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IB Maths Applications & Interpretation · HL · Topic 2

Functions

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 16 October 2026 · new questions added continuouslyFree access until 31 December 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-031-04-r2-01Transformations of graphs and logarithmic scales Paper 1Easy5 marks

The profile of a water slide at a theme park is modelled by $f(x)=0.04x^2-1.6x+20$, for $0\le x\le 20$. Here $x$ is the horizontal distance in metres from the top of the slide, and $f(x)$ is the height in metres above the ground. The slide ends on a raised platform beside a pool.
At a second park the same slide is rebuilt. Its profile is modelled by $g(x)=f(x-4)+2$, for $4\le x\le 24$.

(a)

Describe fully the single transformation that maps the graph of $f$ onto the graph of $g$.

(2)

(b)

Find the coordinates of the vertex of the graph of $f$.

(2)

(c)

Hence write down the height above the ground of the bottom of the new slide.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: translation	✓ 1
A1: by vector $\begin{pmatrix}4\\2\end{pmatrix}$, i.e. 4 m right (horizontally) and 2 m up (vertically)	✓ 1
Part (b)
M1: $x=-\frac{-1.6}{2(0.04)}=20$, or uses the GDC minimum	✓ 1
A1: $(20,\,4)$	✓ 1
Part (c)
A1: the vertex of $g$ is $(24,\,6)$, so the bottom is 6 m above the ground	✓ 1

Exam tip: For $f(x-h)+k$ the graph moves $h$ to the right, not the left. Apply the same translation to any key point, such as a vertex, instead of recalculating it.

+ Add to test✓ In your test2IBMI-031-04-r1-02Transformations of graphs and logarithmic scales Paper 1Easy5 marks

A small company models its weekly profit, P thousand dollars, from producing x hundred items by
$P(x) = -0.5x^2 + 12x - 40$.
After moving to a larger factory, the company's new profit model is $Q(x) = P(x-2) + 5$.

(a)

Describe fully the two transformations that map the graph of P onto the graph of Q.

(2)

(b)

Find the coordinates of the vertex of the graph of Q.

(2)

(c)

Interpret your answer to the previous part in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: horizontal translation 2 units to the right (or translation by vector (2, 0))	✓ 1
A1: vertical translation 5 units up (or translation by vector (0, 5))	✓ 1
Part (b)
M1: finds the vertex of P at (12, 32), e.g. using $x = -\frac{12}{2(-0.5)}$ or a GDC, or finds the vertex of Q directly	✓ 1
A1: vertex of Q is (14, 37)	✓ 1
Part (c)
R1: the maximum weekly profit under the new model is $37 000, achieved when 1400 items are produced	✓ 1

Exam tip: Apply the transformations to the vertex of the original graph rather than expanding Q(x). Replacing x with x − 2 moves points right by 2, not left.

+ Add to test✓ In your test3IBMI3H-2-04-02Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The acidity of a solution is measured by pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol dm⁻³.

(a)

A sample of lemon juice has [H⁺] = 3.2 × 10⁻⁴ mol dm⁻³. Find its pH.

(2)

(b)

Human blood has pH 7.40. Find [H⁺].

(2)

(c)

A solution's pH decreases by 0.3. Find the factor by which [H⁺] changes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −log10(3.2×10⁻⁴)	✓ 1
A1 pH = 3.49	✓ 1
Part (b)
M1 [H⁺] = 10^(−7.40)	✓ 1
A1 3.98 × 10⁻⁸ mol dm⁻³	✓ 1
Part (c)
M1 10^(0.3)	✓ 1
A1 2.00 (the concentration roughly doubles)	✓ 1

Exam tip: A change of 1 on a base-10 logarithmic scale means a factor of 10 in the quantity; a change of 0.3 means a factor of about 2.

+ Add to test✓ In your test4IBMI3H-2-04-10Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The graph of y = h(x) models the height of a wave pulse and has a maximum point at (3, 7).

(a)

Write down the coordinates of the maximum point on the graph of y = h(x + 2) − 1.

(2)

(b)

Write down the coordinates of the maximum point on the graph of y = 3h(2x).

(2)

(c)

The graph of y = −h(x) has a turning point. State its coordinates.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x-coordinate 1	✓ 1
A1 y-coordinate 6, so (1, 6)	✓ 1
Part (b)
A1 x-coordinate 1.5	✓ 1
A1 y-coordinate 21, so (1.5, 21)	✓ 1
Part (c)
A1 (3, −7), a minimum point	✓ 1

Common error: In h(2x) the x-coordinates are halved, not doubled, because the graph is compressed horizontally.

+ Add to test✓ In your test5IBMI3H-2-04-14Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The intensity I W m⁻² of a radio signal at distance d metres from a transmitter satisfies I = k/d². At a distance of 5 m the intensity is 2.4 × 10⁻³ W m⁻².

(a)

Find the value of k.

(2)

(b)

Find the intensity at 12 m.

(2)

(c)

Find the distance at which the intensity falls to 1.0 × 10⁻⁵ W m⁻².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 k = I × d² = 2.4×10⁻³ × 25	✓ 1
A1 k = 0.06	✓ 1
Part (b)
M1 0.06/12²	✓ 1
A1 4.17 × 10⁻⁴ W m⁻²	✓ 1
Part (c)
M1 0.06/d² = 1.0×10⁻⁵	✓ 1
A1 d = 77.5 m	✓ 1

Common error: Treating the relationship as inverse proportion to d instead of to d², which gives the wrong intensity at every other distance.

+ Add to test✓ In your test6IBMI3H-2-04b-04Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

Seawater acidity is measured by pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol/L. The average pH of a bay was 8.18 in 1990 and 8.07 in 2020.

(a)

Find [H⁺] in 1990.

(2)

(b)

Find [H⁺] in 2020.

(1)

(c)

Find the percentage increase in [H⁺] from 1990 to 2020. Use unrounded values.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 [H⁺] = 10^−8.18	✓ 1
A1 6.61 × 10⁻⁹ mol/L	✓ 1
Part (b)
A1 8.51 × 10⁻⁹ mol/L	✓ 1
Part (c)
M1 ratio 10^−8.07 ÷ 10^−8.18 = 10^0.11 = 1.288…	✓ 1
A1 28.8%	✓ 1

Exam tip: A small change on a logarithmic scale is a large change in the underlying quantity. Subtracting the pH values gives the exponent of the ratio.

+ Add to test✓ In your test7IBMI3H-2-04b-10Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The sea temperature T °C at a coastal buoy, t months after 1 January, is modelled by T = 2.4 cos(π(t − 2) ÷ 6) + 14.5. The graph is obtained from the graph of y = cos x by a sequence of transformations.

(a)

Describe the transformations, in order.

(4)

(b)

State the period of the model.

(1)

(c)

Find the lowest temperature predicted by the model.

(1)

(d)

Find the highest temperature predicted by the model.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 vertical stretch with scale factor 2.4	✓ 1
A1 horizontal stretch with scale factor 6 ÷ π	✓ 1
A1 translation 2 units to the right	✓ 1
A1 translation 14.5 units up	✓ 1
Part (b)
A1 12 months	✓ 1
Part (c)
A1 14.5 − 2.4 = 12.1 °C	✓ 1
Part (d)
A1 14.5 + 2.4 = 16.9 °C	✓ 1

Common error: Stating the horizontal stretch factor as π ÷ 6. Writing the argument as (t − 2) ÷ (6 ÷ π) shows the stretch factor is 6 ÷ π.

+ Add to test✓ In your test8IBMI3H-2-03r5-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The curve y = f(x) models the profile of a stage ramp, where x metres is the horizontal distance from the back wall. The curve has a maximum point at (4, 12) and passes through the point (10, 0). The designer changes the ramp so that its profile is the curve y = g(x), where g(x) = 0.5f(x − 2) + 3.

(a)

Write down the coordinates of the maximum point of y = g(x).

(2)

(b)

Write down the coordinates of the image of the point (10, 0) on y = g(x).

(2)

(c)

Describe, in order, the three transformations that map the graph of y = f(x) onto the graph of y = g(x).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x-coordinate 6	✓ 1
A1 y-coordinate 9, so (6, 9)	✓ 1
Part (b)
A1 x-coordinate 12	✓ 1
A1 y-coordinate 3, so (12, 3)	✓ 1
Part (c)
A1 translation 2 units in the positive x direction (right)	✓ 1
A1 vertical stretch, scale factor 0.5	✓ 1
A1 translation 3 units in the positive y direction (up)	✓ 1

Common error: Transformations inside the brackets, such as x − 2, act on x and move the graph the opposite way to the sign; those outside act on y in the expected way.

+ Add to test✓ In your test9IBMI3H-2-03r5-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The value, €V, of a harbour crane t years after it was bought is modelled by V = V₀a^t, where V₀ and a are constants with 0 < a < 1. The crane was worth €34 000 after 1 year and €24 565 after 3 years.

(a)

Find the value of a.

(2)

(b)

Find the value of V₀.

(2)

(c)

Find how many years it takes for the value of the crane to fall to €10 000. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a² = 24 565 ÷ 34 000 = 0.7225	✓ 1
A1 a = 0.85	✓ 1
Part (b)
M1 V₀ × 0.85 = 34 000	✓ 1
A1 V₀ = 40 000	✓ 1
Part (c)
M1 40 000 × 0.85^t = 10 000	✓ 1
M1 takes logarithms: t = ln 0.25 ÷ ln 0.85	✓ 1
A1 t = 8.53 years	✓ 1

Exam tip: Two values at different times give the constants of an exponential model by division: the starting value cancels and leaves a power of the base.

+ Add to test✓ In your test10IBMI3H-2-03r5-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The sound intensity level, L decibels (dB), of a sound with intensity I W m⁻² is given by L = 10 log₁₀(I ÷ 10⁻¹²).

(a)

A loudspeaker in a theatre produces an intensity of 4.0 × 10⁻⁵ W m⁻² at the back row. Calculate the sound intensity level there, correct to 3 significant figures.

(2)

(b)

A second identical loudspeaker is switched on, so the intensities add. Find the new sound intensity level, correct to 3 significant figures.

(2)

(c)

During a thunder effect the level reaches 92 dB. Find the intensity, in W m⁻², correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 L = 10 log₁₀(4.0 × 10⁻⁵ ÷ 10⁻¹²)	✓ 1
A1 L = 76.0 dB	✓ 1
Part (b)
M1 total intensity 8.0 × 10⁻⁵ W m⁻²	✓ 1
A1 L = 79.0 dB	✓ 1
Part (c)
M1 92 = 10 log₁₀(I ÷ 10⁻¹²), so I = 10⁻¹² × 10^9.2	✓ 1
A1 I = 1.58 × 10⁻³ W m⁻² (= 0.00158)	✓ 1

Common error: Doubling the intensity adds about 3 dB to the level; it does not double the number of decibels.

+ Add to test✓ In your test11IBMI3H-2-03r5-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

A counterweight in a theatre flying system moves up and down so that its height above the stage floor is h = 2.4 sin(πt ÷ 6) + 3 metres, where t is the time in seconds after it is released and the angle is in radians.

(a)

Write down the greatest height of the counterweight and the time for one complete oscillation.

(2)

(b)

Find the first time at which the height is 4.2 m.

(2)

(c)

Find the height of the counterweight when t = 4. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 greatest height 5.4 m	✓ 1
A1 period = 2π ÷ (π ÷ 6) = 12 s	✓ 1
Part (b)
M1 2.4 sin(πt ÷ 6) = 1.2, so sin(πt ÷ 6) = 0.5	✓ 1
A1 t = 1 s	✓ 1
Part (c)
M1 h = 2.4 sin(4π ÷ 6) + 3	✓ 1
A1 h = 5.08 m	✓ 1

Common error: If the answer to a sine model looks absurd, check the calculator is in radian mode: the argument πt ÷ 6 is an angle in radians.

+ Add to test✓ In your test12IBMI3H-2-04r5-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The curve y = f(x) shows the shape of a curved stage backdrop, where f(x) = x² − 4x − 1. The curve has its minimum point at (2, −5) and crosses the y-axis at (0, −1).

(a)

The curve is reflected in the x-axis and then translated 1 unit upwards, giving y = 1 − f(x). Write down the coordinates of the image of the point (2, −5).

(2)

(b)

The curve y = f(x) is horizontally compressed with scale factor 0.5, giving y = f(2x). Write down the coordinates of the image of (2, −5) and of (0, −1).

(2)

(c)

The curve y = f(x) is mapped to y = f(x + 3) − 2. Describe this transformation fully.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x-coordinate 2	✓ 1
A1 y-coordinate 5 + 1 = 6, so (2, 6)	✓ 1
Part (b)
A1 image of (2, −5) is (1, −5)	✓ 1
A1 image of (0, −1) is (0, −1)	✓ 1
Part (c)
A1 horizontal translation 3 units to the left	✓ 1
A1 vertical translation 2 units down (translation by the vector (−3, −2))	✓ 1

Common error: A horizontal stretch or compression changes x-coordinates by the reciprocal of the number inside the bracket, so f(2x) halves them.

+ Add to test✓ In your test13IBMI3H-2-04r5-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A shipping company's fuel cost for one voyage is currently €1.2 million. The cost is expected to rise by 2.4% each year, so after n years it is €C million, where C = 1.2 × 1.024ⁿ.

(a)

Find the fuel cost after 5 years, in € million, to 3 significant figures.

(2)

(b)

Find how many years it takes for the cost to reach €1.5 million. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 C = 1.2 × 1.024⁵	✓ 1
A1 €1.35 million	✓ 1
Part (b)
M1 1.2 × 1.024ⁿ = 1.5	✓ 1
M1 n = ln 1.25 ÷ ln 1.024	✓ 1
A1 n = 9.41 years	✓ 1

Common error: A 2.4% yearly rise is multiplication by 1.024, not by 0.024 or by 2.4.

+ Add to test✓ In your test14IBMI3H-2-04r5-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The acidity of a crater lake is measured by pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol dm⁻³.

(a)

The lake has [H⁺] = 2.5 × 10⁻⁴ mol dm⁻³. Calculate its pH, correct to 3 significant figures.

(2)

(b)

A water sample from the lake has pH 2.8. Find its [H⁺], correct to 3 significant figures.

(2)

(c)

Over a year the pH of the lake falls from 4.1 to 3.4. Find the factor by which [H⁺] increases, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 pH = −log₁₀(2.5 × 10⁻⁴)	✓ 1
A1 pH = 3.60	✓ 1
Part (b)
M1 [H⁺] = 10^−2.8	✓ 1
A1 1.58 × 10⁻³ mol dm⁻³	✓ 1
Part (c)
M1 factor = 10^{4.1 − 3.4} = 10^0.7	✓ 1
A1 5.01	✓ 1

Common error: A lower pH means a higher concentration; the minus sign in the formula reverses the direction.

+ Add to test✓ In your test15IBMI3H-2-04r5-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

Two firms ship costumes for a touring theatre company. Firm X charges €C = 480 + 0.35w and Firm Y charges €C = 320 + 0.55w, where w is the mass of the load in kg.

(a)

Find the cost with Firm X for a load of 600 kg.

(2)

(b)

Find which firm is cheaper for a load of 1000 kg, showing both costs.

(2)

(c)

Find the mass for which the two firms charge the same.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 480 + 0.35 × 600	✓ 1
A1 €690	✓ 1
Part (b)
A1 Firm X: 480 + 350 = €830 and Firm Y: 320 + 550 = €870	✓ 1
R1 Firm X is cheaper	✓ 1
Part (c)
M1 480 + 0.35w = 320 + 0.55w	✓ 1
A1 0.2w = 160	✓ 1
A1 w = 800 kg	✓ 1

Exam tip: The break-even point of two linear costs is where the lines intersect; for greater loads the line with the smaller gradient is cheaper.

+ Add to test✓ In your test16IBMI3H-2-04r6-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

Calculator

A train's braking distance d metres from a speed of v km h⁻¹ on level track is modelled by d = 0.045v² + 0.4v, for v ≥ 0.

(a)

Find the braking distance from a speed of 80 km h⁻¹.

(2)

(b)

A signal is 600 m beyond the point where braking starts. Find the greatest speed from which the train can stop before the signal. Give your answer to the nearest km h⁻¹.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes v = 80	✓ 1
A1 d = 320 m	✓ 1
Part (b)
M1 sets 0.045v² + 0.4v = 600 (or equals 0 after rearranging)	✓ 1
M1 solves the quadratic using a graph, a table or the formula, rejecting the negative root	✓ 1
A1 v = 111 km h⁻¹	✓ 1

Common error: Quadratic equations have two roots; reject the negative speed, because the model only applies for v ≥ 0.

+ Add to test✓ In your test17IBMI3H-2-04r6-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

Calculator

A rail company charges a fixed booking fee plus a charge per kilometre travelled. A journey of 50 km costs €13.50 and a journey of 130 km costs €27.50. The cost C euros for a journey of d km is modelled by a linear function.

(a)

Find the charge per kilometre and the fixed booking fee.

(3)

(b)

A passenger pays €36.25 for a journey. Find the distance travelled.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (27.50 − 13.50) ÷ (130 − 50)	✓ 1
A1 charge per kilometre = €0.175	✓ 1
A1 booking fee = 13.50 − 0.175 × 50 = €4.75	✓ 1
Part (b)
M1 solves 4.75 + 0.175d = 36.25	✓ 1
A1 d = 180 km	✓ 1

Exam tip: Use the unit of the gradient to describe its meaning: euros per kilometre is the rate of change of cost with distance.

+ Add to test✓ In your test18IBMI3H-2-04r6-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

Calculator

The apparent magnitude m of a star is related to the energy flux F that reaches the observer by m = m₀ − 2.5 log₁₀ F, where m₀ is a constant. A smaller magnitude means a brighter star.

(a)

Star P has a flux 12 times that of star Q. Find how much smaller the magnitude of P is than the magnitude of Q. Give your answer to 3 significant figures.

(2)

(b)

Star R has a magnitude 1.7 smaller than star S. Find the value of F_R ÷ F_S. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 magnitude difference = 2.5 log₁₀(F_P ÷ F_Q)	✓ 1
A1 difference = 2.5 log₁₀ 12 = 2.70	✓ 1
Part (b)
M1 writes 1.7 = 2.5 log₁₀(F_R ÷ F_S)	✓ 1
M1 rewrites in exponential form: F_R ÷ F_S = 10^{1.7 ÷ 2.5}	✓ 1
A1 ratio = 4.79	✓ 1

Common error: The scale runs backwards: brighter stars have smaller magnitudes, so the flux ratio is greater than 1 for the brighter star.

+ Add to test✓ In your test19IBMI3H-2-04r6-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

Calculator

Sound level L decibels (dB) is related to intensity I W m⁻² by L = 10 log₁₀(I ÷ 10⁻¹²). A passing train produces a sound level of 92 dB at a platform.

(a)

Find the intensity of the sound from the train at the platform. Give your answer in the form a × 10^k with 1 ≤ a < 10 and a to 3 significant figures.

(2)

(b)

Two identical trains pass the platform together, so the total intensity is doubled. Find the new sound level. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solves 92 = 10 log₁₀(I ÷ 10⁻¹²), so I ÷ 10⁻¹² = 10^9.2	✓ 1
A1 I = 1.58 × 10⁻³ W m⁻²	✓ 1
Part (b)
M1 new intensity = 2 × 10^−2.8 W m⁻² (unrounded)	✓ 1
M1 substitutes into the formula for L	✓ 1
A1 L = 95.0 dB	✓ 1

Common error: Doubling the intensity does not double the decibel level; it adds about 3 dB because the scale is logarithmic.

+ Add to test✓ In your test20IBMI3H-2-03r10-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

A wind farm engineer models the electrical power output P kW of one turbine at wind speed v m s⁻¹ by P = 0.35v³, for 4 ≤ v ≤ 12.

(a)

Show that ln P = 3 ln v + ln 0.35.

(2)

(b)

A graph of ln P against ln v is drawn. Write down its gradient and, correct to 3 significant figures, its vertical intercept.

(2)

(c)

Calculate the power output when v = 12 m s⁻¹. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
ln P = ln(0.35v³) = ln 0.35 + ln v³	✓ 1
uses the power law of logarithms to obtain ln P = 3 ln v + ln 0.35 (AG)	✓ 1
Part (b)
gradient = 3	✓ 1
vertical intercept = ln 0.35 = −1.05	✓ 1
Part (c)
P = 0.35 × 12³ = 604.8	✓ 1
P = 605 kW	✓ 1

Exam tip: When a model has the form y = a xⁿ, taking logarithms of both sides gives a straight line in ln y against ln x with gradient n.

+ Add to test✓ In your test21IBMI3H-2-03r10-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

A diver leaves a platform. Her height h metres above the pool surface at time t seconds after leaving the platform is modelled by h = −4.9t² + 4.2t + 3, for t ≥ 0 until she enters the water.

(a)

Find the maximum height reached by the diver and the time at which it occurs.

(2)

(b)

Find the time at which the diver enters the water. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
t = 4.2 / (2 × 4.9) = 0.429 s (0.4286...)	✓ 1
maximum height = 3.9 m	✓ 1
Part (b)
−4.9t² + 4.2t + 3 = 0 solved (formula or GDC), rejecting the negative root	✓ 1
t = 1.32 s	✓ 1

Exam tip: A height of zero means the object reaches the surface, so solve h = 0 and keep only the root with t > 0.

+ Add to test✓ In your test22IBMI3H-2-03r10-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The number of digital loans L made by a library in month t is modelled by L = 500 × 1.06^t, where t = 0 corresponds to the start of the year.

(a)

State the meaning of the value 500 and of the value 1.06 in this context.

(2)

(b)

Find the number of loans predicted when t = 6. Give your answer to the nearest whole number.

(2)

(c)

Find the value of t at which the model predicts 800 loans. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
500 is the number of loans at the start (t = 0)	✓ 1
1.06 means loans increase by 6% each month	✓ 1
Part (b)
500 × 1.06⁶ = 709.26...	✓ 1
709 loans	✓ 1
Part (c)
500 × 1.06^t = 800, so 1.06^t = 1.6	✓ 1
t = ln 1.6 / ln 1.06 = 8.07 months	✓ 1

Exam tip: To solve a × b^t = c for t, divide by a first and then take logarithms of both sides.

+ Add to test✓ In your test23IBMI3H-2-04r13-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

The sound level L in decibels is given by L = 10 log₁₀(I/I₀), where I is the intensity in W m⁻² and I₀ = 1.0 × 10⁻¹² W m⁻². A trumpet produces an intensity of 4.0 × 10⁻⁵ W m⁻² at a listener.

(a)

Calculate the sound level, to 3 significant figures.

(2)

(b)

A second identical trumpet plays at the same place, so the intensities add. Find the new sound level, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute I/I₀ = 4.0 × 10⁷ into the formula	✓ 1
A1 L = 76.0 dB (76.02)	✓ 1
Part (b)
M1 total intensity 8.0 × 10⁻⁵ W m⁻² used in the formula	✓ 1
A1 L = 79.0 dB (79.03)	✓ 1

Common error: Doubling the intensity does not double the decibel level; it adds about 3 dB, because decibels are a logarithmic scale.

+ Add to test✓ In your test24IBMI3H-2-04r13-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The pH of soil is defined by pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³. A field's soil has pH 6.3.

(a)

Find [H⁺] for this soil, giving your answer in the form a × 10^k with a to 3 significant figures.

(2)

(b)

Before liming, the pH was 5.3. State by what factor [H⁺] changed when the pH rose from 5.3 to 6.3.

(1)

(c)

A different field has [H⁺] = 2.5 × 10⁻⁶ mol dm⁻³. Find its pH to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 [H⁺] = 10^−6.3	✓ 1
A1 5.01 × 10⁻⁷ mol dm⁻³	✓ 1
Part (b)
A1 it decreased by a factor of 10 (each pH unit is a factor of 10)	✓ 1
Part (c)
M1 pH = −log₁₀(2.5 × 10⁻⁶)	✓ 1
A1 pH = 5.60	✓ 1

Exam tip: On a log₁₀ scale, a change of 1 unit corresponds to a multiplication by 10, so small differences in pH are large differences in concentration.

+ Add to test✓ In your test25IBMI3H-2-04r13-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

In the lower atmosphere, air pressure p kPa at height h km above sea level is modelled by p = 101e^−h/8.4. An aircraft cruises at 11 km.

(a)

State the pressure at sea level.

(1)

(b)

Find the pressure at the aircraft's cruising height, to 3 significant figures.

(2)

(c)

Find the height at which the pressure is 50 kPa, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 101 kPa	✓ 1
Part (b)
M1 substitute h = 11: 101e^−11/8.4	✓ 1
A1 27.3 kPa (27.26)	✓ 1
Part (c)
M1 set 101e^−h/8.4 = 50 and take natural logs	✓ 1
A1 h = 8.4 ln(101/50) = 5.91 km	✓ 1

Common error: When solving 101e⁻h/⁸.⁴ = 50, divide by 101 first, then take logs; taking logs of 101e… as a single product leads to errors.

+ Add to test✓ In your test26IBMI3H-2-04r13-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

In equal temperament, the frequency of the note n semitones above concert A (440 Hz) is f_n = 440 × 2^n/12 Hz. A negative n means a note below concert A.

(a)

Show that the frequency increases by the same factor, 2^1/12, for each semitone, and write this factor to 4 significant figures.

(2)

(b)

Find the frequency of the note 7 semitones above concert A, to 3 significant figures.

(2)

(c)

Find the frequency of the note 9 semitones below concert A, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f_n+1/f_n = 2^(n+1)/12/2^n/12 = 2^1/12	✓ 1
A1 factor = 1.059	✓ 1
Part (b)
M1 440 × 2^7/12	✓ 1
A1 659 Hz (659.3)	✓ 1
Part (c)
M1 440 × 2^−9/12	✓ 1
A1 262 Hz (261.6)	✓ 1

Exam tip: Musical pitch is exponential in the number of semitones: equal steps in pitch correspond to equal multiplying factors in frequency.

+ Add to test✓ In your test27IBMI3H-2-05r16-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The sound level, L decibels (dB), in a hospital ward is given by L = 10 log₁₀(I/I₀), where I is the sound intensity in W m⁻² and I₀ = 1.0 × 10⁻¹² W m⁻².

(a)

The intensity at night is 3.2 × 10⁻⁷ W m⁻². Calculate the sound level, to 3 significant figures.

(2)

(b)

During a busy period the sound level is 72 dB. Find the intensity, in W m⁻², to 3 significant figures.

(2)

(c)

A noise-reduction programme lowers the sound level by 15 dB. Find the factor by which the intensity is reduced, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 L = 10 log₁₀(3.2 × 10⁻⁷ ÷ 10⁻¹²) = 10 log₁₀(3.2 × 10⁵)	✓ 1
A1 L = 55.05… so 55.1 dB	✓ 1
Part (b)
M1 72 = 10 log₁₀(I/10⁻¹²) so I/10⁻¹² = 10^7.2	✓ 1
A1 I = 1.58 × 10⁻⁵ W m⁻²	✓ 1
Part (c)
M1 intensity ratio satisfies 10 log₁₀(I₁/I₂) = 15, so I₁/I₂ = 10^1.5	✓ 1
A1 factor = 31.62… so 31.6	✓ 1

Exam tip: A change in decibels corresponds to a ratio of intensities, so the I₀ value cancels when comparing two levels.

+ Add to test✓ In your test28IBMI3H-2-05r16-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

The rate at which a patient's body clears a drug, y mL per minute, depends on the patient's mass, x kg. A graph of log₁₀ y against log₁₀ x is a straight line with equation log₁₀ y = 0.75 log₁₀ x + 0.30.

(a)

Show that y = ax^0.75, and find a correct to 3 significant figures.

(2)

(b)

Find the clearance rate for a patient of mass 64 kg. Use the unrounded value of a and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 y = 10^{0.75 log₁₀ x + 0.30} = 10^0.30 × x^0.75	✓ 1
A1 a = 10^0.30 = 1.995… so 2.00	✓ 1
Part (b)
M1 y = 10^0.30 × 64^0.75	✓ 1
A1 45.15… so 45.1 mL per minute	✓ 1

Common error: Writing a = 0.30 from the intercept. The intercept of a log–log line is log a, so a = 10^0.30.

+ Add to test✓ In your test29IBMI3H-2-05r16-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The weekly honey yield, f(x) kg, of a hive with x frames of bees is modelled by f(x) = 12x − 2x². A beekeeper moves the hive to a new site, where the yield is modelled by g(x) = f(x − 2) − 7.

(a)

Find the coordinates of the maximum point of the graph of y = f(x).

(2)

(b)

Describe the transformation that maps the graph of y = f(x) onto y = g(x).

(1)

(c)

Hence find the maximum yield predicted by g, and the number of frames at which it occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 vertex at x = −12 ÷ (2 × (−2)) = 3 (or complete the square / GDC)	✓ 1
A1 (3, 18)	✓ 1
Part (b)
A1 translation by the vector (2, −7)	✓ 1
Part (c)
A1 maximum at (3 + 2, 18 − 7) = (5, 11)	✓ 1
A1 maximum yield 11 kg at 5 frames	✓ 1

Exam tip: Apply a translation to the vertex only; you do not need to expand the new quadratic.

+ Add to test✓ In your test30IBMI3H-2-05r16-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

For a Formula 1 car, the braking distance, d metres, from speed v m s⁻¹ is modelled by d = 0.012v² + 0.30v.

(a)

Calculate the braking distance from 75 m s⁻¹.

(2)

(b)

The braking zone at a corner is 120 m long. Find the greatest speed from which the car can stop within it. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 d = 0.012 × 75² + 0.30 × 75	✓ 1
A1 d = 90.0 m	✓ 1
Part (b)
M1 set 0.012v² + 0.30v = 120 (or ≤ 120)	✓ 1
M1 solve the quadratic (formula or GDC) and reject the negative root	✓ 1
A1 v = 88.28… so 88.3 m s⁻¹	✓ 1

Common error: A quadratic has two roots; the negative speed has no meaning here and must be rejected explicitly.

+ Add to test✓ In your test31IBMI3H-2-03r17-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A hospital pharmacist models the concentration of a drug in a patient's blood after an injection by $C(t)=80e^{-0.173t}$, where $C$ is in mg L⁻¹ and $t$ is the time in hours after the injection.

(a)

Find the concentration of the drug 5 hours after the injection. Give your answer to 3 significant figures.

(2)

(b)

The drug becomes ineffective once the concentration falls below 10 mg L⁻¹. Find the time, in hours, at which this happens. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes $t=5$: $C=80e^{-0.173\times5}$	✓ 1
A1 $C=33.7$ mg L⁻¹	✓ 1
Part (b)
M1 sets $80e^{-0.173t}=10$ (or finds the intersection with $C=10$ on a GDC)	✓ 1
M1 takes natural logarithms: $-0.173t=\ln\frac{1}{8}$ (or equivalent, e.g. $t=\frac{\ln 8}{0.173}$)	✓ 1
A1 $t=12.0$ hours	✓ 1

Exam tip: State the substituted expression before the GDC value so the method mark is secure even if the final value slips.

+ Add to test✓ In your test32IBMI3H-2-03r17-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The volume of water in a storage tank, $V$ litres, is modelled by $V=12000-400t$, where $t$ is the time in hours after the tank's outlet valve is opened.

(a)

Write down the meaning of the number 400 in this model.

(1)

(b)

Find the volume of water in the tank after 7.5 hours.

(2)

(c)

A second tank holds 1500 litres and is being filled at 250 litres per hour, so that its volume is $W=1500+250t$. Find the time, in hours, at which both tanks hold the same volume. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 the water leaves the tank at a constant rate of 400 litres per hour (rate of decrease of volume)	✓ 1
Part (b)
M1 $12000-400\times7.5$	✓ 1
A1 $9000$ litres	✓ 1
Part (c)
M1 equates $12000-400t=1500+250t$ (or intersection on a GDC)	✓ 1
M1 rearranges: $650t=10500$	✓ 1
A1 $t=16.2$ hours	✓ 1

Common error: Mixing up the signs of the two gradients when equating the models; the first volume decreases while the second increases.

+ Add to test✓ In your test33IBMI3H-2-03r17-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The value of a virtual currency in an online game halves every 9 days. The value of one unit at day 0 is 500 gold coins, and its value after $t$ days is modelled by $V=500e^{-kt}$.

(a)

Show that $k=0.0770$ correct to 3 significant figures.

(2)

(b)

Find the value of one unit after 20 days. Give your answer to 3 significant figures.

(2)

(c)

Find the number of days until the value has fallen to 10% of its original value. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $250=500e^{-9k}$, so $e^{-9k}=\frac{1}{2}$	✓ 1
A1 $k=\frac{\ln2}{9}=0.0770$ (AG)	✓ 1
Part (b)
M1 $500e^{-20k}$ with unrounded $k$ (or $500\times0.5^{20/9}$)	✓ 1
A1 107 gold coins	✓ 1
Part (c)
M1 $e^{-kt}=0.1$	✓ 1
M1 $t=\frac{\ln10}{k}$	✓ 1
A1 $29.9$ days	✓ 1

Exam tip: Use the half-life statement to build $k$ exactly (as $\frac{\ln 2}{9}$), then keep it unrounded in later parts.

+ Add to test✓ In your test34IBMI3H-2-03r19-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A wind farm engineer models the power output P kW of a turbine at wind speed v m s⁻¹ by P = kvⁿ. A graph of log₁₀ P against log₁₀ v is a straight line with equation log₁₀ P = 3 log₁₀ v − 0.4.

(a)

State the value of n.

(1)

(b)

Find the value of k, giving your answer to 3 significant figures.

(2)

(c)

Using the unrounded value of k, calculate the power output at a wind speed of 9 m s⁻¹. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 n = 3	✓ 1
Part (b)
M1 k = 10^−0.4 (or equivalent use of the intercept)	✓ 1
A1 k = 0.398	✓ 1
Part (c)
M1 substitution of v = 9 into P = kv³	✓ 1
A1 P = 290 kW	✓ 1

Common error: The intercept of the straight line is log₁₀ k, not k; undo the logarithm before using it.

+ Add to test✓ In your test35IBMI3H-2-03r19-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

A library records the number of visitors per hour, n, t hours after it opens, for 0 ≤ t ≤ 12. The model is n = −0.5t² + 6t + 4.

(a)

Find the maximum number of visitors per hour predicted by the model, and the time at which it occurs.

(3)

(b)

Calculate how many hours the model predicts at least 20 visitors per hour.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 t = −6 ÷ (2 × −0.5) (or use of technology to find the vertex)	✓ 1
A1 t = 6 hours after opening	✓ 1
A1 maximum n = 22 visitors per hour	✓ 1
Part (b)
M1 −0.5t² + 6t + 4 = 20 (or t² − 12t + 32 = 0)	✓ 1
A1 t = 4 and t = 8	✓ 1
A1 4 hours	✓ 1

Exam tip: A symmetrical quadratic is at its maximum halfway between its two roots; use this to check the vertex you find.

+ Add to test✓ In your test36IBMI3H-2-03r19-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The installed capacity of the wind farms in a region is modelled by C = 120 × 1.08^t megawatts (MW), where t is the number of years after the start of 2020.

(a)

Calculate the capacity at t = 5, giving your answer to 3 significant figures.

(2)

(b)

Find the value of t at which the capacity first reaches 300 MW. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 120 × 1.08⁵	✓ 1
A1 176 MW	✓ 1
Part (b)
M1 120 × 1.08^t = 300	✓ 1
M1 1.08^t = 2.5 and use of logarithms (or technology)	✓ 1
A1 t = 11.9 years	✓ 1

Common error: Divide both sides by 120 before taking logarithms; taking logarithms of 300 ÷ 120 is not the same as dividing logarithms of 300 and 120.

+ Add to test✓ In your test37IBMI3H-2-03r19-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

A wave machine makes the water depth at one point in a swimming pool vary according to d(t) = 2.10 + 0.15 sin(^πt⁄₂) metres, where t is the time in seconds and the angle is in radians.

(a)

Write down the maximum and minimum depth.

(2)

(b)

Find the period of the wave.

(2)

(c)

Calculate the depth when t = 1.5, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 maximum 2.25 m	✓ 1
A1 minimum 1.95 m	✓ 1
Part (b)
M1 2π ÷ (π/2)	✓ 1
A1 4 seconds	✓ 1
Part (c)
M1 2.10 + 0.15 sin(0.75π)	✓ 1
A1 2.21 m	✓ 1

Common error: Check that the calculator is in radian mode when the angle contains π; degree mode gives a very different value.

+ Add to test✓ In your test38IBMI3H-2-04r19-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A fisheries scientist models the number of fish in a lake as P = ab^t, where t is the time in years since monitoring began. A graph of log₁₀ P against t is a straight line with equation log₁₀ P = 0.04t + 3.2.

(a)

Find the value of a and the value of b, giving each to 3 significant figures.

(3)

(b)

Use the equation of the line to find the number of fish after 10 years. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = 10^3.2 and b = 10^0.04	✓ 1
A1 a = 1580	✓ 1
A1 b = 1.10	✓ 1
Part (b)
M1 log₁₀ P = 0.04 × 10 + 3.2 = 3.6	✓ 1
A1 P = 3980 fish	✓ 1

Common error: The intercept of the line is log₁₀ a, so a = 10³.², not 3.2.

+ Add to test✓ In your test39IBMI3H-2-04r19-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A company leases a warehouse robot under Plan A, where the total cost is a linear function of the hours of use, h. Using it for 3 hours costs €210 and using it for 7 hours costs €330. Plan B has cost C_B = 60 + 45h.

(a)

Find the cost per hour and the fixed charge in Plan A.

(3)

(b)

Find the number of hours for which the two plans cost the same.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (330 − 210) ÷ (7 − 3)	✓ 1
A1 €30 per hour	✓ 1
A1 fixed charge €120 (C_A = 120 + 30h)	✓ 1
Part (b)
M1 120 + 30h = 60 + 45h	✓ 1
A1 h = 4 hours	✓ 1

Exam tip: Find the gradient first, then substitute one point into y = mx + c to get the intercept.

+ Add to test✓ In your test40IBMI3H-2-04r19-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A fashion shop sells q scarves per day at a price of €p each, where q = 120 − 4p, for 0 < p < 30.

(a)

Show that the daily revenue is R = 120p − 4p² euros.

(2)

(b)

Find the price that gives the greatest revenue and calculate this revenue.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 R = p × q	✓ 1
A1 R = p(120 − 4p) = 120p − 4p²	✓ 1
Part (b)
M1 p = −120 ÷ (2 × −4) (or vertex using technology)	✓ 1
A1 p = €15	✓ 1
A1 R = €900	✓ 1

Common error: The greatest revenue is not the price at which the most items are sold; find the vertex of the revenue quadratic.

+ Add to test✓ In your test41IBMI3H-2-05r21-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The sound level L, in dB, of the cooling fans in a data centre hall is modelled by L = 10 log₁₀(I/I₀), where I is the sound intensity in W m⁻² and I₀ = 1.0×10⁻¹² W m⁻².

(a)

The intensity of the sound in the hall is 4.0×10⁻⁵ W m⁻². Calculate the sound level, giving your answer to 3 significant figures.

(2)

(b)

Staff must wear ear protection when the sound level reaches 85.0 dB. Find the intensity at which this happens, giving your answer in the form a×10^k with 1 ≤ a < 10 and a to 3 significant figures.

(3)

(c)

A second identical rack of fans is switched on next to the first, so that the total intensity doubles. Calculate the increase in the sound level, in dB, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes into 10 log₁₀(4.0×10⁻⁵/1.0×10⁻¹²)	✓ 1
A1 L = 76.0 dB	✓ 1
Part (b)
M1 85.0 = 10 log₁₀(I/10⁻¹²)	✓ 1
M1 I/10⁻¹² = 10^8.5	✓ 1
A1 I = 3.16×10⁻⁴ W m⁻²	✓ 1
Part (c)
M1 increase = 10 log₁₀(2I/I) or 10 log₁₀ 2	✓ 1
A1 3.01 dB	✓ 1

Common error: Doubling the intensity does not double the number of decibels; because the scale is logarithmic, equal intensity ratios add equal amounts to the level.

+ Add to test✓ In your test42IBMI3H-2-05r21-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

Two firms hire out cargo bikes. Firm A charges a fixed fee of £15 plus £4.50 per hour. Firm B charges £7 per hour with no fixed fee. Let h be the number of hours of hire.

(a)

Write down expressions for the cost, in £, of hiring a bike for h hours from Firm A and from Firm B.

(2)

(b)

Find the number of hours for which the two firms charge the same, and the cost of that hire.

(3)

(c)

A customer needs a bike for 4.5 hours. Find how much cheaper the cheaper firm is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 C_A = 15 + 4.5h	✓ 1
A1 C_B = 7h	✓ 1
Part (b)
M1 15 + 4.5h = 7h	✓ 1
A1 h = 6	✓ 1
A1 £42.00	✓ 1
Part (c)
M1 evaluates both costs: A = £35.25 and B = £31.50	✓ 1
A1 Firm B is cheaper by £3.75	✓ 1

Exam tip: State which firm is cheaper as well as the difference; a bare number does not answer a comparison question.

+ Add to test✓ In your test43IBMI3H-2-05r21-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy8 marks

An archaeologist marks out a rectangular excavation area against a straight wall using 40 m of tape. The tape is used for the three sides not on the wall. The two sides perpendicular to the wall each have length x metres.

(a)

Show that the area A m² of the excavation is given by A = 40x − 2x².

(2)

(b)

Find the greatest possible area and the value of x for which it occurs.

(3)

(c)

The permit allows an area of exactly 150 m². Find the two possible values of x.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 side parallel to the wall = 40 − 2x	✓ 1
A1 A = x(40 − 2x) = 40x − 2x² (AG)	✓ 1
Part (b)
M1 vertex at x = −40/(2 × −2) or uses a graph	✓ 1
A1 x = 10 m	✓ 1
A1 maximum area = 200 m²	✓ 1
Part (c)
M1 40x − 2x² = 150	✓ 1
A1 x = 5	✓ 1
A1 x = 15 m (the sides parallel to the wall are then 30 m and 10 m respectively)	✓ 1

Common error: A quadratic model can give two valid solutions; check both against the context before rejecting one.

+ Add to test✓ In your test44IBMI3H-2-05r26-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

In a city, the number of reported cases y of an infection in week t is found to satisfy ln y = 2.1 + 0.35t.

(a)

Show that y can be written as y = ae^kt and find the values of a and k. Give a to 3 significant figures.

(3)

(b)

Find the time taken for the number of weekly cases to double, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 exponentiate both sides: y = e^{2.1 + 0.35t} = e^2.1 × e^0.35t	✓ 1
A1 a = e^2.1 = 8.17	✓ 1
A1 k = 0.35	✓ 1
Part (b)
M1 e^0.35T = 2, so T = ln 2 / 0.35	✓ 1
A1 1.98 weeks	✓ 1

Exam tip: A straight line on a graph of ln y against t means y is exponential in t; the gradient is the growth constant k.

+ Add to test✓ In your test45IBMI3H-2-05r26-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The graph of y = f(x) models the footfall, in hundreds of customers per hour, in a shop x hours after opening. The graph has a maximum point at (6, 20). A new model is g(x) = 2f(x − 1) + 3.

(a)

Describe the sequence of transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

State the coordinates of the maximum point of y = g(x) and hence the greatest footfall predicted by the new model, in hundreds of customers per hour.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation by 1 unit in the positive x-direction	✓ 1
A1 vertical stretch with scale factor 2	✓ 1
A1 translation by 3 units in the positive y-direction, applied after the stretch	✓ 1
Part (b)
A1 maximum point at (7, 43)	✓ 1
A1 greatest footfall 2 × 20 + 3 = 43 hundred customers per hour	✓ 1

Common error: In 2f(x − 1) + 3 the stretch is applied before the vertical shift, so the 3 is not multiplied by 2.

+ Add to test✓ In your test46IBMI3H-2-05r26-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The sound intensity level of a wind turbine at a house is given by L = 10 log₁₀(I/I₀) decibels (dB), where I is the intensity in W m⁻² and I₀ = 1 × 10⁻¹² W m⁻². One turbine produces a level of 52 dB at the house.

(a)

Find the intensity of the sound from one turbine at the house.

(2)

(b)

Two identical turbines operate at the same distance from the house. Find the new sound level, to 3 significant figures.

(2)

(c)

Find the number of identical turbines that would produce a level of 62 dB at the house.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 52 = 10 log₁₀(I/10⁻¹²), so I = 10^5.2 × 10⁻¹²	✓ 1
A1 1.58 × 10⁻⁷ W m⁻²	✓ 1
Part (b)
M1 intensity is doubled, so L = 52 + 10 log₁₀ 2	✓ 1
A1 55.0 dB	✓ 1
Part (c)
M1 62 − 52 = 10 log₁₀ n	✓ 1
A1 n = 10 turbines	✓ 1

Common error: Doubling the intensity adds about 3 dB to the level; it does not double the level in dB.

+ Add to test✓ In your test47IBMI3H-2-05r26-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

A workshop's average cost per unit, A euros (€), when it makes n units in a week is modelled by A(n) = 4 + 3600/n, for n > 0.

(a)

Find the average cost per unit when 300 units are made.

(1)

(b)

Write down the equation of the horizontal asymptote of the graph of A and explain what it means in this context.

(2)

(c)

The total weekly cost is nA(n). Write down the total cost as a function of n, and hence state the fixed weekly cost.

(2)

(d)

Find the smallest whole number of units for which the average cost is less than €5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 4 + 3600/300 = €16	✓ 1
Part (b)
A1 A = 4	✓ 1
A1 the average cost per unit approaches €4, the cost of making each extra unit, as the number of units becomes very large	✓ 1
Part (c)
A1 4n + 3600	✓ 1
A1 fixed cost €3600 per week	✓ 1
Part (d)
M1 4 + 3600/n < 5, so n > 3600	✓ 1
A1 3601 units	✓ 1

Exam tip: When the model is a quotient, solve the inequality as an equation first and then check the direction with a test value.

+ Add to test✓ In your test48IBMI3H-2-04r27-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy8 marks

Underwater sound is measured by the sound pressure level L = 20 log₁₀(p/p₀) decibels (dB), where p is the sound pressure and p₀ = 1 μPa is the reference pressure.

(a)

A ship produces a sound pressure of 3.2 × 10⁷ μPa. Find the sound pressure level, to the nearest dB.

(2)

(b)

A whale call has a sound pressure level of 172 dB. Find its sound pressure, in μPa, to 3 significant figures.

(2)

(c)

Show that doubling the sound pressure increases the sound pressure level by 6.0 dB, to 2 significant figures.

(2)

(d)

Find how many times greater the sound pressure of the 172 dB whale call is than that of the 150 dB ship. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 log₁₀(3.2 × 10⁷)	✓ 1
A1 150 dB	✓ 1
Part (b)
M1 10^172/20 = 10^8.6	✓ 1
A1 3.98 × 10⁸ μPa	✓ 1
Part (c)
M1 20 log₁₀(2p/p₀) − 20 log₁₀(p/p₀) = 20 log₁₀ 2	✓ 1
A1 6.02 which is 6.0 dB to 2 s.f.	✓ 1
Part (d)
M1 10^{(172 − 150)/20}	✓ 1
A1 12.6	✓ 1

Exam tip: A difference in decibels corresponds to a ratio of pressures, so the reference pressure cancels.

+ Add to test✓ In your test49IBMI3H-2-04r27-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The temperature of concrete curing in a precast mould is modelled by T(t) = 20 + 55e^−0.12t, where T is in °C and t is the time in hours after pouring.

(a)

Find the temperature of the concrete at the moment of pouring.

(1)

(b)

Write down the temperature the concrete approaches in the long term.

(1)

(c)

Find the time taken for the temperature to fall to 30 °C. Give your answer to 3 significant figures.

(2)

(d)

The concrete is considered to match the surroundings when it is within 1 °C of its long-term temperature. Find the time at which this first happens, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 75 °C	✓ 1
Part (b)
A1 20 °C	✓ 1
Part (c)
M1 55e^−0.12t = 10	✓ 1
A1 14.2 hours	✓ 1
Part (d)
M1 55e^−0.12t = 1	✓ 1
A1 33.4 hours	✓ 1

Common error: Subtract the long-term temperature before using logarithms: the exponential part equals 10 when T = 30, not T itself.

+ Add to test✓ In your test50IBMI3H-2-04r27-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The downward deflection d mm at the end of a timber cantilever beam of length L metres is modelled by d = kL³, where k is a constant. A beam of length 2.0 m has a deflection of 0.80 mm.

(a)

Find the value of k.

(2)

(b)

Find the deflection of a beam of length 3.5 m, to 3 significant figures.

(1)

(c)

State the factor by which the deflection is multiplied when the length of the beam is doubled.

(1)

(d)

A beam has a deflection of 10 mm. Find its length, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.80 = k × 2³	✓ 1
A1 k = 0.100	✓ 1
Part (b)
A1 4.29 mm	✓ 1
Part (c)
A1 8	✓ 1
Part (d)
M1 10 = 0.1L³	✓ 1
A1 4.64 m	✓ 1

Exam tip: In a power model, doubling the input multiplies the output by 2 raised to the power. Use this as a quick check on your answer.

+ Add to test✓ In your test51IBMI3H-2-04r27-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The mass w grams of a sample of kelp is modelled by w = ab^t, where t is the time in weeks. A graph of log₁₀ w against t is a straight line passing through (0, 1.2) and (5, 2.45).

(a)

Find the gradient of the line.

(2)

(b)

Find the values of a and b, to 3 significant figures.

(3)

(c)

Find the time at which the mass of the sample reaches 400 g. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2.45 − 1.2)/(5 − 0)	✓ 1
A1 0.25	✓ 1
Part (b)
M1 log w = log a + t log b	✓ 1
A1 a = 10^1.2 = 15.8	✓ 1
A1 b = 10^0.25 = 1.78	✓ 1
Part (c)
M1 log₁₀ 400 = 1.2 + 0.25t	✓ 1
A1 5.61 weeks	✓ 1

Common error: The intercept of the line is log a, not a. Undo the logarithm before stating the model.

+ Add to test✓ In your test52IBMI3H-2-05r27-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

Fisheries scientists model the mass M grams of a sea bass of length L cm by M = aL^b. For sea bass between 20 cm and 70 cm long, a graph of log₁₀M against log₁₀L is a straight line with gradient 3.10 and vertical intercept −1.90.

(a)

Write down the value of b.

(1)

(b)

Find the value of a, correct to 3 significant figures.

(2)

(c)

Estimate the mass of a sea bass of length 44 cm. Use the unrounded value of a and give your answer to 3 significant figures.

(2)

(d)

A student uses the model to predict the mass of a sea bass of length 5 cm. State why this prediction may be unreliable.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 b = 3.10	✓ 1
Part (b)
M1 log₁₀a = −1.90 (or a = 10^−1.90)	✓ 1
A1 a = 0.0126	✓ 1
Part (c)
M1 substitutes L = 44 into M = 10^−1.90 × L^3.10	✓ 1
A1 M = 1570 g	✓ 1
Part (d)
R1 5 cm is outside the data range of 20 cm to 70 cm (extrapolation), so the model may not hold	✓ 1

Common error: Remember that the intercept of a log–log graph is log₁₀a, not a; undo the logarithm before substituting.

+ Add to test✓ In your test53IBMI3H-2-05r27-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

A fashion designer's software draws the edge of a scarf pattern as the graph of y = f(x). The graph has a maximum point at (2, 6).

(a)

The graph of y = g(x), where g(x) = 2f(x + 1) − 3, is drawn for a second pattern. Find the coordinates of the maximum point on the graph of g.

(2)

(b)

A third pattern uses h(x) = −f(3x). Find the coordinates of the image of the point (2, 6) on the graph of h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x-coordinate 1 (translation 1 unit left)	✓ 1
A1 y-coordinate 2 × 6 − 3 = 9, so (1, 9)	✓ 1
Part (b)
A1 x-coordinate 2/3 (horizontal stretch, scale factor 1/3)	✓ 1
A1 y-coordinate −6, so (2/3, −6) (reflection in the x-axis)	✓ 1

Common error: Changes inside the brackets act on x and work in the opposite way to what the sign suggests: f(x + 1) moves the graph left.

+ Add to test✓ In your test54IBMI3H-2-05r27-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A designer makes silk scarves. The cost of making n scarves in a week is C = 450 + 12.5n (in €), and the income from selling n scarves is R = 27n (in €).

(a)

Find the smallest whole number of scarves that must be sold in a week for the designer to make a profit.

(3)

(b)

Find the profit made in a week in which 100 scarves are sold.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 27n > 450 + 12.5n (or equates income and cost)	✓ 1
A1 n > 31.03… (14.5n = 450)	✓ 1
A1 32 scarves	✓ 1
Part (b)
M1 27 × 100 − (450 + 12.5 × 100)	✓ 1
A1 €1000	✓ 1

Exam tip: When a question asks for a whole number of items, round up if you need to exceed a threshold, and say which way you rounded.

+ Add to test✓ In your test55IBMI3H-2-05r27-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

In a fish-farm pond, the acidity is measured using pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol dm⁻³.

(a)

The concentration of hydrogen ions in the pond is 3.2 × 10⁻⁸ mol dm⁻³. Find the pH, correct to 3 significant figures.

(2)

(b)

State the factor by which [H⁺] changes when the pH falls by 1.

(1)

(c)

After heavy rain the pH falls from 7.8 to 7.2. Find the factor by which [H⁺] increases. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −log₁₀(3.2 × 10⁻⁸)	✓ 1
A1 pH = 7.49	✓ 1
Part (b)
A1 it is multiplied by 10	✓ 1
Part (c)
M1 [H⁺] ratio = 10^{7.8 − 7.2} = 10^0.6	✓ 1
A1 3.98	✓ 1

Common error: A fall in pH means an increase in [H⁺]; the scale is logarithmic and decreasing.

+ Add to test✓ In your test56IBMI3H-2-06r31-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

In pharmacology the potency of a drug solution is described by pC = −log₁₀(C), where C is the concentration in mol dm⁻³.

(a)

Calculate pC for a solution with concentration 4.0 × 10⁻⁶ mol dm⁻³. Give your answer to 3 significant figures.

(2)

(b)

A solution has pC = 3.6. Find its concentration in the form a × 10^k where 1 ≤ a < 10, with a given to 3 significant figures.

