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A-Level Pure 4 · Edexcel IAL WMA14 · Full paper

Full Paper B · Pure 4 ข้อสอบ Pure 4 เต็มฉบับ · ชุด B

11 questions 11 ข้อ 75 marks 75 คะแนน 1 h 50 min 1 ชม. 50 นาที Edexcel IAL WMA14 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. 5 marks

Express $\dfrac{7x+2}{(x-1)(x+2)^2}$ in partial fractions.

Mark schemeshow ▾
M1 $\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$; $7x+2=A(x+2)^2+B(x-1)(x+2)+C(x-1)$  A1 $x=1$: $9=9A$, $A=1$  A1 $x=-2$: $-12=-3C$, $C=4$  M1 compare $x^2$: $0=A+B$  A1 $B=-1$: $\dfrac{1}{x-1}-\dfrac{1}{x+2}+\dfrac{4}{(x+2)^2}$.
Q2. 4 marks

Prove by contradiction that there are no integers $m$ and $n$ for which $4m+6n=1$.

Mark schemeshow ▾
B1 assume integers $m,n$ exist with $4m+6n=1$  M1 LHS $=2(2m+3n)$ is even  A1 but RHS $=1$ is odd — contradiction  A1 so no such integers exist ∎.
Q3. 6 marks

(a) Expand $\sqrt{4-8x}$ in ascending powers of $x$ up to and including the $x^2$ term, and state the range of validity. (b) Use $x=0.05$ to estimate $\sqrt{3.6}$, giving your answer to 4 decimal places.

Mark schemeshow ▾
(a) M1 $\sqrt{4-8x}=2(1-2x)^{1/2}$  A1 $=2\left(1-x-\tfrac{x^2}{2}\right)=2-2x-x^2$  B1 valid $|x|<\tfrac12$.
(b) M1 $2-0.1-0.0025$  A1 $=1.8975$  B1 (true value $1.89737$ — accurate to 3 s.f.).
Q4. 6 marks

A curve has parametric equations $x=2t$, $y=\dfrac4t$, $t\neq0$. (a) Find a cartesian equation of the curve. (b) Find $\dfrac{dy}{dx}$ in terms of $t$. (c) Find the equation of the tangent at the point where $t=2$.

Mark schemeshow ▾
(a) M1 $t=\tfrac{x}{2}$  A1 $y=\dfrac{8}{x}$ (i.e. $xy=8$).
(b) M1 $\dfrac{dy}{dx}=\dfrac{-4/t^2}{2}$  A1 $=-\dfrac{2}{t^2}$.
(c) M1 at $t=2$: point $(4,2)$, slope $-\tfrac12$  A1 $y-2=-\tfrac12(x-4)$, i.e. $x+2y-8=0$.
Q5. 6 marks

The circle $x^2+y^2+4x-6y=12$ passes through the point $(2,0)$. Find the gradient of the curve at this point using implicit differentiation.

Mark schemeshow ▾
B1 check: $4+0+8-0=12$ ✓  M1 $2x+2y\dfrac{dy}{dx}+4-6\dfrac{dy}{dx}=0$  A1 all terms correct  M1 $\dfrac{dy}{dx}=\dfrac{2x+4}{6-2y}$  M1 substitute $(2,0)$: $\dfrac{8}{6}$  A1 $=\dfrac43$.
Q6. 7 marks

(a) Find the exact value of $\displaystyle\int_0^1 xe^{-x}\,dx$. (b) Show that $\displaystyle\int_1^e \ln x\,dx=1$.

Mark schemeshow ▾
(a) M1 parts: $u=x$, $\frac{dv}{dx}=e^{-x}$  A1 $\big[-xe^{-x}\big]_0^1+\int_0^1 e^{-x}dx$  A1 $=-e^{-1}+\big[-e^{-x}\big]_0^1$  A1 $=1-\dfrac2e$.
(b) M1 parts: $x\ln x-x$  A1 $(e-e)-(0-1)$  A1 $=1$ ∎.
Q7. 7 marks

(a) Use the substitution $u=3x^2+1$ to find the exact value of $\displaystyle\int_0^1 x(3x^2+1)^3\,dx$. (b) The region under $y=e^{-x}$ between $x=0$ and $x=1$ is rotated $360^\circ$ about the $x$-axis. Find the exact volume generated.

