278 free, high-quality past-paper-style questions on Functions for IB Maths AA SL, each with a full mark scheme. Covers: Equations of straight lines; Functions, domain, range and inverse; Graphs of functions; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 2 at Higher Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,185 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AASL2-q00012.1 Equations of straight linesPaper 1Easy8 marks
The line \(L_{1}\) passes through the points \(\mathrm{A}(-1,\,7)\) and \(\mathrm{B}(3,\,-1)\).
Diagram not to scale
(a)
Find the gradient of \(L_{1}\).
(2)
(b)
Find the equation of \(L_{1}\), giving your answer in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(2)
(c)
The line \(L_{2}\) is perpendicular to \(L_{1}\) and passes through \(\mathrm{C}(6,\,4)\). Find the equation of \(L_{2}\), giving your answer in the form \(y=mx+c\).
(2)
(d)
Find the coordinates of the point of intersection of \(L_{1}\) and \(L_{2}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{-1-7}{3-(-1)}\)
(M1)
\(-2\)
A1
(b)
\(y-7=-2(x+1)\) or \(y=-2x+5\)
(M1)
FT from their gradient.
\(2x+y-5=0\)
A1
Accept any integer multiple, e.g. \(-2x-y+5=0\).
(c)
gradient \(\tfrac{1}{2}\) and \(y-4=\tfrac{1}{2}(x-6)\)
(M1)
FT from (a): gradient must be \(-\dfrac{1}{m_{1}}\).
\(y=\tfrac{1}{2}x+1\)
A1
(d)
\(2x+\left(\tfrac{1}{2}x+1\right)-5=0\), i.e. \(\tfrac{5}{2}x=4\)
2AASL2-q00022.1 Equations of straight linesPaper 1Easy6 marks
The line \(L_{1}\) has equation \(3x-4y+12=0\).
(a)
Write the equation of \(L_{1}\) in the form \(y=mx+c\).
(2)
(b)
Find the coordinates of the points where \(L_{1}\) meets the \(x\)-axis and the \(y\)-axis.
(2)
(c)
The line \(L_{2}\) is parallel to \(L_{1}\) and passes through the point \((2,\,-5)\). Write down the equation of \(L_{2}\) in the form \(y-y_{1}=m(x-x_{1})\).
(1)
(d)
Write the equation of \(L_{2}\) in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(4y=3x+12\)
(M1)
Rearranging to make y the subject.
\(y=\tfrac{3}{4}x+3\)
A1
(b)
\(x\)-axis: \((-4,\,0)\)
A1
\(y\)-axis: \((0,\,3)\)
A1
Award A1A0 if both are given as single numbers −4 and 3 without coordinates.
(c)
\(y+5=\tfrac{3}{4}(x-2)\)
A1
FT from their gradient in (a). Accept \(y-(-5)=\tfrac34(x-2)\).
(d)
\(3x-4y-26=0\)
A1
Accept any integer multiple, e.g. \(-3x+4y+26=0\).
3AASL2-q00042.1 Equations of straight linesPaper 1Easy7 marks
A candle burns at a constant rate. Its height is 21 cm two hours after it is lit and 15 cm five hours after it is lit. The height of the candle, \(h\) cm, \(t\) hours after it is lit is modelled by \(h=mt+c\).
(a)
Find the value of \(m\) and the value of \(c\).
(2)
(b)
Interpret, in context, the value of \(m\) and the value of \(c\).
(2)
(c)
A second candle is lit at the same time. Its height is modelled by \(h=18-1.2t\). Find the time when the two candles have the same height, and this height.
(2)
(d)
Write the model for the first candle in the form \(at+bh+d=0\), where \(a,b,d\in\mathbb{Z}\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(m=\dfrac{15-21}{5-2}\)
(M1)
\(m=-2,\ c=25\)
A1
\(21=-2(2)+c\).
(b)
\(m\): the candle’s height decreases by 2 cm every hour
A1
Must refer to rate and units.
\(c\): the height of the candle when it is lit is 25 cm
10AASL2-q00312.5 Composite and inverse functionsPaper 1Easy6 marks
The functions \(f\) and \(g\) are defined on the set \(\{1,\,2,\,3,\,4,\,5\}\) by the following table.
\(x\)
1
2
3
4
5
\(f(x)\)
3
5
1
2
4
\(g(x)\)
4
3
5
1
2
(a)
Find \((f\circ g)(2)\).
(2)
(b)
Find \((g\circ f)(4)\).
(1)
(c)
Write down \(f^{-1}(5)\).
(1)
(d)
Solve \((g\circ f)(x)=5\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(g(2)=3\)
(M1)
\(f(3)=1\)
A1
(b)
\(g(f(4))=g(2)=3\)
A1
(c)
\(2\)
A1
(d)
\(g(f(x))=5\Rightarrow f(x)=3\)
(M1)
\(x=1\)
A1
Answers: (a) 1 · (b) 3 · (c) 2 · (d) \(x=1\)
11AASL2-q00382.6 Quadratic functionsPaper 1Easy8 marks
The following diagram shows part of the graph of a quadratic function \(f\). The vertex of the graph is \((3,\,-2)\) and the graph passes through the point \((1,\,6)\).
(a)
The function can be written in the form \(f(x)=a(x-h)^{2}+k\). Find the value of \(a\), of \(h\) and of \(k\).
(3)
(b)
Express \(f(x)\) in the form \(f(x)=ax^{2}+bx+c\) and hence write down the \(y\)-intercept of the graph.
(2)
(c)
Express \(f(x)\) in the form \(f(x)=a(x-p)(x-q)\) and hence write down the \(x\)-intercepts of the graph.
23AASL2-q00872.11 Transformations of graphsPaper 1Easy8 marks
The following diagram shows the graph of \(y=f(x)\). The graph has a local minimum point at \((-1,\,0)\), a local maximum point at \((1,\,4)\), meets the \(y\)-axis at \((0,\,2)\) and has a horizontal asymptote \(y=2\).
(a)(i)
For each of the following graphs, write down the coordinates of the image of the point \((1,\,4)\) and the equation of the horizontal asymptote.
24AASL2-q00882.11 Transformations of graphsPaper 1Easy7 marks
The graph of \(y=\sqrt{x}\), for \(x\ge0\), is transformed by the following sequence of transformations to give the graph of \(y=g(x)\):
a horizontal stretch with scale factor \(\tfrac{1}{4}\);
then a reflection in the \(x\)-axis;
then a translation of 3 units vertically upwards.
(a)
Show that \(g(x)=3-2\sqrt{x}\).
(2)
(b)
Write down the domain and the range of \(g\).
(2)
(c)
Find the \(x\)-intercept of the graph of \(g\).
(2)
(d)
The horizontal stretch in the sequence above can be replaced by a vertical stretch, giving the same graph. Write down the scale factor of this vertical stretch.
25IBMA2S-046-02-012.5 Composite and inverse functionsPaper 1Easy6 marks
No calculator
Let \(f(x)=\sqrt{10-2x}\).
(a)
Find the largest possible domain of \(f\).
(2)
(b)
State the range of \(f\) for this domain.
(1)
(c)
Find \(f^{-1}(x)\), stating its domain.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(10-2x\ge 0\)
✓ 1
A1 \(x\le 5\)
✓ 1
Part (b)
A1 \(f(x)\ge 0\)
✓ 1
Part (c)
M1 interchanges \(x\) and \(y\) and squares, e.g. \(x^2=10-2y\)
✓ 1
A1 \(f^{-1}(x)=5-\frac{x^2}{2}\)
✓ 1
A1 domain \(x\ge 0\)
✓ 1
Common error: The domain of \(f^{-1}\) is the range of \(f\), not the set of all real numbers. Without the restriction \(x\ge 0\), the formula \(5-\frac{x^2}{2}\) describes a function that is not one-to-one.
26IBMA2S-046-02-032.8 Reciprocal and rational functionsPaper 1Easy5 marks
No calculator
Let \(h(x)=\frac{2x+6}{x-3}\), \(x\ne 3\).
(a)
Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of \(h\).
(2)
(b)
Find the coordinates of the points where the graph of \(h\) meets the axes.
(2)
(c)
Write down the range of \(h\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=3\)
✓ 1
A1 \(y=2\)
✓ 1
Part (b)
A1 \(x\)-intercept \((-3,\,0)\)
✓ 1
A1 \(y\)-intercept \((0,\,-2)\)
✓ 1
Part (c)
A1 \(y\in\mathbb{R},\ y\ne 2\)
✓ 1
Exam tip: For \(\frac{ax+b}{cx+d}\), the horizontal asymptote is \(y=\frac{a}{c}\), and this value is the one missing from the range.
27IBMA2S-046-01-012.6 Quadratic functionsPaper 1Easy6 marks
No calculator
Let \(f(x) = 2x^2 - 12x + 13\).
(a)
Express \(f(x)\) in the form \(a(x-h)^2 + k\), where \(a\), \(h\), \(k \in \mathbb{Z}\).
(3)
(b)
Write down the coordinates of the vertex of the graph of \(f\).
(1)
(c)
Hence solve \(f(x) = 0\), giving your answers in the form \(p \pm \frac{\sqrt{q}}{r}\), where \(p, q, r \in \mathbb{Z}^+\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square, e.g. \(2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13\)
M1 attempt to factorise or solve: \((3y - 1)(y - 3) = 0\)
✓ 1
A1 \(y = \frac{1}{3}\) or \(y = 3\)
✓ 1
A1 \(x = -1\) or \(x = 1\)
✓ 1
Common error: Writing \(3^{2x+1}\) as \((3^x)^2 + 3\) or \(3^{2x} \times 1\). Use the index law \(3^{2x+1} = 3^{2x} \times 3^1\).
29IBMA2S-046-03-012.2 Functions, domain, range and inversePaper 1Easy4 marks
No calculator
A smart thermostat converts a temperature \(c\) in degrees Celsius into degrees Fahrenheit using \(F(c)=\frac{9}{5}c+32\). The temperature sensor inside the thermostat outputs a voltage of \(r\) volts, and the temperature in degrees Celsius is given by \(c(r)=25r-10\).
(a)
Find an expression for \((F\circ c)(r)\), giving your answer in the form \(ar+b\).
(2)
(b)
The thermostat displays a temperature of 95 °F. Find the voltage output by the sensor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(c(r)\) into \(F\): \(\frac{9}{5}(25r-10)+32\)
✓ 1
A1 \((F\circ c)(r)=45r+14\)
✓ 1
Part (b)
M1 sets \(45r+14=95\), or equivalent method, e.g. converts 95 °F to 35 °C and sets \(25r-10=35\)
✓ 1
A1 \(r=\frac{9}{5}\) V \(=1.8\) V
✓ 1
Exam tip: In \((F\circ c)(r)\) the function written nearest to the variable, \(c\), acts first. Substitute the whole of \(c(r)\) into \(F\), keeping the brackets.
30IBMA2S-046-03-022.2 Functions, domain, range and inversePaper 1Easy6 marks
No calculator
A ride-sharing app charges a fare of \(F(d)=2d+3\) dollars for a journey of \(d\) kilometres.
(a)
Find \(F^{-1}(x)\). Hence find \(F^{-1}(19)\) and interpret its meaning in this context.
(3)
(b)
During a promotion, every fare is reduced by 20%, so a fare of \(x\) dollars becomes \(D(x)=0.8x\). Find \((D\circ F)(d)\) and hence find the length of a journey that costs 20 dollars during the promotion.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearranges \(x=2d+3\) to make \(d\) the subject
✓ 1
A1 \(F^{-1}(x)=\frac{x-3}{2}\)
✓ 1
A1 \(F^{-1}(19)=8\): a fare of 19 dollars pays for an 8 km journey
✓ 1
Part (b)
M1 forms \(D(F(d))=0.8(2d+3)\)
✓ 1
A1 \((D\circ F)(d)=1.6d+2.4\)
✓ 1
A1 \(1.6d+2.4=20 \Rightarrow d=11\) km
✓ 1
Common error: \(F^{-1}(19)\) is not \(\frac{1}{F(19)}\). The inverse undoes the function: it takes a fare as input and gives back a distance.
31IBMA2S-046-03-032.5 Composite and inverse functionsPaper 1Easy5 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=\sqrt{x-2}\) for \(x\ge 2\) and \(g(x)=2x-4\) for \(x\in\mathbb{R}\).
(a)
Find an expression for \((f\circ g)(x)\) and state the largest possible domain of \(f\circ g\).
(3)
(b)
Solve \((f\circ g)(x)=4\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(g(x)\) into \(f\): \(\sqrt{(2x-4)-2}\)
✓ 1
A1 \((f\circ g)(x)=\sqrt{2x-6}\)
✓ 1
A1 domain \(x\ge 3\)
✓ 1
Part (b)
M1 squares both sides: \(2x-6=16\)
✓ 1
A1 \(x=11\)
✓ 1
Exam tip: The domain of a composite is found by requiring the inner output to lie in the outer function's domain: here \(2x-4\ge 2\), giving \(x\ge 3\), which is not the same as the domain of \(g\).
32IBMA2S-046-04-012.10 Solving equationsPaper 1Easy6 marks
No calculator
A patient is given an anaesthetic by a continuous intravenous infusion. The concentration of the anaesthetic in the patient's bloodstream is modelled by \(C(t)=\frac{8t}{t+4}\), where \(C\) is measured in mg L−1 and \(t\) is the time in hours after the infusion starts, \(t\ge 0\).
(a)
Find the concentration 4 hours after the infusion starts.
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(C\) and interpret it in context.
(2)
(c)
Find the time at which the concentration reaches 6 mg L−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(t=4\): \(\frac{8\times4}{4+4}\)
✓ 1
A1 \(C=4\) mg L−1
✓ 1
Part (b)
A1 \(C=8\)
✓ 1
R1 in the long term the concentration approaches, but never reaches, 8 mg L−1
✓ 1
Part (c)
M1 sets \(\frac{8t}{t+4}=6\) and rearranges, e.g. \(8t=6t+24\)
✓ 1
A1 \(t=12\) hours
✓ 1
Exam tip: For \(\frac{at+b}{ct+d}\) the horizontal asymptote is \(y=\frac{a}{c}\). In a context, always say what the limiting value means for the real quantity.
33IBMA2S-046-04-022.8 Reciprocal and rational functionsPaper 1Easy5 marks
No calculator
Let \(f(x)=\frac{2x+5}{x-3}\), \(x\ne 3\).
(a)
Write down the equations of the vertical and horizontal asymptotes of the graph of \(f\).
(2)
(b)
Find the coordinates of the points where the graph of \(f\) meets the axes.
Common error: The \(y\)-intercept is \(f(0)=\frac{5}{-3}\), not \(\frac{5}{3}\): keep track of the sign of the denominator.
34IBMA2S-046-04-032.5 Composite and inverse functionsPaper 1Easy6 marks
No calculator
Let \(g(x)=\frac{4}{x+1}-2\), \(x\ne -1\).
(a)
Find \(g^{-1}(x)\).
(3)
(b)
Hence find the value of \(x\) for which \(g(x)=6\).
(2)
(c)
Write down the range of \(g^{-1}\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 interchanges \(x\) and \(y\) or attempts to make \(x\) the subject of \(y=\frac{4}{x+1}-2\)
✓ 1
A1 correct intermediate step, e.g. \(x+1=\frac{4}{y+2}\)
✓ 1
A1 \(g^{-1}(x)=\frac{4}{x+2}-1\) (accept equivalent forms such as \(\frac{2-x}{x+2}\); stating \(x\ne -2\) is not required)
✓ 1
Part (b)
M1 substitutes 6 into their \(g^{-1}\), e.g. \(\frac{4}{6+2}-1\), or solves \(\frac{4}{x+1}-2=6\) directly
✓ 1
A1 \(x=-\frac{1}{2}\)
✓ 1
Part (c)
A1 \(g^{-1}(x)\ne -1\), i.e. all real numbers except \(-1\) (accept \(y\ne -1\))
✓ 1
Exam tip: The range of \(g^{-1}\) is the domain of \(g\), so the excluded value \(-1\) can be read off straight from the original function.
35IBMA2S-046-04-042.10 Solving equationsPaper 1Easy5 marks
No calculator
A triathlete cycles a 30 km stage at a constant speed of \(v\) km h−1, and then spends 0.5 hours in the transition area. The total time, \(T\) hours, for the stage and transition is modelled by \(T(v)=\frac{30}{v}+0.5\), \(v>0\).
(a)
Find \(T(25)\).
(1)
(b)
Write down the equation of the horizontal asymptote of the graph of \(T\) and explain its meaning in context.
(2)
(c)
Find the speed needed for the total time to be 2 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(T=\frac{30}{25}+0.5=1.2+0.5=1.7\) hours
✓ 1
Part (b)
A1 \(T=0.5\)
✓ 1
R1 as the cycling speed increases, the total time approaches (but never reaches) 0.5 hours, the transition time (accept equivalent wording)
✓ 1
Part (c)
M1 sets \(\frac{30}{v}+0.5=2\), so \(\frac{30}{v}=1.5\)
✓ 1
A1 \(v=20\) km h−1
✓ 1
Common error: Subtract the 0.5 h before inverting; writing \(v=\frac{30}{2}\) ignores the transition time.
36IBMA2S-046-07-012.1 Equations of straight linesPaper 1Easy7 marks
No calculator
A delivery drone flies in a straight line on a coordinate grid centred on its control tower. Distances are in kilometres. The drone passes through the points P(−3, 7) and Q(5, −5).
Diagram not to scale
(a)
Find the gradient of the drone's flight path.
(2)
(b)
Find the equation of the flight path, giving your answer in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(3)
(c)
The drone crosses the x-axis at the point R. Find the exact x-coordinate of R.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\frac{y_2-y_1}{x_2-x_1}\), e.g. \(\frac{-5-7}{5-(-3)}\)
✓ 1
A1 gradient \(=-\frac{3}{2}\)
✓ 1
Part (b)
M1 substitutes a point and the gradient into \(y-y_1=m(x-x_1)\), e.g. \(y-7=-\frac{3}{2}(x+3)\)
✓ 1
M1 rearranges and clears fractions, e.g. \(2y-14=-3x-9\)
✓ 1
A1 \(3x+2y-5=0\) (or any integer multiple)
✓ 1
Part (c)
M1 sets \(y=0\) in their equation, e.g. \(3x-5=0\)
✓ 1
A1 \(x=\frac{5}{3}\)
✓ 1
Common error: Subtracting a negative coordinate is where marks slip: 5 − (−3) = 8, not 2. Write the brackets in before simplifying.
37IBMA2S-046-07-032.1 Equations of straight linesPaper 1Easy5 marks
No calculator
On a map grid measured in kilometres, a railway track lies along the line \(L_1\) with equation \(4x-6y+9=0\). A new access road is to be built along a line \(L_2\) that crosses the track at right angles and passes through the station S(4, −1).
(a)
Find the gradient of \(L_1\).
(2)
(b)
Find the equation of \(L_2\) in the form \(y=mx+c\).
Common error: Reading the gradient straight off 4x − 6y + 9 = 0 as 4 or −4/6 is wrong: make y the subject first, which gives +2/3.
38IBMA2S-046-06-022.11 Transformations of graphsPaper 1Easy5 marks
No calculator
A fitness watch displays a curve \(y = f(x)\). After a software update, the watch displays the recalibrated curve \(y = 3f(x-1) + 4\). The point \(P(6,\ 19)\) lies on the recalibrated curve.
(a)
Find the value of \(f(5)\).
(2)
(b)
Write down the coordinates of the corresponding point on the graph of \(y = f(x)\).
(1)
(c)
Find the coordinates of the image of this point on the graph of \(y = -f(2x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(19 = 3f(6-1) + 4\)
✓ 1
A1 \(f(5) = 5\)
✓ 1
Part (b)
A1 \((5,\ 5)\)
✓ 1
Part (c)
M1 recognises that \(2x = 5\), i.e. the \(x\)-coordinate is halved and the \(y\)-coordinate changes sign
✓ 1
A1 \((2.5,\ -5)\)
✓ 1
Exam tip: To undo a transformation, reverse the operations: subtract 4, then divide by 3 for the output, and add 1 to the input.
39IBMA2S-046-06-092.2 Functions, domain, range and inversePaper 1Easy4 marks
No calculator
A training app models an athlete's power output by a function \(f\) with domain \(-2 \le x \le 6\) and range \(1 \le f(x) \le 9\). A comparison curve is given by \(g(x) = 3 - 2f(x-4)\).
(a)
Find the domain of \(g\).
(2)
(b)
Find the range of \(g\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(-2 \le x - 4 \le 6\)
✓ 1
A1 \(2 \le x \le 10\)
✓ 1
Part (b)
M1 \(-18 \le -2f(x-4) \le -2\), or evaluates \(3 - 2 \times 1\) and \(3 - 2 \times 9\)
✓ 1
A1 \(-15 \le g(x) \le 1\)
✓ 1
Common error: Multiplying by \(-2\) reverses the inequality. The largest value of \(f\) gives the smallest value of \(g\).
40IBMA2S-046-05-012.11 Transformations of graphsPaper 1Easy6 marks
No calculator
The function f is defined by f(x) = ln(x − 2) + 1, for x > 2.
(a)
Write down the equation of the vertical asymptote of the graph of f.
(1)
(b)
Find the exact value of the x-intercept of the graph of f.
M1 interchanging x and y and rearranging, e.g. x − 1 = ln(y − 2) so y − 2 = ex−1 ; A1 f−1(x) = ex−1 + 2
✓ 2
Part (d)
A1 x ∈ ℝ (the range of f is all real numbers)
✓ 1
Exam tip: The domain of an inverse function is the range of the original function. A logarithm with domain x > 2 still takes every real value, so the inverse is defined for all real x.
41IBMA2S-046-05-032.2 Functions, domain, range and inversePaper 1Easy4 marks
No calculator
Solve the equation log2x + log2(x − 2) = 3.
(a)
Solve the equation, justifying the rejection of any value.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 using the product law: log2(x(x − 2)) = 3 ; M1 removing logarithms: x(x − 2) = 23 ; A1 x2 − 2x − 8 = 0, so (x − 4)(x + 2) = 0 ; R1 x = −2 rejected because log2 of a negative number is undefined, so x = 4 only
✓ 4
Exam tip: Always check solutions in the original equation: every logarithm's argument must be positive. A brief reason for rejecting a value is needed for the reasoning mark.
42IBMA2S-046-00-012.8 Reciprocal and rational functionsPaper 1Easy5 marks
No calculator
The control software for a robotic arm chooses a joint setting \(x\) by solving the equation \(2x^2-4x+c=0\), where \(c\) is a constant. The software works correctly only when the equation has exactly one solution.
(a)
Find the value of \(c\) for which the equation has exactly one solution.
(3)
(b)
Hence find the joint setting \(x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of discriminant \(b^2-4ac=0\)
✓ 1
A1 \(16-8c=0\)
✓ 1
A1 \(c=2\)
✓ 1
Part (b)
M1 substitutes \(c=2\) and solves, e.g. \(2(x-1)^2=0\) or \(x=\frac{4}{2\times2}\)
✓ 1
A1 \(x=1\)
✓ 1
Exam tip: "Exactly one solution" for a quadratic means a repeated root, so set the discriminant equal to zero, not greater than zero.
43IBMA2S-046-00-032.8 Reciprocal and rational functionsPaper 1Easy4 marks
No calculator
In a computer game, a laser beam travels along the line \(y=2x+1\) and a curved force-field follows the curve \(y=x^2-2x+4\). Find the coordinates of the points where the laser beam meets the force-field.
(a)
Find the coordinates of the points of intersection.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates \(x^2-2x+4=2x+1\)
✓ 1
A1 \(x^2-4x+3=0\)
✓ 1
A1 \(x=1\), \(x=3\)
✓ 1
A1 \((1,\,3)\) and \((3,\,7)\)
✓ 1
Exam tip: After finding the x-values, substitute into the line (the simpler equation) to get the y-coordinates, and give the answer as coordinate pairs.
44IBMA2S-046-00-042.6 Quadratic functionsPaper 1Easy5 marks
No calculator
The graph of a quadratic function \(f\) has its vertex at \((3,\,-8)\) and passes through the point \((5,\,0)\). The function can be written as \(f(x)=a(x-3)^2-8\).
(a)
Find the value of \(a\).
(2)
(b)
Write down the other \(x\)-intercept of the graph of \(f\).
(1)
(c)
Find the \(y\)-intercept of the graph of \(f\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \((5,0)\): \(0=a(5-3)^2-8\)
✓ 1
A1 \(a=2\)
✓ 1
Part (b)
A1 \(x=1\) (by symmetry about \(x=3\))
✓ 1
Part (c)
M1 evaluates \(f(0)=2(0-3)^2-8\)
✓ 1
A1 \(y=10\)
✓ 1
Exam tip: The axis of symmetry passes through the vertex, so the x-intercepts are equal distances either side of it; this saves solving an equation.
45IBMA3S-2-17r11-012.8 Reciprocal and rational functionsPaper 1Easy6 marks
No calculator
The signal-timing correction of a space probe, in seconds, is modelled by \(f(x)=\frac{3x+2}{x-4}\), \(x\neq 4\), where \(x\) is the probe's distance from Earth in suitable units.
(a)
Write down the equation of (i) the vertical asymptote and (ii) the horizontal asymptote of the graph of \(y=f(x)\).
(2)
(b)
Find \(f^{-1}(x)\).
(3)
(c)
Find \(f^{-1}(5)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical asymptote \(x=4\)
✓ 1
A1 horizontal asymptote \(y=3\)
✓ 1
Part (b)
M1 swap variables and multiply out: \(x(y-4)=3y+2\)
✓ 1
M1 collect \(y\) terms: \(y(x-3)=4x+2\)
✓ 1
A1 \(f^{-1}(x)=\frac{4x+2}{x-3}\)
✓ 1
Part (c)
A1 \(f^{-1}(5)=\frac{4(5)+2}{5-3}=11\) (ECF from (b))
✓ 1
Common error: The horizontal asymptote of a rational function comes from the ratio of the leading coefficients, not from the constants.
46IBMA3S-2-17r11-022.7 Quadratic equations and the discriminantPaper 1Easy4 marks
No calculator
The profile of a footbridge support beam is modelled by \(y=x^2-kx+9\), where \(x\) and \(y\) are in metres and \(k>0\) is a constant. The beam just touches the horizontal deck line \(y=0\) at exactly one point.
(a)
Find the value of \(k\).
(3)
(b)
Find the \(x\)-coordinate of the point where the beam touches the deck line.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use discriminant \(=0\): \((-k)^2-4\times1\times9=0\)
✓ 1
A1 \(k^2=36\)
✓ 1
A1 \(k=6\) since \(k>0\)
✓ 1
Part (b)
A1 \(x^2-6x+9=(x-3)^2=0\) so \(x=3\) (or \(x=-\frac{b}{2a}=3\))
✓ 1
Exam tip: One repeated root means the discriminant equals zero; remember to use the extra condition (here \(k>0\)) to reject the negative value.
47IBMA3S-2-17r11-042.6 Quadratic functionsPaper 1Easy5 marks
No calculator
The height above the deck, in metres, of a bridge arch at a horizontal distance \(x\) metres from the left pier is \(h(x)=9-0.25(x-6)^2\) for \(0\leq x\leq 12\).
(a)
Write down the maximum height of the arch and the value of \(x\) at which it occurs.
(2)
(b)
Write down the range of \(h\).
(1)
(c)
Find the value of \(x\), greater than 6, at which the height is 5 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum height 9 m
✓ 1
A1 at \(x=6\)
✓ 1
Part (b)
A1 \(0\leq h(x)\leq 9\) (since \(h(0)=h(12)=0\))
✓ 1
Part (c)
M1 \(0.25(x-6)^2=4\) so \((x-6)^2=16\)
✓ 1
A1 \(x=10\) (reject \(x=2\) as it is less than 6)
✓ 1
Common error: When solving \((x-6)^2=16\) both \(x=2\) and \(x=10\) satisfy it; read the question to decide which one is wanted.
48IBMA3S-2-16r17-012.5 Composite and inverse functionsPaper 1Easy7 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=3x-5\) and \(g(x)=x^2+1\), for \(x\in\mathbb{R}\).
(a)
Find \(f^{-1}(x)\).
(2)
(b)
Find \((g\circ f)(x)\), giving your answer in the form \(ax^2+bx+c\).
(2)
(c)
Given that \((f\circ g)(x)=25\), find the positive value of \(x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to swap \(x\) and \(y\) and rearrange, or write \(y=3x-5 \Rightarrow x=\frac{y+5}{3}\)
✓ 1
A1 \(f^{-1}(x)=\frac{x+5}{3}\)
✓ 1
Part (b)
M1 \(g(f(x))=(3x-5)^2+1\)
✓ 1
A1 \(9x^2-30x+26\)
✓ 1
Part (c)
M1 \(f(g(x))=3(x^2+1)-5=3x^2-2\)
✓ 1
M1 \(3x^2-2=25 \Rightarrow x^2=9\)
✓ 1
A1 \(x=3\)
✓ 1
Common error: \((f\circ g)(x)\) means apply \(g\) first, then \(f\); reversing the order gives \((g\circ f)(x)\), a different function.
49IBMA3S-2-16r17-022.5 Composite and inverse functionsPaper 1Easy7 marks
No calculator
Solve the following equations.
(a)
Solve \(3^{2x-1}=81\).
(3)
(b)
Solve \(\log_2(x+3)+\log_2(x-3)=4\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(81=3^4\)
✓ 1
M1 \(2x-1=4\)
✓ 1
A1 \(x=\frac{5}{2}\)
✓ 1
Part (b)
M1 uses \(\log a+\log b=\log ab\) to obtain \(\log_2\big((x+3)(x-3)\big)=4\)
✓ 1
M1 \(x^2-9=2^4=16\)
✓ 1
A1 \(x=\pm5\)
✓ 1
R1 rejects \(x=-5\) since \(\log_2(-2)\) is undefined, so \(x=5\)
✓ 1
Exam tip: After solving a logarithmic equation, check every solution in the original equation, because the argument of a log must be positive.
50IBMA3S-2-16r17-132.9 Exponential and logarithmic functionsPaper 1Easy4 marks
No calculator
Answer the following question.
(a)
Solve \(\ln(x+2)=2\ln3+\ln4-\ln6\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2\ln3=\ln9\)
✓ 1
A1 \(\ln\frac{9\times4}{6}=\ln6\)
✓ 1
M1 \(x+2=6\)
✓ 1
A1 \(x=4\)
✓ 1
Common error: \(2\ln3\) equals \(\ln9\), not \(\ln6\); the coefficient becomes a power of the argument.
51IBMA3S-2-17r17-012.2 Functions, domain, range and inversePaper 1Easy6 marks
No calculator
A pharmacist estimates the dose, d mg, of a medicine for a patient of mass m kg using the rule d(m) = 0.5m + 20, for 40 ≤ m ≤ 100.
(a)
Find the range of d.
(2)
(b)
Find an expression for the inverse function d−1(x).
(2)
(c)
A patient is prescribed a dose of 55 mg. Use the inverse function to find the mass of this patient.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evaluates d(40) and d(100): 40 and 70
✓ 1
A1 range is 40 ≤ d ≤ 70
✓ 1
Part (b)
M1 writes x = 0.5y + 20 and attempts to make y the subject
✓ 1
A1 d−1(x) = 2x − 40
✓ 1
Part (c)
M1 substitutes 55 into the inverse function: 2 × 55 − 40
✓ 1
A1 70 kg
✓ 1
Common error: When finding an inverse, swap x and y before rearranging, and check the answer by substituting back into the original rule.
52IBMA3S-2-17r17-032.4 Key features and intersectionsPaper 1Easy6 marks
No calculator
The depth of water, h metres, in a reservoir t hours after a valve is opened is modelled by h(t) = −0.5t2 + 3t + 3.5, for t ≥ 0.
(a)
Write h(t) in the form a(t − p)2 + q.
(3)
(b)
State the maximum depth of the water and the time at which it occurs.
(1)
(c)
The reservoir is empty when h = 0. Find the time at which this occurs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts to complete the square or finds the vertex using t = −b/2a
✓ 1
A1 a = −0.5 and p = 3
✓ 1
A1 q = 8, so h = −0.5(t − 3)2 + 8
✓ 1
Part (b)
A1 maximum depth 8 m at t = 3 hours (ECF from (a))
Common error: Take care with signs when completing the square on a quadratic whose leading coefficient is negative: factor out −0.5 from both the t² and t terms first.
53IBMA3S-2-17r26-012.1 Equations of straight linesPaper 1Easy7 marks
No calculator
On a wind farm map, distances are in kilometres. Turbine A is at (2, 5) and turbine B is at (8, 17). A buried cable runs in a straight line along AB. The substation is at S(10, 1).
Diagram not to scale
(a)
Find the gradient of the line AB.
(2)
(b)
Find the equation of the line AB, giving your answer in the form y = mx + c.
(2)
(c)
A second cable runs from S to meet AB at right angles. Find the y-intercept of the line containing this second cable.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(\frac{y_2-y_1}{x_2-x_1}\)
✓ 1
A1 gradient = 2
✓ 1
Part (b)
M1 substitute a point and the gradient into a correct line form
✓ 1
A1 y = 2x + 1
✓ 1
Part (c)
A1 perpendicular gradient = −1⁄2
✓ 1
M1 substitute S(10, 1) into y = −1⁄2x + c or equivalent
✓ 1
A1 y-intercept = 6
✓ 1
Exam tip: Perpendicular gradients multiply to −1, so invert the gradient and change its sign.
54IBMA3S-2-17r26-032.5 Composite and inverse functionsPaper 1Easy6 marks
No calculator
A library runs a volunteer scheme. A volunteer who works x hours earns f(x) = 2x − 3 points. A number of points p can be exchanged for a voucher worth g(p) = p2 + 1 pence.
(a)
Find (g ∘ f)(4).
(2)
(b)
Find (f ∘ g)(x), giving your answer in the form ax2 + b.
(2)
(c)
Solve (f ∘ g)(x) = 17, given that x > 0.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find f(4) = 5 and then g(5)
✓ 1
A1 26
✓ 1
Part (b)
M1 substitute g(x) into f, e.g. 2(x2 + 1) − 3
✓ 1
A1 2x2 − 1
✓ 1
Part (c)
M1 2x2 − 1 = 17 leading to x2 = 9
✓ 1
A1 x = 3
✓ 1
Common error: In (g ∘ f)(x) the function f is applied first, so work from the inside outwards.
55IBMA3S-2-19r28-012.4 Key features and intersectionsPaper 1Easy5 marks
No calculator
A shop applies a 20% discount to the marked price of a jacket and then takes €6 off the discounted price using a voucher. Let p be the marked price in euros. The discount is modelled by \(d(p)=0.8p\) and the voucher by \(v(x)=x-6\).
(a)
Find an expression for \((v\circ d)(p)\). [2]
(2)
(b)
Find \((v\circ d)^{-1}(x)\). [2]
(2)
(c)
A customer pays €34 after both reductions. Find the marked price of the jacket. [1]
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(d(p)\) into \(v\), i.e. \(v(0.8p)\)
Common error: In \(v\circ d\) the function nearest the variable, \(d\), is applied first. Writing \(d(v(p))\) instead gives \(0.8(p-6)\), which is a different price.
56IBMA3S-2-19r28-032.8 Reciprocal and rational functionsPaper 1Easy5 marks
No calculator
Consider the function \(f(x)=\frac{2x+1}{x-3}\), where \(x\neq 3\).
(a)
Write down the equation of the vertical asymptote of the graph of \(y=f(x)\). [1]
(1)
(b)
Write down the equation of the horizontal asymptote. [1]
(1)
(c)
Find the coordinates of the y-intercept. [1]
(1)
(d)
Write down the range of \(f\). [1]
(1)
(e)
Find the x-coordinate of the point where the graph crosses the x-axis. [1]
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=3\)
✓ 1
Part (b)
A1 \(y=2\)
✓ 1
Part (c)
A1 \(\left(0,-\frac{1}{3}\right)\)
✓ 1
Part (d)
A1 \(f(x)\in\mathbb{R}\), \(f(x)\neq 2\)
✓ 1
Part (e)
A1 \(2x+1=0\), so \(x=-\frac{1}{2}\)
✓ 1
Exam tip: For \(\frac{ax+b}{cx+d}\) the horizontal asymptote is \(y=\frac{a}{c}\), and the range excludes exactly this value.
57IBMA3S-2-19r28-042.7 Quadratic equations and the discriminantPaper 1Easy6 marks
No calculator
Consider the equation \(\log_2 x+\log_2(x-2)=3\).