(2)

(c)

Solution X has pC lower than solution Y by 2. Find the ratio of the concentration of X to the concentration of Y.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −log₁₀(4.0 × 10⁻⁶)	✓ 1
A1 5.40	✓ 1
Part (b)
M1 C = 10^−3.6	✓ 1
A1 2.51 × 10⁻⁴ mol dm⁻³	✓ 1
Part (c)
M1 C_X ÷ C_Y = 10^{−(pC_X − pC_Y)} = 10²	✓ 1
A1 100 (X is 100 times more concentrated)	✓ 1

Exam tip: On a logarithmic scale, equal steps in the scale mean equal multiplicative factors in the quantity. A fall of 1 in pC means the concentration is multiplied by 10.

+ Add to test✓ In your test57IBMI3H-2-06r31-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The volume of water in a reservoir, V megalitres (ML), t days after a pump is switched on is modelled by V = 800e^−0.04t.

(a)

Write down the initial volume of water.

(1)

(b)

Find the volume of water after 10 days, to 3 significant figures.

(2)

(c)

Find the time taken for the volume to halve, giving your answer in days to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 800 ML	✓ 1
Part (b)
M1 800e^−0.4	✓ 1
A1 536 ML	✓ 1
Part (c)
M1 800e^−0.04t = 400, so e^−0.04t = 0.5	✓ 1
M1 t = ln 2 ÷ 0.04	✓ 1
A1 17.3 days	✓ 1

Exam tip: For exponential decay the half-life does not depend on the starting amount; it is ln 2 divided by the decay constant.

+ Add to test✓ In your test58IBMI3H-2-06r31-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

A game company models its monthly profit R (in € thousands) from a premium pass priced at €x by R = −0.5x² + 12x − 20.

(a)

Find the price that gives the greatest profit.

(2)

(b)

Find the greatest profit.

(2)

(c)

Find the range of prices for which the monthly profit is at least €40 thousand. Give the endpoints to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x = −12 ÷ (2 × (−0.5)) (or complete the square or use technology)	✓ 1
A1 €12	✓ 1
Part (b)
M1 substitute x = 12 (ECF from (a))	✓ 1
A1 €52 thousand	✓ 1
Part (c)
M1 −0.5x² + 12x − 20 = 40, which gives x² − 24x + 120 = 0	✓ 1
A1 lower endpoint €7.10	✓ 1
A1 upper endpoint €16.9, so the range is 7.10 ≤ x ≤ 16.9	✓ 1

Common error: The vertex of a quadratic model gives the best price and the maximum value, and the endpoints of a range where the value is at least a target are the roots of the equation set equal to that target.

+ Add to test✓ In your test59IBMI3H-2-06r33-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

After a cooling failure in a data centre, the air temperature T °C in a server hall is monitored for t minutes. A graph of ln T against t is a straight line with equation ln T = 0.0318t + 2.95.

(a)

Show that T can be written as $T=Ae^{kt}$, and find the values of A and k to three significant figures.

(2)

(b)

Find the temperature of the hall 25 minutes after the failure, correct to 3 significant figures.

(2)

(c)

The servers shut down automatically when the temperature reaches 60 °C. Find the time, in minutes to 3 significant figures, at which this happens.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes exponentials of both sides: $T=e^{0.0318t+2.95}=e^{2.95}e^{0.0318t}$	✓ 1
A1 A = 19.1 and k = 0.0318	✓ 1
Part (b)
M1 substitutes t = 25	✓ 1
A1 42.3 °C	✓ 1
Part (c)
M1 solves 19.1e^0.0318t = 60, or ln 60 = 0.0318t + 2.95	✓ 1
A1 36.0 minutes	✓ 1

Exam tip: A straight line on a graph of ln T against t means the original relationship is exponential: the gradient is the growth constant and the intercept is ln A.

+ Add to test✓ In your test60IBMI3H-2-06r33-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

The cooling power P kW needed for a data hall with x racks of servers (in hundreds) is such that a graph of log₁₀ P against log₁₀ x is a straight line with equation log₁₀ P = 1.5 log₁₀ x + 0.6.

(a)

Show that P = axⁿ, and find the values of a and n, giving a to three significant figures.

(2)

(b)

Find the cooling power needed when there are 1600 racks, correct to the nearest kW.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewrites as P = 10^0.6 × x^1.5 using the laws of logarithms	✓ 1
A1 a = 3.98 and n = 1.5	✓ 1
Part (b)
M1 x = 16 and P = 3.98 × 16^1.5	✓ 1
A1 255 kW	✓ 1

Common error: The 0.6 is the intercept of the log graph, so it equals log₁₀ a; the constant in the power model is 10⁰.⁶, not 0.6.

+ Add to test✓ In your test61IBMI3H-2-06r33-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The profile of a hill climb on a cycling route is modelled by the graph of $y=f(x)$, which passes through the point (4, −2). A second graph is defined by $g(x)=2f(x-3)+1$.

(a)

Describe a sequence of transformations that maps the graph of $y=f(x)$ onto the graph of $y=g(x)$.

(3)

(b)

Find the coordinates of the image of the point (4, −2) on the graph of y = g(x).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 3 units in the positive x-direction, i.e. by the vector (3, 0)	✓ 1
A1 vertical stretch with scale factor 2	✓ 1
A1 translation 1 unit in the positive y-direction, i.e. by (0, 1), after the stretch (the horizontal translation may be done at any stage)	✓ 1
Part (b)
M1 x = 4 + 3 = 7 or y = 2 × (−2) + 1	✓ 1
A1 (7, −3)	✓ 1

Common error: The vertical stretch must be applied before the vertical translation: stretching after adding 1 would give 2f(x) + 2 instead.

+ Add to test✓ In your test62IBMI3H-2-06r33-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

A bicycle wheel of radius 0.34 m rolls without slipping along level ground at 5 m s⁻¹. A valve on the rim is at its highest point when t = 0. The height of the valve above the ground at time t seconds is modelled by $h=a\cos(bt)+d$.

(a)

Find the values of a, b and d, giving b correct to three significant figures.

(3)

(b)

Find the height of the valve when t = 0.05, correct to three significant figures. Use b = 14.7.

(2)

(c)

Find the first time at which the valve is 0.5 m above the ground, correct to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one revolution takes 2π × 0.34 ÷ 5 = 0.427 s, so b = 2π ÷ 0.427…	✓ 1
A1 b = 14.7	✓ 1
A1 a = 0.34 and d = 0.34	✓ 1
Part (b)
M1 h = 0.34 cos(14.7 × 0.05) + 0.34, with the calculator in radian mode	✓ 1
A1 0.592 m	✓ 1
Part (c)
M1 0.34 cos(14.7t) + 0.34 = 0.5 giving 14.7t = cos⁻¹(0.4706…)	✓ 1
A1 t = 0.0735 s	✓ 1

Common error: The period is the time for one turn of the wheel, so b = 2π ÷ period; check that your calculator is in radian mode before evaluating.

+ Add to test✓ In your test63IBMI3H-2-05r35-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The power P (mW) produced by a solar cell on a space probe depends on its distance d (AU) from the Sun and is modelled by P = ad^b, where a and b are constants. At d = 1.5 the cell produces 40.0 mW and at d = 6.0 it produces 2.50 mW.

(a)

Show that ln P = b ln d + ln a.

(2)

(b)

Find the value of b and the value of a.

(3)

(c)

Estimate the power produced at d = 9.0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 take natural logarithms of both sides of P = ad^b	✓ 1
A1 use ln(xy) = ln x + ln y and ln d^b = b ln d to reach the given form	✓ 1
Part (b)
M1 gradient = (ln 2.50 − ln 40.0) ÷ (ln 6.0 − ln 1.5), or two simultaneous equations in ln a and b	✓ 1
A1 b = −2	✓ 1
A1 a = 90.0 (from 40.0 = a × 1.5⁻²)	✓ 1
Part (c)
M1 substitute d = 9.0 into P = 90.0d⁻²	✓ 1
A1 P = 1.11 mW	✓ 1

Common error: When ln y against ln x is a straight line, the gradient gives the power b but the intercept gives ln a, not a.

+ Add to test✓ In your test64IBMI3H-2-05r35-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The height h metres of a bridge arch above the water, at a horizontal distance x metres from its left base, is modelled by h = −0.02x² + 1.6x, for 0 ≤ x ≤ 80.

(a)

Find the span of the arch at water level.

(2)

(b)

Find the maximum height of the arch.

(2)

(c)

A walkway deck is at a height of 24 m. Find the horizontal width of the arch at this height.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set h = 0, so x(−0.02x + 1.6) = 0	✓ 1
A1 span = 80 m	✓ 1
Part (b)
M1 vertex at x = −1.6 ÷ (2 × −0.02) = 40 (or GDC maximum)	✓ 1
A1 h = 32 m	✓ 1
Part (c)
M1 −0.02x² + 1.6x = 24	✓ 1
M1 solve: x = 20 and x = 60	✓ 1
A1 width = 40 m	✓ 1

Common error: Writing the quadratic as h = 24 and solving gives two x-values; the width is their difference, not either value alone.

+ Add to test✓ In your test65IBMI3H-2-05r35-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy8 marks

Freshly roasted coffee beans leave a roaster at 205 °C and cool in a room at 21 °C. The temperature T (°C) of the beans after t minutes is modelled by T = 21 + 184e^−0.12t.

(a)

Verify that the initial temperature of the beans is 205 °C.

(1)

(b)

Find the temperature of the beans after 5 minutes.

(2)

(c)

State the horizontal asymptote of the graph of T against t and explain its meaning.

(2)

(d)

The beans are safe to pack when their temperature is 40 °C. Find the time at which this occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 at t = 0, T = 21 + 184 = 205 °C	✓ 1
Part (b)
M1 21 + 184e^−0.6	✓ 1
A1 T = 122 °C	✓ 1
Part (c)
A1 T = 21	✓ 1
A1 the beans cool towards, but never go below, room temperature	✓ 1
Part (d)
M1 21 + 184e^−0.12t = 40, so e^−0.12t = 19 ÷ 184	✓ 1
M1 t = ln(184 ÷ 19) ÷ 0.12	✓ 1
A1 t = 18.9 minutes	✓ 1

Common error: Subtract the asymptote value before taking logarithms: it is 184e⁻⁰.¹²t that must equal 19, not the whole expression.

+ Add to test✓ In your test66IBMI3H-2-06r38-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

The gain of a robot's sensor amplifier is defined as G = 20 log₁₀(V_out / V_in) decibels (dB), where V_in and V_out are the input and output voltages.

(a)

An amplifier turns an input of 0.020 V into an output of 0.90 V. Find its gain, to 3 significant figures.

(2)

(b)

Another amplifier has a gain of 28 dB. Find the ratio V_out / V_in, to 3 significant figures.

(2)

(c)

Two amplifiers, each with a gain of 28 dB, are connected in series, so that the output of the first is the input of the second. Find the overall ratio V_out / V_in to the nearest whole number, using unrounded intermediate values.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute the ratio 0.90 / 0.020 = 45	✓ 1
A1 33.1 dB	✓ 1
Part (b)
M1 28 = 20 log₁₀(ratio), so ratio = 10^28/20	✓ 1
A1 25.1	✓ 1
Part (c)
M1 overall gain 56 dB (or ratios multiplied)	✓ 1
M1 10^56/20 (or 25.118...²)	✓ 1
A1 631	✓ 1

Exam tip: On a decibel scale, gains in series add, while the corresponding ratios multiply.

+ Add to test✓ In your test67IBMI3H-2-06r38-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The number of people N following a new fashion trend in a city after t weeks is modelled by N(t) = 2000 / (1 + 19e^−0.6t).

(a)

Find the number of people following the trend at t = 0.

(2)

(b)

Write down the value that N approaches as t becomes very large.

(1)

(c)

Find the time at which 1000 people are following the trend, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0, 2000 / (1 + 19)	✓ 1
A1 100	✓ 1
Part (b)
A1 2000	✓ 1
Part (c)
M1 set 2000 / (1 + 19e^−0.6t) = 1000	✓ 1
M1 19e^−0.6t = 1	✓ 1
A1 t = 4.91 weeks	✓ 1

Exam tip: As t becomes large the exponential term tends to zero, so only the numerator of a logistic function remains.

+ Add to test✓ In your test68IBMI3H-2-06r38-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy7 marks

A fishing company finds that its weekly profit P (in £thousand) when it sends out x boats is modelled by P = −2x² + 36x − 90.

(a)

Find the numbers of boats for which the company breaks even.

(3)

(b)

Find the number of boats that gives the greatest profit.

(2)

(c)

Find the greatest weekly profit, in £thousand.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set P = 0, x² − 18x + 45 = 0	✓ 1
A1 3 boats	✓ 1
A1 15 boats	✓ 1
Part (b)
M1 use of x = −b / 2a (or midpoint of the roots)	✓ 1
A1 9 boats	✓ 1
Part (c)
M1 substitute x = 9	✓ 1
A1 72	✓ 1

Exam tip: The vertex of a quadratic lies halfway between its roots, which gives a quick check on the axis of symmetry.

+ Add to test✓ In your test69IBMI3H-2-05r41-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

Underwater sound is measured using the intensity level L = 10 log₁₀(I / I₀) decibels (dB), where I is the sound intensity in W m⁻² and I₀ = 1.0 × 10⁻¹² W m⁻². A hydrophone on a research vessel records the engine noise of a passing fishing boat.

(a)

The intensity of the noise is 4.0 × 10⁻⁷ W m⁻². Calculate the intensity level in dB, giving your answer to 3 significant figures.

(2)

(b)

A second boat produces a noise with an intensity level 7.0 dB greater than that of the first boat. Find the factor by which the intensity increases, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitution of I / I₀ = 4.0 × 10⁵ into the formula	✓ 1
A1 L = 56.0 dB	✓ 1
Part (b)
M1 10 log₁₀(k) = 7 or k = 10^0.7	✓ 1
A1 factor = 5.01 (accept 5.0)	✓ 1

Common error: An increase of 7 dB does not mean the intensity increases by 7 times; decibels are a logarithmic scale, so equal dB steps correspond to equal multiplying factors.

+ Add to test✓ In your test70IBMI3H-2-05r41-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The number of fish in a farmed tank is modelled by N(t) = 2400 / (1 + 9e^−0.45t), where t is the time in weeks after the tank is stocked.

(a)

Find the number of fish at the start.

(2)

(b)

State the value that N approaches as t becomes very large, and give its meaning in context.

(1)

(c)

Find the time taken for the number of fish to reach 1800. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 N(0) = 2400 / (1 + 9)	✓ 1
A1 240 fish	✓ 1
Part (b)
A1 2400, the maximum number of fish the tank can support	✓ 1
Part (c)
M1 2400 / (1 + 9e^−0.45t) = 1800	✓ 1
M1 e^−0.45t = 1/27 and taking logarithms	✓ 1
A1 t = 7.32 weeks	✓ 1

Common error: When solving for time in a logistic model, rearrange to isolate the exponential term first (here e⁻⁰.⁴⁵t) before taking natural logarithms.

+ Add to test✓ In your test71IBMI3H-2-05r41-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

A sonar survey models a cross-section of a seamount by the graph of y = f(x), where x is the horizontal distance in kilometres and y is the height in hundreds of metres above the surrounding sea floor. The highest point of the graph is (3, 8) and the graph meets the x-axis at the point (5, 0).

(a)

A revised model is g(x) = 2f(x − 1) + 3. Write down the coordinates of the highest point of the graph of y = g(x).

(2)

(b)

Find the coordinates of the image of the point (5, 0) on the graph of y = g(x).

(2)

(c)

A third model is h(x) = −f(2x). Write down the image of the point (3, 8) on the graph of y = h(x).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x-coordinate 4	✓ 1
A1 y-coordinate 19	✓ 1
Part (b)
A1 x-coordinate 6	✓ 1
A1 y-coordinate 3	✓ 1
Part (c)
A1 x-coordinate 1.5	✓ 1
A1 y-coordinate −8	✓ 1

Common error: Changes inside the brackets act on x and work in the opposite way to what they look like: f(x − 1) moves the graph right and f(2x) halves the x-coordinates.

+ Add to test✓ In your test72IBMI3H-2-05r42-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

The power P μW of a radio signal received from a space probe depends on the distance d million km between the probe and the receiving dish. The relationship is of the form P = a d^b. A graph of log₁₀ P against log₁₀ d is a straight line with equation
log₁₀ P = 2.255 − 1.90 log₁₀ d.

(a)

Write down the value of b and find the value of a, correct to 3 significant figures.

(3)

(b)

Find the received power when d = 8.5. Use the unrounded value of a and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 b = −1.90	✓ 1
M1 a = 10^2.255 (or equivalent use of the intercept)	✓ 1
A1 a = 180	✓ 1
Part (b)
M1 P = 10^2.255 × 8.5^−1.90 (or 10^{2.255 − 1.90 log 8.5})	✓ 1
A1 P = 3.08 μW	✓ 1

Common error: To undo a logarithm of base 10, raise 10 to the power of both sides; the intercept of the line is log₁₀ a, not a.

+ Add to test✓ In your test73IBMI3H-2-05r42-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

During roasting, the temperature T °C of a coffee bean after t minutes is modelled by T = 230 − Ae^−kt, where A and k are positive constants. The bean enters the roaster at a temperature of 20 °C. After 3 minutes its temperature is 110 °C.

(a)

Find the value of A.

(2)

(b)

Find the value of k, correct to 3 significant figures.

(2)

(c)

The first crack occurs when the bean reaches 200 °C. Using the unrounded value of k, find the time taken to reach this temperature, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 = 230 − Ae⁰	✓ 1
A1 A = 210	✓ 1
Part (b)
M1 110 = 230 − 210e^−3k, so e^−3k = 120/210	✓ 1
A1 k = 0.187	✓ 1
Part (c)
M1 200 = 230 − 210e^−kt, so e^−kt = 30/210 = 1/7	✓ 1
A1 t = 10.4 minutes	✓ 1

Exam tip: Use the starting condition first to fix the constant that is not a rate, then use the second data point to find the rate.

+ Add to test✓ In your test74IBMI3H-2-05r42-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A small coffee roaster models its weekly profit P euros when it roasts a batch of n kg of beans by P = −0.5n² + 14n − 40, for 0 ≤ n ≤ 28.

(a)

Find the profit when n = 6.

(1)

(b)

Find the batch size that gives the greatest profit, and state this greatest profit.

(2)

(c)

Find the range of values of n for which the weekly profit is positive. Give the limits correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 −0.5(36) + 84 − 40 = €26	✓ 1
Part (b)
M1 n = −14/(2 × −0.5) (or use of GDC maximum)	✓ 1
A1 n = 14 kg and P = €58	✓ 1
Part (c)
M1 solves −0.5n² + 14n − 40 = 0 (or n² − 28n + 80 = 0)	✓ 1
A1 3.23 < n < 24.8	✓ 1

Exam tip: Positive profit means the graph is above the horizontal axis, so the answer to a range question is an interval between the two roots, not the roots themselves.

+ Add to test✓ In your test75IBMI3H-2-05r42-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

A fatigue crack in a bridge girder has length L = 2.5 × 1.07^t millimetres, where t is the time in years since the crack was first detected.

(a)

Find the length of the crack after 10 years, correct to 3 significant figures.

(2)

(b)

The girder must be repaired when the crack reaches 12 mm. Find the time at which this happens, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.5 × 1.07¹⁰	✓ 1
A1 4.92 mm	✓ 1
Part (b)
M1 2.5 × 1.07^t = 12	✓ 1
A1 t = 23.2 years	✓ 1

Exam tip: To find a time in an exponential model, isolate the exponential term first, then use logarithms or the GDC solver.

+ Add to test✓ In your test76IBMI3H-2-06r42-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

Noise in a hospital ward is measured by the sound level L = 10 log₁₀(I/I₀) decibels (dB), where I W m⁻² is the sound intensity and I₀ = 1.0×10⁻¹² W m⁻².

(a)

Find the sound level when the intensity is 4.0×10⁻⁷ W m⁻², correct to 3 significant figures.

(2)

(b)

The ambient level in a quiet ward is 45 dB and a surgical drill produces 76 dB. Find the value of (intensity of the drill) ÷ (ambient intensity), correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10 log₁₀(4.0×10⁻⁷/1.0×10⁻¹²)	✓ 1
A1 56.0 dB	✓ 1
Part (b)
M1 10 log₁₀(ratio) = 76 − 45 = 31, so ratio = 10^3.1	✓ 1
A1 1260	✓ 1

Exam tip: A difference of levels in decibels is 10 times the logarithm of the ratio of intensities, so the reference value cancels.

+ Add to test✓ In your test77IBMI3H-2-06r42-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

After an injection, the concentration of a drug in a patient's blood is C = 120 × 0.82^t mg per litre, where t is the time in hours after the injection.

(a)

Find the concentration after 5 hours, correct to 3 significant figures.

(2)

(b)

The drug is no longer effective when the concentration falls below 10 mg per litre. Find the time at which this happens, correct to 3 significant figures.

(2)

(c)

Find the half-life of the drug, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 120 × 0.82⁵	✓ 1
A1 44.5 mg per litre	✓ 1
Part (b)
M1 120 × 0.82^t = 10	✓ 1
A1 t = 12.5 hours	✓ 1
Part (c)
M1 0.82^t = 0.5	✓ 1
A1 3.49 hours	✓ 1

Exam tip: The half-life does not depend on the starting concentration, so you can solve 0.82t = 0.5 without using 120.

+ Add to test✓ In your test78IBMI3H-2-06r42-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy5 marks

A Formula 1 team models its lap time T seconds as a function of tyre pressure p psi by T = 0.8p² − 34.4p + 451, for 18 ≤ p ≤ 26.

(a)

Find the tyre pressure that gives the shortest lap time.

(2)

(b)

Find the shortest lap time.

(1)

(c)

The team wants a lap time of at most 82.0 s. Find the range of pressures that meets this requirement.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 p = 34.4/(2 × 0.8) (or GDC minimum)	✓ 1
A1 21.5 psi	✓ 1
Part (b)
A1 81.2 s (81.2 exactly)	✓ 1
Part (c)
M1 solves 0.8p² − 34.4p + 451 = 82	✓ 1
A1 20.5 ≤ p ≤ 22.5	✓ 1

Exam tip: A quadratic with a positive leading coefficient has a minimum at its vertex, which lies halfway between the two solutions of any equation T = constant.

+ Add to test✓ In your test79IBMI3H-2-06r42-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

A beekeeper sells jars of honey for €4.50 each. The cost of producing x jars in a year is C = 1200 + 1.85x euros.

(a)

Write down an expression for the annual profit P euros from selling x jars.

(1)

(b)

Find the smallest whole number of jars that must be sold for the beekeeper to make a profit.

(2)

(c)

Find the profit when 800 jars are sold.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 P = 4.50x − 1200 − 1.85x = 2.65x − 1200	✓ 1
Part (b)
M1 2.65x − 1200 > 0, so x > 452.8…	✓ 1
A1 453 jars	✓ 1
Part (c)
A1 €920	✓ 1

Exam tip: A profit requires revenue to exceed cost, and for a whole number of jars you should round up, not to the nearest integer.

+ Add to test✓ In your test80IBMI3H-2-06r44-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

pH is defined by pH = −log₁₀[H⁺], where [H⁺] is the hydrogen-ion concentration in mol dm⁻³. A batch of yoghurt has pH 4.30.

(a)

Calculate the hydrogen-ion concentration of the yoghurt. Give your answer in the form a × 10^k, where 1 ≤ a < 10 and a is given to 3 significant figures.

(2)

(b)

Further fermentation lowers the pH by 0.60. Find the factor by which the hydrogen-ion concentration increases, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 [H⁺] = 10^−4.30	✓ 1
A1 5.01 × 10⁻⁵ mol dm⁻³	✓ 1
Part (b)
M1 factor = 10^0.60 (the new pH is 3.70, so 10^−3.70 ÷ 10^−4.30)	✓ 1
A1 3.98	✓ 1

Common error: A fall of 0.60 in pH is not a 60% change in concentration: each unit of pH is a factor of 10, so the change is 10⁰.⁶⁰.

+ Add to test✓ In your test81IBMI3H-2-06r44-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

The height h (in km) of a ridge above the valley floor at horizontal position x km is modelled by h = f(x). The highest point of the ridge is P(2, 5).

(a)

A second ridge has height h = 2f(x − 3) + 1. Write down the coordinates of the highest point of the second ridge.

(2)

(b)

A third ridge has height h = −f(½x). Write down the coordinates of the image of P.

(2)

(c)

The graph of h = f(x) is reflected in the x-axis and then translated by the vector (−2, 6). Write down the equation of the resulting graph in terms of f.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal translation of 3 to the right gives x = 5	✓ 1
A1 vertical stretch factor 2 then translation up 1 gives height 2 × 5 + 1 = 11, so (5, 11)	✓ 1
Part (b)
A1 horizontal stretch factor 2 gives x = 4	✓ 1
A1 reflection in the x-axis gives (4, −5)	✓ 1
Part (c)
A1 reflection gives h = −f(x)	✓ 1
A1 h = −f(x + 2) + 6	✓ 1

Common error: Inside the brackets, f(x − 3) moves the graph 3 units to the right and f(½x) stretches it horizontally by factor 2; both are the opposite of what the numbers suggest.

+ Add to test✓ In your test82IBMI3H-2-06r44-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy4 marks

A food stall sells jars of preserve at a price of €x per jar. The daily profit €P is modelled by P = −2x² + 36x − 90.

(a)

Find the range of prices for which the stall makes a profit.

(2)

(b)

Find the maximum daily profit predicted by the model.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solve −2x² + 36x − 90 = 0 (or use the GDC)	✓ 1
A1 roots x = 3 and x = 15, so the stall makes a profit for 3 < x < 15	✓ 1
Part (b)
M1 vertex at x = −36 ÷ (2 × −2) = 9	✓ 1
A1 maximum profit €72	✓ 1

Exam tip: The maximum of a quadratic lies halfway between its roots, which provides a quick check on the value found from the vertex formula.

+ Add to test✓ In your test83IBMI3H-2-06r44-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Easy6 marks

On a mountaineering route the temperature falls approximately linearly with height. At Camp 1, at height 2400 m, the temperature is 6.3 °C, and at Camp 3, at height 4100 m, it is −3.9 °C. Let T °C be the temperature at height h metres.

(a)

Find the gradient of the line through the two data points and interpret its value in context.

(3)

(b)

Find the equation of the line in the form T = mh + c.

(1)

(c)

Use the model to estimate the height at which the temperature is 0 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (−3.9 − 6.3) ÷ (4100 − 2400)	✓ 1
A1 −0.006	✓ 1
R1 the temperature falls by 0.006 °C for each additional metre of height (6 °C per km)	✓ 1
Part (b)
A1 T = −0.006h + 20.7	✓ 1
Part (c)
M1 solve −0.006h + 20.7 = 0 (ECF from (b))	✓ 1
A1 3450 m	✓ 1

Exam tip: State the gradient with units and in context: 'per metre' makes the interpretation precise and earns the reasoning mark.

+ Add to test✓ In your test84IBMI-031-00-09Linear models and piecewise models Paper 1Medium8 marks

A print shop charges P(x) dollars to print x posters, where
\(P(x)=\begin{cases}40+2x, & 0The following year all prices increase by 8% and then a fixed $5 administration fee is added to every order. The new cost is given by $Q(x)=1.08P(x)+5$.

(a)

Find the new cost of printing 30 posters.

(2)

(b)

Describe fully the sequence of transformations that maps the graph of P onto the graph of Q.

(2)

(c)

Write down Q(x) as a piecewise function.

(2)

(d)

A customer's bill under the new prices was $221.00. Find the number of posters printed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P(30) = 100 and 1.08 × 100 + 5	✓ 1
A1 $113.00	✓ 1
Part (b)
A1 vertical stretch, scale factor 1.08	✓ 1
A1 followed by a vertical translation of 5 units up	✓ 1
Part (c)
A1 Q(x) = 48.2 + 2.16x for 0 < x ≤ 50	✓ 1
A1 Q(x) = 102.2 + 1.08x for 50 < x ≤ 200	✓ 1
Part (d)
M1 102.2 + 1.08x = 221 (second piece as Q(50) = 156.2 < 221)	✓ 1
A1 110 posters	✓ 1

Common error: The order matters: stretching then translating gives 1.08P(x) + 5, whereas translating first would give 1.08(P(x) + 5), a different function.

+ Add to test✓ In your test85IBMI-031-07-06Exponential and sinusoidal models Paper 1Medium8 marks

A rumour spreads through a town of 4000 people. The number of people, N, who have heard the rumour t days after it starts is modelled by
$N(t) = \frac{4000}{1 + 79e^{-0.6t}}$.

(a)

Find the number of people who knew the rumour at the start.

(1)

(b)

Find the number of people who have heard the rumour after 5 days.

(2)

(c)

Find the time taken for 3000 people to have heard the rumour.

(3)

(d)

Write down the long-term number of people who will have heard the rumour according to this model, and interpret this value in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $N(0) = \frac{4000}{80} = 50$ people	✓ 1
Part (b)
M1: $\frac{4000}{1 + 79e^{-3}}$	✓ 1
A1: 811 people	✓ 1
Part (c)
M1: $\frac{4000}{1 + 79e^{-0.6t}} = 3000$	✓ 1
M1: rearranging to $e^{-0.6t} = \frac{1}{237}$ or using GDC solver/graph intersection	✓ 1
A1: 9.11 days	✓ 1
Part (d)
A1: 4000	✓ 1
R1: eventually (almost) everyone in the town hears the rumour, but the number never exceeds the town population	✓ 1

Exam tip: For a logistic model L/(1 + Ce^{-kt}), the numerator L is the carrying capacity; recognise it directly rather than recalculating.

+ Add to test✓ In your test86IBMI-031-07-09Exponential and sinusoidal models Paper 1Medium7 marks

A pharmacologist measures the mass, M mg, of a drug remaining in a patient's bloodstream t hours after it is given. A graph of ln M against t is approximately a straight line with equation
$\ln M = 4.38 - 0.0275t$.

(a)

Show that M can be written in the form $M = Ae^{kt}$, and find the values of A and k.

(3)

(b)

Find the time taken for the mass of drug in the bloodstream to halve.

(2)

(c)

Find the mass of drug remaining 48 hours after it is given.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $M = e^{4.38 - 0.0275t} = e^{4.38}\times e^{-0.0275t}$	✓ 1
A1: $A = 79.8$ (79.838…)	✓ 1
A1: $k = -0.0275$	✓ 1
Part (b)
M1: $e^{-0.0275t} = 0.5$	✓ 1
A1: 25.2 hours	✓ 1
Part (c)
M1: $79.838\ldots \times e^{-0.0275\times 48}$ or $e^{4.38 - 1.32}$	✓ 1
A1: 21.3 mg	✓ 1

Exam tip: A straight line on a graph of ln y against x signals an exponential model: the gradient is k and the intercept is ln A.

+ Add to test✓ In your test87IBMI-031-04-r1-04Transformations of graphs and logarithmic scales Paper 1Medium8 marks

An ecologist studies the resting metabolic rate, R watts, of mammals with body mass M kg, for masses between 0.02 kg and 4000 kg. A graph of log₁₀ R against log₁₀ M is approximately linear with equation
$\log_{10} R = 0.74\log_{10} M + 0.53$.

(a)

Show that R can be written in the form $R = aM^{b}$, stating the values of a and b.

(3)

(b)

Estimate the resting metabolic rate of a 65 kg mammal.

(2)

(c)

Find the factor by which the body mass of a mammal must increase for its resting metabolic rate to double.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: writes $R = 10^{0.74\log_{10}M + 0.53} = 10^{0.53}\times10^{0.74\log_{10}M}$	✓ 1
M1: uses $10^{0.74\log_{10}M} = M^{0.74}$	✓ 1
A1: a = 3.39 (3.388…), b = 0.74	✓ 1
Part (b)
M1: substitutes, $R = 3.388\times65^{0.74}$	✓ 1
A1: R = 74.4 W	✓ 1
Part (c)
M1: sets up $a(kM)^{0.74} = 2aM^{0.74}$	✓ 1
M1: simplifies to $k^{0.74} = 2$	✓ 1
A1: k = 2.55	✓ 1

Common error: Doubling the rate does not mean doubling the mass. Because R is proportional to M⁰.⁷⁴, the mass must be multiplied by 2¹/⁰.⁷⁴.

+ Add to test✓ In your test88IBMI-031-08-r1-05Choosing and evaluating models Paper 1Medium7 marks

A biologist studies the resting metabolic rate $P$ watts of mammals of mass $M$ kg, for masses between 0.02 kg and 4000 kg. Plotting $\log_{10}P$ against $\log_{10}M$ gives points lying close to the straight line
$\log_{10}P = 0.75\log_{10}M + 0.55$.

(a)

Show that the relationship can be written in the form $P = kM^{n}$, stating the values of $n$ and $k$.

(3)

(b)

Use the model to estimate the resting metabolic rate of a 70 kg mammal.

(2)

(c)

A student uses the model to estimate the metabolic rate of a 0.002 kg insect. Give two reasons why this estimate may be unreliable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses laws of logs: $\log_{10}P = \log_{10}(M^{0.75}) + \log_{10}(10^{0.55})$, so $P = 10^{0.55}M^{0.75}$ M1	✓ 1
$n = 0.75$ A1	✓ 1
$k = 10^{0.55} = 3.55$ A1	✓ 1
Part (b)
Substitutes $M = 70$: $3.548 \times 70^{0.75}$ M1	✓ 1
$P = 85.9$ W A1	✓ 1
Part (c)
Extrapolation: 0.002 kg lies outside the data range 0.02–4000 kg R1	✓ 1
Insects are not mammals, so the relationship found for mammals may not apply R1	✓ 1

Exam tip: On a log–log graph the gradient gives the power $n$ and the intercept gives $\log_{10}k$, not $k$ itself.

+ Add to test✓ In your test89IBMI-031-04-r2-04Transformations of graphs and logarithmic scales Paper 1Medium6 marks

The depth of water at a harbour entrance is modelled by $f(t)=2.4\sin(30t)+5$, where $t$ is the time in hours after midnight and $f(t)$ is the depth in metres. The argument of sine is in degrees.
At a second harbour the depth is modelled by $g(t)=1.25\,f(t-1.5)$.

(a)

Describe fully the two transformations that map the graph of $f$ onto the graph of $g$.

(2)

(b)

Write $g(t)$ in the form $a\sin(b(t-c))+d$.

(2)

(c)

Find the maximum depth at the second harbour and the first time after midnight at which it occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: horizontal translation 1.5 units to the right, i.e. 1.5 hours later	✓ 1
A1: vertical stretch with scale factor 1.25	✓ 1
Part (b)
M1: $1.25\times2.4$ and $1.25\times5$ seen	✓ 1
A1: $g(t)=3\sin(30(t-1.5))+6.25$	✓ 1
Part (c)
A1: maximum depth $3+6.25=9.25$ m	✓ 1
A1: $30(t-1.5)=90$, so $t=4.5$, i.e. 04:30	✓ 1

Common error: A vertical stretch multiplies the principal axis $d$ as well as the amplitude. Only the amplitude changing would give 8 m as the maximum, which is wrong.

+ Add to test✓ In your test90IBMI-031-05-r1-05Linear models and piecewise models Paper 1Medium8 marks

A biologist studies the resting metabolic rate, P watts, of mammals with mass m kg. For mammals with masses between 0.02 kg and 100 kg, a plot of $\ln P$ against $\ln m$ is a straight line with equation
$\ln P = 0.75\ln m + 1.2$.

(a)

Show that the model can be written as $P=am^n$, and find the values of a and n.

(3)

(b)

Use the model to predict the metabolic rate of a mammal of mass 65 kg.

(2)

(c)

Find the mass of a mammal predicted to have a metabolic rate of 500 W.

(2)

(d)

Comment on the reliability of the mass you found for a metabolic rate of 500 W.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Exponentiates both sides, e.g. $P=e^{0.75\ln m+1.2}=e^{1.2}m^{0.75}$ M1 ; $n=0.75$ A1 ; $a=e^{1.2}=3.32$ A1	✓ 3
Part (b)
$P=3.3201\times65^{0.75}$ (or $\ln P=0.75\ln65+1.2$) M1 ; $P=76.0$ W A1	✓ 2
Part (c)
$\ln m=\frac{\ln500-1.2}{0.75}$ (or $500=3.3201m^{0.75}$) M1 ; $m=801$ kg A1	✓ 2
Part (d)
Unreliable, as 801 kg lies outside the 0.02–100 kg range of the data (extrapolation) R1	✓ 1

Common error: The intercept of the log–log line is $\ln a$, not $a$. Leaving $a=1.2$ gives predictions that are wrong by a factor of about 3.

+ Add to test✓ In your test91IBMI-031-04-r1-05Transformations of graphs and logarithmic scales Paper 1Medium8 marks

The depth of water, D metres, at the entrance to a harbour t hours after midnight is modelled by a function of the form $D(t) = a\sin\left(b(t-c)\right) + d$, where angles are measured in degrees.
The graph of D is obtained from the graph of $y = \sin t$ by a vertical stretch with scale factor 1.8, a horizontal stretch with scale factor $\frac{1}{30}$, a horizontal translation of 2 units to the right and a vertical translation of 5.5 units up.

(a)

Write down an expression for D(t).

(3)

(b)

Find the maximum depth of water and the first time after midnight at which it occurs.

(2)

(c)

A cargo boat can only enter the harbour when the depth of water is at least 6.5 m. Find the length of time between 02:00 and 14:00 for which the boat can enter. Give your answer in hours to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: a = 1.8 and d = 5.5	✓ 1
A1: b = 30	✓ 1
A1: c = 2, giving $D(t) = 1.8\sin(30(t-2)) + 5.5$	✓ 1
Part (b)
A1: maximum depth 7.3 m	✓ 1
A1: occurs when 30(t − 2) = 90, i.e. t = 5 (05:00)	✓ 1
Part (c)
M1: sets $1.8\sin(30(t-2)) + 5.5 = 6.5$ or uses GDC intersection	✓ 1
A1: t = 3.12 and t = 6.88 (3.12497… and 6.87503…)	✓ 1
A1: duration = 6.87503… − 3.12497… = 3.75 hours . Accept 3.76 if it comes from subtracting the rounded values 6.88 and 3.12.	✓ 1

Common error: Subtracting rounded intersection times (6.88 − 3.12) gives 3.76. Keep the full GDC values until the final step to get 3.75 hours.

+ Add to test✓ In your test92IBMI-031-03-r1-05Choosing and evaluating models Paper 1Medium8 marks

A biologist records the number of bacteria, $N$, in a culture every hour for $0 \le t \le 8$, where $t$ is the time in hours. She plots $\ln N$ against $t$ and finds that the points lie close to a straight line. The regression line is
$\ln N = 0.231t + 4.62$.
She wants to model the population with $N = Ae^{kt}$.

(a)

Find the value of A and the value of k.

(3)

(b)

Use the model to estimate the number of bacteria when t = 12.

(2)

(c)

Find the time it takes for the number of bacteria to double, according to the model.

(2)

(d)

Comment on the reliability of your estimate in part (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognising ln N = ln A + kt, so ln A = 4.62	✓ 1
A1 A = e^4.62 = 101	✓ 1
A1 k = 0.231	✓ 1
Part (b)
M1 substituting t = 12, e.g. e^{0.231×12 + 4.62} or 101e^0.231×12	✓ 1
A1 N = 1620 (accept answers from 1610 to 1630, e.g. 1615 from using A = 101)	✓ 1
Part (c)
M1 setting up e^0.231t = 2 or t = ln 2 / 0.231	✓ 1
A1 3.00 hours	✓ 1
Part (d)
R1 unreliable, with either of these reasons: t = 12 is outside the data range 0 ≤ t ≤ 8 (extrapolation), or the growth may not stay exponential because nutrients or space run out	✓ 1

Exam tip: A straight line on a graph of ln N against t shows exponential growth: the gradient gives k and the intercept gives ln A, not A.

+ Add to test✓ In your test93IBMI-031-04-r1-03Transformations of graphs and logarithmic scales Paper 1Medium8 marks

A biologist records the number of bacteria, N, in a culture every hour for the first 12 hours of an experiment. She plots ln N against time t hours and finds that the points lie close to the straight line
$\ln N = 0.231t + 4.62$.

(a)

Show that this relationship can be written in the form $N = a\mathrm{e}^{kt}$, and state the values of a and k.

(3)

(b)

Use the model to find the time at which the number of bacteria reaches 1000.

(2)

(c)

Find the time taken for the number of bacteria to double.

(2)

(d)

Comment on the reliability of using the model to predict the number of bacteria after 60 hours.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: exponentiates both sides, $N = \mathrm{e}^{0.231t + 4.62}$	✓ 1
M1: uses laws of indices, $N = \mathrm{e}^{4.62}\times\mathrm{e}^{0.231t}$	✓ 1
A1: a = 101 (101.49…), k = 0.231	✓ 1
Part (b)
M1: sets up $0.231t + 4.62 = \ln 1000$ or $101.49\mathrm{e}^{0.231t} = 1000$	✓ 1
A1: t = 9.90 hours (9.904…)	✓ 1
Part (c)
M1: uses $\mathrm{e}^{0.231T} = 2$ or equivalent, e.g. $T = \frac{\ln 2}{0.231}$	✓ 1
A1: T = 3.00 hours (3.0006…)	✓ 1
Part (d)
R1: states that the prediction is unreliable, with a valid reason: either 60 hours is far outside the 12-hour data range (extrapolation), or limited nutrients and space would slow growth so exponential growth cannot continue	✓ 1

Exam tip: The vertical intercept of a ln N against t line is ln a, not a. Remember to exponentiate it to find the initial value.

+ Add to test✓ In your test94IBMI-031-08-r2-05Choosing and evaluating models Paper 1Medium8 marks

A biologist studies the resting metabolic rate, $P$ watts, of mammals with mass $m$ kg. Data were collected for mammals with masses between 0.02 kg and 60 kg. When $\log_{10}P$ is plotted against $\log_{10}m$ the points lie close to a straight line, and the regression line is
$\log_{10}P=0.75\log_{10}m+0.55$.

(a)

Show that the relationship can be written in the form $P=km^n$, and find the values of $k$ and $n$.

(3)

(b)

According to the model, find the factor by which the metabolic rate increases when the mass of a mammal doubles.

(2)

(c)

Use the model to estimate the metabolic rate of a 250 kg mammal.

(2)

(d)

Comment on the reliability of your estimate for the 250 kg mammal.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $P=10^{0.75\log_{10}m+0.55}=10^{0.55}\times(10^{\log_{10}m})^{0.75}=10^{0.55}\times m^{0.75}$	✓ 1
A1: $n=0.75$	✓ 1
A1: $k=10^{0.55}=3.55$ (3.548…)	✓ 1
Part (b)
M1: $\frac{k(2m)^{0.75}}{km^{0.75}}=2^{0.75}$	✓ 1
A1: 1.68	✓ 1
Part (c)
M1: $3.548\times250^{0.75}$	✓ 1
A1: $P=223$ W	✓ 1
Part (d)
R1: unreliable because 250 kg is outside the data range (0.02–60 kg), so this is extrapolation	✓ 1

Exam tip: A straight line on a log–log graph means a power model $y=kx^n$; a straight line on a semi-log graph (log y against x) means an exponential model. Identify which before rearranging.

+ Add to test✓ In your test95IBMI-031-02-r2-05Exponential and sinusoidal models Paper 1Medium8 marks

A biologist measures the number of bacteria, $N$, in a culture at regular intervals over the first 10 hours of an experiment. She plots $\ln N$ against time $t$ hours and finds that the points lie close to the straight line
$\ln N = 0.231t + 6.12$.

(a)

Show that the data can be modelled by $N = Ae^{kt}$, and find the values of $A$ and $k$.

(3)

(b)

Find the time taken for the number of bacteria to double.

(2)

(c)

Use the model to estimate the time at which the number of bacteria reaches 1.0 × 10⁶.

(2)

(d)

Comment on the reliability of your estimate of the time at which the number of bacteria reaches 1.0 × 10⁶.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Exponentiates both sides: N = e^{0.231t + 6.12} = e^6.12 × e^0.231t, which is of the form Ae^kt	✓ 1
A = e^6.12 = 455 (454.86…)	✓ 1
k = 0.231	✓ 1
Part (b)
Solves e^0.231t = 2, i.e. t = ln 2 ÷ 0.231	✓ 1
t = 3.00 hours	✓ 1
Part (c)
Sets up 0.231t + 6.12 = ln(10⁶) or 455e^0.231t = 10⁶	✓ 1
t = 33.3 hours	✓ 1
Part (d)
Unreliable, with either valid reason: 33.3 h is far outside the 0–10 h data range (extrapolation), OR growth is likely to slow as nutrients or space run out	✓ 1

Exam tip: The gradient of a ln N against t graph gives k directly, but the intercept is ln A, not A. Exponentiate the intercept before writing down A.

+ Add to test✓ In your test96IBMI3H-2-04-03Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The number of views, V, of a video d days after it is posted is modelled by V = ab^d. A plot of ln V against d is a straight line with equation ln V = 1.9 + 0.42d.

(a)

Find the value of a and the value of b.

(3)

(b)

Interpret the value of b in this context.

(1)

(c)

Find the value of d for which the model predicts 10 000 views.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 V = e^(1.9 + 0.42d) = e^1.9 × (e^0.42)^d	✓ 1
A1 a = 6.69	✓ 1
A1 b = 1.52	✓ 1
Part (b)
R1 each day the number of views is multiplied by 1.52, an increase of about 52% per day	✓ 1
Part (c)
M1 ln 10000 = 1.9 + 0.42d	✓ 1
M1 d = (ln 10000 − 1.9)/0.42	✓ 1
A1 d = 17.4 days	✓ 1

Common error: Writing a = 1.9 instead of a = e¹.⁹; the intercept of the line is ln a, not a.

+ Add to test✓ In your test97IBMI3H-2-04-04Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The metabolic rate R watts of a mammal of mass M kg is modelled by R = kM^p. A graph of log₁₀ R against log₁₀ M is a straight line through the points (0.30, 1.68) and (1.70, 2.73).

(a)

Find the gradient of the line.

(2)

(b)

Find the value of p and the value of k.

(4)

(c)

Estimate the metabolic rate of a mammal of mass 70 kg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2.73 − 1.68)/(1.70 − 0.30)	✓ 1
A1 0.75	✓ 1
Part (b)
M1 log10 R = 0.75 log10 M + c, then substitute a point	✓ 1
A1 c = 1.455	✓ 1
M1 R = 10^c × M^0.75	✓ 1
A1 p = 0.75 and k = 28.5	✓ 1
Part (c)
M1 28.5 × 70^0.75	✓ 1
A1 690 W	✓ 1

Exam tip: On a log-log graph the gradient is the power p and the intercept is log₁₀ k; convert back with 10 to the power of the intercept.

+ Add to test✓ In your test98IBMI3H-2-04-05Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The graph of y = f(x) models the cross-section height (in metres) of a skateboard bowl. The graph of y = g(x), where g(x) = 0.5f(x − 2) + 4, models a scaled version of the bowl on a raised platform.

(a)

Describe, in order, the transformations that map the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

The point (6, 10) lies on y = f(x). Find the coordinates of its image on y = g(x).

(2)

(c)

The point (10, 7) lies on y = g(x). Find the coordinates of the corresponding point on y = f(x).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 vertical stretch with scale factor 0.5	✓ 1
A1 translation of 2 units to the right	✓ 1
A1 translation of 4 units up	✓ 1
Part (b)
A1 x-coordinate 8	✓ 1
A1 y-coordinate 0.5 × 10 + 4 = 9, so (8, 9)	✓ 1
Part (c)
M1 reverse the transformations: x − 2 = 8 and (7 − 4)/0.5	✓ 1
A1 (8, 6)	✓ 1

Common error: Treating f(x − 2) as a shift to the left; replacing x by x − 2 moves the graph right.

+ Add to test✓ In your test99IBMI3H-2-04-08Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The number of users, N, of a messaging app m months after launch is modelled by N = ab^m. A graph of log₁₀ N against m is a straight line with gradient 0.06 and vertical intercept 3.2.

(a)

Find the value of a and the value of b.

(3)

(b)

Find the time taken for the number of users to double.

(2)

(c)

Find when the model predicts 1 000 000 users.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 log10 N = 3.2 + 0.06m, so N = 10^3.2 × (10^0.06)^m	✓ 1
A1 a = 1580	✓ 1
A1 b = 1.15	✓ 1
Part (b)
M1 0.06m = log10 2	✓ 1
A1 5.02 months	✓ 1
Part (c)
M1 3.2 + 0.06m = 6	✓ 1
A1 m = 46.7 months	✓ 1

Common error: Using ln instead of log₁₀ when undoing a base-10 graph axis; the base must match the one given in the question.

+ Add to test✓ In your test100IBMI3H-2-04-09Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The energy E joules released by an earthquake of magnitude M satisfies log₁₀ E = 4.8 + 1.5M.

(a)

Find the energy released by an earthquake of magnitude 5.5.

(2)

(b)

Find the factor by which the energy increases when the magnitude increases by 1.

(2)

(c)

An earthquake releases 3.0 × 10¹⁵ J. Find its magnitude.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 log10 E = 4.8 + 1.5 × 5.5 = 13.05	✓ 1
A1 1.12 × 10¹³ J	✓ 1
Part (b)
M1 10^1.5	✓ 1
A1 31.6	✓ 1
Part (c)
M1 log10(3.0×10¹⁵) = 4.8 + 1.5M	✓ 1
A1 M = 7.12	✓ 1

Exam tip: A fixed increase on a logarithmic scale corresponds to a fixed multiplication factor, not a fixed addition, in the original quantity.

+ Add to test✓ In your test101IBMI3H-2-04b-02Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The metabolic rate R watts of marine mammals of mass m kg follows a power model R = am^k. A graph of ln R against ln m is a straight line passing through the points (2.0, 1.1) and (5.0, 3.35).

(a)

Find the gradient and the vertical intercept of the line.

(2)

(b)

Hence write down the values of k and a.

(2)

(c)

Estimate the metabolic rate of a marine mammal of mass 350 kg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient (3.35 − 1.1) ÷ (5.0 − 2.0) = 0.75	✓ 1
A1 intercept 1.1 − 0.75 × 2.0 = −0.4	✓ 1
Part (b)
A1 k = 0.75	✓ 1
A1 a = e^−0.4 = 0.670	✓ 1
Part (c)
M1 substitutes m = 350 into R = e^−0.4m^0.75	✓ 1
A1 R = 54.2 watts	✓ 1

Exam tip: For ln R against ln m the gradient is the power k, and the intercept is ln a. Do not mix these up with the semi-log case.

+ Add to test✓ In your test102IBMI3H-2-04b-03Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

An architect models the cross-section of a curved canopy by the graph of y = f(x), where f(x) = 8x − x² for 0 ≤ x ≤ 8, with lengths in metres. The vertex of the graph of f is (4, 16). A second canopy has the graph of g(x) = 0.5f(x − 3) + 2.

(a)

Describe the sequence of transformations that maps the graph of f onto the graph of g.

(3)

(b)

Find the coordinates of the vertex of the graph of g.

(2)

(c)

A third canopy has the graph of h(x) = f(x ÷ 2). Find the width of its base, which is the distance between the two x-intercepts.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 3 units in the positive x direction	✓ 1
A1 vertical stretch with scale factor 0.5	✓ 1
A1 translation 2 units in the positive y direction	✓ 1
Part (b)
M1 applies the transformations to (4, 16), giving (7, 0.5 × 16 + 2)	✓ 1
A1 (7, 10)	✓ 1
Part (c)
M1 recognises a horizontal stretch with scale factor 2, so the intercepts (0, 0) and (8, 0) move to (0, 0) and (16, 0)	✓ 1
A1 width 16 m	✓ 1

Common error: Applying a horizontal transformation the wrong way round. f(x − 3) moves the graph right, and f(x ÷ 2) stretches it away from the y-axis.

+ Add to test✓ In your test103IBMI3H-2-04b-08Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The number of fish P in a protected bay is modelled by P = 5000 ÷ (1 + Ce^−0.42t), where t is the time in years since the bay was stocked with 400 fish.

(a)

Find the value of C.

(2)

(b)

State the value that the population approaches in the long term, and what it represents.

(2)

(c)

Find the time at which the model predicts the population reaches 2500.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5000 ÷ (1 + C) = 400	✓ 1
A1 C = 11.5	✓ 1
Part (b)
A1 5000 fish	✓ 1
R1 the carrying capacity: the largest population the bay can sustain	✓ 1
Part (c)
M1 sets 5000 ÷ (1 + 11.5e^−0.42t) = 2500	✓ 1
M1 11.5e^−0.42t = 1	✓ 1
A1 t = 5.82 years	✓ 1

Exam tip: For a logistic model the denominator equals 2 when the output is half the limit. This often shortens the algebra.

+ Add to test✓ In your test104IBMI3H-2-04b-11Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A seawater sample at 70 °C is left to cool in a laboratory. Its temperature T °C after t minutes is modelled by T = 18 + 52e^−0.11t.

(a)

Verify that the model gives 70 °C at t = 0.

(1)

(b)

Write down the equation of the horizontal asymptote of the graph and interpret it.

(2)

(c)

Find the time taken for the sample to cool to 30 °C.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 T(0) = 18 + 52 = 70	✓ 1
Part (b)
A1 T = 18	✓ 1
R1 the sample cools towards the room temperature of 18 °C but never goes below it	✓ 1
Part (c)
M1 sets 18 + 52e^−0.11t = 30	✓ 1
M1 e^−0.11t = 12 ÷ 52	✓ 1
A1 t = 13.3 minutes	✓ 1

Common error: Taking logarithms of both sides before isolating the exponential term. First subtract 18, then divide by 52.

+ Add to test✓ In your test105IBMI3H-2-04b-13Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A ferry company runs extra sailings when daylight lasts longer than 15 hours. On day t of the year (t = 1 on 1 January) the hours of daylight at a northern port are modelled by D = 12.2 + 5.1 sin(2π(t − 80) ÷ 365).

(a)

Write down the greatest number of hours of daylight predicted by the model.

(1)

(b)

Find the hours of daylight on day 200.

