Solve the Following Question.(4 Marks)

Question 514 Marks

Differentiate the following w.r.t. x:

$\sqrt[3]{\frac{4 x-1}{(2 x+3)(5-2 x)^2}}$

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Question 524 Marks

Differentiate the following w.r.t. x:

$\frac{(x+1)^2}{(x+2)^3(x+3)^4}$

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Question 534 Marks

Diffrentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right)$

Answer

$\text { Let } y=\cot ^{-1}\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right]$
$1+\sin \left(\frac{4 x}{3}\right)=1+\cos \left(\frac{\pi}{2}-\frac{4 x}{3}\right)=2 \cos ^2\left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
$\therefore \sqrt{1+\sin \left(\frac{4 x}{3}\right)}=\sqrt{2} \cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
Also, $1-\sin \left(\frac{4 x}{3}\right)=1-\cos \left(\frac{\pi}{2}-\frac{4 x}{3}\right)=2 \sin ^2\left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
$\therefore \sqrt{1-\sin \left(\frac{4 x}{3}\right)}=\sqrt{2} \sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)$
$\therefore \frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}$
$=\frac{\sqrt{2} \cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)+\sqrt{2} \sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{\sqrt{2} \cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)-\sqrt{2} \sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}$
$=\frac{\cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)+\sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{\cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)-\sin \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}$
$=\frac{1+\tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{1-\tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)} \quad \cdots\left[\right.$ Dividing by $\left.\cos \left(\frac{\pi}{4}-\frac{2 x}{3}\right)\right]$
$=\frac{\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)}{1-\tan \frac{\pi}{4} \cdot \tan \left(\frac{\pi}{4}-\frac{2 x}{3}\right)} \quad \cdots\left[\because \tan \frac{\pi}{4}=1\right]$
$=\tan \left[\frac{\pi}{4}+\frac{\pi}{4}-\frac{2 x}{3}\right]=\tan \left(\frac{\pi}{2}-\frac{2 x}{3}\right)$
$=\cot \left(\frac{2 x}{3}\right)$
$\therefore y=\cot ^{-1}\left[\cot \left(\frac{2 x}{3}\right)\right]=\frac{2 x}{3}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x} =\frac{d}{d x}\left(\frac{2 x}{3}\right)=\frac{2}{3} \frac{d}{d x}(x)$
$=\frac{2}{3} \times 1=\frac{2}{3} .$

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Question 544 Marks

Find the x co-ordinates of all the points on the curve y = sin 2x – 2 sin x, 0 ≤ x < 2π where

$\frac{d y}{d x}=0$

Answer

$\begin{aligned} & y=\sin 2 x-2 \sin x, 0 \leq x<2 \pi \\ & \begin{aligned} \therefore \frac{d y}{d x} & =\frac{d}{d x}(\sin 2 x-2 \sin x) \\ & =\frac{d}{d x}(\sin 2 x)-2 \frac{d}{d x}(\sin x) \\ & =\cos 2 x \cdot \frac{d}{d x}(2 x)-2 \cos x\end{aligned}\end{aligned}$

$\begin{aligned} & =\cos 2 x \times 2-2 \cos x \\ & =2\left(2 \cos ^2 x-1\right)-2 \cos x \\ & =4 \cos ^2 x-2-2 \cos x \\ & =4 \cos ^2 x-2 \cos x-2\end{aligned}$

$\begin{aligned} & \text { If } \frac{d y}{d x}=0 \text {, then } 4 \cos ^2 x-2 \cos x-2=0 \\ & \therefore 4 \cos ^2 x-4 \cos x+2 \cos x-2=0 \\ & \therefore 4 \cos x(\cos x-1)+2(\cos x-1)=0 \\ & \therefore(\cos x-1)(4 \cos x+2)=0 \\ & \therefore \cos x-1=0 \text { or } 4 \cos x+2=0 \\ & \therefore \cos x=1 \text { or } \cos x=-\frac{1}{2} \\ & \therefore \cos x=\cos 0\end{aligned}$

$\begin{aligned} & \text { or } \cos x=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\frac{\cos 2 \pi}{3} \\ & \text { or } \cos x=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{4 \pi}{3}\end{aligned}$

$\ldots[\because 0 \leqslant x<2 \pi]$

$\begin{aligned} & \therefore x=0 \text { or } x=\frac{2 \pi}{3} \text { or } x=\frac{4 \pi}{3} \text {. } \\ & \therefore x=0 \text { or } \frac{2 \pi}{3} \text { or } \frac{4 \pi}{3} .\end{aligned}$

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Maths Solve the Following Question.(4 Marks)