Edexcel A-Level Mathematics A2 Unit P4 Pure Mathematics 4 Questions

$y=\frac{2}{t(3-t)}=1 \Rightarrow 2=3 t-t^{2} \Rightarrow t=\ldots$$\left(t^{2}-3 t+2=(t-1)(t-2)=0 \Rightarrow\right) a=1, b=2$
(2)

Area under curve $=\int_{t=11}^{t=" 2 "} y \mathrm{~d} x=\int_{1}^{2} y \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t$$t=1 \Rightarrow x=3, \quad t=2 \Rightarrow x=8$
So area of $R=1 \times(8-3)-\int_{1}^{2} \frac{2}{t(3-t)} \times(2 t+2) \mathrm{d} t$$=5-4 \int_{1}^{2} \frac{t+1}{t(3-t)} \mathrm{d} t$
(5)

(ii)

$\frac{t+1}{t(3-t)}=\frac{A}{t}+\frac{B}{3-t}$$$\Rightarrow t+1=A(3-t)+B t$$

E.g. $t=0 \Rightarrow 1=3 A \Rightarrow A=\frac{1}{3} ; t=3 \Rightarrow 4=3 B \Rightarrow B=\frac{4}{3}$$\frac{t+1}{t(3-t)}=\frac{1}{3 t}+\frac{4}{3(3-t)}$

$$\int\frac{t+1}{t(3-t)}\,dt =\int\left(\frac1{3t}+\frac4{3(3-t)}\right)\,dt =\frac13\ln t-\frac43\ln(3-t)$$$$\begin{aligned} \text{Area} &=5-4\left[\frac13\ln t-\frac43\ln(3-t)\right]_1^2\\ &=5-4\left(\frac13\ln2+\frac43\ln2\right) \end{aligned}$$$$=5-\frac{20}{3}\ln2$$

$\frac{\mathrm{d} x}{\mathrm{~d} t}=\sec t \tan t$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}=\sqrt{3} \sec ^{2}\left(t+\frac{\pi}{3}\right)$$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}=\right) \frac{\mathrm{d} y}{\mathrm{~d} t} \div \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\sqrt{3} \sec ^{2}\left(t+\frac{\pi}{3}\right)}{\sec t \tan t}$ oe
(3)

(a) Alt

$x=\sec t \quad y=\sqrt{3} \frac{\tan t+\sqrt{3}}{1-\sqrt{3} \tan t}$$\frac{\mathrm{d} x}{\mathrm{~d} t}=\sec t \tan t$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}=\sqrt{3} \frac{(1-\sqrt{3} \tan t) \sec ^{2} t-(\tan t+\sqrt{3})\left(-\sqrt{3} \sec ^{2} t\right)}{(1-\sqrt{3} \tan t)^{2}}($ oe )
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} t} \div \frac{\mathrm{d} x}{\mathrm{~d} t}=\sqrt{3} \frac{(1-\sqrt{3} \tan t) \sec ^{2} t-(\tan t+\sqrt{3})\left(-\sqrt{3} \sec ^{2} t\right)}{\sec t \tan t(1-\sqrt{3} \tan t)^{2}}$ oe

$$t=\frac{\pi}{3}\Rightarrow x=2,\quad y=-3$$$$\left.\frac{dy}{dx}\right|_{t=\pi/3} =\frac{\sqrt3\sec^2(2\pi/3)}{\sec(\pi/3)\tan(\pi/3)} =2$$y+3=2(x-2)y=2x-7

$y=\sqrt{3} \frac{\tan t \pm \tan \frac{\pi}{3}}{1 \pm \tan t \tan \frac{\pi}{3}}$$x^{2}=\sec ^{2} t=1+\tan ^{2} t \Rightarrow \tan t=\sqrt{x^{2}-1} \Rightarrow y=\sqrt{3} \frac{\sqrt{x^{2}-1}+\sqrt{3}}{1-\sqrt{3} \sqrt{x^{2}-1}}$$=\sqrt{3} \frac{\sqrt{x^{2}-1}+\sqrt{3}}{1-\sqrt{3} \sqrt{x^{2}-1}} \times \frac{1+\sqrt{3} \sqrt{x^{2}-1}}{1+\sqrt{3} \sqrt{x^{2}-1}}=.$.

$=\sqrt{3} \frac{\sqrt{x^{2}-1}+\sqrt{3}\left(x^{2}-1\right)+\sqrt{3}+3 \sqrt{x^{2}-1}}{1-\left(3 x^{2}-3\right)}=\frac{. . \sqrt{x^{2}-1}+. . x^{2}}{4-3 x^{2}}$$=\frac{3 x^{2}+4 \sqrt{3 x^{2}-3}}{4-3 x^{2}}$
(5)