Question

$\mathrm{If}\mathrm{y}={\sec }^{-1}\left(\frac{\left(2\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}\right)+{\sin }^{-1}\frac{[\mathrm{x}-1]}{[\mathrm{x}+1]},\mathrm{then}\frac{\mathrm{dy}}{\mathrm{dx}}=$

• A. $1$
• B. $\frac{x+1}{x-1}$
• C. does not exist
• D. none of these

Answer 1

The correct option is C

does not exist

Explanation for the correct option:

Step : Find the value of $\frac{dy}{dx}$

Given that $\mathrm{y}={\sec }^{-1}\left(\frac{\left(2\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}\right)+{\sin }^{-1}\frac{[\mathrm{x}-1]}{[\mathrm{x}+1]}$

Differentiation by substitution

$$\begin{gathered} \mathrm{since}\mathrm{y}={\sec }^{-1}\left(\frac{\left(2\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}\right)+{\sin }^{-1}\frac{[\mathrm{x}-1]}{[\mathrm{x}+1]} \\ \mathrm{substitute}\mathrm{x}=\mathrm{tan\theta }\mathrm{then}\frac{\left(2\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}=\frac{2\mathrm{tan\theta }}{(1+{\tan }^2\mathrm{\theta })} \\ =\sin 2\mathrm{\theta } \\ \mathrm{Since}-1\le \mathrm{sin\theta }\le 1\Rightarrow -1\le \frac{\left(2\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}\le 1 \\ \mathrm{But}{\sec }^{-1}(\frac{\left(2\mathrm{x}\right)}{(1+{\mathrm{x}}^2)}\mathrm{is}\mathrm{defined}\mathrm{only}\mathrm{at}\mathrm{x}=-1\mathrm{and}1 \\ \mathrm{So}\mathrm{solution}\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$

Hence, option (C) is correct.