Edexcel International A Level Maths · P3 · Pure Mathematics 3
Differentiation: chain, product and quotient rules
28 free, high-quality past-paper-style questions on Differentiation: chain, product and quotient rules 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 · 235 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-6-B-01P3 · A2Easy6 marks
In this question x is measured in radians.
(a)
Differentiate 5 sin 4x − 3 cos (x/2) with respect to x.
(2)
(b)
Differentiate 2e−3x + ln 7x, x > 0, with respect to x.
(2)
(c)
The curve C has equation y = 2 tan 3x, 0 < x < π/6. Find the exact gradient of C at the point where x = π/9.
(2)
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Answer
Mark
Notes
Part (a)
one term of the form k cos 4x or k sin (x/2)
M1
20 cos 4x + (3/2) sin (x/2)
A1
Part (b)
−6e−3x
B1
+ 1/x
B1
Part (c)
dy/dx = 6 sec2 3x
M1
sec2(π/3) = 1/(½)2 = 4, so the gradient is 24
A1
Common error: d/dx(ln 7x) is 1/x, not 7/x: ln 7x = ln 7 + ln x, and ln 7 is a constant.
Figure 1 shows the graph of y = a sin kx, 0 ≤ x ≤ π, where a and k are positive constants and x is in radians. The graph passes through the origin, and its first maximum point M lies on the gridlines.
(a)
Use Figure 1 to write down the value of a and the value of k.
(2)
(b)
Find dy/dx, and hence find the exact gradient of the curve at the point where x = π/6.
(2)
(c)
Find the values of x, 0 < x < π, at which the gradient of the curve is −4.
(2)
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Answer
Mark
Notes
Part (a)
a = 4, the greatest value of y
B1
one complete cycle has length π, so 2π/k = π and k = 2
B1
Part (b)
dy/dx = 8 cos 2x ft their a and k
B1
at x = π/6: 8 cos (π/3) = 4
A1
Part (c)
8 cos 2x = −4 ⇒ cos 2x = −½
M1
2x = 2π/3 or 4π/3, so x = π/3 and x = 2π/3
A1
Common error: Reading k as the length of a cycle. The period of sin kx is 2π/k, so a full cycle of length π gives k = 2. Check it with M: sin (2 × π/4) = 1, a maximum.
Figure 1 shows part of the curve C with equation y = 3 ln 2x, x > 0. The curve meets the x-axis at the point A. The dashed line is the tangent to C at the point P, and it passes through the origin O.
(a)
Write down the exact coordinates of A.
(1)
(b)
Find dy/dx.
(1)
(c)
Find an equation of the tangent to C at A, in the form y = mx + c.
(2)
(d)
The point P has x-coordinate p. Show that ln 2p = 1, and hence find the exact coordinates of P.
(3)
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Answer
Mark
Notes
Part (a)
ln 2x = 0 ⇒ 2x = 1: A(½, 0)
B1
Part (b)
dy/dx = 3/x
B1
Part (c)
gradient at A is 3 ÷ ½ = 6
M1
y = 6x − 3
A1
Part (d)
tangent at P: y − 3 ln 2p = (3/p)(x − p)
M1
it passes through O, so −3 ln 2p = −3, giving ln 2p = 1
A1
2p = e: P(e/2, 3)
A1
Common error: Writing d/dx(ln 2x) = 2/x. Since ln 2x = ln 2 + ln x, and ln 2 is a constant, the derivative is 1/x; with the factor 3 the gradient function is 3/x.
Figure 1 shows part of the curve with equation y = (x2 − 4)3. The points A, B and C on the curve are its three stationary points.
(a)
Use the chain rule to find dy/dx.
(2)
(b)
Find the coordinates of A, B and C.
(3)
(c)
By considering the sign of dy/dx on each side of x = 2, show that C is not a turning point.
