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Learn/ A Level Mathematics/ Past Papers/ 2020/ Feb/March 2020 /Paper 1 (Variant 2)

Pure Mathematics 1 — Variant 2

A Level Mathematics · 9709/12 · Feb/March 2020

Question Paper

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Mark Scheme

PDF · 319 KB · 14 pages

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Questions

9 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 [3]
The function ff is defined by f(x)=13x+2+x2f(x) = \frac{1}{3x + 2} + x^{2} for x<−1x < -1. Determine whether ff is an increasing function, a decreasing function or neither.
2 [4]
The graph of y=f(x)y = f(x) is transformed to the graph of y=1+f(12x)y = 1 + f(\frac{1}{2}x). Describe fully the two single transformations which have been combined to give the resulting transformation.
4 [4]
A curve has equation y=x2−2x−3y = x^{2} - 2x - 3. A point is moving along the curve in such a way that at PP the yy-coordinate is increasing at 44 units per second and the xx-coordinate is increasing at 66 units per second. Find the xx-coordinate of PP.
5 [5]
Solve the equation tan⁡θ+3sin⁡θ+2tan⁡θ−3sin⁡θ+1=2\frac{\tan \theta + 3\sin \theta + 2}{\tan \theta - 3\sin \theta + 1} = 2 for 0∘⩽θ⩽90∘0^\circ \leqslant \theta \leqslant 90^\circ.
6
The coefficient of 1x\frac{1}{x} in the expansion of (2x+ax2)5(2x + \frac{a}{x^{2}})^{5} is 720720.
6(a) [3]
Find the possible values of the constant aa.
6(b) [2]
Hence find the coefficient of 1x7\frac{1}{x^{7}} in the expansion.
8
A woman's basic salary for her first year with a particular company is $30000\$30 000 and at the end of the year she also gets a bonus of $600\$600.
8(a) [1]
For her first year, express her bonus as a percentage of her basic salary. At the end of each complete year, the woman's basic salary will increase by 3%3\% and her bonus will increase by $100\$100.
8(b) [5]
Express the bonus she will be paid at the end of her 24th year as a percentage of the basic salary paid during that year.
9
9(a) [2]
Express 2x2+12x+112x^{2} + 12x + 11 in the form 2(x+a)2+b2(x + a)^{2} + b, where aa and bb are constants. The function ff is defined by f(x)=2x2+12x+11f(x) = 2x^{2} + 12x + 11 for x⩽−4x \leqslant -4.
9(b) [3]
Find an expression for f−1(x)f^{-1}(x) and state the domain of f−1f^{-1}. The function gg is defined by g(x)=2x−3g(x) = 2x - 3 for x⩽kx \leqslant k.
9(c) [2]
For the case where k=−1k = -1, solve the equation fg(x)=193fg(x) = 193.
9(d) [1]
State the largest value of kk possible for the composition fgfg to be defined.
10
The gradient of a curve at the point (x,y)(x, y) is given by dydx=2(x+3)12−x\frac{\mathrm{d}y}{\mathrm{d}x} = 2(x + 3)^{\frac{1}{2}} - x. The curve has a stationary point at (a,14)(a, 14), where aa is a positive constant.
10(a) [3]
Find the value of aa.
10(b) [3]
Determine the nature of the stationary point.
10(c) [4]
Find the equation of the curve.
11
11(a) [4]
Solve the equation 3tan⁡2x−5tan⁡x−2=03\tan^{2}x - 5\tan x - 2 = 0 for 0∘⩽x⩽180∘0^\circ \leqslant x \leqslant 180^\circ.
11(b) [2]
Find the set of values of kk for which the equation 3tan⁡2x−5tan⁡x+k=03\tan^{2}x - 5\tan x + k = 0 has no solutions.
11(c) [3]
For the equation 3tan⁡2x−5tan⁡x+k=03\tan^{2}x - 5\tan x + k = 0, state the value of kk for which there are three solutions in the interval 0∘⩽x⩽180∘0^\circ \leqslant x \leqslant 180^\circ, and find these solutions.

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