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/ May/June 2021 /Paper 2 (Variant 1)

Pure Mathematics 2 — Variant 1

A Level Mathematics · 9709/21 · May/June 2021

Question Paper

PDF · 116 KB · 16 pages

Mark Scheme

PDF · 269 KB · 12 pages

This browser cannot show the PDF inline.

Open the question paper · Open the mark scheme

Questions

5 of 7 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 ∣3x−7∣<∣4x+5∣|3x - 7| < |4x + 5|.
2 [6]
By first expanding sin⁡(θ+30∘)\sin(\theta + 30^\circ), solve the equation sin⁡(θ+30∘)cosec⁡θ=2\sin(\theta + 30^\circ) \operatorname{cosec} \theta = 2 for 0∘<θ<360∘0^\circ < \theta < 360^\circ.
3
3(a) [2]
Show that (sec⁡x+cos⁡x)2(\sec x + \cos x)^{2} can be expressed as sec⁡2x+a+bcos⁡2x\sec^{2}x + a + b\cos 2x, where aa and bb are constants to be determined.
3(b) [4]
Hence find the exact value of ∫014π(sec⁡x+cos⁡x)2 dx\int_{0}^{\frac{1}{4}\pi} (\sec x + \cos x)^{2} \, \mathrm{d}x.
4
A curve has parametric equations x=ln⁡(2t+6)−ln⁡tx = \ln(2t + 6) - \ln t, y=tln⁡ty = t \ln t.
4(a) [3]
Find the value of tt at the point PP on the curve for which x=ln⁡4x = \ln 4.
4(b) [5]
Find the exact gradient of the curve at PP.
7
The polynomial p(x)p(x) is defined by p(x)=ax3−11x2−19x−ap(x) = ax^{3} - 11x^{2} - 19x - a, where aa is a constant. It is given that (x−3)(x - 3) is a factor of p(x)p(x).
7(a) [2]
Find the value of aa.
7(b) [3]
When aa has this value, factorise p(x)p(x) completely.
7(c) [4]
Hence find the exact values of yy that satisfy the equation p(ey+e−y)=0p(e^{y} + e^{-y}) = 0.

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