(2)
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Answer
Mark
Notes
Part (a)
3(x2 − 4)2 × 2x
M1
dy/dx = 6x(x2 − 4)2
A1
Part (b)
6x(x2 − 4)2 = 0 ⇒ x = 0 or x = ±2
M1
B(0, −64)
A1
A(−2, 0) and C(2, 0)
A1
Part (c)
(x2 − 4)2 ≥ 0, so near x = 2 the sign of dy/dx is the sign of 6x
M1
dy/dx > 0 just to the left and just to the right of x = 2, so the gradient does not change sign: C is a stationary point of inflection, not a turning point
A1
Exam tip: Leave a chain-rule derivative factorised. 6x(x2 − 4)2 shows at once where it is zero and which factor decides its sign. Expanding it would hide both.
Figure 1 shows the curve C with equation x = tan 2y, −π/4 < y < π/4. The lines y = π/4 and y = −π/4 are asymptotes to C. The point P on C has x-coordinate 1.
(a)
Find dx/dy in terms of y.
(1)
(b)
Hence show that dy/dx = 1/(2(1 + x2)).
(3)
(c)
Find the exact y-coordinate of P, and find an equation of the tangent to C at P in the form y = mx + c, where m and c are exact constants.
(3)
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Answer
Mark
Notes
Part (a)
dx/dy = 2 sec2 2y
B1
Part (b)
dy/dx = 1/(2 sec2 2y), using dy/dx = 1 ÷ (dx/dy)
M1
sec2 2y = 1 + tan2 2y
M1
= 1 + x2, so dy/dx = 1/(2(1 + x2)), as required
A1
Part (c)
tan 2y = 1 with −π/2 < 2y < π/2 ⇒ y = π/8
B1
gradient 1/(2 × 2) = ¼
M1
y = ¼x + π/8 − ¼
A1
Exam tip: When the answer is wanted in terms of x, look for the original equation inside your derivative: here tan 2y is just x, and the identity sec2 θ = 1 + tan2 θ links them.
The curve C has equation y = ln(sec x + tan x), 0 < x < π/2.
(a)
Show that dy/dx = sec x.
(4)
(b)
Find the exact value of x at which the gradient of C is 2.
(2)
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Answer
Mark
Notes
Part (a)
chain rule: (derivative of sec x + tan x) ÷ (sec x + tan x)
M1
derivative of sec x + tan x is sec x tan x + sec2x
B1
numerator factorised as sec x (tan x + sec x)
M1
cancels sec x + tan x to give sec x, as required
A1
Part (b)
sec x = 2 ⇒ cos x = ½
M1
x = π/3
A1
Exam tip: The derivatives of sec x and tan x are in the formulae booklet, but in a “show that” you must still display the factorisation and the cancelling step.
Verify that the point (e, 1) lies on the curve, and find an equation of the tangent to the curve at this point in the form x + py + q = 0, where p and q are constants to be found in terms of e.
(3)
(c)
Find the exact coordinates of the point on the curve at which the tangent is parallel to the y-axis.
(2)
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Answer
Mark
Notes
Part (a)
product rule: dx/dy = ey + yey
M1
= ey(1 + y)
A1
dy/dx = 1/(ey(1 + y))
A1
Part (b)
1 × e1 = e ✓
B1
gradient 1/(e × 2) = 1/(2e) and y − 1 = (x − e)/(2e)
M1
x − 2ey + e = 0 (p = −2e, q = e)
A1
Part (c)
dx/dy = 0 with ey ≠ 0 ⇒ y = −1
M1
(−e−1, −1)
A1
Model sketch
Figure: the tangent at (e, 1), and the vertical tangent at (−1/e, −1)
Exam tip: A tangent parallel to the y-axis is vertical: dy/dx is undefined there, which happens where dx/dy = 0.
The depth of water, h metres, at the entrance to a harbour t hours after midnight is modelled by h = 5 + 3 sin (πt/6), 0 ≤ t ≤ 12.
(a)
Find dh/dt.
