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Learn/ A Level Mathematics/ Past Papers/ 2021/ May/June 2021 /Paper 3 (Variant 2)

Pure Mathematics 3 — Variant 2

A Level Mathematics · 9709/32 · May/June 2021

Question Paper

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Mark Scheme

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Questions

10 of 11 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 ∣2x−1∣<3∣x+1∣|2x - 1| < 3|x + 1|.
2 [4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z+1−i∣⩽1|z + 1 - i| \leqslant 1 and arg⁡(z−1)⩽34π\arg(z - 1) \leqslant \frac{3}{4}\pi.
3
The variables xx and yy satisfy the equation x=A(3−y)x = A(3^{-y}), where AA is a constant.
3(a) [3]
Explain why the graph of yy against ln⁡x\ln x is a straight line and state the exact value of the gradient of the line. It is given that the line intersects the yy-axis at the point where y=1.3y = 1.3.
3(b) [2]
Calculate the value of AA, giving your answer correct to 2 decimal places.
4 [5]
Using integration by parts, find the exact value of ∫02tan⁡−1(12x) dx\int_{0}^{2} \tan^{-1}(\frac{1}{2}x) \, \mathrm{d}x.
5 [5]
The complex number uu is given by u=10−46iu = 10 - 4\sqrt{6}i. Find the two square roots of uu, giving your answers in the form a+iba + ib, where aa and bb are real and exact.
6
6(a) [3]
Prove that cosec⁡2θ−cot⁡2θ≡tan⁡θ\operatorname{cosec} 2\theta - \cot 2\theta \equiv \tan \theta.
6(b) [4]
Hence show that ∫14π13π(cosec⁡2θ−cot⁡2θ) dθ=12ln⁡2\int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} (\operatorname{cosec} 2\theta - \cot 2\theta) \, \mathrm{d}\theta = \frac{1}{2}\ln 2.
7 [7]
A curve is such that the gradient at a general point with coordinates (x,y)(x, y) is proportional to yx+1\frac{y}{\sqrt{x + 1}}. The curve passes through the points with coordinates (0,1)(0, 1) and (3,e)(3, e). By setting up and solving a differential equation, find the equation of the curve, expressing yy in terms of xx.
8 [8]
The equation of a curve is y=e−5xtan⁡2xy = e^{-5x} \tan^{2}x for −12π<x<12π-\frac{1}{2}\pi < x < \frac{1}{2}\pi. Find the xx-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.
9
Let f(x)=14−3x+2x2(2+x)(3+x2)f(x) = \frac{14 - 3x + 2x^{2}}{(2 + x)(3 + x^{2})}.
9(a) [5]
Express f(x)f(x) in partial fractions.
9(b) [5]
Hence obtain the expansion of f(x)f(x) in ascending powers of xx, up to and including the term in x2x^{2}.
11
With respect to the origin OO, the points AA and BB have position vectors given by OA→=2i−j\overrightarrow{OA} = 2\mathbf{i} - \mathbf{j} and OB→=j−2k\overrightarrow{OB} = \mathbf{j} - 2\mathbf{k}.
11(a) [4]
Show that OA=OBOA = OB and use a scalar product to calculate angle AOBAOB in degrees. The midpoint of ABAB is MM. The point PP on the line through OO and MM is such that PA:OA=7:1PA : OA = \sqrt{7} : 1.
11(b) [6]
Find the possible position vectors of PP.

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