Edexcel International A Level Maths · P1 · Pure Mathematics 1
Integration
28 free, high-quality past-paper-style questions on Integration for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 236 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P1-7-B-01P1 · ASEasy5 marks
Find each of the following integrals, simplifying each answer.
(a)
∫(10x4 + 6x2 − 7) dx
(3)
(b)
∫x(3x − 2) dx
(2)
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Answer
Mark
Notes
Part (a)
raises the power of at least one term by one
M1
two of 2x5, 2x3, −7x correct
A1
2x5 + 2x3 − 7x + c, all correct including + c
A1
Part (b)
multiplies out first: 3x2 − 2x
M1
x3 − x2 + c
A1
Common error: Integrating x and (3x − 2) separately and multiplying the results. Expand the bracket first; there is no product rule for integrals at this level.
Figure 1 shows the graph of y = f′(x). The graph is a straight line through the points A and B, whose coordinates are shown. The curve with equation y = f(x) passes through the point (2, 1).
(a)
Use Figure 1 to find an expression for f′(x).
(2)
(b)
Find f(x).
(3)
(c)
Use the point A in Figure 1 to find the equation of the tangent to the curve y = f(x) at the point where x = 0.
(2)
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Answer
Mark
Notes
Part (a)
gradient of the line = (0 − (−6))/(3 − 0) = 2
M1
f′(x) = 2x − 6
A1
Part (b)
integrates, raising each power by one: x2 − 6x + c
M1
substitutes (2, 1): 4 − 12 + c = 1
M1
f(x) = x2 − 6x + 9 ft from their f′(x)
A1
Part (c)
gradient f′(0) = −6 read from A, and f(0) = 9, so the point of contact is (0, 9)
M1
y = 9 − 6x ft from their f(x)
A1
Exam tip: A graph of y = f′(x) shows gradients, not heights. The point A(0, −6) tells you that the curve y = f(x) has gradient −6 where x = 0; it does not tell you that the curve passes through (0, −6).
Figure 1 is a flow diagram that starts from the expression g(x) in the top box. P is obtained by differentiating g(x) with respect to x, and Q is obtained by integrating g(x) with respect to x. R follows from Q as described in the diagram.
(a)
Find the expression P.
(2)
(b)
Find the expression Q.
(3)
(c)
Find the expression R.
(2)
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Answer
Mark
Notes
Part (a)
lowers the power of at least one term by one
M1
P = 12x − 8
A1
Part (b)
raises the power of at least one term by one
M1
two of 2x3, −4x2, x correct
A1
Q = 2x3 − 4x2 + x + c, including + c
A1
Part (c)
substitutes x = 2, y = 5 into y = Q: 16 − 16 + 2 + c = 5
M1
R = 2x3 − 4x2 + x + 3 ft from their Q
A1
Common error: Mixing up the two directions. Differentiating multiplies by the power and then lowers it by one; integrating raises the power by one and then divides by the new power. Differentiate Q to check: you should get back g(x).
Figure 1 shows part of the curve with equation y = f(x), x > 0. The curve passes through the point A, whose coordinates are shown, and through the point B, where x = 1. Given that f′(x) = 6√x + 8/x2,
(a)
write f′(x) in the form 6xm + 8xn, stating the values of m and n.
(1)
(b)
find ∫(6√x + 8/x2) dx, simplifying each term,
(3)
(c)
use the point A to find f(x),
(2)
(d)
find the y-coordinate of B.
(1)
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Answer
Mark
Notes
Part (a)
m = ½ and n = −2
B1
Part (b)
raises the power of at least one term by one
M1
4x3/2
A1
4x3/2 − 8x−1 + c, all correct including + c
A1
Part (c)
substitutes (4, 35) with 43/2 = 8: 32 − 2 + c = 35
M1
f(x) = 4x3/2 − 8/x + 5
A1
Part (d)
4 − 8 + 5 = 1 ft from their f(x)
B1
Exam tip: For a negative power the new power is still found by adding one: x−2 becomes x−1, and you divide by −1. So 8x−2 integrates to −8x−1; the sign change is easy to miss.
Figure 1 shows the curve C with equation y = f(x) and the straight line ℓ. The line ℓ is the tangent to C at the point P and crosses the y-axis at the point T. The coordinates of P and T are shown in Figure 1. Given that f′(x) = 3x2 − kx, where k is a constant,
(a)
find the gradient of ℓ,
(1)
(b)
show that k = 4,
(2)
(c)
find f(x).