(a)
Show that \(x^2-2x-8=0\). [3]
(3)
(b)
Hence find the value of \(x\) that satisfies the original equation. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\log_2 x+\log_2(x-2)=\log_2\big(x(x-2)\big)\)
✓ 1
M1 write \(x(x-2)=2^3\)
✓ 1
A1 \(x^2-2x-8=0\) (AG)
✓ 1
Part (b)
M1 solve \((x-4)(x+2)=0\) to get \(x=4\) or \(x=-2\)
✓ 1
R1 reject \(x=-2\) because \(\log_2(-2)\) is undefined
✓ 1
A1 \(x=4\)
✓ 1
Common error: Always check solutions of a logarithmic equation against the domain of the original logarithms. A root that makes an argument negative must be rejected.
58IBMA3S-2-17r39-012.6 Quadratic functionsPaper 1Easy6 marks
No calculator
A volcanologist lowers a probe into a gas vent and records the temperature \(T\) °C at a depth of \(d\) metres below the rim. At a depth of 20 m the temperature is 330 °C and at a depth of 60 m it is 410 °C. The relationship between \(T\) and \(d\) is linear.
(a)
Find the gradient of the line, stating its units.
(2)
(b)
Find an equation for \(T\) in terms of \(d\).
(2)
(c)
Find the depth at which the temperature is 530 °C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(\frac{410-330}{60-20}\)
✓ 1
A1 gradient \(=2\) °C per metre
✓ 1
Part (b)
M1 substitute a point and the gradient into \(T=2d+c\)
✓ 1
A1 \(T=2d+290\)
✓ 1
Part (c)
M1 \(530=2d+290\)
✓ 1
A1 \(d=120\) m
✓ 1
Exam tip: Check your line by substituting the other data point: here \(2(20)+290=330\).
59IBMA3S-2-17r39-042.5 Composite and inverse functionsPaper 1Easy7 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=3x-2\) and \(g(x)=x^2+1\), for \(x\in\mathbb{R}\).
(a)
Find \((g\circ f)(2)\).
(2)
(b)
Find an expression for \((f\circ g)(x)\).
(2)
(c)
Solve \((g\circ f)(x)=10\), given that \(x>0\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f(2)=4\)
✓ 1
A1 \(g(4)=17\)
✓ 1
Part (b)
M1 \(3(x^2+1)-2\)
✓ 1
A1 \(3x^2+1\)
✓ 1
Part (c)
M1 \((3x-2)^2+1=10\), so \((3x-2)^2=9\)
✓ 1
M1 \(3x-2=\pm3\)
✓ 1
A1 \(x=\frac53\) (the value \(x=-\frac13\) is rejected as \(x>0\))
✓ 1
Common error: \((g\circ f)(x)\) means apply \(f\) first, then \(g\); writing it the other way round gives \(3x^2+1\) instead.
60IBMA3S-2-19r40-012.1 Equations of straight linesPaper 1Easy4 marks
No calculator
Two marker buoys, A and B, are anchored in a bay. Relative to the harbour, and using units of kilometres, A has coordinates (−2, 7) and B has coordinates (4, −5). The line L passes through A and B.
Diagram not to scale
(a)
Find the gradient of L.
(2)
(b)
Find the equation of L in the form \(ax+by+d=0\), where \(a\), \(b\) and \(d\) are integers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\frac{y_2-y_1}{x_2-x_1}\) with correct substitution, e.g. \(\frac{-5-7}{4-(-2)}\)
✓ 1
A1 gradient = −2
✓ 1
Part (b)
M1 substitutes a point and the gradient into \(y-y_1=m(x-x_1)\) or \(y=mx+c\), e.g. \(y-7=-2(x+2)\) or \(7=-2(-2)+c\)
✓ 1
A1 \(2x+y-3=0\) (or any integer multiple)
✓ 1
Common error: When the gradient uses coordinates in the wrong order, e.g. (−2 − 4) in the denominator but (−5 − 7) in the numerator, the sign of the gradient is reversed. Subtract in the same order top and bottom.
61IBMA3S-2-19r40-052.11 Transformations of graphsPaper 1Easy4 marks
No calculator
A computer model of a curved roof uses the graph of \(y=f(x)\). The point P(4, −6) lies on this graph.
(a)
The roof is redesigned so that its profile is the graph of \(y=f(x-3)+2\). Find the coordinates of the image of P.
(2)
(b)
A second design uses the graph of \(y=\frac{1}{2}f(2x)\). Find the coordinates of the image of P.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 one coordinate correct (7 or −4)
✓ 1
A1 (7, −4)
✓ 1
Part (b)
M1 one coordinate correct (2 or −3)
✓ 1
A1 (2, −3)
✓ 1
Common error: A horizontal stretch or compression acts on x by the reciprocal of the factor inside the bracket: \(f(2x)\) halves the x-coordinate, it does not double it.
62IBMA3S-2-19r40-092.6 Quadratic functionsPaper 1Easy4 marks
No calculator
The profile of a parabolic arch over a walkway is modelled by \(h=-x^2+8x\), where \(h\) is the height of the arch in metres above the ground and \(x\) is the horizontal distance in metres from the left foot of the arch, for \(0\le x\le8\).
(a)
Find the width of the arch at ground level.
(2)
(b)
Find the maximum height of the arch.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(h=0\), so \(x(8-x)=0\) (or uses symmetry about the vertex)
✓ 1
A1 width = 8 m
✓ 1
Part (b)
M1 uses the axis of symmetry \(x=4\) (or completes the square, \(h=16-(x-4)^2\))
✓ 1
A1 maximum height = 16 m
✓ 1
Exam tip: The vertex of a quadratic with two roots lies halfway between them. Use that to find the axis of symmetry without differentiating.
63AASL2-q00032.1 Equations of straight linesPaper 1Medium8 marks
The points \(\mathrm{P}(-3,\,2)\) and \(\mathrm{Q}(5,\,-4)\) are shown on a coordinate plane. The line \(L\) is the perpendicular bisector of \([\mathrm{PQ}]\).
Diagram not to scale
(a)
Find the coordinates of the midpoint of \([\mathrm{PQ}]\).
(1)
(b)
Find the gradient of \([\mathrm{PQ}]\).
(1)
(c)
Find the equation of \(L\), giving your answer in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(3)
(d)
Find the \(y\)-intercept of \(L\).
(1)
(e)
Show that the point \(\mathrm{R}(4,\,3)\) lies on \(L\), and hence explain why \(\mathrm{RP}=\mathrm{RQ}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\((1,\,-1)\)
A1
(b)
\(\dfrac{-4-2}{5-(-3)}=-\dfrac{3}{4}\)
A1
(c)
perpendicular gradient \(\dfrac{4}{3}\)
(M1)
FT: \(-\dfrac{1}{m}\) for their \(m\) in (b).
\(y+1=\tfrac{4}{3}(x-1)\)
(M1)
Substituting their midpoint into a line equation with their perpendicular gradient.
\(4x-3y-7=0\)
A1
Accept any integer multiple.
(d)
\(-\dfrac{7}{3}\)
A1
Accept \(\left(0,\,-\tfrac73\right)\).
(e)
\(4(4)-3(3)-7=16-9-7=0\)
A1
every point on the perpendicular bisector of \([\mathrm{PQ}]\) is the same distance from P and from Q
R1
Accept a calculation showing \(\mathrm{RP}=\mathrm{RQ}=\sqrt{50}\).
64AASL2-q00092.2 Functions, domain, range and inversePaper 1Medium8 marks
The following diagram shows the graph of \(y=f(x)\), for \(-2\le x\le3\). The graph passes through the points \((-2,\,-2.75)\), \((0,\,-2)\), \((2,\,1)\) and \((3,\,5)\). The line \(y=x\) is also shown.
(a)
Write down the value of \(f(2)\).
(1)
(b)
Write down the value of \(f^{-1}(-2)\).
(1)
(c)
Explain why \(f\) has an inverse.
(1)
(d)(i)
Write down the domain of \(f^{-1}\).
(1)
(d)(ii)
Write down the range of \(f^{-1}\).
(1)
(e)
On the same grid, sketch the graph of \(y=f^{-1}(x)\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(1\)
A1
(b)
\(0\)
A1
(c)
\(f\) is one-to-one (each \(y\)-value comes from exactly one \(x\)-value; the graph is increasing)
R1
Accept a correct horizontal-line-test argument.
(d)(i)
\(-2.75\le x\le5\)
A1
(d)(ii)
\(-2\le y\le3\)
A1
Accept \(-2\le f^{-1}(x)\le3\).
(e)
attempt to reflect the graph of \(f\) in the line \(y=x\)
M1
curve passes through \((-2,\,0)\) and \((1,\,2)\), increasing and concave down
65AASL2-q00112.2 Functions, domain, range and inversePaper 1Medium9 marks
The following diagram shows the graph of \(y=f(x)\), for \(-4\le x\le5\). The graph consists of three line segments joining the points \((-4,\,1)\), \((-2,\,-3)\), \((1,\,3)\) and \((5,\,1)\).
(a)
Write down the value of \(f(-2)\).
(1)
(b)
Write down the range of \(f\).
(2)
(c)
Solve \(f(x)=1\).
(2)
(d)
Explain why \(f\) does not have an inverse.
(1)
(e)(i)
The function \(g\) is defined by \(g(x)=f(x)\), for \(-2\le x\le1\).
Write down the domain of \(g^{-1}\).
(1)
(e)(ii)
Write down the range of \(g^{-1}\).
(1)
(f)
Find the value of \(g^{-1}(2)\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(-3\)
A1
(b)
end values \(-3\) and \(3\)
A1
\(-3\le f(x)\le3\)
A1
(c)
drawing \(y=1\), or using the middle segment \(y=2x+1\)
(M1)
\(x=-4,\ x=0,\ x=5\)
A1
All three values required.
(d)
\(f\) is not one-to-one, e.g. \(f(-4)=f(0)=f(5)=1\)
71AASL2-q00402.6 Quadratic functionsPaper 1Medium8 marks
The following diagram shows part of the graph of \(f(x)=x^{2}+bx+c\). The graph crosses the \(x\)-axis at \(\mathrm{A}(1,\,0)\) and \(\mathrm{B}(5,\,0)\) and has vertex V. The line \(L\) passes through A and V.
Diagram not to scale
(a)
Find the value of \(b\) and the value of \(c\).
(2)
(b)
Find the coordinates of V.
(2)
(c)
Find the equation of \(L\).
(2)
(d)
The line \(M\) is parallel to \(L\) and passes through B. Find the \(y\)-intercept of \(M\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f(x)=(x-1)(x-5)\)
(M1)
Or substituting both points into f(x) = x² + bx + c.
72AASL2-q00412.6 Quadratic functionsPaper 1Medium7 marks
The graph of a quadratic function \(f\) has axis of symmetry \(x=3\) and passes through the points \((6,\,15)\) and \((4,\,-9)\). Let \(f(x)=a(x-3)^{2}+k\).
(a)
Find the value of \(a\) and the value of \(k\).
(3)
(b)
Write \(f(x)\) in the form \(a(x-p)(x-q)\).
(2)
(c)
Write \(f(x)\) in the form \(ax^{2}+bx+c\).
(1)
(d)
Write down the \(y\)-intercept of the graph of \(f\).
77AASL2-q00602.8 Reciprocal and rational functionsPaper 1Medium9 marks
The following diagram shows the graph of \(f(x)=\dfrac{x+p}{q-x}\), for \(x\neq q\), where \(p,q\in\mathbb{R}\). The graph has asymptotes \(x=2\) and \(y=-1\), and passes through the points \((-2,\,0)\) and \((0,\,1)\).
(a)
Find the value of \(p\) and the value of \(q\).
(2)
(b)
Solve \(f(x)=3\).
(2)
(c)
Hence, using the graph, solve \(f(x)\gt 3\).
(2)
(d)
Find \(f^{-1}(x)\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(q=2\)
A1
From the vertical asymptote.
\(p=2\)
A1
From \(f(-2)=0\); check \(f(0)=\tfrac{2}{2}=1\).
(b)
\(x+2=3(2-x)\)
M1
\(x=1\)
A1
(c)
recognizing the graph lies above \(y=3\) between \(x=1\) and the vertical asymptote
78AASL2-q00612.11 Transformations of graphsPaper 1Medium7 marks
Let \(f(x)=\dfrac{2x+1}{x-3}\), for \(x\neq3\).
(a)
Write down the equations of the asymptotes of the graph of \(f\).
(2)
(b)
Show that \(f(x)=2+\dfrac{7}{x-3}\).
(2)
(c)
Hence describe a sequence of transformations that maps the graph of \(y=\dfrac{1}{x}\) onto the graph of \(y=f(x)\).
(2)
(d)
Write down the value of \(c\) for which the equation \(f(x)=c\) has no solution.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(x=3\)
A1
\(y=2\)
A1
(b)
\(2+\dfrac{7}{x-3}=\dfrac{2(x-3)+7}{x-3}\)
M1
Or \(2x+1=2(x-3)+7\).
\(=\dfrac{2x-6+7}{x-3}\)
A1
\(=\dfrac{2x+1}{x-3}\)
AG
(c)
vertical stretch with scale factor 7
A1
translation by \(\begin{pmatrix}3\\2\end{pmatrix}\)
A1
The vertical translation must follow the stretch; the horizontal translation may come before or after it.
(d)
\(c=2\)
A1
The horizontal asymptote; FT from (a).
Answers: (a) \(x=3,\ y=2\) · (c) vertical stretch, scale factor 7, then translation by \(\binom{3}{2}\) · (d) \(c=2\)
79AASL2-q00652.9 Exponential and logarithmic functionsPaper 1Medium9 marks
The following diagram shows the graph of \(f(x)=a\times2^{x}+b\), where \(a,b\in\mathbb{R}\). The graph passes through the points \((0,\,-2)\) and \((2,\,4)\) and has a horizontal asymptote \(y=-4\).
(a)
Write down the value of \(b\).
(1)
(b)
Find the value of \(a\).
(2)
(c)
Find the \(x\)-intercept of the graph of \(f\).
(2)
(d)
Find \(f^{-1}(x)\), giving your answer in the form \(\log_{2}(x+p)+q\).
(3)
(e)
Write down the equation of the vertical asymptote of the graph of \(f^{-1}\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(b=-4\)
A1
(b)
\(a\times2^{0}-4=-2\) or \(4a-4=4\)
(M1)
\(a=2\)
A1
(c)
\(2\times2^{x}=4\), so \(2^{x}=2\)
(M1)
\((1,\,0)\)
A1
Accept \(x=1\).
(d)
\(x=2\times2^{y}-4\)
(M1)
Interchanging variables, at any stage.
\(2^{y}=\dfrac{x+4}{2}\), so \(y=\log_{2}\left(\dfrac{x+4}{2}\right)\)
84AASL2-q00852.11 Transformations of graphsPaper 1Medium7 marks
The following diagram shows the graph of \(y=f(x)\), for \(-3\le x\le3\) (solid), which consists of three line segments joining \((-3,\,0)\), \((-1,\,2)\), \((1,\,-1)\) and \((3,\,1)\). The graph of \(y=g(x)\) (dashed) is the image of the graph of \(f\) under two transformations.
(a)(i)
Describe fully the two transformations.
(2)
(a)(ii)
Hence write down \(g(x)\) in the form \(pf(x-q)\), where \(p,q\in\mathbb{Z}\).
(1)
(b)
Find \((g\circ f)(-1)\).
(2)
(c)
Find all solutions of the equation \(f(x)=\dfrac{1}{2}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)(i)
vertical stretch, scale factor 2
A1
translation of 1 unit to the right, \(\begin{pmatrix}1\\0\end{pmatrix}\)
A1
Either order is acceptable here.
(a)(ii)
\(g(x)=2f(x-1)\)
A1
(b)
\(f(-1)=2\), then \(g(2)\)
(M1)
\(-2\)
A1
\(g(2)=2f(1)=-2\), also readable from the graph of \(g\).
(c)
line \(y=\frac{1}{2}\) meets the graph of \(f\) three times, or equation of at least one segment
86AASL2-q00902.11 Transformations of graphsPaper 1Medium8 marks
The points \(\mathrm{P}(2,\,5)\) and \(\mathrm{Q}(-1,\,1)\) lie on the graph of \(y=f(x)\).
(a)
The graph of \(y=f(x)\) is stretched vertically with scale factor 3 and then translated 4 units vertically downwards. Write down the equation of the new graph and the coordinates of the image of P.
(2)
(b)
The same two transformations are applied in the reverse order. Write down the equation of the new graph and the coordinates of the image of P.
(2)
(c)
The graph of \(y=pf(x)+q\) passes through the images of P and Q, which are \((2,\,7)\) and \((-1,\,-5)\) respectively. Find the value of \(p\) and the value of \(q\).
(3)
(d)
Hence describe a sequence of two transformations that maps the graph of \(y=f(x)\) onto the graph of \(y=pf(x)+q\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(y=3f(x)-4\)
A1
\((2,\,11)\)
A1
(b)
\(y=3\big(f(x)-4\big)=3f(x)-12\)
A1
\((2,\,3)\)
A1
(c)
\(5p+q=7\) and \(p+q=-5\)
M1
\(p=3\)
A1
\(q=-8\)
A1
(d)
vertical stretch with scale factor 3, followed by a translation of 8 units downwards
A1
FT from (c). Accept a translation of \(\tfrac{8}{3}\) units down followed by the stretch.
87IBMA2S-046-02-052.5 Composite and inverse functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=\frac{4x-1}{x+2}\), \(x\ne -2\).
(a)
Find an expression for \(f^{-1}(x)\).
(3)
(b)
Write down the domain and the range of \(f^{-1}\).
(2)
(c)
Hence or otherwise, find the value of \(a\) such that \(f^{-1}(a)=3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 interchanges \(x\) and \(y\) (before or after rearranging) and multiplies out, e.g. \(xy+2x=4y-1\)
✓ 1
M1 collects the terms in \(y\), e.g. \(y(4-x)=2x+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{2x+1}{4-x}\)
✓ 1
Part (b)
A1 domain \(x\ne 4\)
✓ 1
A1 range \(y\ne -2\)
✓ 1
Part (c)
M1 uses \(a=f(3)=\frac{4(3)-1}{3+2}\) or solves \(\frac{2a+1}{4-a}=3\)
✓ 1
A1 \(a=\frac{11}{5}\)
✓ 1
Exam tip: \(f^{-1}(a)=3\) means \(a=f(3)\). When a question says 'hence or otherwise', evaluating \(f\) directly is often quicker than solving an equation involving the inverse.
88IBMA2S-046-02-062.6 Quadratic functionsPaper 1Medium10 marks
No calculator
The main cable of a small suspension footbridge hangs between two towers 200 m apart. The height of the cable above the deck, \(h\) metres, at a horizontal distance \(x\) metres from the left tower is modelled by \(h(x)=0.004x^2-0.8x+50\), \(0\le x\le 200\).
(a)
Show that \(h(x)\) can be written as \(0.004(x-100)^2+10\).
(2)
(b)
Find the range of \(h\).
(2)
(c)
The function \(g\) is defined by \(g(x)=h(x)\) for \(100\le x\le 200\). Find \(g^{-1}(x)\), stating its domain.
(4)
(d)
Find the exact horizontal distance from the left tower, beyond the lowest point of the cable, at which the cable is 30 m above the deck.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands \(0.004(x^2-200x+10000)+10\) or completes the square
M1 rearranges to \((y-100)^2=250(x-10)\) and takes the positive square root because \(y\ge 100\)
✓ 1
A1 \(g^{-1}(x)=100+\sqrt{250(x-10)}\)
✓ 1
A1 domain \(10\le x\le 50\)
✓ 1
Part (d)
M1 evaluates \(g^{-1}(30)=100+\sqrt{5000}\), or solves \(0.004(x-100)^2+10=30\) to get \((x-100)^2=5000\) with \(x>100\)
✓ 1
A1 \(100+50\sqrt{2}\) m
✓ 1
Common error: On the domain \(100\le x\le 200\), the inverse must give values of at least 100, so take the positive square root. The negative root gives the point on the other side of the lowest point.
89IBMA2S-046-02-082.7 Quadratic equations and the discriminantPaper 1Medium10 marks
No calculator
A diver jumps from a platform. Her height above the water, \(h\) metres, \(t\) seconds after leaving the platform is modelled by \(h(t)=-5t^2+bt+c\). She leaves the platform at a height of 10 m and reaches her maximum height when \(t=0.4\). She enters the water at time \(T\) seconds.
(a)
Write down the value of \(c\).
(1)
(b)
Find the value of \(b\).
(2)
(c)
Find her maximum height above the water.
(2)
(d)
Write down the range of \(h\) for \(0\le t\le T\).
(2)
(e)
Find the exact value of \(T\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(c=10\)
✓ 1
Part (b)
M1 uses the axis of symmetry: \(-\frac{b}{2(-5)}=0.4\)
✓ 1
A1 \(b=4\)
✓ 1
Part (c)
M1 substitutes \(t=0.4\): \(-5(0.16)+4(0.4)+10\)
✓ 1
A1 \(10.8\) m
✓ 1
Part (d)
A1 lower bound \(0\)
✓ 1
A1 \(0\le h\le 10.8\)
✓ 1
Part (e)
M1 sets \(h(t)=0\): \(5t^2-4t-10=0\)
✓ 1
M1 uses the quadratic formula: \(t=\frac{4\pm\sqrt{216}}{10}\)
✓ 1
A1 \(T=\frac{2+3\sqrt{6}}{5}\), rejecting the negative root
✓ 1
Common error: The range is not \(10\le h\le 10.8\). The diver falls below the platform height before reaching the water, so the lower bound is 0.
90IBMA2S-046-02-102.5 Composite and inverse functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=3+\ln(2x-4)\).
(a)
Write down the largest possible domain of \(f\).
(1)
(b)
Find \(f^{-1}(x)\).
(3)
(c)
Write down the range of \(f^{-1}\).
(1)
(d)
Solve \(f^{-1}(x)=4\), giving your answer in the form \(a+\ln b\), where \(a,b\in\mathbb{Z}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x>2\)
✓ 1
Part (b)
M1 interchanges variables: \(x-3=\ln(2y-4)\)
✓ 1
M1 exponentiates: \(2y-4=e^{x-3}\)
✓ 1
A1 \(f^{-1}(x)=\frac{1}{2}e^{x-3}+2\)
✓ 1
Part (c)
A1 \(y>2\)
✓ 1
Part (d)
M1 obtains \(e^{x-3}=4\), or uses \(x=f(4)\)
✓ 1
A1 \(x=3+\ln 4\)
✓ 1
Exam tip: Check your inverse with a single value. For example, \(f\left(\frac{5}{2}\right)=3+\ln 1=3\), so \(f^{-1}(3)\) should equal \(\frac{5}{2}\).
91IBMA2S-046-01-032.7 Quadratic equations and the discriminantPaper 1Medium6 marks
No calculator
Consider the function \(f(x) = kx^2 - 4x + (k - 3)\), where \(k \in \mathbb{R}\).
(a)
Show that the discriminant of \(f(x)\) can be written as \(-4(k^2 - 3k - 4)\).
(2)
(b)
Find the set of values of \(k\) for which the equation \(f(x) = 0\) has two distinct real roots.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitution into \(b^2 - 4ac\): \((-4)^2 - 4k(k-3)\)
A1 critical values \(k = -1\) and \(k = 4\) from \((k-4)(k+1) = 0\)
✓ 1
A1 \(-1 < k < 4\)
✓ 1
R1 exclude \(k = 0\), because then \(f(x) = -4x - 3\) is linear with only one root. Final answer: \(-1 < k < 0\) or \(0 < k < 4\)
✓ 1
Common error: Forgetting to exclude \(k = 0\). The discriminant condition applies only when the coefficient of \(x^2\) is non-zero, so give the final set as \(-1 < k < 0\) or \(0 < k < 4\).
92IBMA2S-046-01-082.6 Quadratic functionsPaper 1Medium11 marks
No calculator
The entrance to a road tunnel has a parabolic cross-section. The ground is the x-axis and one edge of the entrance is at the origin O. The highest point of the arch is 5.4 m above the ground, at a horizontal distance of 3 m from O. The arch is modelled by \(y = a(x - h)^2 + k\), where x and y are measured in metres.
Diagram not to scale
(a)
Write down the value of h and the value of k.
(2)
(b)
Find the value of a and hence show that the arch can be modelled by \(y = -0.6x^2 + 3.6x\).
(3)
(c)
Find the width of the tunnel entrance at a height of 3 m above the ground.
(3)
(d)
A lorry with a rectangular cross-section is 2.4 m wide. It drives through the tunnel with its centre line directly below the highest point of the arch. Find the greatest height the lorry can have.
A1 \(x^2 - 6x + 5 = 0\), so \(x = 1\) and \(x = 5\)
✓ 1
A1 width \(= 4\) m
✓ 1
Part (d)
M1 the lorry's top corners are at \(x = 3 \pm 1.2\), i.e. \(x = 1.8\) or \(x = 4.2\)
✓ 1
M1 substituting into the model, e.g. \(5.4 - 0.6(1.2)^2\)
✓ 1
A1 \(4.536\) m
✓ 1
Common error: Using the height at the centre of the arch (5.4 m). The top corners of the lorry, not its centre, are the points that limit its height.
93IBMA2S-046-03-062.5 Composite and inverse functionsPaper 1Medium6 marks
No calculator
The function \(f\) is defined by \(f(x)=\dfrac{2x+1}{x-3}\), for \(x\in\mathbb{R},\ x\ne 3\).
(a)
Find \(f^{-1}(x)\), stating any value of \(x\) that must be excluded.
(3)
(b)
Hence state the range of \(f\).
(1)
(c)
Find the exact value of \((f\circ f)(4)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(x=\frac{2y+1}{y-3}\) (or equivalent) and multiplies through: \(xy-3x=2y+1\)
✓ 1
M1 collects the terms in \(y\): \(y(x-2)=3x+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{3x+1}{x-2}\), \(x\ne 2\)
✓ 1
Part (b)
A1 \(f(x)\in\mathbb{R},\ f(x)\ne 2\)
✓ 1
Part (c)
M1 \(f(4)=\frac{9}{1}=9\)
✓ 1
A1 \(f(9)=\frac{19}{6}\)
✓ 1
Exam tip: For a rational function, multiply out, then gather every term containing the new subject on one side before factorising it out.
94IBMA2S-046-03-082.5 Composite and inverse functionsPaper 1Medium6 marks
No calculator
The functions \(f\) and \(g\) are defined for \(x\in\mathbb{R}\) by \(f(x)=x+k\) and \(g(x)=x^2\), where \(k\) is a non-zero constant. It is given that \((g\circ f)(2)=(f\circ g)(2)\).
(a)
Find the value of \(k\).
(4)
(b)
Using your value of \(k\), solve \((g\circ f)(x)=(f\circ g)(x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((g\circ f)(2)=(2+k)^2\)
✓ 1
M1 \((f\circ g)(2)=4+k\)
✓ 1
M1 equates and simplifies: \(k^2+3k=0\)
✓ 1
A1 \(k=-3\), rejecting \(k=0\)
✓ 1
Part (b)
M1 \((x-3)^2=x^2-3 \Rightarrow -6x+9=-3\)
✓ 1
A1 \(x=2\) (the only solution)
✓ 1
Common error: \((g\circ f)(x)\) and \((f\circ g)(x)\) are generally different. Here \((x-3)^2\) and \(x^2-3\) agree at just one value of \(x\).
95IBMA2S-046-04-062.8 Reciprocal and rational functionsPaper 1Medium6 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{ax+b}{x+c}\), \(x\ne -c\), where \(a,b,c\in\mathbb{R}\). The graph of \(f\) has a vertical asymptote \(x=-4\) and a horizontal asymptote \(y=3\), and passes through the point \((-2,5)\).
(a)
Write down the value of \(a\) and the value of \(c\).
(2)
(b)
Find the value of \(b\).
(2)
(c)
Find the coordinates of the points where the graph of \(f\) meets the axes.
Common error: \(f\circ g\) means apply \(g\) first. Computing \(g(f(x))=\frac{2}{x-1}+3\) instead gives a different function.
97IBMA2S-046-07-052.1 Equations of straight linesPaper 1Medium7 marks
No calculator
Two mobile phone masts stand at A(−2, 5) and B(6, 1) on a coordinate grid measured in kilometres. An engineer needs the set of points that are the same distance from both masts; this is the perpendicular bisector of [AB].
Diagram not to scale
(a)
Find the midpoint of [AB].
(1)
(b)
Find the gradient of [AB].
(1)
(c)
Find the equation of the perpendicular bisector of [AB] in the form \(y=mx+c\).
(3)
(d)
A signal repeater must be placed on the line \(y=7\) at a point equidistant from both masts. Find its coordinates.
M1 uses midpoint with their gradient, e.g. \(y-3=2(x-2)\)
✓ 1
A1 \(y=2x-1\)
✓ 1
Part (d)
M1 substitutes \(y=7\) into their bisector, e.g. \(2x-1=7\), or equates squared distances to A and B with \(y=7\)
✓ 1
A1 (4, 7)
✓ 1
Exam tip: A quick check: the distance from (4, 7) to each mast is \(\sqrt{40}=2\sqrt{10}\) km, confirming the point is equidistant.
98IBMA2S-046-07-082.5 Composite and inverse functionsPaper 1Medium7 marks
No calculator
At a concert, two laser beams are projected across a vertical backdrop. With the stage floor as the x-axis and units in metres, the beams lie along the lines \(L_1: y=2x+1\) and \(L_2: x+3y=17\).
(a)
Find the coordinates of the point where the two beams cross.
(3)
(b)
Find the x-intercept of each line.
(2)
(c)
The two beams and the stage floor enclose a triangle. Find its exact area.
Common error: The base of the triangle is the distance between the intercepts, 17 − (−1/2) = 35/2, not 17 − 1/2.
99IBMA2S-046-07-102.1 Equations of straight linesPaper 1Medium11 marks
No calculator
A solar farm is laid out on a coordinate grid measured in metres. Three corners of a rectangular panel array are at A(1, 2), B(7, 5) and C(5, 9). The fourth corner is D.
Diagram not to scale
(a)
Find the gradient of [AB] and the gradient of [BC].
(2)
(b)
Hence show that angle ABC is a right angle.
(1)
(c)
Find the coordinates of D.
(2)
(d)
A cable runs along the diagonal [AC]. Find the equation of the line AC in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(3)
(e)
Find the area of the panel array.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 gradient AB \(=\frac{1}{2}\)
✓ 1
A1 gradient BC \(=-2\)
✓ 1
Part (b)
R1 \(\frac{1}{2}\times(-2)=-1\), so AB is perpendicular to BC
✓ 1
Part (c)
M1 uses parallel sides or equal midpoints of diagonals, e.g. D = A + (C − B)
✓ 1
A1 D(−1, 6)
✓ 1
Part (d)
M1 gradient AC \(=\frac{7}{4}\)
✓ 1
M1 substitutes a point, e.g. \(y-2=\frac{7}{4}(x-1)\)
✓ 1
A1 \(7x-4y+1=0\)
✓ 1
Part (e)
M1 uses distance formula for AB or BC
✓ 1
A1 AB \(=\sqrt{45}=3\sqrt{5}\) and BC \(=\sqrt{20}=2\sqrt{5}\)
✓ 1
A1 area \(=30\) m2
✓ 1
Common error: Corners must be taken in order A, B, C, D around the rectangle. Using D = A + B − C puts the fourth corner in the wrong place.
100IBMA2S-046-06-042.11 Transformations of graphsPaper 1Medium7 marks
No calculator
The path of a water jet from a fountain nozzle is modelled by \(y = g(x)\), where \(x\) is the horizontal distance from the nozzle and \(y\) is the height above the ground, both in metres. The graph of \(y = g(x)\) is obtained from the graph of \(y = x^2\) by a vertical stretch (with reflection) of factor \(a\), followed by a translation. The jet leaves the nozzle at \((0,\ 2)\) and reaches its maximum height at \((4,\ 6)\).
Diagram not to scale
(a)
Find \(g(x)\) in the form \(a(x-h)^2 + k\).
(3)
(b)
The nozzle is adjusted so that the new path is modelled by \(y = g\left(\frac{x}{2}\right)\). Describe the geometric transformation that maps \(y = g(x)\) onto \(y = g\left(\frac{x}{2}\right)\).
(1)
(c)
Find the exact horizontal distance from the nozzle at which the adjusted jet hits the ground.
A1 \(x = 8 + 4\sqrt{6}\) m (negative root rejected)
✓ 1
Common error: Replacing \(x\) with \(\frac{x}{2}\) stretches the graph by a factor of 2, not \(\frac{1}{2}\). Every horizontal distance doubles.
101IBMA2S-046-06-062.11 Transformations of graphsPaper 1Medium7 marks
No calculator
In a graphic design app, a user applies the following transformations, in this order, to the curve \(y = \sqrt{x}\): 1. a translation of 3 units to the left; 2. a horizontal stretch with scale factor \(\frac{1}{2}\); 3. a reflection in the \(x\)-axis. The resulting curve is \(y = g(x)\).
(a)
Find an expression for \(g(x)\).
(3)
(b)
State the domain and range of \(g\).
(2)
(c)
The point \((9,\ 3)\) lies on \(y = \sqrt{x}\). Find the coordinates of its image on \(y = g(x)\).
(2)
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Mark
Notes
Part (a)
A1 after step 1: \(y = \sqrt{x+3}\)
✓ 1
M1 replaces \(x\) by \(2x\): \(y = \sqrt{2x+3}\)
✓ 1
A1 \(g(x) = -\sqrt{2x+3}\)
✓ 1
Part (b)
A1 domain \(x \ge -\frac{3}{2}\)
✓ 1
A1 range \(g(x) \le 0\)
✓ 1
Part (c)
M1 applies the steps in order: \((9,3) \to (6,3) \to (3,3) \to \ldots\)
✓ 1
A1 \((3,\ -3)\)
✓ 1
Common error: After the translation, the stretch replaces \(x\) by \(2x\) to give \(\sqrt{2x+3}\), not \(\sqrt{2(x+3)}\). Writing \(\sqrt{2(x+3)}\) means the stretch was applied first.
102IBMA2S-046-05-052.6 Quadratic functionsPaper 1Medium5 marks
No calculator
Consider the equation 32x+1 − 10 × 3x + 3 = 0.
(a)
Show that, using the substitution y = 3x, the equation can be written as 3y2 − 10y + 3 = 0.
(2)
(b)
Hence find the exact solutions of the original equation.
M1 p + q = loga 10 = 3, so a3 = 10 ; A1 a = ∛10 = 101/3
✓ 2
Exam tip: Rewrite awkward numbers as products and quotients of the given bases first: decimals such as 0.08 become fractions like 2/25 before any log law is applied.
104IBMA2S-046-00-052.7 Quadratic equations and the discriminantPaper 1Medium6 marks
No calculator
In an optics simulation, a light ray travels along the line \(y=mx-1\), where \(m\) is a constant. The ray just grazes a parabolic mirror whose cross-section is \(y=x^2+3\), touching it at exactly one point.
(a)
Find the possible values of \(m\).
(4)
(b)
For the positive value of \(m\), find the coordinates of the point where the ray touches the mirror.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates \(x^2+3=mx-1\)
✓ 1
A1 \(x^2-mx+4=0\)
✓ 1
M1 sets discriminant \(m^2-16=0\)
✓ 1
A1 \(m=\pm4\)
✓ 1
Part (b)
M1 solves \(x^2-4x+4=0\), giving \(x=2\)
✓ 1
A1 \((2,\,7)\)
✓ 1
Common error: Rearrange to the form ax² + bx + c = 0 before identifying a, b and c; here the constant term is 3 + 1 = 4, not 3 − 1.
105IBMA2S-046-00-072.7 Quadratic equations and the discriminantPaper 1Medium7 marks
No calculator
An engineer designing a signal filter requires that the equation \(x^2+(k-2)x+(k+1)=0\), where \(k\) is a real constant, has no real solutions.
(a)
Show that the discriminant of the equation is \(k^2-8k\).
(2)
(b)
Find the range of values of \(k\) for which the equation has no real solutions.
(3)
(c)
For one boundary value of \(k\) the equation has a repeated positive root. Find this root.
R1 the probe reaches a speed of 31 km s−1 4 hours after the manoeuvre began
✓ 1
Common error: When inverting a function with a restricted domain, the choice between the + and − square root must be justified from the original domain.
108IBMA3S-2-17r11-082.5 Composite and inverse functionsPaper 1Medium7 marks
No calculator
A bridge monitoring system converts a raw strain reading \(x\) using \(f(x)=2x-1\), and then estimates stored energy using \(g(x)=x^2+3\), for all real \(x\).
(a)
Find \((f\circ g)(x)\).