(2)

(c)

Find the number of days in a year on which the ferry company runs extra sailings.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 12.2 + 5.1 = 17.3 hours	✓ 1
Part (b)
M1 substitutes t = 200 with the calculator in radian mode	✓ 1
A1 16.7 hours	✓ 1
Part (c)
M1 sets 12.2 + 5.1 sin(2π(t − 80) ÷ 365) = 15	✓ 1
A1 sin(…) = 2.8 ÷ 5.1, giving 2π(t − 80) ÷ 365 = 0.581 and π − 0.581	✓ 1
M1 uses the difference of the two solutions, or the symmetry of the sine curve	✓ 1
A1 115 days	✓ 1

Common error: Stopping after the first solution of the sine equation. The interval above 15 hours lies between the two solutions in each cycle.

+ Add to test✓ In your test106IBMI3H-2-04b-14Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The power output P kW of a wind turbine at wind speed v m/s is modelled by
$P(v)=\begin{cases}0.45v^{3}, & 3\le v\le 11\\ 600, & 11< v\le 25\end{cases}$

(a)

Calculate the power output at a wind speed of 8 m/s.

(2)

(b)

Show that the two pieces of the model are very nearly equal at v = 11.

(2)

(c)

Find the range of wind speeds for which the power output is at least 300 kW.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes v = 8 into 0.45v³	✓ 1
A1 230 kW (230.4)	✓ 1
Part (b)
M1 0.45 × 11³ = 598.95	✓ 1
A1 598.95 ≈ 600, a difference of about 1.05 kW	✓ 1
Part (c)
M1 sets 0.45v³ = 300	✓ 1
A1 v = 8.74 m/s	✓ 1
A1 8.74 ≤ v ≤ 25, as the output is 600 kW from 11 m/s upward	✓ 1

Exam tip: For a piecewise model, test each piece in turn and check that the answer lies in the correct interval before combining the results.

+ Add to test✓ In your test107IBMI3H-2-03r5-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

Seismic tremors beneath a volcano are rated on a logarithmic scale. A tremor with maximum ground amplitude A millimetres is given the magnitude M = log₁₀(A) + 3.

(a)

A tremor has a maximum amplitude of 2.6 mm. Calculate its magnitude, correct to 3 significant figures.

(2)

(b)

A second tremor has magnitude 4.7. Find its maximum amplitude, in mm, correct to 3 significant figures.

(3)

(c)

A third tremor has magnitude 6.1. Find how many times greater its amplitude is than that of the tremor in part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes A = 2.6 into the formula	✓ 1
A1 M = log₁₀(2.6) + 3 = 3.41	✓ 1
Part (b)
M1 4.7 = log₁₀(A) + 3	✓ 1
M1 log₁₀(A) = 1.7, so A = 10^1.7	✓ 1
A1 A = 50.1 mm	✓ 1
Part (c)
M1 ratio = 10^6.1−3 ÷ 10^4.7−3 = 10^1.4	✓ 1
A1 25.1	✓ 1

Exam tip: A difference of 1 on a base-10 scale means a factor of 10 in the measured quantity, so subtract the magnitudes first and then raise 10 to that power.

+ Add to test✓ In your test108IBMI3H-2-03r5-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The thickness, T mm, of volcanic ash on the ground at a distance d km from a vent is recorded at several sites. A graph of ln T against d is a straight line with equation ln T = 5.20 − 0.082d.

(a)

Show that T can be written in the form T = ae^−kd, and find the values of a and k. Give a to 3 significant figures.

(3)

(b)

State what the value of a represents in this context.

(1)

(c)

Calculate the predicted thickness at a site 30 km from the vent. Give your answer to 3 significant figures.

(2)

(d)

Find the distance from the vent at which the predicted thickness is 1 mm. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 T = e^{5.20 − 0.082d} = e^5.20 × e^−0.082d	✓ 1
A1 a = e^5.20 = 181	✓ 1
A1 k = 0.082	✓ 1
Part (b)
R1 the predicted thickness (in mm) at the vent, where d = 0	✓ 1
Part (c)
M1 ln T = 5.20 − 0.082 × 30 = 2.74	✓ 1
A1 T = 15.5 mm	✓ 1
Part (d)
M1 ln 1 = 0, so 0 = 5.20 − 0.082d	✓ 1
A1 d = 63.4 km	✓ 1

Exam tip: On a semi-log graph the intercept is the logarithm of the starting value, so you must take the exponential of it before quoting the thickness at d = 0.

+ Add to test✓ In your test109IBMI3H-2-03r5-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The depth of water, D metres, at a harbour berth t hours after a high tide is modelled by D = a cos(bt) + c, where b > 0 and the angle is in radians. At high tide the depth is 9.4 m and at low tide it is 3.0 m. Successive high tides are 12.4 hours apart.

(a)

Find the value of a and the value of c.

(2)

(b)

Find the value of b, correct to 3 significant figures.

(2)

(c)

A ship needs a depth greater than 8.0 m to pass the berth safely. Find the length of time, in hours, in one tidal cycle during which the ship can pass. Use the unrounded value of b and give your answer to 3 significant figures. (ECF from (a))

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (9.4 − 3.0) ÷ 2 = 3.2	✓ 1
A1 c = (9.4 + 3.0) ÷ 2 = 6.2	✓ 1
Part (b)
M1 period = 2π ÷ b = 12.4	✓ 1
A1 b = 0.507	✓ 1
Part (c)
M1 3.2 cos(bt) + 6.2 = 8, so cos(bt) = 0.5625	✓ 1
A1 first time t = 1.92 (1.921…) hours after high tide	✓ 1
M1 the depth exceeds 8.0 m from −1.921 to +1.921 hours (symmetry about high tide)	✓ 1
A1 3.84 hours	✓ 1

Common error: Check that the calculator is in radian mode when the model uses b in radians; degree mode gives a completely different time.

+ Add to test✓ In your test110IBMI3H-2-03r5-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A theatre sells N = 420 − 12p tickets per night when the ticket price is £p, for 0 < p < 35.

(a)

Show that the nightly revenue, £R, is given by R = 420p − 12p².

(1)

(b)

Find the two ticket prices for which the nightly revenue is £3000.

(3)

(c)

Find the ticket price that gives the greatest nightly revenue, and the value of that revenue.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 R = p × N = p(420 − 12p) = 420p − 12p² (AG)	✓ 1
Part (b)
M1 420p − 12p² = 3000	✓ 1
M1 solves the quadratic (graphing calculator or p² − 35p + 250 = 0)	✓ 1
A1 p = 10 and p = 25	✓ 1
Part (c)
M1 vertex at p = 420 ÷ (2 × 12) (or midpoint of 10 and 25)	✓ 1
A1 p = £17.50	✓ 1
A1 maximum revenue £3675	✓ 1

Exam tip: For a quadratic model with two prices giving the same value, the maximum lies exactly halfway between them.

+ Add to test✓ In your test111IBMI3H-2-03r5-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The graph of y = f(x), where f(x) = 2^x, models the growth of marine fouling on a ship's hull. After a hull treatment the model becomes y = g(x), where g(x) = 5 − 2^x/2.

(a)

Describe, in order, a sequence of three transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

The point (3, 8) lies on y = f(x). Find the coordinates of its image on y = g(x).

(2)

(c)

Write down the equation of the horizontal asymptote of y = g(x).

(1)

(d)

Solve g(x) = 0. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal stretch with scale factor 2	✓ 1
A1 reflection in the x-axis	✓ 1
A1 translation 5 units in the positive y direction	✓ 1
Part (b)
A1 x-coordinate 6	✓ 1
A1 y-coordinate −8 + 5 = −3, so (6, −3)	✓ 1
Part (c)
A1 y = 5	✓ 1
Part (d)
M1 2^x/2 = 5, so x ÷ 2 = log₂ 5	✓ 1
A1 x = 4.64	✓ 1

Exam tip: Apply transformations to a point step by step in the same order as the equation is built; the asymptote is moved only by vertical translations and reflections.

+ Add to test✓ In your test112IBMI3H-2-04r5-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The size of a volcanic eruption is rated by the magnitude m = log₁₀(M) − 7, where M kg is the mass of material erupted.

(a)

An eruption releases 8.0 × 10¹⁰ kg of material. Calculate its magnitude, correct to 3 significant figures.

(2)

(b)

Another eruption has magnitude 5.2. Find the mass erupted, in kg, correct to 3 significant figures.

(2)

(c)

A third eruption has magnitude 6.0. Find how many times greater its erupted mass is than that of the eruption in part (b), correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 m = log₁₀(8.0 × 10¹⁰) − 7	✓ 1
A1 m = 3.90	✓ 1
Part (b)
M1 5.2 = log₁₀(M) − 7, so log₁₀(M) = 12.2	✓ 1
A1 M = 10^12.2 = 1.58 × 10¹² kg	✓ 1
Part (c)
M1 ratio = 10^{6.0 + 7} ÷ 10^{5.2 + 7} = 10^0.8	✓ 1
A1 6.31	✓ 1

Common error: A magnitude difference of 0.8 does not mean 0.8 times the mass; it means a factor of 10⁰.⁸.

+ Add to test✓ In your test113IBMI3H-2-04r5-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The cargo handled by a port is T thousand tonnes in year t, where t = 0 is the start of 2020. The values are T = 52 when t = 1 and T = 73 when t = 6. A graph of ln T against t is a straight line, so T = ae^kt.

(a)

Show that ln T = ln a + kt.

(2)

(b)

Find the value of k, correct to 3 significant figures.

(3)

(c)

Find the time, in years, for the cargo handled to double. Use the unrounded value of k and give your answer to 3 significant figures.

(2)

(d)

State one assumption made by this model.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes natural logarithms of both sides	✓ 1
A1 ln T = ln a + ln(e^kt) = ln a + kt (AG)	✓ 1
Part (b)
M1 gradient = (ln 73 − ln 52) ÷ (6 − 1)	✓ 1
M1 = 0.3392… ÷ 5	✓ 1
A1 k = 0.0678	✓ 1
Part (c)
M1 e^kt = 2, so t = ln 2 ÷ k	✓ 1
A1 10.2 years	✓ 1
Part (d)
R1 the percentage growth in cargo per year stays constant (no limit on capacity)	✓ 1

Exam tip: On a semi-log graph the gradient is the continuous growth rate k; the doubling time does not depend on the starting value.

+ Add to test✓ In your test114IBMI3H-2-04r5-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The function f models the cross-section of a crater floor, with domain 0 ≤ x ≤ 6 and range −2 ≤ y ≤ 5. Here x and y are measured in hundreds of metres. The function g is defined by g(x) = 3f(x ÷ 2) − 4 and the function h by h(x) = 2 − f(x + 1).

(a)

Write down the domain and the range of g.

(3)

(b)

Write down the domain and the range of h.

(3)

(c)

Given that f(4) = 1, find the coordinates of the point on the graph of g that corresponds to this point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 domain 0 ≤ x ≤ 12	✓ 1
A1 least value 3 × (−2) − 4 = −10	✓ 1
A1 greatest value 3 × 5 − 4 = 11, so range −10 ≤ y ≤ 11	✓ 1
Part (b)
A1 domain −1 ≤ x ≤ 5	✓ 1
A1 greatest value 2 − (−2) = 4	✓ 1
A1 least value 2 − 5 = −3, so range −3 ≤ y ≤ 4	✓ 1
Part (c)
M1 x ÷ 2 = 4, so x = 8	✓ 1
A1 g(8) = 3 × 1 − 4 = −1, so (8, −1)	✓ 1

Common error: A reflection swaps the greatest and least values: for h the greatest value comes from the least value of f.

+ Add to test✓ In your test115IBMI3H-2-04r5-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A lava fragment is thrown from a vent that is 120 m above sea level. Its height above sea level is modelled by h = −4.9t² + 45t + 120 metres, where t is the time in seconds after it is thrown.

(a)

Find the greatest height reached and the time at which it is reached. Give both to 3 significant figures.

(3)

(b)

Find the time at which the fragment lands in the sea. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 vertex at t = 45 ÷ (2 × 4.9)	✓ 1
A1 t = 4.59 s	✓ 1
A1 greatest height 223 m	✓ 1
Part (b)
M1 sets h = 0	✓ 1
M1 solves using the quadratic formula or a graphing calculator, giving t = 11.3… or t = −2.16…	✓ 1
R1 rejects the negative root as the model applies only for t ≥ 0	✓ 1
A1 t = 11.3 s	✓ 1

Exam tip: Always say why one root is rejected: a negative time lies before the fragment was thrown.

+ Add to test✓ In your test116IBMI3H-2-04r5-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The depth, T cm, of ash on the ground at a distance d km from a volcano is modelled by T = a + b ln d. The depth is 12.1 cm at 2 km and 7.3 cm at 10 km.

(a)

Find the value of b, correct to 4 significant figures.

(3)

(b)

Find the value of a, correct to 4 significant figures.

(2)

(c)

Using a = 14.17 and b = −2.982, find the predicted depth at 40 km. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12.1 = a + b ln 2 and 7.3 = a + b ln 10	✓ 1
M1 subtracts: −4.8 = b(ln 10 − ln 2) = b ln 5	✓ 1
A1 b = −2.982	✓ 1
Part (b)
M1 a = 12.1 − b ln 2	✓ 1
A1 a = 14.17	✓ 1
Part (c)
M1 T = 14.17 − 2.982 ln 40	✓ 1
A1 T = 3.17 cm	✓ 1

Exam tip: Subtracting two equations removes the constant a, and ln 10 − ln 2 combines to ln 5, which keeps the working exact.

+ Add to test✓ In your test117IBMI3H-2-04r6-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

Calculator

The apparent magnitude m of a variable star is modelled by m(t) = a cos(bt) + d, where t is the time in days since the star was at its dimmest point (a larger magnitude means a dimmer star). The magnitude varies between 8.2 and 9.6, and the cycle repeats every 6.4 days.

(a)

Find the values of a, b and d. Give b to 3 significant figures.

(3)

(b)

In each cycle the star is dimmer than magnitude 9.4 for a period of time. Find the total length of this period. Give your answer in days to 3 significant figures, using the unrounded value of b.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = 0.7 (amplitude (9.6 − 8.2) ÷ 2)	✓ 1
A1 d = 8.9 (midline (9.6 + 8.2) ÷ 2)	✓ 1
A1 b = 2π ÷ 6.4 = 0.982	✓ 1
Part (b)
M1 sets up 0.7 cos(bt) + 8.9 > 9.4	✓ 1
M1 solves cos(bt) > 5/7 to find one boundary time at t = 0.790… days (by graph or inverse cosine)	✓ 1
A1 symmetry about t = 0 and t = 6.4, so total period = 2 × 0.790… = 1.58 days	✓ 1

Exam tip: Sketch one cycle of the graph and shade where the inequality holds, then use the symmetry of the cosine curve to find the total time.

+ Add to test✓ In your test118IBMI3H-2-04r6-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

Calculator

After a sprint, an athlete's heart rate H beats per minute at t minutes after stopping is modelled by H(t) = 70 + 110e^−kt, where k is a positive constant. Two minutes after stopping the heart rate is 100 beats per minute.

(a)

Write down the heart rate at the moment the athlete stops.

(1)

(b)

Find the value of k. Give your answer to 3 significant figures.

(3)

(c)

Find the time taken for the heart rate to fall to 75 beats per minute. Use the unrounded value of k and give your answer in minutes to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 H(0) = 70 + 110 = 180 beats per minute	✓ 1
Part (b)
M1 forms 100 = 70 + 110e^−2k	✓ 1
M1 isolates the exponential: e^−2k = 30/110 and takes natural logarithms	✓ 1
A1 k = 0.650	✓ 1
Part (c)
M1 sets 75 = 70 + 110e^−kt, so e^−kt = 5/110	✓ 1
A1 t = ln 22 ÷ k = 4.76 minutes	✓ 1

Common error: Subtract the horizontal asymptote (70) before taking logarithms; taking logs of 100 directly gives a wrong equation.

+ Add to test✓ In your test119IBMI3H-2-04r6-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

Calculator

A javelin is thrown from a point above the ground. Its height h metres when it has travelled x metres horizontally is modelled by h = −0.05x² + 1.1x + 1.5, for x ≥ 0.

(a)

Write down the height from which the javelin is thrown.

(1)

(b)

Find the maximum height of the javelin and the horizontal distance at which it occurs.

(2)

(c)

Find the horizontal distance travelled when the javelin lands. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 1.5 m (the value of h when x = 0)	✓ 1
Part (b)
A1 horizontal distance = 11 m (using x = −b ÷ 2a or a graph)	✓ 1
A1 maximum height = 7.55 m	✓ 1
Part (c)
M1 sets h = 0	✓ 1
M1 solves the quadratic using a graph or the formula	✓ 1
A1 x = 23.3 m (the positive root only)	✓ 1

Common error: The negative root of the quadratic is outside the domain x ≥ 0, so it must be rejected in context.

+ Add to test✓ In your test120IBMI3H-2-04r6-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The graph of y = f(x) models a force–time curve for a sprinter's start. It has a minimum point at (4, −3) and passes through the point (0, 5). A second curve is defined by g(x) = −2f(x ÷ 3) + 1.

(a)

Describe, in order, a sequence of transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

Write down the coordinates of the turning point of the graph of g, and state whether it is a maximum or a minimum.

(2)

(c)

Find the y-intercept of the graph of g.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal stretch with scale factor 3 (parallel to the x-axis, from x ÷ 3)	✓ 1
A1 vertical stretch with scale factor 2 and a reflection in the x-axis	✓ 1
A1 translation of 1 unit up, by the vector (0, 1)	✓ 1
Part (b)
A1 (12, 7)	✓ 1
R1 the reflection in the x-axis turns the minimum into a maximum	✓ 1
Part (c)
A1 g(0) = −2 × 5 + 1 = −9	✓ 1

Common error: Inside the bracket, f(x ÷ 3) stretches the graph horizontally by 3 (not 1/3), and the order of the vertical steps matters: stretch and reflect before translating.

+ Add to test✓ In your test121IBMI3H-2-03r10-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The sound level L decibels (dB) of a noise of intensity I W m⁻² is given by L = 10 log₁₀(I / I₀), where I₀ = 1.0 × 10⁻¹² W m⁻². A library measures a sound level of 42 dB in its reading room and 61 dB in its group-discussion area.

(a)

Find the intensity of the sound in the reading room. Give your answer in the form a × 10^k where 1 ≤ a < 10 and a is given to 3 significant figures.

(2)

(b)

Find how many times greater the intensity in the group-discussion area is than in the reading room. Give your answer to 3 significant figures.

(2)

(c)

A third area has an intensity 5 times that of the group-discussion area. Find its sound level, in dB, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
42 = 10 log₁₀(I / 10⁻¹²), so I / 10⁻¹² = 10^4.2	✓ 1
I = 1.58 × 10⁻⁸ W m⁻²	✓ 1
Part (b)
difference in levels = 19 dB, so 10 log₁₀(ratio) = 19	✓ 1
ratio = 10^1.9 = 79.4	✓ 1
Part (c)
increase in level = 10 log₁₀ 5 = 6.99 (6.9897...)	✓ 1
L = 61 + 6.99 (or 10 log₁₀(5I/I₀))	✓ 1
L = 68.0 dB	✓ 1

Exam tip: Multiplying the intensity by a fixed factor adds a fixed number of decibels, whatever the starting level.

+ Add to test✓ In your test122IBMI3H-2-03r10-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The concentration C mg L⁻¹ of chlorine in a swimming pool is modelled by C = C₀e^−kt, where t is the time in hours after the pool is treated and C₀ and k are positive constants. A graph of ln C against t is a straight line passing through the points (2, 2.84) and (6, 2.12).

(a)

Find the gradient of the line and its vertical intercept.

(2)

(b)

Hence find the value of C₀ and the value of k. Give C₀ to 3 significant figures.

(2)

(c)

The pool must be treated again when the concentration falls to 5 mg L⁻¹. Find the time, in hours, after treatment at which this happens. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
gradient = (2.12 − 2.84)/(6 − 2) = −0.18	✓ 1
intercept: 2.84 + 0.18 × 2 = 3.2	✓ 1
Part (b)
ln C = ln C₀ − kt, so C₀ = e^3.2 = 24.5 mg L⁻¹	✓ 1
k = 0.18	✓ 1
Part (c)
3.2 − 0.18t = ln 5 (or 5 = 24.53e^−0.18t)	✓ 1
t = 8.84 hours	✓ 1

Common error: In ln C = ln C₀ − kt the gradient is −k, so a negative gradient gives a positive decay constant.

+ Add to test✓ In your test123IBMI3H-2-03r10-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The curve y = f(x) for 0 ≤ x ≤ 10 models the depth, in metres, of the floor of a swimming pool at a horizontal distance x metres from one end. The deepest point of the pool is at (4, 3.5). A designer considers three new curves based on f.

(a)

For the curve y = f(x − 3), describe the transformation and write down the coordinates of the image of the deepest point.

(2)

(b)

For the curve y = 2f(x) + 1, describe the sequence of transformations and write down the coordinates of the image of the deepest point.

(2)

(c)

For the curve y = f(x/2), describe the transformation and write down the coordinates of the image of the deepest point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
translation 3 units in the positive x-direction (vector (3, 0))	✓ 1
image point (7, 3.5)	✓ 1
Part (b)
vertical stretch with scale factor 2, then translation 1 unit in the positive y-direction	✓ 1
image point (4, 8)	✓ 1
Part (c)
horizontal stretch with scale factor 2 (parallel to the x-axis)	✓ 1
image point (8, 3.5)	✓ 1

Common error: A change inside the brackets acts on x and does the opposite of what it seems: f(x − 3) moves right and f(x/2) stretches by 2.

+ Add to test✓ In your test124IBMI3H-2-03r10-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium4 marks

The acidity of pool water is measured by pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol L⁻¹. Pool A has pH 7.2 and pool B has pH 7.8.

(a)

Find the value of [H⁺] in pool A as a multiple of that in pool B. Give your answer to 3 significant figures.

(2)

(b)

Acid is added to pool B so that its [H⁺] triples. Find the new pH of pool B, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
difference in pH = 0.6	✓ 1
ratio = 10^0.6 = 3.98	✓ 1
Part (b)
new pH = −log₁₀(3[H⁺]) = 7.8 − log₁₀ 3 (using log laws)	✓ 1
7.8 − 0.4771 = 7.32	✓ 1

Exam tip: A lower pH means a higher concentration. Each drop of 1 on the pH scale multiplies [H+] by 10.

+ Add to test✓ In your test125IBMI3H-2-03r10-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The total number N of turbines installed in a region t years after 2010 is modelled by N = ab^t. A graph of log₁₀ N against t is a straight line with gradient 0.047 and vertical intercept 1.2.

(a)

Find a and b, each to 3 significant figures.

(3)

(b)

Interpret the value of b in this context.

(1)

(c)

Use the model to predict the number of turbines in 2025. Give your answer to the nearest whole number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
log₁₀ N = log₁₀ a + t log₁₀ b, so log₁₀ a = 1.2 and log₁₀ b = 0.047	✓ 1
a = 15.8	✓ 1
b = 1.11	✓ 1
Part (b)
the number of turbines increases by about 11% each year	✓ 1
Part (c)
log₁₀ N = 1.2 + 0.047 × 15 = 1.905	✓ 1
N = 10^1.905 = 80.35... so 80 turbines	✓ 1

Common error: Always read the unit of the gradient: here it is a change in log₁₀ N per year, not a change in N.

+ Add to test✓ In your test126IBMI3H-2-03r10-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A hot plunge pool at a leisure centre cools after the heater is switched off. Its temperature T °C, t minutes later, is modelled by T = 18 + 22e^−0.05t, t ≥ 0. The graph of T against t can be obtained from the graph of y = e^x by a sequence of transformations.

(a)

Write down the equation of the horizontal asymptote of the graph of T and state what it represents.

(2)

(b)

Describe a sequence of transformations that maps y = e^x onto T = 18 + 22e^−0.05t.

(3)

(c)

Find the time taken for the pool to cool to 28 °C. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
T = 18	✓ 1
the temperature the pool water approaches (the room temperature) in the long term	✓ 1
Part (b)
horizontal stretch with scale factor 20 and reflection in the vertical axis (e^−t/20)	✓ 1
vertical stretch with scale factor 22	✓ 1
translation 18 units up	✓ 1
Part (c)
28 = 18 + 22e^−0.05t, so e^−0.05t = 10/22	✓ 1
t = 20 ln 2.2 = 15.8 minutes	✓ 1

Common error: Subtract the asymptote value before taking logarithms, otherwise the exponential term is not isolated.

+ Add to test✓ In your test127IBMI3H-2-03r10-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A wind farm operator owns a gearbox whose resale value is modelled by V = 24 000 × 0.82^t pounds (£), where t is its age in years. The total cost of repairs over the gearbox's life is modelled by R = 1500 + 1800t pounds (£).

(a)

Find the value of the gearbox when it is 5 years old, to the nearest £.

(2)

(b)

Find the age at which the total repair cost first equals the resale value. Give your answer to 3 significant figures.

(2)

(c)

Suggest, with a reason, what the operator should do at this age.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
24 000 × 0.82⁵ = 8897.76...	✓ 1
£8898	✓ 1
Part (b)
equation 24 000 × 0.82^t = 1500 + 1800t solved (GDC intersection)	✓ 1
t = 4.56 years	✓ 1
Part (c)
consider replacing the gearbox	✓ 1
after this age the repair costs exceed its resale value (accept equivalent reasoning)	✓ 1

Exam tip: When two models are set equal, use the graph intersection on your calculator and state the x-coordinate with units.

+ Add to test✓ In your test128IBMI3H-2-04r13-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The power P (kW) a light aircraft needs for level flight at speed v (m s⁻¹) is modelled by P = kvⁿ. A graph of log₁₀P against log₁₀v is a straight line passing through the points (1.60, 0.60) and (2.00, 1.80).

(a)

Find the gradient of the line.

(2)

(b)

Hence show that n = 3 and find k, giving k to 3 significant figures.

(3)

(c)

Estimate the power needed at 120 m s⁻¹, to 3 significant figures. Use the unrounded value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use (1.80 − 0.60)/(2.00 − 1.60)	✓ 1
A1 gradient = 3.0	✓ 1
Part (b)
M1 log₁₀P = log₁₀k + nlog₁₀v, so n equals the gradient, 3	✓ 1
A1 intercept: 0.60 = 3(1.60) + log₁₀k, so log₁₀k = −4.2	✓ 1
A1 k = 10^−4.2 = 6.31 × 10⁻⁵	✓ 1
Part (c)
M1 substitute into P = k × 120³	✓ 1
A1 P = 109 kW (109.03)	✓ 1

Exam tip: For a power law y = kxⁿ, the gradient of a log–log graph is n and the vertical intercept is log k.

+ Add to test✓ In your test129IBMI3H-2-04r13-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium5 marks

The area N m² covered by an invasive weed in a field is recorded each week. A plot of ln N against the time t (weeks) is a straight line with equation ln N = 0.35t + 2.1.

(a)

Show that N = ae^bt and find a and b to 3 significant figures.

(3)

(b)

Find the time for the area to double, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 take exponentials of both sides: N = e^{0.35t + 2.1} = e^2.1e^0.35t	✓ 1
A1 a = e^2.1 = 8.17 (8.166…)	✓ 1
A1 b = 0.35	✓ 1
Part (b)
M1 set e^0.35t = 2 and take natural logs	✓ 1
A1 t = ln2/0.35 = 1.98 weeks (1.9804)	✓ 1

Common error: A straight line on a semi-log graph (ln N against t) means exponential growth. The intercept is ln a, not a itself.

+ Add to test✓ In your test130IBMI3H-2-04r13-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A drone's height above the ground during a crop-spraying pass is given by y = f(x) (metres), where x is the horizontal distance (m) travelled. The graph of f has a maximum point at (4, 6) and passes through (8, 2).
The function g is defined by g(x) = 2f(x − 3) + 1, and h by h(x) = f(2x).

(a)

Describe the sequence of transformations that maps the graph of y = f(x) onto y = g(x).

(3)

(b)

Find the coordinates of the point on the graph of h that is the image of (8, 2).

(2)

(c)

Write down the coordinates of the maximum point of g, and state the maximum value of g.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 3 units in the positive x direction	✓ 1
A1 vertical stretch with scale factor 2	✓ 1
A1 translation 1 unit in the positive y direction (valid orders of the transformations accepted)	✓ 1
Part (b)
M1 horizontal stretch with scale factor ½ (the x-coordinate is halved)	✓ 1
A1 (4, 2)	✓ 1
Part (c)
A1 maximum point at (7, 13)	✓ 1
A1 maximum value of g is 13 (m)	✓ 1

Common error: For f(x − 3) the graph moves 3 units to the right, and for f(2x) the x-coordinates are halved; horizontal changes act in the opposite sense to what the number suggests.

+ Add to test✓ In your test131IBMI3H-2-04r13-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

A fungal infection spreads in an orchard of 1200 trees. The number of infected trees after t days is modelled by N(t) = 1200/(1 + 19e^−0.45t).

(a)

Find the number of trees infected at t = 0.

(2)

(b)

State the value that N(t) approaches for large t, and what it means in context.

(1)

(c)

Find the number of trees infected after 8 days, to the nearest whole tree.

(2)

(d)

Find the time when 900 trees are infected, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0: 1200/(1 + 19)	✓ 1
A1 60 trees	✓ 1
Part (b)
R1 N → 1200 as t → ∞, meaning every tree in the orchard eventually becomes infected	✓ 1
Part (c)
M1 evaluate 1200/(1 + 19e^−3.6)	✓ 1
A1 790 trees (789.9)	✓ 1
Part (d)
M1 set 1200/(1 + 19e^−0.45t) = 900, so 1 + 19e^−0.45t = 4/3	✓ 1
M1 e^−0.45t = 1/57 and take natural logs	✓ 1
A1 t = ln57/0.45 = 8.98 days (8.985)	✓ 1

Exam tip: In a logistic model the limiting value is the numerator constant, and the model is steepest at half that value.

+ Add to test✓ In your test132IBMI3H-2-04r13-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The hours of daylight D at a farm on day t of the year are modelled by D(t) = acos(b(t − 172)) + d, where b is in radians. The maximum is 15.3 hours (on day 172), the minimum is 9.1 hours and the period is 365 days.

(a)

Find a and d.

(2)

(b)

Find the exact value of b in terms of π.

(1)

(c)

Find the hours of daylight on day 100, to 3 significant figures. Use the unrounded value of b.

(2)

(d)

Find the two days in the year (0 ≤ t ≤ 365) on which there are exactly 14 hours of daylight. Give each answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (15.3 − 9.1)/2 = 3.1	✓ 1
A1 d = (15.3 + 9.1)/2 = 12.2	✓ 1
Part (b)
A1 b = 2π/365	✓ 1
Part (c)
M1 substitute t = 100: 12.2 + 3.1cos(2π(−72)/365)	✓ 1
A1 D = 13.2 hours (13.21)	✓ 1
Part (d)
M1 set 3.1cos(b(t − 172)) + 12.2 = 14, so cos(b(t − 172)) = 1.8/3.1	✓ 1
A1 t = 116.7	✓ 1
A1 t = 227.3	✓ 1

Exam tip: A cosine equation has a second solution symmetric about the line of the maximum; here the two days are equally spaced either side of day 172.

+ Add to test✓ In your test133IBMI3H-2-04r13-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The wheat yield Y (tonnes per hectare) for nitrogen fertiliser applied at x kg per hectare is modelled by Y = −0.00032x² + 0.064x + 2.8, for 0 ≤ x ≤ 200.

(a)

State the yield when no fertiliser is used.

(1)

(b)

Find the amount of fertiliser that gives the maximum yield, and the maximum yield.

(3)

(c)

Find the range of values of x for which the yield is at least 5.0 tonnes per hectare. Give the end-points to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 2.8 tonnes per hectare	✓ 1
Part (b)
M1 use x = −b/(2a) = −0.064/(2 × (−0.00032)) or technology	✓ 1
A1 x = 100 kg per hectare	✓ 1
A1 maximum yield 6.0 tonnes per hectare	✓ 1
Part (c)
M1 solve −0.00032x² + 0.064x + 2.8 = 5, i.e. x² − 200x + 6875 = 0	✓ 1
A1 lower end-point 44.1	✓ 1
A1 upper end-point 156 (155.9…)	✓ 1

Common error: 'At least 5' means the yield is on or above 5, so the answer is the interval between the two roots of Y = 5, not the values outside them.

+ Add to test✓ In your test134IBMI3H-2-04r13-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The concentration of a pesticide in soil decays exponentially: C = C₀e^−kt, where t is in days. After 12 days, 64% of the original concentration remains.

(a)

Find k to 3 significant figures.

(2)

(b)

Find the time for the concentration to fall to 10% of its original value. Use the unrounded value of k, and give your answer to 3 significant figures.

(2)

(c)

Find the percentage of the original concentration remaining after 30 days, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set e^−12k = 0.64	✓ 1
A1 k = −ln(0.64)/12 = 0.0372 (day⁻¹)	✓ 1
Part (b)
M1 set e^−kt = 0.1, so t = ln10/k	✓ 1
A1 t = 61.9 days (61.91)	✓ 1
Part (c)
M1 0.64^30/12 or e^−30k	✓ 1
A1 32.8% (32.768)	✓ 1

Exam tip: If a fixed fraction remains after a given time, use that fraction directly to find the decay constant instead of choosing an arbitrary starting value.

+ Add to test✓ In your test135IBMI3H-2-04r13-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A synthesiser's output voltage (V) at time x (in suitable units) is modelled by y = 2sin(3(x − π/6)) + 1. This is obtained from the graph of y = sin x by a sequence of transformations.

(a)

Describe the four transformations that map y = sin x onto this graph.

(4)

(b)

State the range of the function.

(1)

(c)

Find the smallest positive value of x at which y is a maximum. Give your answer in radians to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 vertical stretch with scale factor 2	✓ 1
A1 horizontal stretch with scale factor 1/3	✓ 1
A1 translation of π/6 to the right (positive x direction)	✓ 1
A1 translation of 1 unit up (positive y direction)	✓ 1
Part (b)
A1 −1 ≤ y ≤ 3	✓ 1
Part (c)
M1 3(x − π/6) = π/2	✓ 1
A1 x = π/3 = 1.05	✓ 1

Common error: In y = sin(3(x − π/6)) the factor 3 compresses the graph horizontally and the translation is π/6, not π/18; factorise inside the bracket before reading off the shift.

+ Add to test✓ In your test136IBMI3H-2-05r16-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A beekeeper counts the number of bees, P, in a hive t days after a new queen is introduced. A plot of ln P against t is a straight line with equation ln P = 0.032t + 7.40.

(a)

Show that the population can be modelled by P = Ae^kt, and state the value of k and the value of A correct to 3 significant figures.

(2)

(b)

Calculate the population after 25 days. Give your answer to the nearest 10 (use ln P = 8.2 directly).

(3)

(c)

Find the number of days for the population to double. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 take exponentials of both sides, P = e^{0.032t + 7.40}	✓ 1
A1 P = e^7.40e^0.032t, so k = 0.032 and A = e^7.40 = 1640 (3 s.f.)	✓ 1
Part (b)
M1 substitute t = 25 to obtain ln P = 0.032 × 25 + 7.40 = 8.2	✓ 1
A1 P = e^8.2 = 3641.0…	✓ 1
A1 P = 3640 bees	✓ 1
Part (c)
M1 doubling: e^0.032T = 2, so T = ln 2 ÷ 0.032 = 21.66…	✓ 1
A1 T = 21.7 days	✓ 1

Exam tip: When a graph of ln y against x is a straight line, the gradient is the exponent k and the intercept is ln A, not A.

+ Add to test✓ In your test137IBMI3H-2-05r16-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The power, P watts, needed to overcome air resistance on a Formula 1 car travelling at speed v m s⁻¹ is modelled by P = kvⁿ. A graph of ln P against ln v is a straight line passing through the points (4.00, 11.60) and (4.40, 12.80).

(a)

Find the gradient of the line and hence write down the value of n.

(2)

(b)

Show that ln P = 3 ln v − 0.40, and find the value of k correct to 4 significant figures.

(2)

(c)

Use k = e^−0.40 (unrounded) to find the power needed at 82 m s⁻¹. Give your answer in kilowatts to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (12.80 − 11.60) ÷ (4.40 − 4.00)	✓ 1
A1 gradient = 3, so n = 3	✓ 1
Part (b)
M1 substitute a point: 11.60 = 3(4.00) + c	✓ 1
A1 c = −0.40 and k = e^−0.40 = 0.6703	✓ 1
Part (c)
M1 P = e^−0.40 × 82³	✓ 1
A1 P = 369593 W	✓ 1
A1 P = 370 kW (3 s.f.)	✓ 1

Common error: In a log–log graph both axes are logarithms, so the intercept is ln k; forgetting to exponentiate it gives k = −0.40.

+ Add to test✓ In your test138IBMI3H-2-05r16-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

During a Formula 1 race, the lap time, T seconds, on lap n for a particular set of tyres is modelled by T = 91.2 + 0.18e^0.11n. The term 0.18e^0.11n represents the time lost to tyre wear.

(a)

Calculate the time lost to tyre wear on lap 10. Give your answer to 3 significant figures.

(2)

(b)

The team will change tyres at the first lap on which the time lost to tyre wear exceeds 2.5 s. Find the lap on which this first happens.

(4)

(c)

State one reason why this model is unlikely to be valid for very large values of n.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.18e^{0.11 × 10} = 0.18e^1.1	✓ 1
A1 0.541… so 0.541 s	✓ 1
Part (b)
M1 set up 0.18e^0.11n = 2.5 (or inequality)	✓ 1
M1 e^0.11n = 13.88… and take natural logs	✓ 1
A1 n = 23.92… so 23.9	✓ 1
A1 the wear exceeds 2.5 s first on lap 24 (whole number of laps)	✓ 1
Part (c)
R1 exponential growth is unbounded, so lap times would increase without limit (or the tyres would have been changed / car would stop)	✓ 1

Common error: When an inequality gives n > 23.9, the answer is the next whole lap, 24, not the rounded value 23.9 or 23.

+ Add to test✓ In your test139IBMI3H-2-05r16-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

Surgical instruments are removed from a steriliser at 100 °C and left to cool in a room at 22 °C. Their temperature after t minutes is T = 22 + Be^−kt. After 6 minutes the temperature is 58 °C.

(a)

Write down the value of B.

(1)

(b)

Find the value of k correct to 3 significant figures.

(3)

(c)

The instruments can be handled safely once they reach 37 °C. Find the time this takes, using the unrounded value of k. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 B = 78 (since T(0) = 100)	✓ 1
Part (b)
M1 58 = 22 + 78e^−6k so e^−6k = 36/78	✓ 1
M1 take natural logs: −6k = ln(6/13)	✓ 1
A1 k = 0.1289… so 0.129	✓ 1
Part (c)
M1 37 = 22 + 78e^−kt so e^−kt = 15/78	✓ 1
M1 t = ln(5.2) ÷ k	✓ 1
A1 t = 12.79… so 12.8 minutes	✓ 1

Exam tip: Newton cooling graphs approach the room temperature, not zero; subtract the room temperature before taking logs.

+ Add to test✓ In your test140IBMI3H-2-03r17-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The demand for water in a town, $D$ megalitres per hour, is modelled by $$D(t)=42+15\cos\left(\frac{\pi}{12}(t-17)\right),\quad 0\le t<24,$$ where $t$ is the number of hours after midnight. (Use radians.)

(a)

Write down the maximum and minimum values of $D$.

(2)

(b)

Find the period of the model.

(1)

(c)

Find the two times in the day when the demand is exactly 50 megalitres per hour. Give your answers in hours to 3 significant figures.

(3)

(d)

Hence find the length of time in each day for which the demand exceeds 50 megalitres per hour. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 maximum $57$ megalitres per hour	✓ 1
A1 minimum $27$ megalitres per hour	✓ 1
Part (b)
A1 $\frac{2\pi}{\pi/12}=24$ hours	✓ 1
Part (c)
M1 sets $42+15\cos\left(\frac{\pi}{12}(t-17)\right)=50$ (or intersection on a GDC)	✓ 1
A1 $t=13.1$	✓ 1
A1 $t=20.9$	✓ 1
Part (d)
M1 subtracts the two times: $20.85\ldots-13.14\ldots$ (unrounded values)	✓ 1
A1 $7.70$ hours	✓ 1

Exam tip: Set the GDC window to radian mode and a domain of 0 to 24 so only the two relevant intersections appear.

+ Add to test✓ In your test141IBMI3H-2-03r17-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

A games company sells a cosmetic pack for €$x$ each. The number sold per day is modelled by $n=400-25x$ and the cost to the company of each pack is €4.

(a)

Find the number of packs sold per day when the price is €6.

(1)

(b)

Show that the daily profit, €$P$, is given by $P=-25x^2+500x-1600$.

(2)

(c)

Find the range of prices for which the company makes a profit.

(2)

(d)

Find the price that gives the greatest daily profit.

(1)

(e)

Find the greatest daily profit.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $n=400-150=250$	✓ 1
Part (b)
M1 $P=(x-4)(400-25x)$	✓ 1
A1 $=400x-25x^2-1600+100x=-25x^2+500x-1600$ (AG)	✓ 1
Part (c)
M1 solves $P=0$ (or $(x-4)(400-25x)=0$)	✓ 1
A1 \(4	✓ 1
Part (d)
A1 $x=-\frac{500}{2(-25)}=10$, so €10	✓ 1
Part (e)
M1 $P(10)=-25(100)+5000-1600$	✓ 1
A1 €900	✓ 1

Common error: Revenue is not profit: forgetting to subtract the €4 cost per pack produces the wrong maximum.

+ Add to test✓ In your test142IBMI3H-2-03r19-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

After the heating is switched off, the water in a swimming pool cools towards the room temperature of 22 °C. At time t hours, the water temperature is T °C and D = T − 22. A graph of ln D against t is a straight line passing through the points (2, 1.9972) and (8, 1.3972).

(a)

Calculate the gradient of the line.

(2)

(b)

The relationship can be written as D = Ae^kt. Find the value of k and the value of A, giving A to 3 significant figures.

(3)

(c)

Use the model, with A = 9 and k = −0.1, to find the water temperature after 5 hours. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1.3972 − 1.9972) ÷ (8 − 2)	✓ 1
A1 gradient = −0.1	✓ 1
Part (b)
A1 k = −0.1 (ECF from (a))	✓ 1
M1 intercept = 1.9972 + 0.1 × 2 = 2.1972, then A = e^2.1972	✓ 1
A1 A = 9.00	✓ 1
Part (c)
M1 T = 22 + 9e^−0.5	✓ 1
A1 T = 27.5 °C	✓ 1

Exam tip: A straight line on a graph of ln D against t means D is exponential in t; the gradient is the exponent and the intercept is ln A.

+ Add to test✓ In your test143IBMI3H-2-03r19-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The depth of water, d metres, in a swimming pool at a distance x metres from the shallow end is modelled by d = f(x). The point (6, 0.9) lies on the graph of y = f(x). A longer, deeper pool has depth modelled by d = g(x), where g(x) = 2f(^x⁄₂) + 0.3.

(a)

Describe the sequence of transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

Write down the coordinates of the image of (6, 0.9) on the graph of y = g(x).

(2)

(c)

The new pool has depth 2.5 m at a distance of 40 m from the shallow end. Find f(20).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal stretch, scale factor 2 (from the y-axis)	✓ 1
A1 vertical stretch, scale factor 2 (from the x-axis)	✓ 1
A1 translation 0.3 units up (stretches before the translation)	✓ 1
Part (b)
A1 x-coordinate 12	✓ 1
A1 y-coordinate 2 × 0.9 + 0.3 = 2.1	✓ 1
Part (c)
M1 2f(20) + 0.3 = 2.5	✓ 1
A1 f(20) = 1.1	✓ 1

Common error: A horizontal stretch of scale factor 2 corresponds to f(x/2), not f(2x); the x-coordinates are multiplied by the factor.

+ Add to test✓ In your test144IBMI3H-2-03r19-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

Sound level, L decibels (dB), is defined by L = 10 log₁₀(^I⁄_I₀), where I is the sound intensity in W m⁻² and I₀ = 1.0 × 10⁻¹² W m⁻².
A library reading room is quiet at 35 dB and busy at 62 dB.

(a)

Find the intensity at 62 dB, giving your answer to 3 significant figures.

(3)

(b)

Calculate the ratio of the intensity when busy to the intensity when quiet, to 3 significant figures.

(2)

(c)

A wind farm has n identical turbines. At a nearby house each turbine alone produces 38 dB, and the intensities add. Find n if all turbines together produce 48 dB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 62 = 10 log₁₀(I/10⁻¹²)	✓ 1
M1 I = 10⁻¹² × 10^6.2	✓ 1
A1 I = 1.58 × 10⁻⁶ W m⁻²	✓ 1
Part (b)
M1 10^{(62 − 35)/10} = 10^2.7	✓ 1
A1 501	✓ 1
Part (c)
M1 10 log₁₀ n = 48 − 38 (or n = 10^{(48 − 38)/10})	✓ 1
A1 n = 10	✓ 1

Common error: Decibels are logarithmic: adding the levels of two sources does not add the decibels; the intensities are added first.

+ Add to test✓ In your test145IBMI3H-2-03r19-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The number of digital loans per month at a library, L thousand, is modelled by L = ab^t, where t is the number of years since the start of 2020. A graph of ln L against t is a straight line with equation ln L = 0.18t + 2.3.

(a)

Find the value of a and the value of b, giving each to 3 significant figures.

(3)

(b)

Using the equation of the line, find the time at which loans reach 50 thousand per month. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = e^2.3 and b = e^0.18	✓ 1
A1 a = 9.97	✓ 1
A1 b = 1.20	✓ 1
Part (b)
M1 ln 50 = 0.18t + 2.3	✓ 1
A1 (ln 50 − 2.3) ÷ 0.18	✓ 1
A1 t = 8.96 years	✓ 1

Exam tip: Both parameters of L = abt come from the line: the intercept gives ln a and the gradient gives ln b.

+ Add to test✓ In your test146IBMI3H-2-03r19-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The annual profit, P thousand euros (€), of a wind farm with x turbines is modelled by P = −0.05x² + 4x − 35, for 0 ≤ x ≤ 80.

(a)

Find the numbers of turbines for which the farm breaks even.

(3)

(b)

Find the number of turbines that gives the greatest profit, and calculate this profit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −0.05x² + 4x − 35 = 0	✓ 1
A1 x = 10	✓ 1
A1 x = 70	✓ 1
Part (b)
M1 x = 4 ÷ (2 × 0.05) or the midpoint of the roots	✓ 1
A1 x = 40	✓ 1
A1 P = €45 thousand	✓ 1

Exam tip: Check the vertex lies halfway between the break-even values (10 and 70) before substituting for the maximum.

+ Add to test✓ In your test147IBMI3H-2-03r19-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium5 marks

Race times for swimmers of a club are modelled by T = kdⁿ, where T is the time in seconds for a race of distance d metres. A swimmer takes 50.0 s for 100 m and 232 s for 400 m.

(a)

Show that n = 1.11 to 3 significant figures.

(2)

(b)

Using the unrounded value of n, predict the time for 200 m. Give your answer to 3 significant figures.

(2)

(c)

Since n > 1, state what the model predicts about the swimmer's average speed as the distance increases.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ln(232/50) = n ln(400/100) (or gradient of ln T against ln d)	✓ 1
A1 n = 1.107, which rounds to 1.11	✓ 1
Part (b)
M1 T = 50.0 × 2ⁿ	✓ 1
A1 108 s	✓ 1
Part (c)
R1 average speed d/T = d¹⁻ⁿ/k decreases as d increases	✓ 1

Common error: Take logarithms of both sides of T = kdⁿ to obtain ln T = n ln d + ln k; the exponent n is the gradient, not the intercept.

+ Add to test✓ In your test148IBMI3H-2-03r19-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

In a swimming pool, the concentration of chlorine, C = f(t) mg per litre, t hours after the pool is first dosed, is modelled by f(t) = 5 × 0.85^t. A second pool is dosed with the same amount of chlorine 3 hours later, so its concentration is modelled by C_B = f(t − 3), for t ≥ 3.

(a)

Describe the transformation that maps the graph of C = f(t) onto the graph of C_B = f(t − 3).

(2)

(b)

Find the concentration in the second pool when t = 8, giving your answer to 3 significant figures.

(2)

(c)

Find the value of t at which the concentration in the second pool falls to 1.0 mg per litre. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation (horizontal shift)	✓ 1
A1 3 units in the positive t direction (to the right)	✓ 1
Part (b)
M1 C_B = 5 × 0.85⁵	✓ 1
A1 2.22 mg per litre	✓ 1
Part (c)
M1 5 × 0.85^t−3 = 1	✓ 1
M1 0.85^t−3 = 0.2 and use of logarithms	✓ 1
A1 t = 12.9 hours	✓ 1

Common error: Replacing t by t − 3 shifts the graph 3 units to the right, not to the left.

+ Add to test✓ In your test149IBMI3H-2-04r19-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The cost per garment, C euros (€), of producing x hundred garments in a fashion workshop is modelled by C = f(x) = 0.2(x − 5)² + 2.4. The profit per garment from a new line is modelled by h(x) = 12 − f(x − 1).

(a)

Describe a sequence of three transformations that maps the graph of y = f(x) onto the graph of y = h(x).

(3)

(b)

The graph of y = f(x) has its minimum at (5, 2.4). Write down the coordinates of the maximum point of y = h(x).

(2)

(c)

Find the profit per garment, in euros, when x = 9.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 1 unit to the right	✓ 1
A1 reflection in the x-axis	✓ 1
A1 translation 12 units up (after the reflection)	✓ 1
Part (b)
A1 x = 6	✓ 1
A1 h = 12 − 2.4 = 9.6, so (6, 9.6)	✓ 1
Part (c)
M1 h(9) = 12 − f(8) = 12 − (0.2 × 3² + 2.4)	✓ 1
A1 €7.80	✓ 1

Common error: f(x − 1) moves the graph right; −f(x) reflects in the x-axis, so a minimum becomes a maximum.

+ Add to test✓ In your test150IBMI3H-2-04r19-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The mass of a fish stock, P tonnes, after t years of protection is modelled by P = ^{12 000}⁄_{1 + 7e^−0.3t}.

(a)

State the initial mass of the stock and the carrying capacity predicted by the model.

(2)

(b)

Find the time for the stock to reach half of the carrying capacity, giving your answer to 3 significant figures.

(3)

(c)

Fishing is permitted once the stock exceeds 9000 tonnes. Find the number of years after which this first happens, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 initial mass 12 000 ÷ 8 = 1500 tonnes	✓ 1
A1 carrying capacity 12 000 tonnes	✓ 1
Part (b)
M1 1 + 7e^−0.3t = 2	✓ 1
M1 e^−0.3t = ¹⁄₇ and take natural logarithms	✓ 1
A1 t = 6.49 years	✓ 1
Part (c)
M1 1 + 7e^−0.3t = ⁴⁄₃, so e^−0.3t = ¹⁄₂₁	✓ 1
A1 t = 10.1 years	✓ 1

Exam tip: When solving for t in a logistic model, first rearrange to isolate the exponential term, then take natural logarithms.

+ Add to test✓ In your test151IBMI3H-2-04r19-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The vibration level of a robot arm is L = 20 log₁₀(^a⁄_a₀) decibels, where a is the vibration amplitude in metres and a₀ = 1.0 × 10⁻⁶ m.

(a)

Calculate L when a = 4.5 × 10⁻⁴ m, giving your answer to 3 significant figures.

(2)

(b)

Find the amplitude when L = 72 dB. Give your answer to 3 significant figures.

(2)

(c)

Dampers reduce the amplitude to one third of its value. Find the decrease in L, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 log₁₀(4.5 × 10⁻⁴ ÷ 10⁻⁶) = 20 log₁₀ 450	✓ 1
A1 53.1 dB	✓ 1
Part (b)
M1 log₁₀(a/a₀) = 3.6, so a = 10⁻⁶ × 10^3.6	✓ 1
A1 a = 3.98 × 10⁻³ m	✓ 1
Part (c)
M1 decrease = 20 log₁₀ 3 (the logarithm of the ratio, independent of a)	✓ 1
A1 9.54 dB	✓ 1

Exam tip: A change in a logarithmic scale depends only on the ratio of the quantities, so a₀ cancels.

+ Add to test✓ In your test152IBMI3H-2-04r19-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The mass m grams of a fish of length L cm is modelled by m = aL^b. A graph of ln m against ln L is a straight line with equation ln m = 3.1 ln L − 4.2.

(a)

Write down the value of b and find the value of a, to 3 significant figures.

(3)

(b)

Use the unrounded value of a to find the mass of a fish of length 30 cm. Give your answer to 3 significant figures.

(2)

(c)

Find the percentage increase in length needed for the mass of a fish to double. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 b = 3.1	✓ 1
M1 a = e^−4.2	✓ 1
A1 a = 0.0150, so 0.0150 to 3 s.f.	✓ 1
Part (b)
M1 ln m = 3.1 ln 30 − 4.2 (or m = a × 30^3.1)	✓ 1
A1 569 g	✓ 1
Part (c)
M1 (L₂/L₁)^3.1 = 2, so L₂/L₁ = 2^1/3.1	✓ 1
A1 25.1%	✓ 1

Exam tip: In a power model m = aLb, multiplying L by a factor k multiplies m by kb; the constant a cancels.

+ Add to test✓ In your test153IBMI3H-2-04r19-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The number of items, f(x) hundred, sold by a fashion website x weeks after a launch is given below, and a second site has sales modelled by g(x) = 2f(^x⁄₂) − 1.
For x = 0, 1, 2, 3, 4 the values of f(x) are 2, 5, 9, 14, 20 respectively.

(a)

Describe the transformations that map the graph of y = f(x) onto the graph of y = g(x).

(2)

(b)

Find the values of g(2), g(4) and g(8).

(3)

(c)

Find the value of x for which g(x) = 27.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal stretch, scale factor 2	✓ 1
A1 vertical stretch, scale factor 2, then translation 1 unit down	✓ 1
Part (b)
A1 g(2) = 2 × 5 − 1 = 9	✓ 1
A1 g(4) = 2 × 9 − 1 = 17	✓ 1
A1 g(8) = 2 × 20 − 1 = 39	✓ 1
Part (c)
M1 2f(x/2) − 1 = 27, so f(x/2) = 14 and x/2 = 3	✓ 1
A1 x = 6	✓ 1

Common error: In f(x/2) the input is halved, so each x-value of the table is doubled for g; the outputs are changed separately.

