Q169 / 240 · KCET 2026 · Mathematics
If $f(x) = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$, then $f'\left(\dfrac{1}{2}\right) = $
- Correct answerA$\dfrac{8}{5}$
- B$\dfrac{5}{8}$
- C$\dfrac{4}{5}$
- D$0$
Solution
Given $f(x) = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$
Let $x = \tan \theta$, then $\theta = \tan^{-1}x$.
Substituting $x = \tan \theta$ in the given function:
$f(x) = \sin^{-1}\left(\dfrac{2\tan\theta}{1 + \tan^2\theta}\right)$
$f(x) = \sin^{-1}(\sin 2\theta)$
For $x = \dfrac{1}{2}$, $\theta = \tan^{-1}\left(\dfrac{1}{2}\right)$, which means $2\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
Therefore, $\sin^{-1}(\sin 2\theta) = 2\theta$.
$f(x) = 2\tan^{-1}x$
Differentiating with respect to $x$:
$f'(x) = \dfrac{2}{1 + x^2}$
Substituting $x = \dfrac{1}{2}$:
$f'\left(\dfrac{1}{2}\right) = \dfrac{2}{1 + \left(\dfrac{1}{2}\right)^2} = \dfrac{2}{1 + \dfrac{1}{4}} = \dfrac{2}{\dfrac{5}{4}} = \dfrac{8}{5}$
Answer: $\dfrac{8}{5}$
Let $x = \tan \theta$, then $\theta = \tan^{-1}x$.
Substituting $x = \tan \theta$ in the given function:
$f(x) = \sin^{-1}\left(\dfrac{2\tan\theta}{1 + \tan^2\theta}\right)$
$f(x) = \sin^{-1}(\sin 2\theta)$
For $x = \dfrac{1}{2}$, $\theta = \tan^{-1}\left(\dfrac{1}{2}\right)$, which means $2\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
Therefore, $\sin^{-1}(\sin 2\theta) = 2\theta$.
$f(x) = 2\tan^{-1}x$
Differentiating with respect to $x$:
$f'(x) = \dfrac{2}{1 + x^2}$
Substituting $x = \dfrac{1}{2}$:
$f'\left(\dfrac{1}{2}\right) = \dfrac{2}{1 + \left(\dfrac{1}{2}\right)^2} = \dfrac{2}{1 + \dfrac{1}{4}} = \dfrac{2}{\dfrac{5}{4}} = \dfrac{8}{5}$
Answer: $\dfrac{8}{5}$
Hence, the correct answer is option: 'A'.