(2)
(b)
Find \((g\circ f)(x)\), giving your answer in the form \(ax^2+bx+c\).
(2)
(c)
Find the positive solution of \((f\circ g)(x)=(g\circ f)(x)\), giving your answer in the form \(\frac{a+\sqrt{b}}{c}\) where \(a,b,c\in\mathbb{Z}^+\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f(x^2+3)=2(x^2+3)-1\)
✓ 1
A1 \((f\circ g)(x)=2x^2+5\)
✓ 1
Part (b)
M1 \((2x-1)^2+3\)
✓ 1
A1 \(4x^2-4x+4\)
✓ 1
Part (c)
M1 \(2x^2+5=4x^2-4x+4\) rearranged to \(2x^2-4x-1=0\)
✓ 1
M1 correct use of the quadratic formula: \(x=\frac{4\pm\sqrt{16+8}}{4}\)
✓ 1
A1 \(x=\frac{2+\sqrt{6}}{2}\) (the other root \(\frac{2-\sqrt{6}}{2}\) is negative)
✓ 1
Common error: \((f\circ g)(x)\) and \((g\circ f)(x)\) are generally different; always apply the inner function first.
109IBMA3S-2-16r17-042.8 Reciprocal and rational functionsPaper 1Medium8 marks
No calculator
Let \(f(x)=\frac{2x+1}{x-3}\), for \(x\in\mathbb{R}\), \(x\neq3\).
(a)
Write down the equation of (i) the vertical asymptote and (ii) the horizontal asymptote of the graph of \(f\).
(2)
(b)
Find the coordinates of the points where the graph of \(f\) crosses the axes.
(2)
(c)
Find \(f^{-1}(x)\).
(3)
(d)
Write down the domain of \(f^{-1}\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical asymptote \(x=3\)
✓ 1
A1 horizontal asymptote \(y=2\)
✓ 1
Part (b)
A1 \(y\)-intercept: \(f(0)=-\frac{1}{3}\), point \(\left(0,-\frac{1}{3}\right)\)
✓ 1
A1 \(x\)-intercept: \(2x+1=0\), point \(\left(-\frac{1}{2},0\right)\)
✓ 1
Part (c)
M1 \(x=\frac{2y+1}{y-3}\) after swapping variables
✓ 1
M1 \(xy-3x=2y+1 \Rightarrow y(x-2)=3x+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{3x+1}{x-2}\)
✓ 1
Part (d)
A1 \(x\in\mathbb{R}\), \(x\neq2\)
✓ 1
Exam tip: The domain of the inverse equals the range of the original function, and for \(\frac{ax+b}{cx+d}\) that excludes the value of the horizontal asymptote.
110IBMA3S-2-16r17-092.5 Composite and inverse functionsPaper 1Medium7 marks
No calculator
The function \(f\) is defined by \(f(x)=e^{2x}+1\), for \(x\in\mathbb{R}\). The function \(g\) is defined by \(g(x)=\ln x\), for \(x>0\).
(a)
Find \(f^{-1}(x)\) and state its domain.
(3)
(b)
Find \((f\circ g)(x)\), giving your answer in a form that does not contain \(e\) or \(\ln\).
(2)
(c)
Solve \(f^{-1}(x)=\ln2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x=e^{2y}+1\) after swapping variables, then \(e^{2y}=x-1\)
✓ 1
A1 \(f^{-1}(x)=\frac{1}{2}\ln(x-1)\)
✓ 1
A1 domain \(x>1\)
✓ 1
Part (b)
M1 \(f(g(x))=e^{2\ln x}+1\)
✓ 1
A1 \(e^{\ln x^2}+1=x^2+1\)
✓ 1
Part (c)
M1 \(\ln(x-1)=2\ln2=\ln4\)
✓ 1
A1 \(x=5\)
✓ 1
Exam tip: The domain of \(f^{-1}\) is the range of \(f\); since \(e^{2x}>0\), the range of \(f\) is \(y>1\).
111IBMA3S-2-17r17-052.5 Composite and inverse functionsPaper 1Medium6 marks
No calculator
The functions f and g are defined by f(x) = 2x + 1 and g(x) = x2 − 3x, for x ∈ ℝ.
(a)
Find (g ∘ f)(x), giving your answer in the form ax2 + bx + c.
(2)
(b)
Solve (g ∘ f)(x) = 0.
(2)
(c)
Find f−1(g(−1)).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes: (2x + 1)2 − 3(2x + 1)
✓ 1
A1 4x2 − 2x − 2
✓ 1
Part (b)
M1 attempts to factorise or solve, e.g. 2x2 − x − 1 = 0 or (2x + 1)(x − 1) = 0
✓ 1
A1 x = −0.5 and x = 1
✓ 1
Part (c)
M1 finds g(−1) = 4 and f−1(x) = (x − 1)/2
✓ 1
A1 1.5
✓ 1
Common error: (g ∘ f)(x) means apply f first, then g. Writing f(g(x)) instead is a very common slip.
112IBMA3S-2-17r17-082.8 Reciprocal and rational functionsPaper 1Medium7 marks
No calculator
A water treatment worker adds x litres of additive to a tank, for x ≥ 0. The concentration of disinfectant, C mg L−1, is modelled by C(x) = (20x + 30)/(x + 5).
(a)
Write down the equation of the horizontal asymptote of the graph of C, and state what it represents in this context.
(2)
(b)
Find an expression for the inverse function C−1(x).
(3)
(c)
Use your inverse function to find the amount of additive needed for a concentration of 15 mg L−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 C = 20
✓ 1
A1 the concentration approaches, but never reaches, 20 mg L−1 however much additive is added
✓ 1
Part (b)
M1 writes y = (20x + 30)/(x + 5) and clears the fraction: yx + 5y = 20x + 30
✓ 1
M1 collects terms in x and factorises: x(y − 20) = 30 − 5y
Exam tip: Check an inverse with one value: C(9) should give 15, which it does, since (180 + 30)/14 = 15.
113IBMA3S-2-17r17-092.2 Functions, domain, range and inversePaper 1Medium7 marks
No calculator
The speed of water, v m s−1, leaving a nozzle at gauge pressure p kPa is modelled by v(p) = √(2p − 5) + 1, for the largest possible domain of real values of p.
(a)
Write down the domain of v.
(1)
(b)
Write down the range of v.
(1)
(c)
Find an expression for the inverse function v−1(x).
(3)
(d)
Find the pressure needed for the water to leave the nozzle at 4.2 m s−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 p ≥ 2.5
✓ 1
Part (b)
A1 v ≥ 1
✓ 1
Part (c)
M1 writes x = √(2y − 5) + 1 and isolates the root: x − 1 = √(2y − 5)
✓ 1
M1 squares both sides and rearranges for y
✓ 1
A1 v−1(x) = ((x − 1)2 + 5)/2
✓ 1
Part (d)
M1 substitutes 4.2: ((4.2 − 1)2 + 5)/2
✓ 1
A1 7.62 kPa (ECF from (c))
✓ 1
Common error: The domain of the inverse is the range of the original function, so v⁻¹ only applies for x ≥ 1.
114IBMA3S-2-17r17-102.10 Solving equationsPaper 1Medium7 marks
M1 converts: (x + 5)/(x − 1) = 23 = 8, so x + 5 = 8x − 8
✓ 1
A1 x = 13/7 (satisfies x > 1)
✓ 1
Part (b)
M1 uses 9x = (3x)2 and sets u = 3x: u2 − 4u − 45 = 0
✓ 1
M1 factorises or solves: (u − 9)(u + 5) = 0
✓ 1
R1 rejects u = −5 because 3x > 0
✓ 1
A1 x = 2
✓ 1
Common error: After solving a logarithmic or exponential equation, check each candidate against the domain: 3x can never be negative.
115IBMA3S-2-17r26-052.11 Transformations of graphsPaper 1Medium6 marks
No calculator
The point P(2, 6) lies on the graph of y = f(x). Each of the following transformations is applied to the graph of y = f(x).
(a)
Find the coordinates of the image of P on the graph of y = f(x + 3).
(1)
(b)
Find the coordinates of the image of P on the graph of y = 2f(x) − 1.
(2)
(c)
Find the coordinates of the image of P on the graph of y = f(x⁄2).
(1)
(d)
The graph of y = f(x) is reflected in the x-axis and then translated 4 units to the right and 1 unit down. Write down the equation of the resulting graph.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 (−1, 6)
✓ 1
Part (b)
A1 x-coordinate unchanged, 2
✓ 1
A1 y-coordinate 2 × 6 − 1 = 11, so (2, 11)
✓ 1
Part (c)
A1 (4, 6)
✓ 1
Part (d)
M1 reflection gives y = −f(x), translation replaces x by x − 4 and subtracts 1
✓ 1
A1 y = −f(x − 4) − 1
✓ 1
Common error: Changes inside the brackets act on x and work 'backwards'; changes outside act on y.
116IBMA3S-2-17r26-072.8 Reciprocal and rational functionsPaper 1Medium7 marks
No calculator
The function f is defined by f(x) = (2x + 1)/(x − 3), for x ≠ 3.
(a)
Write down the equation of (i) the vertical asymptote and (ii) the horizontal asymptote of the graph of y = f(x).
(2)
(b)
Find f−1(x).
(3)
(c)
State the domain of f−1.
(1)
(d)
Find the value of x for which f(x) = 3.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 x = 3
✓ 1
A1 y = 2
✓ 1
Part (b)
M1 swap x and y (or make x the subject) and multiply out the denominator
✓ 1
M1 collect terms in y and factorise
✓ 1
A1 f−1(x) = (3x + 1)/(x − 2)
✓ 1
Part (c)
A1 x ≠ 2
✓ 1
Part (d)
A1 x = 10 (accept f−1(3) = 10)
✓ 1
Exam tip: The domain of an inverse function equals the range of the original, which here excludes the horizontal asymptote value.
117IBMA3S-2-19r28-062.5 Composite and inverse functionsPaper 1Medium8 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=3x-2\) for \(x\in\mathbb{R}\), and \(g(x)=x^2+1\) for \(x\geq 0\).
(a)
Find the value of \((g\circ f)(2)\). [2]
(2)
(b)
Find \(g^{-1}(x)\), and state its domain. [3]
(3)
(c)
Solve \((f\circ g^{-1})(x)=7\). [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f(2)=4\)
✓ 1
A1 \(g(4)=17\)
✓ 1
Part (b)
M1 set \(x=y^2+1\) and rearrange for \(y\)
✓ 1
A1 \(g^{-1}(x)=\sqrt{x-1}\)
✓ 1
A1 domain \(x\geq 1\) (the range of \(g\))
✓ 1
Part (c)
M1 form \(3\sqrt{x-1}-2=7\)
✓ 1
A1 \(\sqrt{x-1}=3\)
✓ 1
A1 \(x=10\)
✓ 1
Common error: The inverse of a function with a restricted domain takes the positive square root only. The domain of \(g^{-1}\) is the range of \(g\), not the domain of \(g\).
118IBMA3S-2-19r28-082.8 Reciprocal and rational functionsPaper 1Medium9 marks
No calculator
Consider \(f(x)=\frac{2x+3}{x-1}\), where \(x\neq 1\).
(a)
Find \(f^{-1}(x)\). [4]
(4)
(b)
State the range of \(f\), and explain how it relates to \(f^{-1}\). [2]
(2)
(c)
The graph of \(y=f(x)\) meets the line \(y=x\) at two points. Find the exact x-coordinate of the point with the larger x-coordinate. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 set \(y=\frac{2x+3}{x-1}\) and swap \(x\) and \(y\) (or rearrange for \(x\))
✓ 1
M1 multiply out: \(x(y-1)=2y+3\)
✓ 1
M1 collect terms in \(y\): \(y(x-2)=x+3\)
✓ 1
A1 \(f^{-1}(x)=\frac{x+3}{x-2}\)
✓ 1
Part (b)
A1 range of \(f\): \(f(x)\neq 2\)
✓ 1
R1 the range of \(f\) is the domain of \(f^{-1}\), so \(x\neq 2\) in \(f^{-1}\)
✓ 1
Part (c)
M1 \(x(x-1)=2x+3\)
✓ 1
A1 \(x^2-3x-3=0\)
✓ 1
A1 \(x=\frac{3+\sqrt{21}}{2}\)
✓ 1
Exam tip: To find an inverse, swap \(x\) and \(y\) and then collect every term containing \(y\) on one side so that \(y\) can be factorised out.
119IBMA3S-2-19r28-102.11 Transformations of graphsPaper 1Medium7 marks
No calculator
A manufacturer models the average cost, in euros per unit, of producing n thousand units by \(A(n)=\frac{an+b}{n+c}\) for \(n\geq 0\), where a, b and c are constants. The graph of \(y=A(n)\) has vertical asymptote \(n=-2\) and horizontal asymptote \(A=3\), and passes through the point \((0,4)\).
(a)
Write down the value of c and the value of a. [2]
(2)
(b)
Find the value of b. [2]
(2)
(c)
Explain what the horizontal asymptote means for the cost of production. [1]
(1)
(d)
Find the number of units, in thousands, for which the average cost is €3.50 per unit. [2]
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(c=2\)
✓ 1
A1 \(a=3\)
✓ 1
Part (b)
M1 \(A(0)=\frac{b}{2}=4\)
✓ 1
A1 \(b=8\)
✓ 1
Part (c)
R1 as production increases, the average cost per unit gets closer to €3 (but never falls below it)
✓ 1
Part (d)
M1 \(3n+8=3.5(n+2)\)
✓ 1
A1 \(n=2\) (that is, 2000 units)
✓ 1
Exam tip: Read the asymptote equations as information about the constants: the vertical asymptote is where the denominator is zero, and the horizontal asymptote is the ratio of the coefficients of \(n\).
120IBMA3S-2-17r39-022.1 Equations of straight linesPaper 1Medium6 marks
No calculator
On a theatre stage plan, with lengths in metres, a safety rail runs from \(A(-2,1)\) to \(B(4,4)\). A second rail runs from \(B\), perpendicular to \(AB\), until it meets the \(x\)-axis at \(C\).
Diagram not to scale
(a)
Find the gradient of \(AB\).
(1)
(b)
Find the equation of line \(BC\), giving your answer in the form \(y=mx+c\).
(2)
(c)
Find the coordinates of \(C\).
(1)
(d)
Find the area of triangle \(ABC\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{4-1}{4-(-2)}=\frac{1}{2}\)
✓ 1
Part (b)
M1 perpendicular gradient \(-2\) used with point \(B\): \(y-4=-2(x-4)\)
✓ 1
A1 \(y=-2x+12\)
✓ 1
Part (c)
A1 \(C=(6,0)\)
✓ 1
Part (d)
M1 area \(=\frac{1}{2}\times AB\times BC\) with \(AB=\sqrt{45}\), \(BC=\sqrt{20}\) (right angle at \(B\))
✓ 1
A1 area \(=15\) m2
✓ 1
Common error: The gradient of a perpendicular line is the negative reciprocal, not just the negative: \(\frac12\) gives \(-2\).
121IBMA3S-2-17r39-052.5 Composite and inverse functionsPaper 1Medium6 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{2x+1}{x-3}\), for \(x\in\mathbb{R}\), \(x\ne3\).
(a)
Write down the range of \(f\).
(1)
(b)
Find \(f^{-1}(x)\).
(3)
(c)
Solve \(f^{-1}(x)=4\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f(x)\in\mathbb{R}\), \(f(x)\ne2\)
✓ 1
Part (b)
M1 let \(y=\frac{2x+1}{x-3}\) and rearrange: \(xy-3y=2x+1\)
✓ 1
M1 collect \(x\) terms: \(x(y-2)=3y+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{3x+1}{x-2}\)
✓ 1
Part (c)
M1 \(3x+1=4(x-2)\) (or \(x=f(4)\))
✓ 1
A1 \(x=9\)
✓ 1
Exam tip: Solving \(f^{-1}(x)=k\) is the same as evaluating \(f(k)\), which is a useful check.
122IBMA3S-2-17r39-062.11 Transformations of graphsPaper 1Medium7 marks
No calculator
The graph of \(y=f(x)\) has a maximum point \(P(2,7)\) and passes through \(Q(-1,1)\).
(a)
Find the coordinates of the image of \(P\) on the graph of \(y=3f(x-4)+1\).
(2)
(b)
Find the coordinates of the image of \(Q\) on the graph of \(y=-f(2x)\).
(2)
(c)
The graph of \(y=f(3-x)\) is obtained from \(y=f(x)\) by a reflection followed by a translation, or a translation followed by a reflection. Describe both transformations fully.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x\)-coordinate \(2+4=6\)
✓ 1
A1 \(y\)-coordinate \(3(7)+1=22\), so \((6,22)\)
✓ 1
Part (b)
A1 \(x\)-coordinate \(-\frac12\)
✓ 1
A1 \(y\)-coordinate \(-1\), so \((-0.5,-1)\)
✓ 1
Part (c)
M1 write \(f(3-x)=f(-(x-3))\) or equivalent
✓ 1
A1 reflection in the \(y\)-axis
✓ 1
A1 horizontal translation of 3 units: to the right if the reflection is done first, to the left if the translation is done first
✓ 1
Common error: Horizontal changes act in the opposite way to what they seem: \(f(x-4)\) moves the graph 4 to the right, and \(f(2x)\) halves the \(x\)-coordinates.
123IBMA3S-2-17r39-082.1 Equations of straight linesPaper 1Medium7 marks
No calculator
On a chart with distances in kilometres, a ship's route \(R_1\) lies along the line \(3x+2y=12\). A port is at \(P(-2,-4)\). A second route \(R_2\) runs from \(P\) at right angles to \(R_1\) and meets \(R_1\) at a point \(Q\).
Diagram not to scale
(a)
Find the equation of \(R_2\) in the form \(ax+by=c\), where \(a\), \(b\) and \(c\) are integers with no common factor.
(3)
(b)
Find the coordinates of \(Q\).
(2)
(c)
Find the exact distance \(PQ\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 gradient of \(R_1\) is \(-\frac32\)
✓ 1
M1 perpendicular gradient \(\frac23\) used with \(P\): \(y+4=\frac23(x+2)\)
✓ 1
A1 \(2x-3y=8\)
✓ 1
Part (b)
M1 solve \(3x+2y=12\) and \(2x-3y=8\) simultaneously
✓ 1
A1 \(Q=(4,0)\)
✓ 1
Part (c)
M1 \(\sqrt{(4-(-2))^2+(0-(-4))^2}=\sqrt{52}\)
✓ 1
A1 \(PQ=2\sqrt{13}\) km (which is 7.2111... km)
✓ 1
Exam tip: Check that \(Q\) lies on both lines by substituting its coordinates back into each equation.
124IBMA3S-2-17r39-142.5 Composite and inverse functionsPaper 1Medium7 marks
No calculator
The function \(f\) is defined by \(f(x)=ax+b\), where \(a\) and \(b\) are constants and \(a>0\). It is given that \((f\circ f)(x)=9x+8\).
(a)
Find the value of \(a\) and the value of \(b\).
(3)
(b)
Find \(f^{-1}(x)\).
(2)
(c)
Solve \(f(x)=f^{-1}(x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((f\circ f)(x)=a(ax+b)+b=a^2x+ab+b\)
✓ 1
A1 \(a^2=9\) with \(a>0\) gives \(a=3\)
✓ 1
A1 \(3b+b=8\) gives \(b=2\)
✓ 1
Part (b)
M1 \(y=3x+2\) and rearrange for \(x\)
✓ 1
A1 \(f^{-1}(x)=\frac{x-2}{3}\)
✓ 1
Part (c)
M1 \(3x+2=\frac{x-2}{3}\), so \(9x+6=x-2\)
✓ 1
A1 \(x=-1\)
✓ 1
Exam tip: Compare coefficients of \(x\) and constants separately when two linear expressions are equal for all \(x\).
125IBMA3S-2-19r40-042.5 Composite and inverse functionsPaper 1Medium6 marks
No calculator
The function f is defined by \(f(x)=\ln(2x-4)\) for \(x>2\).
(a)
Solve the equation \(f(x)=0\).
(2)
(b)
Find an expression for \(f^{-1}(x)\).
(3)
(c)
Write down the range of \(f^{-1}\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2x-4=e^0=1\)
✓ 1
A1 \(x=\frac{5}{2}\)
✓ 1
Part (b)
M1 writes \(y=\ln(2x-4)\) and swaps \(x\) and \(y\) (or makes \(x\) the subject)
✓ 1
M1 \(e^{x}=2y-4\) (exponentiates)
✓ 1
A1 \(f^{-1}(x)=\frac{e^{x}+4}{2}\)
✓ 1
Part (c)
A1 \(f^{-1}(x)>2\)
✓ 1
Exam tip: The range of an inverse function is the domain of the original function. State it as an inequality in terms of \(f^{-1}(x)\) (or y), not x.
126IBMA3S-2-19r40-062.11 Transformations of graphsPaper 1Medium7 marks
No calculator
The graph of \(y=x^2\) is transformed to the graph of \(y=g(x)\), where \(g(x)=2(x+3)^2-5\).
(a)
Describe a sequence of three transformations that maps \(y=x^2\) onto \(y=g(x)\).
(3)
(b)
Write down the coordinates of the vertex of the graph of \(y=g(x)\).
(1)
(c)
Solve \(g(x)=45\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 translation 3 units to the left, i.e. by \(\begin{pmatrix}-3\\0\end{pmatrix}\)
✓ 1
A1 vertical stretch with scale factor 2 (parallel to the y-axis)
✓ 1
A1 translation 5 units down, i.e. by \(\begin{pmatrix}0\\-5\end{pmatrix}\), performed after the stretch (any valid order is accepted provided the stretch comes before the vertical translation)
✓ 1
Part (b)
A1 (−3, −5)
✓ 1
Part (c)
M1 \(2(x+3)^2=50\)
✓ 1
A1 \((x+3)^2=25\) so \(x+3=\pm5\)
✓ 1
A1 \(x=2\) or \(x=-8\)
✓ 1
Exam tip: If a vertical stretch and a vertical translation are both needed, the order matters; the horizontal translation can go anywhere. Test your order on the vertex.
127IBMA3S-2-19r40-112.11 Transformations of graphsPaper 1Medium6 marks
No calculator
A marine biologist models part of a species' depth range using the function \(g(x)=\ln(x+2)-1\) for \(x>-2\).
(a)
Write down the equation of the vertical asymptote of the graph of \(y=g(x)\).
(1)
(b)
Find the exact x-intercept of the graph.
(2)
(c)
Find the exact y-coordinate of the y-intercept.
(1)
(d)
Describe the single transformation that maps the graph of \(y=\ln x\) onto the graph of \(y=g(x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=-2\)
✓ 1
Part (b)
M1 sets \(\ln(x+2)=1\)
✓ 1
A1 \(x=e-2\)
✓ 1
Part (c)
A1 \(y=\ln 2-1\) (≈ -0.307)
✓ 1
Part (d)
M1 identifies a translation
✓ 1
A1 translation by \(\begin{pmatrix}-2\\-1\end{pmatrix}\) (2 units left, 1 unit down)
✓ 1
Exam tip: For \(y=\ln(x+a)\) the asymptote moves with the graph: shift the original asymptote \(x=0\) by the same horizontal translation.
128IBMA3S-2-19r40-122.6 Quadratic functionsPaper 1Medium8 marks
No calculator
The graph of \(f(x)=x^2-4x+1\) is reflected in the x-axis and then translated by the vector \(\begin{pmatrix}2\\-1\end{pmatrix}\). The resulting graph is \(y=g(x)\).
(a)
Show that \(g(x)=-x^2+8x-14\).
(3)
(b)
Solve \(g(x)=0\), giving your answers in exact form.
(3)
(c)
Find the set of values of k for which the equation \(g(x)=k\) has two distinct real solutions.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 reflection gives \(-f(x)=-x^2+4x-1\)
✓ 1
M1 replaces \(x\) by \(x-2\) and subtracts 1
✓ 1
A1 expands to obtain \(g(x)=-x^2+8x-14\) (AG)
✓ 1
Part (b)
M1 \(x^2-8x+14=0\) and uses the quadratic formula or completes the square
✓ 1
A1 \(x=\frac{8\pm\sqrt{8}}{2}\)
✓ 1
A1 \(x=4\pm\sqrt{2}\) (≈ 5.41 or 2.59)
✓ 1
Part (c)
M1 vertex of \(y=g(x)\) is (4, 2) (or uses the discriminant)
✓ 1
A1 \(k<2\)
✓ 1
Common error: Translating by \((2,-1)\) means replacing \(x\) by \(x-2\) (not \(x+2\)), and the reflection must be applied before the translation.
129AASL2-q00052.1 Equations of straight linesPaper 1Hard9 marks
The points \(\mathrm{A}(-2,\,1)\) and \(\mathrm{B}(4,\,4)\) lie on the line \(L_{1}\). The line \(L_{2}\) passes through B and is perpendicular to \(L_{1}\). The line \(L_{2}\) meets the \(x\)-axis at C.
Diagram not to scale
(a)
Find the gradient of \(L_{1}\).
(1)
(b)
Find the equation of \(L_{2}\), giving your answer in the form \(y=mx+c\).
(2)
(c)
Find the coordinates of C.
(1)
(d)
Find the area of triangle ABC.
(3)
(e)
The point D is such that ABCD is a rectangle. Find the coordinates of D.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{4-1}{4-(-2)}=\dfrac{1}{2}\)
A1
(b)
gradient \(-2\) and \(y-4=-2(x-4)\)
(M1)
FT from (a).
\(y=-2x+12\)
A1
(c)
\((6,\,0)\)
A1
FT from (b).
(d)
recognizing the right angle at B and attempting \(\mathrm{AB}\) and \(\mathrm{BC}\)
\(\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BC}}\), i.e. moving from A by \(+2\) in \(x\) and \(-4\) in \(y\); or intersecting the line through A parallel to \(L_{2}\) with the line through C parallel to \(L_{1}\)
136AASL2-q00522.7 Quadratic equations and the discriminantPaper 1Hard8 marks
Consider the quadratic expression \((m-1)x^{2}+4x+m+2\), where \(m\in\mathbb{R}\), \(m\neq1\). The expression is positive for all real values of \(x\).
(a)
Explain why \(m\gt 1\).
(1)
(b)
Show that \(m^{2}+m-6\gt 0\).
(2)
(c)
Hence find the set of possible values of \(m\).
(3)
(d)
For a different value of \(m\), the graph of \(y=(m-1)x^{2}+4x+m+2\) touches the \(x\)-axis at exactly one point and lies on or above it. Find this value of \(m\) and the coordinates of the point.
(2)
Show mark scheme
Working
Mark
Notes
(a)
if \(m-1\lt 0\) the parabola opens downwards, so the expression is negative for large \(|x|\); hence \(m-1\gt 0\)
R1
(b)
no real roots: \(\Delta=16-4(m-1)(m+2)\lt 0\)
M1
\(16-4\left(m^{2}+m-2\right)\lt 0\), so \(24-4m^{2}-4m\lt 0\)
A1
\(m^{2}+m-6\gt 0\)
AG
(c)
\((m+3)(m-2)\gt 0\)
M1
\(m\lt -3\) or \(m\gt 2\)
A1
combining with \(m\gt 1\): \(m\gt 2\)
A1
(d)
\(m=2\)
A1
\(\Delta=0\) gives \(m=2\) or \(m=-3\); \(m=-3\) is rejected as the graph would lie on or below the axis.
\(x^{2}+4x+4=(x+2)^{2}\), point \((-2,\,0)\)
A1
Answers: (c) \(m\gt 2\) · (d) \(m=2\), \((-2,0)\)
137AASL2-q00552.7 Quadratic equations and the discriminantPaper 1Hard15 marks
A gardener has 60 m of fencing to enclose a rectangular vegetable plot against a straight wall. The wall forms one side of the rectangle and is not fenced. The two sides perpendicular to the wall each have length \(x\) metres, where \(0\lt x\lt 30\).
Diagram not to scale
(a)
Show that the area of the plot, \(A\) m2, is given by \(A=60x-2x^{2}\).
(1)
(b)
Find the values of \(x\) for which the area is 400 m2.
(3)
(c)
Express \(A\) in the form \(-2(x-h)^{2}+k\).
(3)
(d)
Hence write down the maximum area and the dimensions of the plot that give this area.
(2)
(e)
Find the set of values of \(x\) for which the area is at least 352 m2.
(3)
(f)
Using the discriminant, find the set of values of \(c\) for which there are two different plots each with area \(c\) m2.
(3)
Show mark scheme
Working
Mark
Notes
(a)
side parallel to the wall \(=60-2x\), so \(A=x(60-2x)\)
A1
\(A=60x-2x^{2}\)
AG
(b)
\(2x^{2}-60x+400=0\), i.e. \(x^{2}-30x+200=0\)
M1
\((x-10)(x-20)=0\)
A1
\(x=10\) or \(x=20\)
A1
(c)
\(-2\left(x^{2}-30x\right)\) and attempt to complete the square
M1
\(-2\big((x-15)^{2}-225\big)\)
A1
\(A=-2(x-15)^{2}+450\)
A1
(d)
maximum area \(450\) m2
A1
FT from (c).
15 m by 30 m
A1
\(x=15\); side parallel to wall \(60-30=30\).
(e)
\(60x-2x^{2}\ge352\), i.e. \(x^{2}-30x+176\le0\)
M1
\((x-8)(x-22)\le0\)
A1
\(8\le x\le22\)
A1
(f)
\(2x^{2}-60x+c=0\) with \(\Delta=3600-8c\gt 0\)
M1
\(c\lt 450\)
A1
\(0\lt c\lt 450\)
A1
Area must be positive. Both roots \(x=15\pm\sqrt{225-\tfrac{c}{2}}\) then lie in \(0\lt x\lt 30\).
Answers: (b) \(x=10,\ 20\) · (c) \(-2(x-15)^{2}+450\) · (d) 450 m2, 15 m by 30 m · (e) \(8\le x\le22\) · (f) \(0\lt c\lt 450\)
138AASL2-q00582.8 Reciprocal and rational functionsPaper 1Hard8 marks
Let \(f(x)=\dfrac{3x+4}{x-3}\), for \(x\neq3\).
(a)
Show that \((f\circ f)(x)=x\).
(3)
(b)
Hence write down \(f^{-1}(x)\).
(1)
(c)
State what the result in part (b) shows about the graph of \(f\).
139AASL2-q00622.8 Reciprocal and rational functionsPaper 1Hard9 marks
Let \(f(x)=\dfrac{x+3}{x-1}\), for \(x\neq1\). The line \(L\) has equation \(y=x-3\).
(a)
Find the coordinates of the points where \(L\) meets the graph of \(f\).
(3)
(b)
Hence solve the inequality \(\dfrac{x+3}{x-1}\gt x-3\).
(3)
(c)
The line with equation \(y=x+c\), where \(c\in\mathbb{R}\), meets the graph of \(f\) at the points A and B. Show that the midpoint of AB lies on the line \(x+y=2\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(x+3=(x-3)(x-1)\)
M1
\(x^{2}-5x=0\), so \(x=0\) or \(x=5\)
A1
\((0,\,-3)\) and \((5,\,2)\)
A1
(b)
considering the graphs: the curve is above \(L\) for \(x\lt 0\) and between the asymptote \(x=1\) and \(x=5\)
M1
Or a sign diagram for \(\dfrac{x(5-x)}{x-1}\) with critical values 0, 1, 5.
\(x\lt 0\)
A1
\(1\lt x\lt 5\)
A1
Award A0 if \(x=1\) is included.
(c)
\(x+3=(x+c)(x-1)\), i.e. \(x^{2}+(c-2)x-(c+3)=0\)
M1
\(x=\dfrac{2-c\pm\sqrt{c^{2}+16}}{2}\), so the midpoint has \(x=\dfrac{2-c}{2}\)
A1
The discriminant \(c^{2}+16\gt 0\), so A and B always exist.
each value \(u\gt -\tfrac{9}{4}\) gives two values of \(x\) and \(u=-\tfrac{9}{4}\) gives one, so one root of \(u^{2}-2u=m\) must equal \(-\tfrac{9}{4}\)
R1
The other root, \(2-\left(-\tfrac{9}{4}\right)=\tfrac{17}{4}\), then gives two values of \(x\).
141AASL2-q00862.11 Transformations of graphsPaper 1Hard8 marks
The graph of \(y=\ln x\) is transformed by the following sequence of transformations to give the graph of \(y=g(x)\):
a reflection in the \(y\)-axis;
then a translation of 2 units vertically upwards;
then a vertical stretch with scale factor 3.
(a)
Show that \(g(x)=3\ln(-x)+6\).
(2)
(b)
Write down the domain of \(g\) and the equation of the vertical asymptote of its graph.
(2)
(c)
Find the exact \(x\)-intercept of the graph of \(g\).
(2)
(d)
The same three transformations are applied to \(y=\ln x\), but with the vertical stretch first, then the reflection, then the translation. Write down the equation of the resulting graph, and explain why it is different from \(y=g(x)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
after reflection \(y=\ln(-x)\); after translation \(y=\ln(-x)+2\)
A1
stretch applied to the whole expression: \(y=3\big(\ln(-x)+2\big)\)
M1
\(g(x)=3\ln(-x)+6\)
AG
(b)
domain \(x\lt 0\)
A1
\(x=0\)
A1
(c)
\(\ln(-x)=-2\)
(M1)
\(x=-\mathrm{e}^{-2}\)
A1
Accept \(\left(-\mathrm{e}^{-2},\,0\right)\).
(d)
\(y=3\ln(-x)+2\)
A1
in \(g\) the translation is applied before the stretch, so the upward shift of 2 is also stretched (to 6); here it is not
142AASL2-q00912.11 Transformations of graphsPaper 1Hard10 marks
Let \(f(x)=\dfrac{x+4}{x-2}\), for \(x\neq2\). The graph of \(y=f(x)\) is stretched horizontally with scale factor \(\tfrac{1}{2}\) and then reflected in the \(y\)-axis to give the graph of \(y=g(x)\).
(a)
Show that \(g(x)=\dfrac{x-2}{x+1}\).
(3)
(b)
Write down the equations of the asymptotes of the graph of \(g\).
(2)
(c)
Explain why the same graph is obtained if the reflection is applied before the stretch.
(2)
(d)
Find the coordinates of the point where the graphs of \(f\) and \(g\) intersect, and explain why this point is unchanged by the two transformations.
(3)
Show mark scheme
Working
Mark
Notes
(a)
stretch: \(f(2x)=\dfrac{2x+4}{2x-2}\)
M1
reflection: replacing \(x\) by \(-x\): \(\dfrac{-2x+4}{-2x-2}\)
M1
\(=\dfrac{-2(x-2)}{-2(x+1)}\)
A1
\(g(x)=\dfrac{x-2}{x+1}\)
AG
(b)
\(x=-1\)
A1
\(y=1\)
A1
(c)
reflection first gives \(f(-x)\); then the stretch replaces \(x\) by \(2x\): \(f(-2x)\)
M1
both orders replace \(x\) by \(-2x\) in \(f\), so they give the same function
R1
Award R1 for a valid general reason, e.g. both transformations act only on \(x\)-coordinates, multiplying them by \(\tfrac{1}{2}\) and by \(-1\), and multiplication is commutative.
(d)
\(\dfrac{x+4}{x-2}=\dfrac{x-2}{x+1}\), so \((x+4)(x+1)=(x-2)^{2}\)
M1
\(9x=0\), point \((0,\,-2)\)
A1
both transformations change only \(x\)-coordinates, by multiplying them, so a point with \(x=0\) (on the \(y\)-axis) is mapped onto itself
R1
Answers: (b) \(x=-1,\ y=1\) · (d) \((0,-2)\)
143IBMA2S-046-01-102.7 Quadratic equations and the discriminantPaper 1Hard7 marks
No calculator
Consider the equation \(x^2 - px + 2p - 3 = 0\), where \(p \in \mathbb{R}\).
(a)
Show that the discriminant of the equation is \(p^2 - 8p + 12\).
(2)
(b)
Find the values of p for which the equation has two distinct real roots.
(2)
(c)
The two roots of the equation differ by exactly 2. Find the possible values of p, giving your answers in the form \(a \pm b\sqrt{2}\), where \(a, b \in \mathbb{Z}\).
Common error: Setting the discriminant itself equal to 2. The difference between the roots is \(\frac{\sqrt{\Delta}}{|a|}\), so here you need \(\sqrt{\Delta} = 2\), i.e. \(\Delta = 4\).
144IBMA2S-046-03-112.5 Composite and inverse functionsPaper 1Hard14 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=2+\ln(x+1)\) for \(x>-1\), and \(g(x)=x^2+2x\) for \(x>-1\).
