Predicted Paper · Pure 2 mixed mock ข้อสอบรวม Pure 2 · ชุดจำลอง
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
Express $2\log_3 x+\log_3 4-\log_3 2x$ as a single logarithm, and hence solve $2\log_3 x+\log_3 4-\log_3 2x=2$.
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M1 $\log_3 2x=2 \Rightarrow 2x=9$ A1 $x=\dfrac92$.
$f(x)=6x^3+x^2-19x+6$. Show that $(x+2)$ is a factor of $f(x)$ and factorise $f(x)$ completely.
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M1 $f(x)=(x+2)(6x^2-11x+3)$ M1 factorise quotient A1 $(x+2)(3x-1)(2x-3)$.
The circle $C$ has equation $x^2+y^2-4x+6y-3=0$. (a) Find the centre and radius of $C$. (b) The point $P(5,2)$ lies outside $C$. Find the length of a tangent from $P$ to $C$.
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(b) M1 $CP^2=(5-2)^2+(2+3)^2=34$ M1 tangent² $=CP^2-r^2=34-16=18$ A1 $=3\sqrt2$.
(a) Find the first three terms, in ascending powers of $x$, of the expansion of $(1+2x)^{10}$. (b) A geometric series has first term 27 and common ratio $\tfrac13$. Find its sum to infinity, and the least $n$ for which $S_\infty-S_n<0.01$.
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(b) A1 $S_\infty=\dfrac{27}{1-\frac13}=40.5$. M1 $S_\infty-S_n=\dfrac{a r^n}{1-r}=40.5\left(\tfrac13\right)^n<0.01$ M1 $n>\dfrac{\ln 4050}{\ln 3}=7.56\ldots$ A1 $n=8$.
The diagram shows a sector $OAB$ of a circle, centre $O$, radius 10 cm, with $\angle AOB=0.8$ rad. (a) Find the area of the sector. (b) Find the area of the segment cut off by the chord $AB$. (c) Find the perimeter of the segment.
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(b) M1 triangle $=\tfrac12(100)\sin 0.8=35.867\ldots$ A1 segment $=4.13$ cm² (3 s.f.).
(c) M1 arc $=8$, chord $=20\sin 0.4=7.788\ldots$ A1 perimeter $=15.8$ cm (3 s.f.).
Solve, for $0\le\theta<360^\circ$: (a) $5\sin\theta=4\cos\theta$ (1 d.p.); (b) $2\cos^2\theta+\sin\theta=1$ (exact).
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(b) M1 $2(1-\sin^2\theta)+\sin\theta=1 \Rightarrow 2\sin^2\theta-\sin\theta-1=0$ M1 $(2\sin\theta+1)(\sin\theta-1)=0$ A1 $\sin\theta=1: \theta=90^\circ$ A1 $\sin\theta=-\tfrac12: \theta=210^\circ, 330^\circ$.
Solve (a) $3^{2x+1}=17$ (3 s.f.); (b) $\log_2(x+7)-\log_2 x=3$.
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(b) M1 $\log_2\dfrac{x+7}{x}=3 \Rightarrow \dfrac{x+7}{x}=8$ A1 $x=1$ B1 check $x>0$ valid.
The curve $y=x^3-3x^2-9x+11$ has two stationary points. (a) Find their coordinates and determine their nature. (b) State the range of $x$ for which $y$ is increasing.
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(b) M1 $y'>0$ A2 $x<-1$ or $x>3$.
The region $R$ is bounded by the curve $y=x^2-4x+5$ and the line $y=2x-3$. (a) Find the $x$-coordinates of the points of intersection. (b) Find the exact area of $R$.
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(b) M1 $\displaystyle\int_2^4\big[(2x-3)-(x^2-4x+5)\big]dx=\int_2^4(-x^2+6x-8)dx$ A1 $=\left[-\dfrac{x^3}{3}+3x^2-8x\right]_2^4$ M1 $=\left(-\dfrac{64}{3}+48-32\right)-\left(-\dfrac{8}{3}+12-16\right)$ A1 $=-\dfrac{16}{3}-\left(-\dfrac{20}{3}\right)$ A1 $=\dfrac{4}{3}$.
B1 exact form quoted.
A closed cylindrical can has volume $330$ cm³. (a) Show that its surface area is $S=2\pi r^2+\dfrac{660}{r}$. (b) Find, to 3 s.f., the radius that minimises $S$, and verify it is a minimum. (c) Use the trapezium rule with 2 strips to estimate $\displaystyle\int_1^3 \dfrac{660}{r}\,dr$ (1 d.p.), and state whether it over- or under-estimates.
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(b) M1 $S'=4\pi r-\dfrac{660}{r^2}=0 \Rightarrow r^3=\dfrac{660}{4\pi}=52.52\ldots$ A1 $r=3.74$ cm (3 s.f.) A1 $S''=4\pi+\dfrac{1320}{r^3}>0$ → minimum.
(c) M1 $h=1$; values $660, 330, 220$ A1 $\tfrac12[660+220+2(330)]=770.0$ B1 curve is convex → overestimate.