(3)
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Answer
Mark
Notes
Part (a)
(3 − (−5))/(2 − 0) = 4
B1
Part (b)
the gradient of C at P equals the gradient of ℓ: f′(2) = 12 − 2k = 4
M1
2k = 8, so k = 4
A1
Part (c)
integrates 3x2 − 4x, raising each power by one: x3 − 2x2 + c
M1
substitutes P(2, 3): 8 − 8 + c = 3
M1
f(x) = x3 − 2x2 + 3
A1
Exam tip: The tangent at P has the same gradient as the curve at P, so the gradient of ℓ read from the figure is the value of f′(2). The coordinates of P then fix the constant of integration.
Water flows into a tank. The volume of water in the tank is V litres at time t minutes after a valve is opened. The rate of flow is modelled by dV/dt = 12√t + 3, t ≥ 0. When the valve is opened the tank already contains 50 litres of water.
(a)
Find V in terms of t.
(4)
(b)
Find the volume of water in the tank 9 minutes after the valve is opened.
(1)
(c)
Find the time at which water is flowing into the tank at 27 litres per minute.
(2)
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Answer
Mark
Notes
Part (a)
raises the power of √t = t½ by one
M1
8t3/2 + 3t
A1
uses V = 50 when t = 0 to find the constant
M1
V = 8t3/2 + 3t + 50
A1
Part (b)
8 × 27 + 27 + 50 = 293 litres ft from their V
B1
Part (c)
12√t + 3 = 27, so √t = 2
M1
t = 4 minutes
A1
Common error: Part (c) asks about the rate, so it uses dV/dt, not V. Read which quantity a question gives you before choosing the equation.
The curve y = f(x), x > 0, has f′(x) = k√x − 6, where k is a constant. The gradient of the curve is zero at the point where x = 4, and the curve passes through the point (1, 3).
(a)
Find the value of k.
(2)
(b)
Find f(x).
(4)
(c)
Find the coordinates of the point on the curve where the gradient is zero.
(1)
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Answer
Mark
Notes
Part (a)
f′(4) = 0: 2k − 6 = 0
M1
k = 3
A1
Part (b)
integrates 3x½ − 6, raising a power by one
M1
2x3/2 − 6x + c
A1
substitutes (1, 3): 2 − 6 + c = 3
M1
f(x) = 2x3/2 − 6x + 7
A1
Part (c)
f(4) = 16 − 24 + 7 = −1, so (4, −1) ft from their f
B1
Exam tip: “The gradient is zero where x = 4” is information about the gradient: it gives f′(4) = 0. The point on the curve gives information about f(x).
The gradient of a curve is given by dy/dx = 4x3 + kx, where k is a constant. The curve passes through the points (1, 3) and (2, 6).
(a)
Find y in terms of x, k and a constant of integration c.
(2)
(b)
Find the value of k and the value of c.
(4)
(c)
The gradient of the curve is zero at three points, one of which has x = 0. Find the exact x-coordinates of the other two points and show that they have the same y-coordinate.
(2)
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Answer
Mark
Notes
Part (a)
raises the power of at least one term by one
M1
y = x4 + ½kx2 + c
A1
Part (b)
(1, 3): 1 + ½k + c = 3
M1
(2, 6): 16 + 2k + c = 6
M1
subtracts: 15 + 1.5k = 3, so k = −8
A1
c = 6
A1
Part (c)
4x3 − 8x = 4x(x2 − 2) = 0, so x = ±√2
B1
y = 4 − 8 + 6 = 2 for both, since y depends only on x2 = 2
B1
Exam tip: Two unknown constants need two conditions. Write one equation for each point, then solve them simultaneously.
A phone is plugged in to charge. Its battery charge is E per cent at time t minutes after it is plugged in. Figure 1 shows the graph of dE/dt against t for 0 ≤ t ≤ 40; the graph is a straight line. When the phone is plugged in, its battery charge is 12%.
(a)
Use Figure 1 to show that dE/dt = 4 − 0.1t.
(2)
(b)
Find E in terms of t.
(3)
(c)
Find the battery charge 40 minutes after the phone is plugged in.
(1)
(d)
Find the time at which the battery charge reaches 72%.