(a)
Find \(f^{-1}(x)\) and state its domain.
(3)
(b)
Show that \((f\circ g)(x)=2+2\ln(x+1)\) for \(x>-1\).
(3)
(c)
Solve \((f\circ g)(x)=6\), giving your answer in exact form.
(2)
(d)
Let \(h=(f\circ g)^{-1}\). Find \(h(x)\).
(3)
(e)
Solve \(h(x)=f^{-1}(x)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 interchanges \(x\) and \(y\) and rearranges \(x=2+\ln(y+1)\) (or equivalent)
✓ 1
A1 \(f^{-1}(x)=e^{x-2}-1\)
✓ 1
A1 domain \(x\in\mathbb{R}\)
✓ 1
Part (b)
M1 forms \(f(x^2+2x)=2+\ln(x^2+2x+1)\)
✓ 1
M1 recognises \(x^2+2x+1=(x+1)^2\)
✓ 1
R1 uses \(\ln\big((x+1)^2\big)=2\ln(x+1)\), which is valid since \(x+1>0\)
✓ 1
Part (c)
M1 \(2\ln(x+1)=4 \Rightarrow \ln(x+1)=2\)
✓ 1
A1 \(x=e^2-1\)
✓ 1
Part (d)
M1 rearranges \(x=2+2\ln(y+1)\) to \(\ln(y+1)=\frac{x-2}{2}\)
✓ 1
M1 exponentiates both sides: \(y+1=e^{\frac{x-2}{2}}\)
✓ 1
A1 \(h(x)=e^{\frac{x-2}{2}}-1\)
✓ 1
Part (e)
M1 equates: \(e^{\frac{x-2}{2}}-1=e^{x-2}-1\)
✓ 1
M1 takes natural logarithms of both sides (or uses the substitution \(u=e^{(x-2)/2}\)) to obtain \(\frac{x-2}{2}=x-2\)
✓ 1
A1 \(x=2\)
✓ 1
Exam tip: On a 'show that' part, the reason matters. To earn the R mark, say why \(\ln\big((x+1)^2\big)=2\ln(x+1)\) is allowed: it is because \(x+1>0\).
145IBMA2S-046-04-112.11 Transformations of graphsPaper 1Hard15 marks
No calculator
Let \(f(x)=\frac{4x+1}{x-2}\), \(x\ne2\).
(a)
Show that \(f(x)=4+\frac{9}{x-2}\).
(2)
(b)
Hence describe a sequence of transformations that maps the graph of \(y=\frac{1}{x}\) onto the graph of \(y=f(x)\).
(3)
(c)
Write down the equations of the asymptotes of the graph of \(f\), and find the coordinates of its axis intercepts.
(4)
(d)
Find \(f^{-1}(x)\), stating its domain.
(3)
(e)
Solve \(f(x)=x\), giving your answers in exact form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(4x+1=4(x-2)+9\), or combines \(4+\frac{9}{x-2}\) over a common denominator
✓ 1
A1 correct working leading to the given expression
✓ 1
Part (b)
A1 horizontal translation 2 units to the right, i.e. by \(\binom{2}{0}\)
✓ 1
A1 vertical stretch with scale factor 9
✓ 1
A1 vertical translation 4 units up, applied after the stretch
✓ 1
Accept any order that produces \(f\). The horizontal translation may come at any point in the sequence. A vertical translation of \(\frac{4}{9}\) up applied before the stretch also earns the final mark.
—
Part (c)
A1 \(x=2\)
✓ 1
A1 \(y=4\)
✓ 1
A1 \(x\)-intercept \(\left(-\frac14,0\right)\)
✓ 1
A1 \(y\)-intercept \(\left(0,-\frac12\right)\)
✓ 1
Part (d)
M1 attempts to rearrange \(y(x-2)=4x+1\) to make \(x\) the subject
✓ 1
A1 \(x(y-4)=2y+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{2x+1}{x-4}\), with domain \(x\in\mathbb{R}\), \(x\ne4\) (accept \(x\ne4\))
✓ 1
Part (e)
M1 multiplies both sides by \((x-2)\): \(4x+1=x^2-2x\)
✓ 1
A1 \(x^2-6x-1=0\)
✓ 1
A1 \(x=3\pm\sqrt{10}\)
✓ 1
Common error: Order matters for the vertical transformations. Translating up 4 and then stretching by 9 moves the horizontal asymptote to \(y=36\). Either stretch first and then translate up 4, or translate up \(\frac{4}{9}\) and then stretch.
146IBMA2S-046-07-122.7 Quadratic equations and the discriminantPaper 1Hard9 marks
No calculator
A straight support beam is fixed between a vertical wall (the y-axis) and a horizontal floor (the x-axis). Units are metres. The beam must pass through a mounting bracket at the point (2, 3). The line of the beam has gradient \(-m\), where \(m>0\).
Diagram not to scale
(a)
Show that the beam meets the wall at height \(3+2m\) and meets the floor at a distance \(2+\frac{3}{m}\) from the wall.
(3)
(b)
The beam, wall and floor enclose a triangle of area 12 m2. Show that \(4m^2-12m+9=0\).
(2)
(c)
Solve the equation to find \(m\), and hence write down the equation of the beam.
M1 endpoints (4, 0) and (0, 6), \(\sqrt{4^2+6^2}\)
✓ 1
A1 \(2\sqrt{13}\) m
✓ 1
Exam tip: In a 'show that' part, every algebraic step must be visible. Multiply out (3 + 2m)(2m + 3)/m fully before stating the given quadratic.
147IBMA2S-046-06-082.11 Transformations of graphsPaper 1Hard13 marks
No calculator
Let \(f(x) = x^2 - 6x + 5\). An engineer designs a tunnel entrance whose arch profile is \(y = g(x)\), where \(g(x) = -2f(x+1) + 8\), with \(x\) and \(y\) in metres.
(a)
Write \(f(x)\) in the form \((x-h)^2 + k\).
(2)
(b)
Show that \(g(x) = -2(x-2)^2 + 16\).
(3)
(c)
Describe a sequence of transformations that maps the graph of \(y = f(x)\) onto the graph of \(y = g(x)\).
(3)
(d)
Find the exact \(x\)-intercepts of the graph of \(y = g(x)\).
(2)
(e)
The design is moved along the road. The new profile is the graph of \(y = g(x)\) translated by \(\begin{pmatrix}c\\0\end{pmatrix}\), where \(c > 0\), and it passes through the origin. Find the exact value of \(c\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square
✓ 1
A1 \(f(x) = (x-3)^2 - 4\)
✓ 1
Part (b)
M1 \(f(x+1) = (x+1-3)^2 - 4 = (x-2)^2 - 4\)
✓ 1
M1 \(g(x) = -2\left[(x-2)^2 - 4\right] + 8\)
✓ 1
A1 \(= -2(x-2)^2 + 8 + 8 = -2(x-2)^2 + 16\) AG
✓ 1
Part (c)
A1 translation 1 unit left
✓ 1
A1 reflection in the \(x\)-axis and vertical stretch with scale factor 2
✓ 1
A1 translation 8 units up, applied after the stretch and reflection
✓ 1
Part (d)
M1 \((x-2)^2 = 8\)
✓ 1
A1 \(x = 2 \pm 2\sqrt{2}\)
✓ 1
Part (e)
M1 the new profile is \(y = g(x-c)\), so \(g(-c) = 0\)
Exam tip: A translation of \(c\) to the right moves every \(x\)-intercept right by \(c\). Moving the left intercept to the origin gives \(c\) directly.
148IBMA2S-046-05-112.11 Transformations of graphsPaper 1Hard12 marks
No calculator
The functions f and g are defined by f(x) = 2 + log3(x − 1), for x > 1, and g(x) = 3x + 1, for x ∈ ℝ.
(a)
Describe fully the transformation that maps the graph of y = log3x onto the graph of y = f(x).
(2)
(b)
Find f−1(x), stating its domain.
(3)
(c)
Find (f ∘ g)(x), giving your answer in its simplest form.
(2)
(d)
Solve the equation f(x) = log3(x + 5) + log3(x − 3), for x > 3. Give your answer in exact form.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 translation of 1 unit in the positive x-direction ; A1 and 2 units in the positive y-direction, i.e. by the vector (1, 2)
✓ 2
Part (b)
M1 interchanging x and y and rearranging: x − 2 = log3(y − 1) ; A1 f−1(x) = 3x−2 + 1 ; A1 domain x ∈ ℝ
Exam tip: Turn a lone constant into a logarithm of the same base (here 2 = log₃ 9) so that every term can be combined into a single log on each side.
149IBMA2S-046-00-122.7 Quadratic equations and the discriminantPaper 1Hard14 marks
No calculator
Let \(f(x)=x^2-4x+7\). A straight line \(L\) passes through the point \((0,\,3)\) and has gradient \(m\), so its equation is \(y=mx+3\).
(a)
Write \(f(x)\) in the form \((x-h)^2+k\) and hence write down the coordinates of the vertex of the graph of \(f\).
(3)
(b)
Show that the \(x\)-coordinates of any points of intersection of \(L\) and the graph of \(f\) satisfy \(x^2-(4+m)x+4=0\).
(2)
(c)
Find the values of \(m\) for which \(L\) is a tangent to the graph of \(f\).
(4)
(d)
Find the set of values of \(m\) for which \(L\) meets the graph of \(f\) at two distinct points.
(2)
(e)
For \(m=1\), find the exact distance between the two points of intersection.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square
✓ 1
A1 \((x-2)^2+3\)
✓ 1
A1 vertex \((2,\,3)\)
✓ 1
Part (b)
M1 equates \(x^2-4x+7=mx+3\)
✓ 1
A1 rearranges to \(x^2-(4+m)x+4=0\)
✓ 1
Part (c)
M1 uses discriminant \(=0\)
✓ 1
A1 \((4+m)^2-16=0\) or \(m^2+8m=0\)
✓ 1
A1 \(m=0\)
✓ 1
A1 \(m=-8\)
✓ 1
Part (d)
M1 \(m(m+8)>0\)
✓ 1
A1 \(m<-8\) or \(m>0\)
✓ 1
Part (e)
M1 solves \(x^2-5x+4=0\) to get \(x=1\), \(x=4\)
✓ 1
A1 points \((1,\,4)\) and \((4,\,7)\)
✓ 1
A1 distance \(=\sqrt{3^2+3^2}=3\sqrt{2}\)
✓ 1
Common error: The tangent with m = 0 is horizontal and touches at the vertex; check this against part (a) rather than dismissing m = 0 as giving no line.
150IBMA3S-2-17r11-122.8 Reciprocal and rational functionsPaper 1Hard11 marks
No calculator
A probe's guidance computer converts a reading using \(f(x)=\frac{3x+4}{x-3}\), \(x\neq 3\).
(a)
Show that \(f(f(x))=x\) for all \(x\neq 3\) in the domain of the composite.
(4)
(b)
Hence write down \(f^{-1}(x)\) and state its domain.
(2)
(c)
Use your answer to (b) to find the value of \(x\) for which \(f(x)=2\).
(2)
(d)
Find the positive value of \(x\) for which \(f(x)=x\).
A1 \(x=3+\sqrt{13}\) (the other root \(3-\sqrt{13}\) is negative)
✓ 1
Exam tip: If \(f(f(x))=x\), the function is its own inverse, so \(f^{-1}(a)\) can be found simply by evaluating \(f(a)\).
151IBMA3S-2-16r17-112.8 Reciprocal and rational functionsPaper 1Hard12 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{ax+b}{x+c}\), for \(x\in\mathbb{R}\), \(x\neq-c\). The graph of \(y=f(x)\) has vertical asymptote \(x=-2\) and horizontal asymptote \(y=3\), and passes through the point \((1,5)\).
(a)
Write down the value of \(c\) and the value of \(a\).
(2)
(b)
Find the value of \(b\).
(2)
(c)
Find the coordinates of the \(x\)-intercept and the \(y\)-intercept of the graph.
(2)
(d)
Find \(f^{-1}(x)\).
(3)
(e)
The line \(y=x\) meets the graph of \(y=f(x)\) at two points. Find the \(x\)-coordinate of the point in the first quadrant.
Exam tip: Asymptotes of \(\frac{ax+b}{x+c}\) are \(x=-c\) and \(y=a\); read off \(a\) and \(c\) before using the point to find \(b\).
152IBMA3S-2-17r17-122.7 Quadratic equations and the discriminantPaper 1Hard10 marks
No calculator
The curve C has equation y = 2x2 + kx + 8, where k is a constant. The line L has equation y = x.
(a)
Show that the x-coordinates of any points of intersection of C and L satisfy 2x2 + (k − 1)x + 8 = 0.
(1)
(b)
The line L is a tangent to the curve C. Find the two possible values of k.
(4)
(c)
Find the set of values of k for which L and C intersect at two distinct points.
(2)
(d)
For the positive value of k found in (b), the line is tangent to the curve at the point P. Find the y-coordinate of P.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates 2x2 + kx + 8 = x and rearranges to the given form (AG)
✓ 1
Part (b)
M1 states that the discriminant equals zero
✓ 1
A1 (k − 1)2 − 4(2)(8) = 0, i.e. (k − 1)2 = 64
✓ 1
M1 solves: k − 1 = ±8
✓ 1
A1 k = 9 or k = −7
✓ 1
Part (c)
M1 sets the discriminant greater than zero: (k − 1)2 > 64
✓ 1
A1 k < −7 or k > 9
✓ 1
Part (d)
M1 substitutes k = 9: 2x2 + 8x + 8 = 0, i.e. (x + 2)2 = 0
✓ 1
A1 x = −2
✓ 1
A1 y = −2 (since P lies on y = x)
✓ 1
Exam tip: Tangency means a repeated root, so the discriminant is zero; two distinct intersections need a discriminant that is strictly greater than zero.
153IBMA3S-2-17r26-122.5 Composite and inverse functionsPaper 1Hard11 marks
No calculator
The function f is defined by f(x) = x2 − 6x + 10, for x ≥ 3. The function g is defined by g(x) = 2x − 5, for x ∈ ℝ.
(a)
Write f(x) in the form (x − a)2 + b.
(2)
(b)
Write down the range of f.
(1)
(c)
Find f−1(x) and state its domain.
(4)
(d)
Solve (f ∘ g)(x) = 5. Justify your answer.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (x − 3)2 seen, or a = 3
✓ 1
A1 b = 1
✓ 1
Part (b)
A1 f(x) ≥ 1
✓ 1
Part (c)
M1 swap x and y in y = (x − 3)2 + 1
✓ 1
M1 rearrange, e.g. x − 3 = ±√(y − 1)
✓ 1
A1 f−1(x) = 3 + √(x − 1) (positive root chosen since x ≥ 3)
✓ 1
A1 domain x ≥ 1
✓ 1
Part (d)
M1 (f ∘ g)(x) = (2x − 8)2 + 1
✓ 1
M1 (2x − 8)2 = 4
✓ 1
A1 x = 5 or x = 3
✓ 1
R1 g(x) must be in the domain of f, so 2x − 5 ≥ 3, x ≥ 4; reject x = 3, so x = 5
✓ 1
Common error: A composite is only defined where the output of the inner function lies in the domain of the outer function.
154IBMA3S-2-19r28-122.8 Reciprocal and rational functionsPaper 1Hard8 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{x+3}{x-1}\), where \(x\neq 1\).
(a)
Show that \((f\circ f)(x)=x\). [4]
(4)
(b)
Hence write down \(f^{-1}(x)\). [1]
(1)
(c)
The graphs of \(y=f(x)\) and \(y=x\) meet at two points. Find the x-coordinate of the point in the first quadrant. [3]
A1 numerator \(\frac{4x}{x-1}\) and denominator \(\frac{4}{x-1}\)
✓ 1
R1 quotient is \(x\) (AG)
✓ 1
Part (b)
A1 \(f^{-1}(x)=\frac{x+3}{x-1}\) (since \(f\circ f\) is the identity, \(f\) is its own inverse)
✓ 1
Part (c)
M1 \(x+3=x(x-1)\)
✓ 1
A1 \(x^2-2x-3=0\), so \(x=3\) or \(x=-1\)
✓ 1
A1 \(x=3\) (first quadrant)
✓ 1
Exam tip: If \(f(f(x))=x\) then \(f\) is its own inverse, so its graph is symmetric about the line \(y=x\). This gives a quick check of the answer to the last part.
155IBMA3S-2-19r28-132.8 Reciprocal and rational functionsPaper 1Hard8 marks
No calculator
Consider the equation \(9^{x}-4\times3^{x+1}+27=0\). Let \(u=3^{x}\).
(a)
Show that the equation can be written as \(u^2-12u+27=0\). [3]
(3)
(b)
Solve \(u^2-12u+27=0\). [2]
(2)
(c)
Hence find the two solutions of the original equation. State which is the larger. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(9^{x}=(3^{x})^2=u^2\)
✓ 1
M1 \(3^{x+1}=3\times3^{x}=3u\), so \(4\times3^{x+1}=12u\)
✓ 1
A1 \(u^2-12u+27=0\) (AG)
✓ 1
Part (b)
M1 factorise \((u-3)(u-9)=0\)
✓ 1
A1 \(u=3\) or \(u=9\)
✓ 1
Part (c)
M1 \(3^{x}=3\) or \(3^{x}=9\)
✓ 1
A1 \(x=1\)
✓ 1
A1 \(x=2\), the larger solution
✓ 1
Common error: After finding \(u\), remember to return to \(x\) by solving \(3^{x}=u\). Quoting \(u=3\) and \(u=9\) as the solutions leaves the question unanswered.
156IBMA3S-2-17r39-102.5 Composite and inverse functionsPaper 1Hard14 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=\frac{x+1}{x-2}\), \(x\ne2\), and \(g(x)=3x-1\), \(x\in\mathbb{R}\).
(a)
Show that \((f\circ g)(x)=\frac{x}{x-1}\).
(3)
(b)
State the largest possible domain of \(f\circ g\).
(1)
(c)
Show that \(f\circ g\) is a self-inverse function; that is, \((f\circ g)^{-1}(x)=(f\circ g)(x)\).
(3)
(d)
Find \(f^{-1}(x)\).
(3)
(e)
Find \(g^{-1}(x)\).
(1)
(f)
Show that \((g^{-1}\circ f^{-1})(x)=(f\circ g)(x)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(g(x)\) into \(f\): \(\frac{(3x-1)+1}{(3x-1)-2}\)
✓ 1
A1 \(\frac{3x}{3x-3}\)
✓ 1
A1 divide numerator and denominator by 3 to obtain \(\frac{x}{x-1}\)
✓ 1
Part (b)
A1 \(x\in\mathbb{R}\), \(x\ne1\)
✓ 1
Part (c)
M1 let \(y=\frac{x}{x-1}\) and rearrange: \(y(x-1)=x\)
✓ 1
M1 collect \(x\) terms: \(x(y-1)=y\)
✓ 1
A1 \(x=\frac{y}{y-1}\), so \((f\circ g)^{-1}(x)=\frac{x}{x-1}=(f\circ g)(x)\)
✓ 1
Part (d)
M1 let \(y=\frac{x+1}{x-2}\) and rearrange: \(y(x-2)=x+1\)
✓ 1
M1 collect \(x\) terms: \(x(y-1)=2y+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{2x+1}{x-1}\)
✓ 1
Part (e)
A1 \(g^{-1}(x)=\frac{x+1}{3}\)
✓ 1
Part (f)
M1 substitute \(f^{-1}\) into \(g^{-1}\): \(\frac13\left(\frac{2x+1}{x-1}+1\right)\)
✓ 1
A1 \(\frac13\times\frac{3x}{x-1}\)
✓ 1
A1 \(\frac{x}{x-1}=(f\circ g)(x)\)
✓ 1
Exam tip: The inverse of a composite reverses the order: \((f\circ g)^{-1}=g^{-1}\circ f^{-1}\). Part (c) tells you that here it also equals \(f\circ g\) itself, which is a useful check on part (f).
157IBMA3S-2-19r40-142.9 Exponential and logarithmic functionsPaper 1Hard12 marks
No calculator
This question is about exponential and logarithmic equations.
(a)
Solve \(4^x-5\times2^x+4=0\).
(5)
(b)
Solve \(\ln x+\ln(x-2)=\ln 3\).
(4)
(c)
Hence, or otherwise, solve \(4^x-5\times2^x+6=0\), giving your answers in exact form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(4^x=(2^x)^2\) and lets \(u=2^x\)
✓ 1
A1 \(u^2-5u+4=0\)
✓ 1
M1 \((u-1)(u-4)=0\) so \(u=1\) or \(u=4\)
✓ 1
A1 \(x=0\)
✓ 1
A1 \(x=2\)
✓ 1
Part (b)
M1 combines the logarithms, \(\ln\left(x(x-2)\right)=\ln 3\)
✓ 1
A1 \(x^2-2x-3=0\)
✓ 1
M1 \((x-3)(x+1)=0\)
✓ 1
A1 \(x=3\), with \(x=-1\) rejected because \(\ln x\) is undefined for negative \(x\)
Common error: Always check solutions of logarithmic equations in the original equation: a root that makes the argument of a logarithm negative must be rejected.
158AASL2-q00062.1 Equations of straight linesPaper 2Easy8 marks
A straight mountain road climbs from a village at a height of 312 m above sea level to a pass at a height of 396 m. The horizontal distance from the village to the pass is 1200 m. Let \(x\) be the horizontal distance, in metres, from the village and \(h\) the height, in metres, of the road above sea level.
Diagram not to scale
(a)
Find the gradient of the road.
(2)
(b)
Write down an equation for \(h\) in terms of \(x\).
(1)
(c)
A viewpoint on the road is 350 m above sea level. Find the horizontal distance of the viewpoint from the village.
(2)
(d)
A straight cable-car line is parallel to the road and passes through the point \((0,\,340)\). Write down the equation of the cable-car line.
(1)
(e)
Building regulations state that an access ramp must have a gradient no greater than \(\dfrac{1}{12}\). A ramp must rise 0.65 m. Find the least horizontal length of the ramp.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{396-312}{1200}\)
(M1)
\(0.07\)
A1
Accept \(\tfrac{7}{100}\) or 7%.
(b)
\(h=0.07x+312\)
A1
FT from (a).
(c)
\(0.07x+312=350\)
(M1)
\(543\) m
A1
\(542.857\ldots\) FT from (b).
(d)
\(h=0.07x+340\)
A1
Parallel lines have equal gradients, \(m_{1}=m_{2}\). FT from (a).
(e)
\(\dfrac{0.65}{\ell}\le\dfrac{1}{12}\) or \(0.65\times12\)
(M1)
\(7.8\) m
A1
Answers: (a) 0.07 · (b) \(h=0.07x+312\) · (c) 543 m · (d) \(h=0.07x+340\) · (e) 7.8 m
159AASL2-q00162.2 Functions, domain, range and inversePaper 2Easy8 marks
The fare, \(F\) dollars, for a taxi journey of \(d\) kilometres is modelled by \(F(d)=3.20+1.45d\), for \(0\lt d\le40\).
(a)
Find the fare for a journey of 12 km.
(2)
(b)
Interpret the value 1.45 in the context of the model.
(1)
(c)
Find an expression for \(F^{-1}(x)\).
(2)
(d)
Find \(F^{-1}(30.75)\) and interpret your answer in context.
(2)
(e)
Write down the range of \(F\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(3.20+1.45(12)\)
(M1)
$20.60
A1
(b)
the cost, in dollars, per kilometre travelled
A1
Accept the fare increases by $1.45 for each extra km.
(c)
interchanging variables: \(x=3.20+1.45d\)
(M1)
Or attempt to make \(d\) the subject.
\(F^{-1}(x)=\dfrac{x-3.20}{1.45}\)
A1
(d)
\(19\)
A1
a fare of $30.75 corresponds to a journey of 19 km
160AASL2-q00272.4 Key features and intersectionsPaper 2Easy6 marks
A point O lies at the foot of a straight hillside. A ball is thrown from a point 1.8 m vertically above O, up the hillside. In a vertical cross-section, the path of the ball is modelled by \(y=1.8+0.6x-0.025x^{2}\) and the hillside is modelled by \(y=0.1x\), for \(x\ge0\), where \(x\) metres is the horizontal distance from O and \(y\) metres is the height above the level of O.
Diagram not to scale
(a)
Find the horizontal distance from O to the point where the ball lands on the hillside.
(2)
(b)
Find the maximum height of the ball above the level of O.
(1)
(c)
Find the greatest vertical distance between the ball and the hillside, and the value of \(x\) at which it occurs.
(2)
(d)
Explain why the answer to (c) occurs at a smaller value of \(x\) than the answer to (b).
(1)
Show mark scheme
Working
Mark
Notes
(a)
intersecting \(y=1.8+0.6x-0.025x^{2}\) with \(y=0.1x\) on a GDC
(M1)
\(23.1\) m
A1
\(23.1148\ldots\) (the other intersection is negative and is rejected).
(b)
\(5.4\) m
A1
At \(x=12\).
(c)
graphing \(d(x)=1.8+0.6x-0.025x^{2}-0.1x\), or using its vertex
(M1)
\(4.3\) m at \(x=10\)
A1
Both values required.
(d)
the hillside rises as \(x\) increases, so beyond \(x=10\) the ground rises faster than the ball does; the gap is largest before the ball reaches its highest point
R1
Accept: \(d(x)=1.8+0.5x-0.025x^{2}\) has its vertex at \(x=10\lt12\).
Answers: (a) 23.1 m · (b) 5.4 m · (c) 4.3 m at \(x=10\)
161IBMA2S-046-02-022.2 Functions, domain, range and inversePaper 2Easy6 marks
Calculator
A ride-share company charges a fare of \(C\) dollars for a journey of \(d\) kilometres, where \(C(d)=2.80+1.35d\), \(d\ge 0\).
(a)
Find the fare for a journey of 12.5 km.
(2)
(b)
Find \(C^{-1}(x)\).
(2)
(c)
Find \(C^{-1}(20)\) and interpret its meaning in this context.
M1 interchanges variables or rearranges, e.g. \(x=2.80+1.35y\)
✓ 1
A1 \(C^{-1}(x)=\frac{x-2.80}{1.35}\)
✓ 1
Part (c)
A1 \(C^{-1}(20)=12.7\) (12.74…)
✓ 1
R1 a fare of $20 corresponds to a journey of about 12.7 km
✓ 1
Exam tip: When asked to interpret an inverse value, state both quantities with their units: the input of \(C^{-1}\) is a fare in dollars and the output is a distance in kilometres.
162IBMA2S-046-02-092.10 Solving equationsPaper 2Easy9 marks
Calculator
The sound level \(L\) decibels (dB) produced by a pair of headphones is related to the sound intensity \(I\) W m−2 by \(L(I)=10\log_{10}\left(\frac{I}{10^{-12}}\right)\), where \(10^{-12}\le I\le 10\).
(a)
Find the sound level when \(I=2.5\times10^{-6}\).
(2)
(b)
Find the range of \(L\).
(2)
(c)
Find an expression for \(L^{-1}(x)\).
(3)
(d)
A safety setting limits the sound level to 85 dB. Find the corresponding intensity.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes: \(10\log_{10}(2.5\times10^{6})\)
✓ 1
A1 \(64.0\) dB
✓ 1
Part (b)
A1 \(L(10^{-12})=0\)
✓ 1
A1 \(L(10)=130\), so \(0\le L\le 130\)
✓ 1
Part (c)
M1 divides by 10: \(\frac{x}{10}=\log_{10}\left(\frac{y}{10^{-12}}\right)\)
✓ 1
M1 writes in exponential form: \(\frac{y}{10^{-12}}=10^{x/10}\)
✓ 1
A1 \(L^{-1}(x)=10^{\frac{x}{10}-12}\)
✓ 1
Part (d)
M1 evaluates \(L^{-1}(85)=10^{8.5-12}\)
✓ 1
A1 \(3.16\times10^{-4}\) W m−2
✓ 1
Exam tip: Before converting to exponential form, divide by the coefficient of the logarithm. Writing \(10^{x}\) instead of \(10^{x/10}\) is a frequent slip.
163IBMA2S-046-01-022.7 Quadratic equations and the discriminantPaper 2Easy7 marks
Calculator
A delivery drone is launched from the roof of a building. Its height, h metres, above the ground t seconds after launch is modelled by \(h(t) = -0.8t^2 + 9.6t + 1.5\), for \(t \geq 0\), until the drone reaches the ground.
(a)
Find the maximum height of the drone and the time at which it occurs.
(3)
(b)
Find the time at which the model predicts the drone reaches the ground.
(2)
(c)
Find the length of time for which the drone is more than 20 m above the ground.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(t = -\frac{b}{2a} = \frac{9.6}{1.6}\) or GDC maximum
✓ 1
A1 \(t = 6\) s
✓ 1
A1 maximum height \(= 30.3\) m
✓ 1
Part (b)
M1 attempt to solve \(-0.8t^2 + 9.6t + 1.5 = 0\)
✓ 1
A1 \(t = 12.2\) s (12.154…), rejecting the negative root
✓ 1
Part (c)
M1 solving \(h(t) = 20\) to obtain \(t = 2.41\) and \(t = 9.59\)
✓ 1
A1 \(9.588 - 2.412 = 7.18\) s
✓ 1
Exam tip: When a model only applies for \(t \geq 0\), state clearly that the negative root is rejected rather than silently ignoring it.
164IBMA2S-046-01-112.10 Solving equationsPaper 2Easy5 marks
Calculator
Let \(f(x) = x^2 - 3x - 1\) and \(g(x) = 4\ln x\), for \(x > 0\).
(a)
Solve the equation \(f(x) = g(x)\).
(3)
(b)
Hence solve the inequality \(f(x) < g(x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find intersections of the graphs of f and g on GDC
✓ 1
A1 \(x = 0.555\)
✓ 1
A1 \(x = 4.55\)
✓ 1
Part (b)
M1 recognising that the graph of f lies below the graph of g between the intersections
✓ 1
A1 \(0.555 < x < 4.55\)
✓ 1
Exam tip: A standard GDC window may hide the intersection near \(x = 0.5\). Zoom in close to the y-axis, since \(\ln x\) changes rapidly there.
165IBMA2S-046-03-042.2 Functions, domain, range and inversePaper 2Easy6 marks
Calculator
A children's liquid painkiller is dosed according to body mass. For a child of mass \(w\) kg, the recommended dose in millilitres is \(D(w)=0.4w+1.5\), for \(5\le w\le 40\).
(a)
Find the range of \(D\).
(2)
(b)
Find \(D^{-1}(x)\) and state its domain.
(3)
(c)
A nurse gives a child a dose of 9.3 ml. Find the mass of the child for whom this is the recommended dose.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evaluates \(D(5)=3.5\) and \(D(40)=17.5\)
✓ 1
A1 \(3.5\le D(w)\le 17.5\)
✓ 1
Part (b)
M1 rearranges \(x=0.4w+1.5\) to make \(w\) the subject
Exam tip: The domain of an inverse function is the range of the original function. Find the range first and you get the inverse's domain for free.
166IBMA2S-046-07-022.10 Solving equationsPaper 2Easy6 marks
Calculator
A hospital uses a linear rule to set the loading dose D mg of a drug for a patient of body mass m kg. A patient of mass 60 kg receives 420 mg and a patient of mass 85 kg receives 545 mg.
(a)
Find the gradient of the line relating D to m, and state its units.
(2)
(b)
Find an equation for D in terms of m.
(2)
(c)
A patient is given a loading dose of 500 mg. Find the patient's body mass.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{545-420}{85-60}\)
✓ 1
A1 5 mg per kg
✓ 1
Part (b)
M1 substitutes a data point, e.g. \(420=5(60)+c\)
✓ 1
A1 \(D=5m+120\)
✓ 1
Part (c)
M1 \(5m+120=500\)
✓ 1
A1 \(m=76\) kg
✓ 1
Exam tip: When the unknown is the input rather than the output, set your equation equal to the given value and solve; do not substitute 500 for m.
167IBMA2S-046-07-042.10 Solving equationsPaper 2Easy5 marks
Calculator
The battery level of a smartphone decreases linearly during a working day. At 08:00 (t = 0 hours) the battery level is 92%. At 11:30 the battery level is 64%.
(a)
Find the gradient of the line giving battery level against t.
(2)
(b)
Interpret the gradient in context.
(1)
(c)
The phone shows a low-battery warning when the level reaches 15%. Find the time of day at which the warning appears.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{64-92}{3.5-0}\)
✓ 1
A1 \(-8\) (% per hour; units not required)
✓ 1
Part (b)
A1 the battery level falls by 8 percentage points every hour
✓ 1
Part (c)
M1 \(92-8t=15\)
✓ 1
A1 \(t=9.625\) hours, so at 17:37:30 (accept 17:37 or 17:38)
✓ 1
Common error: 11:30 is 3.5 hours after 08:00, not 3.3 hours. Convert minutes to decimal hours before finding the gradient, and convert 0.625 h back to minutes (37.5 min) at the end.
168IBMA2S-046-06-032.10 Solving equationsPaper 2Easy7 marks
Calculator
After a single dose, the concentration of a drug in the blood of a first patient is modelled by \(C(t) = 20e^{-0.3t}\), \(t \ge 0\), where \(C\) is in mg L−1 and \(t\) is the time in hours after a clock is started. A second patient receives an identical dose 2 hours later. Their concentration is modelled by \(D(t) = C(t-2)\), \(t \ge 2\).
(a)
Describe fully the transformation that maps the graph of \(y = C(t)\) onto the graph of \(y = D(t)\).
(2)
(b)
Find \(D(5)\).
(2)
(c)
A third patient receives a larger dose 1 hour after the clock is started. Their concentration is modelled by \(E(t) = 1.5\,C(t-1)\), \(t \ge 1\). Find the value of \(t\) at which \(E(t) = 10\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 horizontal translation
✓ 1
A1 2 units in the positive \(t\)-direction (to the right), i.e. by vector \(\binom{2}{0}\)
✓ 1
Part (b)
M1 \(D(5) = C(3) = 20e^{-0.3 \times 3}\)
✓ 1
A1 \(8.13\) mg L−1
✓ 1
Part (c)
M1 \(30e^{-0.3(t-1)} = 10\)
✓ 1
M1 \(-0.3(t-1) = \ln\frac{1}{3}\), or an equivalent method such as solving on a GDC
✓ 1
A1 \(t = 4.66\) hours
✓ 1
Exam tip: Write out \(E(t)\) in full before solving: substitute \(t-1\) for every \(t\) in \(C\), then multiply the whole expression by 1.5.
169IBMA2S-046-05-022.9 Exponential and logarithmic functionsPaper 2Easy6 marks
Calculator
A fully charged smartphone is left running a navigation app. Its battery charge, C%, after t hours is modelled by C(t) = 100e−kt, where k is a positive constant. After 5 hours the charge is 64%.
(a)
Show that k = 0.0893, correct to three significant figures.
(2)
(b)
The phone displays a low-battery warning when the charge falls to 15%. Find how many hours after being fully charged the warning appears.
(2)
(c)
Find the percentage of its current charge that the battery loses during any one-hour period.
M1 15 = 100e−kt, giving t = ln(100/15)/k ; A1 t = 21.3 hours using the unrounded k, or 21.2 hours using k = 0.0893; accept either
✓ 2
Part (c)
M1 the ratio of the charges one hour apart is e−k = 0.9146… ; A1 1 − 0.9146… gives a loss of 8.54%
✓ 2
Common error: Losing 8.54% of the current charge each hour is not the same as losing 8.54 percentage points each hour. The exponential model removes a constant proportion, not a constant amount.
170IBMA2S-046-00-022.4 Key features and intersectionsPaper 2Easy6 marks
Calculator
The path of a basketball shot is modelled by \(h(x)=-0.12x^2+1.08x+2.1\), where \(h\) is the height of the ball above the ground in metres and \(x\) is its horizontal distance in metres from the point of release.
(a)
Find the maximum height reached by the ball.
(3)
(b)
The ball passes through the hoop, which is 3.05 m above the ground, while it is descending. Find the horizontal distance from the point of release to the hoop.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(x=-\frac{b}{2a}\) or calculator maximum
✓ 1
A1 \(x=4.5\)
✓ 1
A1 maximum height \(=4.53\) m
✓ 1
Part (b)
M1 sets \(-0.12x^2+1.08x+2.1=3.05\)
✓ 1
A1 \(x=0.988\) or \(x=8.01\)
✓ 1
A1 \(x=8.01\) m, as the ball is descending only after \(x=4.5\)
✓ 1
Common error: Both solutions of h(x) = 3.05 are valid points on the path. Choose the one after the vertex, because the ball is descending there.