(2)
(b)
Find the exact rate at which the depth is changing at 2 a.m., stating whether the depth is increasing or decreasing.
(2)
(c)
Find the two times at which dh/dt = 0, and the depth of water at each of these times.
(3)
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Answer
Mark
Notes
Part (a)
k cos (πt/6)
M1
dh/dt = (π/2) cos (πt/6)
A1
Part (b)
(π/2) cos (π/3) = π/4
M1
increasing at π/4 m per hour (≈ 0.785 m per hour)
A1
Part (c)
cos (πt/6) = 0 ⇒ πt/6 = π/2 or 3π/2
M1
t = 3 and t = 9 (3 a.m. and 9 a.m.)
A1
depths 8 m and 2 m
A1
Model sketch
Figure: depth of water: dh/dt = 0 at t = 3 and t = 9
Exam tip: The chain rule brings out the factor π/6 from inside the sine. Leave it as an exact multiple of π unless a decimal is asked for, and give units with every rate.
The curve C has equation y = x2e−x/2, x ≥ 0. Figure 1 shows part of C, which has a stationary point at the origin O and a maximum point B. The point P on C has x-coordinate 2, and the dashed line is the tangent to C at P.
(a)
Show that dy/dx = ½x(4 − x)e−x/2.
(3)
(b)
Find the exact coordinates of B.
(2)
(c)
Show that the tangent to C at P passes through O.
(3)
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Answer
Mark
Notes
Part (a)
product rule: 2xe−x/2 + x2 × d/dx(e−x/2)
M1
d/dx(e−x/2) = −½e−x/2
B1
2xe−x/2 − ½x2e−x/2 = ½x(4 − x)e−x/2, as required
A1
Part (b)
e−x/2 ≠ 0 and x ≠ 0 at B, so 4 − x = 0 and x = 4
M1
B(4, 16e−2)
A1
Part (c)
P is (2, 4e−1)
B1
gradient at P: ½ × 2 × 2 × e−1 = 2e−1, and y − 4e−1 = 2e−1(x − 2)
M1
y = 2e−1x, which has no constant term, so the tangent passes through O
A1
Exam tip: Take out the common factor xe−x/2 as soon as the product rule is done. The bracket that is left, (4 − x), gives the stationary points and makes the signs easy to read.
Figure 1 shows part of the curve C with equation y = e2x/(x + 1), x > −1. The line x = −1 is an asymptote to C. The curve has a minimum point M and crosses the y-axis at A.
(a)
Show that dy/dx = (2x + 1)e2x/(x + 1)2.
(3)
(b)
Find the exact coordinates of M.
(2)
(c)
The tangent to C at A meets the x-axis at the point T. Find the coordinates of T.
(3)
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Answer
Mark
Notes
Part (a)
quotient rule with u = e2x, v = x + 1
M1
numerator 2e2x(x + 1) − e2x × 1
A1
= e2x(2x + 2 − 1) = (2x + 1)e2x, as required
A1
Part (b)
e2x > 0, so 2x + 1 = 0 and x = −½
M1
M(−½, 2e−1)
A1
Part (c)
A is (0, 1)
B1
gradient at A is (0 + 1) × 1 ÷ 1 = 1, so the tangent is y = x + 1
M1
T(−1, 0)
A1
Common error: Writing the quotient-rule numerator the wrong way round, as uv′ − vu′. That changes the sign of dy/dx, and here it would give a gradient of −1 at A, which Figure 1 shows cannot be right.
Figure 1 shows the curve C with equation y = 2 sec x − tan x, −π/2 < x < π/2. The lines x = −π/2 and x = π/2 are asymptotes to C, and M is the minimum point of C.
(a)
Show that dy/dx = (2 sin x − 1)/cos2x.
(3)
(b)
Find the exact coordinates of M.
(3)
(c)
Find an equation of the tangent to C at the point where C crosses the y-axis.
