Worked solutions · Pure 2 Paper A เฉลยละเอียด ข้อสอบ Pure 2 ชุด A
Every question worked end to end, with the mark codes showing exactly where each mark is earned. Use it after sitting the paper — not instead of sitting it. เฉลยครบทุกขั้นพร้อมจุดให้คะแนน M1/A1 — ใช้หลังลองทำข้อสอบเอง ไม่ใช่แทนการทำ
Find, in ascending powers of $x$, the first three terms of the expansion of $(3-2x)^5$.
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$f(x)=4x^3-8x^2-x+2$. (a) Show that $(2x-1)$ is a factor of $f(x)$. (b) Hence factorise $f(x)$ completely.
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(b) M1 divide: $f(x)=(2x-1)(2x^2-3x-2)$ A1 quotient correct A1 $f(x)=(2x-1)(2x+1)(x-2)$.
(a) Solve $\log_5(x+4)+\log_5 x=1$. (b) Solve $2^{3x-1}=10$, giving your answer to 3 significant figures.
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(b) M1 $(3x-1)\ln 2=\ln 10$ A1 $x=\tfrac13\!\left(1+\dfrac{\ln 10}{\ln 2}\right)$ A1 $=1.44$ (3 s.f.).
The circle $C$ has equation $x^2+y^2-8x+2y+8=0$. (a) Find the centre and radius of $C$. (b) Show that the point $P(8,2)$ lies outside $C$, and find the length of a tangent from $P$ to $C$.
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(b) M1 $CP^2=(8-4)^2+(2+1)^2=25$, $CP=5>3$ → outside M1 tangent$^2=CP^2-r^2=25-9$ A1 tangent $=4$.
A geometric series has first term 18 and sum to infinity 27. (a) Find the common ratio. (b) Write down the third term. (c) Find $S_5$, correct to 3 significant figures.
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(b) B1 $u_3=18\left(\tfrac13\right)^2=2$.
(c) M1 $S_5=\dfrac{18\left(1-(1/3)^5\right)}{1-\tfrac13}=27\left(1-\tfrac{1}{243}\right)$ A1 $=\dfrac{242\cdot 27}{243}=26.888\ldots$ A1 $=26.9$ (3 s.f.).
A sector of a circle has perimeter 20 cm and area 24 cm². Find the two possible pairs of values of the radius $r$ and the angle $\theta$ (in radians).
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Solve, for $0\le\theta<360^\circ$: (a) $2\sin\theta=5\cos\theta$, to 1 decimal place; (b) $4\sin^2\theta+8\cos\theta=7$, giving exact answers.
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(b) M1 $4(1-\cos^2\theta)+8\cos\theta=7$ A1 $4\cos^2\theta-8\cos\theta+3=0$ M1 $(2\cos\theta-1)(2\cos\theta-3)=0$, $\cos\theta=\tfrac32$ rejected A1 $\cos\theta=\tfrac12:\ \theta=60^\circ,\ 300^\circ$.
The curve $y=x^3-4x^2+4x$ has two stationary points. Find their coordinates and determine the nature of each.
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The number of users of an app is modelled by $N=200e^{0.15t}$, where $t$ is the time in years since launch. (a) Write down the number of users at launch. (b) Find the time for the user base to reach 500, to 3 s.f. (c) Show that the model predicts growth of about 16.2% per year. (d) Find when $N$ reaches 1000, and give one reason the model may be unrealistic for large $t$.
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(b) M1 $e^{0.15t}=2.5 \Rightarrow t=\dfrac{\ln 2.5}{0.15}$ A1 $=6.11$ years.
(c) M1 yearly factor $e^{0.15}=1.1618\ldots$ A1 $\approx 16.2\%$ per year ∎.
(d) M1 $e^{0.15t}=5$ A1 $t=\dfrac{\ln 5}{0.15}=10.7$ years B1 e.g. exponential growth is unbounded — real user growth slows as the market saturates.
(a) Use the trapezium rule with 4 strips to estimate $\displaystyle\int_0^2\sqrt{4+x^3}\,dx$, giving your answer to 3 decimal places. (b) State one way to improve the accuracy of the estimate. (c) Find the exact value of $\displaystyle\int_1^9\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx$.
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(b) B1 use more strips (smaller $h$).
(c) M1 $\int\left(x^{1/2}+x^{-1/2}\right)dx=\tfrac23x^{3/2}+2x^{1/2}$ A1 $(18+6)-\left(\tfrac23+2\right)$ A1 $=\dfrac{64}{3}$.
The curve $y=x^3-6x^2+9x$ and the line $y=x$ intersect at three points. (a) Show that the $x$-coordinates of the intersections are $x=0$, $x=2$ and $x=4$. (b) Find the total area of the two regions enclosed between the curve and the line.
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(b) M1 $\displaystyle\int(x^3-6x^2+8x)\,dx$ A1 $=\dfrac{x^4}{4}-2x^3+4x^2$ A1 $\big[\cdot\big]_0^2=4-16+16=4$ M1 $\big[\cdot\big]_2^4=(64-128+64)-4$ A1 $=-4$, so area $=4$ M1 total $=4+|-4|$ A1 $=8$ square units.