171IBMA3S-2-17r11-032.4 Key features and intersectionsPaper 2Easy7 marks
Calculator
In a coffee roaster the bean temperature, in °C, after \(t\) minutes is \(f(t)=20+16t\) for \(0\leq t\leq 10\). A darkness score is calculated from the temperature \(T\) using \(g(T)=0.35T-6\).
(a)
Find \((g\circ f)(t)\), giving your answer in the form \(a+bt\).
(2)
(b)
Hence find the time at which the darkness score equals 29.
(2)
(c)
Find \(g^{-1}(x)\).
(2)
(d)
Hence find the bean temperature when the darkness score is 29.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(g(20+16t)=0.35(20+16t)-6\)
✓ 1
A1 \((g\circ f)(t)=1+5.6t\)
✓ 1
Part (b)
M1 \(1+5.6t=29\)
✓ 1
A1 \(t=5\) minutes
✓ 1
Part (c)
M1 \(y=0.35x-6\) rearranged for \(x\) (or swap and rearrange)
✓ 1
A1 \(g^{-1}(x)=\frac{x+6}{0.35}\)
✓ 1
Part (d)
A1 \(g^{-1}(29)=\frac{35}{0.35}=100\) °C (consistent with \(f(5)=100\))
✓ 1
Exam tip: In \((g\circ f)(t)\) the inner function \(f\) is applied first, so substitute \(f(t)\) into \(g\).
172IBMA3S-2-16r17-032.4 Key features and intersectionsPaper 2Easy5 marks
Calculator
The air pressure \(P\) kilopascals (kPa) at an altitude of \(h\) kilometres above sea level is modelled by \(P=101e^{-0.125h}\), for \(0\le h\le 12\). An aircraft cruises at an altitude of 10.5 km.
(a)
Find the air pressure at the cruising altitude. Give your answer to 3 significant figures.
(2)
(b)
Find the altitude at which the air pressure is 50 kPa. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P=101e^{-0.125\times10.5}\)
✓ 1
A1 \(P=27.2\) kPa
✓ 1
Part (b)
M1 \(101e^{-0.125h}=50\)
✓ 1
M1 \(-0.125h=\ln\frac{50}{101}\), or \(h=\frac{\ln(101/50)}{0.125}\)
✓ 1
A1 \(h=5.62\) km
✓ 1
Exam tip: To undo \(e^{kx}\) take natural logarithms of both sides; keep full calculator values until the final rounding.
173IBMA3S-2-17r17-022.9 Exponential and logarithmic functionsPaper 2Easy5 marks
Calculator
The number of active players of a new online game, t weeks after its launch, is modelled by P(t) = 800 × 1.12t.
(a)
Find the number of active players predicted after 5 weeks. Give your answer to the nearest whole number.
(2)
(b)
Find the value of t for which the model predicts 2000 active players. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes t = 5: 800 × 1.125 (= 1409.87…)
✓ 1
A1 1410
✓ 1
Part (b)
M1 forms 1.12t = 2.5
✓ 1
M1 takes logarithms: t = ln 2.5 / ln 1.12 (accept a correct GDC method)
✓ 1
A1 t = 8.09 weeks
✓ 1
Exam tip: Isolate the exponential term first (divide by 800) and only then take logarithms of both sides.
174IBMA3S-2-17r17-042.9 Exponential and logarithmic functionsPaper 2Easy6 marks
Calculator
The pH of a sample of water is given by pH = −log10(H), where H mol L−1 is the concentration of hydrogen ions. Sample A has H = 4.0 × 10−8. Sample B has a pH of 6.8.
(a)
Find the pH of sample A. Give your answer to 3 significant figures.
(2)
(b)
Find the hydrogen ion concentration of sample B. Give your answer in the form a × 10k, where 1 ≤ a < 10 and k is an integer, with a to 3 significant figures.
(2)
(c)
Calculate how many times greater the hydrogen ion concentration of sample B is than that of sample A. Use the unrounded value of H for sample B and give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 −log10(4.0 × 10−8)
✓ 1
A1 pH = 7.40
✓ 1
Part (b)
M1 H = 10−6.8
✓ 1
A1 1.58 × 10−7 mol L−1
✓ 1
Part (c)
M1 divides: 10−6.8 / (4.0 × 10−8)
✓ 1
A1 3.96
✓ 1
Exam tip: A lower pH means a higher concentration: use the inverse function H = 10⁻pH and keep the full calculator value until the final line.
175IBMA3S-2-17r26-022.7 Quadratic equations and the discriminantPaper 2Easy6 marks
Calculator
A swimming lake has a measured depth D metres at a distance x metres from the shore along a marked line, modelled by D(x) = −0.02x2 + 0.4x + 1, for 0 ≤ x ≤ 20.
(a)
Write down the depth of the lake at the shore.
(1)
(b)
Find the maximum depth of the lake along this line and the value of x at which it occurs.
(2)
(c)
Swimmers must be kept to the part of the line where the depth is at least 2.5 m. Find the length of this part of the line.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 D(0) = 1 m
✓ 1
Part (b)
A1 x = 10
✓ 1
A1 maximum depth = 3 m
✓ 1
Part (c)
M1 set up D(x) = 2.5 or an inequality, or sketch/GDC solve
✓ 1
A1 x = 5 and x = 15
✓ 1
A1 length = 10 m
✓ 1
Exam tip: A quadratic graph is symmetrical about its vertex, so the two solutions of D = 2.5 lie the same distance either side of x = 10.
176IBMA3S-2-17r26-042.6 Quadratic functionsPaper 2Easy5 marks
Calculator
A household's monthly electricity bill is C = f(u) = 0.15u + 12, where C is the cost in pounds (£) and u is the number of units used.
(a)
Find f−1(C).
(2)
(b)
Explain what f−1(C) represents in this context.
(1)
(c)
A bill is £51. Find the number of units used.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to rearrange C = 0.15u + 12 for u
✓ 1
A1 f−1(C) = (C − 12)/0.15
✓ 1
Part (b)
R1 the number of units used for a bill of £C
✓ 1
Part (c)
M1 substitute C = 51 into the inverse
✓ 1
A1 260 units
✓ 1
Exam tip: Check an inverse by substituting your answer back into the original function.
177IBMA3S-2-19r28-022.9 Exponential and logarithmic functionsPaper 2Easy6 marks
Calculator
The number of customers using a shop's loyalty app is modelled by \(N(t)=40\,e^{0.15t}\), where t is the number of months after the app was launched.
(a)
Write down the number of customers at launch. [1]
(1)
(b)
Find the number of customers 10 months after launch. Give your answer to the nearest whole number. [2]
(2)
(c)
Find the time taken for the number of customers to reach 500. Give your answer in months to 3 significant figures. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 40
✓ 1
Part (b)
M1 \(40e^{0.15\times10}=40e^{1.5}\)
✓ 1
A1 179
✓ 1
Part (c)
M1 \(40e^{0.15t}=500\), so \(e^{0.15t}=12.5\)
✓ 1
M1 \(0.15t=\ln 12.5\)
✓ 1
A1 \(t=16.8\) months
✓ 1
Exam tip: Isolate the exponential first by dividing by 40, then take natural logarithms of both sides. Do not take logarithms of 500 and 40 separately and divide them.
178IBMA3S-2-17r39-032.6 Quadratic functionsPaper 2Easy6 marks
Calculator
A shipping company charges €\(C\) to carry one container of mass \(w\) tonnes, where \(C(w)=12+0.8w\) for \(5\le w\le 40\).
(a)
Find \(C(25)\).
(1)
(b)
Find the range of \(C\).
(2)
(c)
Find an expression for \(C^{-1}(x)\).
(2)
(d)
A container costs €30.40 to carry. Find its mass.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(C(25)=32\)
✓ 1
Part (b)
M1 evaluate \(C(5)\) and \(C(40)\): 16 and 44
✓ 1
A1 \(16\le C\le 44\)
✓ 1
Part (c)
M1 \(x=12+0.8w\) and rearrange for \(w\)
✓ 1
A1 \(C^{-1}(x)=\frac{x-12}{0.8}=1.25x-15\)
✓ 1
Part (d)
A1 \(C^{-1}(30.4)=23\) tonnes
✓ 1
Exam tip: The inverse swaps inputs and outputs: here it takes a cost and returns a mass.
179IBMA3S-2-17r39-092.6 Quadratic functionsPaper 2Easy6 marks
Calculator
A theatre sells \(N=600-15p\) tickets for each performance when the ticket price is €\(p\), where \(0<p<40\). The revenue from a performance is €\(R\).
(a)
Show that \(R=600p-15p^2\).
(1)
(b)
Find the ticket price that gives the greatest revenue.
(2)
(c)
Find the greatest revenue.
(1)
(d)
Find the greatest ticket price for which the revenue is at least €5400, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(R=pN=p(600-15p)=600p-15p^2\)
✓ 1
Part (b)
M1 axis of symmetry \(p=-\frac{600}{2(-15)}\) or \(\frac{dR}{dp}\) method not required
✓ 1
A1 \(p=20\)
✓ 1
Part (c)
A1 \(R(20)=600(20)-15(400)=6000\), i.e. €6000
✓ 1
Part (d)
M1 solve \(600p-15p^2=5400\), i.e. \(p^2-40p+360=0\)
✓ 1
A1 \(p=20+\sqrt{40}=26.3\), i.e. €26.3 (the other root is 13.7)
✓ 1
Common error: Both roots of the equation are needed to describe where \(R\ge5400\); the greatest price is the larger root, not the smaller.
180AASL2-q00072.1 Equations of straight linesPaper 2Medium6 marks
A funicular railway runs in a straight line up a hillside. The track is 1500 m long, measured along the slope, and the upper station is 270 m higher than the lower station.
Diagram not to scale
(a)
Find the horizontal distance between the two stations.
(2)
(b)
Find the gradient of the track. Give your answer as a decimal and as a percentage.
(2)
(c)
A second funicular is to be built on another hillside. Its track must rise 300 m and, for safety, its gradient must not exceed 0.25. Find the least possible length of this track, measured along the slope.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\sqrt{1500^{2}-270^{2}}\)
(M1)
Use of Pythagoras with the track as the hypotenuse.
\(1480\) m
A1
\(1475.49\ldots\) Accept 1475 or 1475.5.
(b)
\(\dfrac{270}{1475.49\ldots}=0.183\)
A1
\(0.182988\ldots\) FT from (a). Award A0 for \(\tfrac{270}{1500}=0.18\) (slope length used).
\(18.3\%\)
A1
FT from their decimal.
(c)
least horizontal distance \(\dfrac{300}{0.25}=1200\) m, and attempt to use Pythagoras
(M1)
\(\sqrt{1200^{2}+300^{2}}=1240\) m
A1
\(1236.93\ldots\) Accept 1237. Award M1A0 for 1200 (horizontal distance only).
Answers: (a) 1480 m (1475.5) · (b) 0.183, 18.3% · (c) 1240 m (1236.9)
181AASL2-q00192.3 Graphs of functionsPaper 2Medium7 marks
Let \(h(x)=\mathrm{e}^{\frac{x}{2}}-x^{2}+3\), for \(-3\le x\le6\).
(a)
Sketch the graph of \(y=h(x)\) on the given domain, showing the coordinates of any intercepts with the axes, the local maximum point and the endpoints.
(3)
(b)
Find the range of \(h\).
(2)
(c)
Find the set of values of \(k\) for which the equation \(h(x)=k\) has exactly two solutions.
(2)
Show mark scheme
Working
Mark
Notes
(a)
correct shape: increases to a single maximum, then decreases throughout to the right endpoint
A1
\(x\)-intercepts \((-1.84,\,0)\) and \((2.57,\,0)\); \(y\)-intercept \((0,\,4)\)
A1
maximum \((0.289,\,4.07)\); endpoints \((-3,\,-5.78)\) and \((6,\,-12.9)\)
A1
Award for at least the maximum and both endpoints in approximately correct positions.
(b)
recognizing the least value occurs at the endpoint \(x=6\)
(M1)
\(h(6)=-12.914\ldots\)
\(-12.9\le h(x)\le4.07\)
A1
(c)
horizontal line meets the graph twice between \(y=h(-3)\) and the maximum
(M1)
\(-5.78\le k\lt 4.07\)
A1
Both ends must be correct, including the inclusive left end: \(h(-3)=-5.7768\ldots\), \(h_{\max}=4.0719\ldots\)
182AASL2-q00202.3 Graphs of functionsPaper 2Medium8 marks
Let \(f(x)=x\) and \(g(x)=\mathrm{e}^{-x}\). The function \(h\) is defined by \(h(x)=f(x)+g(x)\), for \(-2\le x\le3\).
(a)
On the same set of axes, sketch the graphs of \(y=f(x)\) and \(y=h(x)\), for \(-2\le x\le3\).
(3)
(b)
Write down the coordinates of the minimum point of the graph of \(h\).
(2)
(c)
Explain why the graph of \(h\) lies above the graph of \(f\) for all values of \(x\).
(1)
(d)
Solve \(h(x)=2\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
straight line \(y=x\) through the origin
A1
graph of \(h\) with a single minimum near \((0,\,1)\), decreasing then increasing
A1
endpoints of \(h\) approximately \((-2,\,5.39)\) and \((3,\,3.05)\), with \(h\) above \(y=x\) and approaching it as \(x\) increases
A1
(b)
\(x=0\)
A1
\(y=1\)
A1
Award A1A0 if only the x-coordinate is given.
(c)
\(h(x)-f(x)=\mathrm{e}^{-x}\gt 0\) for all \(x\)
R1
(d)
\(x=-1.15\)
A1
\(-1.14619\ldots\)
\(x=1.84\)
A1
\(1.84140\ldots\)
Answers: (b) \((0,\,1)\) · (d) \(x=-1.15,\ 1.84\)
Model sketch
Diagram not to scale
183AASL2-q00222.4 Key features and intersectionsPaper 2Medium8 marks
The concentration of a drug in a patient’s blood, \(C\) mg l−1, \(t\) hours after an injection is modelled by \(C(t)=12t\,\mathrm{e}^{-0.4t}\), for \(t\ge0\).
(a)
Find the maximum concentration and the time at which it occurs.
(2)
(b)
The drug is effective while the concentration is at least 8 mg l−1. Find the length of time for which the drug is effective.
(3)
(c)
Find the time after which the concentration remains below 0.5 mg l−1.
(2)
(d)
Write down the equation of the horizontal asymptote of the graph of \(C\) and interpret it in context.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(t=2.5\) hours
A1
Using the maximum feature of the GDC.
\(11.0\) mg l−1
A1
\(11.036\ldots\)
(b)
intersecting \(y=C(t)\) with \(y=8\)
(M1)
Or using the GDC equation solver twice.
\(t=0.991\) and \(t=5.07\)
(A1)
\(4.08\) hours
A1
\(4.0830\ldots\) Accept 4.07–4.08.
(c)
\(12t\,\mathrm{e}^{-0.4t}=0.5\), with \(t\gt 2.5\)
(M1)
\(14.7\) hours
A1
\(14.657\ldots\)
(d)
\(C=0\): in the long term the drug is eliminated from the blood
A1
Both equation and interpretation required.
Answers: (a) 11.0 mg l−1 at 2.5 h · (b) 4.08 h · (c) 14.7 h · (d) \(C=0\)
184AASL2-q00232.4 Key features and intersectionsPaper 2Medium6 marks
Let \(f(x)=4-x^{2}\) and \(g(x)=2^{x}\), for \(x\in\mathbb{R}\). The graphs of \(f\) and \(g\) intersect at two points, A and B, where A has the smaller \(x\)-coordinate.
(a)
Find the coordinates of A and of B.
(3)
(b)
Write down the set of values of \(x\) for which \(f(x)\gt g(x)\).
(1)
(c)
Let \(d(x)=f(x)-g(x)\). Find the maximum value of \(d(x)\) for values of \(x\) between A and B, and the value of \(x\) at which it occurs.
(2)
Show mark scheme
Working
Mark
Notes
(a)
intersecting the two graphs on a GDC, or solving \(4-x^{2}=2^{x}\)
(M1)
\(\mathrm{A}(-1.93,\,0.262)\)
A1
\(\mathrm{B}(1.26,\,2.40)\)
A1
(b)
\(-1.93\lt x\lt 1.26\)
A1
FT from (a).
(c)
graphing \(y=4-x^{2}-2^{x}\) and finding its maximum
185AASL2-q00242.8 Reciprocal and rational functionsPaper 2Medium7 marks
Let \(f(x)=\dfrac{2x^{2}+1}{x^{2}-9}\), for \(x\in\mathbb{R}\), \(x\neq\pm3\).
(a)
Write down the equations of the vertical and horizontal asymptotes of the graph of \(f\).
(3)
(b)
Write down the coordinates of the local maximum point of the graph of \(f\).
(1)
(c)
Find the range of \(f\).
(2)
(d)
Describe the symmetry of the graph of \(f\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(x=3\)
A1
\(x=-3\)
A1
\(y=2\)
A1
(b)
\((0,\,-0.111)\)
A1
Accept \(\left(0,\,-\tfrac19\right)\).
(c)
\(f(x)\le-0.111\)
A1
Accept \(f(x)\le-\tfrac19\).
or \(f(x)\gt2\)
A1
The graph approaches y = 2 from above for |x| > 3.
(d)
symmetrical about the \(y\)-axis (the line \(x=0\))
A1
Answers: (a) \(x=\pm3\), \(y=2\) · (b) \((0,-0.111)\) · (c) \(f(x)\le-0.111\) or \(f(x)\gt2\) · (d) about the \(y\)-axis
186AASL2-q00252.4 Key features and intersectionsPaper 2Medium6 marks
For a particular product, the price, \(p\) dollars, at which consumers will buy \(q\) thousand units per week is modelled by the demand function \(p=50-0.4q\). The price at which producers will supply \(q\) thousand units per week is modelled by the supply function \(p=5+0.02q^{2}\). Both models apply for \(0\le q\le120\).
(a)
Market equilibrium occurs where the graphs of the two functions intersect. Find the equilibrium quantity and price.
(2)
(b)
A tax raises the supply function to \(p=11+0.02q^{2}\). Find the new equilibrium quantity and price.
(2)
(c)
Find the increase in the equilibrium price, and express it as a percentage of the tax of $6.
(2)
Show mark scheme
Working
Mark
Notes
(a)
intersecting the graphs on a GDC, or solving \(50-0.4q=5+0.02q^{2}\)
187AASL2-q00362.5 Composite and inverse functionsPaper 2Medium9 marks
A bank converts euros (€) to US dollars ($). The function \(f(x)=1.08x\) gives the number of dollars equivalent to \(x\) euros. The bank then deducts a commission of 2% of the dollar amount and a fixed fee of $3, so that a dollar amount \(x\) becomes \(g(x)=0.98x-3\) dollars.
(a)
Find \((g\circ f)(500)\) and interpret your answer in context.
(2)
(b)
Find \((g\circ f)(x)\).
(2)
(c)
Find \((g\circ f)^{-1}(x)\).
(2)
(d)
Hence find the number of euros that a customer must exchange to receive exactly $1000. Give your answer correct to the nearest cent.
(1)
(e)
Find \((f\circ g)(x)\) and show that \((f\circ g)(x)-(g\circ f)(x)\) is constant.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\$526.20\)
A1
\(0.98\times540-3\)
a customer exchanging €500 receives $526.20
A1
(b)
\(0.98(1.08x)-3\)
(M1)
\(1.0584x-3\)
A1
(c)
\(x=1.0584y-3\)
(M1)
\((g\circ f)^{-1}(x)=\dfrac{x+3}{1.0584}\)
A1
(d)
\(\dfrac{1003}{1.0584}=947.66\), i.e. €947.66
A1
\(947.656\ldots\)
(e)
\((f\circ g)(x)=1.08(0.98x-3)=1.0584x-3.24\)
A1
difference \(=-0.24\), which does not depend on \(x\)
188AASL2-q00422.6 Quadratic functionsPaper 2Medium7 marks
The cross-section of a tunnel entrance is modelled by the arch \(y=h(x)\), where \(h(x)=ax(x-40)\), for \(0\le x\le40\). Distances are in metres, the ground is the \(x\)-axis and the highest point of the arch is \((20,\,12)\).
Diagram not to scale
(a)
Show that \(a=-0.03\).
(2)
(b)
Write down \(h(x)\) in the form \(a(x-p)^{2}+q\).
(1)
(c)
Find the height of the arch at a horizontal distance of 5 m from O.
(1)
(d)
A lorry with a rectangular cross-section is 10.5 m high. Find the maximum width of the lorry that can pass through the arch.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(12=a(20)(20-40)\)
M1
\(12=-400a\)
A1
\(a=-0.03\)
AG
(b)
\(h(x)=-0.03(x-20)^{2}+12\)
A1
(c)
\(5.25\) m
A1
(d)
\(-0.03(x-20)^{2}+12=10.5\)
(M1)
\(x=12.93,\ 27.07\)
(A1)
\(20\pm\sqrt{50}\)
\(14.1\) m
A1
\(2\sqrt{50}=14.142\ldots\)
Answers: (b) \(-0.03(x-20)^{2}+12\) · (c) 5.25 m · (d) 14.1 m
189AASL2-q00532.7 Quadratic equations and the discriminantPaper 2Medium8 marks
A ball is thrown upwards from the edge of a platform. Its height above the ground, \(h\) metres, \(t\) seconds after it is thrown is modelled by \(h(t)=1.6+12t-4.9t^{2}\), for \(t\ge0\), until the ball reaches the ground.
(a)
Write down the height from which the ball is thrown.
(1)
(b)
Find the maximum height of the ball and the time at which it occurs.
(3)
(c)
Find the time at which the ball reaches the ground.
(2)
(d)
Find the length of time for which the ball is more than 6 m above the ground.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(1.6\) m
A1
(b)
\(t=-\dfrac{12}{2(-4.9)}\), or maximum on GDC
(M1)
\(t=1.22\) s
A1
\(1.2244\ldots\)
\(8.95\) m
A1
\(8.9469\ldots\)
(c)
\(1.6+12t-4.9t^{2}=0\)
(M1)
\(t=2.58\) s
A1
\(2.5757\ldots\) Award A0 if the negative root is also given as an answer.
(d)
\(1.6+12t-4.9t^{2}=6\), giving \(t=0.449\) and \(t=2\)
(M1)
\(1.55\) s
A1
\(1.5510\ldots\)
Answers: (a) 1.6 m · (b) 8.95 m at 1.22 s · (c) 2.58 s · (d) 1.55 s
190AASL2-q00542.7 Quadratic equations and the discriminantPaper 2Medium8 marks
On a dry road, the stopping distance, \(d\) metres, of a car travelling at \(v\) km h−1 is modelled by \(d=0.2v+0.006v^{2}\), for \(v\ge0\).
(a)
Find the stopping distance of the car when it travels at 80 km h−1.
(1)
(b)
Find the greatest speed at which the car can stop within 60 m.
(3)
(c)
On a wet road the model becomes \(d=0.2v+kv^{2}\). At 50 km h−1 the stopping distance on the wet road is 32.5 m. Find the value of \(k\).
(2)
(d)
Find the speed on the wet road that gives the same stopping distance as 80 km h−1 on the dry road.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(54.4\) m
A1
(b)
\(0.006v^{2}+0.2v-60=0\) or \(0.006v^{2}+0.2v\le60\)
(M1)
\(v=\dfrac{-0.2+\sqrt{0.04+1.44}}{0.012}\), or GDC
(A1)
\(84.7\) km h−1
A1
\(84.712\ldots\) Award A0 if the negative root is also given as a final answer.
(c)
\(10+2500k=32.5\)
(M1)
\(k=0.009\)
A1
(d)
\(0.2v+0.009v^{2}=54.4\)
(M1)
FT from their (a) and (c).
\(67.4\) km h−1
A1
\(67.424\ldots\)
Answers: (a) 54.4 m · (b) 84.7 km h−1 · (c) \(k=0.009\) · (d) 67.4 km h−1
191AASL2-q00642.8 Reciprocal and rational functionsPaper 2Medium9 marks
A tank initially contains 200 litres of pure water. Salt water with a concentration of 40 grams per litre is pumped into the tank at a rate of 5 litres per minute, and mixes instantly. No liquid leaves the tank. The tank can hold at most 1000 litres.
(a)
Show that the concentration of salt in the tank, \(C\) grams per litre, \(t\) minutes after pumping starts is \(C(t)=\dfrac{40t}{t+40}\).
(2)
(b)
Find the concentration after 20 minutes.
(1)
(c)
Find the time at which the concentration reaches 30 grams per litre.
(2)
(d)
Write down the domain of \(C\) in this context, and find the greatest concentration that can be reached.
(2)
(e)
If the tank had no maximum capacity, the graph of \(C\) would have a horizontal asymptote. Write down its equation and interpret it in context.
(2)
Show mark scheme
Working
Mark
Notes
(a)
mass of salt \(=200t\) g and volume \(=200+5t\) L
M1
\(C=\dfrac{200t}{200+5t}\)
A1
\(C(t)=\dfrac{40t}{t+40}\)
AG
Dividing numerator and denominator by 5.
(b)
\(13.3\) g L−1
A1
\(\tfrac{40}{3}=13.33\ldots\)
(c)
\(40t=30(t+40)\)
(M1)
\(t=120\) minutes
A1
(d)
\(0\le t\le160\)
A1
From \(200+5t=1000\).
\(C(160)=32\) g L−1
A1
(e)
\(C=40\)
A1
the concentration approaches (but never reaches) that of the incoming salt water, 40 g L−1
R1
Answers: (b) 13.3 g L−1 · (c) 120 min · (d) \(0\le t\le160\); 32 g L−1 · (e) \(C=40\)
192AASL2-q00702.9 Exponential and logarithmic functionsPaper 2Medium8 marks
A radioactive sample initially has a mass of 80 mg. Its mass, \(m\) mg, after \(t\) years is modelled by \(m(t)=80\mathrm{e}^{-kt}\), where \(k\) is a positive constant. The half-life of the substance is 5.3 years.
(a)
Show that \(k=0.131\), correct to three significant figures.
(2)
(b)
Find the mass of the sample after 12 years.
(2)
(c)
Find the mass of the substance that decays during the third year.
(2)
(d)
Find the time it takes for the mass of the sample to fall below 1 mg.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(80\mathrm{e}^{-5.3k}=40\), so \(\mathrm{e}^{-5.3k}=\dfrac{1}{2}\)
M1
\(k=\dfrac{\ln2}{5.3}=0.13078\ldots\)
A1
Must see a value to more than 3 s.f., or the exact form.
\(k=0.131\)
AG
(b)
\(80\mathrm{e}^{-12k}\)
(M1)
\(16.7\) mg
A1
\(16.653\ldots\) Accept 16.6 from \(k=0.131\).
(c)
\(m(2)-m(3)\)
(M1)
Award M0 for \(m(3)-m(4)\) or for \(80-m(3)\).
\(7.55\) mg
A1
\(7.5501\ldots\) Accept 7.56 from \(k=0.131\).
(d)
\(80\mathrm{e}^{-kt}=1\), or \(t=\dfrac{\ln80}{k}\)
193AASL2-q00712.9 Exponential and logarithmic functionsPaper 2Medium8 marks
The number of birds, \(N\), in a protected wetland \(t\) years after the start of a conservation programme is modelled by \(N(t)=A\mathrm{e}^{kt}\), where \(A\) and \(k\) are positive constants. After 2 years there are 3000 birds and after 6 years there are 4800 birds.
(a)
Show that \(k=\dfrac{1}{4}\ln1.6\).
(2)
(b)
Find the value of \(A\).
(2)
(c)
The aim of the programme is to have at least 6000 birds after 8 years. Determine whether the model predicts that this aim is met.
(2)
(d)
Find the value of \(t\) when the number of birds first reaches 10 000.
196AASL2-q00802.10 Solving equationsPaper 2Medium7 marks
Car A is bought for $30 000 and its value decreases by 20% each year. Car B is bought for $24 000 at the same time and its value decreases by 15% each year. The value of each car after \(t\) years is modelled by a continuous function of \(t\).
(a)
Find the value of car A after 3 years.
(2)
(b)
Show that the two cars have the same value when \(\left(\dfrac{16}{17}\right)^{t}=0.8\).
(2)
(c)
Hence find the value of \(t\) when the two cars have the same value.
(1)
(d)
Find the number of complete years after which the value of car B is first less than one third of its purchase price.
197AASL2-q00812.10 Solving equationsPaper 2Medium10 marks
Two students start a typing course at the same time. Student A’s typing speed, in words per minute, \(t\) weeks after the start of the course is modelled by \(W(t)=20+12\ln(t+1)\). Student B’s typing speed is modelled by \(V(t)=60-45\mathrm{e}^{-0.15t}\). Both models apply for \(t\ge0\).
(a)
Write down the typing speed of each student at the start of the course.
(1)
(b)
Find the typing speed of student A after 10 weeks.
(1)
(c)
Find the exact time at which student A’s typing speed reaches 50 words per minute.
(2)
(d)
Write down the equation of the horizontal asymptote of the graph of \(V\), and interpret it in context.
(2)
(e)
Find the set of values of \(t\) for which student B types faster than student A.
(3)
(f)
Suggest one reason why the model for student A is not appropriate for large values of \(t\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
A: 20 and B: 15 words per minute
A1
Both values required.
(b)
\(48.8\) words per minute
A1
\(48.774\ldots\)
(c)
\(12\ln(t+1)=30\), so \(\ln(t+1)=2.5\)
(M1)
\(t=\mathrm{e}^{2.5}-1\) weeks
A1
\((=11.2)\). Award A0 for 11.2 only (exact answer required).
(d)
\(V=60\)
A1
student B’s speed approaches (but never reaches) 60 words per minute
R1
(e)
intersections of the graphs of \(W\) and \(V\)
(M1)
\(t=7.89\) and \(t=24.5\)
A1
\(7.8877\ldots\), \(24.4838\ldots\)
\(7.89\lt t\lt 24.5\)
A1
(f)
\(W(t)\to\infty\) as \(t\to\infty\), but a typing speed cannot increase without limit
198IBMA2S-046-02-042.4 Key features and intersectionsPaper 2Medium7 marks
Calculator
After a tablet is swallowed, the concentration of a drug in a patient's blood is modelled by \(C(t)=12te^{-0.4t}\), where \(C\) is measured in mg L−1 and \(t\) is the time in hours, \(0\le t\le 10\). The drug is effective while its concentration is at least 6 mg L−1.
(a)
Find the length of time for which the drug is effective.
(3)
(b)
Find the maximum concentration and the time at which it occurs.
(2)
(c)
Write down the range of \(C\) for the given domain.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 solves \(12te^{-0.4t}=6\) graphically or numerically
✓ 1
A1 \(t=0.648\) and \(t=6.36\)
✓ 1
A1 5.71 hours
✓ 1
Part (b)
A1 \(t=2.5\) hours
✓ 1
A1 maximum concentration \(11.0\) mg L−1 (11.036…)
✓ 1
Part (c)
A1 lower bound \(0\)
✓ 1
A1 \(0\le C\le 11.0\)
✓ 1
Common error: The range is not found from the endpoints alone. Here \(C(10)\approx 2.20\), but the greatest value occurs at the turning point inside the domain.
199IBMA2S-046-02-072.4 Key features and intersectionsPaper 2Medium9 marks
Calculator
The orbital period, \(T\) hours, of a satellite in a circular orbit around the Earth is modelled by \(T(r)=0.0877r^{1.5}\), where \(r\) is the distance from the centre of the Earth in thousands of kilometres. The radius of the Earth is 6.37 thousand km.
(a)
Explain why a suitable domain for \(T\) is \(r\ge 6.37\), and find the corresponding range of \(T\).
(3)
(b)
Find an expression for \(T^{-1}(t)\).
(3)
(c)
A navigation satellite has an orbital period of 12 hours. Find its height above the Earth's surface, in thousands of kilometres.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 a satellite cannot orbit closer to the centre than the Earth's surface
Exam tip: Read the definition of the variable carefully. Here \(r\) is measured from the Earth's centre, so subtract the Earth's radius to obtain the height above the surface.
200IBMA2S-046-01-042.4 Key features and intersectionsPaper 2Medium7 marks
Calculator
After a patient takes a tablet, the concentration of a drug in the bloodstream, C mg L−1, t hours later is modelled by \(C(t) = 12te^{-0.4t}\), for \(t \geq 0\). The drug is effective while its concentration is at least 8 mg L−1.
(a)
Find the maximum concentration of the drug and the time at which it occurs.
(2)
(b)
Solve the equation \(C(t) = 8\).
(3)
(c)
Hence find the length of time for which the drug is effective.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(t = 2.5\) hours
✓ 1
A1 maximum \(C = 11.0\) mg L−1 (11.036…)
✓ 1
Part (b)
M1 attempt to find intersection of \(y = 12te^{-0.4t}\) and \(y = 8\) on GDC
✓ 1
A1 \(t = 0.991\)
✓ 1
A1 \(t = 5.07\)
✓ 1
Part (c)
M1 subtracting their two values, \(5.0740 - 0.9910\)
✓ 1
A1 \(4.08\) hours
✓ 1
Exam tip: Keep the full GDC values of both roots when subtracting; rounding each to 3 s.f. first can change the final digit.
201IBMA2S-046-01-062.4 Key features and intersectionsPaper 2Medium5 marks
Calculator
A basketball player releases a ball at a height of 2.1 m. The path of the ball is modelled by \(y = -0.45x^2 + bx + 2.1\), where y metres is the height of the ball and x metres is the horizontal distance travelled from the point of release. The ball passes through the centre of the hoop, which is 3.05 m above the floor and 4.2 m horizontally from the point of release.
M1 substituting their unrounded x into the model, or using the GDC maximum, or using \(y_{max} = 2.1 + \frac{b^2}{4 \times 0.45}\)
✓ 1
A1 \(4.59\) m (4.5879…)
✓ 1
Note: accept 4.60 m if it comes from correctly using the rounded value \(b = 2.12\).
—
Exam tip: Store the unrounded value of b in your GDC before finding the maximum. This avoids rounding errors, for example 4.60 m instead of 4.59 m.
202IBMA2S-046-01-072.10 Solving equationsPaper 2Medium13 marks
Calculator
Two mobile phones are plugged in to charge at the same time. The charge in each battery, as a percentage of full capacity, t minutes after being plugged in is modelled by Phone A: \(B_A(t) = 100 - 85e^{-0.025t}\), for \(t \geq 0\) Phone B: \(B_B(t) = 20 + 0.9t\), for \(0 \leq t \leq \frac{800}{9}\), after which Phone B remains at 100%.
(a)
Write down the charge in each phone at the moment they are plugged in.
(2)
(b)
Write down the value that \(B_A(t)\) approaches as \(t\) increases, and comment on what this means for Phone A.
(2)
(c)
Solve the equation \(B_A(t) = B_B(t)\).
(3)
(d)
Hence write down the interval of time during which Phone A has the greater charge.
(2)
(e)
Determine which phone reaches 95% charge first, and by how many minutes.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Phone A: 15%
✓ 1
A1 Phone B: 20%
✓ 1
Part (b)
A1 100
✓ 1
R1 the model predicts Phone A gets ever closer to full charge but never reaches exactly 100%
✓ 1
Part (c)
M1 attempt to find points of intersection of the two graphs on GDC
✓ 1
A1 \(t = 4.51\)
✓ 1
A1 \(t = 74.1\)
✓ 1
Part (d)
R1 Phone A starts with less charge than Phone B, so it is ahead between the two intersections
Exam tip: For equations mixing exponentials and linear terms, there is no algebraic method at SL: sketch both graphs on a suitable window so you do not miss the second intersection.
203IBMA2S-046-03-052.10 Solving equationsPaper 2Medium7 marks
Calculator
A cup of coffee is poured into a reusable travel mug. Its temperature, in °C, \(t\) minutes after pouring is modelled by \(T(t)=20+70e^{-0.05t}\), for \(t\ge 0\).
(a)
Find an expression for \(T^{-1}(x)\).
(3)
(b)
State the domain of \(T^{-1}\).
(2)
(c)
Find \(T^{-1}(45)\) and interpret your answer in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 isolates the exponential: \(\frac{x-20}{70}=e^{-0.05t}\)
A1 lower bound \(x>20\) (strict, because the exponential term is never zero)
✓ 1
A1 upper bound \(x\le 90\), so the domain is \(20<x\le 90\)
✓ 1
Part (c)
A1 \(T^{-1}(45)=20\ln 2.8=20.6\)
✓ 1
A1 the coffee takes about 20.6 minutes after pouring to cool to 45 °C
✓ 1
Common error: A frequent slip is to divide only the \(x\) by 70 and write \(e^{-0.05t}=\frac{x}{70}-20\). Subtract 20 from both sides first. Then divide the whole of \(x-20\) by 70 to get \(\frac{x-20}{70}\).
