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A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Full paper

Full Paper A · Pure 2 ข้อสอบ Pure 2 เต็มฉบับ · ชุด A

11 questions 11 ข้อ 75 marks 75 คะแนน 1 h 50 min 1 ชม. 50 นาที Edexcel IAL WMA12 format

Sit it like the real thing: timer on, no notes, calculator allowed. Open the mark schemes only when you finish, then self-mark honestly. ทำเหมือนสอบจริง: เปิดตัวจับเวลา ไม่เปิดโน้ต ใช้เครื่องคิดเลขได้ ทำเสร็จค่อยเปิดมาร์คสกีมแล้วให้คะแนนตัวเองตามจริง

⏱ 110:00 Scoreคะแนนรวม: –
Q1. 4 marks

Find, in ascending powers of $x$, the first three terms of the expansion of $(3-2x)^5$.

Mark schemeshow ▾
M1 binomial: $3^5+\binom51 3^4(-2x)+\binom52 3^3(-2x)^2$  A1 $243$  A1 $-810x$  A1 $+1080x^2$.
Q2. 5 marks

$f(x)=4x^3-8x^2-x+2$. (a) Show that $(2x-1)$ is a factor of $f(x)$. (b) Hence factorise $f(x)$ completely.

Mark schemeshow ▾
(a) M1 $f\!\left(\tfrac12\right)=\tfrac12-2-\tfrac12+2$  A1 $=0$, so $(2x-1)$ is a factor (conclusion stated).
(b) M1 divide: $f(x)=(2x-1)(2x^2-3x-2)$  A1 quotient correct  A1 $f(x)=(2x-1)(2x+1)(x-2)$.
Q3. 6 marks

(a) Solve $\log_5(x+4)+\log_5 x=1$. (b) Solve $2^{3x-1}=10$, giving your answer to 3 significant figures.

Mark schemeshow ▾
(a) M1 $\log_5\big(x(x+4)\big)=1 \Rightarrow x^2+4x=5$  A1 $(x+5)(x-1)=0$  A1 $x=1$ only ($x=-5$ rejected: $\log_5 x$ undefined).
(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.).
Q4. 6 marks

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 $C$.

Mark schemeshow ▾
(a) M1 $(x-4)^2+(y+1)^2=9$  A1 centre $(4,-1)$  A1 $r=3$.
(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 $=4$.
Q5. 6 marks

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.

Mark schemeshow ▾
(a) M1 $\dfrac{18}{1-r}=27$  A1 $r=\tfrac13$.
(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.).
Q6. 7 marks

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).

Mark schemeshow ▾
M1 $2r+r\theta=20$  M1 $\tfrac12 r^2\theta=24$  M1 eliminate: $\theta=\dfrac{20-2r}{r}$, so $\tfrac12 r(20-2r)=24$  A1 $r^2-10r+24=0$  A1 $r=4$ or $r=6$  A1 $r=4:\ \theta=3$  A1 $r=6:\ \theta=\tfrac43$.
Q7. 7 marks

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.

Mark schemeshow ▾
(a) M1 $\tan\theta=2.5$  A1 $\theta=68.2^\circ$  A1 $248.2^\circ$.
(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 $\cos\theta=\tfrac12:\ \theta=60^\circ,\ 300^\circ$.
Q8. 7 marks

The curve $y=x^3-4x^2+4x$ has two stationary points. Find their coordinates and determine the nature of each.

Mark schemeshow ▾
M1 $y'=3x^2-8x+4$  A1 correct  M1 $(3x-2)(x-2)=0$  A1 $x=\tfrac23,\ 2$  A1 points $\left(\tfrac23,\tfrac{32}{27}\right)$ and $(2,0)$  M1 $y''=6x-8$  A1 at $x=\tfrac23$: $y''=-4<0$ maximum; at $x=2$: $y''=4>0$ minimum.
Q9. 8 marks

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 $t$.

Mark schemeshow ▾
(a) B1 $200$.
(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.
Q10. 9 marks

(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 $\displaystyle\int_1^9\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx$.

Mark schemeshow ▾
(a) B2 $h=0.5$; $y$-values $2,\ 2.031,\ 2.236,\ 2.716,\ 3.464$ (3 d.p.)  M1 $\tfrac{0.5}{2}\big[2+3.464+2(2.031+2.236+2.716)\big]$  A1 $=\tfrac14(5.464+13.966)$  A1 $=4.858$ (3 d.p.).
(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 $=\dfrac{64}{3}$.
Q11. 10 marks

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.

Mark schemeshow ▾
(a) M1 $x^3-6x^2+9x=x \Rightarrow x^3-6x^2+8x=0$  A1 $x(x-2)(x-4)=0$  A1 $x=0,2,4$ ∎.
(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.