+ Add to test✓ In your test154IBMI3H-2-04r19-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The acidity of seawater is measured by pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol per litre. A fishery's seawater has pH 8.15, and a long-term study predicts that this will fall to 8.05.

(a)

Find [H⁺] at pH 8.15, giving your answer to 3 significant figures.

(2)

(b)

Find the percentage increase in [H⁺] when the pH falls from 8.15 to 8.05, to 3 significant figures.

(3)

(c)

Find by how much the pH must fall for [H⁺] to double, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 [H⁺] = 10^−8.15	✓ 1
A1 7.08 × 10⁻⁹ mol per litre	✓ 1
Part (b)
M1 ratio = 10^{8.15 − 8.05} = 10^0.1	✓ 1
M1 percentage increase = (ratio − 1) × 100	✓ 1
A1 25.9%	✓ 1
Part (c)
M1 pH change = log₁₀ 2	✓ 1
A1 0.301	✓ 1

Common error: A lower pH means a higher concentration; the pH scale is logarithmic so a fall of 0.1 is a multiplication, not an addition, of [H⁺].

+ Add to test✓ In your test155IBMI3H-2-04r19-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

In a harbour, the tide height h metres follows a model h(t) = a cos(bt) + d, where t is in hours after high tide and the angle is in radians. High tide is 4.2 m and low tide is 1.8 m, and low tide occurs 6 hours after high tide.

(a)

Find the values of a, b and d.

(3)

(b)

A fishing boat needs a depth of at least 2.5 m to leave the harbour. Find the total time in one 12-hour cycle for which h ≥ 2.5, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (4.2 − 1.8) ÷ 2 = 1.2	✓ 1
A1 d = (4.2 + 1.8) ÷ 2 = 3.0	✓ 1
A1 b = 2π ÷ 12 = π/6	✓ 1
Part (b)
M1 3 + 1.2 cos(πt/6) = 2.5, so cos(πt/6) = −⁵⁄₁₂	✓ 1
A1 t = 3.82 hours after high tide	✓ 1
A1 by symmetry, 2 × 3.8208 = 7.64 hours	✓ 1

Exam tip: Sketch one period to see which interval satisfies the inequality; a cosine model starting at a maximum is symmetric about t = 0.

+ Add to test✓ In your test156IBMI3H-2-04r19-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The speed s m s⁻¹ of a robot carrying a payload of mass m kg is modelled by s = km^−0.5. When m = 4, s = 2.4.

(a)

Show that k = 4.8.

(2)

(b)

A graph of log₁₀ s against log₁₀ m is a straight line. State its gradient and its vertical intercept, to 3 significant figures.

(2)

(c)

Find the payload mass for which the speed is 1.5 m s⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.4 = k × 4^−0.5	✓ 1
A1 k = 2.4 × 2 = 4.8	✓ 1
Part (b)
A1 gradient = −0.5	✓ 1
A1 intercept = log₁₀ 4.8 = 0.681	✓ 1
Part (c)
M1 1.5 = 4.8m^−0.5, so m^0.5 = ^4.8⁄_1.5 = 3.2	✓ 1
A1 m = 10.24 kg	✓ 1

Common error: Take logarithms of s = kmⁿ to get log s = n log m + log k; the intercept is log k, not k.

+ Add to test✓ In your test157IBMI3H-2-05r21-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The daily traffic N (terabytes) through a data centre network is modelled by N = ab^t, where t is the number of years since 2020. A graph of log₁₀ N against t is a straight line passing through the points (0, 2.40) and (8, 3.20).

(a)

Find the gradient of the line.

(2)

(b)

Hence find the values of a and b, each to 3 significant figures.

(3)

(c)

Find the time, to 3 significant figures, for the traffic to double. Use b = 10^0.1 exactly.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (3.20 − 2.40)/(8 − 0)	✓ 1
A1 0.100	✓ 1
Part (b)
M1 log₁₀ N = 2.4 + 0.1t, so N = 10^2.4 × (10^0.1)^t	✓ 1
A1 a = 251	✓ 1
A1 b = 1.26	✓ 1
Part (c)
M1 b^T = 2 or 0.1T = log₁₀ 2	✓ 1
A1 T = 3.01 years	✓ 1

Common error: On a semi-logarithmic graph the gradient is log₁₀ b, not b; remember to undo the logarithm to recover the growth factor.

+ Add to test✓ In your test158IBMI3H-2-05r21-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The magnitude M of an earthquake is related to the energy E joules released by log₁₀ E = 1.5M + 4.8. An earthquake is thought to have destroyed part of an ancient settlement.

(a)

Calculate the energy released by an earthquake of magnitude 5.4. Give your answer in the form a×10^k, where 1 ≤ a < 10 and a is given to 3 significant figures.

(2)

(b)

A different earthquake releases 3.0×10¹⁵ J. Find its magnitude, to 3 significant figures.

(2)

(c)

Show that an earthquake of magnitude 6.8 releases about 63 times as much energy as one of magnitude 5.6.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 log₁₀ E = 1.5 × 5.4 + 4.8 = 12.9	✓ 1
A1 E = 10^12.9 = 7.94×10¹² J	✓ 1
Part (b)
M1 M = (log₁₀(3.0×10¹⁵) − 4.8)/1.5	✓ 1
A1 M = 7.12	✓ 1
Part (c)
M1 log₁₀(E₂/E₁) = 1.5 × (6.8 − 5.6) = 1.8	✓ 1
A1 E₂/E₁ = 10^1.8 = 63.1 (AG)	✓ 1

Common error: A difference of 1 in magnitude corresponds to a factor of 10¹.⁵ in energy, so differences of magnitude must be converted using a power of ten, not multiplied.

+ Add to test✓ In your test159IBMI3H-2-05r21-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The brightness B watts of a bicycle lamp after it is switched on is modelled by B(t) = 2 + 4e^−0.5t, t ≥ 0, where t is the time in seconds.

(a)

Find the initial brightness.

(1)

(b)

Write down the equation of the horizontal asymptote of the graph of B and interpret it in context.

(1)

(c)

Find the time at which the brightness is 3 W, to 3 significant figures.

(2)

(d)

The graph of y = e^x is transformed to give the graph of B. Describe a sequence of transformations that achieves this.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 B(0) = 6 W	✓ 1
Part (b)
A1 B = 2, the brightness the lamp settles to in the long run	✓ 1
Part (c)
M1 3 = 2 + 4e^−0.5t, so e^−0.5t = 0.25	✓ 1
A1 t = 2 ln 4 = 2.77 s	✓ 1
Part (d)
A1 reflection in the y-axis and horizontal stretch with scale factor 2 (together giving x → −0.5x)	✓ 1
A1 vertical stretch with scale factor 4	✓ 1
A1 translation 2 units up (B = 2 + …)	✓ 1

Exam tip: Reflections and stretches can be combined: replacing x by −0.5x is a reflection in the y-axis together with a horizontal stretch of scale factor 2.

+ Add to test✓ In your test160IBMI3H-2-05r26-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The power output P kW of a wind turbine is modelled by P = kvⁿ, where v m s⁻¹ is the wind speed. The turbine produces 32 kW when v = 4 and 500 kW when v = 10.

(a)

Explain why a graph of ln P against ln v is a straight line, and state its gradient.

(2)

(b)

Find the values of n and k.

(3)

(c)

Find the wind speed at which the power output is 2000 kW, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ln P = ln k + n ln v, which has the form Y = c + nX	✓ 1
A1 gradient is n (the intercept is ln k)	✓ 1
Part (b)
M1 n = (ln 500 − ln 32)/(ln 10 − ln 4)	✓ 1
A1 n = 3	✓ 1
A1 k = 32/4³ = 0.5	✓ 1
Part (c)
M1 0.5v³ = 2000, so v³ = 4000	✓ 1
A1 v = 15.9 m s⁻¹	✓ 1

Common error: Do not take logarithms of the sum; ln(kvⁿ) = ln k + n ln v, so the gradient is n and not k.

+ Add to test✓ In your test161IBMI3H-2-05r26-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The number of people infected in an outbreak is modelled by N(t) = 2000 / (1 + 19e^−0.6t), where t is the time in days since the first case was recorded.

(a)

Find the number of people infected at t = 0.

(1)

(b)

State the value that N approaches as t becomes large.

(1)

(c)

Find the time at which 1000 people are infected, to 3 significant figures.

(3)

(d)

This model satisfies dN/dt = 0.6N(1 − N/2000). The rate of increase is greatest when N = 1000. Find the greatest rate of increase.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 N = 2000/20 = 100	✓ 1
Part (b)
A1 2000	✓ 1
Part (c)
M1 1 + 19e^−0.6t = 2	✓ 1
A1 e^−0.6t = 1/19	✓ 1
A1 t = ln 19 / 0.6 = 4.91 days	✓ 1
Part (d)
M1 substitute N = 1000: 0.6 × 1000 × (1 − 1000/2000)	✓ 1
A1 300 people per day	✓ 1

Exam tip: For a logistic model, the limit as t → ∞ is the numerator of the fraction, because the exponential term in the denominator tends to zero.

+ Add to test✓ In your test162IBMI3H-2-05r26-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The capacity of a home battery after t years is modelled in two ways, as a percentage of its original capacity. Model A is exponential: C = 100 × 0.97^t. Model B is linear: C = 100 − 2.4t.

(a)

Interpret the value 0.97 in Model A.

(1)

(b)

Find which model predicts the higher capacity after 20 years, showing both values to 3 significant figures.

(2)

(c)

Find the value of t greater than 0 for which the two models give the same capacity, to 3 significant figures.

(2)

(d)

Using Model A, find how long it takes for the capacity to fall to 80%, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 the capacity falls to 97% of its previous value each year (a 3% yearly decrease)	✓ 1
Part (b)
A1 Model A: 54.4% and Model B: 52.0%	✓ 1
A1 Model A predicts the higher capacity	✓ 1
Part (c)
M1 solve 100 × 0.97^t = 100 − 2.4t using technology	✓ 1
A1 t = 16.3 years	✓ 1
Part (d)
M1 0.97^t = 0.8, so t = ln 0.8 / ln 0.97	✓ 1
A1 7.33 years	✓ 1

Exam tip: When two models are equated, an intersection at t = 0 is always present; read the other intersection from the graph on your calculator.

+ Add to test✓ In your test163IBMI3H-2-05r26-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A company models the satisfaction score S of customers x months after joining a loyalty scheme by S(x) = 2 ln(x + 3) − 1, for x ≥ 0.

(a)

Describe a sequence of three transformations that maps the graph of y = ln x onto the graph of y = S(x).

(3)

(b)

Write down the equation of the vertical asymptote of the graph of y = S(x).

(1)

(c)

Find the number of months after which the score first reaches 4, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 3 units in the negative x-direction	✓ 1
A1 vertical stretch with scale factor 2	✓ 1
A1 translation 1 unit downwards (applied after the stretch)	✓ 1
Part (b)
A1 x = −3	✓ 1
Part (c)
M1 2 ln(x + 3) − 1 = 4, so ln(x + 3) = 2.5	✓ 1
A1 x = e^2.5 − 3 = 9.18 months	✓ 1

Common error: After the vertical stretch by 2, the shift of −1 is not doubled: 2 ln(x + 3) − 1 is not 2 ln(x + 3 − 1).

+ Add to test✓ In your test164IBMI3H-2-05r26-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

A study of towns finds that a graph of log₁₀ E against log₁₀ P is a straight line with gradient 0.85 and vertical intercept −1, where E is the annual electricity use in GWh and P is the population.

(a)

Show that E = kPⁿ and find the values of n and k.

(3)

(b)

Find the annual electricity use of a town of population 20 000, to the nearest GWh.

(2)

(c)

The population of a town increases by 40%. Find the percentage increase in its annual electricity use, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 log₁₀ E = 0.85 log₁₀ P − 1, so E = 10⁻¹ × P^0.85	✓ 1
A1 n = 0.85	✓ 1
A1 k = 0.1	✓ 1
Part (b)
M1 0.1 × 20000^0.85	✓ 1
A1 453 GWh (452.8…)	✓ 1
Part (c)
M1 new E = 0.1(1.4P)^0.85 = 1.4^0.85 × E	✓ 1
A1 1.4^0.85 = 1.331…	✓ 1
A1 increase of 33.1%	✓ 1

Common error: A gradient below 1 on the log–log graph means electricity use grows more slowly than population; do not assume a 40% increase in population gives a 40% increase in use.

+ Add to test✓ In your test165IBMI3H-2-04r27-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A logistics company models the average cost per parcel, C euros, of a delivery round with n parcels by C = a/n + b, where a and b are constants. When 50 parcels are delivered the average cost is €21.00, and when 100 parcels are delivered the average cost is €16.50.

(a)

Find the values of a and b.

(3)

(b)

Write down the equation of the horizontal asymptote of the graph of C against n, and explain what it means in this context.

(2)

(c)

Find the least number of parcels needed for the average cost to be at most €14.00.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 forming a/50 + b = 21 and a/100 + b = 16.5	✓ 1
A1 a = 450	✓ 1
A1 b = 12	✓ 1
Part (b)
A1 C = 12	✓ 1
R1 however many parcels are delivered, the average cost per parcel approaches €12 but never goes below it	✓ 1
Part (c)
M1 450/n + 12 ≤ 14	✓ 1
A1 225 parcels	✓ 1

Common error: Do not round the answer down: the number of parcels must be a whole number that satisfies the inequality. Check by substituting.

+ Add to test✓ In your test166IBMI3H-2-04r27-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The height y metres of a curved facade at horizontal distance x metres is modelled by g(x) = 3 ln(x − 2) + 1, for x > 2. The graph is obtained from the graph of y = ln x by a sequence of transformations.

(a)

Describe the transformations, in the order they are applied.

(3)

(b)

Write down the equation of the vertical asymptote of the graph of y = g(x).

(1)

(c)

Find the x-coordinate where the graph crosses the x-axis, to 3 significant figures.

(2)

(d)

The graph of y = g(x) is reflected in the x-axis to give the graph of y = h(x). Find h(5), to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal translation of 2 units to the right	✓ 1
A1 vertical stretch, scale factor 3	✓ 1
A1 vertical translation of 1 unit up	✓ 1
Part (b)
A1 x = 2	✓ 1
Part (c)
M1 3 ln(x − 2) + 1 = 0	✓ 1
A1 x = 2 + e^−1/3 = 2.72	✓ 1
Part (d)
M1 h(5) = −g(5) = −(3 ln 3 + 1)	✓ 1
A1 −4.30	✓ 1

Exam tip: A reflection in the x-axis changes the sign of the whole function: h(x) = −g(x), not g(−x).

+ Add to test✓ In your test167IBMI3H-2-05r27-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The population P of a species of fish in a managed lake, t years after stocking began, is modelled by P = 2400 / (1 + 11e^−0.35t).

(a)

Find the initial population.

(2)

(b)

State the value that the population approaches in the long term.

(1)

(c)

Find the time taken for the population to reach 1500. Give your answer to 3 significant figures.

(3)

(d)

The population grows fastest when it is half of its long-term value. Find, to 3 significant figures, the value of t at which this occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes t = 0 : 2400/(1 + 11)	✓ 1
A1 200 fish	✓ 1
Part (b)
A1 2400 fish	✓ 1
Part (c)
M1 2400/(1 + 11e^−0.35t) = 1500, so 1 + 11e^−0.35t = 1.6	✓ 1
M1 e^−0.35t = 0.6/11 and takes natural logarithms	✓ 1
A1 t = 8.31 years	✓ 1
Part (d)
M1 sets P = 1200 : 1 + 11e^−0.35t = 2 so e^−0.35t = 1/11	✓ 1
A1 t = ln(11)/0.35 = 6.85 years	✓ 1

Exam tip: For a logistic model the point of fastest growth is where the population is half of the carrying capacity; use this to avoid differentiating.

+ Add to test✓ In your test168IBMI3H-2-05r27-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The signal power P mW received by an underwater robot from a beacon t minutes after the beacon is switched on is modelled by P = Ae^kt. A graph of ln P against t is a straight line passing through (2, 3.10) and (8, 1.60).

(a)

Find the gradient of the line.

(2)

(b)

Hence find the values of k and A. Give A to 3 significant figures.

(3)

(c)

The robot can only decode signals above 5 mW. Using A = e^3.60 and k = −0.25, find the time after which the signal can no longer be decoded. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1.60 − 3.10)/(8 − 2)	✓ 1
A1 −0.25	✓ 1
Part (b)
A1 k = −0.25 (ECF from (a))	✓ 1
M1 ln A = 3.10 + 0.25 × 2 (substitutes a point into ln P = ln A + kt)	✓ 1
A1 A = e^3.60 = 36.6	✓ 1
Part (c)
M1 e^3.60e^−0.25t = 5 (or 3.60 − 0.25t = ln 5)	✓ 1
M1 rearranges using natural logarithms: t = (3.60 − ln 5)/0.25	✓ 1
A1 t = 7.96 minutes	✓ 1

Common error: On a graph of ln P against t, the intercept is ln A; forgetting to apply the exponential to it gives a wrong value of A.

+ Add to test✓ In your test169IBMI3H-2-05r27-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A boutique sells hand-dyed silk scarves. When the selling price is €x, the weekly profit is €P, where P(x) = −3x² + 84x − 300, for 0 < x < 28.

(a)

Find the selling price that gives the greatest weekly profit.

(2)

(b)

Find the greatest weekly profit.

(2)

(c)

The owner wants a weekly profit of at least €150. Find the range of selling prices that achieves this. Give your values to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses x = −84/(2 × −3) (or completes the square / uses the GDC)	✓ 1
A1 €14	✓ 1
Part (b)
M1 substitutes x = 14	✓ 1
A1 €288	✓ 1
Part (c)
M1 sets −3x² + 84x − 300 = 150	✓ 1
M1 solves the quadratic, e.g. x = 14 ± √46	✓ 1
A1 7.22 ≤ x ≤ 20.8	✓ 1

Exam tip: For "at least" or "no more than" questions, solve the equation first, then use the shape of the graph to decide which side of the roots is required.

+ Add to test✓ In your test170IBMI3H-2-05r27-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The height H metres above the floor of the gripper of a robotic arm, t seconds after the start of a task, is modelled by H = a cos(b(t − 0.5)) + d, where a > 0 and b > 0. The gripper's greatest height is 1.8 m, first reached when t = 0.5. Its lowest height is 0.6 m, first reached when t = 1.7.

(a)

Find the value of a and the value of d.

(2)

(b)

Find the value of b, correct to 3 significant figures.

(2)

(c)

In each cycle, the gripper is above 1.5 m for a time of T seconds. Using the exact value of b, find T.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (1.8 − 0.6)/2 = 0.6	✓ 1
A1 d = (1.8 + 0.6)/2 = 1.2	✓ 1
Part (b)
M1 half a period is 1.7 − 0.5 = 1.2, so period = 2.4 s (and b = 2π/period)	✓ 1
A1 b = 2.62 (exact value 5π/6)	✓ 1
Part (c)
M1 0.6 cos(b(t − 0.5)) + 1.2 = 1.5	✓ 1
A1 cos(b(t − 0.5)) = 0.5	✓ 1
M1 uses symmetry: \|b(t − 0.5)\| < π/3, so interval length = 2π/(3b)	✓ 1
A1 T = 0.8 s	✓ 1

Exam tip: Half a period separates a maximum and the next minimum; use this rather than counting full cycles.

+ Add to test✓ In your test171IBMI3H-2-05r27-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium5 marks

The sound intensity level L decibels (dB) of a sound of intensity I W m⁻² is given by L = 10 log₁₀(I/I₀), where I₀ is a constant reference intensity.

(a)

An echo-sounder on a fishing vessel produces a sound level of 63 dB at a point in the water. A second sounder produces 84 dB at the same point. Find the ratio of the intensity of the second sound to that of the first. Give your answer to 3 significant figures.

(2)

(b)

Two engines run at the same time. Alone, the first produces a sound level of 60 dB and the second 66 dB at a hydrophone. Intensities add when sounds combine. Find the combined sound level, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 log₁₀(I₂/I₁) = (84 − 63)/10 = 2.1	✓ 1
A1 ratio = 10^2.1 = 126	✓ 1
Part (b)
M1 I/I₀ = 10⁶ and 10^6.6 for the two engines	✓ 1
M1 adds the intensities: 10⁶ + 10^6.6 and applies 10 log₁₀(…)	✓ 1
A1 67.0 dB	✓ 1

Common error: Decibel levels cannot be added directly; convert to intensities, add, then convert back.

+ Add to test✓ In your test172IBMI3H-2-06r31-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A games programmer measures how the frame-processing time T (milliseconds) depends on the number of objects N in a scene. She models the relationship as T = aNⁿ. When ln T is plotted against ln N, the points lie close to the straight line ln T = 1.5 ln N + 0.4.

(a)

Find the values of n and a. Give a to 3 significant figures.

(3)

(b)

Predict the processing time for a scene with 250 objects. Use the unrounded value of a and give your answer to 3 significant figures.

(2)

(c)

Find the factor by which the processing time increases when the number of objects doubles. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 take logarithms: ln T = ln a + n ln N	✓ 1
A1 n = 1.5	✓ 1
A1 a = e^0.4 = 1.49	✓ 1
Part (b)
M1 T = e^0.4 × 250^1.5	✓ 1
A1 5.90 × 10³ ms	✓ 1
Part (c)
M1 (2N)ⁿ ÷ Nⁿ = 2^1.5	✓ 1
A1 2.83	✓ 1

Exam tip: For a power model y = axⁿ, the gradient of the ln y against ln x line is the power n, and the intercept is ln a, not a.

+ Add to test✓ In your test173IBMI3H-2-06r31-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The concentration C (mg dm⁻³) of a drug in a patient's blood is measured at times t hours after a dose. A graph of log₁₀ C against t is a straight line with intercept 2.3 on the vertical axis and gradient −0.1.

(a)

Show that C can be written as C = ab^t, and find the values of a and b to 3 significant figures.

(3)

(b)

Interpret the value of b in this context.

(1)

(c)

Find the half-life of the drug, giving your answer in hours to 3 significant figures.

(2)

(d)

The patient must not receive a second dose until the concentration has fallen to 25 mg dm⁻³. Find, to 3 significant figures, how many hours after the first dose this happens.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 log₁₀ C = 2.3 − 0.1t so C = 10^2.3 × (10^−0.1)^t	✓ 1
A1 a = 200	✓ 1
A1 b = 0.794	✓ 1
Part (b)
R1 each hour the concentration is multiplied by 0.794, a decrease of about 20.6% per hour	✓ 1
Part (c)
M1 10^−0.1t = 0.5 (or log₁₀ 0.5 = −0.1t)	✓ 1
A1 3.01 hours	✓ 1
Part (d)
M1 2.3 − 0.1t = log₁₀ 25	✓ 1
A1 9.02 hours	✓ 1

Exam tip: A straight line on a semi-log graph (log of the quantity against time) indicates exponential behaviour. Raise 10 to both sides of the line equation to recover the model.

+ Add to test✓ In your test174IBMI3H-2-06r31-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

A water company is connecting households to a new supply network. The number of connected households t months after the project starts is modelled by N(t) = 1200 ÷ (1 + 9e^−0.4t).

(a)

Find the number of households connected at the start of the project.

(2)

(b)

Find the value that N(t) approaches for large t, and explain why.

(2)

(c)

Find the time at which 600 households are connected. Give your answer in months to 3 significant figures.

(3)

(d)

Explain why the number of new connections per month is greatest at about this time.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0: 1200 ÷ (1 + 9)	✓ 1
A1 120	✓ 1
Part (b)
M1 e^−0.4t → 0 as t → ∞	✓ 1
A1 N → 1200 households (the capacity of the network)	✓ 1
Part (c)
M1 1200 ÷ (1 + 9e^−0.4t) = 600, so 9e^−0.4t = 1	✓ 1
M1 e^−0.4t = 1/9, so t = ln 9 ÷ 0.4	✓ 1
A1 5.49 months	✓ 1
Part (d)
R1 the growth rate of a logistic model is greatest when N is half of the limiting value (600 of 1200); before this the growth is still speeding up, and after it the network nears capacity so growth slows	✓ 1

Exam tip: A logistic curve has its steepest point where the value is half of the limiting value. At that point the model changes from speeding up to slowing down.

+ Add to test✓ In your test175IBMI3H-2-06r31-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The water level in a reservoir, h metres, varies with time t hours after midnight and is modelled by h(t) = a cos(b(t − 3)) + d, where b is in radians per hour. The maximum level is 14.2 m at t = 3, and the minimum level is 9.8 m at t = 9.

(a)

Find the values of a, b and d.

(3)

(b)

For how many hours, in the interval 0 ≤ t ≤ 6, is the water level at least 13 m? Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (14.2 − 9.8) ÷ 2 = 2.2	✓ 1
A1 d = (14.2 + 9.8) ÷ 2 = 12	✓ 1
A1 half a period is 6 hours, so the period is 12 and b = 2π ÷ 12 = π/6 (0.524)	✓ 1
Part (b)
M1 2.2 cos(π(t − 3)/6) + 12 = 13, so cos(π(t − 3)/6) = 1 ÷ 2.2	✓ 1
A1 the boundary times are t = 0.901 and t = 5.10	✓ 1
A1 duration 4.20 hours	✓ 1

Exam tip: Amplitude is half the difference between maximum and minimum; the midline is their mean. For a period of 12 use b = 2π ÷ 12, and make sure your calculator is in radian mode.

+ Add to test✓ In your test176IBMI3H-2-06r31-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The effect of a power boost in a game is described by the function f(x) = ln x, for x > 0. A designer defines a second function g(x) = ln(5x).

(a)

Show that the graph of y = g(x) is a translation of the graph of y = f(x), and write down the translation vector. Give any decimal to 3 significant figures.

(2)

(b)

Describe the single stretch that also maps the graph of y = f(x) onto the graph of y = g(x).

(1)

(c)

Find the x-coordinate of the point where the graph of y = g(x) crosses the x-axis.

(2)

(d)

A third function is k(x) = ln(x ÷ 4). Write k(x) in the form ln x + c and find c to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 ln(5x) = ln 5 + ln x (laws of logarithms)	✓ 1
A1 translation vector (0, ln 5) = (0, 1.61)	✓ 1
Part (b)
A1 horizontal stretch (parallel to the x-axis) with scale factor 1/5	✓ 1
Part (c)
M1 ln(5x) = 0 so 5x = 1	✓ 1
A1 x = 1/5	✓ 1
Part (d)
M1 ln(x ÷ 4) = ln x − ln 4	✓ 1
A1 c = −1.39	✓ 1

Exam tip: A change inside a logarithm can be rewritten as a vertical translation using ln(ab) = ln a + ln b, so a horizontal stretch and a vertical translation can give the same graph.

+ Add to test✓ In your test177IBMI3H-2-06r33-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A cyclist’s power output P watts needed to ride at constant speed v m s⁻¹ on level ground is modelled by $P=kv^n$. At 8 m s⁻¹ the power needed is 110 W and at 12 m s⁻¹ it is 371 W.

(a)

By taking natural logarithms of the model, find the value of n correct to three significant figures.

(3)

(b)

Take n = 3. Find the value of k, correct to three significant figures.

(2)

(c)

Use the model with n = 3 and k = 110 ÷ 512 to find the speed at which the power needed is 600 W. Give your answer to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\ln P=n\ln v+\ln k$, and forms two equations or the gradient between (ln 8, ln 110) and (ln 12, ln 371)	✓ 1
M1 $n=\dfrac{\ln 371-\ln 110}{\ln 12-\ln 8}$	✓ 1
A1 n = 3.00	✓ 1
Part (b)
M1 110 = k × 8³	✓ 1
A1 k = 0.215	✓ 1
Part (c)
M1 600 = (110/512)v³ leading to v³ = 2792.7…	✓ 1
A1 14.1 m s⁻¹	✓ 1

Common error: A power model gives a straight line on a graph of ln P against ln v, whose gradient is n; do not plot ln P against v, which is the test for an exponential model.

+ Add to test✓ In your test178IBMI3H-2-06r33-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The sound level L decibels (dB) of a source of intensity I W m⁻² is given by $L=10\log_{10}\left(\dfrac{I}{I_0}\right)$, where $I_0=1\times10^{-12}$ W m⁻². In a data centre the cooling units together produce a sound level of 78 dB.

(a)

Find the intensity of the sound from the cooling units. Give your answer in the form a × 10^k with 1 ≤ a < 10 and a to three significant figures.

(2)

(b)

A second identical set of cooling units is switched on. Find the new sound level, correct to the nearest 0.1 dB.

(2)

(c)

Each set of units alone produces 78 dB. Find the smallest whole number of identical sets that together produce a sound level of at least 90 dB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $78=10\log_{10}(I/10^{-12})$ giving $I=10^{7.8}\times10^{-12}$	✓ 1
A1 I = 6.31 × 10⁻⁵ W m⁻²	✓ 1
Part (b)
M1 intensity doubles, so L = 10 log₁₀(2I/I₀) = 78 + 10 log₁₀ 2	✓ 1
A1 81.0 dB	✓ 1
Part (c)
M1 10 log₁₀ n ≥ 12 giving n ≥ 15.85…	✓ 1
A1 16 sets	✓ 1

Common error: Sound levels in decibels do not add: doubling the intensity raises the level by about 3 dB, not by a factor of two.

+ Add to test✓ In your test179IBMI3H-2-06r33-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

Archaeologists date a wooden beam using carbon-14. The amount of carbon-14 remaining after t years is $N=N_0\left(\tfrac12\right)^{t/5730}$, where N₀ is the original amount.

(a)

Show that this model can be written as $N=N_0e^{-\lambda t}$ and find the value of $\lambda$ to three significant figures.

(2)

(b)

The beam contains 62% of its original carbon-14. Find its age, correct to the nearest 10 years.

(3)

(c)

Find the percentage of the original carbon-14 remaining in a sample 8000 years old, correct to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\left(\tfrac12\right)^{t/5730}=e^{-t\ln 2/5730}$, so $\lambda=\dfrac{\ln2}{5730}$	✓ 1
A1 λ = 1.21 × 10⁻⁴	✓ 1
Part (b)
M1 0.62 = (1/2)^t/5730	✓ 1
A1 t = 5730 × ln 0.62 ÷ ln 0.5 = 3951.8…	✓ 1
A1 3950 years	✓ 1
Part (c)
M1 (1/2)^8000/5730	✓ 1
A1 38.0%	✓ 1

Exam tip: Half-life models can be written with base 1/2 or with base e; both give the same values, so use whichever makes the unknown easier to isolate.

+ Add to test✓ In your test180IBMI3H-2-06r33-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The monthly data traffic V petabytes (PB) of a data centre is recorded each year for 5 years, where t is the number of years after the first record.
t: 0, 1, 2, 3, 4, 5
V: 13.0, 15.7, 25.7, 35.8, 52.1, 76.3
The data are modelled by $V=ab^t$.

(a)

Complete a table of values of ln V for the six data points, to three decimal places.

(2)

(b)

Use linear regression on the six points (t, ln V) to find the equation of the regression line of ln V on t, with coefficients to three significant figures.

(2)

(c)

Hence find the values of a and b, correct to three significant figures.

(2)

(d)

Use the unrounded regression line to predict the monthly data traffic when t = 8, correct to the nearest PB. Comment on the reliability of this prediction.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to find ln of the V values	✓ 1
A1 2.565, 2.754, 3.246, 3.578, 3.953, 4.335	✓ 1
Part (b)
M1 attempt at the regression line using the calculator	✓ 1
A1 ln V = 0.365t + 2.49	✓ 1
Part (c)
M1 a = e^2.49… and b = e^0.365…	✓ 1
A1 a = 12.1 and b = 1.44	✓ 1
Part (d)
A1 224 PB	✓ 1
R1 this is extrapolation beyond the data range (t = 0 to 5), so the model may not remain valid; any equivalent comment	✓ 1

Exam tip: If ln V is linear in t, the gradient is ln b and the intercept is ln a; always convert back before giving the model in the form abt.

+ Add to test✓ In your test181IBMI3H-2-06r33-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The drag force F newtons on a cyclist riding at speed v m s⁻¹ is measured in a wind tunnel.
v (m s⁻¹): 4, 6, 8, 10, 12
F (N): 6.1, 13.7, 24.5, 38.2, 55.0
The data are modelled by $F=kv^n$.

(a)

Find the regression line of ln F on ln v, with coefficients to four significant figures.

(3)

(b)

Hence find the values of n and k, correct to three significant figures.

(2)

(c)

Use the unrounded values to predict the drag force at 15 m s⁻¹, correct to three significant figures.

(1)

(d)

Interpret the value of n in this context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to find ln v and ln F for the five points	✓ 1
M1 uses linear regression on (ln v, ln F)	✓ 1
A1 ln F = 2.003 ln v − 0.9688	✓ 1
Part (b)
A1 n = 2.00	✓ 1
A1 k = e^−0.9688 = 0.380	✓ 1
Part (c)
A1 86.1 N	✓ 1
Part (d)
R1 n ≈ 2, so doubling the speed roughly multiplies the drag force by four	✓ 1

Common error: Regression for a power model uses ln v and ln F for both variables; taking logs of only the output variable fits an exponential model instead.

+ Add to test✓ In your test182IBMI3H-2-06r33-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium5 marks

The energy E joules released by an earthquake of magnitude M satisfies log₁₀ E = 4.8 + 1.5M. Two earthquakes of magnitudes 5.2 and 6.4 are recorded near an ancient temple site.

(a)

Find the ratio of the energy released by the larger earthquake to the energy released by the smaller one, correct to three significant figures.

(3)

(b)

By how much must the magnitude increase for the energy released to be multiplied by 1000?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 subtracts the equations: log₁₀(E₂/E₁) = 1.5 × (6.4 − 5.2)	✓ 1
A1 log₁₀(E₂/E₁) = 1.8	✓ 1
A1 ratio = 10^1.8 = 63.1	✓ 1
Part (b)
M1 1.5 × ΔM = log₁₀ 1000 = 3	✓ 1
A1 ΔM = 2	✓ 1

Common error: Each unit of magnitude multiplies the energy by 10¹.⁵ ≈ 31.6, not by 10; the factor depends on the coefficient 1.5 in the scale.

+ Add to test✓ In your test183IBMI3H-2-06r33-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

A geologist models the age A (thousands of years) of a buried layer at depth d metres, for d > 2, by $A=3\ln(d-2)+1$ at an archaeological dig.

(a)

Describe a sequence of transformations that maps the graph of $y=\ln x$ onto the graph of $A=3\ln(d-2)+1$.

(3)

(b)

Write down the equation of the vertical asymptote of the graph, and state what this means for the model.

(1)

(c)

Find the depth at which the model gives an age of zero, correct to three significant figures.

(2)

(d)

Find the depth at which the age is 4 thousand years, correct to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 2 units in the positive x-direction, by the vector (2, 0)	✓ 1
A1 vertical stretch with scale factor 3	✓ 1
A1 translation 1 unit up, by (0, 1), after the stretch (the horizontal translation may be done at any stage)	✓ 1
Part (b)
A1 d = 2; the model is only valid below a depth of 2 m (any equivalent interpretation)	✓ 1
Part (c)
M1 3 ln(d − 2) + 1 = 0 giving d − 2 = e^−1/3	✓ 1
A1 2.72 m	✓ 1
Part (d)
M1 3 ln(d − 2) = 3 giving d − 2 = e	✓ 1
A1 4.72 m	✓ 1

Exam tip: A horizontal translation changes the asymptote of a logarithmic graph: replacing x with x − 2 moves the asymptote from x = 0 to x = 2.

+ Add to test✓ In your test184IBMI3H-2-05r35-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The natural frequency f (Hz) of a bridge cable of length L (m) is modelled by f = aL^b. A graph of ln f against ln L is a straight line with equation ln f = 4.25 − 0.600 ln L.

(a)

Find the value of b and the value of a, giving a to 3 significant figures.

(3)

(b)

Use the equation of the line to predict the natural frequency of a cable of length 50 m.

(2)

(c)

Find the length of cable for which the natural frequency is 10.0 Hz.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 exponentiate: f = e^4.25 × L^−0.600	✓ 1
A1 b = −0.600	✓ 1
A1 a = e^4.25 = 70.1	✓ 1
Part (b)
M1 ln f = 4.25 − 0.600 ln 50 (= 1.9027…)	✓ 1
A1 f = 6.70 Hz	✓ 1
Part (c)
M1 ln 10 = 4.25 − 0.600 ln L	✓ 1
M1 ln L = (4.25 − ln 10) ÷ 0.600 (= 3.2457…)	✓ 1
A1 L = 25.7 m	✓ 1

Common error: Do not forget to undo the logarithm at the end: the line gives ln L, so use ex to find L itself.

+ Add to test✓ In your test185IBMI3H-2-05r35-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

The water level h (m above chart datum) in a harbour is modelled by h(t) = 3.4 + 1.8 cos(πt ÷ 6), where t is the time in hours after high tide, and t is measured in the domain 0 ≤ t ≤ 12. The underside of a bridge over the harbour is 7.2 m above chart datum.

(a)

Write down the maximum and the minimum water level.

(2)

(b)

Write down the period of the tide.

(1)

(c)

Find the minimum clearance between the water and the underside of the bridge.

(2)

(d)

A boat needs a water level of at least 2.5 m. Find the length of time, in the first 12 hours, for which the water level is below 2.5 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 maximum = 3.4 + 1.8 = 5.2 m	✓ 1
A1 minimum = 3.4 − 1.8 = 1.6 m	✓ 1
Part (b)
A1 period = 2π ÷ (π ÷ 6) = 12 hours	✓ 1
Part (c)
M1 7.2 − 5.2	✓ 1
A1 2.0 m	✓ 1
Part (d)
M1 3.4 + 1.8 cos(πt ÷ 6) = 2.5, so cos(πt ÷ 6) = −0.5	✓ 1
A1 t = 4 and t = 8	✓ 1
A1 duration = 4.00 hours	✓ 1

Exam tip: Sketch the cosine curve first so you can see that the water is below the target between two solutions, not just after the first one.

+ Add to test✓ In your test186IBMI3H-2-05r35-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The charge in a space probe's battery is predicted by two models, where t is the time in hours after the battery is disconnected from its charger and charge is given as a percentage. Model A (linear): L(t) = 95 − 4.5t. Model B (exponential): E(t) = 95 × 0.94^t.

(a)

Find the charge predicted by each model after 8 hours.

(2)

(b)

Use your GDC to find the positive value of t at which the two models predict the same charge.

(2)

(c)

Find the time at which Model A predicts that the charge reaches 0 %, and state why Model B never predicts a charge of exactly 0 %.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 Model A: 95 − 4.5 × 8 = 59.0 %	✓ 1
A1 Model B: 95 × 0.94⁸ = 57.9 %	✓ 1
Part (b)
M1 solve 95 − 4.5t = 95 × 0.94^t (graphs intersect)	✓ 1
A1 t = 9.06 hours	✓ 1
Part (c)
M1 95 − 4.5t = 0	✓ 1
A1 t = 21.1 hours	✓ 1
R1 0.94^t > 0 for all t, so Model B only approaches 0 and never reaches it	✓ 1

Common error: The graphs also meet at t = 0, so check that your GDC intersection is the positive one.

+ Add to test✓ In your test187IBMI3H-2-06r38-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The sound intensity level of a robot motor is given by L = 10 log₁₀(I / I₀) dB, where I is the intensity in W m⁻² and I₀ = 10⁻¹² W m⁻².

(a)

One motor has a level of 72 dB and another has a level of 65 dB. Find the ratio of the intensity of the first to that of the second, to 3 significant figures.

(2)

(b)

Show that two identical 65 dB motors running together produce a level of approximately 68 dB.

(2)

(c)

The 72 dB motor and the 65 dB motor run together. Find the combined level, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 difference of 7 dB gives 7 = 10 log₁₀(ratio)	✓ 1
A1 10^0.7 = 5.01	✓ 1
Part (b)
M1 intensity doubles, so level increases by 10 log₁₀2 = 3.01 dB	✓ 1
R1 65 + 3.01 = 68.01 ≈ 68 dB	✓ 1
Part (c)
M1 intensities proportional to 10^7.2 and 10^6.5 (in units of I₀)	✓ 1
M1 add intensities and apply 10 log₁₀	✓ 1
A1 72.8 dB	✓ 1

Common error: Decibel levels cannot be added directly: convert to intensities, add those, then convert back.

+ Add to test✓ In your test188IBMI3H-2-06r38-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The graph of y = f(x) passes through the points A(−2, 4), B(1, −3) and C(6, 0), and the maximum value of f(x) is 7. A second graph has equation y = g(x) where g(x) = 3f(x/2) − 2.

(a)

Describe fully the three transformations that map the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

Find the coordinates of the images of A and C on the graph of y = g(x).

(2)

(c)

Find the maximum value of g(x).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal stretch (parallel to the x-axis) with scale factor 2	✓ 1
A1 vertical stretch (parallel to the y-axis) with scale factor 3	✓ 1
A1 translation by (0, −2), applied after the vertical stretch	✓ 1
Part (b)
M1 (x, y) → (2x, 3y − 2)	✓ 1
A1 (−4, 10) and (12, −2)	✓ 1
Part (c)
M1 3 × 7 − 2	✓ 1
A1 19	✓ 1

Common error: A horizontal stretch by factor 2 appears as x/2 inside the function, so the x-coordinates are multiplied by 2, not halved.

+ Add to test✓ In your test189IBMI3H-2-06r38-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The stiffness S (N mm⁻¹) of a robot leg is modelled by S = kLⁿ, where L is the length of the leg in metres and k and n are constants. When the length of a leg is doubled, its stiffness is multiplied by 6.5.

(a)

Find the value of n, to 3 significant figures.

(2)

(b)

A leg of length 0.4 m has stiffness 12 N mm⁻¹. Find the value of k to the nearest whole number, using the unrounded value of n.

(2)

(c)

The length of a leg is increased by 30%. Find the percentage increase in stiffness, to 3 significant figures, using the unrounded value of n.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2ⁿ = 6.5	✓ 1
A1 n = log 6.5 / log 2 = 2.70	✓ 1
Part (b)
M1 12 = k × 0.4ⁿ	✓ 1
A1 k = 142	✓ 1
Part (c)
M1 stiffness multiplied by 1.3ⁿ	✓ 1
A1 2.03	✓ 1
A1 103%	✓ 1

Exam tip: A percentage change in length of 30% is a multiplier of 1.3, so the stiffness multiplier is 1.3 raised to the power n.

+ Add to test✓ In your test190IBMI3H-2-05r41-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium8 marks

Biologists model the resting metabolic rate R watts of a marine mammal of mass M kg by the linear relationship log₁₀ R = 0.75 log₁₀ M + 0.62.

(a)

Show that the relationship can be written in the form R = a M^k, and find the value of k and the value of a to 3 significant figures.

(3)

(b)

An orca has a mass of 4000 kg. Use the model to estimate its metabolic rate, giving your answer to 3 significant figures. Use the unrounded value of a.

(2)

(c)

A seal has a metabolic rate of 95 W. Find its mass to the nearest kilogram. Use the unrounded value of a.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 writing log₁₀ R = log₁₀(10^0.62 M^0.75) or taking powers of 10 on both sides	✓ 1
A1 k = 0.75	✓ 1
A1 a = 10^0.62 = 4.17	✓ 1
Part (b)
M1 substitution of M = 4000 into R = a M^0.75	✓ 1
A1 R = 2100 W (accept 2097 W)	✓ 1
Part (c)
M1 95 = a M^0.75 and rearranging to M^0.75 = 95 / a	✓ 1
M1 raising both sides to the power 1/0.75	✓ 1
A1 M = 65 kg	✓ 1

Exam tip: When a linear relationship is given between logarithms, take powers of the base on both sides to recover the original model; the intercept becomes the multiplying constant.

+ Add to test✓ In your test191IBMI3H-2-05r41-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

Ballast water taken on by a cargo ship contains bacteria. A technician records the number of cells per millilitre, N, after t hours and finds that the points (t, log₁₀ N) lie on the straight line log₁₀ N = 2.4 + 0.045t.

(a)

Find the initial number of cells per millilitre, giving your answer to 3 significant figures.

(2)

(b)

Show that the number of cells increases by approximately 10.9 % each hour.

(2)

(c)

Find the time, in hours, for the concentration to reach 5000 cells per millilitre. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 N = 10^2.4 (when t = 0)	✓ 1
A1 N = 251	✓ 1
Part (b)
M1 N = 10^2.4 × (10^0.045)^t identifying the hourly growth factor 10^0.045	✓ 1
A1 10^0.045 = 1.1092, so the increase is 10.9 % (to 3 s.f.)	✓ 1
Part (c)
M1 log₁₀ 5000 = 2.4 + 0.045t	✓ 1
M1 rearranging for t	✓ 1
A1 t = 28.9 hours	✓ 1

Exam tip: A straight line on a graph of log N against t means N is exponential in t; the gradient gives log₁₀ of the growth factor, not the growth factor itself.

+ Add to test✓ In your test192IBMI3H-2-05r41-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The acidity of seawater is measured by pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³.

(a)

A sample of surface seawater has [H⁺] = 7.9 × 10⁻⁹ mol dm⁻³. Calculate its pH to 3 significant figures.

(2)

(b)

Dissolved carbon dioxide increases the hydrogen ion concentration of this sample by 26 %. Find the new pH to 3 significant figures. Use the unrounded pH from part (a) if you need it.

(2)

(c)

In another experiment the pH of a water sample falls by 0.25. Find the factor by which [H⁺] is multiplied, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −log₁₀(7.9 × 10⁻⁹)	✓ 1
A1 pH = 8.10	✓ 1
Part (b)
M1 new [H⁺] = 1.26 × 7.9 × 10⁻⁹ (accept pH_new = pH_old − log₁₀ 1.26)	✓ 1
A1 pH = 8.00	✓ 1
Part (c)
M1 −log₁₀(f) = −0.25 or [H⁺]_new / [H⁺]_old = 10^0.25	✓ 1
A1 factor = 1.78	✓ 1

Common error: A fall in pH means an increase in [H⁺]; the minus sign in the definition reverses the direction of the change.

+ Add to test✓ In your test193IBMI3H-2-05r41-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A shipping firm finds that the cost per container, C pounds, falls as the number of containers x in an order increases. It models this by C = a + b ln x. An order of 10 containers costs £640 per container and an order of 60 containers costs £520 per container.

(a)

Find the values of a and b, each to 3 significant figures.

(4)

(b)

Use the model to estimate the cost per container for an order of 200 containers. Use unrounded values of a and b and give your answer to the nearest pound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 640 = a + b ln 10 and 520 = a + b ln 60	✓ 1
M1 subtracting to eliminate a: −120 = b ln 6	✓ 1
A1 b = -67.0	✓ 1
A1 a = 794	✓ 1
Part (b)
M1 substitution of x = 200	✓ 1
A1 £439	✓ 1

Exam tip: Use two data points to set up a pair of simultaneous equations; subtracting them eliminates the constant term and leaves a single logarithm, such as ln 6.

+ Add to test✓ In your test194IBMI3H-2-05r41-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The weekly profit, P pounds, of a delivery company that runs n vans is modelled by P(n) = −25n² + 900n − 5000, where n is a whole number.

(a)

Find the number of vans that gives the greatest weekly profit, and state this profit.

(3)

(b)

Interpret the value −5000 in the model.

(1)

(c)

Find how many different fleet sizes give a positive weekly profit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 axis of symmetry n = −900 / (2 × −25) (or use of a graphic display calculator)	✓ 1
A1 n = 18	✓ 1
A1 maximum profit £3100	✓ 1
Part (b)
R1 with no vans the company makes a loss of £5000 per week (fixed weekly costs)	✓ 1
Part (c)
M1 solving −25n² + 900n − 5000 > 0 (roots 6.86… and 29.1…)	✓ 1
A1 7 ≤ n ≤ 29	✓ 1
A1 23 different fleet sizes	✓ 1

Common error: Roots of a quadratic model are rarely whole numbers; round inwards so that the stated condition (profit positive) is still true for every whole number in the range.

+ Add to test✓ In your test195IBMI3H-2-05r41-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A sample of water is taken from a depth of 400 m at a temperature of 4 °C and left in a laboratory where the air temperature is 21 °C. The sample temperature T °C after t minutes is modelled by T = 21 − 17e^−kt, where k is a positive constant.

(a)

State the temperature that the model predicts the sample will approach in the long term.

(1)

(b)

After 10 minutes the temperature of the sample is 15 °C. Find the value of k to 3 significant figures.

(3)

(c)

Find the time for the sample to reach 18 °C. Use the unrounded value of k and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 21 °C	✓ 1
Part (b)
M1 15 = 21 − 17e^−10k	✓ 1
M1 e^−10k = 6/17 and taking natural logarithms	✓ 1
A1 k = 0.104	✓ 1
Part (c)
M1 18 = 21 − 17e^−kt	✓ 1
M1 e^−kt = 3/17 and taking natural logarithms	✓ 1
A1 t = 16.7 minutes	✓ 1

Exam tip: The long-term value in a model of the form a − b e⁻kt is the constant a, because the exponential term tends to zero.

+ Add to test✓ In your test196IBMI3H-2-05r42-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The main cable of a footbridge is modelled by the curve y = f(x), where f(x) = 0.05x², with x and y in metres. A second, wider cable B is the image of this curve after a horizontal stretch with scale factor 3 (from the y-axis), followed by a translation with vector
(−10, 4).

(a)

The point (6, 1.8) lies on y = f(x). Find the coordinates of its image on cable B.

(2)

(b)

Write down the equation of cable B in the form y = g(x).

(2)

(c)

Write down the coordinates of the lowest point of cable B.

(1)

(d)

Find the value of g(20).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x multiplied by 3 then 10 subtracted (and 4 added to y)	✓ 1
A1 (8, 5.8)	✓ 1
Part (b)
M1 replaces x with x/3 and x + 10 (in either order of working) and adds 4	✓ 1
A1 g(x) = 0.05((x + 10)/3)² + 4	✓ 1
Part (c)
A1 (−10, 4)	✓ 1
Part (d)
M1 0.05((20 + 10)/3)² + 4	✓ 1
A1 9 m (9.00)	✓ 1

Common error: A stretch with scale factor 3 parallel to the x-axis replaces x by x/3, not by 3x; apply a translation after the stretch by substituting for x in the stretched function.

+ Add to test✓ In your test197IBMI3H-2-05r42-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The depth y mm of corrosion on a steel bridge girder is modelled by y = a b^x, where x is the time in years since the girder was installed. A graph of ln y against x is a straight line passing through the points (4, 1.28) and (14, 2.48).

(a)

Find the equation of the line in the form ln y = mx + c.

(2)

(b)

Find the values of a and b, correct to 3 significant figures.

(3)

(c)

Find the time taken for the depth of corrosion to double, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (2.48 − 1.28)/(14 − 4)	✓ 1
A1 ln y = 0.12x + 0.8 (m = 0.12, c = 0.80)	✓ 1
Part (b)
M1 y = e^{0.12x + 0.8} = e^0.8(e^0.12)^x	✓ 1
A1 a = 2.23	✓ 1
A1 b = 1.13	✓ 1
Part (c)
M1 b^x = 2 (or 0.12x = ln 2)	✓ 1
A1 x = 5.78 years	✓ 1

Exam tip: When ln y is plotted against x, the gradient is ln b and the intercept is ln a; take e to each of them to recover b and a.

+ Add to test✓ In your test198IBMI3H-2-05r42-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

In a coffee roaster the temperature excess of a cooling tray above its surroundings, in °C, is modelled by f(x) = 20e^−0.3x, where x is the time in minutes. A second tray is modelled by g(x) = 0.5f(x − 2) + 6.

(a)

Describe a sequence of three transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

Write down the equation of the horizontal asymptote of y = g(x).

(1)

(c)

Find g(0), correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation by vector (2, 0) (2 units to the right)	✓ 1
A1 vertical stretch with scale factor 0.5	✓ 1
A1 translation by vector (0, 6) (accept the translation (2, 0) at any stage, and the stretch before the final translation)	✓ 1
Part (b)
A1 y = 6	✓ 1
Part (c)
M1 0.5 × 20e^{−0.3(0 − 2)} + 6	✓ 1
A1 24.2	✓ 1

Common error: To find a horizontal translation read the bracket carefully: x − 2 shifts the graph 2 units to the right, which is the opposite of what the sign might suggest.

+ Add to test✓ In your test199IBMI3H-2-05r42-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A network of sensors is installed on a suspension bridge. The number of sensors that are online t hours after the network is switched on is modelled by N(t) = 240 / (1 + 19e^−0.4t), t ≥ 0.

(a)

Find the number of sensors online when the network is switched on.

(2)

(b)

Find the time at which exactly half of the 240 sensors are online. Give your answer to 3 significant figures.

(2)

(c)

Find the rate at which sensors are coming online when t = 5. Give your answer to 3 significant figures.

(2)

(d)

Explain why this model cannot be used to predict the number of sensors online after a very long time if more sensors are added to the bridge.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 240/(1 + 19)	✓ 1
A1 12	✓ 1
Part (b)
M1 120 = 240/(1 + 19e^−0.4t), so e^−0.4t = 1/19	✓ 1
A1 t = 7.36 hours	✓ 1
Part (c)
M1 gradient of N at t = 5 (GDC derivative)	✓ 1
A1 19.4 sensors per hour	✓ 1
Part (d)
R1 the model has a limiting value (horizontal asymptote) of 240 sensors, so it cannot show more than 240 online	✓ 1

Exam tip: For a logistic model the greatest rate of growth occurs when the quantity reaches half of its limiting value.

+ Add to test✓ In your test200IBMI3H-2-05r42-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The arch of a road bridge is modelled by a quadratic curve y = ax² + bx + c, where y is the height in metres above the water and x is the horizontal distance in metres from the left end of the arch. The curve passes through the points (0, 3), (20, 11.4) and (60, 18.6).

(a)

Find the values of a, b and c.

(3)

(b)

Find the greatest height of the arch above the water, correct to 3 significant figures.

(2)

(c)

The road deck is 10 m above the water. Find the horizontal width of the arch that lies above the level of the road deck, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 c = 3	✓ 1
M1 two simultaneous equations from the other two points (or quadratic regression on GDC)	✓ 1
A1 a = −0.004 and b = 0.5	✓ 1
Part (b)
M1 vertex at x = −b/(2a) = 62.5 (or GDC maximum)	✓ 1
A1 18.6 m	✓ 1
Part (c)
M1 solves −0.004x² + 0.5x + 3 = 10 (roots 16.06… and 108.9…)	✓ 1
A1 92.9 m	✓ 1

Exam tip: Three points on a quadratic give three linear equations; the point on the vertical axis gives the constant term immediately.