204IBMA2S-046-03-072.9 Exponential and logarithmic functionsPaper 2Medium13 marks
Calculator
A portable speaker emits sound with power 0.08 W. At a distance of \(r\) metres from the speaker, the sound intensity is \(I(r)=\dfrac{0.08}{4\pi r^2}\) W m−2, for \(r>0\). The sound level in decibels corresponding to an intensity \(I\) is \(L(I)=10\log_{10}\left(\dfrac{I}{10^{-12}}\right)\).
(a)
Find \(I(5)\).
(2)
(b)
Find \((L\circ I)(5)\).
(2)
(c)
Show that \((L\circ I)(r)\) can be written as \(a-20\log_{10} r\), and find the value of \(a\) correct to 3 significant figures.
(3)
(d)
Find an expression for \(L^{-1}(x)\).
(3)
(e)
A noise regulation requires the sound level to be no more than 70 dB. Find the minimum distance from the speaker at which the regulation is met.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(r=5\): \(\frac{0.08}{4\pi(25)}\)
✓ 1
A1 \(2.55\times10^{-4}\) W m−2
✓ 1
Part (b)
M1 substitutes their \(I(5)\) into \(L\)
✓ 1
A1 84.1 dB
✓ 1
Part (c)
M1 forms \(10\log_{10}\left(\frac{0.08}{4\pi\times10^{-12}\,r^2}\right)\)
✓ 1
M1 uses a law of logarithms to separate the term \(-10\log_{10}r^2=-20\log_{10}r\)
M1 writes \(x=10\log_{10}\left(\frac{I}{10^{-12}}\right)\), or interchanges the variables, so that \(\frac{x}{10}=\log_{10}\left(\frac{I}{10^{-12}}\right)\)
✓ 1
M1 uses the inverse relationship between log and exponential: \(\frac{I}{10^{-12}}=10^{x/10}\)
✓ 1
A1 \(L^{-1}(x)=10^{\frac{x}{10}-12}\) (or equivalent, such as \(10^{-12}\times10^{x/10}\))
✓ 1
Part (e)
M1 finds the intensity at 70 dB: \(L^{-1}(70)=10^{-5}\) W m−2, OR sets up \(a-20\log_{10}r=70\) using the result of (c)
✓ 1
M1 solves for \(r\): \(\frac{0.08}{4\pi r^2}=10^{-5}\), OR \(\log_{10}r=\frac{a-70}{20}\)
✓ 1
A1 \(r=25.2\) m
✓ 1
Exam tip: Keep full calculator accuracy between parts. If you use the rounded value \(a=98.0\) in (e), you get \(r\approx25.1\) m. Use the unrounded value of \(a\) or the intensity method instead.
205IBMA2S-046-03-092.5 Composite and inverse functionsPaper 2Medium6 marks
Calculator
A traveller converts British pounds into euros at a kiosk. For \(p\) pounds, the kiosk gives \(E(p)=1.16p-2.5\) euros, after a fixed fee. The traveller then converts all of these euros into Japanese yen, receiving \(Y(e)=162e\) yen for \(e\) euros.
(a)
Find \((Y\circ E)(p)\).
(2)
(b)
Find \((Y\circ E)^{-1}(x)\).
(2)
(c)
The traveller needs 50 000 yen. Find how many pounds they must exchange, giving your answer to the nearest penny.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms \(162(1.16p-2.5)\)
✓ 1
A1 \((Y\circ E)(p)=187.92p-405\)
✓ 1
Part (b)
M1 rearranges \(x=187.92p-405\) to make \(p\) the subject
✓ 1
A1 \((Y\circ E)^{-1}(x)=\frac{x+405}{187.92}\)
✓ 1
Part (c)
M1 substitutes \(x=50000\) into their inverse
✓ 1
A1 £268.23
✓ 1
Exam tip: As a check, \((Y\circ E)^{-1}=E^{-1}\circ Y^{-1}\): convert yen back to euros, then euros back to pounds. Both routes must give the same answer.
206IBMA2S-046-03-102.7 Quadratic equations and the discriminantPaper 2Medium13 marks
Calculator
During the final stage of a delivery flight, the height of a drone above the ground, in metres, is modelled by \(h(t)=20-1.25(t-4)^2\), where \(t\) is the time in seconds and \(4\le t\le 8\).
(a)
Find \(h(4)\) and \(h(8)\). Hence write down the range of \(h\).
(2)
(b)
Find an expression for \(h^{-1}(x)\).
(4)
(c)
State the domain and range of \(h^{-1}\).
(2)
(d)
Find the time at which the drone is 12 m above the ground.
(2)
(e)
The drone's altimeter displays height in feet, where a height of \(h\) metres is displayed as \(F(h)=3.28h\) feet. Find the time at which the altimeter displays 30 feet.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(h(4)=20\) and \(h(8)=0\)
✓ 1
A1 \(0\le h(t)\le 20\)
✓ 1
Part (b)
M1 rearranges \(x=20-1.25(t-4)^2\)
✓ 1
A1 \((t-4)^2=0.8(20-x)\)
✓ 1
M1 takes the square root
✓ 1
R1 takes the positive root because \(t\ge 4\), giving \(h^{-1}(x)=4+\sqrt{0.8(20-x)}\)
✓ 1
Part (c)
A1 domain \(0\le x\le 20\)
✓ 1
A1 range \(4\le h^{-1}(x)\le 8\)
✓ 1
Part (d)
M1 evaluates \(h^{-1}(12)=4+\sqrt{6.4}\)
✓ 1
A1 \(t=6.53\) s
✓ 1
Part (e)
M1 \(F^{-1}(30)=\frac{30}{3.28}=9.146\) m
✓ 1
M1 evaluates \(h^{-1}(9.146...)\), i.e. \((h^{-1}\circ F^{-1})(30)\)
✓ 1
A1 \(t=6.95\) s
✓ 1
Common error: Writing \(\pm\sqrt{\;}\) in the inverse loses the reasoning mark. The restricted domain \(t\ge 4\) makes \(h\) one-to-one and fixes the sign of the root.
207IBMA2S-046-04-052.1 Equations of straight linesPaper 2Medium6 marks
Calculator
In a camera, the distance \(v\) cm from a lens to a sharp image of an object placed \(u\) cm from the lens is given by \(v=\frac{5u}{u-5}\), \(u>5\).
(a)
Find the image distance when the object is 12 cm from the lens.
(2)
(b)
Write down the equations of the asymptotes of the graph of \(v\) against \(u\).
(2)
(c)
Find the object distance for which the image distance equals the object distance.
M1 sets \(\frac{5u}{u-5}=u\) (or equivalent, e.g. \(5u=u(u-5)\) or \(u^2-10u=0\), or uses the intersection of the graph with \(v=u\))
✓ 1
A1 \(u=10\) cm (rejecting \(u=0\) because \(u>5\))
✓ 1
Exam tip: If you expand \(5u=u(u-5)\) to \(u^2-10u=0\), you get \(u=0\) or \(u=10\). Use the domain \(u>5\) to reject \(u=0\), and say so in your answer.
208IBMA2S-046-04-072.11 Transformations of graphsPaper 2Medium7 marks
Calculator
A mobile data plan charges a fixed $15 per month plus $2 per gigabyte (GB) used. The average cost per GB, in dollars, when \(n\) GB are used in a month is \(C(n)=\frac{15+2n}{n}\), \(n>0\).
(a)
Find the average cost per GB when 6 GB are used.
(1)
(b)
(i) Write \(C(n)\) in the form \(a+\frac{b}{n}\). (ii) Hence describe a sequence of two transformations that maps the graph of \(y=\frac{1}{n}\) onto the graph of \(y=C(n)\).
(4)
(c)
Find the least amount of data that must be used for the average cost to be at most $3 per GB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(C(6)=\frac{27}{6}=4.5\) dollars per GB
✓ 1
Part (b)
(i) A1 \(a=2\)
✓ 1
A1 \(b=15\)
✓ 1
(ii) A1 a stretch with scale factor 15, either vertical (parallel to the \(y\)-axis) or horizontal (parallel to the \(x\)-axis)
✓ 1
A1 followed by a vertical translation of 2 units up, i.e. by vector \(\binom{0}{2}\)
✓ 1
Also accept, for both A1 marks: a vertical translation of \(\frac{2}{15}\) up, followed by a vertical stretch with scale factor 15.
Exam tip: Splitting the fraction into \(2+\frac{15}{n}\) makes both the transformations and the inequality immediate. State the order of the transformations clearly, because changing the order changes the result.
209IBMA2S-046-04-092.10 Solving equationsPaper 2Medium6 marks
Calculator
An engineer connects a variable resistor of resistance \(R\) Ω in parallel with a fixed 6 Ω resistor. The combined resistance, \(R_T\) Ω, is given by \(R_T=\frac{6R}{R+6}\), \(R>0\).
(a)
Find \(R_T\) when \(R=3\).
(1)
(b)
Write down the equation of the horizontal asymptote of the graph of \(R_T\) against \(R\) and interpret it in context.
(2)
(c)
Find the value of \(R\) that gives a combined resistance of 4.5 Ω.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(R_T=\frac{18}{9}=2\) Ω
✓ 1
Part (b)
A1 \(R_T=6\)
✓ 1
R1 the combined resistance is always less than 6 Ω however large \(R\) becomes
✓ 1
Part (c)
M1 sets \(\frac{6R}{R+6}=4.5\)
✓ 1
M1 rearranges to \(6R=4.5R+27\)
✓ 1
A1 \(R=18\) Ω
✓ 1
Exam tip: Multiply through by the denominator first; this turns the rational equation into a linear one.
210IBMA2S-046-04-102.10 Solving equationsPaper 2Medium11 marks
Calculator
During a basketball season, Amara has scored 30 of her 50 free throws. She then scores every one of her next \(n\) free throws. Her overall success rate is \(P(n)=\frac{30+n}{50+n}\), \(n\ge0\).
(a)
Write down \(P(0)\).
(1)
(b)
Find the number of consecutive successful free throws needed for her success rate to reach 0.8.
(3)
(c)
Write down the equation of the horizontal asymptote of the graph of \(P\) and interpret it in context.
(2)
(d)
Find \(P^{-1}(x)\) and explain what \(P^{-1}(x)\) represents in this context.
(3)
(e)
Her teammate Ben has scored 18 of his 40 free throws. Find how many consecutive successful free throws he needs for his success rate to equal 0.6.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(0)=0.6\)
✓ 1
Part (b)
M1 sets \(\frac{30+n}{50+n}=0.8\)
✓ 1
M1 rearranges to \(0.2n=10\)
✓ 1
A1 \(n=50\)
✓ 1
Part (c)
A1 \(P=1\)
✓ 1
R1 her success rate gets ever closer to 100% but never reaches it, since she has already missed 20
✓ 1
Part (d)
M1 attempts to make \(n\) the subject of \(x(50+n)=30+n\)
✓ 1
A1 \(P^{-1}(x)=\frac{50x-30}{1-x}\)
✓ 1
R1 the number of consecutive successes needed to reach a success rate of \(x\)
✓ 1
Part (e)
M1 sets \(\frac{18+n}{40+n}=0.6\), so \(0.4n=6\)
✓ 1
A1 \(n=15\)
✓ 1
Exam tip: Check an inverse function against an earlier answer: \(P^{-1}(0.8)=\frac{40-30}{0.2}=50\) confirms part (b).
211IBMA2S-046-07-062.1 Equations of straight linesPaper 2Medium7 marks
Calculator
A straight ski slope descends at an angle of 28° below the horizontal. Using coordinates in metres, x is the horizontal distance from the top of the slope and y is the height above the base. The top of the slope is at (0, 450) and the slope ends where it meets the base, \(y=0\).
Diagram not to scale
(a)
Find the gradient of the slope.
(2)
(b)
Write down the equation of the line of the slope.
(1)
(c)
Find the horizontal distance from the top to the base of the slope.
(2)
(d)
Find the length of the slope measured along its surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 gradient \(=\tan 28^\circ\) in magnitude
✓ 1
A1 \(-0.532\) (negative sign required)
✓ 1
Part (b)
A1 \(y=-0.532x+450\)
✓ 1
Part (c)
M1 \(0=-\tan 28^\circ\,x+450\)
✓ 1
A1 \(x=846\) m
✓ 1
Part (d)
M1 \(\frac{450}{\sin 28^\circ}\) or \(\sqrt{450^2+846.3^2}\)
✓ 1
A1 959 m
✓ 1
Common error: The gradient is tan of the angle with the horizontal, not sin or cos. A downhill line also needs a negative gradient.
212IBMA2S-046-07-072.10 Solving equationsPaper 2Medium10 marks
Calculator
Two electric cars, P and Q, are plugged into chargers at the same moment, t = 0 minutes. The charge level C% of each car increases linearly with time. Car P starts at 18% and reaches 66% after 40 minutes. Car Q starts at 45% and gains 0.5 percentage points per minute.
(a)
Find an equation for the charge level of car P in terms of t.
(2)
(b)
Write down an equation for the charge level of car Q in terms of t.
(1)
(c)
Find the time at which both cars have the same charge level, and that charge level.
(3)
(d)
Each charger switches off when the car reaches 80%. Find the time at which car P's charger switches off.
(2)
(e)
Find how much longer car Q's charger stays on than car P's.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 gradient \(\frac{66-18}{40}=1.2\)
✓ 1
A1 \(C=1.2t+18\)
✓ 1
Part (b)
A1 \(C=0.5t+45\)
✓ 1
Part (c)
M1 equates, e.g. \(1.2t+18=0.5t+45\)
✓ 1
A1 \(t=38.6\) minutes
✓ 1
A1 \(C=64.3\%\)
✓ 1
Part (d)
M1 \(1.2t+18=80\)
✓ 1
A1 \(t=51.7\) minutes
✓ 1
Part (e)
M1 Q: \(0.5t+45=80\) gives \(t=70\)
✓ 1
A1 \(70-51.7=18.3\) minutes
✓ 1
Exam tip: Keep the unrounded value t = 51.666… in your calculator for part (e); subtracting a rounded value can shift the final 3 s.f. answer.
213IBMA2S-046-07-092.10 Solving equationsPaper 2Medium7 marks
Calculator
The speed of sound in air, v m s−1, is modelled as a linear function of air temperature T °C. At 0 °C the speed is 331 m s−1 and at 30 °C it is 349 m s−1.
(a)
Find the gradient of the model and interpret it in context.
(2)
(b)
Write down the equation of the model.
(1)
(c)
On a winter night the speed of sound is measured as 325 m s−1. Find the air temperature.
(2)
(d)
A sports starter's pistol is fired 500 m from a timing sensor on a day when the temperature is 25 °C. Find the time taken for the sound to reach the sensor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0.6
✓ 1
R1 speed increases by 0.6 m s−1 for each 1 °C rise in temperature
✓ 1
Part (b)
A1 \(v=0.6T+331\)
✓ 1
Part (c)
M1 \(0.6T+331=325\)
✓ 1
A1 \(T=-10\) °C
✓ 1
Part (d)
M1 \(v=346\) and time \(=\frac{500}{346}\)
✓ 1
A1 1.45 s
✓ 1
Exam tip: An interpretation of a gradient earns credit only if it states the change in the output per unit change in the input, with units.
214IBMA2S-046-06-052.11 Transformations of graphsPaper 2Medium6 marks
Calculator
A calibration curve for a light sensor is modelled by \(g(x) = 3\ln(x-2) + 1\), \(x > 2\). The graph of \(y = g(x)\) is obtained from the graph of \(y = \ln x\) by a sequence of transformations.
(a)
Describe, in order, a sequence of three transformations that maps the graph of \(y = \ln x\) onto the graph of \(y = g(x)\).
(3)
(b)
Write down the equation of the vertical asymptote of the graph of \(y = g(x)\).
(1)
(c)
Find the \(x\)-intercept of the graph of \(y = g(x)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 translation 2 units right, i.e. by \(\begin{pmatrix}2\\0\end{pmatrix}\)
✓ 1
A1 vertical stretch with scale factor 3
✓ 1
A1 translation 1 unit up, i.e. by \(\begin{pmatrix}0\\1\end{pmatrix}\), in an order that gives \(g\)
✓ 1
Accept any valid order. The horizontal translation may come before or after the stretch, with the translation up 1 applied after the stretch. Alternatively, accept a translation up \(\frac{1}{3}\) applied before the stretch, scale factor 3.
Exam tip: The vertical asymptote moves with horizontal transformations only. Vertical stretches and translations leave it unchanged. When ordering transformations, check that your sequence really rebuilds \(g(x)\): a vertical translation applied before a stretch is also scaled.
215IBMA2S-046-06-072.11 Transformations of graphsPaper 2Medium13 marks
Calculator
A smart mug records the temperature of the coffee inside it. The temperature, in °C, \(t\) minutes after the coffee is poured is modelled by \(T(t) = 70e^{-0.05t} + 20\), \(t \ge 0\).
(a)
Describe a sequence of transformations that maps the graph of \(y = e^t\) onto the graph of \(y = T(t)\).
(4)
(b)
Write down the equation of the horizontal asymptote of the graph of \(y = T(t)\) and interpret it in context.
(2)
(c)
A heater raises the room temperature, and the model becomes \(W(t) = T(t) + 5\). Find the time at which \(W(t) = 50\).
(3)
(d)
In the original room, a mug made of a different material gives the model \(V(t) = T(kt)\), where \(k\) is a positive constant. The coffee in this mug reaches 50 °C after 12 minutes. Find \(k\) and state, with a reason, whether this mug keeps coffee hot for longer than the original mug.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 reflection in the vertical axis
✓ 1
A1 horizontal stretch with scale factor 20
✓ 1
A1 vertical stretch with scale factor 70
✓ 1
A1 translation 20 units up, applied after the vertical stretch
✓ 1
Part (b)
A1 \(T = 20\)
✓ 1
R1 the coffee cools towards the room temperature of 20 °C
R1 no: \(k > 1\) is a horizontal stretch of factor \(\frac{1}{k} \approx 0.708 < 1\), so each temperature is reached sooner and this mug cools faster
✓ 1
Exam tip: Write \(e^{-0.05t}\) as \(e^{-t/20}\). This shows the horizontal stretch factor is 20 and that a reflection in the vertical axis is also needed.
216IBMA2S-046-06-122.4 Key features and intersectionsPaper 2Medium10 marks
Calculator
The height, \(h\) metres, of a golf ball above level ground is modelled by \(h(x) = -0.004x^2 + 0.6x\), where \(x\) is the horizontal distance in metres from the point where it is struck.
(a)
Find the horizontal distance at which the ball lands.
(2)
(b)
Find the maximum height of the ball.
(2)
(c)
At a high-altitude course, the same shot travels 12% further horizontally and reaches the same maximum height. Write down the model \(k(x)\) for this shot in terms of \(h\), and find where the ball lands.
(3)
(d)
On the original course, the ball is instead struck from a raised tee 3 m above the level ground, so its height is modelled by \(m(x) = h(x) + 3\). Find the horizontal distance at which the ball lands.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x(0.6 - 0.004x) = 0\)
✓ 1
A1 \(x = 150\) m (rejecting \(x = 0\), the point of striking)
✓ 1
Part (b)
M1 vertex at \(x = \frac{-0.6}{2(-0.004)} = 75\)
✓ 1
A1 \(h(75) = 22.5\) m
✓ 1
Part (c)
M1 recognises a horizontal stretch with scale factor 1.12
M1 uses the quadratic formula or technology and takes the positive root, e.g. \(x = \frac{0.6 + \sqrt{0.36 + 0.048}}{0.008}\)
✓ 1
A1 \(x = 155\) m (3 s.f.) (accept 154.8)
✓ 1
Common error: A vertical translation does not move the landing point by the same amount. Find the new \(x\)-intercept by solving \(m(x) = 0\) again, and reject the negative root because it lies behind the tee.
217IBMA2S-046-05-062.10 Solving equationsPaper 2Medium12 marks
Calculator
A patient receives an injection of an antibiotic. The concentration of the antibiotic in the patient's blood, C mg L−1, t hours after the injection is modelled by C(t) = 18e−0.23t. The antibiotic is effective only while C ≥ 4.
(a)
Write down the concentration immediately after the injection.
(1)
(b)
Find the time taken for the concentration to halve.
(3)
(c)
Find the length of time for which the first dose is effective.
(3)
(d)
At the moment the first dose stops being effective, a second identical injection is given, which immediately raises the concentration by 18 mg L−1. The concentration then decays at the same rate. Find the concentration 2 hours after the second injection.
(2)
(e)
Find how long after the second injection the concentration falls back to 4 mg L−1.
M1 4 = 18e−0.23t ; M1 t = ln(4.5)/0.23 ; A1 t = 6.54 hours
✓ 3
Part (d)
M1 concentration after second dose is 22, so C = 22e−0.23×2 ; A1 13.9 mg L−1
✓ 2
Part (e)
M1 4 = 22e−0.23s ; M1 s = ln(5.5)/0.23 ; A1 s = 7.41 hours
✓ 3
Exam tip: After the second dose the starting value is 4 + 18 = 22, not 18. Restarting the clock at the second injection keeps the model simple.
218IBMA2S-046-05-072.10 Solving equationsPaper 2Medium10 marks
Calculator
After a racing car stops, the temperature of one of its brake discs, T °C, t minutes later is modelled by T(t) = 25 + Ae−kt, where A and k are positive constants. When the car stops the disc is at 625 °C, and 4 minutes later it is at 385 °C.
(a)
Find the value of A.
(2)
(b)
Show that k = 0.128, correct to three significant figures.
(3)
(c)
Write down the equation of the horizontal asymptote of the graph of T, and interpret it in context.
(2)
(d)
A mechanic can safely touch the disc once its temperature has fallen to 100 °C. Find how long this takes.
A1 T = 25 ; R1 the disc cools towards the surrounding (room or air) temperature of 25 °C but never quite reaches it
✓ 2
Part (d)
M1 100 = 25 + 600e−kt ; A1 e−kt = 0.125 ; A1 t = 16.3 minutes (accept 16.2 minutes from using k = 0.128)
✓ 3
Exam tip: Store the unrounded value k = ln(5/3)/4 in your calculator and use it in later parts. Using the rounded value 0.128 shifts the final answer slightly, from 16.3 to 16.2 minutes.
219IBMA2S-046-05-092.9 Exponential and logarithmic functionsPaper 2Medium7 marks
Calculator
The change in velocity, Δv m s−1, that a rocket stage can produce is given by Δv = ve ln(m0/mf), where m0 is the mass at ignition, mf is the mass when the fuel is used up, and ve = 3100 m s−1 for this engine.
(a)
A stage has mass 550 tonnes at ignition and 180 tonnes when its fuel is used up. Find Δv.
(2)
(b)
To reach low Earth orbit a single stage would need Δv = 9400 m s−1. Find the minimum value of the ratio m0/mf.
(3)
(c)
Hence find the minimum percentage of the ignition mass that must be fuel.
Exam tip: Undo ln with the exponential: if ln R = 3.03, then R = e³.⁰³. Keep the full value of R in your calculator before using it in the final part.
220IBMA2S-046-05-102.10 Solving equationsPaper 2Medium11 marks
Calculator
The speed of a sprinter, v m s−1, t seconds after leaving the blocks is modelled by v(t) = 11.2(1 − e−0.8t), t ≥ 0.
(a)
Find the sprinter's speed after 2 seconds.
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of v, and interpret it in context.
(2)
(c)
Find the time taken for the sprinter to reach 95% of this limiting speed.
(3)
(d)
Show that the time taken to reach a speed v is t = −1.25 ln(1 − v/11.2).
(2)
(e)
Hence find the time at which the sprinter reaches 10 m s−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 v(2) = 11.2(1 − e−1.6) ; A1 8.94 m s−1
✓ 2
Part (b)
A1 v = 11.2 ; R1 the sprinter's speed approaches a maximum of 11.2 m s−1 but never exceeds it
✓ 2
Part (c)
M1 0.95 × 11.2 = 11.2(1 − e−0.8t) ; A1 e−0.8t = 0.05 ; A1 t = 3.74 s
✓ 3
Part (d)
M1 rearranging to e−0.8t = 1 − v/11.2 ; A1 taking natural logarithms and dividing by −0.8 to obtain t = −1.25 ln(1 − v/11.2)
✓ 2
Part (e)
M1 t = −1.25 ln(1 − 10/11.2) ; A1 t = 2.79 s
✓ 2
Common error: The result in the fourth part is the inverse of v(t); it is only defined for 0 ≤ v < 11.2, which matches the asymptote found earlier.
221IBMA2S-046-00-062.7 Quadratic equations and the discriminantPaper 2Medium7 marks
Calculator
A small probe is launched vertically from a platform on Mars. Its height, \(h\) metres, above the ground \(t\) seconds after launch is modelled by \(h(t)=-1.85t^2+vt+2\), where \(v\) is the launch speed in m s−1.
(a)
Show that the maximum height of the probe is \(2+\dfrac{v^2}{7.4}\) metres.
(2)
(b)
The probe reaches a maximum height of 60 m. Find \(v\).
Exam tip: Keep the unrounded value of v in your calculator memory when using it in the next part to avoid accumulated rounding errors.
222IBMA2S-046-00-082.4 Key features and intersectionsPaper 2Medium6 marks
Calculator
In a warehouse design, a curved roof beam is modelled by \(f(x)=0.5x^2-3x+7\) and a straight overhead cable is modelled by \(g(x)=12-0.8x\), where \(x\) and the heights are measured in metres. Vertical brackets join the cable to the beam wherever the cable lies above the beam.
(a)
Find the \(x\)-coordinates of the points where the cable meets the beam.
(3)
(b)
Find the length of the longest vertical bracket.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates \(0.5x^2-3x+7=12-0.8x\)
✓ 1
A1 \(0.5x^2-2.2x-5=0\)
✓ 1
A1 \(x=-1.65\), \(x=6.05\)
✓ 1
Part (b)
M1 forms \(d(x)=g(x)-f(x)=-0.5x^2+2.2x+5\)
✓ 1
M1 maximum at \(x=2.2\)
✓ 1
A1 \(7.42\) m
✓ 1
Exam tip: The vertical gap between two graphs is itself a quadratic here, so its maximum is at the vertex of g(x) − f(x), not at the vertex of f.
223IBMA2S-046-00-092.10 Solving equationsPaper 2Medium11 marks
Calculator
After a single dose, the concentration of a drug in a patient's blood is modelled by \(C(t)=-0.6t^2+4.8t\), where \(C\) is measured in mg L−1 and \(t\) is the time in hours after the dose, \(C\geq0\).
(a)
Find the time at which the concentration is greatest, and the greatest concentration.
(3)
(b)
Find the time at which the model predicts the drug has been completely eliminated.
(2)
(c)
The drug is effective when its concentration is above 6 mg L−1. Find the length of time for which the drug is effective.
(3)
(d)
For a different patient the model is \(C(t)=-0.6t^2+kt\), where \(k>0\). The concentration must never exceed 12 mg L−1. Find the largest possible value of \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(t=-\frac{4.8}{2(-0.6)}\), or finds the vertex by another valid method (e.g. GDC, completing the square)
✓ 1
A1 \(t=4\) hours
✓ 1
A1 \(C=9.6\) mg L−1
✓ 1
Part (b)
M1 solves \(-0.6t(t-8)=0\), or uses a GDC to find the positive zero
✓ 1
A1 \(t=8\) hours
✓ 1
Part (c)
M1 sets \(-0.6t^2+4.8t=6\), or finds the intersections with \(C=6\) on a GDC
✓ 1
A1 \(t=1.55\) and \(t=6.45\) (i.e. \(4\pm\sqrt{6}\))
✓ 1
A1 \(4.90\) hours (accept \(2\sqrt{6}\))
✓ 1
Part (d)
M1 maximum concentration \(=\frac{k^2}{4(0.6)}=\frac{k^2}{2.4}\), from the vertex at \(t=\frac{k}{1.2}\)
Exam tip: "Never exceeds" is a condition on the maximum value, so work with the vertex of the quadratic rather than solving C(t) = 12 for t.
224IBMA3S-2-17r11-062.10 Solving equationsPaper 2Medium8 marks
Calculator
During the early stage of a roast the bean temperature, in °C, is modelled by \(T(t)=-2t^2+28t+60\) for \(0\leq t\leq 14\), where \(t\) is in minutes.
(a)
Use the discriminant to show that the bean temperature never reaches 200 °C.
(4)
(b)
By considering the discriminant of \(T(t)=k\), find the greatest value of \(k\) for which this equation has real solutions.
(3)
(c)
Interpret your answer to (b) in context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 set \(-2t^2+28t+60=200\) and rearrange to \(t^2-14t+70=0\) (or equivalent)
✓ 1
M1 attempt to find the discriminant \(b^2-4ac\)
✓ 1
A1 \(\Delta=(-14)^2-4\times1\times70=-84\) (or \(-336\) if not divided by \(-2\))
✓ 1
R1 \(\Delta<0\) so there are no real solutions
✓ 1
Part (b)
M1 rearrange to \(-2t^2+28t+(60-k)=0\) and use \(\Delta\geq 0\)
✓ 1
A1 \(\Delta=28^2+8(60-k)=1264-8k\)
✓ 1
A1 \(k\leq 158\) so the greatest value is 158
✓ 1
Part (c)
R1 the maximum bean temperature during this stage is 158 °C
✓ 1
Exam tip: To find the greatest or least value of a quadratic without calculus, set it equal to \(k\) and require the discriminant to be at least zero.
225IBMA3S-2-17r11-072.2 Functions, domain, range and inversePaper 2Medium8 marks
Calculator
The quality reading of a probe's radio link, in suitable units, is modelled by \(R(t)=\frac{6t+9}{t+3}\) for \(t\geq 0\), where \(t\) is the time in hours since a signal check began.
(a)
Show that \(R(t)\) can be written in the form \(6-\frac{9}{t+3}\).
(2)
(b)
Write down the equation of the horizontal asymptote and state what it means for the reading.
(2)
(c)
Find the range of \(R\) for \(t\geq 0\).
(2)
(d)
Find the time at which the reading equals 5.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to divide or write \(6t+9=6(t+3)-9\)
✓ 1
A1 \(R(t)=\frac{6(t+3)-9}{t+3}=6-\frac{9}{t+3}\)
✓ 1
Part (b)
A1 \(R=6\)
✓ 1
R1 the reading gets closer to 6 as time increases but never reaches it
✓ 1
Part (c)
M1 \(R(0)=3\) and \(R\) is increasing towards the asymptote
✓ 1
A1 \(3\leq R<6\)
✓ 1
Part (d)
M1 \(6-\frac{9}{t+3}=5\) so \(t+3=9\)
✓ 1
A1 \(t=6\) hours
✓ 1
Exam tip: Rewriting a rational function as a constant plus a fraction makes the asymptote and the range easy to read off.
226IBMA3S-2-17r11-092.7 Quadratic equations and the discriminantPaper 2Medium7 marks
Calculator
A space probe's path near a planet is modelled by the parabola \(y=x^2-4x+7\), and a radio beam is modelled by the line \(y=2x+k\), where \(x\) and \(y\) are in km and \(k\) is a constant.
(a)
Show that the \(x\)-coordinates of the intersections satisfy \(x^2-6x+7-k=0\) and find the discriminant in terms of \(k\).
(2)
(b)
Find the value of \(k\) for which the beam is a tangent to the path.
(2)
(c)
Find the range of values of \(k\) for which the beam meets the path at two distinct points.
(1)
(d)
For the tangent case, find the \(y\)-coordinate of the point of contact.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equate \(x^2-4x+7=2x+k\) and rearrange
✓ 1
A1 \(\Delta=36-4(7-k)=8+4k\)
✓ 1
Part (b)
M1 \(\Delta=0\)
✓ 1
A1 \(k=-2\)
✓ 1
Part (c)
A1 \(k>-2\) (from \(8+4k>0\))
✓ 1
Part (d)
M1 repeated root \(x=3\) from \(x^2-6x+9=0\)
✓ 1
A1 \(y=2(3)-2=4\) km (check: \(9-12+7=4\))
✓ 1
Exam tip: Tangency to a curve at a single point corresponds to a repeated root, so the discriminant is zero; two intersections need a positive discriminant.
227IBMA3S-2-17r11-102.6 Quadratic functionsPaper 2Medium12 marks
Calculator
The arch of a footbridge has height \(h(x)=-0.05x^2+1.6x\) metres above the deck at a horizontal distance \(x\) metres from the left pier, for \(0\leq x\leq 32\).
(a)
Find the values of \(x\) at which the arch meets the deck.
(2)
(b)
Find the maximum height of the arch.
(2)
(c)
Write down the range of \(h\).
(1)
(d)
By considering the discriminant of \(-0.05x^2+1.6x-k=0\), show that \(h(x)=k\) has real solutions only when \(k\leq 12.8\).
(3)
(e)
Find the horizontal length of the part of the span where the arch is at least 10 m high. Give your answer to 4 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(h(x)=0\) and factorise: \(-0.05x(x-32)=0\)
✓ 1
A1 \(x=0\) and \(x=32\)
✓ 1
Part (b)
M1 axis of symmetry \(x=-\frac{1.6}{2(-0.05)}=16\)
✓ 1
A1 \(h(16)=12.8\) m
✓ 1
Part (c)
A1 \(0\leq h(x)\leq 12.8\)
✓ 1
Part (d)
M1 use \(\Delta\geq 0\) with \(a=-0.05\), \(b=1.6\), \(c=-k\)
✓ 1
A1 \(\Delta=2.56-0.2k\)
✓ 1
R1 \(2.56-0.2k\geq 0\) gives \(k\leq 12.8\)
✓ 1
Part (e)
M1 \(-0.05x^2+1.6x=10\) rearranged to \(x^2-32x+200=0\)
✓ 1
M1 solve: \(x=16\pm\sqrt{56}\)
✓ 1
A1 \(x=8.517\) and \(x=23.48\)
✓ 1
A1 length \(=2\sqrt{56}=14.97\) m
✓ 1
Exam tip: For a quadratic with known zeros, the axis of symmetry lies midway between them, which gives a quick check on the vertex.
228IBMA3S-2-17r11-112.10 Solving equationsPaper 2Medium7 marks
Calculator
A cooling tray lowers the temperature of roasted beans according to \(T(t)=22+78e^{-0.2t}\) degrees Celsius, \(t\geq 0\), where \(t\) is in minutes.
(a)
Write down the range of \(T\).
(2)
(b)
Find an expression for the time \(t\) in terms of \(T\).
(3)
(c)
Find the time taken to cool to 40 °C. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
A1 \(22<T\leq 100\) (initial value \(T(0)=100\))
✓ 1
A1 correct strict inequality at 22 (asymptote not reached)
✓ 1
Part (b)
M1 \(\frac{T-22}{78}=e^{-0.2t}\)
✓ 1
M1 take natural logarithms: \(-0.2t=\ln\frac{T-22}{78}\)
✓ 1
A1 \(t=-5\ln\frac{T-22}{78}\)
✓ 1
Part (c)
M1 \(t=-5\ln\frac{18}{78}\)
✓ 1
A1 \(t=7.33\) minutes
✓ 1
Common error: Take logarithms only after the exponential term has been isolated; the constant 22 must be subtracted first.
229IBMA3S-2-16r17-052.10 Solving equationsPaper 2Medium8 marks
Calculator
After a fertiliser is applied, the concentration of nitrate in a field, \(C\) mg per kg of soil, decreases exponentially. It is modelled by \(C=48e^{-kt}\), where \(t\) is the time in days after application and \(k>0\). After 10 days the concentration is 30 mg per kg. In parts (b) and (c) use the unrounded value of \(k\) from part (a).
(a)
Find the value of \(k\). Give your answer to 3 significant figures.
(3)
(b)
Find the time taken for the concentration to halve. Give your answer to 3 significant figures.
(2)
(c)
Find the time taken for the concentration to fall to 10 mg per kg. Give your answer to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 \(30=48e^{-10k}\)
✓ 1
M1 \(-10k=\ln\frac{30}{48}\)
✓ 1
A1 \(k=0.0470\)
✓ 1
Part (b)
M1 \(e^{-kt}=\frac{1}{2}\), so \(t=\frac{\ln2}{k}\)
✓ 1
A1 \(t=14.7\) days
✓ 1
Part (c)
M1 \(10=48e^{-kt}\)
✓ 1
M1 \(t=\frac{\ln4.8}{k}\)
✓ 1
A1 \(t=33.4\) days
✓ 1
Common error: Writing \(\ln\frac{30}{48}\) as \(\frac{\ln30}{\ln48}\) is wrong; the quotient must stay inside the logarithm.