(2)
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Answer
Mark
Notes
Part (a)
uses d/dx(sec x) = sec x tan x and d/dx(tan x) = sec2x
M1
dy/dx = 2 sec x tan x − sec2x
A1
= 2 sin x/cos2x − 1/cos2x = (2 sin x − 1)/cos2x, as required
A1
Part (b)
2 sin x − 1 = 0 ⇒ sin x = ½
M1
x = π/6, the only solution in the interval
A1
y = 2 × 2/√3 − 1/√3 = 3/√3: M(π/6, √3)
A1
Part (c)
point (0, 2) and gradient (0 − 1)/1 = −1
M1
y = 2 − x
A1
Exam tip: Writing sec x and tan x in terms of sin x and cos x puts everything over cos2 x. That is positive on this interval, so the sign of dy/dx is the sign of 2 sin x − 1, which confirms that M is a minimum.
The curve C has equation x = ln(y2 + 1). Figure 1 shows C, which is symmetrical about the x-axis and passes through the origin O. The point P on C has y-coordinate 2.
(a)
Find dy/dx in terms of y.
(2)
(b)
Find an equation of the normal to C at P, in the form ax + by + c = 0, where a and b are integers and c is an exact constant.
(3)
(c)
Find the exact coordinates of the point on C at which the gradient is 1.
(2)
(d)
Use dx/dy to explain why the tangent to C at O is the y-axis.
(1)
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Answer
Mark
Notes
Part (a)
chain rule: dx/dy = 2y/(y2 + 1)
M1
dy/dx = (y2 + 1)/(2y)
A1
Part (b)
P is (ln 5, 2)
B1
gradient of C at P is 5/4, so the normal has gradient −4/5 and y − 2 = −⅘(x − ln 5)
M1
4x + 5y − 10 − 4 ln 5 = 0
A1
Part (c)
y2 + 1 = 2y ⇒ (y − 1)2 = 0, so y = 1
M1
(ln 2, 1)
A1
Part (d)
at O, y = 0 so dx/dy = 0: dy/dx is undefined, the tangent is vertical, and through O it is the line x = 0
B1
Exam tip: With x given in terms of y, differentiate with respect to y and then take the reciprocal. Check against Figure 1: near O the curve is almost vertical, and (y2 + 1)/(2y) is very large when y is small.
The temperature, T °C, of a cup of tea t minutes after it is poured is modelled by T = 20 + Ae−kt, t ≥ 0, where A and k are positive constants. Figure 1 shows the graph of T against t. The two marked points lie on gridline intersections.
(a)
Use the marked points in Figure 1 to show that A = 70 and k = (ln 2)/10.
(3)
(b)
Find dT/dt, and hence find the rate at which the temperature is falling when t = 10. Give your answer to 3 significant figures.
(2)
(c)
Find the time at which the temperature is falling at 1 °C per minute. Give your answer to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
at t = 0: 90 = 20 + A, so A = 70
B1
at t = 10: 55 = 20 + 70e−10k ⇒ e−10k = ½
M1
−10k = ln ½ = −ln 2, so k = (ln 2)/10, as required
A1
Part (b)
dT/dt = −70ke−kt = −7 ln 2 e−kt
M1
at t = 10: −7 ln 2 × ½ = −2.426…, so falling at 2.43 °C per minute
A1
Part (c)
7 ln 2 e−kt = 1
M1
e−kt = 1/(7 ln 2), so t = 10 ln (7 ln 2)/ln 2
M1
t = 22.8 minutes
A1
Exam tip: “Falling at 1 °C per minute” means dT/dt = −1. Either keep the minus sign or work with the size of the rate, 7 ln 2 e−kt = 1, but say which you are doing.
Figure 1 shows the curve C with equation y = esin 2x, 0 ≤ x ≤ π. The curve has a maximum point A and a minimum point B, and it meets the y-axis at D. The dashed lines are the tangents to C at D and at A; they meet at the point E.
(a)
Find dy/dx.
