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Learn/ A Level Mathematics/ Past Papers/ 2020/ May/June 2020 /Paper 2 (Variant 1)

Pure Mathematics 2 — Variant 1

A Level Mathematics · 9709/21 · May/June 2020

Question Paper

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Mark Scheme

PDF · 262 KB · 10 pages

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Questions

6 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 [3]
Solve the equation ln⁡(x+1)−ln⁡x=2ln⁡2\ln(x + 1) - \ln x = 2 \ln 2.
2 [5]
The polynomial p(x)p(x) is defined by p(x)=6x3+ax2+9x+b,p(x) = 6x^{3} + ax^{2} + 9x + b, where aa and bb are constants. It is given that (x−2)(x - 2) and (2x+1)(2x + 1) are factors of p(x)p(x). Find the values of aa and bb.
3 [5]
A curve has parametric equations x=et−2e−t,y=3e2t+1.x = e^{t} - 2e^{-t}, \quad y = 3e^{2t} + 1. Find the equation of the tangent to the curve at the point for which t=0t = 0.
4
4(a) [3]
Sketch, on the same diagram, the graphs of y=∣3x+2a∣y = |3x + 2a| and y=∣3x−4a∣y = |3x - 4a|, where aa is a positive constant. Give the coordinates of the points where each graph meets the axes.
4(b) [3]
Find the coordinates of the point of intersection of the two graphs.
4(c) [1]
Deduce the solution of the inequality ∣3x+2a∣<∣3x−4a∣|3x + 2a| < |3x - 4a|.
6
6(a) [5]
Prove that sin⁡2θ(cosec⁡θ−sec⁡θ)≡8cos⁡(θ+14π).\sin 2\theta(\operatorname{cosec} \theta - \sec \theta) \equiv \sqrt{8} \cos\left(\theta + \frac{1}{4}\pi\right).
6(b) [2]
Solve the equation sin⁡2θ(cosec⁡θ−sec⁡θ)=1\sin 2\theta(\operatorname{cosec} \theta - \sec \theta) = 1 for 0<θ<12π0 < \theta < \frac{1}{2}\pi. Give the answer correct to 3 significant figures.
6(c) [3]
Find ∫sin⁡x(cosec⁡12x−sec⁡12x) dx.\int \sin x\left(\operatorname{cosec} \frac{1}{2}x - \sec \frac{1}{2}x\right) \, \mathrm{d}x.
7
7(a) [3]
Find the quotient when 9x3−6x2−20x+19x^{3} - 6x^{2} - 20x + 1 is divided by (3x+2)(3x + 2), and show that the remainder is 99.
7(b) [5]
Hence find ∫169x3−6x2−20x+13x+2 dx\int_{1}^{6} \frac{9x^{3} - 6x^{2} - 20x + 1}{3x + 2} \, \mathrm{d}x, giving the answer in the form a+ln⁡ba + \ln b where aa and bb are integers.
7(c) [4]
Find the exact root of the equation 9e9y−6e6y−20e3y−8=09e^{9y} - 6e^{6y} - 20e^{3y} - 8 = 0.

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