Pure Mathematics 3 — Variant 2

A Level Mathematics · 9709/32 · Feb/March 2020

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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
1(a) [1]
Sketch the graph of y=∣x−2∣y = |x - 2|.
1(b) [3]
Solve the inequality ∣x−2∣<3x−4|x - 2| < 3x - 4.
2 [4]
Solve the equation ln⁡3+ln⁡(2x+5)=2ln⁡(x+2)\ln 3 + \ln(2x + 5) = 2\ln(x + 2). Give your answer in a simplified exact form.
3
3(a) [2]
By sketching a suitable pair of graphs, show that the equation sec⁡x=2−12x\sec x = 2 - \frac{1}{2}x has exactly one root in the interval 0⩽x<12π0 \leqslant x < \frac{1}{2}\pi.
3(b) [2]
Verify by calculation that this root lies between 0.8 and 1.
3(c) [3]
Use the iterative formula xn+1=cos⁡−1(24−xn)x_{n+1} = \cos^{-1}\left(\frac{2}{4 - x_{n}}\right) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
4 [7]
Find ∫16π13πxsec⁡2x dx\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} x \sec^{2}x \, \mathrm{d}x. Give your answer in a simplified exact form.
5
5(a) [4]
Show that cos⁡3xsin⁡x+sin⁡3xcos⁡x=2cot⁡2x\dfrac{\cos 3x}{\sin x} + \dfrac{\sin 3x}{\cos x} = 2\cot 2x.
5(b) [3]
Hence solve the equation cos⁡3xsin⁡x+sin⁡3xcos⁡x=4\dfrac{\cos 3x}{\sin x} + \dfrac{\sin 3x}{\cos x} = 4, for 0<x<π0 < x < \pi.
6
The variables xx and yy satisfy the differential equation dydx=1+4y2ex\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1 + 4y^{2}}{\mathrm{e}^{x}}. It is given that y=0y = 0 when x=1x = 1.
6(a) [7]
Solve the differential equation, obtaining an expression for yy in terms of xx.
6(b) [1]
State what happens to the value of yy as xx tends to infinity.
7
The equation of a curve is x3+3xy2−y3=5x^{3} + 3xy^{2} - y^{3} = 5.
7(a) [4]
Show that dydx=x2+y2y2−2xy\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x^{2} + y^{2}}{y^{2} - 2xy}.
7(b) [5]
Find the coordinates of the points on the curve where the tangent is parallel to the yy-axis.
9
Let f(x)=2+11x−10x2(1+2x)(1−2x)(2+x)\mathrm{f}(x) = \dfrac{2 + 11x - 10x^{2}}{(1 + 2x)(1 - 2x)(2 + x)}.
9(a) [5]
Express f(x)\mathrm{f}(x) in partial fractions.
9(b) [5]
Hence obtain the expansion of f(x)\mathrm{f}(x) in ascending powers of xx, up to and including the term in x2x^{2}.

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