(2)
(b)
Find the exact coordinates of A and of B.
(3)
(c)
Find the exact x-coordinate of E.
(2)
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Answer
Mark
Notes
Part (a)
k cos 2x esin 2x
M1
dy/dx = 2 cos 2x esin 2x
A1
Part (b)
esin 2x > 0, so cos 2x = 0
M1
2x = π/2 or 3π/2, so x = π/4 or 3π/4
A1
A(π/4, e) and B(3π/4, e−1)
A1
Part (c)
D is (0, 1) with gradient 2 cos 0 × e0 = 2, so the tangent at D is y = 2x + 1
M1
the tangent at A is y = e, so 2x + 1 = e and x = (e − 1)/2
A1
Common error: Writing the derivative of esin 2x as esin 2x or as e2 cos 2x. The power stays as it is; the chain rule multiplies by the derivative of the power, 2 cos 2x.
The curve C has equation y = sin x/(2 + cos x), 0 ≤ x ≤ 2π.
(a)
Show that dy/dx = (2 cos x + 1)/(2 + cos x)2.
(4)
(b)
Find the exact coordinates of the stationary points of C.
(4)
(c)
Find an equation of the tangent to C at the point where x = π/2, giving your answer in the form y = mx + c, where m and c are exact constants.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
quotient rule with u′ = cos x and v′ = −sin x
M1
numerator cos x(2 + cos x) − sin x(−sin x)
A1
= 2 cos x + cos2x + sin2x, using cos2x + sin2x = 1
M1
= 2 cos x + 1, as required
A1
Part (b)
2 cos x + 1 = 0 ⇒ cos x = −½
M1
x = 2π/3 and x = 4π/3
A1
y = (√3/2) ÷ (3/2) at x = 2π/3
M1
(2π/3, √3/3) and (4π/3, −√3/3)
A1
Part (c)
point (π/2, ½)
B1
gradient (0 + 1)/22 = ¼ and y − ½ = ¼(x − π/2)
M1
y = ¼x + ½ − π/8
A1
Model sketch
Figure: C: y = sin x/(2 + cos x) with its stationary points and the tangent at x = π/2 (red)
Exam tip: After the quotient rule, look for cos2 x + sin2 x in the numerator: it collapses to 1 and is usually the step that makes a printed answer appear.
A straight line passes through the point P(8, 125). It meets the positive x-axis at A and the positive y-axis at B, and P lies between A and B. The acute angle OAB is θ radians, where O is the origin. The length AB is L.
Figure: the line through P meeting the axes at A and B; not to scale
(a)
Show that L = 125 cosec θ + 8 sec θ.
(2)
(b)
Find dL/dθ and hence show that dL/dθ = 0 when tan3 θ = 125/8.
(4)
(c)
Find the exact minimum value of L, giving your answer in the form k√29.
(3)
(d)
Justify that this value of L is a minimum.
(2)
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Answer
Mark
Notes
Part (a)
P is 125 above the x-axis, so AP = 125/sin θ = 125 cosec θ
M1
P is 8 from the y-axis, so PB = 8/cos θ = 8 sec θ, and L = AP + PB
A1
Part (b)
derivatives of cosec θ and sec θ used
M1
dL/dθ = −125 cosec θ cot θ + 8 sec θ tan θ
A1
8 sin θ/cos2 θ = 125 cos θ/sin2 θ, written in sin and cos
M1
sin3 θ/cos3 θ = 125/8, so tan3 θ = 125/8, as required
A1
Part (c)
tan θ = 5/2
B1
sin θ = 5/√29 and cos θ = 2/√29, from a 2, 5, √29 triangle
every term is positive for 0 < θ < π/2, so d2L/dθ2 > 0 and the stationary value is a minimum
A1
Exam tip: When an equation contains several trigonometric functions, rewrite everything in sin θ and cos θ, clear the fractions, then look for tan θ = sin θ/cos θ.