(2)
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Answer
Mark
Notes
Part (a)
reads the intercept 4 and the gradient −4/40 = −0.1 from the graph
t = 20 minutes; t = 60 is rejected because the model only applies for 0 ≤ t ≤ 40
A1
Common error: Reading the 4 on the vertical axis as a charge of 4%. The graph shows the rate of charging, so 4 means 4% per minute; the starting charge of 12% comes from the stem and gives the constant.
A curve has equation y = f(x), x > 0, where dy/dx = (4 + x)/√x. A student is asked whether the curve can pass through the point (9, 30). Figure 1 shows the student’s attempt.
(a)
Identify the error in Line 1, and write dy/dx correctly as a sum of powers of x.
(2)
(b)
Line 2 integrates the expression in Line 1 correctly but contains a different error. State this error and explain why the conclusion in Line 3 is not valid.
(2)
(c)
Find f(x) correctly, given that the curve passes through (9, 30).
(3)
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Answer
Mark
Notes
Part (a)
dividing by √x = x½ should lower each power by ½, but Line 1 multiplies by √x instead
B1
dy/dx = 4x−½ + x½
B1
Part (b)
the constant of integration + c is missing
B1
with + c included, the value of y at x = 9 depends on c, so the point (9, 30) can always be reached by choosing c; it determines c and cannot be ruled out
B1
Part (c)
raises a fractional power by one, e.g. x−½ → x½
M1
8x½ + (2/3)x3/2 + c
A1
24 + 18 + c = 30: f(x) = 8x½ + (2/3)x3/2 − 12
A1
Exam tip: Before integrating a fraction with √x underneath, write √x as x½ and divide each term of the numerator by it. The powers go down by ½, not up.
Figure 1 shows the graph of y = f′(x), x > 0, where f′(x) = ax + b/x2 and a and b are constants. The graph passes through the points A and B, whose coordinates are shown.
(a)
Show that a = 2 and b = −8.
(3)
(b)
Given that the curve y = f(x) passes through the point (4, 20), find f(x).
(4)
(c)
Use the point B in Figure 1 to write down the gradient of the curve y = f(x) at the point where x = 2, and find the equation of the tangent to the curve at this point.
(2)
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Answer
Mark
Notes
Part (a)
uses A: a + b = −6
M1
uses B: 2a + b/4 = 2
M1
solves simultaneously, e.g. 8a + b = 8 so 7a = 14: a = 2, b = −8
A1
Part (b)
writes −8/x2 as −8x−2 and raises a power by one
M1
x2 + 8x−1 + c
A1
substitutes (4, 20): 16 + 2 + c = 20
M1
f(x) = x2 + 8/x + 2
A1
Part (c)
gradient = f′(2) = 2, read from B, and f(2) = 4 + 4 + 2 = 10
B1
y − 10 = 2(x − 2), so y = 2x + 6 ft from their f(x)
B1
Exam tip: A point on the graph of y = f′(x) gives an equation for f′, such as f′(2) = 2. It says nothing about the value of f itself, which needs a separate point on the curve y = f(x).
Figure 1 shows a direction field for the gradient function dy/dx = 4x − x2, 0 ≤ x ≤ 5. Each short segment is drawn with the gradient given by this function at its midpoint, so any curve with this gradient function follows the directions of the segments. The points P and Q are marked.
(a)
Explain why all the segments in the same vertical column are parallel.
(1)
(b)
Find the equation of the curve with this gradient function that passes through P.
(3)
(c)
Show that the curve in part (b) does not pass through Q, and find the equation of the curve with this gradient function that does pass through Q.
(3)
(d)
Explain why the curves in parts (b) and (c) never meet, and state the vertical distance between them for every value of x.
(2)
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Answer
Mark
Notes
Part (a)
dy/dx depends only on x, not on y, so all points with the same x-coordinate have the same gradient
B1
Part (b)
raises the power of at least one term by one
M1
2x2 − x3/3 + c
A1
P(0, 2) gives c = 2: y = 2x2 − x3/3 + 2
A1
Part (c)
at x = 3 the curve in (b) has y = 18 − 9 + 2 = 11, not 4
B1
18 − 9 + c = 4
M1
y = 2x2 − x3/3 − 5
A1
Part (d)
the curves have the same gradient function and differ only in the constant of integration, so one is the other translated vertically and they cannot meet
B1
vertical distance = 2 − (−5) = 7
B1
Exam tip: Every curve with the same gradient function has the same shape; the constant of integration only moves it up or down. A direction field shows the whole family at once, and one point picks out one curve.
