JEE PYQ: Differentiation Question 13
Question 13 - 2019 (08 Apr Shift 1)
If $2y = \left(\cot^{-1}\left(\frac{\sqrt{3}\cos x + \sin x}{\cos x - \sqrt{3}\sin x}\right)\right)^2$, $x \in \left(0, \frac{\pi}{6}\right)$, then $\frac{dy}{dx}$ is equal to:
(1) $\frac{\pi}{6} - x$ (2) $x - \frac{\pi}{6}$ (3) $\frac{\pi}{3} - x$ (4) $2x - \frac{\pi}{3}$
Show Answer
Answer: (2) $x - \frac{\pi}{6}$
Solution
Simplifying: $2y = \left(\frac{\pi}{6} - x\right)^2$. So $\frac{dy}{dx} = x - \frac{\pi}{6}$.
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