JEE PYQ: Differentiation Question 8
Question 8 - 2020 (07 Jan Shift 1)
If $y(\alpha) = \sqrt{2\left(\frac{\tan\alpha + \cot\alpha}{1+\tan^2\alpha}\right) + \frac{1}{\sin^2\alpha}}$, $\alpha \in \left(\frac{3\pi}{4}, \pi\right)$, then $\frac{dy}{d\alpha}$ at $\alpha = \frac{5\pi}{6}$ is:
(1) $4$ (2) $\frac{4}{3}$ (3) $-4$ (4) $-\frac{1}{4}$
Show Answer
Answer: (1) $4$
Solution
Simplifying: $y = |1 + \cot\alpha| = -1 - \cot\alpha$. So $\frac{dy}{d\alpha} = \csc^2\alpha = 4$ at $\alpha = \frac{5\pi}{6}$.
BETA