230IBMA3S-2-16r17-062.3 Graphs of functionsPaper 2Medium6 marks
Calculator
The cross-section of a ridge on a farm is modelled by \(f(x)=x^3-7x+3\), where \(x\) is the horizontal distance in hundreds of metres. Use your graphic display calculator to answer the following.
(a)
Find the zeros of \(f\). Give your answers to 3 significant figures.
(2)
(b)
The line \(y=2x-1\) meets the graph of \(y=f(x)\) at three points. Find the \(x\)-coordinate of the point with the greatest \(x\)-coordinate. Give your answer to 3 significant figures.
(2)
(c)
Find the coordinates of the local minimum point of the graph of \(f\). Give your answers to 3 significant figures.
(2)
Show mark scheme
Marking point
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Notes
Part (a)
M1 attempt to solve \(f(x)=0\) using technology
✓ 1
A1 \(x=-2.84,\ 0.441,\ 2.40\)
✓ 1
Part (b)
M1 attempt to solve \(x^3-7x+3=2x-1\), or equivalent, using technology
✓ 1
A1 \(x=2.75\)
✓ 1
Part (c)
M1 attempt to find the minimum using technology
✓ 1
A1 \((1.53,\,-4.13)\)
✓ 1
Exam tip: Quote the technology result to 3 significant figures in the final line; show the equation you entered so a method mark is available.
231IBMA3S-2-16r17-072.7 Quadratic equations and the discriminantPaper 2Medium5 marks
Calculator
The area of weeds in a field is \(W=4e^{0.3t}\) square metres, where \(t\) is the time in days after the weeds are first measured. A treated area increases at a steady rate and is modelled by \(A=6t+10\) square metres. For \(t>0\) the curve and the line meet once.
(a)
Find the value of \(t\) at which \(W=A\). Give your answer to 3 significant figures.
(2)
(b)
Find the area at this time. Give your answer to 3 significant figures.
(1)
(c)
Find the time taken for the area of weeds to double. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to solve \(4e^{0.3t}=6t+10\) using technology
✓ 1
A1 \(t=9.35\) days
✓ 1
Part (b)
A1 \(66.1\) m\(^2\)
✓ 1
Part (c)
M1 \(e^{0.3t}=2\), so \(0.3t=\ln2\)
✓ 1
A1 \(t=2.31\) days
✓ 1
Common error: The doubling time of \(e^{kt}\) does not depend on the starting value, so there is no need to use the 4 m\(^2\).
232IBMA3S-2-16r17-082.9 Exponential and logarithmic functionsPaper 2Medium7 marks
Calculator
In equal temperament, the frequency \(f\) Hz of a note \(n\) semitones above the note A4 (440 Hz) is given by \(f=440\times2^{\frac{n}{12}}\).
(a)
Find the frequency of the note 7 semitones above A4. Give your answer to the nearest Hz.
(2)
(b)
Show that the note 12 semitones above A4 has a frequency twice that of A4.
(2)
(c)
A synthesiser produces a tone of frequency 1200 Hz. Find the value of \(n\) for this frequency. Give your answer to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 \(440\times2^{\frac{7}{12}}\)
✓ 1
A1 \(659\) Hz
✓ 1
Part (b)
M1 \(f=440\times2^{\frac{12}{12}}\)
✓ 1
A1 \(=440\times2=880\), which is \(2\times440\)
✓ 1
Part (c)
M1 \(1200=440\times2^{\frac{n}{12}}\)
✓ 1
M1 \(\frac{n}{12}=\log_2\frac{1200}{440}\), or uses natural logarithms correctly
✓ 1
A1 \(n=17.4\)
✓ 1
Exam tip: To solve \(a^{x}=b\) use \(x=\log_a b=\frac{\ln b}{\ln a}\) and keep the full calculator value until the end.
233IBMA3S-2-16r17-102.8 Reciprocal and rational functionsPaper 2Medium7 marks
Calculator
A farmer mixes \(x\) litres of a concentrated herbicide with 20 litres of water. The percentage concentration \(c\) of the mixture is modelled by \(c(x)=\frac{100x}{x+20}\), for \(x\ge0\).
(a)
Find the concentration when 5 litres of herbicide are added.
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(c\), and interpret it in this context.
(2)
(c)
Find the volume of herbicide needed for a concentration of 60%.
(3)
Show mark scheme
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Notes
Part (a)
M1 \(c(5)=\frac{100\times5}{5+20}\)
✓ 1
A1 \(20\%\)
✓ 1
Part (b)
A1 \(c=100\)
✓ 1
R1 the concentration approaches but never reaches 100% however much herbicide is added
✓ 1
Part (c)
M1 \(\frac{100x}{x+20}=60\)
✓ 1
M1 \(100x=60x+1200\)
✓ 1
A1 \(x=30\) litres
✓ 1
Common error: When interpreting an asymptote, relate it to the context: say what happens to the quantity as \(x\) becomes very large.
234IBMA3S-2-17r17-062.10 Solving equationsPaper 2Medium7 marks
Calculator
The concentration of a drug in a patient's blood, C mg L−1, t hours after it is administered, is modelled by C(t) = Ae−kt, where A and k are positive constants. The concentration is 24.0 mg L−1 when t = 2 and 15.0 mg L−1 when t = 5. In parts (b) and (c) use the unrounded value of k and give your answers to 3 significant figures.
(a)
Find the value of k.
(3)
(b)
Find the value of A.
(2)
(c)
Find the time at which the concentration falls to 5.00 mg L−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms e−3k = 15/24 (dividing the two equations)
✓ 1
M1 takes natural logarithms: −3k = ln 0.625
✓ 1
A1 k = 0.157 (0.156678…)
✓ 1
Part (b)
M1 substitutes into 24.0 = Ae−2k
✓ 1
A1 A = 32.8
✓ 1
Part (c)
M1 solves Ae−kt = 5, e.g. t = ln(A/5)/k
✓ 1
A1 t = 12.0 hours
✓ 1
Exam tip: Dividing the two given equations eliminates A and leaves an equation in k alone.
235IBMA3S-2-17r17-072.7 Quadratic equations and the discriminantPaper 2Medium6 marks
Calculator
In a video game, a character jumps from a platform. The height of the character above the ground, h metres, t seconds after the jump begins is modelled by h(t) = 1.5 + 7.2t − 4.9t2, for t ≥ 0, until the character lands.
(a)
Find the time at which the character lands on the ground. Give your answer to 3 significant figures.
(2)
(b)
Find the length of time for which the character is more than 3 m above the ground. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 solves 1.5 + 7.2t − 4.9t2 = 0 (quadratic formula or GDC)
✓ 1
A1 t = 1.65 s (rejects the negative root)
✓ 1
Part (b)
M1 forms the equation or inequality h(t) = 3, i.e. 4.9t2 − 7.2t + 1.5 = 0
✓ 1
A1 roots t = 0.251… and t = 1.22…
✓ 1
M1 subtracts the roots (unrounded)
✓ 1
A1 0.967 s
✓ 1
Common error: Read the question for what is being asked: the answer is the difference between the two times at which h = 3, not either root on its own.
236IBMA3S-2-17r17-132.8 Reciprocal and rational functionsPaper 2Medium5 marks
Calculator
A water company pumps water from a borehole at a constant rate. The volume pumped in t hours is V(t) = 45t m3. The cost, in €, of treating and pumping V m3 is C(V) = 0.35V + 120.
(a)
Find (C ∘ V)(t).
(2)
(b)
State what (C ∘ V)(t) represents in this context.
(1)
(c)
The total cost is €1500. Find the number of hours of pumping, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes V(t) into C: 0.35 × 45t + 120
✓ 1
A1 15.75t + 120
✓ 1
Part (b)
A1 the total cost in € of pumping and treating water for t hours
✓ 1
Part (c)
M1 solves 15.75t + 120 = 1500
✓ 1
A1 t = 87.6 hours
✓ 1
Exam tip: Include units and meaning when asked to interpret a composite function: the input is time, the output is cost.
237IBMA3S-2-17r26-062.1 Equations of straight linesPaper 2Medium7 marks
Calculator
Two straight service roads on a wind farm are drawn on a grid with units in kilometres. Road 1 passes through A(1, 2) and B(9, 8). A turbine is at T(10, −2). A cable is to be laid from T to the nearest point on Road 1.
Diagram not to scale
(a)
Find the equation of Road 1 in the form ax + by + d = 0, where a, b and d are integers.
(2)
(b)
Find the equation of the line through T that is perpendicular to Road 1.
(2)
(c)
Find the length of the shortest cable from T to Road 1.
(3)
Show mark scheme
Marking point
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Notes
Part (a)
M1 gradient = 6⁄8 = 3⁄4 and substitute a point
✓ 1
A1 3x − 4y + 5 = 0 (or any integer multiple)
✓ 1
Part (b)
A1 perpendicular gradient = −4⁄3
✓ 1
A1 4x + 3y = 34 (or equivalent)
✓ 1
Part (c)
M1 solve the two line equations simultaneously
✓ 1
A1 meeting point (4.84, 4.88)
✓ 1
A1 distance = 8.6 km
✓ 1
Exam tip: The shortest distance from a point to a line is along the perpendicular from the point.
238IBMA3S-2-17r26-082.7 Quadratic equations and the discriminantPaper 2Medium11 marks
Calculator
A wind speed v m s−1 at a height h metres above ground on a small wind farm is modelled by v = f(h) = 2 + 0.1h, for 0 ≤ h ≤ 100. The power output P kW of a turbine at wind speed v is modelled by P = g(v) = 0.5v2, for 0 ≤ v ≤ 12.
(a)
Find the wind speed at a height of 80 m and the power output of a turbine in that wind.
(3)
(b)
Find an expression for (g ∘ f)(h) in the form a + bh + ch2.
(2)
(c)
Find f−1(v) and state what it represents.
(3)
(d)
Determine the height at which a turbine produces 72 kW.
(3)
Show mark scheme
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Notes
Part (a)
A1 f(80) = 10 m s−1
✓ 1
M1 g(10)
✓ 1
A1 50 kW
✓ 1
Part (b)
M1 0.5(2 + 0.1h)2 with an attempt to expand
✓ 1
A1 2 + 0.2h + 0.005h2
✓ 1
Part (c)
M1 rearrange v = 2 + 0.1h for h
✓ 1
A1 f−1(v) = 10v − 20
✓ 1
R1 the height at which the wind speed is v m s−1
✓ 1
Part (d)
M1 0.5v2 = 72
✓ 1
A1 v = 12 (reject −12 as v ≥ 0)
✓ 1
A1 h = 100 m (accept solving 2 + 0.2h + 0.005h2 = 72)
✓ 1
Exam tip: When a question asks for a height from a power, work backwards through the composite one function at a time.
239IBMA3S-2-17r26-102.1 Equations of straight linesPaper 2Medium7 marks
Calculator
A swimming pool plan uses a grid with units in metres. A lane rope lies along line L1 with equation 2x − 3y = 12.
(a)
Find the coordinates of the points where L1 crosses the x-axis and the y-axis.
(2)
(b)
Hence find the area of the triangle bounded by L1 and the two axes.
(1)
(c)
A second rope lies along line L2, parallel to L1, passing through P(3, 5). Find the equation of L2.
(2)
(d)
A third rope lies along L3 with equation kx + 6y = 1 and is perpendicular to L1. Find k.
(2)
Show mark scheme
Marking point
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Notes
Part (a)
A1 (6, 0)
✓ 1
A1 (0, −4)
✓ 1
Part (b)
A1 1⁄2 × 6 × 4 = 12 m2
✓ 1
Part (c)
M1 same gradient 2⁄3 used with P, or 2x − 3y = c with substitution
✓ 1
A1 2x − 3y = −9 (or equivalent)
✓ 1
Part (d)
M1 gradient of L3 is −k/6, set equal to −3⁄2
✓ 1
A1 k = 9
✓ 1
Exam tip: Parallel lines have equal gradients; to compare gradients, rewrite each line as y = mx + c.
240IBMA3S-2-17r26-112.11 Transformations of graphsPaper 2Medium7 marks
Calculator
The graph of y = x2 is transformed into the graph of y = g(x) by the following transformations, applied in this order: a reflection in the x-axis, a vertical stretch with scale factor 3, then a translation by the vector (2, 7).
(a)
Write down g(x) in the form a(x − h)2 + k.
(3)
(b)
Expand g(x) to the form ax2 + bx + c.
(2)
(c)
Solve g(x) = 0, giving your answers to 3 significant figures.
(2)
Show mark scheme
Marking point
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Notes
Part (a)
A1 reflection gives −x2
✓ 1
A1 stretch gives −3x2
✓ 1
A1 g(x) = −3(x − 2)2 + 7
✓ 1
Part (b)
M1 expand (x − 2)2 = x2 − 4x + 4
✓ 1
A1 −3x2 + 12x − 5
✓ 1
Part (c)
M1 use the quadratic formula or GDC on −3x2 + 12x − 5 = 0
✓ 1
A1 x = 0.472 and x = 3.53
✓ 1
Common error: Apply transformations in the order stated: a stretch before a translation changes the stretched value, not the shifted one.
241IBMA3S-2-19r28-052.4 Key features and intersectionsPaper 2Medium7 marks
Calculator
The power output, P megawatts (MW), of a solar farm is modelled by \(P(t)=-0.4t^2+5.6t-12.8\), where t is the number of hours after 04:00. The model applies while the farm is generating power.
(a)
Find the two values of t for which \(P(t)=0\). Give your answers to 3 significant figures. [2]
(2)
(b)
Find the maximum power output and the time at which it occurs, in hours after 04:00. [2]
(2)
(c)
The farm can sell power to the grid only while \(P>4\). Find the length of time, in hours to 3 significant figures, for which this is true. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 solve \(-0.4t^2+5.6t-12.8=0\)
✓ 1
A1 \(t=2.88\) and \(t=11.1\)
✓ 1
Part (b)
M1 axis of symmetry \(t=\frac{-5.6}{2(-0.4)}=7\)
✓ 1
A1 maximum output 6.8 MW at \(t=7\)
✓ 1
Part (c)
M1 solve \(-0.4t^2+5.6t-12.8=4\), i.e. \(t^2-14t+42=0\)
✓ 1
A1 \(t=4.35\) and \(t=9.65\)
✓ 1
A1 duration \(=5.29\) hours
✓ 1
Exam tip: For \"above a value\" questions, solve the equation first, then use the shape of the graph to decide whether the required values lie between the roots or outside them.
242IBMA3S-2-19r28-072.9 Exponential and logarithmic functionsPaper 2Medium7 marks
Calculator
During an outbreak, the number of reported cases is modelled by \(I(t)=120\times1.18^{t}\), where t is the number of days since the outbreak was first detected.
(a)
Find the smallest whole number of days after which more than 1000 cases have been reported. [3]
(3)
(b)
Find the time, in days to 3 significant figures, for the number of cases to double. [2]
(2)
(c)
Find an expression for t in terms of I. [2]
(2)
Show mark scheme
Marking point
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Notes
Part (a)
M1 \(120\times1.18^{t}=1000\)
✓ 1
A1 \(t=12.8\)
✓ 1
A1 13 days
✓ 1
Part (b)
M1 \(1.18^{t}=2\)
✓ 1
A1 \(t=\frac{\ln 2}{\ln 1.18}=4.19\) days
✓ 1
Part (c)
M1 \(\frac{I}{120}=1.18^{t}\)
✓ 1
A1 \(t=\frac{\ln(I/120)}{\ln 1.18}\)
✓ 1
Common error: The doubling time does not depend on the starting value: solve \(1.18^t=2\), not \(120\times1.18^t=2\).
243IBMA3S-2-19r28-092.9 Exponential and logarithmic functionsPaper 2Medium8 marks
Calculator
A grid battery is charged from a partially full state. The energy stored, E MWh, after t hours of charging is modelled by \(E(t)=2000-1500e^{-0.3t}\), for \(t\geq 0\).
(a)
Find the energy stored when charging begins. [1]
(1)
(b)
Write down the equation of the horizontal asymptote of the graph of \(E\), and state what it represents in this context. [2]
(2)
(c)
Find the time taken for the stored energy to reach 1500 MWh. Give your answer in hours to 3 significant figures. [3]
(3)
(d)
Find an expression for the inverse function \(E^{-1}(x)\). [2]
(2)
Show mark scheme
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Notes
Part (a)
A1 500 MWh
✓ 1
Part (b)
A1 \(E=2000\)
✓ 1
R1 the maximum energy the battery can hold (the stored energy can never exceed 2000 MWh)
✓ 1
Part (c)
M1 \(2000-1500e^{-0.3t}=1500\), so \(e^{-0.3t}=\frac{1}{3}\)
✓ 1
M1 \(-0.3t=\ln\frac{1}{3}\)
✓ 1
A1 \(t=3.66\) hours
✓ 1
Part (d)
M1 \(x=2000-1500e^{-0.3t}\), so \(e^{-0.3t}=\frac{2000-x}{1500}\)
Exam tip: An exponential model of the form \(a-be^{-kt}\) levels off at \(a\). Use this to check that your answers to any \"time to reach\" question are less than the asymptote.
244IBMA3S-2-19r28-112.4 Key features and intersectionsPaper 2Medium8 marks
Calculator
Two electricity suppliers have unit prices, in pence per kWh, t years after the start of 2025 given by: Supplier A: \(A(t)=22+1.4t\) Supplier B: \(B(t)=18\times1.07^{t}\)
(a)
Write down the price charged by each supplier at \(t=0\). [2]
(2)
(b)
State the annual percentage increase in the price charged by supplier B. [1]
(1)
(c)
Find the value of t, for \(t>0\), at which the two suppliers charge the same price. Give your answer to 3 significant figures. [3]
(3)
(d)
Find the common price at this time. Give your answer in pence to 3 significant figures, using the unrounded value of t. [2]
(2)
Show mark scheme
Marking point
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Notes
Part (a)
A1 A: 22 pence
✓ 1
A1 B: 18 pence
✓ 1
Part (b)
A1 7%
✓ 1
Part (c)
M1 equate \(22+1.4t=18\times1.07^{t}\)
✓ 1
M1 solve using the intersection of the graphs or an equation solver
✓ 1
A1 \(t=10.6\) years
✓ 1
Part (d)
M1 substitute into \(A(t)=22+1.4t\) (or into \(B(t)\))
✓ 1
A1 36.8 pence
✓ 1
Exam tip: When an equation mixes a linear and an exponential expression there is no algebraic method. Use the intersection feature of your graphing calculator, and note the window needed to see the point.
245IBMA3S-2-17r39-112.10 Solving equationsPaper 2Medium7 marks
Calculator
Lava from a vent cools after it stops flowing. Its temperature in °C, \(t\) minutes after it stops, is modelled by \(\theta(t)=20+980e^{-0.04t}\), for \(t\ge0\).
(a)
Find the temperature of the lava at the moment it stops flowing.
(1)
(b)
Find an expression for \(\theta^{-1}(x)\) in terms of \(x\).
(3)
(c)
Find how long the lava takes to cool to 500 °C, giving your answer to the nearest tenth of a minute.
(2)
(d)
Write down the domain of \(\theta^{-1}\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\theta(0)=1000\) °C
✓ 1
Part (b)
M1 \(x-20=980e^{-0.04t}\) so \(e^{-0.04t}=\frac{x-20}{980}\)
✓ 1
M1 take natural logarithms: \(-0.04t=\ln\frac{x-20}{980}\)
Exam tip: The domain of an inverse is the range of the original function: \(\theta\) decreases from 1000 towards, but never reaches, 20.
246IBMA3S-2-17r39-122.5 Composite and inverse functionsPaper 2Medium10 marks
Calculator
The function \(f\) is defined by \(f(x)=\sqrt{2x-6}+1\) for the largest possible domain.
(a)
Write down the domain of \(f\).
(1)
(b)
Write down the range of \(f\).
(1)
(c)
Find \(f^{-1}(x)\).
(3)
(d)
Write down the domain of \(f^{-1}\).
(1)
(e)
Solve \(f^{-1}(x)=11\).
(3)
(f)
Hence write down the value of \(f(11)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x\ge3\)
✓ 1
Part (b)
A1 \(f(x)\ge1\)
✓ 1
Part (c)
M1 \(y-1=\sqrt{2x-6}\) (isolate the root)
✓ 1
M1 square both sides: \((y-1)^2=2x-6\)
✓ 1
A1 \(f^{-1}(x)=\frac{(x-1)^2+6}{2}\)
✓ 1
Part (d)
A1 \(x\ge1\)
✓ 1
Part (e)
M1 \((x-1)^2+6=22\), so \((x-1)^2=16\)
✓ 1
A1 \(x=5\) or \(x=-3\)
✓ 1
R1 reject \(x=-3\) because it lies outside the domain \(x\ge1\)
✓ 1
Part (f)
A1 \(f(11)=5\)
✓ 1
Common error: Squaring can introduce a solution that does not belong to the function: always check answers against the domain of the inverse.
247IBMA3S-2-19r40-032.1 Equations of straight linesPaper 2Medium6 marks
Calculator
A bloom of algae on the surface of a lagoon covers an area of A m2 at time t days after it is first measured. The area is modelled by \(A=12e^{0.35t}\).
(a)
Write down the area of the bloom when it is first measured.
(1)
(b)
Find the area of the bloom after 5 days. Give your answer to 3 significant figures.
(2)
(c)
Find the time taken for the area to reach 100 m2. Give your answer in days to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 12 m2
✓ 1
Part (b)
M1 substitutes \(t=5\) into \(12e^{0.35t}\)
✓ 1
A1 \(12e^{1.75}\) = 69.1 m2 (69.055…)
✓ 1
Part (c)
M1 sets \(12e^{0.35t}=100\) and isolates the exponential, \(e^{0.35t}=\frac{100}{12}\)
✓ 1
M1 takes natural logarithms, \(0.35t=\ln\frac{100}{12}\) (or finds the intersection with a graphing tool)
✓ 1
A1 \(t\) = 6.06 days
✓ 1
Common error: Taking logarithms before isolating the exponential (e.g. writing \(0.35t=\ln 100-12\)) is not valid. Divide by the coefficient first.
248IBMA3S-2-19r40-072.10 Solving equationsPaper 2Medium5 marks
Calculator
A logistics firm compares the storage charge for a pallet at two depots, \(t\) weeks after the pallet arrives, where \(t\ge0\). Depot A charges £\(S\), where \(S=250+18t\). Depot B charges £\(E\), where \(E=90e^{0.3t}\).
(a)
Find the value of \(t\) at which the two depots charge the same amount. Give your answer to 3 significant figures.
(2)
(b)
Write down the values of \(t\) for which Depot A charges less than Depot B.
(1)
(c)
Find the charge at the time found in part (a). Use the unrounded value of \(t\) and give your answer in pounds to 2 decimal places.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates \(250+18t=90e^{0.3t}\) (or sketches both graphs)
✓ 1
A1 \(t\) = 4.31 weeks
✓ 1
Part (b)
A1 \(t>4.31\) (ECF from (a))
✓ 1
Part (c)
M1 substitutes the unrounded \(t\) (4.3056…) into \(250+18t\) or \(90e^{0.3t}\)
✓ 1
A1 £327.50
✓ 1
Exam tip: Store the intersection value on your calculator and reuse it. Re-typing a rounded value can shift the final pence.
249IBMA3S-2-19r40-102.1 Equations of straight linesPaper 2Medium9 marks
Calculator
A courier firm has depots at A(1, 2) and B(7, 6), where the coordinates are in kilometres. A new distribution centre, C, is to be the same distance from A and from B, and it must lie on a straight road with equation \(y=0.5x+3\).
Diagram not to scale
(a)
Find the coordinates of the midpoint M of AB.
(1)
(b)
Show that the perpendicular bisector of AB has equation \(3x+2y=20\).
(3)
(c)
Find the coordinates of C.
(3)
(d)
Find the distance AC. Give your answer in kilometres to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 M(4, 4)
✓ 1
Part (b)
A1 gradient of AB = \(\frac{6-2}{7-1}=\frac{2}{3}\)
✓ 1
M1 perpendicular gradient = \(-\frac{3}{2}\), and uses the line through M(4, 4)
✓ 1
A1 \(y-4=-\frac{3}{2}(x-4)\) leading to \(3x+2y=20\) (AG)
✓ 1
Part (c)
M1 substitutes \(y=0.5x+3\) into \(3x+2y=20\) (or solves the two equations simultaneously)
✓ 1
A1 \(x=3.5\)
✓ 1
A1 \(y=4.75\), C(3.5, 4.75)
✓ 1
Part (d)
M1 \(\sqrt{(3.5-1)^2+(4.75-2)^2}\)
✓ 1
A1 AC = 3.72 km
✓ 1
Exam tip: After finding C, check it by calculating BC as well: if AC and BC are not equal, there is an error in the bisector or the intersection.
250AASL2-q00212.3 Graphs of functionsPaper 2Hard9 marks
Let \(s(x)=\tfrac{1}{2}x^{2}+4\mathrm{e}^{-x^{2}}\), for \(-3\le x\le3\).
(a)
Sketch the graph of \(y=s(x)\), for \(-3\le x\le3\).
(3)
(b)
Show that the graph is symmetrical about the \(y\)-axis.
(1)
(c)
Find the coordinates of the two minimum points of the graph.
(2)
(d)
Find the set of values of \(k\) for which the equation \(s(x)=k\) has exactly four solutions.
(2)
(e)
Write down the number of solutions of \(s(x)=4\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
local maximum at \((0,\,4)\)
A1
two minimum points, symmetrically placed either side of the \(y\)-axis
A1
endpoints approximately \((\pm3,\,4.50)\), above the level of the maximum
251AASL2-q00262.4 Key features and intersectionsPaper 2Hard7 marks
During a chemical process, the temperature of a liquid, \(T\,{}^{\circ}\mathrm{C}\), \(h\) minutes after the start is modelled by \(T(h)=0.02h^{3}-0.75h^{2}+7.2h+8\), for \(0\le h\le24\).
(a)
Find the maximum temperature of the liquid during the process, and the time at which it occurs.
(2)
(b)
Find the minimum temperature of the liquid during the process.
(2)
(c)
A cooling fan is switched off whenever the temperature is below \(20\,{}^{\circ}\mathrm{C}\). Find the total length of time for which the fan is switched off.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(28.6\,{}^{\circ}\mathrm{C}\)
A1
\(28.6051\ldots\) The value at \(h=24\) is \(25.28\), which is less.
\(h=6.48\) minutes
A1
\(6.47920\ldots\)
(b)
comparing the local minimum \(T(18.5)=11.1\) with the endpoint values \(T(0)=8\) and \(T(24)=25.28\)
252AASL2-q00282.4 Key features and intersectionsPaper 2Hard14 marks
A cable hangs between the tops of two towers that stand 100 m apart on opposite banks of a river. The height of the cable above the river, \(h\) metres, at a horizontal distance \(x\) metres from the midpoint between the towers, is modelled by \(h(x)=10\left(\mathrm{e}^{\frac{x}{40}}+\mathrm{e}^{-\frac{x}{40}}\right)-12\), for \(-50\le x\le50\).
(a)
Write down the height of the cable at the midpoint between the towers.
(1)
(b)
Find the height of each tower.
(1)
(c)
Find the total horizontal distance for which the cable is less than 10 m above the river.
(2)
(d)
A straight walkway below the cable is modelled by \(w(x)=4+0.04x\), for \(-50\le x\le50\). Vertical hangers join the cable to the walkway. Find the length of the shortest hanger, and the value of \(x\) at which it occurs.
(3)
(e)
Find the greatest vertical distance between the cable and the walkway.
(2)
(f)
Hangers longer than 5 m need extra bracing. Find the total horizontal length of walkway for which the hangers need extra bracing.
(3)
(g)
The walkway is to be replaced by a horizontal walkway, \(y=c\). The shortest vertical distance between the cable and the new walkway must be 3 m. Find the value of \(c\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(h(0)=8\) m
A1
(b)
\(h(\pm50)=25.8\) m
A1
\(25.7684\ldots\)
(c)
solving \(h(x)=10\), e.g. \(x=\pm17.74\)
(M1)
\(\pm17.7427\ldots\)
\(35.5\) m
A1
\(35.4854\ldots\)
(d)
graphing \(h(x)-w(x)\) and finding its minimum
(M1)
\(3.94\) m
A1
\(3.93603\ldots\)
\(x=3.20\)
A1
\(3.19659\ldots\) Accept 3.2.
(e)
considering the end values \(h(-50)-w(-50)\) and \(h(50)-w(50)\)
(M1)
Evidence of checking the endpoints of the domain.
\(23.8\) m (at \(x=-50\))
A1
\(23.7684\ldots\) Award M1A0 for 19.8 (at x = 50).
(f)
solving \(h(x)-w(x)=5\)
(M1)
\(x=-9.83\) and \(x=16.1\)
A1
\(-9.82892\ldots,\ 16.1116\ldots\)
\((-9.83+50)+(50-16.11)=74.1\) m
A1
\(74.0594\ldots\) FT from their roots.
(g)
the least value of \(h\) is \(8\), at \(x=0\)
(M1)
FT from (a).
\(c=5\)
A1
Answers: (a) 8 m · (b) 25.8 m · (c) 35.5 m · (d) 3.94 m at \(x=3.20\) · (e) 23.8 m · (f) 74.1 m · (g) \(c=5\)
253AASL2-q00442.6 Quadratic functionsPaper 2Hard13 marks
A workshop sells handmade lamps. When the selling price is \(x\) dollars, the weekly profit, \(P\) dollars, is modelled by \(P(x)=ax^{2}+bx+c\). The workshop has fixed weekly costs of $200, so that \(P(0)=-200\). It records a weekly profit of $150 when the price is $10 and a weekly profit of $400 when the price is $20.
(a)
Find the value of \(a\), of \(b\) and of \(c\).
(4)
(b)
Express \(P(x)\) in the form \(a(x-h)^{2}+k\).
(2)
(c)
Write down the maximum weekly profit and the price at which it occurs.
(2)
(d)
Find the range of selling prices for which the workshop makes a profit.
(3)
(e)
Find the range of selling prices for which the weekly profit is at least $450.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(c=-200\)
A1
\(100a+10b-200=150\) and \(400a+20b-200=400\)
(M1)
Substituting both points to form two equations in a and b.
\(a=-0.5\)
A1
\(b=40\)
A1
(b)
\(-0.5\left(x^{2}-80x\right)-200\), or vertex at \(x=-\dfrac{40}{2(-0.5)}=40\)
(M1)
\(P(x)=-0.5(x-40)^{2}+600\)
A1
(c)
maximum profit \(\$600\)
A1
FT from (b).
at a price of \(\$40\)
A1
(d)
solving \(P(x)=0\)
(M1)
\(x=5.36\) and \(x=74.6\)
A1
\(40\pm\sqrt{1200}=5.3589\ldots,\ 74.641\ldots\)
\(5.36\lt x\lt74.6\)
A1
Accept \(\$5.36\lt x\lt\$74.64\).
(e)
\(-0.5(x-40)^{2}+600\ge450\), i.e. \((x-40)^{2}\le300\)
254AASL2-q00722.9 Exponential and logarithmic functionsPaper 2Hard8 marks
Alex invests $5000 in an account in which the value after \(t\) years is \(A(t)=5000\mathrm{e}^{0.04t}\). Beth invests $6000 in an account that pays 3% per year, compounded annually, so the value after \(t\) complete years is \(B(t)=6000(1.03)^{t}\).
(a)
Find the value of Alex’s investment after 10 years. Give your answer correct to the nearest cent.
(2)
(b)
Show that Alex’s account is equivalent to an annual percentage increase of 4.08%, correct to three significant figures.
(2)
(c)
Show that \(A(t)=B(t)\) when \(t=\dfrac{\ln1.2}{0.04-\ln1.03}\).
(2)
(d)
Find the least number of complete years after which Alex’s investment is worth more than Beth’s.
\(t=17.46\ldots\), or comparing values: \(A(17)=9869.39\lt B(17)=9917.09\), \(A(18)=10\,272.17\gt B(18)=10\,214.60\)
(M1)
18 years
A1
Answers: (a) $7459.12 · (d) 18 years
255AASL2-q00732.9 Exponential and logarithmic functionsPaper 2Hard13 marks
Two colonies of bacteria are being studied. The size of colony A, \(t\) hours after the start of the study, is modelled by \(A(t)=500\mathrm{e}^{0.12t}\). The size of colony B is modelled by \(B(t)=2000+150t\). Both models apply for \(t\ge0\).
(a)
Write down the initial size of colony A.
(1)
(b)
Show that, according to the model, colony A increases by approximately 12.7% each hour.
(2)
(c)
Find the time taken for the size of colony A to quadruple.
(2)
(d)
Find the value of \(t\) when the two colonies are the same size.
(2)
(e)
For values of \(t\) between 0 and the value found in part (d), find the greatest value of \(B(t)-A(t)\), and the time at which it occurs.
(3)
(f)
Find the time at which colony A is twice the size of colony B.
(2)
(g)
Suggest one reason why the model for colony A may not be valid for large values of \(t\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
500
A1
(b)
\(\dfrac{A(t+1)}{A(t)}=\mathrm{e}^{0.12}\)
M1
\(\mathrm{e}^{0.12}=1.1275\ldots\)
A1
Must see a value to at least 4 s.f.
increase of approximately 12.7% per hour
AG
(c)
\(\mathrm{e}^{0.12t}=4\), so \(t=\dfrac{\ln4}{0.12}\)
(M1)
\(11.6\) hours
A1
\(11.552\ldots\)
(d)
intersection of \(y=500\mathrm{e}^{0.12t}\) and \(y=2000+150t\)
(M1)
\(t=18.9\)
A1
\(18.911\ldots\)
(e)
graph of \(y=2000+150t-500\mathrm{e}^{0.12t}\) and its maximum
(M1)
\(1895\) (\(1.90\times10^{3}\))
A1
\(1894.6\ldots\) Accept 1890–1900.
at \(t=7.64\) hours
A1
\(7.6357\ldots\)
(f)
\(500\mathrm{e}^{0.12t}=2(2000+150t)\)
(M1)
\(t=26.4\) hours
A1
\(26.435\ldots\)
(g)
exponential growth would continue without limit, but in reality space and nutrients are limited
R1
Accept any reasonable limiting factor.
Answers: (a) 500 · (c) 11.6 h · (d) \(t=18.9\) · (e) 1895 at \(t=7.64\) · (f) 26.4 h
256AASL2-q00822.10 Solving equationsPaper 2Hard14 marks
A cup of tea is left to cool in a room. Its temperature, \(T\,{}^{\circ}\mathrm{C}\), \(t\) minutes after it is poured is modelled by \(T(t)=22+68\mathrm{e}^{-kt}\), for \(t\ge0\), where \(k\) is a positive constant.
(a)
Write down the temperature of the tea when it is poured.
(1)
(b)
After 10 minutes the temperature of the tea is \(60\,{}^{\circ}\mathrm{C}\). Find the value of \(k\).
(3)
(c)
Find the temperature of the tea after 25 minutes.
(2)
(d)
Find the time taken for the temperature of the tea to fall to \(40\,{}^{\circ}\mathrm{C}\).
(2)
(e)
Write down the equation of the horizontal asymptote of the graph of \(T\), and interpret it in context.
(2)
(f)
At \(t=0\), a glass of cold juice is placed in the same room. Its temperature is modelled by \(U(t)=22-18\mathrm{e}^{-0.08t}\). Write down the initial temperature of the juice.
(1)
(g)
Find the time at which the temperature of the tea is \(5\,{}^{\circ}\mathrm{C}\) more than the temperature of the juice.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(90\,{}^{\circ}\mathrm{C}\)
A1
(b)
\(22+68\mathrm{e}^{-10k}=60\)
(M1)
\(\mathrm{e}^{-10k}=\dfrac{38}{68}\), or GDC solver
the tea approaches room temperature, \(22\,{}^{\circ}\mathrm{C}\), but never falls below it
R1
(f)
\(4\,{}^{\circ}\mathrm{C}\)
A1
(g)
\(T(t)-U(t)=5\), i.e. \(68\mathrm{e}^{-kt}+18\mathrm{e}^{-0.08t}=5\)
M1
graphing \(y=T(t)-U(t)\) and \(y=5\), or GDC solver
(M1)
Must use technology.
\(46.4\) minutes
A1
\(46.430\ldots\) (46.42 with \(k=0.0582\)).
Answers: (a) 90 °C · (b) \(k=0.0582\) · (c) 37.9 °C · (d) 22.8 min · (e) \(T=22\) · (f) 4 °C · (g) 46.4 min
257AASL2-q00832.10 Solving equationsPaper 2Hard15 marks
The concentration of a drug in a patient’s blood, \(C\) mg L−1, \(t\) hours after a single dose is taken is modelled by \(C(t)=20\left(\mathrm{e}^{-0.2t}-\mathrm{e}^{-0.9t}\right)\), for \(t\ge0\).