The curve C has equation y = f(x), where f(x) = ln(x2 + 3) − x, x ∈ ℝ.
(a)
Find f′(x).
(2)
(b)
Show that C has no stationary points and that f is a decreasing function.
(3)
(c)
Find an equation of the normal to C at the point where x = 1, giving your answer in the form y = 2x + a + ln b, where a and b are integers.
(3)
(d)
Use the quotient rule to find f″(x). Hence find the exact x-coordinate of the point on C at which the gradient is most negative, and state this gradient.
f″(x) = 0 ⇒ x = ±√3; f′(√3) = √3/3 − 1 and f′(−√3) = −√3/3 − 1, while f′(x) → −1 as x → ±∞
M1
most negative at x = −√3, gradient −1 − √3/3
A1
Exam tip: To prove a quadratic expression is always positive, complete the square or show its discriminant is negative, and then say what that means for the sign of the derivative.
Figure 1 shows a straight road through the points F and J, where FJ = 800 m. A power station S is 300 m from the road, and SF is perpendicular to the road. A cable is to be laid in a straight line from S to a point P on the road between F and J, and then along the road from P to J. The angle FSP is θ radians, where 0 ≤ θ ≤ α and α is the angle FSJ. Laying the cable costs £50 per metre from S to P and £30 per metre along the road. The total cost is £C.
(a)
Show that C = 15 000 sec θ − 9000 tan θ + 24 000.
(2)
(b)
Find dC/dθ.
(2)
(c)
Show that dC/dθ = 0 when sin θ = 3/5.
(3)
(d)
Find the minimum total cost and the corresponding distance FP.
(3)
(e)
Use the second derivative to show that this cost is a minimum.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
SP = 300 sec θ and FP = 300 tan θ, so PJ = 800 − 300 tan θ
M1
C = 50 × 300 sec θ + 30(800 − 300 tan θ) = 15 000 sec θ − 9000 tan θ + 24 000, as required
A1
Part (b)
uses the derivatives of sec θ and tan θ
M1
dC/dθ = 15 000 sec θ tan θ − 9000 sec2 θ
A1
Part (c)
3000 sec θ(5 tan θ − 3 sec θ) = 0
M1
sec θ ≠ 0, so 5 sin θ/cos θ = 3/cos θ
M1
cos θ ≠ 0, so 5 sin θ = 3 and sin θ = 3/5, as required
at sec θ = 5/4, tan θ = 3/4: 15 000 × 170/64 − 18 000 × 75/64 = 18 750 > 0, so a minimum
A1
Exam tip: Start from the right-angled triangle SFP in Figure 1. With θ at S, SP is the hypotenuse and SF is adjacent to θ, so SP = 300 sec θ and FP = 300 tan θ.
Figure 1 shows part of the curve C with equation y = e−x/3. The point P(p, e−p/3), where p ≥ 0, lies on C. The tangent to C at P meets the x-axis at A and the y-axis at B, and O is the origin. The shaded triangle OAB has area R. Figure 1 is drawn for one particular value of p.
(a)
Show that the tangent to C at P has equation x + 3ep/3y = p + 3.
(3)
(b)
Hence show that R = (p + 3)2e−p/3/6.
(2)
(c)
Show that dR/dp = (p + 3)(3 − p)e−p/3/18.
(3)
(d)
Find the exact maximum value of R, justifying that it is a maximum.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
dy/dx = −⅓e−x/3
B1
y − e−p/3 = −⅓e−p/3(x − p)
M1
multiplying by 3ep/3: 3ep/3y − 3 = −x + p, so x + 3ep/3y = p + 3, as required
A1
Part (b)
A(p + 3, 0) and B(0, ⅓(p + 3)e−p/3)
M1
R = ½ × (p + 3) × ⅓(p + 3)e−p/3 = (p + 3)2e−p/3/6, as required
p + 3 > 0 and e−p/3 > 0, so dR/dp = 0 only when p = 3
M1
R = 36e−1/6 = 6/e
A1
dR/dp > 0 for 0 ≤ p < 3 and dR/dp < 0 for p > 3, so this is a maximum
A1
Exam tip: Once R is written in terms of p, treat p as the variable. The factors p + 3 and e−p/3 are positive, so the sign of dR/dp is the sign of 3 − p.