Mark schemeshow ▾
(a) M1 $du=6x\,dx$; limits $u:1\to4$  A1 $\tfrac16\int_1^4 u^3du=\tfrac16\left[\tfrac{u^4}{4}\right]_1^4$  A1 $=\dfrac{256-1}{24}=\dfrac{85}{8}$.
(b) M1 $V=\pi\displaystyle\int_0^1 e^{-2x}dx$  A1 $=\pi\left[-\tfrac12e^{-2x}\right]_0^1$  A1 $=\dfrac{\pi}{2}\left(1-e^{-2}\right)$  B1 exact form.
Q8. 7 marks

Solve the differential equation $y\dfrac{dy}{dx}=2x+1$, given that $y=2$ when $x=0$. Give $y^2$ in terms of $x$, and find the exact value of $y$ when $x=1$, given $y>0$.

Mark schemeshow ▾
M1 separate: $\int y\,dy=\int(2x+1)dx$  A1 $\dfrac{y^2}{2}=x^2+x+c$  M1 $y(0)=2$: $c=2$  A1 $y^2=2x^2+2x+4$  M1 $x=1$: $y^2=8$  A1 $y=2\sqrt2$  B1 positive root chosen with reason.
Q9. 8 marks

Lines $\ell_1: \mathbf{r}=\begin{pmatrix}2\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}$ and $\ell_2: \mathbf{r}=\begin{pmatrix}5\\-2\\6\end{pmatrix}+\mu\begin{pmatrix}1\\0\\1\end{pmatrix}$. (a) Show that $\ell_1$ and $\ell_2$ intersect, and find the point of intersection. (b) Find the acute angle between the two lines.

Mark schemeshow ▾
(a) M1 equate components: $1-\lambda=-2 \Rightarrow \lambda=3$  M1 $2+3=5+\mu \Rightarrow \mu=0$  M1 check third: $2(3)=6$ and $6+0=6$ ✓  A1 intersect at $(5,-2,6)$.
(b) M1 $\mathbf{d_1}\cdot\mathbf{d_2}=1+0+2=3$  M1 $|\mathbf{d_1}|=\sqrt6$, $|\mathbf{d_2}|=\sqrt2$  A1 $\cos\theta=\dfrac{3}{\sqrt{12}}=\dfrac{\sqrt3}{2}$  A1 $\theta=30^\circ$.
Q10. 9 marks

$g(x)=\dfrac{1}{(1+x)(1+3x)}$. (a) Express $g(x)$ in partial fractions. (b) Hence find the expansion of $g(x)$ in ascending powers of $x$ up to and including the $x^2$ term, and state the range of validity.

Mark schemeshow ▾
(a) M1 $1=A(1+3x)+B(1+x)$  A1 $x=-1$: $1=-2A$, $A=-\tfrac12$  A1 $x=-\tfrac13$: $1=\tfrac23B$, $B=\tfrac32$. So $g(x)=\dfrac{3/2}{1+3x}-\dfrac{1/2}{1+x}$.
(b) M1 $\tfrac32(1-3x+9x^2)$  A1 correct  M1 $-\tfrac12(1-x+x^2)$  A1 correct  A1 $g(x)=1-4x+13x^2$  B1 valid $|x|<\tfrac13$.
Q11. 10 marks

Water drains from a tank so that $\dfrac{dV}{dt}=-k\sqrt{V}$, where $V$ m³ is the volume after $t$ minutes and $k$ is a positive constant. Initially $V=400$, and after 10 minutes $V=100$. (a) Show that $\sqrt{V}=20-\dfrac{kt}{2}$. (b) Find the value of $k$. (c) Show that $V=(20-t)^2$ for $0\le t\le20$, and find the time at which the tank is empty.

Mark schemeshow ▾
(a) M1 separate: $\displaystyle\int V^{-1/2}dV=-\int k\,dt$  A1 $2\sqrt V=-kt+c$  M1 $V(0)=400$: $c=40$  A1 $\sqrt V=20-\dfrac{kt}{2}$ ∎.
(b) M1 $V(10)=100$: $10=20-5k$  A1 $k=2$.
(c) M1 $\sqrt V=20-t$  A1 $V=(20-t)^2$, valid while $\sqrt V\ge0$  A1 empty when $V=0$: $t=20$ minutes  B1 domain $0\le t\le 20$ stated.