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Learn/ A Level Mathematics/ Past Papers/ 2021/ May/June 2021 /Paper 3 (Variant 1)

Pure Mathematics 3 — Variant 1

A Level Mathematics · 9709/31 · May/June 2021

Question Paper

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Mark Scheme

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Questions

8 of 10 questions shown here, checked against the paper. The others are in the PDF above — they contain diagrams or maths we could not yet verify, and we would rather show you the original than a transcription we are not sure of.

1 [4]
Solve the inequality 2∣3x−1∣<∣x+1∣2|3x - 1| < |x + 1|.
2 [5]
Find the real root of the equation 2ex+e−x2+ex=3\frac{2e^{x} + e^{-x}}{2 + e^{x}} = 3, giving your answer correct to 3 decimal places. Your working should show clearly that the equation has only one real root.
3
3(a) [4]
Given that cos⁡(x−30∘)=2sin⁡(x+30∘)\cos(x - 30^\circ) = 2\sin(x + 30^\circ), show that tan⁡x=2−31−23\tan x = \frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}.
3(b) [2]
Hence solve the equation cos⁡(x−30∘)=2sin⁡(x+30∘)\cos(x - 30^\circ) = 2\sin(x + 30^\circ), for 0∘<x<360∘0^\circ < x < 360^\circ.
4
4(a) [2]
Prove that 1−cos⁡2θ1+cos⁡2θ≡tan⁡2θ\frac{1 - \cos 2\theta}{1 + \cos 2\theta} \equiv \tan^{2}\theta.
4(b) [4]
Hence find the exact value of ∫16π13π1−cos⁡2θ1+cos⁡2θ dθ\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \frac{1 - \cos 2\theta}{1 + \cos 2\theta} \, \mathrm{d}\theta.
6
The parametric equations of a curve are x=ln⁡(2+3t)x = \ln(2 + 3t), y=t2+3ty = \frac{t}{2 + 3t}.
6(a) [5]
Show that the gradient of the curve is always positive.
6(b) [3]
Find the equation of the tangent to the curve at the point where it intersects the yy-axis.
8
With respect to the origin OO, the points AA and BB have position vectors given by OA→=(121)\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} and OB→=(31−2)\overrightarrow{OB} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}. The line ll has equation r=(231)+λ(1−21)\mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.
8(a) [4]
Find the acute angle between the directions of ABAB and ll.
8(b) [5]
Find the position vector of the point PP on ll such that AP=BPAP = BP.
9
The equation of a curve is y=x−23ln⁡xy = x^{-\frac{2}{3}} \ln x for x>0x > 0. The curve has one stationary point.
9(a) [5]
Find the exact coordinates of the stationary point.
9(b) [5]
Show that ∫18y dx=18ln⁡2−9\int_{1}^{8} y \, \mathrm{d}x = 18\ln 2 - 9.
10 [11]
The variables xx and tt satisfy the differential equation dxdt=x2(1+2x)\frac{\mathrm{d}x}{\mathrm{d}t} = x^{2}(1 + 2x), and x=1x = 1 when t=0t = 0. Using partial fractions, solve the differential equation, obtaining an expression for tt in terms of xx.

Published by Cambridge Assessment International Education. Reproduced for study use.