JEE PYQ: Differentiation Question 12
Question 12 - 2020 (09 Jan Shift 2)
If $x = 2\sin\theta - \sin 2\theta$ and $y = 2\cos\theta - \cos 2\theta$, $\theta \in [0, 2\pi]$, then $\frac{d^2y}{dx^2}$ at $\theta = \pi$ is:
(1) $\frac{3}{4}$ (2) $\frac{3}{8}$ (3) $\frac{3}{2}$ (4) $-\frac{3}{4}$
Show Answer
Answer: (2) $\frac{3}{8}$
Solution
$\frac{dy}{dx} = \cot\frac{3\theta}{2}$. $\frac{d^2y}{dx^2} = \frac{-\frac{3}{2}\csc^2\frac{3\theta}{2}}{2(\cos\theta - \cos 2\theta)}$. At $\theta = \pi$: $\frac{3}{8}$.
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