Question

If $y = \sec^{-1} \frac{2x}{1+x^2} + \sin^{-1} \frac{x -1}{x+1} $ ,then $ \frac{dy}{dx}$ is equal to :

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

Answer 1

Answer: (C)

Let $ y = \sec^{-1} \left(\frac{2x}{ 1+x^2}\right) + \sin^{-1} \left(\frac{x - 1}{ x + 1}\right) $
Put $x = \tan \:\, \theta$ , we get
$\frac{2x}{ 1+x^2} = \frac{2 \:\tan \theta}{1 + \tan^2 \theta} = \sin \: 2 \theta$
Since, $ - 1 \leq \: \sin \theta \leq 1$
$ \therefore \:\: - 1 \leq \frac{2x}{1 + x^2} \leq 1$
$ \Rightarrow \:\: \sec^{-1} \left(\frac{2x}{ 1+x^2} \right)$ is defined only at
$x = 1, - 1$
$ \therefore \:\: \frac{dy}{dx}$ does not exist.