+ Add to test✓ In your test201IBMI3H-2-06r42-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A designer models the cross-section of a track kerb by y = f(x), where f(x) = x² − 4x, with x and y in centimetres. A redesigned kerb is modelled by y = g(x), where g(x) = −2f(x + 3).

(a)

Describe a sequence of three transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(3)

(b)

The graph of y = f(x) has its vertex at (2, −4). Write down the coordinates of the vertex of y = g(x).

(2)

(c)

Find the value of g(0).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation by vector (−3, 0)	✓ 1
A1 vertical stretch with scale factor 2	✓ 1
A1 reflection in the x-axis (accept the stretch and reflection combined, and in either order)	✓ 1
Part (b)
M1 applies the transformations to the vertex (2 → −1, −4 → 8)	✓ 1
A1 (−1, 8)	✓ 1
Part (c)
A1 −2 × (3² − 12) = 6	✓ 1

Exam tip: Describe each transformation separately; a horizontal translation affects only x, while the factor −2 acts on all y-values.

+ Add to test✓ In your test202IBMI3H-2-06r42-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The number of bacteria N on a hospital surface t hours after cleaning is modelled by N = a b^t. A graph of log₁₀ N against t is a straight line passing through (0, 3.0) and (6, 4.2).

(a)

Find the equation of the line in the form log₁₀ N = mt + c.

(2)

(b)

Find the values of a and b, giving b correct to 3 significant figures.

(3)

(c)

Find the time at which the number of bacteria reaches 1.0×10⁶.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (4.2 − 3.0)/6	✓ 1
A1 log₁₀ N = 0.2t + 3	✓ 1
Part (b)
M1 N = 10^{0.2t + 3} = 10³(10^0.2)^t	✓ 1
A1 a = 1000	✓ 1
A1 b = 1.58	✓ 1
Part (c)
M1 0.2t + 3 = 6	✓ 1
A1 t = 15.0 hours	✓ 1

Exam tip: On a semi-log graph, the intercept is log₁₀ a and the gradient is log₁₀ b; use powers of 10 to return to the original model.

+ Add to test✓ In your test203IBMI3H-2-06r42-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The blood pressure P₁ mmHg of a patient t seconds after a heartbeat is modelled by P₁(t) = 95 + 20 sin(2.5πt), where the angle is in radians. A second patient has pressure P₂(t) = P₁(t/1.5) − 10.

(a)

Find the period of P₁ and hence the heart rate of the first patient in beats per minute.

(2)

(b)

Describe the two transformations that map the graph of P₁ onto the graph of P₂.

(2)

(c)

Write down the greatest and least values of P₂.

(1)

(d)

Find the heart rate of the second patient in beats per minute.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 period = 2π/(2.5π) = 0.8 s	✓ 1
A1 60/0.8 = 75 beats per minute	✓ 1
Part (b)
A1 horizontal stretch with scale factor 1.5	✓ 1
A1 vertical translation by −10 (down 10 units)	✓ 1
Part (c)
A1 greatest 105 mmHg and least 65 mmHg	✓ 1
Part (d)
A1 60/(0.8 × 1.5) = 50 beats per minute	✓ 1

Exam tip: A horizontal stretch with scale factor s multiplies the period by s, so the rate of beats is divided by s.

+ Add to test✓ In your test204IBMI3H-2-06r42-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

The number of patients I infected on a hospital ward t days after the first case is reported is modelled by I(t) = 150 / (1 + 24e^−0.6t), t ≥ 0. The ward has 150 beds.

(a)

Find the number of patients infected when the first case is reported.

(2)

(b)

Find the time at which 90% of the beds on the ward contain infected patients. Give your answer to 3 significant figures.

(2)

(c)

State what value I(t) approaches as t becomes large, and explain this in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 150/(1 + 24)	✓ 1
A1 6	✓ 1
Part (b)
M1 150/(1 + 24e^−0.6t) = 135	✓ 1
A1 t = 8.96 days	✓ 1
Part (c)
A1 150	✓ 1
R1 every bed on the ward eventually holds an infected patient, so the number cannot exceed the number of beds	✓ 1

Exam tip: For logistic models always check what the limiting value represents in context; here it is fixed by the number of beds.

+ Add to test✓ In your test205IBMI3H-2-06r42-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A Formula 1 car's speed v m s⁻¹, t seconds after a standing start, is modelled by
v(t) = 12t for 0 ≤ t ≤ 2.5,
v(t) = −0.8t² + bt + 10 for 2.5 < t ≤ 6,
where b is a constant. The model is continuous at t = 2.5.

(a)

Find the value of b.

(2)

(b)

Find the time at which the speed is 35 m s⁻¹. Give your answer to 3 significant figures.

(3)

(c)

Find the speed at t = 6.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 2.5 = −0.8(2.5)² + 2.5b + 10	✓ 1
A1 b = 10	✓ 1
Part (b)
M1 −0.8t² + 10t + 10 = 35 (the first piece cannot reach 35 in its domain: 12 × 2.5 = 30)	✓ 1
A1 roots 3.45 and 9.05	✓ 1
A1 t = 3.45 s (the other root is outside the domain)	✓ 1
Part (c)
A1 −0.8(36) + 60 + 10 = 41.2 m s⁻¹	✓ 1

Exam tip: For a piecewise function, solve in the piece whose domain contains the answer, and reject roots outside that domain.

+ Add to test✓ In your test206IBMI3H-2-06r44-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

On a mountaineering expedition, atmospheric pressure p kPa is measured at several heights h km above sea level. A plot of ln p against h is a straight line with gradient −0.125 and vertical intercept 4.615. The relationship has the form p = Ae^−kh.

(a)

Show that ln p = ln A − kh.

(2)

(b)

Find the values of A and k. Give A to the nearest integer.

(2)

(c)

The summit camp is at a height where the pressure is 40 kPa. Find this height in km, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 take natural logarithms of both sides: ln p = ln A + ln e^−kh	✓ 1
A1 use ln e^−kh = −kh to reach ln A − kh (AG)	✓ 1
Part (b)
A1 k = 0.125 (the negative of the gradient)	✓ 1
A1 ln A = 4.615, so A = 101	✓ 1
Part (c)
M1 solve 40 = 101e^−0.125h (or 4.615 − 0.125h = ln 40) (ECF from (b))	✓ 1
A1 7.41 km	✓ 1

Exam tip: When a straight line is obtained by plotting ln y against x, the gradient is the exponent coefficient and the intercept is ln of the multiplier.

+ Add to test✓ In your test207IBMI3H-2-06r44-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

The temperature T °C at a mountain hut t hours after midnight is modelled by T = 4 + 7 sin((π/12)(t − 9)), for 0 ≤ t ≤ 24. The graph of this model is obtained from the graph of y = sin t by a sequence of transformations.

(a)

Describe the transformations that map the graph of y = sin t onto the graph of the model. The horizontal and vertical stretches should be applied before the translation.

(3)

(b)

Write down the maximum temperature predicted by the model.

(1)

(c)

Find the first time after midnight at which the model predicts a temperature of 8 °C. Give your answer in hours after midnight to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 vertical stretch with scale factor 7	✓ 1
A1 horizontal stretch with scale factor 12/π	✓ 1
A1 translation by the vector (9, 4)	✓ 1
Part (b)
A1 11 °C	✓ 1
Part (c)
M1 set 4 + 7 sin((π/12)(t − 9)) = 8 and rearrange to sin((π/12)(t − 9)) = 4/7	✓ 1
M1 inverse sine: (π/12)(t − 9) = 0.608 (the temperature is rising between t = 3 and t = 15)	✓ 1
A1 t = 11.3	✓ 1

Exam tip: Set your GDC to radians for a sinusoidal model, and check the solution lies on the rising part of the graph if the question asks for the first time the temperature is reached.

+ Add to test✓ In your test208IBMI3H-2-06r44-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

A cake is removed from an oven at 180 °C and left to cool in a kitchen at a constant 22 °C. Its temperature T °C after t minutes is modelled by T = 22 + 158e^−kt. After 10 minutes the temperature of the cake is 112 °C.

(a)

Find the value of k, giving your answer to 3 significant figures.

(3)

(b)

The cake can be decorated once its temperature falls to 40 °C. Using k = 0.0563, find how many minutes after leaving the oven this happens, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 10 and T = 112: 90 = 158e^−10k	✓ 1
A1 e^−10k = 90/158	✓ 1
A1 k = 0.0563	✓ 1
Part (b)
M1 set 22 + 158e^−0.0563t = 40	✓ 1
M1 rearrange to e^−0.0563t = 18/158 and take natural logarithms	✓ 1
A1 t = 38.6 minutes	✓ 1

Common error: The exponential term models the difference between the cake and the room, so subtract the room temperature of 22 before dividing.

+ Add to test✓ In your test209IBMI3H-2-06r44-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium7 marks

A snow bridge between two ice walls has a cross-section modelled by y = g(x), where g(x) = −2f(x − 3) + 5, f(x) = x², and x and y are measured in metres.

(a)

The graph of y = x² is transformed to give the graph of y = g(x). Describe the three transformations involved, in the order that gives g.

(3)

(b)

Write down the coordinates of the highest point of the bridge and the range of g.

(2)

(c)

The bridge meets the level ground at y = 0. Find the width of the bridge at ground level, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 3 units to the right (replace x by x − 3)	✓ 1
A1 vertical stretch with scale factor 2 and reflection in the x-axis (multiplication by −2)	✓ 1
A1 translation 5 units up	✓ 1
Part (b)
A1 highest point (3, 5)	✓ 1
A1 range y ≤ 5	✓ 1
Part (c)
M1 solve −2(x − 3)² + 5 = 0, giving x = 1.42 and x = 4.58	✓ 1
A1 width = 3.16 m	✓ 1

Common error: A stretch factor of −2 combines a vertical stretch of factor 2 and a reflection in the x-axis; the order of the vertical steps matters, so apply the stretch and reflection before the upward translation.

+ Add to test✓ In your test210IBMI3H-2-06r44-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

In a telecoms link the signal-to-noise ratio, in decibels, is SNR = 10 log₁₀(P_s ÷ P_n), where P_s is the signal power and P_n is the noise power, both in watts.

(a)

The noise power is 3.5 × 10⁻¹⁰ W and the SNR is 28 dB. Find the signal power, giving your answer in the form a × 10^k with 1 ≤ a < 10 and a to 3 significant figures.

(2)

(b)

With the same noise power, the engineers want the SNR to rise from 28 dB to 40 dB. Find the factor by which the signal power must be multiplied, to 3 significant figures.

(2)

(c)

With the signal power unchanged, the noise power is halved. Find the increase in SNR, in dB, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10^2.8 = P_s ÷ (3.5 × 10⁻¹⁰)	✓ 1
A1 2.21 × 10⁻⁷ W	✓ 1
Part (b)
M1 an increase of 12 dB corresponds to a factor 10^12/10	✓ 1
A1 15.8	✓ 1
Part (c)
M1 increase = 10 log₁₀(2)	✓ 1
A1 3.01 dB	✓ 1

Exam tip: An increase of 10 dB always means the power ratio is multiplied by 10, so any change in decibels depends only on the ratio of the powers, not on their individual sizes.

+ Add to test✓ In your test211IBMI3H-2-06r44-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Medium6 marks

At a mountain base camp the number of hours of daylight d on day t of the year (with t = 1 on 1 January) is modelled by d = a cos(b(t − 172)) + k. The longest day has 14.6 hours of daylight and the shortest has 9.4 hours. The pattern repeats every 365 days, and angles are in radians.

(a)

Find the values of a and k.

(2)

(b)

Find the value of b, to 3 significant figures.

(1)

(c)

Find the number of days in the year on which there are more than 13 hours of daylight, giving your answer to the nearest whole number. Use b = 2π ÷ 365.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (14.6 − 9.4) ÷ 2 = 2.6	✓ 1
A1 k = (14.6 + 9.4) ÷ 2 = 12	✓ 1
Part (b)
A1 b = 2π ÷ 365 = 0.0172	✓ 1
Part (c)
M1 solve 2.6cos(b(t − 172)) + 12 = 13	✓ 1
A1 t = 103.7 and t = 240.3 (or t − 172 = ±68.3)	✓ 1
A1 137 days	✓ 1

Common error: With a cosine model the maximum occurs when the bracket is zero, so the horizontal shift equals the day of the maximum, not the day of the minimum.

+ Add to test✓ In your test212IBMI3H-2-03r10-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard8 marks

A library network introduces tablet loans. The number N of tablets on loan, t weeks after the scheme starts, is modelled by N = 800 / (1 + Ae^−kt), where A and k are positive constants. When t = 0, N = 100. When t = 3, N = 250.

(a)

Show that A = 7.

(2)

(b)

Find the value of k to 3 significant figures.

(3)

(c)

The scheme is judged a success when 90% of the 800 tablets are on loan. Using k = 0.386, find the time at which this happens. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
100 = 800 / (1 + A)	✓ 1
1 + A = 8, so A = 7 (AG)	✓ 1
Part (b)
250 = 800 / (1 + 7e^−3k), so 7e^−3k = 2.2	✓ 1
k = ln(7/2.2) / 3	✓ 1
k = 0.386	✓ 1
Part (c)
720 = 800 / (1 + 7e^−0.386t), so 7e^−0.386t = 1/9	✓ 1
t = ln 63 / 0.386	✓ 1
t = 10.7 weeks	✓ 1

Exam tip: Treat each given point as one equation; with A known, the second point leaves a single unknown, k.

+ Add to test✓ In your test213IBMI3H-2-04r13-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard6 marks

The number N of downloads of a new song after t days follows the logistic model N = 50 000/(1 + 49e^−0.8t), which satisfies dN/dt = 0.8N(1 − N/50 000). Initially N = 1000.

(a)

Show that the rate of download is greatest when N = 25 000.

(2)

(b)

Find the greatest rate of download.

(2)

(c)

Find the time at which the rate of download is greatest, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dN/dt = 0.8N − 0.8N²/50 000 is a downward quadratic in N	✓ 1
A1 its maximum is at the vertex N = 0.8/(2 × 0.8/50 000) = 25 000	✓ 1
Part (b)
M1 substitute N = 25 000: 0.8 × 25 000 × (1 − 0.5)	✓ 1
A1 10 000 downloads per day	✓ 1
Part (c)
M1 solve 50 000/(1 + 49e^−0.8t) = 25 000, so 49e^−0.8t = 1	✓ 1
A1 t = ln49/0.8 = 4.86 days (4.865)	✓ 1

Exam tip: For a logistic curve the fastest growth occurs at half the carrying capacity; the rate there equals one quarter of (growth constant × carrying capacity).

+ Add to test✓ In your test214IBMI3H-2-04r13-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard6 marks

An aircraft's sound intensity falls with the inverse square of the distance from it. At 100 m the sound level is 110 dB, where the sound level is L = 10 log₁₀(I/I₀).

(a)

Find the sound level at 400 m, to 3 significant figures.

(3)

(b)

Find the distance at which the sound level is 85 dB, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 intensity at 400 m is (100/400)² = 1/16 of that at 100 m	✓ 1
M1 change in level = 10 log₁₀(1/16) = −12.04…	✓ 1
A1 L = 98.0 dB (97.96)	✓ 1
Part (b)
M1 change in level of −25 dB: 10 log₁₀((100/d)²) = −25	✓ 1
M1 log₁₀(d/100) = 1.25	✓ 1
A1 d = 1780 m (1778.3…)	✓ 1

Common error: Because intensity is proportional to 1/d², the level changes by 20 log₁₀ of the distance ratio, not 10 log₁₀.

+ Add to test✓ In your test215IBMI3H-2-05r35-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard8 marks

The curve y = f(x) models the top edge of a bridge arch. Its highest point is M(4, 7) and it passes through A(1, 2), with x and y in metres. A hanging cable follows the curve y = g(x), where g(x) = −2f(x/3) + 5.

(a)

Find the coordinates of the lowest point of the cable, which is the image of M.

(2)

(b)

Describe fully a sequence of transformations that maps the graph of y = f(x) onto the graph of y = g(x).

(4)

(c)

Find the vertical distance between the images of A and M under the transformation from f to g.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 x-coordinate: 3 × 4 = 12	✓ 1
A1 y-coordinate: −2 × 7 + 5 = −9, so (12, −9)	✓ 1
Part (b)
A1 stretch parallel to the x-axis, scale factor 3	✓ 1
A1 stretch parallel to the y-axis, scale factor 2	✓ 1
A1 reflection in the x-axis	✓ 1
A1 translation by the vector (0, 5), applied after the vertical stretch and the reflection (the horizontal stretch may be done at any stage; the vertical stretch and reflection may be in either order)	✓ 1
Part (c)
M1 image of A has y-coordinate −2 × 2 + 5 = 1	✓ 1
A1 distance = 1 − (−9) = 10 m	✓ 1

Common error: Horizontal changes act on x inside the brackets and work in the opposite way to what you might expect: f(x/3) stretches by a factor of 3.

+ Add to test✓ In your test216IBMI3H-2-05r35-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard8 marks

The temperature set-point of a coffee roaster oscillates according to T(t) = 185 + 12 sin(πt ÷ 4), where T is in °C and t is in minutes. A second programme uses g(t) = 200 + 6 sin(πt ÷ 8).

(a)

Write down the period and the amplitude of g.

(2)

(b)

Describe a sequence of transformations that maps the graph of y = T(t) onto the graph of y = g(t).

(3)

(c)

Find the first time after t = 0 at which g(t) = 204.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 period = 2π ÷ (π ÷ 8) = 16 minutes	✓ 1
A1 amplitude = 6 °C	✓ 1
Part (b)
A1 stretch parallel to the t-axis, scale factor 2	✓ 1
A1 stretch parallel to the vertical axis, scale factor 1/2	✓ 1
A1 translation by the vector (0, 107.5), applied after the vertical stretch (since g(t) = ½T(t/2) + 107.5; the horizontal stretch may be done at any stage)	✓ 1
Part (c)
M1 200 + 6 sin(πt ÷ 8) = 204, so sin(πt ÷ 8) = 2/3	✓ 1
M1 πt ÷ 8 = sin⁻¹(2/3) (smallest positive solution)	✓ 1
A1 t = 1.86 minutes	✓ 1

Exam tip: When a sine graph is stretched and shifted, work out the vertical stretch first, then the translation, to see the order that gives the new midline.

+ Add to test✓ In your test217IBMI3H-2-05r41-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard6 marks

The intensity level of a sound is L = 10 log₁₀(I / I₀) dB, where I₀ = 1.0 × 10⁻¹² W m⁻². When two sources sound together, their intensities add. A hydrophone records noise from two ships. Ship A alone gives an intensity level of 72 dB and ship B alone gives 75 dB.

(a)

Calculate the intensity level when both ships are sounding together. Give your answer to 3 significant figures.

(3)

(b)

A third ship, C, then joins them, and the combined intensity level of all three ships is 78.0 dB. Find the intensity level that ship C would give on its own, to 3 significant figures. Use unrounded values.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 I_A / I₀ = 10^7.2 and I_B / I₀ = 10^7.5	✓ 1
M1 10 log₁₀(10^7.2 + 10^7.5)	✓ 1
A1 L = 76.8 dB	✓ 1
Part (b)
M1 total I / I₀ = 10^7.8	✓ 1
M1 I_C / I₀ = 10^7.8 − 10^7.2 − 10^7.5	✓ 1
A1 L_C = 71.9 dB	✓ 1

Common error: Decibel values cannot be added directly; convert each level back to an intensity, add the intensities, and then convert the total back to decibels.

+ Add to test✓ In your test218IBMI3H-2-05r42-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard7 marks

The level of vibration of a bridge deck is measured as L = 10 log₁₀(I/I₀) dB, where I is the intensity of the vibration and I₀ is a fixed reference intensity. When two independent sources act together, their intensities add. Heavy lorries alone produce a level of 62 dB and wind alone produces a level of 65 dB.

(a)

Find the value of I/I₀ for the lorries, correct to 3 significant figures.

(2)

(b)

Find the level of vibration when the lorries and the wind act together, correct to 3 significant figures.

(3)

(c)

The lorry traffic doubles, so the intensity from the lorries doubles while the wind is unchanged. Find the new combined level, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10^62/10	✓ 1
A1 1.58×10⁶	✓ 1
Part (b)
M1 I/I₀ for wind = 10^6.5 and adds the two ratios	✓ 1
M1 10 log₁₀(sum)	✓ 1
A1 66.8 dB	✓ 1
Part (c)
M1 10 log₁₀(2 × 10^6.2 + 10^6.5)	✓ 1
A1 68.0 dB	✓ 1

Common error: Levels in decibels cannot be added directly; convert each to an intensity ratio, add the ratios, then convert back.

+ Add to test✓ In your test219IBMI3H-2-06r42-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard6 marks

The pH of a solution is pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol per litre. A hospital technician mixes 0.20 litres of cleaning solution of pH 2.5 with 0.80 litres of solution of pH 5.0.

(a)

Find the value of [H⁺] in the solution of pH 2.5, correct to 3 significant figures.

(1)

(b)

The amounts of hydrogen ions (concentration × volume) add when the solutions are mixed. Find the pH of the 1.00 litre mixture, correct to 3 significant figures.

(3)

(c)

The technician dilutes the 1.00 litre mixture with water of negligible acidity until its pH is 4.0. Find the volume of water to be added, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 10^−2.5 = 3.16×10⁻³ mol per litre	✓ 1
Part (b)
M1 amount of H⁺ = 0.20 × 10^−2.5 + 0.80 × 10⁻⁵	✓ 1
M1 [H⁺] = 6.40×10⁻⁴ mol per litre (volume 1.00 litre)	✓ 1
A1 pH = 3.19	✓ 1
Part (c)
M1 (amount of H⁺)/(1 + V) = 10⁻⁴, using the unrounded amount from (b)	✓ 1
A1 V = 5.40 litres	✓ 1

Exam tip: Hydrogen-ion amounts add, but pH values do not; convert to concentrations, combine, then convert back with the logarithm.

+ Add to test✓ In your test220IBMI3H-2-06r44-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 1Hard7 marks

The viscosity η (in Pa s) of a sauce depends on the shear rate γ (in s⁻¹) according to η = aγⁿ. Two measurements are made: when γ = 1.0, η = 12.0, and when γ = 25, η = 1.74. A graph of ln η against ln γ is a straight line through the two corresponding points.

(a)

Find the gradient of this line, giving your answer to 3 significant figures.

(3)

(b)

Hence write down the values of n and a.

(1)

(c)

Using n = −0.600 and a = 12.0, find the shear rate at which the viscosity is 2.5 Pa s. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 points (ln 1, ln 12.0) = (0, 2.485) and (ln 25, ln 1.74) = (3.219, 0.554)	✓ 1
M1 gradient = (ln 1.74 − ln 12.0) ÷ (ln 25 − ln 1)	✓ 1
A1 -0.600	✓ 1
Part (b)
A1 n = −0.600 and a = 12.0 (the intercept is ln a = ln 12.0, or η = a when γ = 1)	✓ 1
Part (c)
M1 set 12.0γ^−0.600 = 2.5	✓ 1
M1 rearrange: γ = (12.0 ÷ 2.5)^1/0.600	✓ 1
A1 γ = 13.7 s⁻¹	✓ 1

Exam tip: A straight line on a ln–ln plot signals a power law y = axⁿ: the gradient is the power n and the intercept is ln a.

+ Add to test✓ In your test221IBMI3H-2-03r10-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Easy9 marks

The tip of one blade of a wind turbine rotates in a vertical circle. Its height h metres above the ground, t seconds after it is at its highest point, is modelled by h = 45 + 30 cos(60t°). Use your calculator in degree mode.

(a)

Write down the maximum and minimum heights of the blade tip.

(2)

(b)

Find the period of the motion.

(2)

(c)

Find the height of the tip when t = 2.

(2)

(d)

During one complete rotation, find the total time for which the tip is higher than 60 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
maximum = 75 m	✓ 1
minimum = 15 m	✓ 1
Part (b)
period = 360 / 60	✓ 1
6 seconds	✓ 1
Part (c)
h = 45 + 30 cos 120°	✓ 1
h = 30 m	✓ 1
Part (d)
45 + 30 cos(60t°) = 60 gives cos(60t°) = 0.5, so t = 1	✓ 1
above 60 m for −1 < t < 1 (symmetry about the highest point)	✓ 1
time = 2 s	✓ 1

Exam tip: Sketch the cosine curve and the horizontal line to see how many intervals lie above the line before solving.

+ Add to test✓ In your test222IBMI3H-2-05r21-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Easy9 marks

The curved cable of a footbridge used by cyclists hangs in the shape of a quadratic curve y = ax² + bx + c, where y is the height in metres of the cable above the deck and x is the horizontal distance in metres from the left tower. The cable is 12 m above the deck at x = 0, 4 m above the deck at x = 20 and 12 m above the deck at x = 40.

(a)

Find the values of a, b and c.

(4)

(b)

Find the least height of the cable above the deck.

(2)

(c)

The deck is only safe for cycling where the cable is at most 6 m above it. Find the horizontal length of the section of cable where the height is at most 6 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 c = 12	✓ 1
M1 forms 400a + 20b + 12 = 4 and 1600a + 40b + 12 = 12	✓ 1
A1 a = 0.02	✓ 1
A1 b = −0.8	✓ 1
Part (b)
M1 vertex at x = −b/(2a) = 20 (or substitutes x = 20)	✓ 1
A1 4 m	✓ 1
Part (c)
M1 solves 0.02x² − 0.8x + 12 = 6	✓ 1
A1 x = 10 and x = 30	✓ 1
A1 length = 20 m	✓ 1

Exam tip: Use the symmetry of a parabola: the vertex lies halfway between two points at the same height.

+ Add to test✓ In your test223IBMI3H-2-06r38-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Easy9 marks

The resale value, V pounds, of a designer coat after t years is modelled by V = Ab^t. The coat is worth £450 after 1 year and £364.50 after 3 years.

(a)

Find the value of b.

(2)

(b)

Find the value of A.

(2)

(c)

Find the value of the coat after 6 years, to the nearest penny.

(2)

(d)

Find how many years it takes for the value to fall to half of its original value, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 b² = 364.50 / 450 = 0.81	✓ 1
A1 b = 0.9	✓ 1
Part (b)
M1 450 = A × 0.9	✓ 1
A1 A = 500	✓ 1
Part (c)
M1 500 × 0.9⁶	✓ 1
A1 £265.72	✓ 1
Part (d)
M1 500 × 0.9^t = 250	✓ 1
M1 take logarithms, t = ln 0.5 / ln 0.9	✓ 1
A1 6.58 years	✓ 1

Exam tip: To solve an exponential equation for the time, isolate the power first and then take logarithms of both sides.

+ Add to test✓ In your test224IBMI-031-03-r2-05Choosing and evaluating models Paper 2Medium10 marks

A biologist measures the resting metabolic rate, $P$ watts, and the body mass, $m$ kg, of mammals ranging from 0.02 kg to 70 kg. She plots $\log_{10} P$ against $\log_{10} m$. The points lie close to the straight line
$\log_{10} P = 0.75\log_{10} m + 0.55$.

(a)

Show that the relationship can be written in the form $P = a m^b$. Find the value of $a$ and the value of $b$.

(3)

(b)

Use the model to estimate the metabolic rate of a 60 kg mammal.

(2)

(c)

A colleague suggests an exponential model, $P = ke^{rm}$. Explain why the straight line on the log–log graph supports the power model rather than the exponential model.

(2)

(d)

An elephant has a mass of 5000 kg. Use the model to estimate its metabolic rate, and comment on the reliability of this estimate.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $P = 10^{0.75\log_{10} m + 0.55} = 10^{0.55} \times m^{0.75}$	✓ 1
A1 $b = 0.75$	✓ 1
A1 $a = 10^{0.55} = 3.55$	✓ 1
Part (b)
M1 $3.548 \times 60^{0.75}$	✓ 1
A1 76.5 W	✓ 1
Part (c)
R1 a power model gives a straight line when $\log P$ is plotted against $\log m$	✓ 1
R1 an exponential model would give a straight line when $\log P$ is plotted against $m$ itself, so a linear log–log graph indicates a power relationship	✓ 1
Part (d)
M1 $3.548 \times 5000^{0.75}$	✓ 1
A1 2110 W	✓ 1
R1 unreliable because it is an extrapolation far beyond the data range of 0.02 kg to 70 kg	✓ 1

Common error: The intercept 0.55 is log₁₀ a, not a itself. You must calculate 10^0.55 to find a.

+ Add to test✓ In your test225IBMI-031-04-r2-03Transformations of graphs and logarithmic scales Paper 2Medium13 marks

A biologist studies the resting metabolic rate $P$ watts of mammals of mass $M$ kg, for masses between 3 kg and 320 kg. She proposes a model of the form $P=aM^{b}$, where $a$ and $b$ are constants.

(a)

Show that $\log_{10}P=b\log_{10}M+\log_{10}a$.

(2)

(b)

She plots $\log_{10}P$ against $\log_{10}M$. Her line of best fit passes through the points $(0.5,\,0.919)$ and $(2.5,\,2.403)$. Find the gradient and the vertical-axis intercept of this line.

(3)

(c)

Hence find the value of $a$ and write down the value of $b$.

(2)

(d)

Use the model to estimate the metabolic rate of a 60 kg mammal.

(2)

(e)

Find the mass of a mammal whose metabolic rate is predicted to be 250 W.

(3)

(f)

Comment on the reliability of using this model to predict the metabolic rate of a 4000 kg elephant.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: takes $\log_{10}$ of both sides: $\log_{10}P=\log_{10}(aM^{b})$	✓ 1
A1: uses product and power laws to get $\log_{10}a+b\log_{10}M$, AG	✓ 1
Part (b)
M1: gradient $=\frac{2.403-0.919}{2.5-0.5}$	✓ 1
A1: gradient $=0.742$	✓ 1
A1: intercept $=0.919-0.742(0.5)=0.548$	✓ 1
Part (c)
M1: $a=10^{0.548}$	✓ 1
A1: $a=3.53$ and $b=0.742$	✓ 1
Part (d)
M1: $3.532\times60^{0.742}$	✓ 1
A1: 73.7 W	✓ 1
Part (e)
M1: $250=3.532M^{0.742}$	✓ 1
M1: $M=\left(\frac{250}{3.532}\right)^{1/0.742}$ or solves with GDC	✓ 1
A1: 311 kg	✓ 1
Part (f)
R1: unreliable, because 4000 kg is far outside the 3–320 kg data range, so it would be extrapolation	✓ 1

Exam tip: On a log–log graph the gradient is the power $b$ directly, but the intercept is $\log_{10}a$. Raise 10 to the intercept to find $a$, and keep unrounded values for later parts.

+ Add to test✓ In your test226IBMI-031-04-r2-05Transformations of graphs and logarithmic scales Paper 2Medium10 marks

The magnitude $M$ of an earthquake is defined by $M=\log_{10}\left(\frac{A}{A_0}\right)$, where $A$ is the amplitude recorded by a seismograph and $A_0$ is a fixed reference amplitude.
The energy $E$ joules released by an earthquake of magnitude $M$ is modelled by $\log_{10}E=4.8+1.5M$.

(a)

Two earthquakes have magnitudes 6.3 and 4.1. Find how many times larger the recorded amplitude of the first is than the amplitude of the second.

(2)

(b)

Find the energy released by the magnitude 6.3 earthquake.

(2)

(c)

Show that an increase of 1 in magnitude multiplies the energy released by approximately 31.6.

(2)

(d)

An earthquake releases $5.0\times10^{15}$ J of energy. Find its magnitude.

(2)

(e)

A poster shows energies on a logarithmic axis. The axis starts at $10^{10}$ J, and each power of 10 is 2 cm long. Find the distance from the start of the axis at which the earthquake releasing $5.0\times10^{15}$ J should be plotted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{A_1}{A_2}=10^{6.3-4.1}=10^{2.2}$	✓ 1
A1: 158 (accept 158.5)	✓ 1
Part (b)
M1: $E=10^{4.8+1.5(6.3)}=10^{14.25}$	✓ 1
A1: $1.78\times10^{14}$ J	✓ 1
Part (c)
M1: $\frac{10^{4.8+1.5(M+1)}}{10^{4.8+1.5M}}$, or subtracting the logs to get a difference of 1.5	✓ 1
A1: $=10^{1.5}=31.62\ldots\approx31.6$, AG	✓ 1
Part (d)
M1: $1.5M=\log_{10}(5.0\times10^{15})-4.8$	✓ 1
A1: $M=7.27$	✓ 1
Part (e)
M1: $(\log_{10}(5.0\times10^{15})-10)\times2$	✓ 1
A1: 11.4 cm	✓ 1

Exam tip: On a logarithmic axis, distance is proportional to the logarithm of the value. Find how many powers of ten the value lies above the start, then multiply by the length of one power.

+ Add to test✓ In your test227IBMI-031-01-r2-05Quadratic and cubic models Paper 2Medium12 marks

The cross-section of a footbridge arch is modelled by $f(x) = -0.05x^2 + 2x$, for 0 ≤ x ≤ 40, where x and f(x) are measured in metres from a point at ground level.
A larger arch for a second bridge is modelled by
$g(x) = 0.75\,f\left(\frac{x-5}{1.5}\right)$.

(a)

Find the coordinates of the highest point of the arch modelled by f.

(2)

(b)

Describe fully a sequence of transformations that maps the graph of f onto the graph of g.

(3)

(c)

Hence find the coordinates of the highest point of the arch modelled by g.

(2)

(d)

Express g(x) in the form $ax^2 + bx + c$.

(3)

(e)

A walkway is placed horizontally inside the larger arch, 9 m above the ground, with each end touching the arch. Find the length of the walkway.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = -\frac{2}{2(-0.05)} = 20$, or use of a GDC (M1)	✓ 1
(20, 20) (A1)	✓ 1
Part (b)
Horizontal stretch, scale factor 1.5 (A1)	✓ 1
Vertical stretch, scale factor 0.75 (A1)	✓ 1
Then a translation of 5 units in the positive x-direction, i.e. by the vector (5, 0), applied after the horizontal stretch (A1)	✓ 1
Accept any correct equivalent sequence, e.g. a translation by $\binom{10/3}{0}$ followed by a horizontal stretch of scale factor 1.5 about the y-axis, with the vertical stretch at any stage.	—
Part (c)
Applies the transformations to (20, 20): x = 1.5 × 20 + 5, y = 0.75 × 20 (M1)	✓ 1
(35, 15) (A1)	✓ 1
Part (d)
Substitutes: $0.75\left[-0.05\frac{(x-5)^2}{2.25} + 2\cdot\frac{x-5}{1.5}\right]$ (M1)	✓ 1
Simplifies to $-\frac{1}{60}(x-5)^2 + (x-5)$ and expands (M1)	✓ 1
$g(x) = -\frac{1}{60}x^2 + \frac{7}{6}x - \frac{65}{12}$ (A1)	✓ 1
Part (e)
Solves g(x) = 9 (e.g. by GDC), giving x = 16.03 and x = 53.97, i.e. $x = 35 \pm \sqrt{360}$ (M1)	✓ 1
Length = $2\sqrt{360}$ = 37.9 m (A1)	✓ 1
Use unrounded roots: subtracting the 3 s.f. values 16.0 and 54.0 gives 38.0, which is inaccurate.	—

Common error: Replacing x by x/1.5 stretches the graph horizontally by factor 1.5, not 1/1.5. Inside the bracket, the effect on the graph is the inverse of the operation applied to x.

+ Add to test✓ In your test228IBMI3H-2-04-06Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Medium12 marks

A deer population in a reserve, P, t years after the reserve opens, is modelled by P = 1200 / (1 + 11e^−0.4t).

(a)

Find the initial population.

(2)

(b)

Find the limiting value of P as t increases, and interpret it.

(2)

(c)

Find the population after 5 years.

(2)

(d)

Find the time at which the population first reaches 900.

(3)

(e)

Find the time at which the population is growing most rapidly.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1200 / (1 + 11e⁰)	✓ 1
A1 100	✓ 1
Part (b)
A1 1200	✓ 1
R1 the carrying capacity: the largest population the reserve can support	✓ 1
Part (c)
M1 1200 / (1 + 11e^(−2))	✓ 1
A1 482	✓ 1
Part (d)
M1 1200 / (1 + 11e^(−0.4t)) = 900	✓ 1
M1 e^(−0.4t) = 1/33 and take logarithms	✓ 1
A1 t = 8.74 years	✓ 1
Part (e)
M1 greatest growth when P is half the carrying capacity, P = 600	✓ 1
M1 1 + 11e^(−0.4t) = 2	✓ 1
A1 t = 5.99 years	✓ 1

Exam tip: A logistic curve grows fastest at half its carrying capacity, so you can find that time without calculus.

+ Add to test✓ In your test229IBMI3H-2-04-11Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Medium12 marks

The depth of water in a harbour is H = 2.4 sin(π(t − 1.5)/6.2) + 3.1 metres, where t is the time in hours after midnight and the angle is in radians.

(a)

Describe the sequence of transformations that maps the graph of y = sin t onto the graph of H.

(4)

(b)

Write down the maximum depth and the minimum depth.

(2)

(c)

Find the first time after midnight at which the depth is 4.5 m.

(3)

(d)

In each cycle, find the time for which the depth is greater than 4.5 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 vertical stretch with scale factor 2.4	✓ 1
A1 horizontal stretch with scale factor 6.2/π	✓ 1
A1 translation 1.5 units to the right	✓ 1
A1 translation 3.1 units up	✓ 1
Part (b)
A1 5.5 m	✓ 1
A1 0.7 m	✓ 1
Part (c)
M1 sin(π(t − 1.5)/6.2) = 1.4/2.4	✓ 1
A1 π(t − 1.5)/6.2 = 0.6228	✓ 1
A1 t = 2.73 hours	✓ 1
Part (d)
M1 second solution π(t − 1.5)/6.2 = π − 0.6228	✓ 1
M1 subtract the two times	✓ 1
A1 3.74 hours	✓ 1

Common error: Stopping at the first solution of a sine equation; a duration question needs both solutions in the cycle.

+ Add to test✓ In your test230IBMI3H-2-04b-06Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The bacterial count C per mL in a tank of treated ballast water is recorded each hour.

t (hours)	1	2	3	4	5
C	64	41	27	17	11

(a)

Complete the values of ln C for the five times, correct to 2 decimal places.

(2)

(b)

Find the equation of the regression line of ln C on t.

(2)

(c)

Hence write the model in the form C = ae^−kt.

(2)

(d)

Find the percentage by which the count falls each hour.

(2)

(e)

Use the model to find when the count is predicted to fall to 2 per mL.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 4.16, 3.71, 3.30 (first three)	✓ 1
A1 2.83, 2.40 (last two)	✓ 1
Part (b)
A1 gradient −0.440	✓ 1
A1 ln C = −0.440t + 4.60	✓ 1
Part (c)
A1 k = 0.440	✓ 1
A1 a = e^4.60 = 99.5	✓ 1
Part (d)
M1 factor e^−0.440 = 0.644 each hour	✓ 1
A1 35.6%	✓ 1
Part (e)
M1 sets 99.5e^−0.44t = 2	✓ 1
A1 t = 8.88 hours	✓ 1

Exam tip: Regression on ln C against t turns an exponential relationship into a straight line. Convert back by exponentiating both intercept and gradient.

+ Add to test✓ In your test231IBMI3H-2-03r5-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The resistance R kN that water exerts on a ship's hull is modelled by R = kvⁿ, where v m s⁻¹ is the speed of the ship and k and n are constants. Measurements give R = 19.2 when v = 4 and R = 145.8 when v = 9.

(a)

Show that ln R = ln k + n ln v.

(2)

(b)

Hence, using the two measurements, find the value of n.

(3)

(c)

Find the value of k.

(2)

(d)

Use the model to predict the resistance, in kN, when v = 12. Give your answer to 3 significant figures. (ECF from (b) and (c))

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes natural logarithms of both sides of R = kvⁿ	✓ 1
A1 ln R = ln k + ln vⁿ = ln k + n ln v (AG)	✓ 1
Part (b)
M1 ln 19.2 = 2.955…, ln 145.8 = 4.982…, ln 4 = 1.386…, ln 9 = 2.197…	✓ 1
M1 gradient = (4.982 − 2.955) ÷ (2.197 − 1.386)	✓ 1
A1 n = 2.5	✓ 1
Part (c)
M1 19.2 = k × 4^2.5, so k = 19.2 ÷ 32	✓ 1
A1 k = 0.6	✓ 1
Part (d)
M1 R = 0.6 × 12^2.5	✓ 1
A1 299 kN	✓ 1

Common error: When finding the gradient of a log–log graph, use the logarithms of both variables; using the original values of R and v gives the wrong exponent.

+ Add to test✓ In your test232IBMI3H-2-04r5-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The illuminance, I lux, of a theatre spotlight at a distance d metres from the lamp follows a power model I = kdⁿ. On a graph of log₁₀ I against log₁₀ d the data lie on a straight line through the points (0.20, 2.62) and (0.80, 1.48).

(a)

Find the gradient of the line.

(2)

(b)

Find the intercept on the vertical axis.

(2)

(c)

Hence write the model in the form I = kdⁿ, stating k and n. (ECF from (a) and (b))

(2)

(d)

Find the distance at which the illuminance is 40 lux. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1.48 − 2.62) ÷ (0.80 − 0.20)	✓ 1
A1 −1.90	✓ 1
Part (b)
M1 2.62 = c + (−1.90)(0.20)	✓ 1
A1 c = 3.00	✓ 1
Part (c)
M1 log₁₀ I = 3.00 − 1.90 log₁₀ d gives I = 10^3.00d^−1.90	✓ 1
A1 k = 1000 and n = −1.90	✓ 1
Part (d)
M1 40 = 1000d^−1.90, so d^−1.90 = 0.04	✓ 1
M1 d = 0.04^{−1 ÷ 1.90}	✓ 1
A1 d = 5.44 m	✓ 1

Exam tip: Read the gradient of a log–log line as the exponent n and the intercept as log₁₀ k; remember to undo the logarithm to get k.

+ Add to test✓ In your test233IBMI3H-2-04r5-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

A buoy's height above the seabed is modelled by h = a cos(b(t − 2)) + d metres, where t is the time in hours after midnight, b > 0 and the angle is in radians. The highest point of the buoy is 4.1 m, reached at t = 2. The next lowest point is 1.3 m, reached at t = 5.5.

(a)

Find the value of a and the value of d.

(2)

(b)

Find the value of b, correct to 3 significant figures.

(2)

(c)

Find the height of the buoy at t = 10, correct to 3 significant figures. Use the unrounded value of b.

(2)

(d)

Find the first time after midnight (t > 0) at which the height is 3.5 m. Use the unrounded value of b and give t to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (4.1 − 1.3) ÷ 2 = 1.4	✓ 1
A1 d = (4.1 + 1.3) ÷ 2 = 2.7	✓ 1
Part (b)
M1 half a period is 3.5 hours, so period 7 and b = 2π ÷ 7	✓ 1
A1 b = 0.898	✓ 1
Part (c)
M1 h = 1.4 cos(b × 8) + 2.7	✓ 1
A1 h = 3.57 m	✓ 1
Part (d)
M1 1.4 cos(b(t − 2)) + 2.7 = 3.5, so cos(b(t − 2)) = 0.5714…	✓ 1
M1 b(t − 2) = −0.9626… (the solution that gives t > 0 closest to 0)	✓ 1
A1 t = 0.928 hours	✓ 1

Common error: A cosine equation has two solutions in every cycle; choose the one that fits the time window asked for, not just the value your calculator returns.

+ Add to test✓ In your test234IBMI3H-2-04r6-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

Calculator

The table gives the orbital radius r (in millions of km) and orbital period T (in days) for five planets orbiting a distant star. It is suggested that T = ar^b.
r = 6, T = 1.4
r = 11, T = 3.6
r = 18, T = 7.3
r = 30, T = 15.6
r = 52, T = 35.5

(a)

Show that this suggestion leads to a linear relationship between ln T and ln r.

(2)

(b)

Find the equation of the regression line of ln T on ln r. Give the gradient and intercept to 3 significant figures.

(3)

(c)

Hence find the values of a and b in the model T = ar^b. Give a to 3 significant figures.

(3)

(d)

Use the unrounded regression values to estimate the orbital period of a planet with r = 40. Give your answer in days to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes natural logarithms of both sides: ln T = ln a + b ln r	✓ 1
R1 this is of the form y = c + bx with gradient b and intercept ln a	✓ 1
Part (b)
M1 calculates ln r and ln T for each planet and uses a regression tool	✓ 1
A1 gradient = 1.49	✓ 1
A1 intercept = -2.32	✓ 1
Part (c)
A1 b = gradient = 1.49	✓ 1
M1 a = e^intercept	✓ 1
A1 a = 0.0983	✓ 1
Part (d)
M1 substitutes r = 40 into T = ar^b	✓ 1
A1 T = 24.1 days	✓ 1

Exam tip: For a power model, plot ln y against ln x; for an exponential model, plot ln y against x. The gradient and intercept then give the parameters.

+ Add to test✓ In your test235IBMI3H-2-04r6-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

Calculator

Daily passenger numbers P on a new rail line, t years after it opens, are modelled by an equation of the form P = A × B^t. A graph of log₁₀ P against t is a straight line with equation log₁₀ P = 0.045t + 3.2.

(a)

Find the values of A and B. Give A to 4 significant figures and B to 3 significant figures.

(4)

(b)

Interpret the value of A in this context.

(1)

(c)

Estimate the daily number of passengers 10 years after opening. Give your answer to the nearest whole number, using unrounded values.

(2)

(d)

Find the time, in years to 3 significant figures, for the daily passenger numbers to double.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewrites P = 10^{0.045t + 3.2}	✓ 1
M1 separates: P = 10^3.2 × (10^0.045)^t	✓ 1
A1 A = 1585	✓ 1
A1 B = 1.11	✓ 1
Part (b)
R1 A is the number of passengers per day when the line opens (t = 0)	✓ 1
Part (c)
M1 substitutes t = 10 into P = 10^{0.045t + 3.2}	✓ 1
A1 P = 4467	✓ 1
Part (d)
M1 sets 2 = B^t or uses the log graph: 0.045t = log₁₀ 2	✓ 1
A1 t = 6.69 years	✓ 1

Common error: A straight line on a log10 graph means an exponential model, and its intercept is the logarithm of A, not A itself.

+ Add to test✓ In your test236IBMI3H-2-04r6-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

Calculator

The time T seconds that a club runner takes to run a race of distance d metres is modelled by T = ad^b. The runner takes 330 seconds to run 1500 m and 2640 seconds to run 10000 m.

(a)

Show that ln T = ln a + b ln d.

(2)

(b)

Find the value of b to 4 significant figures.

(3)

(c)

Find the value of a to 3 significant figures, using the unrounded value of b.

(2)

(d)

Use the model, with unrounded values, to predict the time for a 5000 m race. Give your answer to the nearest second.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes natural logarithms of both sides and uses ln(xy) = ln x + ln y	✓ 1
A1 ln a + ln(d^b) = ln a + b ln d (AG)	✓ 1
Part (b)
M1 forms two equations from the data or finds the gradient between the two points	✓ 1
M1 b = (ln 2640 − ln 330) ÷ (ln 10000 − ln 1500)	✓ 1
A1 b = 1.096	✓ 1
Part (c)
M1 substitutes one data point, for example a = 330 ÷ 1500^b	✓ 1
A1 a = 0.109	✓ 1
Part (d)
M1 substitutes d = 5000	✓ 1
A1 T = 1235 seconds	✓ 1

Exam tip: For power models, use the ratio form T₂ ÷ T₁ = (d₂ ÷ d₁)^b; it avoids calculating a when you only need a prediction.

+ Add to test✓ In your test237IBMI3H-2-05r16-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

A hospital laboratory grows a bacterial culture. The number of bacteria, N, is counted every two hours.

Time t (hours)	0	2	4	6	8
N	300	515	883	1516	2601

It is suggested that N = ae^bt.

(a)

Calculate ln N when t = 4 and when t = 8, to 3 significant figures.

(2)

(b)

Use your graphic display calculator to find the equation of the regression line of ln N on t.

(2)

(c)

Hence write down values of a and b in the model N = ae^bt, giving a to the nearest integer and b to 3 significant figures. ECF from (b).

(2)

(d)

Use the model N = 300e^0.270t to predict the number of bacteria when t = 10. Give your answer to the nearest 10.

(2)

(e)

Interpret the value of e^b for this culture, using e^0.270 = 1.31 (3 s.f.).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 ln 883 = 6.78	✓ 1
A1 ln 2601 = 7.86	✓ 1
Part (b)
M1 linear regression of ln N against t (using all five points)	✓ 1
A1 ln N = 0.270t + 5.70 (accept 0.270t + 5.70)	✓ 1
Part (c)
A1 b = 0.270 (the gradient)	✓ 1
A1 a = e^5.70 ≈ 299 to 300 (accept 298–301)	✓ 1
Part (d)
M1 N = 300e^{0.270 × 10} = 300e^2.7	✓ 1
A1 4463.9 so 4460 bacteria	✓ 1
Part (e)
R1 each hour the number of bacteria is multiplied by 1.31	✓ 1
A1 an increase of about 31% per hour	✓ 1

Exam tip: If you can see ln y against x is linear, use regression on ln y, not on y, and convert back with e.

+ Add to test✓ In your test238IBMI3H-2-05r16-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium11 marks

The number of bees, P, in a colony t days after it is established is modelled by P(t) = 60000 ÷ (1 + 9e^−0.1t), for t ≥ 0.

(a)

Find the number of bees in the colony when it is established.

(2)

(b)

Write down the limiting value of P as t → ∞.

(1)

(c)

Find the number of days for the colony to reach 40000 bees. Give your answer to 3 significant figures.

(3)

(d)

Find the time at which the colony is growing most rapidly, to 3 significant figures.

(3)

(e)

Find the rate of growth, in bees per day, at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P(0) = 60000 ÷ (1 + 9)	✓ 1
A1 6000 bees	✓ 1
Part (b)
A1 60000 (as t → ∞, e^−0.1t → 0)	✓ 1
Part (c)
M1 40000 = 60000 ÷ (1 + 9e^−0.1t) so 9e^−0.1t = 0.5	✓ 1
M1 e^−0.1t = 1/18 so t = 10 ln 18	✓ 1
A1 t = 28.90… so 28.9 days	✓ 1
Part (d)
R1 fastest growth when P = 30000 (half the limiting value), so 9e^−0.1t = 1	✓ 1
M1 t = 10 ln 9 (accept GDC maximum of derivative)	✓ 1
A1 t = 21.972… so 22.0 days	✓ 1
Part (e)
M1 differentiate (GDC) or use dP/dt = 0.1 P(1 − P/60000) at P = 30000	✓ 1
A1 1500 bees per day	✓ 1

Exam tip: For a logistic model with limit L, the fastest growth is when the value is L/2; check by substituting back into the formula.

+ Add to test✓ In your test239IBMI3H-2-03r17-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

A water engineer measures the flow rate $Q$ (m³ per hour) through pipes of internal diameter $d$ metres. A plot of $\ln Q$ against $\ln d$ is a straight line with equation $$\ln Q=2.46\ln d+0.89.$$

(a)

Show that the flow rate can be modelled by $Q=kd^{a}$, where $k$ and $a$ are constants.

(3)

(b)

State the value of $k$ and the value of $a$, correct to 3 significant figures.

(2)

(c)

Use the model to estimate the flow rate through a pipe of diameter 0.35 m. Give your answer to 3 significant figures.

(2)

(d)

Find the diameter of pipe required for a flow rate of 1.5 m³ per hour. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes exponentials (or writes $Q=e^{2.46\ln d+0.89}$)	✓ 1
M1 uses $e^{2.46\ln d}=d^{2.46}$ and $e^{x+y}=e^xe^y$	✓ 1
A1 $Q=e^{0.89}d^{2.46}$, which has the form $kd^a$ (AG)	✓ 1
Part (b)
A1 $a=2.46$	✓ 1
A1 $k=e^{0.89}=2.44$	✓ 1
Part (c)
M1 substitutes $d=0.35$ into $Q=2.4351\ldots\times d^{2.46}$ (unrounded $k$)	✓ 1
A1 $Q=0.184$ m³ per hour	✓ 1
Part (d)
M1 sets $2.4351\ldots d^{2.46}=1.5$	✓ 1
M1 isolates $d=\left(\frac{1.5}{e^{0.89}}\right)^{\frac{1}{2.46}}$	✓ 1
A1 $d=0.821$ m	✓ 1

Common error: A straight line on a log–log plot means a power model, not an exponential one; the gradient is the power and the intercept is $\ln k$, not $k$.

+ Add to test✓ In your test240IBMI3H-2-03r17-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The number of active players, $N$, on a new online game server is recorded $t$ days after launch. A plot of $\log_{10}N$ against $t$ is a straight line with equation $$\log_{10}N=0.12t+3.2.$$

(a)

Find the number of active players at launch ($t=0$). Give your answer to 3 significant figures.

(2)

(b)

Show that the model can be written $N=A\times B^{t}$ and find $B$ to 3 significant figures. Interpret $B$ in context as a percentage daily increase.

(3)

(c)

Find the number of days after launch for the server to reach 100 000 active players.

(2)

(d)

Find the time taken for the number of players to double. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $N=10^{3.2}$	✓ 1
A1 $N=1580$	✓ 1
Part (b)
M1 $N=10^{0.12t+3.2}=10^{3.2}\times(10^{0.12})^t$	✓ 1
A1 $B=10^{0.12}=1.32$	✓ 1
A1 the number of players increases by about 31.8% (each day), since $B=1.318\ldots$	✓ 1
Part (c)
M1 sets $0.12t+3.2=\log_{10}100000=5$	✓ 1
A1 $t=15$ days	✓ 1
Part (d)
M1 uses $B^{t}=2$ (or $0.12t=\log_{10}2$)	✓ 1
M1 $t=\frac{\log_{10}2}{0.12}$	✓ 1
A1 $t=2.51$ days	✓ 1

Exam tip: Doubling time does not depend on the starting value: divide $\log_{10}2$ by the gradient of the semi-log line.

