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Learn/ A Level Mathematics/ Past Papers/ 2021/ May/June 2021 /Paper 1 (Variant 2)

Pure Mathematics 1 — Variant 2

A Level Mathematics · 9709/12 · May/June 2021

Question Paper

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Mark Scheme

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Questions

10 of 12 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 16x2−24x+1016x^{2} - 24x + 10 in the form (4x+a)2+b(4x + a)^{2} + b.
1(b) [2]
It is given that the equation 16x2−24x+10=k16x^{2} - 24x + 10 = k, where kk is a constant, has exactly one root. Find the value of this root.
2
2(a) [3]
The graph of y=f(x)y = \mathrm{f}(x) is transformed to the graph of y=2f(x−1)y = 2\mathrm{f}(x - 1). Describe fully the two single transformations which have been combined to give the resulting transformation.
2(b) [2]
The curve y=sin⁡2x−5xy = \sin 2x - 5x is reflected in the yy-axis and then stretched by scale factor 13\frac{1}{3} in the xx-direction. Write down the equation of the transformed curve.
3
The equation of a curve is y=(x−3)x+1+3y = (x - 3)\sqrt{x + 1} + 3. The following points lie on the curve. Non-exact values are rounded to 4 decimal places. A(2,k)A(2, k) B(2.9,2.8025)B(2.9, 2.8025) C(2.99,2.9800)C(2.99, 2.9800) D(2.999,2.9980)D(2.999, 2.9980) E(3,3)E(3, 3)
3(a) [1]
Find kk, giving your answer correct to 4 decimal places.
3(b) [1]
Find the gradient of AEAE, giving your answer correct to 4 decimal places. The gradients of BEBE, CECE and DEDE, rounded to 4 decimal places, are 1.97481.9748, 1.99751.9975 and 1.99971.9997 respectively.
3(c) [2]
State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at the point EE.
4 [5]
The coefficient of xx in the expansion of (4x+10x)3\left( 4x + \frac{10}{x} \right)^{3} is pp. The coefficient of 1x\frac{1}{x} in the expansion of (2x+kx2)5\left( 2x + \frac{k}{x^{2}} \right)^{5} is qq. Given that p=6qp = 6q, find the possible values of kk.
5
The function f\mathrm{f} is defined by f(x)=2x2+3\mathrm{f}(x) = 2x^{2} + 3 for x⩾0x \geqslant 0.
5(a) [2]
Find and simplify an expression for ff(x)\mathrm{ff}(x).
5(b) [4]
Solve the equation ff(x)=34x2+19\mathrm{ff}(x) = 34x^{2} + 19.
6 [4]
Points AA and BB have coordinates (8,3)(8, 3) and (p,q)(p, q) respectively. The equation of the perpendicular bisector of ABAB is y=−2x+4y = -2x + 4. Find the values of pp and qq.
7
The point AA has coordinates (1,5)(1, 5) and the line ll has gradient −23-\frac{2}{3} and passes through AA. A circle has centre (5,11)(5, 11) and radius 52\sqrt{52}.
7(a) [2]
Show that ll is the tangent to the circle at AA.
7(b) [3]
Find the equation of the other circle of radius 52\sqrt{52} for which ll is also the tangent at AA.
8
The first, second and third terms of an arithmetic progression are aa, 32a\frac{3}{2}a and bb respectively, where aa and bb are positive constants. The first, second and third terms of a geometric progression are aa, 1818 and b+3b + 3 respectively.
8(a) [5]
Find the values of aa and bb.
8(b) [3]
Find the sum of the first 20 terms of the arithmetic progression.
10
10(a) [4]
Prove the identity 1+sin⁡x1−sin⁡x−1−sin⁡x1+sin⁡x≡4tan⁡xcos⁡x\dfrac{1 + \sin x}{1 - \sin x} - \dfrac{1 - \sin x}{1 + \sin x} \equiv \dfrac{4 \tan x}{\cos x}.
10(b) [3]
Hence solve the equation 1+sin⁡x1−sin⁡x−1−sin⁡x1+sin⁡x=8tan⁡x\dfrac{1 + \sin x}{1 - \sin x} - \dfrac{1 - \sin x}{1 + \sin x} = 8 \tan x for 0⩽x⩽12π0 \leqslant x \leqslant \frac{1}{2}\pi.
11
The gradient of a curve is given by dydx=6(3x−5)3−kx2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6(3x - 5)^{3} - kx^{2}, where kk is a constant. The curve has a stationary point at (2,−3.5)(2, -3.5).
11(a) [2]
Find the value of kk.
11(b) [4]
Find the equation of the curve.
11(c) [2]
Find d2ydx2\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}.
11(d) [2]
Determine the nature of the stationary point at (2,−3.5)(2, -3.5).

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