Figure 1 shows the side view of a grass bank. The profile of the bank is modelled by the curve y = f(x), 0 ≤ x ≤ 9, where x metres is the horizontal distance from the top of the bank, A, and y metres is the height above level ground. The coordinates of A, and of B, where the bank meets the ground, are shown in Figure 1. In the model, f′(x) = k√x − 1, where k is a constant.
(a)
Find f(x) in terms of k.
(3)
(b)
Show that k = 1/6.
(2)
(c)
Find the height of the bank at a horizontal distance of 4 m from A, giving your answer to 3 significant figures.
(2)
(d)
Find the coordinates of the point on the profile at which the gradient is −5/6.
(2)
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Answer
Mark
Notes
Part (a)
raises the power of √x = x½ by one
M1
(2k/3)x3/2 − x
A1
+ 6, using A(0, 6)
B1
Part (b)
uses B: f(9) = 0, with 93/2 = 27: 18k − 9 + 6 = 0
M1
18k = 3, so k = 1/6
A1
Part (c)
f(4) = (1/9) × 8 − 4 + 6
M1
= 26/9 = 2.89 m
A1
Part (d)
√x/6 − 1 = −5/6, so √x = 1
M1
(1, 46/9), i.e. (1, 5.11) to 3 s.f.
A1
Exam tip: Evaluate 93/2 as (√9)3 = 27 before multiplying: (2k/3) × 27 = 18k. Exact values keep the equation for k simple.
Figure 1 (top) shows the graph of y = f′(x), where f′(x) is a cubic function. The graph meets the x-axis only at the three points where x = −2, 0 and 2, and passes through the labelled point. Figure 1 (bottom) shows three sketches, A, B and C. The curve y = f(x) passes through the point (2, −1).
(a)
Use the labelled point to write down the gradient of the curve y = f(x) at the point where x = 1. Hence state which one of the sketches A, B and C could be the graph of y = f(x), giving a reason.
(2)
(b)
Write f′(x) in the form px(x − q)(x + q), and hence show that f′(x) = x3 − 4x.
(3)
(c)
Find f(x).
(3)
(d)
Write down the coordinates of the point where the curve y = f(x) meets the y-axis.
(1)
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Answer
Mark
Notes
Part (a)
gradient = value of f′ at x = 1 = −3
B1
A: at x = 1 the curve must slope downwards, which rules out B (sloping upwards at x = 1) and C (horizontal at x = 1)
B1
Part (b)
uses the roots 0 and ±2: f′(x) = px(x − 2)(x + 2)
M1
substitutes the point (1, −3): p × 1 × (−1) × 3 = −3
M1
p = 1, so f′(x) = x(x2 − 4) = x3 − 4x
A1
Part (c)
raises each power by one
M1
x4/4 − 2x2 + c
A1
4 − 8 + c = −1: f(x) = x4/4 − 2x2 + 3
A1
Part (d)
(0, 3) ft from their f(x)
B1
Common error: Choosing the sketch that looks like the graph of f′. The height of the graph of f′ at a value of x is the gradient of y = f(x) there: f′(1) = −3 means the curve slopes downwards at x = 1.
uses the known root x = 2: (x − 2)(x2 − 2x + 2) = 0
M1
discriminant of x2 − 2x + 2 is 4 − 8 = −4 < 0, so x = 2 is the only solution and the normal meets C only at A
A1
Model sketch
Figure: model sketch: C with the normal x + y − 3 = 0 at A
Exam tip: When you already know one intersection, (x − 2) must be a factor of the cubic. Divide it out and check the discriminant of the quadratic that is left.
The side profile of a playground slide is modelled by the curve y = f(x), 0 ≤ x ≤ 4, where x metres is the horizontal distance from the top of the slide and y metres is the height above level ground. In the model, f′(x) = k(x2 − 4x), where k is a positive constant. The top of the slide, where x = 0, is 2.4 m above the ground, and the bottom end, where x = 4, is 0.8 m above the ground.
Figure: side profile of the slide
(a)
Find f(x) in terms of k.
(3)
(b)
Show that k = 3/20.
(3)
(c)
By writing f′(x) in completed-square form, or otherwise, find the value of x at which the slide is steepest, and the gradient of the slide there.
(3)
(d)
Find the angle that the slide makes with the horizontal at its steepest point, giving your answer in degrees to one decimal place.