+ Add to test✓ In your test241IBMI3H-2-03r17-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The number of players of a mobile game is modelled by the logistic function $$N(t)=\frac{50000}{1+Ce^{-kt}},\quad t\ge0,$$ where $t$ is the number of weeks since release and $C,k$ are positive constants. There were 2000 players at release and 9000 players after 3 weeks.

(a)

Show that $C=24$.

(2)

(b)

Find the value of $k$. Give your answer to 3 significant figures.

(3)

(c)

State the value that $N(t)$ approaches as $t$ becomes large, and interpret it in context.

(1)

(d)

Find the time, in weeks, for the game to reach 40 000 players. Use the unrounded value of $k$ and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes $t=0$: $2000=\frac{50000}{1+C}$	✓ 1
A1 $1+C=25$, so $C=24$ (AG)	✓ 1
Part (b)
M1 substitutes $t=3$: $9000=\frac{50000}{1+24e^{-3k}}$	✓ 1
M1 rearranges to $e^{-3k}=\frac{1}{24}\left(\frac{50}{9}-1\right)$ (or solves on a GDC)	✓ 1
A1 $k=0.554$	✓ 1
Part (c)
A1 50 000; the maximum number of players the market can support (carrying capacity)	✓ 1
Part (d)
M1 sets $\frac{50000}{1+24e^{-kt}}=40000$	✓ 1
M1 $24e^{-kt}=0.25$, so $e^{-kt}=\frac{1}{96}$	✓ 1
A1 $t=8.24$ weeks	✓ 1

Common error: Rounding $k$ before the final part can shift the last digit; keep the unrounded value in the calculator memory.

+ Add to test✓ In your test242IBMI3H-2-03r17-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The acidity of water is measured by its pH, where $\text{pH}=-\log_{10}[\text{H}^+]$ and $[\text{H}^+]$ is the hydrogen-ion concentration in mol L⁻¹. A water sample has $[\text{H}^+]=4.5\times10^{-8}$.

(a)

Calculate the pH of the sample. Give your answer to 3 significant figures.

(2)

(b)

After treatment the pH of the water is 6.8. Find the new hydrogen-ion concentration. Give your answer in the form $a\times10^{k}$ with $1\le a<10$ and $a$ to 3 significant figures.

(2)

(c)

Find how many times greater the hydrogen-ion concentration is after treatment than before. Use unrounded values and give your answer to 3 significant figures.

(2)

(d)

A different treatment multiplies the hydrogen-ion concentration by 3. Find the change in pH, giving your answer to 3 significant figures and stating whether the pH rises or falls.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $-\log_{10}(4.5\times10^{-8})$	✓ 1
A1 $7.35$	✓ 1
Part (b)
M1 $[\text{H}^+]=10^{-6.8}$	✓ 1
A1 $1.58\times10^{-7}$ mol L⁻¹	✓ 1
Part (c)
M1 $\frac{10^{-6.8}}{4.5\times10^{-8}}$	✓ 1
A1 $3.52$	✓ 1
Part (d)
M1 $\Delta\text{pH}=-\log_{10}3$	✓ 1
A1 $-0.477$	✓ 1
A1 the pH falls by 0.477	✓ 1

Common error: The pH scale is logarithmic and decreasing: a larger concentration means a smaller pH, and equal pH steps are equal ratios of concentration.

+ Add to test✓ In your test243IBMI3H-2-03r17-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

A cup of tea is left to cool in a room at a constant temperature of 20 °C. Its temperature $T$ °C is recorded at five-minute intervals.

t (minutes)	0	5	10	15	20
T (°C)	84.7	63.4	49.1	39.5	33.1
ln(T − 20)	4.17	3.77		2.97

(a)

Complete the table by finding the two missing values of $\ln(T-20)$, correct to 3 significant figures.

(2)

(b)

A linear regression of $\ln(T-20)$ on $t$ gives $\ln(T-20)=4.17-0.0800t$. Hence show that $T=20+Ae^{-kt}$ and find $A$ and $k$.

(3)

(c)

Explain why the model uses $T-20$ rather than $T$.

(1)

(d)

Use the regression equation to find when the tea reaches 30 °C. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\ln(29.1)=3.37$	✓ 1
A1 $\ln(13.1)=2.57$	✓ 1
Part (b)
M1 $T-20=e^{4.17-0.0800t}$	✓ 1
M1 $=e^{4.17}e^{-0.0800t}$	✓ 1
A1 $A=64.7$ and $k=0.0800$	✓ 1
Part (c)
R1 the temperature tends to the room temperature 20 °C (horizontal asymptote $T=20$), so $T-20$ decays to zero exponentially	✓ 1
Part (d)
M1 sets $4.17-0.0800t=\ln10$	✓ 1
M1 $t=\frac{4.17-\ln10}{0.0800}$	✓ 1
A1 $t=23.3$ minutes	✓ 1

Exam tip: Linearising means rewriting the model so one side is a straight-line function of the other variable; read the gradient as $-k$ and the intercept as $\ln A$.

+ Add to test✓ In your test244IBMI3H-2-05r21-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium11 marks

A cyclist on a flat road at constant speed v (m s⁻¹) needs a power P (W) to overcome air resistance and rolling friction. A graph of log₁₀ P against log₁₀ v is a straight line with equation
log₁₀ P = −0.05 + 2.6 log₁₀ v

(a)

The relationship can be written as P = av^b. Find the values of a and b, giving a to 3 significant figures.

(3)

(b)

Calculate the power needed at 11 m s⁻¹, to the nearest watt.

(2)

(c)

The cyclist's power output is limited to 400 W. Find the greatest speed that can be sustained, to 3 significant figures.

(3)

(d)

Find the percentage increase in the power needed when the speed is increased by 10%, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P = 10^−0.05 × 10^{2.6 log₁₀ v} or equivalent	✓ 1
A1 a = 0.891	✓ 1
A1 b = 2.6	✓ 1
Part (b)
M1 P = 10^−0.05 × 11^2.6	✓ 1
A1 P = 455 W	✓ 1
Part (c)
M1 400 = 0.891v^2.6 (or using unrounded a)	✓ 1
A1 rearranges v = (400/a)^1/2.6	✓ 1
A1 v = 10.5 m s⁻¹	✓ 1
Part (d)
M1 ratio = (1.1v)^2.6/v^2.6 = 1.1^2.6	✓ 1
A1 1.28 (or 1.281)	✓ 1
A1 28.1%	✓ 1

Common error: A 10% rise in speed multiplies power by 1.1b, not by 1.1 and not by 1 + 0.1b; the percentage change depends only on the exponent.

+ Add to test✓ In your test245IBMI3H-2-05r21-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The height profile of a cycling route is modelled by the graph of y = f(x), where x is the horizontal distance in km. The graph has a maximum point at (2, 5), an x-intercept at (−1, 0) and passes through (0, 3). A second route is modelled by y = g(x), where g(x) = 3f(x/2 + 1).

(a)

Write down the coordinates of the maximum point of the graph of g.

(2)

(b)

Find the coordinates of the x-intercept of the graph of g that is the image of (−1, 0).

(2)

(c)

Find the coordinates of the image of the point (0, 3).

(3)

(d)

Describe a sequence of three transformations that maps the graph of f onto the graph of g.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solves x/2 + 1 = 2 for x (or applies transformations to the point)	✓ 1
A1 (2, 15)	✓ 1
Part (b)
M1 solves x/2 + 1 = −1	✓ 1
A1 (−4, 0)	✓ 1
Part (c)
M1 solves x/2 + 1 = 0	✓ 1
A1 x = −2	✓ 1
A1 (−2, 9)	✓ 1
Part (d)
A1 horizontal stretch with scale factor 2 (from the y-axis)	✓ 1
A1 translation of 2 units in the negative x direction (applied after the stretch; accept translation 1 unit left before the stretch)	✓ 1
A1 vertical stretch with scale factor 3	✓ 1

Common error: In f(x/2 + 1) the order of the horizontal transformations matters: translating by 1 unit left and then stretching by 2 is the same as stretching by 2 and then translating 2 units left.

+ Add to test✓ In your test246IBMI3H-2-05r21-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium11 marks

The depth d metres of water above an underwater archaeological site in a harbour is modelled by
d(t) = p + a cos(b(t − 3.5)), where t is the time in hours after midnight and b is in radians.
The depth has a maximum of 6.8 m at t = 3.5 and the next minimum of 1.2 m at t = 9.7.

(a)

Find the values of a and p.

(3)

(b)

Show that b = 0.507 to 3 significant figures.

(2)

(c)

Find the depth at midnight (t = 0), to 3 significant figures. Use the unrounded value of b.

(2)

(d)

Divers can only work when the depth is greater than 5.5 m. Find the length of time, to the nearest minute, for which diving is possible in the first tidal cycle around the maximum at t = 3.5. Use the unrounded value of b.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = (6.8 − 1.2)/2 or p = (6.8 + 1.2)/2	✓ 1
A1 a = 2.8	✓ 1
A1 p = 4.0	✓ 1
Part (b)
M1 period = 2 × (9.7 − 3.5) = 12.4 hours	✓ 1
A1 b = 2π/12.4 = 0.507 (0.50670…)	✓ 1
Part (c)
M1 d(0) = 4 + 2.8 cos(b × (−3.5)) with b in radians	✓ 1
A1 3.44 m	✓ 1
Part (d)
M1 forms 4 + 2.8 cos(b(t − 3.5)) = 5.5	✓ 1
M1 solves for t − 3.5 = ±acos(1.5/2.8)/b (= ±1.984…)	✓ 1
M1 duration = 2 × 1.984… = 3.97 hours	✓ 1
A1 3.97 hours = 238 minutes	✓ 1

Exam tip: Check that your calculator is in radian mode whenever the argument of the trigonometric function contains a value of b found as 2π divided by the period.

+ Add to test✓ In your test247IBMI3H-2-05r26-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The depth of water at the entrance to a tidal-power barrage is modelled by d(t) = 5.2 + 1.8 sin(π(t − 2)/6) metres, where t is the time in hours after midnight, 0 ≤ t ≤ 12.

(a)

Describe a sequence of four transformations that maps the graph of y = sin t onto the graph of y = d(t).

(4)

(b)

State the period of the model.

(1)

(c)

Find the two times when the depth is 6 metres, to 3 significant figures.

(3)

(d)

Hence find, to 3 significant figures, the length of time within this 12-hour interval for which the depth exceeds 6 metres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 stretch parallel to the vertical axis with scale factor 1.8	✓ 1
A1 stretch parallel to the t-axis with scale factor 6/π	✓ 1
A1 translation 2 units in the positive t-direction	✓ 1
A1 translation 5.2 units upwards	✓ 1
Accept any order that gives the same graph.	—
Part (b)
A1 12 hours	✓ 1
Part (c)
M1 sin(π(t − 2)/6) = 0.8/1.8	✓ 1
A1 t = 2.88	✓ 1
A1 t = 7.12	✓ 1
Part (d)
M1 7.1204… − 2.8796…	✓ 1
A1 4.24 hours	✓ 1

Exam tip: Use the unrounded times from the previous part: subtracting the rounded values 7.12 and 2.88 gives 4.24 here but can differ in the last digit in other questions.

+ Add to test✓ In your test248IBMI3H-2-05r26-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

A shop's daily sales, S units, t days after the start of a promotion are modelled by S(t) = a sin(b(t − c)) + d, where b > 0. Sales have a maximum of 480 on day 3 and the next minimum of 120 occurs on day 6.5.

(a)

Find the values of a, b, c and d, giving b in terms of π.

(4)

(b)

Find the total length of time between t = 0 and t = 14 for which daily sales exceed 400, to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (480 − 120)/2 = 180	✓ 1
A1 d = (480 + 120)/2 = 300	✓ 1
A1 half a period is 3.5, so period 7 and b = 2π/7	✓ 1
A1 c = 1.25 (or any equivalent value, e.g. 1.25 + 7m)	✓ 1
Part (b)
M1 180 sin(2π(t − 1.25)/7) + 300 = 400, so the sine equals 5/9	✓ 1
A1 sin⁻¹(5/9) = 0.589…	✓ 1
M1 the sine exceeds 5/9 for an angle interval of π − 2 × 0.589… = 1.964…	✓ 1
A1 time per cycle = 1.964… × 7/(2π) = 2.19 days	✓ 1
A1 two complete cycles in the interval: 4.38 days	✓ 1

Exam tip: Convert a time interval to an angle interval by multiplying by 2π/period, solve, and convert back; sketching one cycle shows the other solution.

+ Add to test✓ In your test249IBMI3H-2-04r27-01Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

A marine biologist monitors an algal bloom. The population density P (thousands of cells per mL) is modelled by P = ab^t, where t is the time in days. The values of ln P are recorded and lie on a straight line.

t (days): 0, 2, 4, 6, 8
ln P: 2.10, 2.80, 3.50, 4.20, 4.90

(a)

Find the gradient and the vertical intercept of the line of ln P against t.

(2)

(b)

Hence find the values of a and b, to 3 significant figures.

(3)

(c)

Predict the population density when t = 12.

(2)

(d)

Find the time taken for the population density to double, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (4.90 − 2.10)/8	✓ 1
A1 gradient 0.35 and intercept 2.10	✓ 1
Part (b)
M1 ln P = ln a + t ln b	✓ 1
A1 a = e^2.10 = 8.17	✓ 1
A1 b = e^0.35 = 1.42	✓ 1
Part (c)
M1 ln P = 2.10 + 0.35 × 12 = 6.3	✓ 1
A1 P = 545 thousand cells per mL	✓ 1
Part (d)
M1 b^T = 2, so T = ln 2 / ln b = ln 2 / 0.35	✓ 1
A1 1.98 days	✓ 1

Exam tip: If ln P is linear in t, the model is exponential: the gradient is ln b and the intercept is ln a.

+ Add to test✓ In your test250IBMI3H-2-04r27-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The metabolic rate R (watts) of a marine mammal of mass M (kg) is modelled by R = kMⁿ. When log₁₀ R is plotted against log₁₀ M, the graph is a straight line through the points (0.6, 0.95) and (2.4, 2.30).

(a)

Find the gradient of the line.

(2)

(b)

Find the values of n and k. Give k to 3 significant figures.

(3)

(c)

A mammal has a metabolic rate of 100 W. Find its mass.

(2)

(d)

Find the factor by which the metabolic rate is multiplied when the mass doubles. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2.30 − 0.95)/(2.4 − 0.6)	✓ 1
A1 0.75	✓ 1
Part (b)
M1 log R = n log M + log k	✓ 1
A1 n = 0.75	✓ 1
A1 log k = 0.95 − 0.75 × 0.6 = 0.5, so k = 3.16	✓ 1
Part (c)
M1 2 = 0.75 log M + 0.5	✓ 1
A1 M = 100 kg	✓ 1
Part (d)
M1 2^0.75	✓ 1
A1 1.68	✓ 1

Common error: On a log-log graph the gradient is the power n, and the intercept is log k, not k itself.

+ Add to test✓ In your test251IBMI3H-2-04r27-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The height of the tide in a harbour is modelled by h(t) = A cos(B(t − C)) + D, where h is in metres and t is the time in hours after midnight. High tide is 4.8 m and occurs at 02:00. Low tide is 0.6 m. The time between consecutive high tides is 12.4 hours.

(a)

Find the values of A and D.

(2)

(b)

Find the values of B and C, and write down the equation of the model. Give B to 3 significant figures.

(2)

(c)

Find the height of the tide at t = 7.

(2)

(d)

A boat needs a depth of at least 3.5 m to enter the harbour. Find the length of time in one tidal cycle during which the boat can enter. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 A = (4.8 − 0.6)/2 or D = (4.8 + 0.6)/2	✓ 1
A1 A = 2.1 and D = 2.7	✓ 1
Part (b)
A1 B = 2π/12.4 = 0.507 and C = 2	✓ 1
A1 h(t) = 2.1 cos(0.507(t − 2)) + 2.7	✓ 1
Part (c)
M1 substitute t = 7	✓ 1
A1 0.98 m	✓ 1
Part (d)
M1 2.1 cos(B(t − 2)) + 2.7 = 3.5	✓ 1
A1 B(t − 2) = ±1.18, so the solutions are t = 2 ± 2.33	✓ 1
A1 duration 4.66 hours	✓ 1

Exam tip: Check your calculator is in radians when the model uses B = 2π/period.

+ Add to test✓ In your test252IBMI3H-2-04r27-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium12 marks

The number N of electric vans in a delivery fleet is modelled by N = 200/(1 + A e^−rt), where t is the time in months after the programme begins, and A and r are positive constants. The following values are recorded.

t (months): 0, 2, 4, 6
N: 20, 46.4, 90.2, 138.1

(a)

Show that ln(200/N − 1) = ln A − rt.

(3)

(b)

Calculate the values of ln(200/N − 1) for the four data points, to 3 decimal places.

(2)

(c)

Use the first and last points to find r and A, each to 3 significant figures.

(3)

(d)

Use A = 9 and r = 0.5 to find the time at which N = 180. Give your answer to 3 significant figures.

(2)

(e)

Explain why the model never predicts more than 200 vans.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 200/N = 1 + A e^−rt	✓ 1
M1 200/N − 1 = A e^−rt	✓ 1
A1 taking natural logarithms: ln(200/N − 1) = ln A − rt	✓ 1
Part (b)
M1 attempt to evaluate ln(200/N − 1)	✓ 1
A1 2.197, 1.197, 0.197, −0.802	✓ 1
Part (c)
M1 gradient = (−0.802 − 2.197)/6	✓ 1
A1 r = 0.500	✓ 1
A1 A = e^2.197 = 9.00	✓ 1
Part (d)
M1 ln(200/180 − 1) = ln 9 − 0.5t	✓ 1
A1 t = 8.79 months	✓ 1
Part (e)
R1 as t increases, e^−rt tends to 0	✓ 1
R1 so N tends to 200 from below, and the denominator is always greater than 1	✓ 1

Exam tip: For a logistic model with known limit K, plot ln(K/N − 1) against t: the result is a straight line with gradient −r.

+ Add to test✓ In your test253IBMI3H-2-05r27-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The electrical power P watts used by a delivery robot when moving at speed v m s⁻¹ is modelled by P = kvⁿ, where k and n are constants. Measurements are shown in the table.

v (m s⁻¹)	0.5	1.0	2.0	4.0
P (W)	0.606	3.20	16.9	89.1

(a)

Find the values of ln v and ln P when v = 4.0. Give your answers to 3 significant figures.

(2)

(b)

A graph of ln P against ln v is a straight line. Use the data for v = 0.5 and v = 4.0 to find the gradient, and hence the value of n correct to 1 decimal place.

(3)

(c)

Find the value of k.

(2)

(d)

Using k = 3.20 and n = 2.4, estimate the power used at 3.0 m s⁻¹. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 ln 4.0 = 1.39	✓ 1
A1 ln 89.1 = 4.49	✓ 1
Part (b)
M1 ln P = ln k + n ln v, so n is the gradient	✓ 1
M1 (ln 89.1 − ln 0.606)/(ln 4 − ln 0.5) = 2.400	✓ 1
A1 n = 2.4	✓ 1
Part (c)
M1 substitutes v = 1.0, P = 3.20 (ln 1 = 0), so ln k = ln 3.20	✓ 1
A1 k = 3.20	✓ 1
Part (d)
M1 3.20 × 3.0^2.4	✓ 1
A1 44.7 W	✓ 1

Exam tip: On a log–log graph the gradient is the power n; use two points far apart to reduce the effect of rounding in the data.

+ Add to test✓ In your test254IBMI3H-2-06r31-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The pH of a solution is defined as pH = −log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol dm⁻³. Tank A of a treatment works has pH 7.8 and tank B has pH 6.8.

(a)

Find [H⁺] in tank A, giving your answer in the form a × 10^k where 1 ≤ a < 10, with a given to 3 significant figures.

(2)

(b)

Find how many times greater [H⁺] is in tank B than in tank A.

(2)

(c)

Equal volumes of water from the two tanks are mixed. Assume that the hydrogen-ion concentration of the mixture is the mean of the two concentrations. Explain why the pH of the mixture is not 7.3.

(1)

(d)

Find the pH of the mixture. Use unrounded values and give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 [H⁺] = 10^−7.8	✓ 1
A1 1.58 × 10⁻⁸ mol dm⁻³	✓ 1
Part (b)
M1 10^{7.8 − 6.8}	✓ 1
A1 10	✓ 1
Part (c)
R1 pH is a logarithmic scale, so the mean of two pH values is not the pH of the mean concentration (the mixture is dominated by the more concentrated tank)	✓ 1
Part (d)
M1 mean [H⁺] = (10^−7.8 + 10^−6.8) ÷ 2	✓ 1
A1 8.72 × 10⁻⁸ mol dm⁻³	✓ 1
M1 pH = −log₁₀(8.72 × 10⁻⁸) using the unrounded concentration	✓ 1
A1 7.06	✓ 1

Common error: Never average values on a logarithmic scale directly. Convert back to the original quantity, average there, then convert to the scale again.

+ Add to test✓ In your test255IBMI3H-2-05r35-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

During roasting, the moisture content M (% of bean mass) of green coffee beans falls with time t minutes. A graph of ln M against t is a straight line through the points (2, 2.15) and (10, 1.35).

(a)

Find the gradient and the vertical intercept of the line, and write down its equation in the form ln M = mt + c.

(3)

(b)

Hence show that M = Ae^−kt, and find A to 3 significant figures and the value of k.

(3)

(c)

State what A represents in this context.

(1)

(d)

Use the equation from part (a) to find the time at which the moisture content reaches 4.0 %.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (1.35 − 2.15) ÷ (10 − 2)	✓ 1
A1 m = −0.1	✓ 1
A1 c = 2.35, so ln M = 2.35 − 0.1t	✓ 1
Part (b)
M1 exponentiate both sides: M = e^2.35 × e^−0.1t	✓ 1
A1 k = 0.100 (0.1)	✓ 1
A1 A = e^2.35 = 10.5	✓ 1
Part (c)
A1 the moisture content (10.5 %) of the beans at the start of roasting, when t = 0	✓ 1
Part (d)
M1 ln 4.0 = 2.35 − 0.1t	✓ 1
M1 t = (2.35 − ln 4.0) ÷ 0.1	✓ 1
A1 t = 9.64 minutes	✓ 1

Exam tip: Always say what the intercept means in context: for an exponential model it is the starting value, found by exponentiating, not read directly from the log graph.

+ Add to test✓ In your test256IBMI3H-2-05r35-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The percentage P of vehicles using an electronic tag on a toll bridge, t years after the system opened, is modelled by
P(t) = 80 ÷ (1 + 7e^−0.5t), t ≥ 0.

(a)

Find the initial value of P and the limiting value of P as t becomes large.

(2)

(b)

Find the time at which 60 % of vehicles use a tag.

(3)

(c)

The rate of increase of P is greatest when P is half of its limiting value. Find the time at which this happens.

(2)

(d)

Use your GDC to find the greatest rate of increase of P.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 initial value P(0) = 80 ÷ 8 = 10.0 %	✓ 1
A1 limiting value 80 % (as e^−0.5t → 0)	✓ 1
Part (b)
M1 60 = 80 ÷ (1 + 7e^−0.5t)	✓ 1
M1 e^−0.5t = 1 ÷ 21, then take logarithms (or use GDC to solve)	✓ 1
A1 t = 6.09 years	✓ 1
Part (c)
M1 80 ÷ (1 + 7e^−0.5t) = 40, so e^−0.5t = 1 ÷ 7	✓ 1
A1 t = 3.89 years	✓ 1
Part (d)
M1 gradient of the graph of P against t at t = 3.89 (maximum of dP/dt)	✓ 1
A1 10.0 % per year	✓ 1

Exam tip: For a logistic curve, the steepest point is at half the carrying capacity, which saves you from differentiating.

+ Add to test✓ In your test257IBMI3H-2-05r35-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The power P (W) from a radioisotope generator on a space probe, t years after launch, is such that the graph of log₁₀ P against t is a straight line through (0, 2.60) and (20, 2.52).

(a)

Find the gradient and the vertical intercept of the line.

(2)

(b)

Show that P can be written as P = a × b^t, and find a and b to 3 significant figures.

(3)

(c)

Use the equation of the line to find the power 30 years after launch.

(2)

(d)

Find the half-life of the power output.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 gradient = (2.52 − 2.60) ÷ 20 = −0.004	✓ 1
A1 intercept = 2.60	✓ 1
Part (b)
M1 P = 10^{2.60 − 0.004t} = 10^2.60 × (10^−0.004)^t	✓ 1
A1 a = 398	✓ 1
A1 b = 0.991	✓ 1
Part (c)
M1 log₁₀ P = 2.60 − 0.004 × 30 = 2.48	✓ 1
A1 P = 302 W	✓ 1
Part (d)
M1 log₁₀(0.5) = −0.004t (or a(b^t) = ½a)	✓ 1
M1 t = log₁₀ 2 ÷ 0.004	✓ 1
A1 t = 75.3 years	✓ 1

Common error: A straight line on a log-linear graph means exponential behaviour: the gradient gives log₁₀ b, not b.

+ Add to test✓ In your test258IBMI3H-2-05r35-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The power P (W) delivered by a probe's solar panel is modelled by P(t) = a sin(b(t + 3)) + d, where t is the time in hours after midnight and b > 0. The maximum power is 420 W at 03:00 and the first minimum power is 60 W at 15:00.

(a)

Find the value of a and the value of d.

(2)

(b)

Find the exact value of b.

(2)

(c)

Find the power at 08:00.

(2)

(d)

Find the first time after midnight at which the power falls to 300 W.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (420 − 60) ÷ 2 = 180	✓ 1
A1 d = (420 + 60) ÷ 2 = 240	✓ 1
Part (b)
M1 period = 2 × (15 − 3) = 24, so 2π ÷ b = 24	✓ 1
A1 b = π ÷ 12	✓ 1
Part (c)
M1 180 sin(11π ÷ 12) + 240	✓ 1
A1 P = 287 W	✓ 1
Part (d)
M1 180 sin(π(t + 3) ÷ 12) + 240 = 300, so sin(π(t + 3) ÷ 12) = 1/3	✓ 1
M1 choose the decreasing solution: π(t + 3) ÷ 12 = π − sin⁻¹(1/3)	✓ 1
A1 t = 7.70 hours (approximately 07:42)	✓ 1

Exam tip: Check that the angle you choose lies on the falling part of the curve: the arcsine value alone gives the rising crossing, which here occurs before midnight.

+ Add to test✓ In your test259IBMI3H-2-06r38-02Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

Fisheries scientists model the mass W grams of a fish of length L cm by W = aL^b. A graph of log₁₀W against log₁₀L is a straight line passing through the points (1.0, 0.9) and (1.5, 2.45).

(a)

Find the value of b.

(2)

(b)

Show that log₁₀W = b log₁₀L + log₁₀a, and hence find a to 3 significant figures.

(3)

(c)

Predict the mass of a fish of length 42 cm, to the nearest gram. Use the unrounded value of a.

(2)

(d)

Find the factor by which the mass is multiplied when the length of a fish is doubled, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = (2.45 − 0.9) / (1.5 − 1.0)	✓ 1
A1 b = 3.1	✓ 1
Part (b)
M1 taking logs of both sides of W = aL^b and using the laws of logarithms	✓ 1
A1 intercept log₁₀a = 0.9 − 3.1 = −2.2	✓ 1
A1 a = 10^−2.2 = 0.00631	✓ 1
Part (c)
M1 substitute L = 42 into W = 10^−2.2 × 42^3.1	✓ 1
A1 679 g	✓ 1
Part (d)
M1 2^b = 2^3.1	✓ 1
A1 8.57	✓ 1

Exam tip: A straight line on a log–log graph signals a power law: the gradient is the power and the intercept is the logarithm of the multiplier.

+ Add to test✓ In your test260IBMI3H-2-06r38-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

A fashion designer models the edge of a fabric border by the curve y = f(x), where f(x) = x² − 4x + 1. The curve is transformed to y = g(x) by a reflection in the x-axis, then a vertical stretch of scale factor 3, then a translation by the vector (4, 1) (that is, 4 to the right and 1 up).

(a)

Write f(x) in the form (x − h)² + k.

(2)

(b)

Write down the coordinates of the vertex of y = g(x).

(2)

(c)

Hence write g(x) in the form a(x − h)² + k.

(2)

(d)

Find the larger x-intercept of y = g(x), to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 complete the square	✓ 1
A1 (x − 2)² − 3	✓ 1
Part (b)
M1 apply the transformations in order to the vertex (2, −3)	✓ 1
A1 (6, 10)	✓ 1
Part (c)
M1 g(x) = −3f(x − 4) + 1 (or equivalent use of the vertex)	✓ 1
A1 −3(x − 6)² + 10	✓ 1
Part (d)
M1 set g(x) = 0	✓ 1
M1 (x − 6)² = 10/3	✓ 1
A1 7.83	✓ 1

Exam tip: When transformations are applied in a stated order, apply them to a key point one at a time, tracking the vertex or other known points.

+ Add to test✓ In your test261IBMI3H-2-06r38-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium9 marks

The depth h metres of water at a harbour entrance is modelled by h(t) = a cos(b(t − 2)) + d, where t is the time in hours after midnight and the angle is in radians. The maximum depth of 6.8 m occurs at t = 2 and the next minimum depth of 1.2 m occurs at t = 8.

(a)

Find the values of a and d.

(3)

(b)

Find the exact value of b.

(2)

(c)

A boat can enter only when the depth is at least 5.2 m. Find the length of time, in each cycle, for which the depth is at least 5.2 m, to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = (6.8 − 1.2) / 2 or d = (6.8 + 1.2) / 2	✓ 1
A1 a = 2.8	✓ 1
A1 d = 4	✓ 1
Part (b)
M1 period = 12 hours, b = 2π / 12	✓ 1
A1 b = π/6	✓ 1
Part (c)
M1 2.8 cos(b(t − 2)) + 4 = 5.2	✓ 1
M1 cos(b(t − 2)) = 1.2/2.8, giving b(t − 2) = ±1.128	✓ 1
M1 time = 2 × 1.1279 / b (symmetry about the maximum)	✓ 1
A1 4.31 hours	✓ 1

Exam tip: The depth is at least a given value on either side of the maximum, so double the time from the maximum to the first crossing.

+ Add to test✓ In your test262IBMI3H-2-05r42-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium12 marks

The memory of a space probe stores observation data. The amount of data stored, S gigabytes (GB), t days after the start of an observation campaign is modelled by
S(t) = 480 / (1 + Ae^−kt), t ≥ 0,
where A and k are positive constants. Initially 12 GB of data are stored. After 5 days, 60 GB are stored.

(a)

Show that A = 39.

(2)

(b)

Find the value of k, correct to 3 significant figures.

(3)

(c)

The memory is considered nearly full when 90% of its 480 GB capacity is used. Using the unrounded value of k, find the time at which this occurs, correct to 3 significant figures.

(3)

(d)

Find the time at which data are being stored at the greatest rate, correct to 3 significant figures.

(2)

(e)

Find the greatest rate at which data are stored, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 = 480/(1 + A)	✓ 1
A1 1 + A = 40, so A = 39 (AG)	✓ 1
Part (b)
M1 60 = 480/(1 + 39e^−5k)	✓ 1
M1 e^−5k = 7/39 (or 1 + 39e^−5k = 8)	✓ 1
A1 k = 0.344 (0.3435 unrounded)	✓ 1
Part (c)
M1 432 = 480/(1 + 39e^−kt)	✓ 1
M1 e^−kt = (480/432 − 1)/39 (or equivalent, e.g. using GDC to solve)	✓ 1
A1 t = 17.1 days	✓ 1
Part (d)
M1 S = 240 (half of the limiting value), so 39e^−kt = 1	✓ 1
A1 t = 10.7 days (ECF from (b))	✓ 1
Part (e)
M1 gradient of S(t) at the time found in (d), or use of kC/4 with C = 480 (ECF from (b))	✓ 1
A1 41.2 GB per day	✓ 1

Exam tip: The steepest part of a logistic curve occurs when the quantity is half of its limiting value; use this to avoid searching for the maximum of the derivative.

+ Add to test✓ In your test263IBMI3H-2-05r42-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The vertical clearance c metres between the underside of a bridge and the water below it varies with the tide. It is modelled by c(t) = a cos(b(t − 1.5)) + d, where t is the time in hours after midnight, a, b and d are positive constants and b is in radians. The greatest clearance is 9.6 m and occurs at t = 1.5. The next least clearance is 5.2 m and occurs at t = 7.7.

(a)

Find the value of a and the value of d.

(3)

(b)

Show that b = 0.507, correct to 3 significant figures.

(2)

(c)

Find the value of c(0), correct to 3 significant figures.

(1)

(d)

A vessel needs a clearance of at least 6.0 m to pass under the bridge. Using the unrounded value of b, find the total length of time, in the 12.4 hours from t = 0, during which the vessel cannot pass. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = (9.6 − 5.2)/2 (or d = (9.6 + 5.2)/2)	✓ 1
A1 a = 2.2	✓ 1
A1 d = 7.4	✓ 1
Part (b)
M1 period = 2 × (7.7 − 1.5) = 12.4 (or b × 6.2 = π)	✓ 1
A1 b = 2π/12.4 = 0.5067 = 0.507 (AG)	✓ 1
Part (c)
A1 c(0) = 8.99 m	✓ 1
Part (d)
M1 solves 2.2 cos(b(t − 1.5)) + 7.4 = 6 (or < 6)	✓ 1
A1 t = 5.96 (first time clearance falls to 6.0 m)	✓ 1
A1 t = 9.44 (clearance rises back to 6.0 m)	✓ 1
A1 duration 3.48 hours	✓ 1

Exam tip: Solve the equation c = 6 on the GDC, then decide from the shape of the graph which side of the two solutions is below the threshold.

+ Add to test✓ In your test264IBMI3H-2-06r42-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium12 marks

The number of bees in a hive, N, t weeks after the start of spring is modelled by N(t) = 50 000 / (1 + Ae^−kt), where A and k are positive constants. At the start of spring there are 8000 bees in the hive. The number of bees is increasing most rapidly when t = 14.

(a)

Find the value of A.

(2)

(b)

Find the value of k, correct to 3 significant figures.

(3)

(c)

Using the unrounded value of k, find the number of bees in the hive after 8 weeks, correct to the nearest hundred.

(2)

(d)

Find the greatest rate of increase in the number of bees, correct to 3 significant figures.

(2)

(e)

Using the unrounded value of k, find the time at which the hive first contains 46 000 bees. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 8000 = 50 000/(1 + A)	✓ 1
A1 A = 5.25	✓ 1
Part (b)
M1 the greatest rate occurs when N = 25 000 (half the limiting value)	✓ 1
M1 5.25e^−14k = 1	✓ 1
A1 k = 0.118 (0.1184… unrounded)	✓ 1
Part (c)
M1 50 000/(1 + 5.25e^−8k)	✓ 1
A1 16 500	✓ 1
Part (d)
M1 gradient of N at t = 14 (GDC) or kC/4 with C = 50 000 (ECF from (b))	✓ 1
A1 1480 bees per week	✓ 1
Part (e)
M1 50 000/(1 + 5.25e^−kt) = 46 000	✓ 1
M1 e^−kt = (50 000/46 000 − 1)/5.25 (or GDC solve)	✓ 1
A1 t = 34.6 weeks	✓ 1

Exam tip: The inflection point of a logistic curve is where the quantity is half of its limiting value, so the given time of greatest growth links A and k.

+ Add to test✓ In your test265IBMI3H-2-06r42-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Medium10 marks

The number of patients admitted to a hospital in week t of the year is modelled by A(t) = a cos(b(t − 3)) + d, where b is in radians and a, b and d are positive constants. The greatest weekly number of admissions is 142 and occurs at t = 3. The next least number is 58 and occurs at t = 29.

(a)

Find the value of a and the value of d.

(3)

(b)

Show that b = 0.121, correct to 3 significant figures.

(2)

(c)

Find the number of admissions predicted for week 0, correct to the nearest whole number.

(1)

(d)

Using the unrounded value of b, find the total length of time in one 52-week year for which the model predicts more than 120 admissions per week. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 a = (142 − 58)/2 (or d = (142 + 58)/2)	✓ 1
A1 a = 42	✓ 1
A1 d = 100	✓ 1
Part (b)
M1 period = 2 × (29 − 3) = 52	✓ 1
A1 b = 2π/52 = 0.1208 = 0.121 (AG)	✓ 1
Part (c)
A1 139	✓ 1
Part (d)
M1 100 + 42 cos(b(t − 3)) > 120, so cos(b(t − 3)) > 20/42	✓ 1
M1 b\|t − 3\| < arccos(20/42) = 1.074	✓ 1
M1 total width = 2 × 1.074/b	✓ 1
A1 17.8 weeks	✓ 1

Exam tip: The inequality gives an interval centred on the time of the maximum; the interval may wrap around the end of the year but its total length is unchanged.

+ Add to test✓ In your test266IBMI-031-00-12Linear models and piecewise models Paper 2Hard12 marks

The battery level, B %, of a phone t hours after it is unplugged is modelled by
$B(t)=\begin{cases}100-12t, & 0\le t\le 5\\ 40+30(t-5), & 5where the phone is in use for the first 5 hours and then placed on charge.
A newer model of phone uses power more efficiently, so that each stage takes 25% longer. Its battery level is modelled by \(N(t)=B\left(\frac{t}{1.25}\right)$.

(a)

Find B(3).

(1)

(b)

Find all the times at which the battery level of the original phone is 70%.

(3)

(c)

Describe the transformation that maps the graph of B onto the graph of N.

(2)

(d)

Write down N(t) as a piecewise function and state its domain.

(4)

(e)

Find how much later the newer phone first reaches a battery level of 52% than the original phone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 64 (%)	✓ 1
Part (b)
M1 equating either piece to 70	✓ 1
A1 t = 2.5 hours	✓ 1
A1 t = 6 hours	✓ 1
Part (c)
A1 horizontal stretch (parallel to the t-axis)	✓ 1
A1 scale factor 1.25	✓ 1
Part (d)
A1 N(t) = 100 − 9.6t	✓ 1
A1 N(t) = 40 + 24(t − 6.25) or 24t − 110	✓ 1
A1 pieces valid for 0 ≤ t ≤ 6.25 and 6.25 < t ≤ 8.75	✓ 1
A1 domain 0 ≤ t ≤ 8.75	✓ 1
Part (e)
M1 100 − 12t = 52 gives t = 4 and 100 − 9.6t = 52 gives t = 5	✓ 1
A1 1 hour later	✓ 1

Common error: Replacing t with t/1.25 stretches the graph by factor 1.25, not 1/1.25; the newer phone takes longer, so its graph is wider.

+ Add to test✓ In your test267IBMI-031-02-r1-06Exponential and sinusoidal models Paper 2Hard12 marks

An ecologist records the number of grey seals, N, on an island, where t is the number of years after 1 January 2000. A plot of ln N against t is approximately linear, and the regression line is
$\ln N = 0.0815t + 6.42$.

(a)

Show that the model can be written in the form $N = a e^{bt}$, and find the values of a and b.

(3)

(b)

Find the time taken for the seal population to double, according to this model.

(2)

(c)

A report requires the model to use s, the number of years after 1 January 2010.
(i) Write the model in the form $N = c\,e^{0.0815s}$, giving the value of c.
(ii) Describe the single transformation that maps the graph of N against t onto the graph of N against s.

(3)

(d)

A second species, harbour seals, has population M on the same island, modelled by $\ln M = -0.046t + 8.10$. Find the value of t when the two populations are predicted to be equal.

(3)

(e)

Give one reason why the model for N is unlikely to remain valid for many decades.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Exponentiates both sides: N = e^{0.0815t + 6.42} = e^6.42 × e^0.0815t (M1)	✓ 1
a = e^6.42 = 614 (A1)	✓ 1
b = 0.0815 (A1)	✓ 1
Part (b)
Sets up e^0.0815t = 2, or equivalent (M1)	✓ 1
t = ln 2 / 0.0815 = 8.50 years (A1)	✓ 1
Part (c)
(i) Substitutes t = s + 10: N = 614e^{0.0815(s + 10)} = 614e^0.815e^0.0815s (M1)	✓ 1
c = 1390 (1387.14…) (A1)	✓ 1
(ii) Horizontal translation of 10 units in the negative direction (to the left), or equivalently a vertical stretch with scale factor e^0.815 ≈ 2.26 (A1)	✓ 1
Part (d)
Equal populations means ln N = ln M (M1)	✓ 1
0.0815t + 6.42 = −0.046t + 8.10, so 0.1275t = 1.68 (A1)	✓ 1
t = 13.2 years (A1)	✓ 1
Part (e)
Valid reason in context, e.g. exponential growth is unbounded, but the island has limited space and food, so growth must slow down (R1)	✓ 1

Exam tip: If ln N is linear in t, then N is exponential in t: the gradient gives b and the intercept gives ln a. Comparing two populations is easier on the log scale, because two straight lines can be equated directly.

+ Add to test✓ In your test268IBMI-031-04-r1-06Transformations of graphs and logarithmic scales Paper 2Hard16 marks

A cup of coffee is left to cool in a room. Its temperature, T °C, t minutes after it is poured is modelled by
$T(t) = 70\mathrm{e}^{-0.05t} + 20$, $t \geq 0$.

(a)

Describe a sequence of transformations that maps the graph of $y = \mathrm{e}^{x}$ onto the graph of T.

(3)

(b)

A café displays temperatures in degrees Fahrenheit, using $F = 1.8T + 32$. Find an expression for the temperature of the coffee, F(t) °F, and describe the transformations that map the graph of T onto the graph of F.

(3)

(c)

The café recommends that coffee is drunk once it has cooled to 140 °F. Find the time taken for the coffee to reach this temperature.

(3)

(d)

Show that a graph of $\ln(T - 20)$ against t is a straight line, and state its gradient and vertical intercept.

(3)

(e)

A second cup of coffee in the same room gives data for which the line of best fit is $\ln(T - 20) = 4.38 - 0.062t$.
(i) Find the initial temperature of the second cup.
(ii) Find the time taken for the second cup to cool to 40 °C.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: reflection in the vertical axis and horizontal stretch with scale factor 20	✓ 1
A1: vertical stretch with scale factor 70	✓ 1
A1: followed by a vertical translation of 20 units up (stretch before translation)	✓ 1
Part (b)
M1: substitutes, $F(t) = 1.8(70\mathrm{e}^{-0.05t} + 20) + 32$	✓ 1
A1: $F(t) = 126\mathrm{e}^{-0.05t} + 68$	✓ 1
A1: vertical stretch scale factor 1.8 followed by vertical translation 32 units up	✓ 1
Part (c)
M1: sets $126\mathrm{e}^{-0.05t} + 68 = 140$ (or equivalent 60 °C in T)	✓ 1
M1: rearranges to $\mathrm{e}^{-0.05t} = \frac{4}{7}$ or uses GDC	✓ 1
A1: t = 11.2 minutes	✓ 1
Part (d)
M1: $T - 20 = 70\mathrm{e}^{-0.05t}$ so $\ln(T-20) = \ln 70 - 0.05t$, which is linear in t	✓ 1
A1: gradient −0.05	✓ 1
A1: vertical intercept ln 70 = 4.25	✓ 1
Part (e)
(i) M1: at t = 0, $T = \mathrm{e}^{4.38} + 20$	✓ 1
A1: T = 99.8 °C	✓ 1
(ii) M1: sets $4.38 - 0.062t = \ln 20$	✓ 1
A1: t = 22.3 minutes	✓ 1

Common error: When linearising, subtract the room temperature before taking logarithms: ln T is not linear in t for this model, but ln(T − 20) is.

+ Add to test✓ In your test269IBMI-031-01-r1-06Quadratic and cubic models Paper 2Hard13 marks

The profile of a children's slide is modelled by $f(x) = ax^3 + bx^2 + cx + d$ for 0 ≤ x ≤ 5, where x is the horizontal distance in metres from the top of the slide and f(x) is the height in metres above the ground.
The slide starts horizontally at a height of 4 m at x = 0, and finishes horizontally at ground level at x = 5.

(a)

Find the values of a, b, c and d.

(4)

(b)

Find the horizontal distance from the top at which the slide is steepest, and the angle the slide makes with the horizontal at this point.

(4)

(c)

A redesigned slide has profile $g(x) = f\left(\frac{x}{1.5}\right)$. Describe fully the single transformation that maps the graph of f onto the graph of g.

(2)

(d)

A safety rule states that a slide must make an angle of no more than 40° with the horizontal at any point. A slide is designed with profile $h(x) = f\left(\frac{x}{k}\right)$, where k > 1. Find the least value of k for which this slide meets the safety rule.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: f(0) = 4 gives d = 4, and f '(0) = 0 gives c = 0	✓ 1
M1: differentiates: f '(x) = 3ax² + 2bx + c	✓ 1
M1: forms 125a + 25b + 4 = 0 and 75a + 10b = 0	✓ 1
A1: a = 0.064, b = −0.48	✓ 1
Part (b)
M1: sets f ''(x) = 0.384x − 0.96 = 0, or finds the minimum of f '(x) on GDC	✓ 1
A1: x = 2.5 m	✓ 1
M1: f '(2.5) = 0.192(2.5)² − 0.96(2.5) = −1.2	✓ 1
A1: angle = tan⁻¹(1.2) = 50.2°	✓ 1
Part (c)
A1: horizontal stretch (parallel to the x-axis)	✓ 1
A1: scale factor 1.5	✓ 1
Part (d)
M1: recognises the steepest gradient of h is $\frac{1}{k}f'(2.5) = -\frac{1.2}{k}$	✓ 1
M1: sets $\frac{1.2}{k} \le \tan 40°$	✓ 1
A1: k = 1.43 (1.4301…)	✓ 1

Exam tip: A horizontal stretch by scale factor k leaves heights unchanged but divides every gradient by k, so the steepest point moves to x = 2.5k while its gradient becomes f′(2.5)/k.

+ Add to test✓ In your test270IBMI-031-02-r2-06Exponential and sinusoidal models Paper 2Hard13 marks

The depth of water, $d$ metres, at the entrance to a harbour is modelled by
$d(t) = 2.6\cos(29t) + 7.4$,
where $t$ is the time in hours after midnight and angles are in degrees. A cargo ship can only pass through the entrance when the depth of water is at least 9.0 m.

(a)

Find the period of the model, giving your answer in hours and minutes to the nearest minute.

(2)

(b)

Find the total length of time, in each tidal cycle, for which the ship can pass through the entrance.

(4)

(c)

A storm surge raises the water level by 0.8 m at all times and delays each high tide by 1.5 hours. Write down the new model, $d_2(t)$, for the depth of water.

(2)

(d)

Using $d_2$, find the first time after midnight at which the depth falls to 9.0 m. Give your answer as a time of day to the nearest minute.

(3)

(e)

Using $d_2$, find the percentage of each tidal cycle during which the ship can pass through the entrance.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Period = 360 ÷ 29 = 12.414 hours (M1)	✓ 1
12 hours 25 minutes (A1)	✓ 1
Part (b)
Sets up 2.6cos(29t) + 7.4 = 9.0, i.e. cos(29t) = 0.615… (M1)	✓ 1
t = 1.79 h (1.7938…) (A1)	✓ 1
t = 10.6 h (10.620…) (A1)	✓ 1
Total time = 2 × 1.7938 = 3.59 hours (A1)	✓ 1
Part (c)
Horizontal translation: cos(29(t − 1.5)) (A1)	✓ 1
d₂(t) = 2.6cos(29(t − 1.5)) + 8.2 (A1)	✓ 1
Part (d)
Sets up 2.6cos(29(t − 1.5)) + 8.2 = 9.0 (M1)	✓ 1
t = 3.99 hours (3.9855…) (A1)	✓ 1
03:59 (A1)	✓ 1
Part (e)
Time per cycle with d ≥ 9.0: 2 × 2.4855 = 4.971 h, divided by period 12.414 h (M1)	✓ 1
40.0% (A1)	✓ 1

Common error: A delay of 1.5 hours is a translation in the positive t-direction, so replace t with (t − 1.5), not (t + 1.5). Check by confirming that the new maximum occurs at t = 1.5.

+ Add to test✓ In your test271IBMI-031-08-r1-06Choosing and evaluating models Paper 2Hard13 marks

A conservation group introduces 400 fish into a lake. Three years later a survey estimates there are 1100 fish. Two models are proposed for the number of fish $N$, $t$ years after introduction.
Model A: $N = 400e^{kt}$
Model B: $N = \dfrac{6000}{1 + 14e^{-rt}}$
Both models are fitted to the survey data at $t = 0$ and $t = 3$.

(a)

Find the value of $k$.

(2)

(b)

Verify that Model B gives $N = 400$ when $t = 0$, and show that $r = 0.382$ to 3 significant figures.

(3)

(c)

Find the number of fish predicted by each model when $t = 8$.

(3)

(d)

A survey at $t = 8$ estimates 3900 fish. State, with a reason, which model is more appropriate.

(2)

(e)

Interpret the value 6000 in Model B in the context of the question.

(1)

(f)

Using Model B, find the time at which the fish population reaches 5000.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$400e^{3k} = 1100$ M1	✓ 1
$k = \frac{1}{3}\ln 2.75 = 0.337$ A1	✓ 1
Part (b)
$\frac{6000}{1+14} = 400$ A1	✓ 1
Sets $\frac{6000}{1 + 14e^{-3r}} = 1100$, giving $e^{-3r} = \frac{7}{22}$ M1	✓ 1
$r = \frac{1}{3}\ln\frac{22}{7} = 0.3817… = 0.382$ A1	✓ 1
Part (c)
Substitutes $t = 8$ into either model M1	✓ 1
Model A: $N = 5940$ (5937.6…) A1	✓ 1
Model B: $N = 3610$ (3613.2… using unrounded $r$; accept 3620 from using $r = 0.382$) A1	✓ 1
Part (d)
Compares both predictions with 3900, e.g. differences of about 2040 and 290 M1	✓ 1
Model B, because its prediction is much closer to the observed value R1	✓ 1
Part (e)
6000 is the carrying capacity: the maximum number of fish the lake can sustain in the long term R1	✓ 1
Part (f)
Solves $\frac{6000}{1 + 14e^{-rt}} = 5000$, giving $e^{-rt} = \frac{1}{70}$ M1	✓ 1
$t = \frac{\ln 70}{r} = 11.1$ years (unrounded $r$ or $r = 0.382$) A1	✓ 1

Exam tip: When choosing a model, quote both predictions next to the observed value and then state the conclusion. A conclusion with no comparison usually earns no reasoning mark. Keep unrounded parameter values in your calculator to avoid rounding drift.

+ Add to test✓ In your test272IBMI-031-06-r2-06Quadratic and cubic models Paper 2Hard12 marks

The population of an insect species in Region A, in thousands, is modelled by
$f(x) = 0.5x^3 - 6x^2 + 18x + 40$, for $0 \leq x \leq 8$,
where x is the time in months after monitoring began.
In Region B the population, in thousands, follows a slower cycle and is modelled by
$g(x) = 1.5\,f(0.5x) - 12$, for $0 \leq x \leq 16$.

(a)

Find the coordinates of the local maximum and the local minimum of the graph of f.

(3)

(b)

Describe fully a sequence of transformations that maps the graph of f onto the graph of g.

(3)

(c)

Hence write down the coordinates of the local maximum and the local minimum of the graph of g.

(2)

(d)

Find the total length of time, within the 16 months, for which the population of Region B exceeds 60 thousand.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Use of GDC, or f′(x) = 1.5x² − 12x + 18 = 0 M1	✓ 1
Local maximum (2, 56) A1	✓ 1
Local minimum (6, 40) A1	✓ 1
Part (b)
Horizontal stretch with scale factor 2 (parallel to the x-axis), in any position in the sequence A1	✓ 1
Vertical stretch with scale factor 1.5 (parallel to the y-axis) A1	✓ 1
Vertical translation by the vector (0, −12) applied after the vertical stretch, OR a vertical translation by (0, −8) applied before the vertical stretch A1	✓ 1
Part (c)
Local maximum (4, 72) A1	✓ 1
Local minimum (12, 48) A1	✓ 1
Part (d)
g(x) > 60 ⇔ f(0.5x) > 48, or solving g(x) = 60 directly with GDC M1	✓ 1
x = 1.07, x = 8 and x = 14.9 (exact values 8 − 4√3, 8, 8 + 4√3) A1	✓ 1
g(x) > 60 for 1.07 < x < 8 and 14.9 < x ≤ 16, using the shape of the graph R1	✓ 1
Total time = 6.93 + 1.07 = 8.00 months A1	✓ 1

Exam tip: In y = a f(bx) + d, the horizontal scale factor is 1/b, not b. The order of the vertical stretch and the vertical translation matters. If you stretch first, translate by d. If you translate first, translate by d/a. State the order clearly.

+ Add to test✓ In your test273IBMI-031-04-r2-06Transformations of graphs and logarithmic scales Paper 2Hard15 marks

The temperature of a cup of coffee $t$ minutes after it is poured is modelled by $f(t)=70e^{-0.05t}+20$, in °C.
A second cup, in an insulated mug in a warmer room, is modelled by $g(t)=f(0.8t)+4$.

(a)

Write down the equation of the horizontal asymptote of the graph of $f$ and interpret it in context.

(2)

(b)

Describe fully the transformations that map the graph of $f$ onto the graph of $g$.

(3)

(c)

Write $g(t)$ in the form $ae^{-kt}+c$.

(2)

(d)

Find the time taken for the second cup to cool to 60 °C, and hence find how much longer the second cup takes than the first cup to cool to 60 °C.

(5)

(e)

Find the maximum value of $g(t)-f(t)$ for $t\ge0$, and the time at which it occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $y=20$ (accept $T=20$ or $f(t)=20$)	✓ 1
R1: the coffee cools towards room temperature, 20 °C, but never falls below it	✓ 1
Part (b)
A1: horizontal stretch (parallel to the $t$-axis)	✓ 1
A1: scale factor $\frac{1}{0.8}=1.25$	✓ 1
A1: vertical translation 4 units up	✓ 1
Part (c)
M1: $70e^{-0.05(0.8t)}+20+4$	✓ 1
A1: $g(t)=70e^{-0.04t}+24$	✓ 1
Part (d)
M1: $70e^{-0.04t}+24=60$	✓ 1
A1: $t=\frac{\ln(70/36)}{0.04}=16.6$ minutes	✓ 1
M1: $70e^{-0.05t}+20=60$	✓ 1
A1: $t=\frac{\ln 1.75}{0.05}=11.2$ minutes	✓ 1
A1: $16.62-11.19=5.43$ minutes	✓ 1
Part (e)
M1: forms $70e^{-0.04t}-70e^{-0.05t}+4$ and uses a GDC maximum or sets the derivative to zero	✓ 1
A1: $t=100\ln1.25=22.3$ minutes	✓ 1
A1: maximum difference 9.73 °C	✓ 1

Common error: $f(0.8t)$ is a horizontal stretch with scale factor $\frac{1}{0.8}=1.25$, not 0.8. A factor of 0.8 would mean the second cup cools faster, which contradicts the insulated mug.

