Choose the correct alternative in the following

Solution:

Let $u=\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$ and $v=\sqrt{1+3 x}$

$\left(\frac{d u}{d v}\right)_{x=-\frac{1}{3}}=?$

Considering $u$,

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

Put $x=\cos \theta$

$\theta=\cos ^{-1} x \cdots(1)$

$u=\sec ^{-1}\left(\frac{1}{2 \cos ^{2} \theta+1}\right)=\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right) \because 2 \cos ^{2} \theta+1=\cos 2 \theta$

$=\sec ^{-1}(\sec 2 \theta)=2 \theta$

$\Rightarrow u=2 \cos ^{-1} x$ From (1)

Differentiating w.r.t $x$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=-\frac{2}{\sqrt{1-\mathrm{x}^{2}}}$

Considering $\mathrm{V}$,

$\mathrm{v}=\sqrt{1+3 \mathrm{x}}$

Differentiating w.r.t $x$

$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{3}{2 \sqrt{1+3 \mathrm{x}}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\mathrm{dv}}=\frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dv}}$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=-\frac{2}{\sqrt{1-\mathrm{x}^{2}}} \cdot\left(\frac{2 \sqrt{1+3 \mathrm{x}}}{3}\right)$

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dv}}=-\frac{4}{3} \cdot\left(\sqrt{\frac{1+3 \mathrm{x}}{1-\mathrm{x}^{2}}}\right)$

$\Rightarrow\left(\frac{\mathrm{du}}{\mathrm{dv}}\right)_{\mathrm{x}=-\frac{1}{3}}=-\frac{4}{3} \cdot\left(\sqrt{\frac{1+3\left(-\frac{1}{3}\right)}{1-\left(-\frac{1}{3}\right)^{2}}}\right)$

$\Rightarrow\left(\frac{\mathrm{du}}{\mathrm{dv}}\right)_{\mathrm{x}=-\frac{1}{3}}=0=(\mathrm{B})$