Q75 / 150 · MHT CET 2020 (14 Oct Shift 1) · Mathematics
If $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$, then $\frac{d u}{d v}$ at $x=0$ is
- Correct answerA$\frac{1}{4}$
- B$\frac{1}{8}$
- C1
- D$\frac{-1}{8}$
Solution
Given $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$
Put $x=\tan \theta$
$u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \quad=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos }\right)$
$\therefore u=\frac{\tan ^{-1} x}{2} \Rightarrow \frac{d u}{d x}=\frac{1}{2\left(1+x^{2}\right)}$
We have, $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$
Put $x=\sin \theta$
$\therefore v=\tan ^{-1}\left(\frac{2 \sin \theta \cdot \cos \theta}{\cos 2 \theta}\right)=\tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right)=\tan ^{-1}(\tan 2 \theta)=2 \theta$
$\therefore v=2 \tan ^{-1} x \Rightarrow \frac{d v}{d x}=\frac{2}{1+x^{2}}$
$\left.\frac{d u}{d v}=\frac{d u}{d x}\right)=\frac{1}{2\left(1+x^{2}\right)} \times \frac{\left(1+x^{2}\right)}{2}=\frac{1}{4}$
Hence, the correct answer is option: 'A'.