JEE PYQ: Differentiation Question 10
Question 10 - 2020 (08 Jan Shift 1)
Let $f(x) = (\sin(\tan^{-1}x) + \sin(\cot^{-1}x))^2 - 1$, $|x| > 1$. If $\frac{dy}{dx} = \frac{1}{2}\frac{d}{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt{3}) = \frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to:
(1) $\frac{2\pi}{3}$ (2) $-\frac{\pi}{6}$ (3) $\frac{5\pi}{6}$ (4) $\frac{\pi}{3}$
Show Answer
Answer: (2) $-\frac{\pi}{6}$
Solution
$2y = \sin^{-1}(\sin(2\tan^{-1}x)) + C$. Using $y(\sqrt{3}) = \frac{\pi}{6}$: $C = 0$. For $x = -\sqrt{3}$: $y = -\frac{\pi}{6}$.
BETA
