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MHT CET 2021 Math Solved Paper

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Question : 41 of 100

Marks: +1, -0

If

\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)

$\frac{1}{2}$
$\frac{1}{4}$
$\frac{3}{2}$
$\frac{3}{4}$

Solution:

Let

u=\tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

Put

x=\tan\theta\Rightarrow\theta=\tan^{-1}x

, then

u=\tan^{-1}\left[\frac{\sqrt{1+\tan^{2}\theta}-1}{\tan\theta}\right]=\tan^{-1}\left[\frac{\sqrt{\sec^{2}\theta}-1}{\tan\theta}\right]

=\tan^{-1}\left[\frac{\sec\theta-1}{\tan\theta}\right]=\tan^{-1}\left[\frac{1-\cos\theta}{\sin\theta}\right]

=\tan^{-1}\left[\frac{2\sin^{2}\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right]=\tan^{-1}\left[\tan\frac{\theta}{2}\right]

\Rightarrow u=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x

r\cdot\tan^{-1}(\tan\theta)=\theta]

On differentiating both sides w.r.t.

x

, we get

\frac{du}{dx}=\frac{1}{2(1+x)^{2}}\left[\because\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^{2}}\right]\ldots\text{(i)}

Also, let

v=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)

Put

x=\tan\theta\Rightarrow\theta=\tan^{-1}x

, then we get

v=\sin^{-1}\left[\frac{2\tan\theta}{1+\tan^{2}\theta}\right]

\Rightarrow v=\sin^{-1}[\sin2\theta]

\Rightarrow v=2\theta\Rightarrow v=2\tan^{-1}x

\text{On differentiating both sides w.r.t. }x,\text{ we get}

\frac{dv}{dx}=\frac{2}{1+x^{2}}

\text{Now, }\frac{du}{dv}=\frac{du}{dx}\times\frac{dx}{dv}=\frac{1}{2(1-x^{2})}\times\frac{1+x^{2}}{2}

\text{[from Eqs. (i) and (ii)]}

\therefore\frac{du}{dv}=\frac{1}{4}

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