+ Add to test✓ In your test274IBMI-031-06-r1-06Quadratic and cubic models Paper 2Hard15 marks

The profile of a section of a children's roller-coaster track is modelled by
$f(x) = -0.004x^3 + 0.12x^2$, for $0 \le x \le 30$,
where $f(x)$ is the height in metres above the ground at a horizontal distance x metres from the start.
A larger version of the ride is designed with profile $g(x) = p\,f(qx)$, where $p, q > 0$. The new track is the image of the whole original section. It must reach its maximum height of 20 m at a horizontal distance of 30 m from the start.

(a)

Find the maximum height of the original track and the horizontal distance at which it occurs.

(2)

(b)

Find the value of p and the value of q.

(3)

(c)

Describe fully the two transformations that map the graph of f onto the graph of g.

(2)

(d)

Find the total horizontal length of the new track section.

(2)

(e)

Show that $g(x) = -\frac{1}{675}x^3 + \frac{1}{15}x^2$.

(3)

(f)

A horizontal safety walkway is to be fixed between the two points on the new track that are 12 m above the ground. Find the length of the walkway.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 20$ m (e.g. from $f'(x) = -0.012x^2 + 0.24x = 0$ or GDC)	✓ 1
Maximum height = 16 m	✓ 1
Part (b)
Recognises the maximum of g occurs where $qx = 20$ and equals $16p$	✓ 1
$p = 1.25$	✓ 1
$q = \frac{2}{3}$	✓ 1
Part (c)
Vertical stretch (parallel to the y-axis), scale factor 1.25	✓ 1
Horizontal stretch (parallel to the x-axis), scale factor 1.5	✓ 1
Part (d)
Original track ends at $x = 30$, so $\frac{2}{3}x = 30$ (or $30 \times 1.5$)	✓ 1
Length = 45 m	✓ 1
Part (e)
Substitutes: $g(x) = 1.25\left[-0.004\left(\frac{2x}{3}\right)^3 + 0.12\left(\frac{2x}{3}\right)^2\right]$	✓ 1
$x^3$ coefficient: $1.25 \times (-0.004) \times \frac{8}{27} = -\frac{1}{675}$	✓ 1
$x^2$ coefficient: $1.25 \times 0.12 \times \frac{4}{9} = \frac{1}{15}$	✓ 1
Part (f)
Solves $g(x) = 12$ by GDC, or solves $f(x) = 9.6$ and scales the solutions by 1.5	✓ 1
$x = 17.0$ and $x = 39.9$	✓ 1
Length = 22.9 m	✓ 1

Common error: Replacing x by qx stretches the graph horizontally by scale factor 1/q, not q. Here q = 2/3 gives a stretch of 1.5, which is why the track lengthens from 30 m to 45 m.

+ Add to test✓ In your test275IBMI3H-2-04-07Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard13 marks

A student measures the drag force F newtons on a cyclist at speed v m s⁻¹.

v (m s⁻¹): 5, 10, 15, 20, 25
F (N): 6.4, 23.8, 51.5, 88.9, 135.9
The student suggests a power model F = av^b.

(a)

Calculate ln v and ln F for each data pair, to 3 decimal places.

(2)

(b)

Show that the points (ln v, ln F) lie close to a straight line.

(2)

(c)

Hence find the values of a and b.

(4)

(d)

Another student suggests an exponential model F = cr^v. Use the ratios of F over equal steps in v to show that this model is unsuitable.

(3)

(e)

Use the power model to estimate the drag force at 35 m s⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 ln v: 1.609, 2.303, 2.708, 2.996, 3.219	✓ 1
A1 ln F: 1.856, 3.170, 3.942, 4.488, 4.912	✓ 1
Part (b)
M1 gradient between first and last points (4.912 − 1.856)/(3.219 − 1.609) and between second and fourth points	✓ 1
A1 both gradients are about 1.90, so the points are close to collinear	✓ 1
Part (c)
M1 ln F = b ln v + ln a	✓ 1
A1 b = 1.90	✓ 1
M1 ln a = 1.856 − 1.90 × 1.609	✓ 1
A1 a = 0.301	✓ 1
Part (d)
M1 ratios of consecutive values of F	✓ 1
A1 3.72, 2.16, 1.73, 1.53	✓ 1
R1 the ratios are not constant, so an exponential model does not fit	✓ 1
Part (e)
M1 0.301 × 35^1.90	✓ 1
A1 258 N	✓ 1

Exam tip: Straight line on log-log axes means a power model; straight line on semi-log axes (ln of the output against the input) means an exponential model.

+ Add to test✓ In your test276IBMI3H-2-04-12Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard14 marks

A rumour spreads among the 800 students of a school. The number of students who have heard it, P after t days, is modelled by P = 800 / (1 + ce^−kt). Initially 60 students know the rumour and after 3 days 200 students know it.

(a)

Show that c = 37/3.

(2)

(b)

Find the value of k.

(3)

(c)

Find the time at which 600 students know the rumour.

(3)

(d)

Find the time at which the rumour is spreading fastest.

(2)

(e)

A rival model is R = 60 × 1.5^t. Find when the rival model predicts that 800 students know the rumour, and the logistic prediction at that time. Comment.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 800/(1 + c) = 60	✓ 1
A1 c = 37/3 (shown)	✓ 1
Part (b)
M1 800/(1 + (37/3)e^(−3k)) = 200	✓ 1
M1 1 + (37/3)e^(−3k) = 4, so (37/3)e^(−3k) = 3	✓ 1
A1 k = 0.471	✓ 1
Part (c)
M1 (37/3)e^(−kt) = 1/3	✓ 1
A1 e^(−kt) = 1/37	✓ 1
A1 t = 7.66 days	✓ 1
Part (d)
M1 P = 400	✓ 1
A1 t = 5.33 days	✓ 1
Part (e)
M1 60 × 1.5^t = 800	✓ 1
A1 t = 6.39 days	✓ 1
A1 logistic model gives P = 498 students at this time	✓ 1
R1 the exponential model ignores the limited number of students, so it overestimates	✓ 1

Exam tip: Substitute the initial condition first to fix the constant c, then use the second data point to find k.

+ Add to test✓ In your test277IBMI3H-2-04-13Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard13 marks

The height H metres of a sapling t years after planting is modelled by H = 1.8 ln(t + 1.5) + 0.4.

(a)

Describe the transformations that map the graph of y = ln t onto the graph of this model.

(3)

(b)

Find the height after 10 years.

(2)

(c)

Find the height predicted after 100 years, and explain why the model is unrealistic for large t.

(2)

(d)

Find the inverse function, expressing t in terms of H.

(3)

(e)

Use your inverse function to find when the sapling reaches a height of 6 m.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 translation 1.5 units to the left	✓ 1
A1 vertical stretch with scale factor 1.8	✓ 1
A1 translation 0.4 units up	✓ 1
Part (b)
M1 1.8 ln(11.5) + 0.4	✓ 1
A1 4.80 m	✓ 1
Part (c)
A1 8.72 m	✓ 1
R1 the model has no upper limit as t increases, while real trees reach a maximum height	✓ 1
Part (d)
M1 ln(t + 1.5) = (H − 0.4)/1.8	✓ 1
M1 t + 1.5 = e^((H − 0.4)/1.8)	✓ 1
A1 t = e^((H − 0.4)/1.8) − 1.5	✓ 1
Part (e)
M1 substitute H = 6	✓ 1
M1 e^(5.6/1.8) − 1.5	✓ 1
A1 t = 20.9 years	✓ 1

Common error: Subtracting 1.5 before applying the exponential when inverting; the 1.5 must be subtracted after undoing the logarithm.

+ Add to test✓ In your test278IBMI3H-2-04b-05Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

Noise at a harbour construction site is measured as a sound level L decibels, where L = 10 log₁₀(I ÷ 10⁻¹²) and I W/m² is the sound intensity. When machines operate together, their intensities add.

(a)

A piling rig produces a sound level of 105 dB at a monitoring point. Find the intensity.

(2)

(b)

Two identical rigs operate together. Show that the sound level increases by 3.0 dB.

(3)

(c)

Five machines each produce 92 dB at the monitoring point. Find the sound level when all five operate together.

(2)

(d)

The sound level at the point must not exceed 100 dB. Find the greatest number of these 92 dB machines that may operate together.

(2)

(e)

One machine produces 95 dB and another produces 91 dB at the point. Find the sound level when both operate together.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 10 log₁₀(I ÷ 10⁻¹²) = 105, so I = 10^10.5−12	✓ 1
A1 I = 3.16 × 10⁻² W/m²	✓ 1
Part (b)
M1 new intensity is 2I	✓ 1
M1 change in level = 10 log₁₀(2I ÷ 10⁻¹²) − 10 log₁₀(I ÷ 10⁻¹²) = 10 log₁₀ 2	✓ 1
A1 = 3.01 dB, which is 3.0 dB to two significant figures	✓ 1
Part (c)
M1 total intensity is 5 times one machine's intensity, so L = 92 + 10 log₁₀ 5	✓ 1
A1 99.0 dB	✓ 1
Part (d)
M1 92 + 10 log₁₀ n ≤ 100, so n ≤ 10^0.8 = 6.31	✓ 1
A1 6 machines	✓ 1
Part (e)
M1 intensities 10^−2.5 and 10^−2.9 W/m²	✓ 1
M1 adds the intensities (4.42 × 10⁻³) and finds 10 log₁₀((4.42 × 10⁻³) ÷ 10⁻¹²) using unrounded values	✓ 1
A1 96.5 dB	✓ 1

Common error: Adding decibel values to combine sources. Add the intensities, then convert back to decibels.

+ Add to test✓ In your test279IBMI3H-2-04b-07Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

The mass W grams and length L cm of five fish of one species are recorded.

L	10	15	20	30	40
W	11.2	37.0	88.5	295	707

The relationship is modelled by W = aL^b.

(a)

Calculate ln L and ln W for the first two fish, correct to 3 significant figures.

(2)

(b)

Use technology to find the regression line ln W = p ln L + q for all five fish.

(2)

(c)

Hence find the values of a and b.

(2)

(d)

Suggest why b is close to 3.

(1)

(e)

Use W = 0.0114L^2.99 to estimate the mass of a fish of length 55 cm.

(2)

(f)

A second species follows W = 0.0021L^3.4. Taking the first species to follow W = 0.011L³, find the length at which fish of the two species have equal mass.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 ln 10 = 2.30 and ln 11.2 = 2.42	✓ 1
A1 ln 15 = 2.71 and ln 37 = 3.61	✓ 1
Part (b)
A1 p = 2.99	✓ 1
A1 q = −4.48	✓ 1
Part (c)
A1 b = 2.99	✓ 1
A1 a = e^−4.477 = 0.0114	✓ 1
Part (d)
R1 for fish of similar shape, mass is proportional to volume, which is proportional to length cubed	✓ 1
Part (e)
M1 substitutes L = 55	✓ 1
A1 1820 g	✓ 1
Part (f)
M1 sets 0.0021L^3.4 = 0.011L³	✓ 1
M1 L^0.4 = 0.011 ÷ 0.0021 (= 5.238)	✓ 1
A1 L = 5.238^2.5 = 62.8 cm	✓ 1

Common error: Taking the regression gradient as a. On a log-log graph the gradient is the power b and the intercept is ln a.

+ Add to test✓ In your test280IBMI3H-2-04b-09Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard14 marks

A freight company is introducing electronic tags on its containers. The number of tagged containers, N, is modelled by N = 1800 ÷ (1 + ae^−bt), where t is the time in months since the start of the scheme. At t = 0 there are 150 tagged containers and at t = 3 there are 450.

(a)

Show that a = 11.

(2)

(b)

Show that b = 0.433 to three significant figures.

(3)

(c)

Find when 90% of the maximum number of containers are tagged.

(3)

(d)

The number of containers tagged each month grows fastest when N is half of 1800. Find the time at which this happens.

(2)

(e)

A second model uses exponential growth, N = 150e^kt, through the same two data points. Find k, then find the number of tagged containers it predicts at t = 10, and comment.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1800 ÷ (1 + a) = 150	✓ 1
A1 1 + a = 12, so a = 11	✓ 1
Part (b)
M1 1800 ÷ (1 + 11e^−3b) = 450	✓ 1
M1 e^−3b = 3 ÷ 11	✓ 1
A1 b = ln(11 ÷ 3) ÷ 3 = 0.433	✓ 1
Part (c)
M1 sets 1800 ÷ (1 + 11e^−0.4331t) = 1620	✓ 1
M1 e^−0.4331t = 1 ÷ 99	✓ 1
A1 t = 10.6 months	✓ 1
Part (d)
M1 sets 11e^−0.4331t = 1	✓ 1
A1 t = 5.54 months	✓ 1
Part (e)
M1 150e^3k = 450, so k = ln 3 ÷ 3 = 0.366	✓ 1
M1 N(10) = 150e^3.662 (using unrounded k)	✓ 1
A1 5840 containers	✓ 1
R1 this exceeds the 1800 maximum, so the exponential model is unrealistic for large t	✓ 1

Exam tip: When asked for a time at a fraction of the limit, rearrange to isolate the exponential term first, then take natural logs of both sides.

+ Add to test✓ In your test281IBMI3H-2-04b-12Logarithmic scales and linearising data (log-log and semi-log); transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

The depth d metres of a lake is measured at distances x metres from the shore along a straight line.

x	0	20	40	60	80	100
d	0	3.9	6.8	8.0	6.1	1.2

(a)

Find the cubic regression model d = ax³ + bx² + cx + e.

(2)

(b)

Use unrounded coefficients to estimate the depth at x = 45.

(2)

(c)

The R² values are 0.977 for the quadratic model and 0.999 for the cubic model. State which model fits better and why.

(2)

(d)

Use the cubic model to find the distances from the shore at which the depth is 5 m.

(3)

(e)

Use the unrounded cubic model to find the predicted depth at x = 100, and compare it with the measured value.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = −1.64 × 10⁻⁵ and b = −3.56 × 10⁻⁴	✓ 1
A1 c = 0.212 and e = −0.0198	✓ 1
Part (b)
M1 substitutes x = 45	✓ 1
A1 7.29 m	✓ 1
Part (c)
A1 the cubic model	✓ 1
R1 its R² is closer to 1, so it explains more of the variation in depth	✓ 1
Part (d)
M1 sets the cubic equal to 5 and solves with technology	✓ 1
A1 x = 26.3 m	✓ 1
A1 x = 86.5 m (the root near −134 is rejected as outside the lake)	✓ 1
Part (e)
M1 substitutes x = 100	✓ 1
A1 1.16 m	✓ 1
R1 the measured depth is 1.2 m, a difference of about 0.04 m, so the model is very close	✓ 1

Exam tip: When a model is evaluated by technology, store the full coefficients. Rounding to three significant figures can change a prediction in the second decimal place.

+ Add to test✓ In your test282IBMI3H-2-03r5-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

A port introduces an automated berthing system. The number of vessels, N, using the system t years after its introduction is modelled by N = 480 ÷ (1 + Ce^−kt), where C and k are positive constants. Initially 60 vessels use the system, and after 3 years 150 vessels use it.

(a)

Find the value of C.

(2)

(b)

Show that k = 0.386, correct to 3 significant figures.

(3)

(c)

Use k = 0.3858 to find the number of years until 400 vessels use the system. Give your answer to 3 significant figures.

(3)

(d)

Find the greatest rate at which the number of vessels using the system increases, in vessels per year. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 = 480 ÷ (1 + C)	✓ 1
A1 C = 7	✓ 1
Part (b)
M1 150 = 480 ÷ (1 + 7e^−3k)	✓ 1
M1 7e^−3k = 2.2, so k = ln(7 ÷ 2.2) ÷ 3	✓ 1
A1 k = 0.3858… = 0.386 (AG)	✓ 1
Part (c)
M1 400 = 480 ÷ (1 + 7e^−0.3858t)	✓ 1
M1 e^−0.3858t = 0.2 ÷ 7, so t = ln 35 ÷ 0.3858	✓ 1
A1 t = 9.22 years	✓ 1
Part (d)
M1 greatest rate occurs when N = 240 (half of the limiting value 480)	✓ 1
M1 uses a graphing calculator (or dN/dt = kN(480 − N)/480) at this point	✓ 1
A1 46.3 vessels per year	✓ 1

Exam tip: In a logistic model the growth is fastest when the quantity reaches half of its limiting value, so you can locate the point of greatest rate before using any calculus.

+ Add to test✓ In your test283IBMI3H-2-03r5-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

Advance bookings for a theatre run are modelled by P = 2400 ÷ (1 + Ce^−kt), where P is the number of tickets sold t days after booking opens, and C and k are positive constants. On day 0, 100 tickets have been sold. Sales are fastest on day 8.

(a)

Find the value of C.

(2)

(b)

Show that k = 0.392, correct to 3 significant figures.

(3)

(c)

Use k = 0.392 to find the number of tickets sold by day 5. Give your answer to the nearest whole number.

(2)

(d)

Find the day on which the number of tickets sold first exceeds 2000. Use k = 0.392 and give the time in days to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100 = 2400 ÷ (1 + C)	✓ 1
A1 C = 23	✓ 1
Part (b)
M1 sales are fastest when P = 1200, so 1 + 23e^−8k = 2	✓ 1
M1 e^8k = 23, so 8k = ln 23	✓ 1
A1 k = 0.3919… = 0.392 (AG)	✓ 1
Part (c)
M1 P = 2400 ÷ (1 + 23e^{−0.392 × 5})	✓ 1
A1 566	✓ 1
Part (d)
M1 2000 = 2400 ÷ (1 + 23e^−0.392t)	✓ 1
M1 e^−0.392t = 0.2 ÷ 23, so t = ln 115 ÷ 0.392	✓ 1
A1 t = 12.1 days	✓ 1

Common error: In a logistic model the maximum sales rate happens at half the limiting value, not at half the time or at the end of the run.

+ Add to test✓ In your test284IBMI3H-2-03r5-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

Lava from a vent cools in the surrounding air, which is at 25 °C. The temperature T °C of a lava sample is recorded t minutes after it leaves the vent.
t: 0, 20, 40, 60
T: 1000, 549.5, 307.1, 176.8
A model of the form T − 25 = Ae^−kt is proposed.

(a)

Show that a graph of ln(T − 25) against t should be a straight line, and state what its gradient represents.

(2)

(b)

Calculate ln(T − 25) for t = 20 and for t = 60, to 3 significant figures.

(2)

(c)

Use your graphing calculator to find the regression line ln(T − 25) = a + bt. Give a to 2 decimal places and b to 3 significant figures.

(2)

(d)

Using a = 6.88 and b = -0.0310, find the time for the sample to cool to 100 °C. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 takes logarithms: ln(T − 25) = ln A + ln(e^−kt)	✓ 1
A1 ln(T − 25) = ln A − kt, linear in t with gradient −k	✓ 1
Part (b)
A1 t = 20: 6.26	✓ 1
A1 t = 60: 5.02	✓ 1
Part (c)
A1 a = 6.88	✓ 1
A1 b = -0.0310	✓ 1
Part (d)
M1 ln(100 − 25) = ln 75 = 4.317…	✓ 1
M1 6.88 − 0.0310t = ln 75 (using the gradient from (c))	✓ 1
A1 t = 82.7 minutes	✓ 1

Exam tip: Subtract the surrounding temperature before taking logarithms; plotting ln T directly does not give a straight line for this model.

+ Add to test✓ In your test285IBMI3H-2-04r5-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The number of visitors, N, walking a volcano trail in a month is modelled by N = 1800 ÷ (1 + 14e^−0.6t), where t is the number of months after the trail opens.

(a)

Find the number of visitors in the first month, t = 0.

(2)

(b)

Write down the value that N approaches as t becomes very large.

(1)

(c)

Find the value of t for which N = 900. Give your answer to 3 significant figures.

(3)

(d)

Find the rate of change of N when t = 2. Give your answer to 3 significant figures.

(2)

(e)

Find the greatest rate of increase of N. (ECF from (c))

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 N = 1800 ÷ (1 + 14)	✓ 1
A1 120	✓ 1
Part (b)
A1 1800	✓ 1
Part (c)
M1 900 = 1800 ÷ (1 + 14e^−0.6t), so 14e^−0.6t = 1	✓ 1
M1 t = ln 14 ÷ 0.6	✓ 1
A1 t = 4.40	✓ 1
Part (d)
M1 uses a graphing calculator to evaluate dN/dt at t = 2	✓ 1
A1 167 visitors per month	✓ 1
Part (e)
M1 the greatest rate occurs when N = 900, that is at t = 4.40	✓ 1
M1 evaluates dN/dt at this t (or uses 0.6 × 1800 ÷ 4)	✓ 1
A1 270 visitors per month	✓ 1

Exam tip: The steepest point of a logistic curve is its midpoint; evaluate the gradient there to get the greatest rate of increase.

+ Add to test✓ In your test286IBMI3H-2-04r5-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

The daily fuel use of a cargo ship is modelled by F = a + bv³ tonnes per day, where v knots is its speed and a and b are constants. At 10 knots the ship uses 80 tonnes per day and at 14 knots it uses 114.88 tonnes per day. A speed of v knots covers 24v nautical miles per day.

(a)

Write down two equations for a and b.

(2)

(b)

Solve these equations to find a and b.

(3)

(c)

The fuel used per nautical mile is f = F ÷ (24v). Use your graphing calculator to find the speed that minimises f, to 3 significant figures. (ECF from (b))

(3)

(d)

Find the least fuel use per nautical mile, in tonnes, to 3 significant figures. (ECF from (c))

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a + 1000b = 80	✓ 1
A1 a + 2744b = 114.88	✓ 1
Part (b)
M1 subtracts or solves simultaneously: 1744b = 34.88	✓ 1
A1 b = 0.02	✓ 1
A1 a = 60	✓ 1
Part (c)
M1 f = (60 + 0.02v³) ÷ (24v)	✓ 1
M1 sketches or uses the minimum function on the graph of f	✓ 1
A1 v = 11.4 knots	✓ 1
Part (d)
M1 substitutes the speed from (c) into f	✓ 1
A1 0.328 tonnes per nautical mile	✓ 1

Exam tip: The best speed for fuel per distance is not the slowest speed: dividing by v creates a minimum where fixed and speed-dependent costs balance.

+ Add to test✓ In your test287IBMI3H-2-04r5-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

The depth of water, D metres, in a shipping channel t hours after midnight is modelled by D = a cos(b(t − 3.2)) + k, where b > 0 and the angle is in radians. High tide is 11.2 m at t = 3.2 and the next low tide is 3.8 m at t = 9.4.

(a)

Find the value of a and the value of k.

(2)

(b)

Find the value of b, correct to 3 significant figures.

(2)

(c)

A ship needs a depth of at least 6.0 m and reaches the channel at t = 8.0. Show that the ship cannot enter at that time. Use the unrounded value of b.

(2)

(d)

Find the earliest time after t = 8.0 at which the ship can enter. Use the unrounded value of b and give t to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = (11.2 − 3.8) ÷ 2 = 3.7	✓ 1
A1 k = (11.2 + 3.8) ÷ 2 = 7.5	✓ 1
Part (b)
M1 half period = 9.4 − 3.2 = 6.2, so period = 12.4 and b = 2π ÷ 12.4	✓ 1
A1 b = 0.507	✓ 1
Part (c)
M1 D = 3.7 cos(b × 4.8) + 7.5	✓ 1
A1 D = 4.69 m, which is less than 6.0 m	✓ 1
Part (d)
M1 3.7 cos(b(t − 3.2)) + 7.5 = 6, so cos(b(t − 3.2)) = −0.4054…	✓ 1
M1 first solution is b(t − 3.2) = 1.988… (t = 7.12…), which is before 8.0, so uses the next rising solution 2π − 1.988…	✓ 1
M1 t = 3.2 + 4.295… ÷ b	✓ 1
A1 t = 11.7 hours after midnight	✓ 1

Common error: Solving the cosine equation once gives the end of one window only; use the symmetry of the cosine curve to find the start of the next window.

+ Add to test✓ In your test288IBMI3H-2-04r6-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

Calculator

The number of passengers P per day using a new tram line, t months after opening, is modelled by P(t) = 24000 ÷ (1 + Ce^−kt), where C and k are positive constants. When the line opens, 3000 passengers per day use it. After 4 months, 9000 passengers per day use it.

(a)

Show that C = 7.

(2)

(b)

Find the value of k. Give your answer to 3 significant figures.

(3)

(c)

Use the unrounded value of k to find the predicted number of passengers per day after 8 months. Give your answer to the nearest 10 passengers.

(2)

(d)

The number of passengers is growing fastest when P = 12000. Find the value of t at which this occurs. Give your answer in months to 3 significant figures.

(3)

(e)

State what the model predicts for the number of passengers per day in the long term, and give one reason why this limit is realistic.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes t = 0 and P = 3000: 3000 = 24000 ÷ (1 + C)	✓ 1
A1 1 + C = 8, so C = 7 (AG)	✓ 1
Part (b)
M1 substitutes t = 4 and P = 9000: 9000 = 24000 ÷ (1 + 7e^−4k)	✓ 1
M1 rearranges to e^−4k = (24000/9000 − 1) ÷ 7 = 0.238…	✓ 1
A1 k = 0.359	✓ 1
Part (c)
M1 substitutes t = 8	✓ 1
A1 P = 17180	✓ 1
Part (d)
M1 sets 24000 ÷ (1 + 7e^−kt) = 12000	✓ 1
M1 e^−kt = 1/7	✓ 1
A1 t = ln 7 ÷ k = 5.42 months	✓ 1
Part (e)
R1 24000 passengers per day (the limit of P as t → ∞), because the line has a maximum capacity or a limited population to serve	✓ 1

Exam tip: In a logistic model L ÷ (1 + Ce^(−kt)), the numerator L is the limiting value and the steepest growth occurs when P is half of L.

+ Add to test✓ In your test289IBMI3H-2-04r6-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

Calculator

During a warm-up, a runner's core temperature T °C at t minutes is modelled by T = 37 − 5e^−0.2t, for t ≥ 0.

(a)

The graph of T is obtained from the graph of y = e^x by a sequence of transformations. Describe each transformation, in order.

(4)

(b)

Write down the equation of the horizontal asymptote of the graph of T.

(1)

(c)

Find the temperature at the start of the warm-up.

(1)

(d)

Find the time at which the temperature reaches 36 °C. Give your answer in minutes to 3 significant figures.

(3)

(e)

The rate of change of temperature is given by dT/dt = e^−0.2t. Find the time at which the temperature is rising at 0.3 °C per minute. Give your answer in minutes to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 horizontal stretch with scale factor 5	✓ 1
A1 reflection in the y-axis	✓ 1
A1 vertical stretch with scale factor 5 and a reflection in the x-axis	✓ 1
A1 translation 37 units up	✓ 1
Part (b)
A1 T = 37	✓ 1
Part (c)
A1 T(0) = 37 − 5 = 32 °C	✓ 1
Part (d)
M1 sets 37 − 5e^−0.2t = 36	✓ 1
A1 e^−0.2t = 0.2	✓ 1
A1 t = 5 ln 5 = 8.05 minutes	✓ 1
Part (e)
M1 sets e^−0.2t = 0.3	✓ 1
M1 takes natural logarithms: −0.2t = ln 0.3	✓ 1
A1 t = 6.02 minutes	✓ 1

Exam tip: Read transformations from the inside of the function outwards for horizontal changes, then follow the order of operations for the vertical changes.

+ Add to test✓ In your test290IBMI3H-2-04r6-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

Calculator

The distance modulus of a star relates its apparent magnitude m, its absolute magnitude M and its distance d parsecs: m − M = 5 log₁₀ d − 5.

(a)

A star has m = 11.2 and M = −3.4. Find its distance. Give your answer to the nearest parsec.

(3)

(b)

A second star has M = 1.2 and appears with m = 9.5. Find its distance. Give your answer to the nearest parsec.

(3)

(c)

A star's distance is multiplied by a factor k, with M unchanged. Show that its apparent magnitude increases by 5 log₁₀ k.

(2)

(d)

A star is at a distance of 125 parsecs. Find the distance at which the same star would appear 2.0 magnitudes fainter. Give your answer to the nearest parsec.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes: 11.2 − (−3.4) = 5 log₁₀ d − 5	✓ 1
A1 log₁₀ d = 3.92	✓ 1
A1 d = 8318 parsecs	✓ 1
Part (b)
M1 substitutes: 9.5 − 1.2 = 5 log₁₀ d − 5	✓ 1
A1 log₁₀ d = 2.66	✓ 1
A1 d = 457 parsecs	✓ 1
Part (c)
M1 new m − M = 5 log₁₀(kd) − 5 and subtracts the original equation	✓ 1
R1 uses log₁₀(kd) = log₁₀ k + log₁₀ d, so the increase in m is 5 log₁₀ k (AG)	✓ 1
Part (d)
M1 sets 5 log₁₀ k = 2.0	✓ 1
A1 k = 10^0.4 = 2.51…	✓ 1
A1 distance = 125 × 10^0.4 = 314 parsecs	✓ 1

Common error: A change of 5 magnitudes corresponds to a factor of 10 in distance (a factor of 100 in flux): the relationship is logarithmic, not linear.

+ Add to test✓ In your test291IBMI3H-2-03r10-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The number of members N of a library's online reading club, t months after it was launched, is modelled by N = 1200 / (1 + 7e^−0.4t), t ≥ 0.

(a)

Find the number of members when the club was launched.

(2)

(b)

Write down the limiting value of N as t → ∞ and give a reason for your answer.

(2)

(c)

Find the time at which the club has 900 members. Give your answer to 3 significant figures.

(3)

(d)

It can be shown that the rate of growth of the membership is dN/dt = 0.4N(1 − N/1200). Find the greatest rate of growth, in members per month.

(2)

(e)

Find the time at which the membership grows at this greatest rate. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
N(0) = 1200 / (1 + 7)	✓ 1
N = 150	✓ 1
Part (b)
limiting value = 1200	✓ 1
e^−0.4t → 0 as t → ∞, so the denominator tends to 1	✓ 1
Part (c)
900 = 1200 / (1 + 7e^−0.4t), so 1 + 7e^−0.4t = 4/3	✓ 1
e^−0.4t = 1/21	✓ 1
t = ln 21 / 0.4 = 7.61 months	✓ 1
Part (d)
rate is greatest when N = 600 (half of the limiting value) (or by sketch/GDC maximum)	✓ 1
0.4 × 600 × (1 − 0.5) = 120 members per month	✓ 1
Part (e)
600 = 1200 / (1 + 7e^−0.4t), so 7e^−0.4t = 1	✓ 1
t = ln 7 / 0.4 = 4.86 months	✓ 1

Exam tip: In a logistic model the growth rate is greatest when the quantity is half of its limiting value.

+ Add to test✓ In your test292IBMI3H-2-05r16-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard15 marks

During an outbreak in a hospital, the number of patients, N, who have been infected t days after the first count is modelled by N = 240 ÷ (1 + Ce^−kt), where C and k are positive constants. The hospital has 240 patients in total. At t = 0, 12 patients are infected, and at t = 4, 60 patients are infected.

(a)

Show that C = 19.

(2)

(b)

Find the value of k correct to 3 significant figures.

(4)

(c)

Use the unrounded value of k to find the time at which 180 patients are infected. Give your answer in days to 3 significant figures.

(3)

(d)

Sketch the graph of N against t for t ≥ 0. Describe your sketch in words, stating any intercept, asymptote and the shape.

(3)

(e)

Find the time, in days to 3 significant figures, at which the number of infections is increasing most rapidly. Use the unrounded value of k.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0 and N = 12: 12 = 240 ÷ (1 + C)	✓ 1
A1 1 + C = 20, so C = 19	✓ 1
Part (b)
M1 60 = 240 ÷ (1 + 19e^−4k) so 1 + 19e^−4k = 4	✓ 1
M1 e^−4k = 3/19	✓ 1
M1 take natural logs: −4k = ln(3/19) (accept GDC solve)	✓ 1
A1 k = 0.461	✓ 1
Part (c)
M1 180 = 240 ÷ (1 + 19e^−kt) so 19e^−kt = 1/3	✓ 1
M1 e^−kt = 1/57 so t = ln 57 ÷ k	✓ 1
A1 t = 8.76 so 8.76 days	✓ 1
Part (d)
A1 starts at (0, 12) on the vertical axis	✓ 1
A1 increasing S-shaped curve (steepening then flattening) approaching the horizontal asymptote N = 240	✓ 1
A1 point of inflection (steepest point) marked at N = 120	✓ 1
Part (e)
R1 fastest growth occurs at the point of inflection, where N = 240 ÷ 2 = 120 (accept GDC maximum gradient)	✓ 1
M1 1 + 19e^−kt = 2, so e^−kt = 1/19	✓ 1
A1 t = ln 19 ÷ k = 6.381… so 6.38 days	✓ 1

Exam tip: For a logistic curve the fastest growth is at half the carrying capacity; using this saves a derivative.

+ Add to test✓ In your test293IBMI3H-2-05r16-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

Two strains of bacteria, A and B, are grown on a hospital culture plate. The numbers of bacteria after t hours are N_A and N_B. When log₁₀ N is plotted against t, both graphs are straight lines:
log₁₀ N_A = 2.90 + 0.020t and log₁₀ N_B = 2.40 + 0.035t.

(a)

Find the initial number of bacteria of each strain, to the nearest whole number.

(2)

(b)

Show that strain A grows by about 4.7% each hour.

(2)

(c)

Find the time, in hours to 3 significant figures, when the two strains are equal in number.

(3)

(d)

Find the number of bacteria of each strain at this time. Use t = 100/3 and give your answer to 3 significant figures.

(2)

(e)

State, with a reason, which strain is more numerous at t = 40 hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 N_A = 10^2.90 = 794	✓ 1
A1 N_B = 10^2.40 = 251	✓ 1
Part (b)
M1 N_A = 10^2.90 × (10^0.020)^t, so hourly multiplier is 10^0.020	✓ 1
A1 10^0.020 = 1.047… so an increase of 4.7%	✓ 1
Part (c)
M1 equate 2.90 + 0.020t = 2.40 + 0.035t	✓ 1
M1 0.50 = 0.015t	✓ 1
A1 t = 33.3 hours (100/3)	✓ 1
Part (d)
M1 log₁₀ N = 2.90 + 0.020 × 100/3 = 3.5666…	✓ 1
A1 N = 3687… so 3690	✓ 1
Part (e)
M1 compute log₁₀ N_A = 3.70 and log₁₀ N_B = 3.80 (or compare gradients)	✓ 1
R1 B is more numerous, because its line has a greater gradient and the lines cross at 33.3 hours	✓ 1

Exam tip: On a semi-log graph, equal values of log N mean equal numbers, so equating the two linear equations finds the crossing time directly.

+ Add to test✓ In your test294IBMI3H-2-03r17-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

Algae spreads across the surface of a reservoir. The area covered, $P$ m², is modelled by $$P(t)=\frac{800}{1+Ae^{-rt}},\quad t\ge0,$$ where $t$ is the time in days since the algae was first detected, and $A$, $r$ are positive constants. Initially 40 m² is covered. The area is growing fastest when $t=6$.

(a)

Find the value of $A$.

(2)

(b)

Given that the growth is fastest when the covered area is half of the maximum area, find the value of $r$. Give your answer to 3 significant figures.

(3)

(c)

Find the area covered when $t=10$. Use the unrounded value of $r$ and give your answer to 3 significant figures.

(2)

(d)

Use your GDC to find the maximum rate of growth of the area covered, giving your answer to 3 significant figures and including units.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $40=\frac{800}{1+A}$	✓ 1
A1 $A=19$	✓ 1
Part (b)
R1 the maximum area is 800, so the fastest growth is at $P=400$	✓ 1
M1 $400=\frac{800}{1+19e^{-6r}}$, so $e^{-6r}=\frac{1}{19}$	✓ 1
A1 $r=0.491$	✓ 1
Part (c)
M1 $P(10)=\frac{800}{1+19e^{-10r}}$	✓ 1
A1 $P(10)=701$ m²	✓ 1
Part (d)
M1 uses a GDC to find the maximum of $P'(t)$ (or the gradient of $P$ at $t=6$)	✓ 1
A1 $98.1$	✓ 1
A1 m² per day	✓ 1

Exam tip: For any logistic curve the steepest point is at half the carrying capacity, which gives a second equation without needing a second data point.

+ Add to test✓ In your test295IBMI3H-2-03r17-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

Pharmacologists model the rate $R$ (mg per hour) at which a mammal clears a drug from its blood by $R=am^{b}$, where $m$ is its body mass in kg and $a$, $b$ are constants. A 4 kg animal clears the drug at 3.0 mg per hour and a 64 kg animal at 24.0 mg per hour.

(a)

Show that $b=0.75$.

(3)

(b)

Find the value of $a$ to 3 significant figures.

(2)

(c)

Estimate the clearance rate for a 70 kg animal. Use the unrounded value of $a$ and give your answer to 3 significant figures.

(2)

(d)

Show that a plot of $\ln R$ against $\ln m$ is a straight line, stating its gradient and its vertical intercept.

(2)

(e)

Find the body mass of an animal that clears the drug at 30 mg per hour. Use the unrounded value of $a$ and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 forms $\frac{24}{3}=\left(\frac{64}{4}\right)^b$, i.e. $8=16^b$	✓ 1
M1 takes logarithms: $b=\frac{\ln8}{\ln16}$	✓ 1
A1 $b=\frac{3}{4}=0.75$ (AG)	✓ 1
Part (b)
M1 $3.0=a\times4^{0.75}$	✓ 1
A1 $a=1.06$	✓ 1
Part (c)
M1 $R=\frac{3}{4^{0.75}}\times70^{0.75}$	✓ 1
A1 $25.7$ mg per hour	✓ 1
Part (d)
M1 $\ln R=\ln a+b\ln m$	✓ 1
A1 gradient $0.75$ and vertical intercept $\ln a=0.0589$	✓ 1
Part (e)
M1 $30=a\,m^{0.75}$, so $m^{0.75}=\frac{30}{a}$	✓ 1
M1 $m=\left(\frac{30}{a}\right)^{\frac{1}{0.75}}$	✓ 1
A1 $m=86.2$ kg	✓ 1

Exam tip: With two data points and a power model, divide the two equations to eliminate $a$ before taking logarithms.

+ Add to test✓ In your test296IBMI3H-2-03r19-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The number of registered members of a new library, N, after t months is modelled by N = ⁴⁰⁰⁰⁄_{1 + 19e^−0.4t}, t ≥ 0.

(a)

Find the number of members when the library opens.

(2)

(b)

Find the number of members after 6 months, giving your answer to 3 significant figures.

(2)

(c)

Find the time taken for the library to reach 2000 members, giving your answer to 3 significant figures.

(3)

(d)

Explain why the model predicts that the number of members never exceeds 4000.

(2)

(e)

Use technology to find the greatest rate at which the number of members increases, in members per month.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0: 4000 ÷ (1 + 19)	✓ 1
A1 200 members	✓ 1
Part (b)
M1 N(6) = 4000 ÷ (1 + 19e^−2.4)	✓ 1
A1 1469, so 1470 members	✓ 1
Part (c)
M1 1 + 19e^−0.4t = 2	✓ 1
M1 e^−0.4t = ¹⁄₁₉, then take natural logarithms	✓ 1
A1 t = 7.36 months	✓ 1
Part (d)
R1 as t → ∞, e^−0.4t → 0 so N → 4000 from below	✓ 1
R1 the denominator is always greater than 1, so N < 4000	✓ 1
Part (e)
M1 attempt to find the maximum gradient of the graph (or use of N = 2000 at the steepest point)	✓ 1
A1 400 members per month	✓ 1

Exam tip: On a logistic curve the steepest point is where N is half of the limiting value, which matches the time found in (c).

+ Add to test✓ In your test297IBMI3H-2-03r19-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The tip of a wind turbine blade has height h metres above the ground at time t seconds, modelled by h(t) = a cos(bt) + d. The hub is 80 m above the ground, the blade is 45 m long, the blade makes 15 complete rotations per minute and the tip is at its highest point when t = 0.
Give answers to 3 significant figures and use unrounded values in later working.

(a)

Write down the values of a, b and d.

(3)

(b)

Find the first time at which the tip is 100 m above the ground.

(3)

(c)

Describe the transformations that map the graph of y = cos t onto the graph of y = h(t).

(3)

(d)

In each rotation, find for how long the tip is higher than 100 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 a = 45	✓ 1
A1 b = ^2π⁄₄ = ^π⁄₂ (period 4 s)	✓ 1
A1 d = 80	✓ 1
Part (b)
M1 80 + 45 cos(πt/2) = 100	✓ 1
M1 cos(πt/2) = ²⁰⁄₄₅ and take the inverse cosine	✓ 1
A1 t = 0.707 s	✓ 1
Part (c)
A1 vertical stretch, scale factor 45	✓ 1
A1 horizontal stretch, scale factor ²⁄_π	✓ 1
A1 translation 80 units up	✓ 1
Part (d)
M1 by symmetry the tip is above 100 m from −t₁ to t₁, where t₁ is the answer to (b)	✓ 1
A1 1.41 s	✓ 1

Exam tip: In a cosine model starting at a maximum, the time spent above a given height in one period is symmetric about the maximum.

+ Add to test✓ In your test298IBMI3H-2-03r19-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

The number of members of a swimming club, M, after t months is modelled by M = ⁶⁰⁰⁄_{1 + ae^−kt}, where a and k are positive constants. At t = 0 there are 50 members and after 4 months there are 200 members.

(a)

Show that a = 11.

(2)

(b)

Find the value of k, giving your answer to 3 significant figures.

(3)

(c)

Using the unrounded value of k, find the time at which the club has 500 members. Give your answer to 3 significant figures.

(3)

(d)

State the largest number of members predicted by the model, and give a reason why this limit is realistic for a club.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 600 ÷ (1 + a) = 50	✓ 1
A1 1 + a = 12, so a = 11	✓ 1
Part (b)
M1 600 ÷ (1 + 11e^−4k) = 200, so e^−4k = ²⁄₁₁	✓ 1
M1 take natural logarithms: 4k = ln 5.5	✓ 1
A1 k = 0.426	✓ 1
Part (c)
M1 1 + 11e^−kt = 1.2	✓ 1
M1 e^−kt = ¹⁄₅₅ and take logarithms	✓ 1
A1 t = 9.40 months	✓ 1
Part (d)
R1 600; any sensible reason such as limited pool space, lane capacity or local population	✓ 1

Exam tip: To solve a logistic equation, rearrange so the exponential term is isolated before taking natural logarithms.

+ Add to test✓ In your test299IBMI3H-2-04r19-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The voltage V volts across a battery in a robot, t minutes after it starts to discharge, is modelled by V = 9 + Ae^−kt, where A and k are positive constants. Initially V = 12.6, and after 5 minutes V = 10.8.

(a)

Find the value of A.

(2)

(b)

Find the value of k, giving your answer to 3 significant figures.

(3)

(c)

Using the unrounded value of k, find the time at which V falls to 9.5 volts. Give your answer to 3 significant figures.

(3)

(d)

A graph of ln(V − 9) against t is plotted. State the gradient of the line and its vertical intercept, to 3 significant figures.

(2)

(e)

State the lowest voltage the model predicts for the battery.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12.6 = 9 + A	✓ 1
A1 A = 3.6	✓ 1
Part (b)
M1 10.8 = 9 + 3.6e^−5k, so e^−5k = 0.5	✓ 1
M1 −5k = ln 0.5	✓ 1
A1 k = 0.139	✓ 1
Part (c)
M1 9 + 3.6e^−kt = 9.5	✓ 1
M1 e^−kt = ^0.5⁄_3.6 and take logarithms	✓ 1
A1 t = 14.2 minutes	✓ 1
Part (d)
A1 gradient = −k = −0.139	✓ 1
A1 intercept = ln 3.6 = 1.28	✓ 1
Part (e)
A1 9 volts (the horizontal asymptote)	✓ 1

Exam tip: Subtract the asymptote before taking logarithms: it is ln(V − 9), not ln V, that is linear against t.

+ Add to test✓ In your test300IBMI3H-2-05r21-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The number of active users N of a cycle-sharing app is modelled by
N(t) = 6000 / (1 + 39e^−0.45t), t ≥ 0,
where t is the time in months since the app was launched.

(a)

Find the number of users at launch.

(1)

(b)

State the limiting value of N as t → ∞.

(1)

(c)

Find the time at which the app has 3000 users, to 3 significant figures.

(3)

(d)

Using your graphic display calculator, find the maximum rate of increase of N and the value of t at which it occurs. Give t to 3 significant figures and the rate to the nearest whole number.

(3)

(e)

Find how long it takes for the number of users to reach 98% of the limiting value. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 N(0) = 6000/40 = 150	✓ 1
Part (b)
A1 6000 users	✓ 1
Part (c)
M1 3000 = 6000/(1 + 39e^−0.45t)	✓ 1
A1 e^−0.45t = 1/39	✓ 1
A1 t = ln 39 / 0.45 = 8.14 months	✓ 1
Part (d)
M1 attempts to find the maximum of N′(t) (graph or numerical derivative)	✓ 1
A1 t = 8.14 months	✓ 1
A1 maximum rate = 675 users per month	✓ 1
Part (e)
M1 5880 = 6000/(1 + 39e^−0.45t)	✓ 1
A1 e^−0.45t = (6000/5880 − 1)/39 = 5.23×10⁻⁴ (or 1/1911)	✓ 1
A1 t = 16.8 months	✓ 1

Exam tip: For a logistic curve the steepest point is at half the limiting value; checking your calculator result against this gives a quick consistency test.

+ Add to test✓ In your test301IBMI3H-2-05r21-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

A server is switched off and its temperature T °C is measured after t minutes. The temperature is modelled by T = A + Be^−kt, where A, B and k are positive constants.
Readings: T = 70 at t = 0; T = 50 at t = 5; T = 38 at t = 10.

(a)

Explain why (50 − A)/(70 − A) = (38 − A)/(50 − A), and use this to find A.

(4)

(b)

Find the value of B and the value of k, to 3 significant figures.

(3)

(c)

Find the time taken for the temperature to fall to 25 °C. Give your answer to 3 significant figures, using the unrounded value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 each side equals e^−5k, because T − A = Be^−kt is multiplied by e^−5k every 5 minutes	✓ 1
M1 cross-multiplies: (50 − A)² = (70 − A)(38 − A)	✓ 1
M1 expands and simplifies to a linear equation (2500 − 100A = 2660 − 108A)	✓ 1
A1 A = 20	✓ 1
Part (b)
A1 B = 50	✓ 1
M1 e^−5k = 30/50 = 0.6	✓ 1
A1 k = 0.102	✓ 1
Part (c)
M1 25 = 20 + 50e^−kt, so e^−kt = 0.1	✓ 1
A1 t = ln 10 / k = 22.5 minutes	✓ 1

Exam tip: In a model of the form A + Be⁻kt, equal time steps multiply the difference from A by the same factor; this is a quick way to find A from three readings.

+ Add to test✓ In your test302IBMI3H-2-05r26-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

The cumulative number of units, S, of a new product sold t months after launch is modelled by S = 5000 / (1 + Ce^−kt), where C and k are positive constants. After launch (t = 0), 500 units had been sold, and after 3 months 1500 units had been sold.

(a)

Show that C = 9.

(2)

(b)

Find the value of k, to 3 significant figures.

(3)

(c)

State what the value 5000 represents in this context.

(1)

(d)

Find the time at which 90% of this maximum has been sold. Use the unrounded value of k and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 500 = 5000/(1 + C)	✓ 1
A1 1 + C = 10, so C = 9 (AG)	✓ 1
Part (b)
M1 1500 = 5000/(1 + 9e^−3k), so 1 + 9e^−3k = 10/3	✓ 1
M1 e^−3k = 7/27 and take logarithms: 3k = ln(27/7)	✓ 1
A1 k = 0.450	✓ 1
Part (c)
A1 the maximum (saturation) number of units that will be sold in total	✓ 1
Part (d)
M1 4500 = 5000/(1 + 9e^−kt), so e^−kt = 1/81	✓ 1
M1 t = ln 81 / k	✓ 1
A1 t = 9.77 months	✓ 1

Exam tip: Keep k unrounded (store it on your calculator) so that the later answer is not affected by rounding.

+ Add to test✓ In your test303IBMI3H-2-05r26-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The concentration C mg per litre of a drug in a patient's blood was measured t hours after the drug was given.
t: 1, 2, 3, 4, 5, 6
C: 46.2, 34.1, 25.0, 18.7, 13.6, 10.2
It is suggested that C = ae^−kt, where a and k are positive constants.

(a)

Show that a graph of ln C against t would be a straight line if this model is correct, and state the gradient and intercept in terms of a and k.

(2)

(b)

Use technology to find the regression line of ln C on t, giving the gradient and intercept to 3 significant figures.

(2)

(c)

Hence find the values of a and k, to 3 significant figures.

(2)

(d)

Find the half-life of the drug, to 3 significant figures.

(2)

(e)

The drug is ineffective once the concentration is below 2 mg per litre. Use unrounded values from the regression to find when this happens, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ln C = ln a − kt	✓ 1
A1 gradient −k, intercept ln a	✓ 1
Part (b)
A1 gradient −0.303	✓ 1
A1 intercept 4.13	✓ 1
Part (c)
A1 k = 0.303	✓ 1
A1 a = e^4.1338… = 62.4	✓ 1
Part (d)
M1 e^−kT = 1/2, so T = ln 2 / k	✓ 1
A1 2.29 hours	✓ 1
Part (e)
M1 62.41…e^{−0.30288…t} = 2	✓ 1
M1 t = (4.1338… − ln 2)/0.30288…	✓ 1
A1 11.4 hours	✓ 1

Common error: Do not take the logarithm of the time values; only C is transformed, because the relationship is exponential in t.

+ Add to test✓ In your test304IBMI3H-2-05r26-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard13 marks

In a town of 800 residents, the number of people P who have caught an infection t days after the first diagnosis is modelled by P = 800 / (1 + Ce^−kt), where C and k are positive constants. A student plots Y = ln(800/P − 1) against t and obtains a straight line passing through the points (2, 1.9) and (8, −0.5).

(a)

Show that Y = ln C − kt.

(3)

(b)

Find the gradient of the line and hence the value of k.

(2)

(c)

Find the value of C, to 3 significant figures.

(3)

(d)

Use the unrounded value of C to estimate the number of infected people after 10 days, to the nearest whole number.

(2)

(e)

Find the time at which 700 people are infected, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 800/P = 1 + Ce^−kt, so 800/P − 1 = Ce^−kt	✓ 1
M1 take natural logarithms of both sides	✓ 1
A1 Y = ln C + ln e^−kt = ln C − kt (AG)	✓ 1
Part (b)
M1 gradient = (−0.5 − 1.9)/(8 − 2)	✓ 1
A1 gradient −0.4, so k = 0.4	✓ 1
Part (c)
M1 use a point, e.g. 1.9 = ln C − 0.4 × 2	✓ 1
A1 ln C = 2.7	✓ 1
A1 C = e^2.7 = 14.9	✓ 1
Part (d)
M1 P = 800/(1 + e^2.7e⁻⁴) = 800/(1 + e^−1.3)	✓ 1
A1 629 people (628.7…)	✓ 1
Part (e)
M1 Y = ln(800/700 − 1) = ln(1/7) = −1.946…	✓ 1
M1 −1.946… = 2.7 − 0.4t	✓ 1
A1 t = 11.6 days	✓ 1

Exam tip: In a linearisation question, name the vertical and horizontal variables first; the gradient and intercept then give the unknown constants directly.

+ Add to test✓ In your test305IBMI3H-2-04r27-03Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

A fish population in a marine reserve is modelled by P(t) = K/(1 + A e^−rt), where P is the number of fish and t is the time in years since the reserve opened. The population cannot exceed 1200 fish, so K = 1200. Initially there are 150 fish, and after 3 years there are 400 fish.

(a)

Find the value of A.

(2)

(b)

Find the value of r, to 3 significant figures.

(3)

(c)

Find the time at which the population reaches 1000 fish. Use the unrounded value of r and give your answer to 3 significant figures.

(3)

(d)

The population grows at the rate dP/dt = rP(1 − P/K) fish per year. The growth rate is greatest when P = 600. Find the time at which P = 600, and the greatest growth rate to 3 significant figures. Use the unrounded value of r.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 150 = 1200/(1 + A)	✓ 1
A1 A = 7	✓ 1
Part (b)
M1 400 = 1200/(1 + 7e^−3r)	✓ 1
M1 e^−3r = 2/7	✓ 1
A1 r = 0.418	✓ 1
Part (c)
M1 1000 = 1200/(1 + 7e^−rt)	✓ 1
A1 e^−rt = 1/35	✓ 1
A1 t = 8.51 years	✓ 1
Part (d)
M1 600 = 1200/(1 + 7e^−rt), so e^−rt = 1/7	✓ 1
A1 t = 4.66 years	✓ 1
A1 greatest growth rate 125 fish per year	✓ 1

Exam tip: Substitute the initial condition first: at t = 0 the exponential equals 1, so A = K/P(0) − 1.

+ Add to test✓ In your test306IBMI3H-2-04r27-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard13 marks

The natural frequency f (Hz) of a steel beam of length L (m) is measured.

L (m): 1.0, 1.5, 2.0, 3.0, 4.0
f (Hz): 118, 66.0, 41.9, 23.5, 15.2

An engineer models the data by f = aL^b. In this question use your calculator's unrounded regression values in later parts.

(a)

Calculate ln L and ln f for each beam, to 3 significant figures.

(2)

(b)

Find the equation of the regression line of ln f on ln L, with coefficients to 3 significant figures.

(2)

(c)

Show that this gives a model of the form f = aL^b, and state the values of a and b to 3 significant figures.

(3)

(d)

Use the model to predict the natural frequency of a beam of length 3.5 m.