Figure 1 shows the curve C with equation x = 4 sin2y, 0 ≤ y ≤ π/2. The point P on C has x-coordinate 1. The point Q is the point on C at which the gradient dy/dx takes its least value.
(a)
Show that dx/dy = 4 sin 2y.
(2)
(b)
Show that, for 0 < x < 4, dy/dx = 1/(2√(4x − x2)).
(4)
(c)
Find the exact y-coordinate of P, and find an equation of the tangent to C at P.
(3)
(d)
Find the least value of the gradient, and the exact coordinates of Q.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
chain rule: 4 × 2 sin y cos y = 8 sin y cos y
M1
= 4 sin 2y, using sin 2y = 2 sin y cos y
A1
Part (b)
dy/dx = 1/(4 sin 2y) = 1/(8 sin y cos y)
M1
sin y = √x/2
B1
cos y = √(1 − x/4) = √(4 − x)/2, taking the positive root as 0 ≤ y ≤ π/2
M1
8 × (√x/2) × (√(4 − x)/2) = 2√(4x − x2), giving the result
A1
Part (c)
sin2y = ¼ with 0 ≤ y ≤ π/2 ⇒ y = π/6
B1
gradient 1/(2√(4 − 1)) = 1/(2√3)
M1
y − π/6 = (x − 1)/(2√3), or y = (√3/6)(x − 1) + π/6
A1
Part (d)
4x − x2 = 4 − (x − 2)2, which is greatest when x = 2
M1
least gradient 1/(2√4) = ¼
A1
4 sin2y = 2 ⇒ sin y = 1/√2: Q(2, π/4)
A1
Common error: Taking cos y = −√(4 − x)/2, or not saying which root is used. The restriction 0 ≤ y ≤ π/2, the part of the curve drawn in Figure 1, is what makes cos y ≥ 0.
A point on a vibrating spring is x cm from its rest position at time t seconds. Its motion is modelled by x = 5e−t/5 sin 2t, t ≥ 0. Figure 1 shows the graph of x against t for 0 ≤ t ≤ 7.2. The points A and B are the first two maximum points of the graph.
(a)
Find dx/dt.
(3)
(b)
Show that at a stationary point, tan 2t = 10.
(2)
(c)
Find the value of t at A and the displacement at A, giving each answer to 3 significant figures.
(3)
(d)
Explain why the value of t at B is π more than the value of t at A, and show that the ratio of the displacement at B to the displacement at A is e−π/5.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
product rule of the form ae−t/5 sin 2t + be−t/5 cos 2t
M1
one term correct, e.g. −e−t/5 sin 2t
A1
dx/dt = e−t/5(10 cos 2t − sin 2t)
A1
Part (b)
e−t/5 ≠ 0, so 10 cos 2t = sin 2t
M1
cos 2t ≠ 0 (otherwise sin 2t would be 0 too), so tan 2t = 10, as required
A1
Part (c)
2t = arctan 10 = 1.4711…
M1
t = 0.736
A1
x = 5e−0.1471… × sin 1.4711… = 4.29 cm
A1
Part (d)
tan 2t = 10 has solutions every π/2 in t, alternately maxima and minima as Figure 1 shows, so B is at tA + π
M1
ratio = 5e−(tA + π)/5 sin(2tA + 2π) ÷ 5e−tA/5 sin 2tA
M1
sin(2tA + 2π) = sin 2tA, so the ratio is e−π/5
A1
Exam tip: Use the figure to choose the right solution of tan 2t = 10. Solutions come every π/2 seconds and alternate between maximum and minimum points, so the next maximum is π seconds later.
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