(2)
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Answer
Mark
Notes
Part (a)
raises the powers by one
M1
k(x3/3 − 2x2)
A1
+ 2.4, using f(0) = 2.4
B1
Part (b)
f(4) = 0.8: k(64/3 − 32) + 2.4 = 0.8
M1
−(32/3)k = −1.6
M1
k = 4.8/32 = 3/20
A1
Part (c)
f′(x) = (3/20)[(x − 2)2 − 4], or solves f″(x) = 0
M1
steepest where x = 2
A1
gradient = (3/20)(−4) = −0.6
A1
Part (d)
tan θ = 0.6
M1
θ = 31.0°
A1
Exam tip: A point given in context, such as “2.4 m above the ground at the top”, is just a coordinate: here (0, 2.4). Translate each sentence into a condition on f or f′.
The curve C has equation y = f(x), x ≥ 0, where dy/dx = 9√x − kx and k is a constant. C passes through the origin O and through the point A(4, 16).
Figure: sketch of the curve C
(a)
Find y in terms of x and k, and show that k = 4.
(5)
(b)
Find the equation of the normal to C at A, giving your answer in the form ax + by + d = 0, where a, b and d are integers.
(4)
(c)
C meets the x-axis at O and at the point B. Find the x-coordinate of B.
(3)
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Answer
Mark
Notes
Part (a)
raises the power of at least one term by one
M1
6x3/2
A1
−½kx2
A1
constant of integration is 0 because C passes through O
B1
at A: 48 − 8k = 16, so k = 4
A1
Part (b)
substitutes x = 4 into dy/dx = 9√x − 4x
M1
gradient = 18 − 16 = 2
A1
normal gradient −½ and uses y − 16 = −½(x − 4)
M1
x + 2y − 36 = 0
A1
Part (c)
6x3/2 − 2x2 = 0, so x3/2(6 − 2x½) = 0
M1
√x = 3
M1
x = 9
A1
Common error: Assuming the constant of integration is zero without a reason. Here it is zero because the curve passes through the origin — say so in your working.
Figure 1 shows a plan of part of a new road. The straight road ℓ1 lies along the x-axis for x ≤ 0. A curved section of road joins ℓ1 at the origin O to the straight road ℓ2 at the point P. The road ℓ2 passes through P and Q. The coordinates of P and Q, in kilometres, are shown in Figure 1. The curved section is modelled by y = f(x), 0 ≤ x ≤ 2, where f′(x) = ax2 + bx and a and b are constants. The curved section meets ℓ2 at P with the same gradient as ℓ2.
(a)
Find f(x) in terms of a and b, explaining why there is no constant term.
(3)
(b)
Find the gradient of ℓ2.
(1)
(c)
Show that a = 3/8, and find the value of b.
(4)
(d)
Show that the curved section also joins ℓ1 without a change of direction.
(1)
(e)
Find the coordinates of the point on the curved section at which the road is heading in the same direction as the line y = x. Give each coordinate to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
raises each power by one
M1
ax3/3 + bx2/2
A1
the curve passes through O, so f(0) = 0 and the constant of integration is 0
B1
Part (b)
(7 − 2)/(4 − 2) = 5/2
B1
Part (c)
P lies on the curve: 8a/3 + 2b = 2
M1
same gradient at P: 4a + 2b = 5/2
M1
subtracts: 4a/3 = 1/2, so a = 3/8
A1
b = ½
A1
Part (d)
f′(0) = 0, which equals the gradient of ℓ1 (the x-axis)
B1
Part (e)
(3/8)x2 + ½x = 1, i.e. 3x2 + 4x − 8 = 0
M1
x = (−4 + √112)/6 = 1.097…, so x = 1.10 (rejecting the negative root)
A1
y = x3/8 + x2/4 = 0.466
A1
Exam tip: “Meets with the same gradient” gives two conditions at P: the curve passes through P, and f′ at P equals the gradient of the line. Two unknown constants need exactly two equations like these.
Figure 1 shows the graph of y = f′(x), where f′(x) is a quadratic function, and below it, on the same x-scale, the graph of y = f(x). The graph of y = f′(x) has its vertex at V and crosses the x-axis at the labelled point (3, 0). The curve y = f(x) passes through the labelled point (0, 2).
(a)
Write f′(x) in the form p(x − 1)2 − 4, and hence show that f′(x) = x2 − 2x − 3.
(3)
(b)
Find f(x).
(3)
(c)
The curve y = f(x) has gradient −3 at two points. Find the exact coordinates of these two points.