(2)

(e)

The correlation coefficient for ln f and ln L is r = −0.9999. Comment on what this indicates about the model.

(2)

(f)

A beam has a natural frequency of 30 Hz. Find its length, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to take natural logarithms of the values	✓ 1
A1 ln L: 0, 0.405, 0.693, 1.10, 1.39; ln f: 4.77, 4.19, 3.74, 3.16, 2.72	✓ 1
Part (b)
M1 use of linear regression on the values of ln L and ln f	✓ 1
A1 ln f = −1.48 ln L + 4.78	✓ 1
Part (c)
M1 ln f = ln a + b ln L	✓ 1
A1 b = −1.48	✓ 1
A1 a = e^4.775... = 119	✓ 1
Part (d)
M1 substitute L = 3.5 into f = aL^b	✓ 1
A1 18.6 Hz	✓ 1
Part (e)
R1 r is very close to −1, so there is a very strong negative linear relationship between ln f and ln L	✓ 1
R1 so the power model fits the data very well for lengths between 1.0 m and 4.0 m	✓ 1
Part (f)
M1 30 = aL^b solved for L	✓ 1
A1 2.53 m	✓ 1

Exam tip: Linearise with logarithms: a straight line on a log-log plot signals a power model, and the gradient of that line is the power.

+ Add to test✓ In your test307IBMI3H-2-04r27-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

A marine biologist models the excess nutrient concentration C (mg/L above a baseline) at a distance x km from a river mouth by the graph obtained from y = 2^x by these transformations, applied in this order: a reflection in the y-axis, a vertical stretch with scale factor 4, then a translation of 3 units down. The resulting function is g(x).

(a)

Write down g(x).

(3)

(b)

Show that the graph of y = g(x) can also be obtained from y = 2^x by a reflection in the y-axis, followed by a horizontal translation, followed by a vertical translation. State both translations.

(3)

(c)

Write down the equation of the horizontal asymptote and the y-intercept of the graph of y = g(x).

(2)

(d)

Find the set of values of x for which g(x) > 1.

(2)

(e)

Find the x-coordinate at which the concentration is zero, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 reflection gives 2^−x	✓ 1
A1 vertical stretch gives 4 × 2^−x	✓ 1
A1 g(x) = 4 × 2^−x − 3	✓ 1
Part (b)
M1 4 × 2^−x = 2² × 2^−x	✓ 1
A1 = 2^2−x = 2^−(x−2)	✓ 1
R1 so the stretch is replaced by a translation of 2 units to the right, followed by a translation of 3 units down	✓ 1
Part (c)
A1 y = −3	✓ 1
A1 (0, 1)	✓ 1
Part (d)
M1 g(0) = 1 and g is decreasing	✓ 1
A1 x < 0	✓ 1
Part (e)
M1 4 × 2^−x = 3, so 2^−x = 0.75	✓ 1
A1 x = 0.415	✓ 1

Common error: The order of transformations matters when a stretch and a translation are combined: stretching after translating gives a different graph.

+ Add to test✓ In your test308IBMI3H-2-05r27-11Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The number N of delivery robots in a swarm running a new software version, t days after release, is modelled by N = K / (1 + Ae^−rt), where K, A and r are positive constants. The swarm has 200 robots in total, so K = 200. When t = 0, N = 5, and when t = 6, N = 50.

(a)

Show that A = 39.

(1)

(b)

Find the value of r, correct to 4 significant figures.

(3)

(c)

Use the unrounded value of r to find the time at which 190 robots are running the new version. Give your answer to 3 significant figures.

(3)

(d)

Show that the time taken for N to rise from 10 % of K to 90 % of K is (ln 81)/r, and evaluate this time to 3 significant figures using the unrounded value of r.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 200/(1 + A) = 5, so 1 + A = 40 and A = 39	✓ 1
Part (b)
M1 200/(1 + 39e^−6r) = 50, so e^−6r = 3/39 = 1/13	✓ 1
M1 −6r = ln(1/13)	✓ 1
A1 r = 0.4275	✓ 1
Part (c)
M1 200/(1 + 39e^−rt) = 190, so e^−rt = (200/190 − 1)/39 = 1/741	✓ 1
M1 takes natural logarithms: t = ln(741)/r	✓ 1
A1 t = 15.5 days	✓ 1
Part (d)
M1 N = 20 gives e^−rt₁ = 9/39; N = 180 gives e^−rt₂ = 1/(9 × 39)	✓ 1
M1 e^{−r(t₂ − t₁)} = (1/351) ÷ (9/39) = 1/81	✓ 1
A1 t₂ − t₁ = (ln 81)/r	✓ 1
A1 10.3 days	✓ 1

Exam tip: In a "show that" question, keep every step of the algebra visible; the final line must match the result given in the question exactly.

+ Add to test✓ In your test309IBMI3H-2-05r27-14Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard13 marks

The monthly catch N tonnes of a fishing fleet, t years after 2020, is modelled by the exponential function N = ab^t, where a and b are constants. The catch was 1800 tonnes when t = 2 and 1200 tonnes when t = 5.

(a)

Find the value of b, correct to 4 significant figures.

(3)

(b)

Find the value of a, correct to 4 significant figures. Use the unrounded value of b.

(2)

(c)

Find the value of t at which the model predicts the monthly catch falls to 500 tonnes. Use the unrounded values of a and b and give your answer to 3 significant figures.

(3)

(d)

An alternative linear model N = mt + c passes through the same two data points. Find m and c.

(2)

(e)

For 2 ≤ t ≤ 5 the linear model gives larger values than the exponential model. Find the greatest difference between the two models in this interval. Use the unrounded values of a and b and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ab⁵ ÷ ab²: b³ = 1200/1800 = 2/3	✓ 1
M1 b = (2/3)^1/3	✓ 1
A1 b = 0.8736	✓ 1
Part (b)
M1 a = 1800/b²	✓ 1
A1 a = 2359	✓ 1
Part (c)
M1 ab^t = 500	✓ 1
M1 t = ln(500/a)/ln b	✓ 1
A1 t = 11.5	✓ 1
Part (d)
M1 m = (1200 − 1800)/(5 − 2)	✓ 1
A1 m = −200 and c = 2200	✓ 1
Part (e)
M1 forms the difference (2200 − 200t) − ab^t and finds its maximum (GDC or setting the derivative to zero)	✓ 1
A1 maximum occurs at t = 3.45	✓ 1
A1 greatest difference = 30.3 tonnes	✓ 1

Exam tip: Keep unrounded values in your calculator memory for later parts, then round only the final answer.

+ Add to test✓ In your test310IBMI3H-2-06r31-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

After an oral dose, the concentration of a drug in the blood is modelled by C(t) = 40te^−0.5t mg dm⁻³, where t ≥ 0 is the time in hours after the dose.

(a)

Show that the maximum concentration occurs at t = 2.

(4)

(b)

The drug is effective while the concentration is at least 10 mg dm⁻³. Find, to 3 significant figures, the length of time for which the drug is effective.

(4)

(c)

Find the maximum concentration, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule: C′(t) = 40e^−0.5t + 40t × (−0.5)e^−0.5t	✓ 1
A1 C′(t) = 40e^−0.5t(1 − 0.5t)	✓ 1
M1 set C′(t) = 0, noting e^−0.5t ≠ 0	✓ 1
A1 1 − 0.5t = 0 so t = 2 (a sign change from positive to negative shows a maximum)	✓ 1
Part (b)
M1 solve 40te^−0.5t = 10 using technology	✓ 1
A1 t = 0.289	✓ 1
A1 t = 6.52	✓ 1
A1 duration 6.23 hours	✓ 1
Part (c)
M1 C(2) = 40 × 2 × e⁻¹	✓ 1
A1 29.4 mg dm⁻³	✓ 1

Exam tip: When an equation has two solutions, a sketch of the curve with the horizontal line shows where they lie. Check that you subtract the earlier time from the later one to get a duration.

+ Add to test✓ In your test311IBMI3H-2-06r31-09Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

The number of players on a game server, P thousand, t days after launch is modelled by P(t) = 60 ÷ (1 + 11e^−0.8t).

(a)

Write down the number of players at launch.

(1)

(b)

Find the time at which the number of players is increasing most rapidly, to 3 significant figures.

(2)

(c)

Find the maximum rate of increase of the number of players.

(2)

(d)

Explain, in context, why the model predicts the rate of increase will fall after this time.

(1)

(e)

Find the time at which the server first reaches 80% of its limiting number of players. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 5 thousand (5000 players)	✓ 1
Part (b)
M1 the rate is greatest when P = 30 (half the limit), so 11e^−0.8t = 1 (or use technology to find the maximum gradient)	✓ 1
A1 t = ln 11 ÷ 0.8 = 3.00 days	✓ 1
Part (c)
M1 gradient at t = 2.997… found using technology, or kL/4 = 0.8 × 60 ÷ 4	✓ 1
A1 12 thousand players per day	✓ 1
Part (d)
R1 the server is approaching its limit of 60 thousand players, so fewer new players remain to join	✓ 1
Part (e)
M1 60 ÷ (1 + 11e^−0.8t) = 48	✓ 1
M1 e^−0.8t = 0.25 ÷ 11 = 1/44, so t = ln 44 ÷ 0.8	✓ 1
A1 4.73 days	✓ 1

Exam tip: The limiting value of a logistic model is the numerator. The point of fastest growth is at half that value, so you can find its time without differentiating.

+ Add to test✓ In your test312IBMI3H-2-06r31-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The rate v (μmol per minute) at which an enzyme converts a drug depends on the drug concentration S (mmol dm⁻³) according to v = VS ÷ (K + S), where V and K are positive constants. Experimental data are:
S: 2.5, 5, 10, 20
v: 4.00, 6.00, 8.00, 9.60

(a)

Show that 1 ÷ v = (K ÷ V)(1 ÷ S) + 1 ÷ V.

(2)

(b)

The data are plotted as 1 ÷ v against 1 ÷ S. Write down the coordinates of the points for S = 2.5 and S = 10.

(2)

(c)

Find the gradient and the vertical intercept of the straight line through these two points, to 3 significant figures.

(3)

(d)

Hence find the values of V and K.

(2)

(e)

Find the concentration S at which the rate is 7.0 μmol per minute. Use V = 12 and K = 5, and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 take reciprocals: 1 ÷ v = (K + S) ÷ (VS)	✓ 1
A1 split the fraction: K ÷ (VS) + S ÷ (VS) = (K ÷ V)(1 ÷ S) + 1 ÷ V (AG)	✓ 1
Part (b)
A1 (0.4, 0.25)	✓ 1
A1 (0.1, 0.125)	✓ 1
Part (c)
M1 gradient = (0.25 − 0.125) ÷ (0.4 − 0.1)	✓ 1
A1 gradient 0.417	✓ 1
A1 intercept 0.0833 (ECF from (b))	✓ 1
Part (d)
A1 V = 1 ÷ 0.0833 = 12.0	✓ 1
A1 K = gradient × V = 5.00	✓ 1
Part (e)
M1 12S ÷ (5 + S) = 7, so 5S = 35	✓ 1
A1 S = 7.00 mmol dm⁻³	✓ 1

Exam tip: When a model is not a straight line in v and S, look for a transformation such as a reciprocal that makes it linear. Read the constants from the gradient and intercept of the transformed graph, and use two well-separated points to find them.

+ Add to test✓ In your test313IBMI3H-2-06r33-05Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

The number of members of a cycling club t years after it was founded is modelled by $N=\dfrac{400}{1+Ce^{-kt}}$, where C and k are positive constants. The club had 25 members when it was founded, and 90 members after 4 years.

(a)

Show that C = 15.

(2)

(b)

Find the value of k, correct to three significant figures.

(3)

(c)

Find the number of members after 10 years, correct to three significant figures. Use k = 0.368.

(2)

(d)

Find how many years after the foundation the club first has 300 members, correct to three significant figures. Use k = 0.368.

(3)

(e)

State what happens to N as t becomes very large, and give one reason why this is reasonable in this context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes t = 0 and N = 25: 25 = 400/(1 + C)	✓ 1
A1 1 + C = 16, so C = 15, shown	✓ 1
Part (b)
M1 90 = 400/(1 + 15e^−4k)	✓ 1
A1 e^−4k = 0.2296…	✓ 1
A1 k = 0.368	✓ 1
Part (c)
M1 N = 400/(1 + 15e^−3.68)	✓ 1
A1 290 members	✓ 1
Part (d)
M1 300 = 400/(1 + 15e^−0.368t)	✓ 1
A1 e^−0.368t = 1/45	✓ 1
A1 t = 10.3 years	✓ 1
Part (e)
A1 N approaches 400 (the carrying capacity)	✓ 1
R1 any valid reason, e.g. limited local population of cyclists, club size or facilities	✓ 1

Exam tip: In a logistic model the numerator is the limiting value, and the initial value N(0) = L/(1 + C) lets you find C without any calculus.

+ Add to test✓ In your test314IBMI3H-2-06r33-06Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

The number of downloads of a cycle-route app, N, t weeks after its release is modelled by $N=\dfrac{12\,000}{1+79e^{-0.9t}}$.

(a)

Find the number of downloads at the moment of release.

(2)

(b)

Find the time at which N reaches half of its limiting value, correct to three significant figures.

(3)

(c)

The rate of increase of downloads is greatest at the time found in part (b). Find the maximum rate of increase, in downloads per week.

(2)

(d)

Find the number of weeks during which the rate of increase exceeds 2000 downloads per week, correct to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes t = 0: 12 000 ÷ (1 + 79)	✓ 1
A1 150 downloads	✓ 1
Part (b)
M1 sets N = 6000 giving 1 + 79e^−0.9t = 2	✓ 1
A1 e^−0.9t = 1/79	✓ 1
A1 t = 4.85 weeks	✓ 1
Part (c)
M1 attempt to find dN/dt at t = 4.85…, using the graphing facility or the derivative	✓ 1
A1 2700 downloads per week	✓ 1
Part (d)
M1 solves dN/dt = 2000 using the graphing facility	✓ 1
A1 t = 3.61 and t = 6.10	✓ 1
A1 duration 2.50 weeks	✓ 1

Exam tip: For a logistic curve the steepest point is at half the limiting value; the sketch of the rate against time is a bell shape, so a rate is exceeded between two solutions.

+ Add to test✓ In your test315IBMI3H-2-06r33-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

After a sprint, the concentration of lactate in a cyclist’s blood is modelled by $L=at\,e^{-bt}$, where L is in mmol per litre, t is the time in minutes after the sprint and a and b are positive constants. At t = 2 the concentration is 19.76 and at t = 8 it is 13.06.

(a)

Show that $\ln\left(\dfrac{L}{t}\right)=\ln a-bt$.

(2)

(b)

Use the two data points to find a and b, each correct to three significant figures.

(4)

(c)

Use a = 18.0 and b = 0.300 to find the time at which the concentration is greatest, and the greatest concentration, each to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\dfrac{L}{t}=ae^{-bt}$	✓ 1
A1 takes natural logs: ln(L/t) = ln a + ln e^−bt = ln a − bt, shown	✓ 1
Part (b)
M1 finds ln(L/t) at both points: ln 9.88 = 2.2905… and ln 1.6325 = 0.4901…	✓ 1
M1 gradient = (0.4901 − 2.2905) ÷ (8 − 2), so b = −gradient	✓ 1
A1 b = 0.300	✓ 1
A1 ln a = 2.2905 + 2b, so a = 18.0	✓ 1
Part (c)
M1 attempt to find the maximum of 18.0te^−0.3t, using the graphing facility or calculus	✓ 1
A1 t = 3.33 minutes	✓ 1
A1 L = 22.1 mmol per litre	✓ 1

Exam tip: When a model mixes a power and an exponential, divide by the power of t before taking logarithms: the quantity ln(L/t) is then linear in t.

+ Add to test✓ In your test316IBMI3H-2-05r35-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard9 marks

The percentage C of a bridge pier covered by lichen, t years after cleaning, is modelled by
C = 60 ÷ (1 + Ae^−kt), where A and k are positive constants. Immediately after cleaning, 5.0 % of the pier is covered. After 4.0 years, 18.0 % is covered.

(a)

Show that A = 11.

(2)

(b)

Find the value of k, correct to 3 significant figures.

(3)

(c)

State the long-term percentage of the pier covered, according to the model.

(1)

(d)

Use k = 0.388 to find the time taken for 50 % of the pier to be covered.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5 = 60 ÷ (1 + A) at t = 0	✓ 1
A1 1 + A = 12, so A = 11	✓ 1
Part (b)
M1 18 = 60 ÷ (1 + 11e^−4k), so 11e^−4k = 7/3	✓ 1
M1 e^−4k = 7/33, so k = −ln(7/33) ÷ 4	✓ 1
A1 k = 0.388	✓ 1
Part (c)
A1 60 % (as t → ∞, e^−kt → 0)	✓ 1
Part (d)
M1 50 = 60 ÷ (1 + 11e^−0.388t), so 11e^−0.388t = 0.2	✓ 1
M1 t = ln 55 ÷ 0.388	✓ 1
A1 t = 10.3 years	✓ 1

Exam tip: In a logistic model the constant A is fixed by the initial value, so use t = 0 first before using any other data.

+ Add to test✓ In your test317IBMI3H-2-06r38-07Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

A fish farm stocks a lake. The number of fish P after t years is modelled by P(t) = 1200 / (1 + Ae^−rt), where A and r are positive constants. Initially there are 150 fish, and after 3 years there are 300 fish.

(a)

Find the value of A.

(2)

(b)

Find the value of r, to 3 significant figures.

(3)

(c)

Find the number of fish after 10 years, to the nearest whole number. Use the unrounded value of r.

(2)

(d)

The population grows fastest when it is half of its limiting value. Find the time at which this happens, to 3 significant figures. Use the unrounded value of r.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0, 150 = 1200 / (1 + A)	✓ 1
A1 A = 7	✓ 1
Part (b)
M1 300 = 1200 / (1 + 7e^−3r)	✓ 1
M1 e^−3r = 3/7	✓ 1
A1 r = 0.282	✓ 1
Part (c)
M1 substitute t = 10	✓ 1
A1 848	✓ 1
Part (d)
M1 set P = 600	✓ 1
M1 7e^−rt = 1	✓ 1
A1 6.89 years	✓ 1

Exam tip: In a logistic model the limiting value is the numerator, and the fastest growth occurs when the population is half of that value.

+ Add to test✓ In your test318IBMI3H-2-05r41-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

A marine biologist monitors a restored reef plot. The area covered by living coral, C m², is modelled by C(t) = 120 / (1 + A e^−kt), where t is the time in months since restoration began and A and k are positive constants. Initially 15 m² is covered, and after 5 months 48 m² is covered.

(a)

Show that A = 7.

(2)

(b)

Find the value of k. Give your answer to 3 significant figures.

(3)

(c)

Find the time at which the coral covers 100 m². Use the unrounded value of k and give your answer to 3 significant figures.

(3)

(d)

Use your graphic display calculator to find the maximum rate of increase of the coral area, and the time at which it occurs. Give the rate to 3 significant figures, in m² per month.

(3)

(e)

State one reason why the logistic model may stop being appropriate for this reef after several years.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 C(0) = 120 / (1 + A) = 15	✓ 1
A1 1 + A = 8, so A = 7 (AG)	✓ 1
Part (b)
M1 120 / (1 + 7e^−5k) = 48	✓ 1
M1 7e^−5k = 1.5, so e^−5k = 3/14 (accept equivalent)	✓ 1
A1 k = 0.308	✓ 1
Part (c)
M1 120 / (1 + 7e^−kt) = 100 (or 1 + 7e^−kt = 1.2)	✓ 1
M1 e^−kt = 0.2/7 and taking natural logarithms	✓ 1
A1 t = 11.5 months	✓ 1
Part (d)
M1 attempt to find the maximum of C′(t) (graph of the gradient function or by sketching)	✓ 1
A1 time of maximum growth t = 6.32 months (when C = 60)	✓ 1
A1 maximum rate = 9.24 m² per month	✓ 1
Part (e)
R1 for a valid reason, e.g. other limiting factors such as bleaching events, storms or changes in water temperature are not included, or the maximum area may itself change	✓ 1

Exam tip: In a logistic model the numerator is the limiting value, and the fastest growth occurs when the quantity is half of that limit.

+ Add to test✓ In your test319IBMI3H-2-05r41-13Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

The depth of water d metres in a harbour entrance is modelled by d(t) = 6.4 + 2.6 sin(b(t − 1.5)), where t is the time in hours after midnight and b is a positive constant. The time between successive high tides is 12.4 hours.

(a)

Find the value of b to 3 significant figures.

(2)

(b)

Write down the minimum depth of water in the harbour entrance.

(2)

(c)

A ferry needs a depth of at least 8.0 m to enter the harbour. During the first tidal cycle (0 ≤ t ≤ 12.4), find the two times, to 3 significant figures, at which the depth is exactly 8.0 m. Use the unrounded value of b.

(4)

(d)

The ferry needs the depth to be at least 8.0 m for a continuous period of at least 3.5 hours. Find the length of time, in hours, for which the depth is at least 8.0 m in this cycle, and state whether the ferry can enter. Give the time to 3 significant figures and use unrounded values.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 period = 2π / b = 12.4	✓ 1
A1 b = 0.507	✓ 1
Part (b)
M1 6.4 − 2.6	✓ 1
A1 3.8 m	✓ 1
Part (c)
M1 6.4 + 2.6 sin(b(t − 1.5)) = 8	✓ 1
M1 sin(b(t − 1.5)) = 1.6/2.6 and use of inverse sine, or graphical solution with a graphic display calculator	✓ 1
A1 t = 2.81 hours	✓ 1
A1 t = 6.39 hours	✓ 1
Part (d)
A1 duration = 3.58 hours	✓ 1
R1 3.58 > 3.5, so the ferry can enter	✓ 1

Exam tip: A sine equation has a second solution in each cycle: after finding one angle with inverse sine, the other is π minus that angle.

+ Add to test✓ In your test320IBMI3H-2-05r42-08Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard14 marks

The strength of a radio signal from a space probe is measured on a logarithmic scale. The signal level is L = 10 log₁₀(P/P₀) decibels (dB), where P is the received power and P₀ is a fixed reference power. The received power is inversely proportional to the square of the distance d from the probe, so P = k/d², where k is a constant depending on the transmitter.
A probe at distance 2.0×10⁸ km produces a signal level of −110 dB.

(a)

Show that, when the distance changes from d₁ to d₂, the signal level changes by −20 log₁₀(d₂/d₁) dB.

(3)

(b)

Find the signal level when the probe is at a distance of 6.0×10⁸ km, correct to 4 significant figures.

(2)

(c)

The receiving dish can only detect signals with a level of at least −140 dB. Find the greatest distance at which the probe can be detected. Give your answer in the form a × 10ⁿ km, with 1 ≤ a < 10 and a given to 3 significant figures.

(3)

(d)

The transmitter power is multiplied by 4. Find the new greatest distance at which the probe can be detected, giving your answer in the form a × 10ⁿ km with a given to 3 significant figures.

(3)

(e)

Mission planners need to detect the original probe at a distance of 1.0×10¹⁰ km. Find the smallest factor by which the transmitter power must be multiplied. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 L₂ − L₁ = 10 log₁₀(P₂/P₁) (the reference power cancels)	✓ 1
M1 P₂/P₁ = d₁²/d₂², then use of log law for powers	✓ 1
A1 change = −20 log₁₀(d₂/d₁) (AG)	✓ 1
Part (b)
M1 −110 − 20 log₁₀(6.0×10⁸/2.0×10⁸)	✓ 1
A1 -119.5 dB	✓ 1
Part (c)
M1 −140 = −110 − 20 log₁₀(d/2.0×10⁸)	✓ 1
M1 log₁₀(d/2.0×10⁸) = 1.5, so d = 2.0×10⁸ × 10^1.5	✓ 1
A1 6.32×10⁹ km	✓ 1
Part (d)
M1 power × 4 increases every signal level by 10 log₁₀ 4 ≈ 6.02 dB (or k becomes 4k)	✓ 1
M1 distance for a given level is multiplied by √4 = 2 (or solves −140 = −110 + 6.02 − 20 log₁₀(d/2.0×10⁸))	✓ 1
A1 1.26×10¹⁰ km (ECF from (c))	✓ 1
Part (e)
M1 level at 1.0×10¹⁰ km with original transmitter = −110 − 20 log₁₀(50) = -143.98 dB	✓ 1
M1 required gain = −140 − (that level), and factor = 10^gain/10	✓ 1
A1 factor = 2.50	✓ 1

Exam tip: A decibel scale turns multiplication of powers into addition of levels, so a power factor f adds 10 log₁₀ f dB.

+ Add to test✓ In your test321IBMI3H-2-06r42-12Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard10 marks

A beekeeper counts the varroa mites y found on sticky boards beneath a hive at the end of week t:
week 0: 35 mites
week 2: 49 mites
week 4: 66 mites
week 6: 93 mites
week 8: 126 mites
The beekeeper models the count by y = a b^t.

(a)

Complete a table of values of t and ln y, giving each value of ln y correct to 3 decimal places.

(2)

(b)

Find the equation of the regression line of ln y on t, in the form ln y = mt + q, giving m and q correct to 3 significant figures.

(3)

(c)

Hence write down the values of a and b, correct to 3 significant figures.

(2)

(d)

The hive must be treated when the model predicts 500 mites. Using the unrounded values of m and q, find the time at which this occurs. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 ln y values for the first three weeks: 3.555, 3.892, 4.190	✓ 1
A1 ln y values for weeks 6 and 8: 4.533, 4.836	✓ 1
Part (b)
M1 linear regression of ln y on t using GDC	✓ 1
A1 m = 0.160	✓ 1
A1 q = 3.56	✓ 1
Part (c)
M1 a = e^q and b = e^m (ECF from (b))	✓ 1
A1 a = 35.2 and b = 1.17	✓ 1
Part (d)
M1 ln 500 = mt + q (ECF from (b))	✓ 1
M1 t = (ln 500 − q)/m	✓ 1
A1 t = 16.6 weeks	✓ 1

Exam tip: Fit the straight line to the transformed data, not the original counts; the regression gradient is ln b and the intercept is ln a.

+ Add to test✓ In your test322IBMI3H-2-06r44-04Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard12 marks

The number of subscribers N (in thousands) to a new mobile service t years after launch is modelled by N = 480 ÷ (1 + Ce^−kt), where C and k are positive constants. At launch there are 40 thousand subscribers, and after 3 years there are 120 thousand.

(a)

Show that C = 11.

(2)

(b)

Find the value of k, giving your answer to 3 significant figures.

(3)

(c)

Use k = 0.433 to find the predicted number of subscribers after 6 years, to the nearest thousand.

(2)

(d)

The rate of growth of the number of subscribers is greatest at the point where N = 240. Find the time at which this occurs, giving your answer in years to 3 significant figures.

(3)

(e)

The company regards the service as mature when the number of subscribers reaches 95% of the upper limit of the model. Find the time at which this happens, to 3 significant figures, using k = 0.433.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute t = 0 and N = 40: 40 = 480 ÷ (1 + C)	✓ 1
A1 1 + C = 12, so C = 11 (AG)	✓ 1
Part (b)
M1 substitute t = 3 and N = 120: 1 + 11e^−3k = 4	✓ 1
A1 e^−3k = 3/11	✓ 1
A1 k = 0.433	✓ 1
Part (c)
M1 substitute t = 6 into the model (ECF from (b))	✓ 1
A1 264 thousand	✓ 1
Part (d)
M1 set 480 ÷ (1 + 11e^−0.433t) = 240	✓ 1
A1 e^−0.433t = 1/11	✓ 1
A1 t = 5.54 years	✓ 1
Part (e)
M1 set 480 ÷ (1 + 11e^−0.433t) = 456 and rearrange to e^−0.433t = 1/209	✓ 1
A1 t = 12.3 years	✓ 1

Exam tip: In a logistic model N = L ÷ (1 + Ce⁻kt) the value L is the limiting value, and growth is fastest when N = L/2.

+ Add to test✓ In your test323IBMI3H-2-06r44-10Logarithmic scales and linearising data; transformations of graphs; logistic and other non-linear models Paper 2Hard11 marks

The population N of bacteria in a sample of yoghurt, in millions per gram, t hours after the sample is stored is modelled by N = 84 ÷ (1 + Ae^−0.64t). When t = 5 the population is half of its limiting value.

(a)

Find the value of A, giving your answer to 3 significant figures.

(3)

(b)

Find the initial population of bacteria, to 3 significant figures.

(2)

(c)

Find the time at which the population first reaches 75 million per gram. Give your answer in hours to 3 significant figures.

(3)

(d)

Find the maximum rate of growth of the population, in millions per gram per hour, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set 84 ÷ (1 + Ae^−3.2) = 42	✓ 1
A1 1 + Ae^−3.2 = 2	✓ 1
A1 A = e^3.2 = 24.5	✓ 1
Part (b)
M1 N(0) = 84 ÷ (1 + 24.5) (ECF from (a))	✓ 1
A1 3.29 million per gram	✓ 1
Part (c)
M1 set 84 ÷ (1 + 24.5e^−0.64t) = 75 and rearrange to 24.5e^−0.64t = 0.12	✓ 1
M1 take natural logarithms: t = ln(24.5 ÷ 0.12) ÷ 0.64	✓ 1
A1 t = 8.31 hours	✓ 1
Part (d)
M1 the growth rate is greatest when N = 42, that is when t = 5	✓ 1
M1 use the GDC to find dN/dt at t = 5, or use rate = 0.64 × 84 ÷ 4	✓ 1
A1 13.4	✓ 1

Exam tip: For a logistic model the greatest growth rate is one quarter of the product of the growth constant and the limiting value, and it occurs at half the limiting value.

Marking point	Mark	Notes
Part (a)
M1 substitutes \(t=5\): \(C=80e^{-0.173\times5}\)	✓ 1
A1 \(C=33.7\) mg L⁻¹	✓ 1
Part (b)
M1 sets \(80e^{-0.173t}=10\) (or finds the intersection with \(C=10\) on a GDC)	✓ 1
M1 takes natural logarithms: \(-0.173t=\ln\frac{1}{8}\) (or equivalent, e.g. \(t=\frac{\ln 8}{0.173}\))	✓ 1
A1 \(t=12.0\) hours	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(250=500e^{-9k}\), so \(e^{-9k}=\frac{1}{2}\)	✓ 1
A1 \(k=\frac{\ln2}{9}=0.0770\) (AG)	✓ 1
Part (b)
M1 \(500e^{-20k}\) with unrounded \(k\) (or \(500\times0.5^{20/9}\))	✓ 1
A1 107 gold coins	✓ 1
Part (c)
M1 \(e^{-kt}=0.1\)	✓ 1
M1 \(t=\frac{\ln10}{k}\)	✓ 1
A1 \(29.9\) days	✓ 1

Marking point	Mark	Notes
Part (a)
M1 takes exponentials of both sides: \(T=e^{0.0318t+2.95}=e^{2.95}e^{0.0318t}\)	✓ 1
A1 A = 19.1 and k = 0.0318	✓ 1
Part (b)
M1 substitutes t = 25	✓ 1
A1 42.3 °C	✓ 1
Part (c)
M1 solves 19.1e^0.0318t = 60, or ln 60 = 0.0318t + 2.95	✓ 1
A1 36.0 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(N(0) = \frac{4000}{80} = 50\) people	✓ 1
Part (b)
M1: \(\frac{4000}{1 + 79e^{-3}}\)	✓ 1
A1: 811 people	✓ 1
Part (c)
M1: \(\frac{4000}{1 + 79e^{-0.6t}} = 3000\)	✓ 1
M1: rearranging to \(e^{-0.6t} = \frac{1}{237}\) or using GDC solver/graph intersection	✓ 1
A1: 9.11 days	✓ 1
Part (d)
A1: 4000	✓ 1
R1: eventually (almost) everyone in the town hears the rumour, but the number never exceeds the town population	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(M = e^{4.38 - 0.0275t} = e^{4.38}\times e^{-0.0275t}\)	✓ 1
A1: \(A = 79.8\) (79.838…)	✓ 1
A1: \(k = -0.0275\)	✓ 1
Part (b)
M1: \(e^{-0.0275t} = 0.5\)	✓ 1
A1: 25.2 hours	✓ 1
Part (c)
M1: \(79.838\ldots \times e^{-0.0275\times 48}\) or \(e^{4.38 - 1.32}\)	✓ 1
A1: 21.3 mg	✓ 1

Marking point	Mark	Notes
Part (a)
M1: writes \(R = 10^{0.74\log_{10}M + 0.53} = 10^{0.53}\times10^{0.74\log_{10}M}\)	✓ 1
M1: uses \(10^{0.74\log_{10}M} = M^{0.74}\)	✓ 1
A1: a = 3.39 (3.388…), b = 0.74	✓ 1
Part (b)
M1: substitutes, \(R = 3.388\times65^{0.74}\)	✓ 1
A1: R = 74.4 W	✓ 1
Part (c)
M1: sets up \(a(kM)^{0.74} = 2aM^{0.74}\)	✓ 1
M1: simplifies to \(k^{0.74} = 2\)	✓ 1
A1: k = 2.55	✓ 1

Marking point	Mark	Notes
Part (a)
Uses laws of logs: \(\log_{10}P = \log_{10}(M^{0.75}) + \log_{10}(10^{0.55})\), so \(P = 10^{0.55}M^{0.75}\) M1	✓ 1
\(n = 0.75\) A1	✓ 1
\(k = 10^{0.55} = 3.55\) A1	✓ 1
Part (b)
Substitutes \(M = 70\): \(3.548 \times 70^{0.75}\) M1	✓ 1
\(P = 85.9\) W A1	✓ 1
Part (c)
Extrapolation: 0.002 kg lies outside the data range 0.02–4000 kg R1	✓ 1
Insects are not mammals, so the relationship found for mammals may not apply R1	✓ 1

Marking point	Mark	Notes
Part (a)
Exponentiates both sides, e.g. \(P=e^{0.75\ln m+1.2}=e^{1.2}m^{0.75}\) M1 ; \(n=0.75\) A1 ; \(a=e^{1.2}=3.32\) A1	✓ 3
Part (b)
\(P=3.3201\times65^{0.75}\) (or \(\ln P=0.75\ln65+1.2\)) M1 ; \(P=76.0\) W A1	✓ 2
Part (c)
\(\ln m=\frac{\ln500-1.2}{0.75}\) (or \(500=3.3201m^{0.75}\)) M1 ; \(m=801\) kg A1	✓ 2
Part (d)
Unreliable, as 801 kg lies outside the 0.02–100 kg range of the data (extrapolation) R1	✓ 1

Marking point	Mark	Notes
Part (a)
M1: exponentiates both sides, \(N = \mathrm{e}^{0.231t + 4.62}\)	✓ 1
M1: uses laws of indices, \(N = \mathrm{e}^{4.62}\times\mathrm{e}^{0.231t}\)	✓ 1
A1: a = 101 (101.49…), k = 0.231	✓ 1
Part (b)
M1: sets up \(0.231t + 4.62 = \ln 1000\) or \(101.49\mathrm{e}^{0.231t} = 1000\)	✓ 1
A1: t = 9.90 hours (9.904…)	✓ 1
Part (c)
M1: uses \(\mathrm{e}^{0.231T} = 2\) or equivalent, e.g. \(T = \frac{\ln 2}{0.231}\)	✓ 1
A1: T = 3.00 hours (3.0006…)	✓ 1
Part (d)
R1: states that the prediction is unreliable, with a valid reason: either 60 hours is far outside the 12-hour data range (extrapolation), or limited nutrients and space would slow growth so exponential growth cannot continue	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(P=10^{0.75\log_{10}m+0.55}=10^{0.55}\times(10^{\log_{10}m})^{0.75}=10^{0.55}\times m^{0.75}\)	✓ 1
A1: \(n=0.75\)	✓ 1
A1: \(k=10^{0.55}=3.55\) (3.548…)	✓ 1
Part (b)
M1: \(\frac{k(2m)^{0.75}}{km^{0.75}}=2^{0.75}\)	✓ 1
A1: 1.68	✓ 1
Part (c)
M1: \(3.548\times250^{0.75}\)	✓ 1
A1: \(P=223\) W	✓ 1
Part (d)
R1: unreliable because 250 kg is outside the data range (0.02–60 kg), so this is extrapolation	✓ 1

Marking point	Mark	Notes
Part (a)
A1 maximum \(57\) megalitres per hour	✓ 1
A1 minimum \(27\) megalitres per hour	✓ 1
Part (b)
A1 \(\frac{2\pi}{\pi/12}=24\) hours	✓ 1
Part (c)
M1 sets \(42+15\cos\left(\frac{\pi}{12}(t-17)\right)=50\) (or intersection on a GDC)	✓ 1
A1 \(t=13.1\)	✓ 1
A1 \(t=20.9\)	✓ 1
Part (d)
M1 subtracts the two times: \(20.85\ldots-13.14\ldots\) (unrounded values)	✓ 1
A1 \(7.70\) hours	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(n=400-150=250\)	✓ 1
Part (b)
M1 \(P=(x-4)(400-25x)\)	✓ 1
A1 \(=400x-25x^2-1600+100x=-25x^2+500x-1600\) (AG)	✓ 1
Part (c)
M1 solves \(P=0\) (or \((x-4)(400-25x)=0\))	✓ 1
A1 \(4	✓ 1
Part (d)
A1 \(x=-\frac{500}{2(-25)}=10\), so €10	✓ 1
Part (e)
M1 \(P(10)=-25(100)+5000-1600\)	✓ 1
A1 €900	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\ln P=n\ln v+\ln k\), and forms two equations or the gradient between (ln 8, ln 110) and (ln 12, ln 371)	✓ 1
M1 \(n=\dfrac{\ln 371-\ln 110}{\ln 12-\ln 8}\)	✓ 1
A1 n = 3.00	✓ 1
Part (b)
M1 110 = k × 8³	✓ 1
A1 k = 0.215	✓ 1
Part (c)
M1 600 = (110/512)v³ leading to v³ = 2792.7…	✓ 1
A1 14.1 m s⁻¹	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(78=10\log_{10}(I/10^{-12})\) giving \(I=10^{7.8}\times10^{-12}\)	✓ 1
A1 I = 6.31 × 10⁻⁵ W m⁻²	✓ 1
Part (b)
M1 intensity doubles, so L = 10 log₁₀(2I/I₀) = 78 + 10 log₁₀ 2	✓ 1
A1 81.0 dB	✓ 1
Part (c)
M1 10 log₁₀ n ≥ 12 giving n ≥ 15.85…	✓ 1
A1 16 sets	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\left(\tfrac12\right)^{t/5730}=e^{-t\ln 2/5730}\), so \(\lambda=\dfrac{\ln2}{5730}\)	✓ 1
A1 λ = 1.21 × 10⁻⁴	✓ 1
Part (b)
M1 0.62 = (1/2)^t/5730	✓ 1
A1 t = 5730 × ln 0.62 ÷ ln 0.5 = 3951.8…	✓ 1
A1 3950 years	✓ 1
Part (c)
M1 (1/2)^8000/5730	✓ 1
A1 38.0%	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(P = 10^{0.75\log_{10} m + 0.55} = 10^{0.55} \times m^{0.75}\)	✓ 1
A1 \(b = 0.75\)	✓ 1
A1 \(a = 10^{0.55} = 3.55\)	✓ 1
Part (b)
M1 \(3.548 \times 60^{0.75}\)	✓ 1
A1 76.5 W	✓ 1
Part (c)
R1 a power model gives a straight line when \(\log P\) is plotted against \(\log m\)	✓ 1
R1 an exponential model would give a straight line when \(\log P\) is plotted against \(m\) itself, so a linear log–log graph indicates a power relationship	✓ 1
Part (d)
M1 \(3.548 \times 5000^{0.75}\)	✓ 1
A1 2110 W	✓ 1
R1 unreliable because it is an extrapolation far beyond the data range of 0.02 kg to 70 kg	✓ 1

Marking point	Mark	Notes
Part (a)
M1: takes \(\log_{10}\) of both sides: \(\log_{10}P=\log_{10}(aM^{b})\)	✓ 1
A1: uses product and power laws to get \(\log_{10}a+b\log_{10}M\), AG	✓ 1
Part (b)
M1: gradient \(=\frac{2.403-0.919}{2.5-0.5}\)	✓ 1
A1: gradient \(=0.742\)	✓ 1
A1: intercept \(=0.919-0.742(0.5)=0.548\)	✓ 1
Part (c)
M1: \(a=10^{0.548}\)	✓ 1
A1: \(a=3.53\) and \(b=0.742\)	✓ 1
Part (d)
M1: \(3.532\times60^{0.742}\)	✓ 1
A1: 73.7 W	✓ 1
Part (e)
M1: \(250=3.532M^{0.742}\)	✓ 1
M1: \(M=\left(\frac{250}{3.532}\right)^{1/0.742}\) or solves with GDC	✓ 1
A1: 311 kg	✓ 1
Part (f)
R1: unreliable, because 4000 kg is far outside the 3–320 kg data range, so it would be extrapolation	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{A_1}{A_2}=10^{6.3-4.1}=10^{2.2}\)	✓ 1
A1: 158 (accept 158.5)	✓ 1
Part (b)
M1: \(E=10^{4.8+1.5(6.3)}=10^{14.25}\)	✓ 1
A1: \(1.78\times10^{14}\) J	✓ 1
Part (c)
M1: \(\frac{10^{4.8+1.5(M+1)}}{10^{4.8+1.5M}}\), or subtracting the logs to get a difference of 1.5	✓ 1
A1: \(=10^{1.5}=31.62\ldots\approx31.6\), AG	✓ 1
Part (d)
M1: \(1.5M=\log_{10}(5.0\times10^{15})-4.8\)	✓ 1
A1: \(M=7.27\)	✓ 1
Part (e)
M1: \((\log_{10}(5.0\times10^{15})-10)\times2\)	✓ 1
A1: 11.4 cm	✓ 1

Marking point	Mark	Notes
Part (a)
\(x = -\frac{2}{2(-0.05)} = 20\), or use of a GDC (M1)	✓ 1
(20, 20) (A1)	✓ 1
Part (b)
Horizontal stretch, scale factor 1.5 (A1)	✓ 1
Vertical stretch, scale factor 0.75 (A1)	✓ 1
Then a translation of 5 units in the positive x-direction, i.e. by the vector (5, 0), applied after the horizontal stretch (A1)	✓ 1
Accept any correct equivalent sequence, e.g. a translation by \(\binom{10/3}{0}\) followed by a horizontal stretch of scale factor 1.5 about the y-axis, with the vertical stretch at any stage.	—
Part (c)
Applies the transformations to (20, 20): x = 1.5 × 20 + 5, y = 0.75 × 20 (M1)	✓ 1
(35, 15) (A1)	✓ 1
Part (d)
Substitutes: \(0.75\left[-0.05\frac{(x-5)^2}{2.25} + 2\cdot\frac{x-5}{1.5}\right]\) (M1)	✓ 1
Simplifies to \(-\frac{1}{60}(x-5)^2 + (x-5)\) and expands (M1)	✓ 1
\(g(x) = -\frac{1}{60}x^2 + \frac{7}{6}x - \frac{65}{12}\) (A1)	✓ 1
Part (e)
Solves g(x) = 9 (e.g. by GDC), giving x = 16.03 and x = 53.97, i.e. \(x = 35 \pm \sqrt{360}\) (M1)	✓ 1
Length = \(2\sqrt{360}\) = 37.9 m (A1)	✓ 1
Use unrounded roots: subtracting the 3 s.f. values 16.0 and 54.0 gives 38.0, which is inaccurate.	—

Marking point	Mark	Notes
Part (a)
M1 takes exponentials (or writes \(Q=e^{2.46\ln d+0.89}\))	✓ 1
M1 uses \(e^{2.46\ln d}=d^{2.46}\) and \(e^{x+y}=e^xe^y\)	✓ 1
A1 \(Q=e^{0.89}d^{2.46}\), which has the form \(kd^a\) (AG)	✓ 1
Part (b)
A1 \(a=2.46\)	✓ 1
A1 \(k=e^{0.89}=2.44\)	✓ 1
Part (c)
M1 substitutes \(d=0.35\) into \(Q=2.4351\ldots\times d^{2.46}\) (unrounded \(k\))	✓ 1
A1 \(Q=0.184\) m³ per hour	✓ 1
Part (d)
M1 sets \(2.4351\ldots d^{2.46}=1.5\)	✓ 1
M1 isolates \(d=\left(\frac{1.5}{e^{0.89}}\right)^{\frac{1}{2.46}}\)	✓ 1
A1 \(d=0.821\) m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(N=10^{3.2}\)	✓ 1
A1 \(N=1580\)	✓ 1
Part (b)
M1 \(N=10^{0.12t+3.2}=10^{3.2}\times(10^{0.12})^t\)	✓ 1
A1 \(B=10^{0.12}=1.32\)	✓ 1
A1 the number of players increases by about 31.8% (each day), since \(B=1.318\ldots\)	✓ 1
Part (c)
M1 sets \(0.12t+3.2=\log_{10}100000=5\)	✓ 1
A1 \(t=15\) days	✓ 1
Part (d)
M1 uses \(B^{t}=2\) (or \(0.12t=\log_{10}2\))	✓ 1
M1 \(t=\frac{\log_{10}2}{0.12}\)	✓ 1
A1 \(t=2.51\) days	✓ 1

Marking point	Mark	Notes
Part (a)
M1 substitutes \(t=0\): \(2000=\frac{50000}{1+C}\)	✓ 1
A1 \(1+C=25\), so \(C=24\) (AG)	✓ 1
Part (b)
M1 substitutes \(t=3\): \(9000=\frac{50000}{1+24e^{-3k}}\)	✓ 1
M1 rearranges to \(e^{-3k}=\frac{1}{24}\left(\frac{50}{9}-1\right)\) (or solves on a GDC)	✓ 1
A1 \(k=0.554\)	✓ 1
Part (c)
A1 50 000; the maximum number of players the market can support (carrying capacity)	✓ 1
Part (d)
M1 sets \(\frac{50000}{1+24e^{-kt}}=40000\)	✓ 1
M1 \(24e^{-kt}=0.25\), so \(e^{-kt}=\frac{1}{96}\)	✓ 1
A1 \(t=8.24\) weeks	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(-\log_{10}(4.5\times10^{-8})\)	✓ 1
A1 \(7.35\)	✓ 1
Part (b)
M1 \([\text{H}^+]=10^{-6.8}\)	✓ 1
A1 \(1.58\times10^{-7}\) mol L⁻¹	✓ 1
Part (c)
M1 \(\frac{10^{-6.8}}{4.5\times10^{-8}}\)	✓ 1
A1 \(3.52\)	✓ 1
Part (d)
M1 \(\Delta\text{pH}=-\log_{10}3\)	✓ 1
A1 \(-0.477\)	✓ 1
A1 the pH falls by 0.477	✓ 1

Marking point	Mark	Notes
Part (a)
A1 \(\ln(29.1)=3.37\)	✓ 1
A1 \(\ln(13.1)=2.57\)	✓ 1
Part (b)
M1 \(T-20=e^{4.17-0.0800t}\)	✓ 1
M1 \(=e^{4.17}e^{-0.0800t}\)	✓ 1
A1 \(A=64.7\) and \(k=0.0800\)	✓ 1
Part (c)
R1 the temperature tends to the room temperature 20 °C (horizontal asymptote \(T=20\)), so \(T-20\) decays to zero exponentially	✓ 1
Part (d)
M1 sets \(4.17-0.0800t=\ln10\)	✓ 1
M1 \(t=\frac{4.17-\ln10}{0.0800}\)	✓ 1
A1 \(t=23.3\) minutes	✓ 1

Marking point	Mark	Notes
Part (a)
\(400e^{3k} = 1100\) M1	✓ 1
\(k = \frac{1}{3}\ln 2.75 = 0.337\) A1	✓ 1
Part (b)
\(\frac{6000}{1+14} = 400\) A1	✓ 1
Sets \(\frac{6000}{1 + 14e^{-3r}} = 1100\), giving \(e^{-3r} = \frac{7}{22}\) M1	✓ 1
\(r = \frac{1}{3}\ln\frac{22}{7} = 0.3817… = 0.382\) A1	✓ 1
Part (c)
Substitutes \(t = 8\) into either model M1	✓ 1
Model A: \(N = 5940\) (5937.6…) A1	✓ 1
Model B: \(N = 3610\) (3613.2… using unrounded \(r\); accept 3620 from using \(r = 0.382\)) A1	✓ 1
Part (d)
Compares both predictions with 3900, e.g. differences of about 2040 and 290 M1	✓ 1
Model B, because its prediction is much closer to the observed value R1	✓ 1
Part (e)
6000 is the carrying capacity: the maximum number of fish the lake can sustain in the long term R1	✓ 1
Part (f)
Solves \(\frac{6000}{1 + 14e^{-rt}} = 5000\), giving \(e^{-rt} = \frac{1}{70}\) M1	✓ 1
\(t = \frac{\ln 70}{r} = 11.1\) years (unrounded \(r\) or \(r = 0.382\)) A1	✓ 1

Marking point	Mark	Notes
Part (a)
A1: \(y=20\) (accept \(T=20\) or \(f(t)=20\))	✓ 1
R1: the coffee cools towards room temperature, 20 °C, but never falls below it	✓ 1
Part (b)
A1: horizontal stretch (parallel to the \(t\)-axis)	✓ 1
A1: scale factor \(\frac{1}{0.8}=1.25\)	✓ 1
A1: vertical translation 4 units up	✓ 1
Part (c)
M1: \(70e^{-0.05(0.8t)}+20+4\)	✓ 1
A1: \(g(t)=70e^{-0.04t}+24\)	✓ 1
Part (d)
M1: \(70e^{-0.04t}+24=60\)	✓ 1
A1: \(t=\frac{\ln(70/36)}{0.04}=16.6\) minutes	✓ 1
M1: \(70e^{-0.05t}+20=60\)	✓ 1
A1: \(t=\frac{\ln 1.75}{0.05}=11.2\) minutes	✓ 1
A1: \(16.62-11.19=5.43\) minutes	✓ 1
Part (e)
M1: forms \(70e^{-0.04t}-70e^{-0.05t}+4\) and uses a GDC maximum or sets the derivative to zero	✓ 1
A1: \(t=100\ln1.25=22.3\) minutes	✓ 1
A1: maximum difference 9.73 °C	✓ 1

Marking point	Mark	Notes
Part (a)
\(x = 20\) m (e.g. from \(f'(x) = -0.012x^2 + 0.24x = 0\) or GDC)	✓ 1
Maximum height = 16 m	✓ 1
Part (b)
Recognises the maximum of g occurs where \(qx = 20\) and equals \(16p\)	✓ 1
\(p = 1.25\)	✓ 1
\(q = \frac{2}{3}\)	✓ 1
Part (c)
Vertical stretch (parallel to the y-axis), scale factor 1.25	✓ 1
Horizontal stretch (parallel to the x-axis), scale factor 1.5	✓ 1
Part (d)
Original track ends at \(x = 30\), so \(\frac{2}{3}x = 30\) (or \(30 \times 1.5\))	✓ 1
Length = 45 m	✓ 1
Part (e)
Substitutes: \(g(x) = 1.25\left[-0.004\left(\frac{2x}{3}\right)^3 + 0.12\left(\frac{2x}{3}\right)^2\right]\)	✓ 1
\(x^3\) coefficient: \(1.25 \times (-0.004) \times \frac{8}{27} = -\frac{1}{675}\)	✓ 1
\(x^2\) coefficient: \(1.25 \times 0.12 \times \frac{4}{9} = \frac{1}{15}\)	✓ 1
Part (f)
Solves \(g(x) = 12\) by GDC, or solves \(f(x) = 9.6\) and scales the solutions by 1.5	✓ 1
\(x = 17.0\) and \(x = 39.9\)	✓ 1
Length = 22.9 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(40=\frac{800}{1+A}\)	✓ 1
A1 \(A=19\)	✓ 1
Part (b)
R1 the maximum area is 800, so the fastest growth is at \(P=400\)	✓ 1
M1 \(400=\frac{800}{1+19e^{-6r}}\), so \(e^{-6r}=\frac{1}{19}\)	✓ 1
A1 \(r=0.491\)	✓ 1
Part (c)
M1 \(P(10)=\frac{800}{1+19e^{-10r}}\)	✓ 1
A1 \(P(10)=701\) m²	✓ 1
Part (d)
M1 uses a GDC to find the maximum of \(P'(t)\) (or the gradient of \(P\) at \(t=6\))	✓ 1
A1 \(98.1\)	✓ 1
A1 m² per day	✓ 1

Marking point	Mark	Notes
Part (a)
M1 forms \(\frac{24}{3}=\left(\frac{64}{4}\right)^b\), i.e. \(8=16^b\)	✓ 1
M1 takes logarithms: \(b=\frac{\ln8}{\ln16}\)	✓ 1
A1 \(b=\frac{3}{4}=0.75\) (AG)	✓ 1
Part (b)
M1 \(3.0=a\times4^{0.75}\)	✓ 1
A1 \(a=1.06\)	✓ 1
Part (c)
M1 \(R=\frac{3}{4^{0.75}}\times70^{0.75}\)	✓ 1
A1 \(25.7\) mg per hour	✓ 1
Part (d)
M1 \(\ln R=\ln a+b\ln m\)	✓ 1
A1 gradient \(0.75\) and vertical intercept \(\ln a=0.0589\)	✓ 1
Part (e)
M1 \(30=a\,m^{0.75}\), so \(m^{0.75}=\frac{30}{a}\)	✓ 1
M1 \(m=\left(\frac{30}{a}\right)^{\frac{1}{0.75}}\)	✓ 1
A1 \(m=86.2\) kg	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\dfrac{L}{t}=ae^{-bt}\)	✓ 1
A1 takes natural logs: ln(L/t) = ln a + ln e^−bt = ln a − bt, shown	✓ 1
Part (b)
M1 finds ln(L/t) at both points: ln 9.88 = 2.2905… and ln 1.6325 = 0.4901…	✓ 1
M1 gradient = (0.4901 − 2.2905) ÷ (8 − 2), so b = −gradient	✓ 1
A1 b = 0.300	✓ 1
A1 ln a = 2.2905 + 2b, so a = 18.0	✓ 1
Part (c)
M1 attempt to find the maximum of 18.0te^−0.3t, using the graphing facility or calculus	✓ 1
A1 t = 3.33 minutes	✓ 1
A1 L = 22.1 mmol per litre	✓ 1