StudentAid
HomeLearn
9th Class10th Class11th Class12th ClassCambridge past papersBrowse everything
Attendance
Class attendanceVisitor register
For schoolsFor driversFor parentsBook a demo
Learn/ A Level Mathematics/ Past Papers/ 2021/ Oct/Nov 2021 /Paper 3 (Variant 2)

Pure Mathematics 3 — Variant 2

A Level Mathematics · 9709/32 · Oct/Nov 2021

Question Paper

PDF · 122 KB · 20 pages

Mark Scheme

PDF · 361 KB · 17 pages

This browser cannot show the PDF inline.

Open the question paper · Open the mark scheme

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]
Find the value of xx for which 3(21−x)=7x3(2^{1-x}) = 7^{x}. Give your answer in the form ln⁡aln⁡b\frac{\ln a}{\ln b}, where aa and bb are integers.
2 [4]
Solve the inequality ∣3x−a∣>2∣x+2a∣|3x - a| > 2|x + 2a|, where aa is a positive constant.
3
3(a) [2]
Given the complex numbers u=a+ibu = a + \mathrm{i}b and w=c+idw = c + \mathrm{i}d, where aa, bb, cc and dd are real, prove that (u+w)∗=u∗+w∗(u + w)^{*} = u^{*} + w^{*}.
3(b) [4]
Solve the equation (z+2+i)∗+(2+i)z=0(z + 2 + \mathrm{i})^{*} + (2 + \mathrm{i})z = 0, giving your answer in the form x+iyx + \mathrm{i}y where xx and yy are real.
4 [5]
Express 4x2−13x+13(2x−1)(x−3)\frac{4x^{2} - 13x + 13}{(2x - 1)(x - 3)} in partial fractions.
5
5(a) [4]
On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z−3−2i∣⩽1|z - 3 - 2\mathrm{i}| \leqslant 1 and Im z⩾2\mathrm{Im}\, z \geqslant 2.
5(b) [3]
Find the greatest value of arg⁡z\arg z for points in the shaded region, giving your answer in degrees.
6
6(a) [3]
Using the expansions of sin⁡(3x+2x)\sin(3x + 2x) and sin⁡(3x−2x)\sin(3x - 2x), show that 12(sin⁡5x+sin⁡x)≡sin⁡3xcos⁡2x\frac{1}{2}(\sin 5x + \sin x) \equiv \sin 3x \cos 2x.
6(b) [3]
Hence show that ∫014πsin⁡3xcos⁡2x dx=15(3−2)\int_{0}^{\frac{1}{4}\pi} \sin 3x \cos 2x \, \mathrm{d}x = \frac{1}{5}(3 - \sqrt{2}).
7 [7]
The variables xx and yy satisfy the differential equation e2xdydx=4xy2\mathrm{e}^{2x} \frac{\mathrm{d}y}{\mathrm{d}x} = 4xy^{2}, and it is given that y=1y = 1 when x=0x = 0. Solve the differential equation, obtaining an expression for yy in terms of xx.
8
8(a) [3]
By first expanding (cos⁡2θ+sin⁡2θ)2(\cos^{2}\theta + \sin^{2}\theta)^{2}, show that cos⁡4θ+sin⁡4θ≡1−12sin⁡22θ\cos^{4}\theta + \sin^{4}\theta \equiv 1 - \frac{1}{2}\sin^{2}2\theta.
8(b) [4]
Hence solve the equation cos⁡4θ+sin⁡4θ=59\cos^{4}\theta + \sin^{4}\theta = \frac{5}{9}, for 0∘<θ<180∘0^\circ < \theta < 180^\circ.
9
The equation of a curve is ye2x−y2ex=2y\mathrm{e}^{2x} - y^{2}\mathrm{e}^{x} = 2.
9(a) [4]
Show that dydx=2yex−y22y−ex\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2y\mathrm{e}^{x} - y^{2}}{2y - \mathrm{e}^{x}}.
9(b) [4]
Find the exact coordinates of the point on the curve where the tangent is parallel to the yy-axis.
10
With respect to the origin OO, the position vectors of the points AA and BB are given by OA→=(12−1)\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} and OB→=(031)\overrightarrow{OB} = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}.
10(a) [3]
Find a vector equation for the line ll through AA and BB.
10(b) [2]
The point CC lies on ll and is such that AC→=3AB→\overrightarrow{AC} = 3\overrightarrow{AB}. Find the position vector of CC.
10(c) [5]
Find the possible position vectors of the point PP on ll such that OP=14OP = \sqrt{14}.

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