(3)
(d)
Use the point V in Figure 1 to write down the gradient of the curve y = f(x) at the point where x = 1. Find the equation of the tangent to y = f(x) at this point, giving your answer in the form rx + sy + d = 0, where r, s and d are integers.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
the vertex V(1, −4) gives f′(x) = p(x − 1)2 − 4
B1
substitutes (3, 0): 4p − 4 = 0, so p = 1
M1
(x − 1)2 − 4 = x2 − 2x + 1 − 4 = x2 − 2x − 3
A1
Part (b)
raises each power by one
M1
x3/3 − x2 − 3x + c
A1
c = 2 from (0, 2): f(x) = x3/3 − x2 − 3x + 2
A1
Part (c)
x2 − 2x − 3 = −3
M1
x(x − 2) = 0, so x = 0 or x = 2
A1
(0, 2) and (2, −16/3)
A1
Part (d)
gradient = −4, the value of f′ at x = 1 read from V
B1
at x = 1, y = 1/3 − 1 − 3 + 2 = −5/3, and y + 5/3 = −4(x − 1)
M1
12x + 3y − 7 = 0
A1
Exam tip: A point on the graph of y = f′(x) gives a gradient, not a point on the curve. V(1, −4) tells you that y = f(x) has gradient −4 where x = 1; the y-coordinate of the curve there comes from f(1).
A hot-air balloon is released from a platform 20 m above the ground. Its height above the ground t minutes after release is h metres. The rate of climb is modelled by dh/dt = a√t − bt, 0 ≤ t ≤ 16, where a and b are constants; a negative value means that the balloon is descending. Figure 1 shows the graph of dh/dt against t, which passes through the points A and B shown.
(a)
Show that a = 6 and b = 2.
(4)
(b)
Find h in terms of t.
(3)
(c)
Find the height of the balloon at the time shown by A, and use Figure 1 to state its rate of climb at that time.
(2)
(d)
Find the value of t at which the balloon is again 20 m above the ground.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
uses A: 2a − 4b = 4
M1
uses B: 3a − 9b = 0
M1
solves simultaneously, e.g. a = 3b so 6b − 4b = 4
M1
b = 2 and a = 6
A1
Part (b)
raises the power of t½ or t by one
M1
4t3/2 − t2
A1
h = 20 when t = 0: h = 4t3/2 − t2 + 20
B1
Part (c)
h = 4 × 8 − 16 + 20 = 36 m
B1
rate of climb = 4 m per minute, read from A(4, 4)
B1
Part (d)
4t3/2 − t2 = 0
M1
t3/2(4 − t½) = 0, so √t = 4
M1
t = 16
A1
Common error: Reading A(4, 4) as a height of 4 m. The vertical axis shows the rate of climb in metres per minute; the height at t = 4 comes from your expression for h, using the starting height of 20 m.
Figure 1 shows part of the curve C with equation y = f(x), x > 0. C crosses the x-axis at the points A and B and has a minimum point M. The coordinates of A, M and B are shown in Figure 1. Given that f′(x) = (p√x + q)/√x, where p and q are constants,
(a)
write f′(x) in the form p + qxn, stating the value of n, and find f(x) in terms of p, q and a constant of integration c,
(3)
(b)
use the fact that the gradient of C is zero at M to show that q = −2p,
(2)
(c)
use the coordinates of A and M to find the values of p, q and c,
(4)
(d)
verify that B lies on C,
(1)
(e)
find the equation of the tangent to C at B, giving your answer in the form rx + sy + d = 0, where r, s and d are integers.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
n = −½
B1
raises the power, x−½ → x½
M1
f(x) = px + 2qx½ + c
A1
Part (b)
f′(4) = 0: p + q/2 = 0
M1
q = −2p
A1
Part (c)
A: p + 2q + c = 0, so c = 3p
M1
M: 4p + 4q + c = −2, so 4p − 8p + 3p = −2
M1
p = 2, q = −4
A1
c = 6
A1
Part (d)
f(9) = 18 − 8 × 3 + 6 = 0
B1
Part (e)
f′(9) = 2 − 4/3 = 2/3
M1
y − 0 = (2/3)(x − 9)
M1
2x − 3y − 18 = 0
A1
Exam tip: Three unknowns (p, q and c) need three conditions. The figure supplies them: the gradient is zero at M, and the coordinates of A and of M each satisfy y = f(x).
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