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Learn/ A Level Mathematics/ Past Papers/ 2020/ Oct/Nov 2020 /Paper 1 (Variant 3)

Pure Mathematics 1 — Variant 3

A Level Mathematics · 9709/13 · Oct/Nov 2020

Question Paper

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Mark Scheme

PDF · 291 KB · 14 pages

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Questions

9 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
1(a) [2]
Express x2+6x+5x^{2} + 6x + 5 in the form (x+a)2+b(x + a)^{2} + b, where aa and bb are constants.
1(b) [2]
The curve with equation y=x2y = x^{2} is transformed to the curve with equation y=x2+6x+5y = x^{2} + 6x + 5. Describe fully the transformation(s) involved.
2
The function f is defined by f(x)=2(x+2)2\mathrm{f}(x) = \frac{2}{(x + 2)^{2}} for x>−2x > -2.
2(a) [3]
Find ∫1∞f(x) dx\int_{1}^{\infty} \mathrm{f}(x) \, \mathrm{d}x.
2(b) [2]
The equation of a curve is such that dydx=f(x)\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{f}(x). It is given that the point (−1,−1)(-1, -1) lies on the curve. Find the equation of the curve.
3 [5]
Solve the equation 3tan⁡2θ+1=2tan⁡2θ3\tan^{2}\theta + 1 = \frac{2}{\tan^{2}\theta} for 0∘<θ<180∘0^\circ < \theta < 180^\circ.
4 [5]
A curve has equation y=3x2−4x+4y = 3x^{2} - 4x + 4 and a straight line has equation y=mx+m−1y = mx + m - 1, where mm is a constant. Find the set of values of mm for which the curve and the line have two distinct points of intersection.
5 [5]
In the expansion of (a+bx)7(a + bx)^{7}, where aa and bb are non-zero constants, the coefficients of xx, x2x^{2} and x4x^{4} are the first, second and third terms respectively of a geometric progression. Find the value of ab\frac{a}{b}.
6
The function f is defined by f(x)=2x3x−1\mathrm{f}(x) = \frac{2x}{3x - 1} for x>13x > \frac{1}{3}.
6(a) [3]
Find an expression for f−1(x)\mathrm{f}^{-1}(x).
6(b) [2]
Show that 23+23(3x−1)\frac{2}{3} + \frac{2}{3(3x - 1)} can be expressed as 2x3x−1\frac{2x}{3x - 1}.
6(c) [1]
State the range of f.
7
The first and second terms of an arithmetic progression are 1cos⁡2θ\frac{1}{\cos^{2}\theta} and −tan⁡2θcos⁡2θ-\frac{\tan^{2}\theta}{\cos^{2}\theta}, respectively, where 0<θ<12π0 < \theta < \frac{1}{2}\pi.
7(a) [4]
Show that the common difference is −1cos⁡4θ-\frac{1}{\cos^{4}\theta}.
7(b) [3]
Find the exact value of the 13th term when θ=16π\theta = \frac{1}{6}\pi.
8
The equation of a curve is y=2x+1+12x+1y = 2x + 1 + \frac{1}{2x + 1} for x>−12x > -\frac{1}{2}.
8(a) [3]
Find dydx\frac{\mathrm{d}y}{\mathrm{d}x} and d2ydx2\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}.
8(b) [5]
Find the coordinates of the stationary point and determine the nature of the stationary point.
10
A curve has equation y=1kx12+x−12+1k2y = \frac{1}{k}x^{\frac{1}{2}} + x^{-\frac{1}{2}} + \frac{1}{k^{2}} where x>0x > 0 and kk is a positive constant.
10(a) [4]
It is given that when x=14x = \frac{1}{4}, the gradient of the curve is 33. Find the value of kk.
10(b) [5]
It is given instead that ∫14k2k2(1kx12+x−12+1k2)dx=1312\int_{\frac{1}{4}k^{2}}^{k^{2}} \left( \frac{1}{k}x^{\frac{1}{2}} + x^{-\frac{1}{2}} + \frac{1}{k^{2}} \right) \mathrm{d}x = \frac{13}{12}. Find the value of kk.

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