(a)
Write down the value of \(C(0)\).
(1)
(b)
Find the maximum concentration and the time at which it occurs.
(3)
(c)
The drug is effective when the concentration is at least 8 mg L−1. Find the length of time for which a single dose is effective.
(3)
(d)
Find the time after which the concentration remains below 0.5 mg L−1.
(2)
(e)(i)
A second, identical dose is taken at \(t=6\). For \(t\ge6\), the concentration is modelled by \(D(t)=C(t)+C(t-6)\).
Describe the transformation that maps the graph of \(y=C(t)\) onto the graph of \(y=C(t-6)\).
(2)
(e)(ii)
Find \(D(6)\).
(1)
(f)
Find the maximum value of \(D(t)\) for \(t\ge6\), and the time at which it occurs.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(0\)
A1
(b)
maximum on the graph of \(C\)
(M1)
\(10.1\) mg L−1
A1
\(10.1216\ldots\)
\(t=2.15\) hours
A1
\(2.14868\ldots\)
(c)
intersections of \(y=C(t)\) and \(y=8\)
(M1)
\(t=0.942\) and \(t=4.33\)
A1
\(0.94233\ldots\), \(4.33498\ldots\)
\(3.39\) hours
A1
\(3.39264\ldots\)
(d)
\(C(t)=0.5\) on the decreasing part of the graph
(M1)
\(t=18.4\) hours
A1
\(18.4443\ldots\)
(e)(i)
translation
A1
6 units in the positive \(t\)-direction, \(\begin{pmatrix}6\\0\end{pmatrix}\)
A1
(e)(ii)
\(D(6)=C(6)+C(0)=5.93\) mg L−1
A1
\(5.93355\ldots\)
(f)
maximum on the graph of \(y=D(t)\) for \(t\ge6\)
(M1)
\(14.2\) mg L−1
A1
\(14.1810\ldots\)
\(t=7.78\) hours
A1
\(7.77900\ldots\)
Answers: (a) 0 · (b) 10.1 mg L−1 at 2.15 h · (c) 3.39 h · (d) 18.4 h · (e)(i) translation by \(\binom{6}{0}\); (ii) 5.93 mg L−1 · (f) 14.2 mg L−1 at 7.78 h
258AASL2-q00922.11 Transformations of graphsPaper 2Hard13 marks
The weekly demand, \(D\) units, for a product sold at a price of \(p\) dollars is modelled by \(D(p)=800\mathrm{e}^{-0.05p}\), for \(0\le p\le60\).
(a)
Find the demand when the price is $10.
(1)
(b)
Find the price at which the weekly demand is 300 units.
(2)
(c)(i)
After an advertising campaign, the graph of the demand function is translated 4 units in the positive \(p\)-direction. The new demand is \(E(p)=D(p-4)\).
Show that \(E(p)=800\mathrm{e}^{0.2}\mathrm{e}^{-0.05p}\).
(1)
(c)(ii)
Hence describe the same change as a single vertical stretch, giving the scale factor to three significant figures.
(2)
(d)
Without solving a further equation, find the price at which the weekly demand after the campaign is 300 units. Justify your answer.
(2)
(e)
The weekly revenue after the campaign is \(R(p)=pE(p)\). Find the price that maximizes the revenue and the maximum revenue.
(3)
(f)
Find the horizontal translation of the original demand graph that would be equivalent to a vertical stretch with scale factor 1.6.
Translation of \(a\) units in the positive \(p\)-direction.
\(a=20\ln1.6=9.40\) units in the positive \(p\)-direction
A1
\(9.4000\ldots\)
Answers: (a) 485 · (b) $19.62 · (c)(ii) vertical stretch, scale factor 1.22 · (d) $23.62 · (e) $20, $7189 · (f) 9.40 to the right
259IBMA2S-046-02-122.10 Solving equationsPaper 2Hard12 marks
Calculator
The effect of a pain-relief drug, \(E\), as a percentage of its maximum possible effect, is modelled by \(E(c)=\frac{100c}{c+k}\), \(c\ge 0\), where \(c\) is the drug concentration in mg L−1 and \(k\) is a positive constant. When \(c=4\), the effect is \(40\%\).
(a)
Show that \(k=6\).
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(E\) and interpret it in context.
(2)
(c)
Find \(E^{-1}(x)\).
(3)
(d)
Write down the range of \(E\).
(1)
(e)
After the drug is given, the concentration \(t\) hours later is modelled by \(c(t)=8e^{-0.2t}\). The patient needs a further dose when the effect falls to \(30\%\). Find the time at which this happens.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms \(\frac{400}{4+k}=40\)
✓ 1
A1 \(4+k=10\), so \(k=6\)
✓ 1
Part (b)
A1 \(E=100\)
✓ 1
R1 as the concentration increases, the effect approaches its maximum but never reaches \(100\%\)
✓ 1
Part (c)
M1 interchanges variables and multiplies out: \(x(y+6)=100y\)
Exam tip: Use the inverse function to turn a condition on the effect into a condition on the concentration. This avoids working with the composite function \(E(c(t))\) directly.
260IBMA2S-046-01-092.7 Quadratic equations and the discriminantPaper 2Hard12 marks
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A footballer kicks a ball from ground level with speed 20 m s−1 at an angle θ above the horizontal. The path of the ball is modelled by \(y = xt - 0.01225(1 + t^2)x^2\), where \(t = \tan\theta\), x metres is the horizontal distance from the kick and y metres is the height of the ball. The player wants the ball to pass through a point 30 m away horizontally and 2.4 m above the ground.
(a)
Show that t must satisfy \(11.025t^2 - 30t + 13.425 = 0\).
(3)
(b)
Hence find the two possible values of θ.
(4)
(c)
The player now wants the ball to pass at a height of 2.4 m at a horizontal distance of x metres, for some angle θ. By considering a quadratic equation in t, find the greatest value of x for which this is possible.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substituting \(x = 30\) and \(y = 2.4\)
✓ 1
A1 \(2.4 = 30t - 11.025(1 + t^2)\)
✓ 1
A1 rearranging to \(11.025t^2 - 30t + 13.425 = 0\)
✓ 1
Part (b)
M1 attempt to solve the quadratic, by formula or GDC
✓ 1
A1 \(t = 0.565\) and \(t = 2.16\)
✓ 1
M1 use of \(\theta = \arctan t\)
✓ 1
A1 \(\theta = 29.5^\circ\) and \(\theta = 65.1^\circ\)
A1 dividing by \(x^2\): \(0.01225x^2 + 2.4 \leq \frac{1}{0.049}\), so \(x^2 \leq 1470\)
✓ 1
A1 greatest \(x = 38.3\) m
✓ 1
Exam tip: When a question asks when something is possible for a quadratic in a parameter, the condition is almost always on the discriminant. Treat t as the variable and x as a constant.
261IBMA2S-046-03-122.5 Composite and inverse functionsPaper 2Hard13 marks
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Mountaineers boil water to cook food. Atmospheric pressure, in kPa, at an altitude of \(h\) km is modelled by \(P(h)=101.3e^{-0.12h}\), for \(0\le h\le 9\). The boiling point of water, in °C, at a pressure of \(P\) kPa is modelled by \(B(P)=100+27.6\ln\left(\dfrac{P}{101.3}\right)\).
(a)
Find an expression for \(P^{-1}(x)\).
(3)
(b)
Show that \((B\circ P)(h)=100-3.312h\).
(3)
(c)
Find the range of \(B\circ P\).
(2)
(d)
Find \((B\circ P)^{-1}(x)\) and state its domain.
(3)
(e)
A dehydrated meal needs water boiling at 88 °C or above. Find the greatest altitude at which the meal can be prepared.
M1 evaluates \((B\circ P)(0)=100\) and \((B\circ P)(9)=70.192\)
✓ 1
A1 \(70.192\le(B\circ P)(h)\le 100\)
✓ 1
Part (d)
M1 rearranges \(x=100-3.312h\) to make \(h\) the subject
✓ 1
A1 \((B\circ P)^{-1}(x)=\frac{100-x}{3.312}\)
✓ 1
A1 domain \(70.192\le x\le 100\) (follow through from their range in (c))
✓ 1
Part (e)
M1 evaluates \((B\circ P)^{-1}(88)\), or solves \(100-3.312h\ge 88\)
✓ 1
A1 \(h=3.62\) km
✓ 1
Exam tip: The domain of an inverse function is the range of the original function. State endpoints exactly rather than rounding them, because a rounded endpoint can cut off part of the true range.
262IBMA2S-046-04-122.8 Reciprocal and rational functionsPaper 2Hard12 marks
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A water-treatment tank initially holds 200 litres of water containing 10 kg of dissolved salt. Brine containing 0.3 kg of salt per litre is pumped in at 5 litres per minute, and nothing leaves the tank. The salt concentration, in kg L−1, after \(t\) minutes is \(c(t)=\frac{10+1.5t}{200+5t}\).
(a)
Find the initial concentration.
(1)
(b)
Show that \(c(t)=0.3-\frac{50}{200+5t}\).
(2)
(c)
Hence write down the equation of the horizontal asymptote of the graph of \(c\), interpret it in context, and explain why \(c\) is increasing for \(t\ge0\).
(3)
(d)
Find the time at which the concentration reaches 0.2 kg L−1.
(2)
(e)
The tank has a capacity of 1000 litres and pumping stops when it is full. Find the domain of \(c\) for this model and hence find the range of \(c\).
(4)
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Notes
Part (a)
A1 \(c(0)=0.05\) kg L−1
✓ 1
Part (b)
M1 writes \(10+1.5t=0.3(200+5t)-50\) or combines the right-hand side over a common denominator
✓ 1
A1 correct working leading to the given expression
✓ 1
Part (c)
A1 \(c=0.3\)
✓ 1
R1 the concentration approaches the concentration of the incoming brine, 0.3 kg L−1, but never reaches it
✓ 1
R1 as \(t\) increases, \(200+5t\) increases so \(\frac{50}{200+5t}\) decreases, hence \(c\) increases
✓ 1
Part (d)
M1 sets \(\frac{50}{200+5t}=0.1\) or equivalent
✓ 1
A1 \(t=60\) minutes
✓ 1
Part (e)
A1 \(200+5t\le1000\) gives \(0\le t\le160\)
✓ 1
M1 evaluates \(c(160)=\frac{10+240}{1000}\)
✓ 1
A1 \(c(160)=0.25\)
✓ 1
A1 range \(0.05\le c\le0.25\)
✓ 1
Exam tip: For an increasing function on a closed interval, the range runs from the value at the left endpoint to the value at the right endpoint, not up to the asymptote.
263IBMA2S-046-07-112.1 Equations of straight linesPaper 2Hard14 marks
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A radar station is at the origin O of a coordinate grid measured in kilometres. Aircraft A flies in a straight line through the points (−40, 10) and (20, 40). Aircraft B flies in a straight line through the point (50, 5), perpendicular to the path of aircraft A.
Diagram not to scale
(a)
Find the equation of the path of aircraft A in the form \(y=mx+c\).
(2)
(b)
Find the equation of the path of aircraft B.
(2)
(c)
Find the coordinates of the point where the two paths cross.
(2)
(d)
The point F on the path of aircraft A is the point closest to O. By considering the line through O perpendicular to the path of A, find the coordinates of F.
(3)
(e)
Hence find the shortest distance from the radar station to the path of aircraft A.
(2)
(f)
The radar detects aircraft within 35 km of O. Find the length of the part of aircraft A's path that lies within detection range.
R1 the section in range is symmetric about F, since OF is perpendicular to the path
✓ 1
M1 Pythagoras: half-length \(=\sqrt{35^2-720}\)
✓ 1
A1 length \(=44.9\) km
✓ 1
Exam tip: The shortest distance from a point to a line is always along the perpendicular. Once you have the foot F, a right-angled triangle gives the rest.
264IBMA2S-046-06-112.10 Solving equationsPaper 2Hard12 marks
Calculator
After a single dose, the concentration of a drug in the blood of patient A is modelled by \(C(t) = 12te^{-0.5t}\), \(t \ge 0\), where \(C\) is in mg L−1 and \(t\) is in hours. The drug is effective when the concentration is at least 6 mg L−1.
(a)
Find the maximum concentration and the time at which it occurs.
(2)
(b)
Find the length of time for which the drug is effective for patient A.
(3)
(c)
Patient B's concentration is modelled by \(D(t) = 1.5\,C(t-3)\), \(t \ge 3\). Describe the two transformations that map the graph of \(y = C(t)\) onto the graph of \(y = D(t)\), and interpret them in context.
(3)
(d)
Without further solving, state the length of time for which \(D(t) \ge 9\), justifying your answer.
(2)
(e)
Write down the coordinates of the maximum point of the graph of \(y = D(t)\).
(2)
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Notes
Part (a)
A1 \(t = 2\) hours
✓ 1
A1 maximum concentration \(8.83\) mg L−1
✓ 1
Part (b)
M1 attempts to solve \(12te^{-0.5t} = 6\), e.g. by finding the intersection on a graph
✓ 1
A1 \(t = 0.7145\) and \(t = 4.3066\)
✓ 1
A1 \(3.59\) hours, found from the unrounded values
✓ 1
Part (c)
A1 horizontal translation of 3 units to the right
✓ 1
A1 vertical stretch with scale factor 1.5
✓ 1
R1 patient B receives the dose 3 hours later and their concentrations are 1.5 times as large
✓ 1
Part (d)
R1 \(1.5\,C(t-3) \ge 9 \iff C(t-3) \ge 6\), and a horizontal translation does not change the length of an interval
✓ 1
A1 \(3.59\) hours
✓ 1
Part (e)
A1 \(t = 5\)
✓ 1
A1 \(D = 13.2\) mg L−1, giving \((5,\ 13.2)\)
✓ 1
Exam tip: Look for ways to reuse earlier answers. A horizontal translation shifts every time by the same amount, so durations stay the same. A vertical stretch multiplies every concentration by the same factor. Keep unrounded values in your calculator when subtracting.
265IBMA2S-046-05-122.9 Exponential and logarithmic functionsPaper 2Hard10 marks
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Two fitness apps are launched at the same time. The numbers of monthly active users, in thousands, t months after launch are modelled by A(t) = 500(1.35)t and B(t) = 2000(1.12)t.
(a)
The model for A can be written as A(t) = 500ekt. Find the value of k.
(2)
(b)
Find the time taken for the number of users of app A to double.
(2)
(c)
Show that A(t) = B(t) when t = ln 4 / ln(135/112), and hence find this value of t.
(4)
(d)
Find the number of monthly active users of each app at this time.
M1 500(1.35)t = 2000(1.12)t ; M1 (1.35/1.12)t = 4 ; A1 taking logarithms: t ln(135/112) = ln 4, giving the result ; A1 t = 7.42 months
✓ 4
Part (d)
M1 substituting the unrounded value of t into A(t) or B(t) ; A1 4640 thousand (about 4.64 million) users
✓ 2
Common error: Do not subtract the bases: 1.35t ÷ 1.12t equals (1.35/1.12)t, not 0.23t. Divide both sides first, then take logarithms.
266IBMA2S-046-00-112.6 Quadratic functionsPaper 2Hard15 marks
Calculator
The arch of a footbridge over a canal is modelled by \(y=-0.04x^2+1.6x\), where \(x\) is the horizontal distance in metres from one base of the arch and \(y\) is the height in metres above the canal path.
(a)
Find the span of the arch at path level and its maximum height.
(3)
(b)
A horizontal walkway is fixed inside the arch at a height of 12 m. Find the length of the walkway.
(3)
(c)
A straight maintenance ramp follows the line \(y=0.4x+c\) and touches the arch at exactly one point. Find \(c\).
(4)
(d)
Find the coordinates of the point where the ramp touches the arch.
(2)
(e)
A maintenance vehicle 8 m wide drives along the canal path, centred under the highest point of the arch. Find the maximum height the vehicle can have if it is to pass under the arch.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factorises \(y=-0.04x(x-40)\) or solves \(y=0\)
✓ 1
A1 span \(=40\) m
✓ 1
A1 maximum height \(16\) m (at \(x=20\))
✓ 1
Part (b)
M1 sets \(-0.04x^2+1.6x=12\)
✓ 1
A1 \(x=10\), \(x=30\)
✓ 1
A1 length \(=20\) m
✓ 1
Part (c)
M1 equates \(-0.04x^2+1.6x=0.4x+c\)
✓ 1
A1 \(0.04x^2-1.2x+c=0\)
✓ 1
M1 sets discriminant to zero, \(1.44-0.16c=0\) (or uses gradient \(-0.08x+1.6=0.4\))
✓ 1
A1 \(c=9\)
✓ 1
Part (d)
M1 \(x=\frac{1.2}{2(0.04)}=15\) (or from \(-0.08x+1.6=0.4\))
✓ 1
A1 \(y=0.4(15)+9=15\), giving \((15,\,15)\)
✓ 1
Part (e)
M1 edges of vehicle at \(x=16\) and \(x=24\)
✓ 1
M1 substitutes, e.g. \(-0.04(16)^2+1.6(16)\)
✓ 1
A1 \(15.36\) m (accept \(15.3\) m rounded down; do not accept \(15.4\) m, which would not fit)
✓ 1
Exam tip: The limiting height for a wide vehicle is at its edges, not its centre, because the arch is lowest at the points furthest from the axis of symmetry. For a clearance limit, give the exact value or round down, never up.
267IBMA3S-2-17r11-132.7 Quadratic equations and the discriminantPaper 2Hard10 marks
Calculator
A thruster-correction curve for a space probe is modelled by \(y=x^2-2kx+3k+4\), where \(k\) is a constant.
(a)
Find the discriminant of \(x^2-2kx+3k+4=0\) in terms of \(k\).
(2)
(b)
Find the values of \(k\) for which the curve touches the \(x\)-axis.
(3)
(c)
Find the range of values of \(k\) for which the curve lies entirely above the \(x\)-axis.
(2)
(d)
The minimum value of \(y\) is 6. Find the largest possible value of \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(b^2-4ac\) with \(b=-2k\), \(c=3k+4\)
✓ 1
A1 \(\Delta=4k^2-12k-16\)
✓ 1
Part (b)
M1 \(4k^2-12k-16=0\) so \(k^2-3k-4=0\)
✓ 1
M1 factorise: \((k-4)(k+1)=0\)
✓ 1
A1 \(k=4\) or \(k=-1\)
✓ 1
Part (c)
M1 \(\Delta<0\)
✓ 1
A1 \(-1<k<4\)
✓ 1
Part (d)
M1 complete the square or use vertex \(x=k\): \(y=(x-k)^2-k^2+3k+4\)
✓ 1
A1 minimum value \(-k^2+3k+4\)
✓ 1
A1 \(-k^2+3k+4=6\) gives \(k^2-3k+2=0\), so \(k=1\) or \(k=2\); the largest is \(k=2\)
✓ 1
Exam tip: The vertex form \((x-k)^2+c\) shows the minimum value of \(y\) immediately without needing the discriminant.
268IBMA3S-2-17r11-142.10 Solving equationsPaper 2Hard12 marks
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The strength of a signal received from a space probe is modelled by \(S(x)=\frac{20x+30}{x+5}\) for \(x\geq 0\), where \(x\) is the time in hours since the probe came into view. All times are to be given exactly or to 3 significant figures.
(a)
Find \(S(0)\) and write down the equation of the horizontal asymptote of the graph of \(S\).
(2)
(b)
Find \(S^{-1}(x)\), and state its domain.
(3)
(c)
A relay converts strength using \(g(s)=\frac{s}{2}+1\). Show that \((g\circ S)(x)=\frac{11x+20}{x+5}\).
(3)
(d)
Find the range of \(g\circ S\) for \(x\geq 0\).
(2)
(e)
Use your answer to (b) to find the time at which the signal strength is 15.
Common error: A reciprocal-type function increasing towards an asymptote has a range that excludes the asymptote value; check the end values separately.
269IBMA3S-2-16r17-122.2 Functions, domain, range and inversePaper 2Hard13 marks
Calculator
After landing, the temperature \(T\) degrees Celsius of an aircraft brake disc, \(t\) minutes after it stops, is modelled by \(T(t)=20+380e^{-0.05t}\), for \(t\ge0\).
(a)
Write down the temperature of the disc when it stops.
(1)
(b)
Find the temperature of the disc after 20 minutes. Give your answer to 1 decimal place.
(2)
(c)
Write down the equation of the horizontal asymptote of the graph of \(T\), and interpret it in this context.
(2)
(d)
Write down the range of \(T\).
(1)
(e)
Show that the inverse function is \(T^{-1}(x)=20\ln\left(\frac{380}{x-20}\right)\), and write down its domain.
(4)
(f)
Find the time at which the temperature of the disc is \(100\,{}^{\circ}\)C. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
A1 \(T(0)=20+380=400\,{}^{\circ}\)C
✓ 1
Part (b)
M1 \(T(20)=20+380e^{-0.05\times20}\)
✓ 1
A1 \(159.8\,{}^{\circ}\)C
✓ 1
Part (c)
A1 \(T=20\)
✓ 1
R1 the disc cools towards the surrounding temperature of \(20\,{}^{\circ}\)C but never falls below it
✓ 1
Part (d)
A1 \(20<T\le400\)
✓ 1
Part (e)
M1 \(x=20+380e^{-0.05y}\) after swapping, so \(e^{-0.05y}=\frac{x-20}{380}\)
M1 \(T^{-1}(100)=20\ln\frac{380}{100-20}\), or solves \(20+380e^{-0.05t}=100\)
✓ 1
M1 \(\ln\frac{380}{80}=\ln4.75\)
✓ 1
A1 \(t=31.2\) minutes
✓ 1
Exam tip: The range of an exponential model with a vertical shift is bounded by the horizontal asymptote; the domain of the inverse equals this range.
270IBMA3S-2-16r17-142.5 Composite and inverse functionsPaper 2Hard10 marks
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The function \(f\) is defined by \(f(x)=x^2-6x+7\), for \(x\ge3\).
(a)
Write \(f(x)\) in the form \((x-h)^2+k\).
(2)
(b)
Write down the range of \(f\).
(1)
(c)
Find \(f^{-1}(x)\).
(3)
(d)
Write down the domain of \(f^{-1}\).
(1)
(e)
Solve \(f(x)=f^{-1}(x)\). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square, \((x-3)^2-9+7\)
✓ 1
A1 \((x-3)^2-2\)
✓ 1
Part (b)
A1 \(f(x)\ge-2\)
✓ 1
Part (c)
M1 \(x=(y-3)^2-2\) after swapping, so \((y-3)^2=x+2\)
✓ 1
M1 \(y=3\pm\sqrt{x+2}\)
✓ 1
A1 \(f^{-1}(x)=3+\sqrt{x+2}\), taking the positive root as \(f^{-1}(x)\ge3\)
✓ 1
Part (d)
A1 \(x\ge-2\)
✓ 1
Part (e)
R1 since \(f\) is increasing, the graphs of \(f\) and \(f^{-1}\) meet on the line \(y=x\), so \(f(x)=x\)
✓ 1
M1 \(x^2-7x+7=0\), so \(x=\frac{7\pm\sqrt{21}}{2}\)
✓ 1
A1 \(x=5.79\) (reject \(x=1.21\) since \(x\ge3\))
✓ 1
Common error: The restricted domain \(x\ge3\) decides which root to keep when finding the inverse and when solving \(f(x)=x\).
271IBMA3S-2-17r17-142.10 Solving equationsPaper 2Hard10 marks
Calculator
Two drugs are given to a patient at the same time. The concentration of drug A in the blood is CA(t) = 30e−0.2t mg L−1 and the concentration of drug B is CB(t) = 12e−0.05t mg L−1, where t is the time in hours after they are given.
(a)
Find the difference between the two concentrations at t = 0.
(1)
(b)
Find the time at which the two concentrations are equal. Give your answer to 3 significant figures.
(4)
(c)
Find the common concentration at the time found in (b). Use the unrounded value of t and give your answer to 3 significant figures.
(2)
(d)
Drug A stops being effective when its concentration falls below 1 mg L−1. Find the time at which this happens, to 3 significant figures.
(3)
Show mark scheme
Marking point
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Notes
Part (a)
A1 30 − 12 = 18 mg L−1
✓ 1
Part (b)
M1 equates: 30e−0.2t = 12e−0.05t
✓ 1
M1 rearranges: e−0.15t = 0.4 (or e0.15t = 2.5)
✓ 1
M1 takes natural logarithms: t = ln 2.5 / 0.15
✓ 1
A1 t = 6.11 hours
✓ 1
Part (c)
M1 substitutes t = 6.1086… into either function
✓ 1
A1 8.84 mg L−1
✓ 1
Part (d)
M1 solves 30e−0.2t = 1, i.e. e−0.2t = 1/30
✓ 1
M1 takes natural logarithms: t = ln 30 / 0.2
✓ 1
A1 t = 17.0 hours
✓ 1
Common error: When equating two exponentials with different rates, divide to collect the powers of e into a single term before taking logarithms; do not take logarithms of each side term by term.
272IBMA3S-2-17r26-132.1 Equations of straight linesPaper 2Hard13 marks
Calculator
A hillside path is modelled by the curve C with equation y = x2 − 4x + 1. A straight boundary fence lies along the line L with equation y = 2x − 8. Both x and y are in metres.
(a)
Find the coordinates of the lowest point of C.
(2)
(b)
Show that L meets C at exactly one point, and find the coordinates of that point.
(4)
(c)
The line M passes through (3, −2) and is perpendicular to L. Find the equation of M in the form y = mx + c.
(2)
(d)
M meets C at (3, −2) and at a second point Q. Find the coordinates of Q.
(3)
(e)
Find the distance between (3, −2) and Q, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 x = 2
✓ 1
A1 y = −3, so (2, −3)
✓ 1
Part (b)
M1 equate x2 − 4x + 1 = 2x − 8
✓ 1
A1 x2 − 6x + 9 = 0
✓ 1
R1 (x − 3)2 = 0 (or discriminant = 0), so one repeated solution
✓ 1
A1 point (3, −2)
✓ 1
Part (c)
A1 gradient of M = −1⁄2
✓ 1
A1 y = −0.5x − 0.5
✓ 1
Part (d)
M1 equate −0.5x − 0.5 = x2 − 4x + 1
✓ 1
A1 2x2 − 7x + 3 = 0, so x = 0.5 or 3
✓ 1
A1 Q = (0.5, −0.75)
✓ 1
Part (e)
M1 √((3 − 0.5)2 + (−2 + 0.75)2)
✓ 1
A1 2.80 m
✓ 1
Exam tip: If a line touches a curve, the equation formed by equating them has a repeated root, so its discriminant is zero.
273IBMA3S-2-17r26-142.3 Graphs of functionsPaper 2Hard13 marks
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The net energy balance of a wind farm, f(x) MWh, is recorded x hours after 06:00 (negative when the farm draws more from the grid than it supplies). It is modelled by a function f for 0 ≤ x ≤ 4. The values are: x: 0, 1, 2, 3, 4 f(x): 3, −1, −2, 1, 6 The graph of y = f(x) has its minimum point at (2, −2). Ignore domain restrictions for the transformed functions below.
(a)
The function g is given by g(x) = f(x − 2) + 4. Find g(5) and g(2).
(2)
(b)
The function k is given by k(x) = f(x⁄2). Find k(6) and k(4).
(2)
(c)
Write down the coordinates of the maximum point of the graph of y = −2f(x).
(2)
(d)
Write down the coordinates of the minimum point of the graph of y = 3f(x + 1) − 1.
(2)
(e)
The graph of y = af(x − b) + c, where a > 0, has its minimum point at (6, −10) and passes through the point (7, 8). Find the values of a, b and c.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 g(5) = f(3) + 4 = 5
✓ 1
A1 g(2) = f(0) + 4 = 7
✓ 1
Part (b)
A1 k(6) = f(3) = 1
✓ 1
A1 k(4) = f(2) = −2
✓ 1
Part (c)
A1 x-coordinate 2
✓ 1
A1 y-coordinate 4, so (2, 4)
✓ 1
Part (d)
A1 x-coordinate 1
✓ 1
A1 y-coordinate 3 × (−2) − 1 = −7, so (1, −7)
✓ 1
Part (e)
A1 minimum moves from (2, −2) to (6, −10), so b = 4
✓ 1
M1 −2a + c = −10
✓ 1
M1 point (7, 8) corresponds to f(3) = 1, so a + c = 8
✓ 1
A1 a = 6
✓ 1
A1 c = 2
✓ 1
Common error: A minimum point of f maps to a minimum of af when a is positive, but to a maximum when a is negative.
274IBMA3S-2-19r28-142.9 Exponential and logarithmic functionsPaper 2Hard8 marks
Calculator
The installed capacity, K gigawatts (GW), of a region's wind farms is modelled by \(K(t)=Ae^{kt}\), where t is the number of years since monitoring began and A and k are constants. The capacity was 60 GW when \(t=3\) and 108 GW when \(t=8\).
(a)
Find the value of k. Give your answer to 3 significant figures. [3]
(3)
(b)
Find the value of A. Give your answer to 3 significant figures. Use the unrounded value of k. [2]
(2)
(c)
Find the value of t at which the capacity first reaches 200 GW. Give your answer to 3 significant figures. Use the unrounded values of k and A. [3]
(3)
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Mark
Notes
Part (a)
M1 \(Ae^{3k}=60\) and \(Ae^{8k}=108\), divide to get \(e^{5k}=1.8\)
✓ 1
M1 \(k=\frac{\ln 1.8}{5}\)
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A1 \(k=0.118\)
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Part (b)
M1 \(A=60e^{-3k}\)
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A1 \(A=42.2\)
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Part (c)
M1 \(200=Ae^{kt}\)
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M1 \(t=\frac{1}{k}\ln\frac{200}{A}\)
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A1 \(t=13.2\) years
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Exam tip: Dividing the two equations eliminates \(A\) and leaves a single exponential equation. Store \(k\) in your calculator so later parts use its unrounded value.
275IBMA3S-2-17r39-072.1 Equations of straight linesPaper 2Hard10 marks
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A lava bomb is thrown from a vent on a volcanic island. Its height above sea level is modelled by \(h(t)=15+39.2t-4.9t^2\) metres, where \(t\) is the time in seconds after launch, until the bomb lands in the sea.
(a)
Find the time at which the bomb is at its greatest height, and find this greatest height.
(3)
(b)
Find the time at which the bomb lands in the sea, giving your answer to 3 significant figures.
(2)
(c)
Write down the domain and the range of \(h\) in this model.
(2)
(d)
Find, to 3 significant figures, the length of time for which the bomb is higher than 80 m. Use unrounded values in your working.
(3)
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Part (a)
M1 use axis of symmetry \(t=-\frac{39.2}{2(-4.9)}\) or complete the square
✓ 1
A1 \(t=4\) s
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A1 greatest height \(=93.4\) m
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Part (b)
M1 solve \(h(t)=0\) with \(t>0\)
✓ 1
A1 \(t=8.37\) s
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Part (c)
A1 domain \(0\le t\le 8.37\) (accept the unrounded value 8.3659...)
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A1 range \(0\le h\le 93.4\)
✓ 1
Part (d)
M1 solve \(h(t)=80\), i.e. \(4.9t^2-39.2t+65=0\)
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A1 \(t=2.35\) and \(t=5.65\) (2.3463... and 5.6537...)
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A1 duration \(=3.31\) s
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Exam tip: Domain and range must come from the physical situation: the model only applies from launch until landing, so the range ends at sea level, not at the vertex of a full parabola.
276IBMA3S-2-17r39-132.7 Quadratic equations and the discriminantPaper 2Hard11 marks
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A theatre spotlight at the point \(S(0,-1)\) shines a straight beam along the line \(y=mx-1\), where lengths are in metres. The top edge of a curved backdrop is modelled by the curve \(y=x^2+3\).
Diagram not to scale
(a)
Show that the \(x\)-coordinates of any points where the beam meets the backdrop satisfy \(x^2-mx+4=0\).
(2)
(b)
Two beams just touch the backdrop, each at a single point. Find the two values of \(m\).
(3)
(c)
Find the coordinates of the two points where these beams touch the backdrop.
(2)
(d)
Find the set of values of \(m\) for which the beam meets the backdrop at two distinct points.
(2)
(e)
Find the angle between the two beams that just touch the backdrop, giving your answer in degrees to 3 significant figures.
M1 half-angle from the vertical \(=\tan^{-1}\frac{2}{8}\) (using \(S\) and a touching point)
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A1 \(28.1^\circ\)
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Exam tip: A line touches a curve exactly when the equation formed by equating them has a repeated root, so set the discriminant equal to zero.
277IBMA3S-2-19r40-082.10 Solving equationsPaper 2Hard10 marks
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Sunlight is absorbed as it passes down through the water of a marine reserve. The intensity of light at a depth of d metres is modelled by \(I(d)=I_0e^{-kd}\), where \(I_0\) is the intensity at the surface and \(k>0\) is a constant. At a depth of 6 m the intensity is 45% of the surface intensity.
(a)
Show that \(k=\frac{1}{6}\ln\frac{20}{9}\).
(3)
(b)
Photosynthesis stops where the intensity falls below 5% of the surface intensity. Find the depth at which the intensity is exactly 5% of the surface intensity. Use the unrounded value of \(k\) and give your answer in metres to 3 significant figures.
(3)
(c)
Show that \(I(d)=I_0\left(\frac{9}{20}\right)^{d/6}\).
(2)
(d)
Find the percentage by which the intensity decreases with each additional metre of depth. Give your answer to 3 significant figures.
(2)
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Part (a)
M1 \(I_0e^{-6k}=0.45I_0\) so \(e^{-6k}=0.45\)
✓ 1
M1 takes logarithms, \(-6k=\ln 0.45\) or \(6k=\ln\frac{1}{0.45}\)
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A1 \(k=\frac{1}{6}\ln\frac{20}{9}\) (AG, reaching this from \(\frac{1}{0.45}=\frac{20}{9}\))
✓ 1
Part (b)
M1 \(e^{-kd}=0.05\)
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M1 \(d=\frac{\ln 20}{k}\)
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A1 \(d\) = 22.5 m (22.51…)
✓ 1
Part (c)
M1 \(e^{-kd}=\left(e^{-6k}\right)^{d/6}\) or \(e^{-\frac{d}{6}\ln\frac{20}{9}}\)
✓ 1
A1 \(=\left(\frac{9}{20}\right)^{d/6}\) (AG)
✓ 1
Part (d)
M1 compares \(I(d+1)\) with \(I(d)\), giving a ratio of \(\left(\frac{9}{20}\right)^{1/6}\) = 0.875
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A1 percentage decrease = 12.5%
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Common error: A 12.5% decrease per metre is not an intensity of 87.5% after 6 m; the factor must be applied once per metre, so it is raised to a power.
278IBMA3S-2-19r40-132.10 Solving equationsPaper 2Hard9 marks
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Two samples of seawater are heated to 30 °C and then left to cool in a laboratory. The temperature, in °C, of sample A after t minutes is \(T_A=12+18e^{-0.08t}\) and of sample B is \(T_B=10+20e^{-0.06t}\).
(a)
Write down the temperature that sample A approaches in the long term.
(1)
(b)
Find the time at which sample A reaches 15 °C. Give your answer in minutes to 3 significant figures.
(3)
(c)
For \(t>0\), find the time at which the two samples are at the same temperature. Give your answer in minutes to 3 significant figures.
(2)
(d)
Find the common temperature at that time. Use the unrounded value of \(t\) and give your answer to 3 significant figures.
(2)
(e)
Explain why sample A is warmer than sample B for all large values of t.
(1)
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Part (a)
A1 12 °C
✓ 1
Part (b)
M1 \(18e^{-0.08t}=3\) so \(e^{-0.08t}=\frac{1}{6}\)
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M1 \(-0.08t=\ln\frac{1}{6}\)
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A1 \(t\) = 22.4 minutes
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Part (c)
M1 equates \(T_A=T_B\), or sketches both graphs and finds the intersection with \(t>0\)
✓ 1
A1 \(t\) = 25.4 minutes
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Part (d)
M1 substitutes the unrounded \(t\) (25.365…) into \(T_A\) or \(T_B\)
✓ 1
A1 14.4 °C
✓ 1
Part (e)
R1 as \(t\to\infty\) sample A approaches 12 °C while sample B approaches 10 °C (so A is warmer once the exponential terms are negligible)
✓ 1
Exam tip: When both graphs start at the same point, the intersection at \(t=0\) is not the one wanted. Use the stated condition \(t>0\) to select the correct crossing.
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