Question

If $x=2\sin \theta -\sin 2\theta$ and $y=2\cos \theta -\cos 2\theta$, $\theta \in [0,2\pi ],$ then $\frac{d^{2}y}{d{x}^2}$ at $\theta =\pi$ is :

• A. $-\frac{3}{8}$
• B. $\frac{3}{4}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{4}$
• E. None of these

Answer 1

The correct option is E None of these

$\frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta$
$\frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta$
$\frac{dy}{dx}=\frac{2\cos \frac{3\theta }{2}\sin \frac{\theta }{2}}{2\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}$
$\frac{dy}{dx}=\cot \frac{3\theta }{2}$
$\frac{d^{2}y}{d{x}^2}=-\frac{3}{2}{\text{cosec}}^2\frac{3\theta }{2}\frac{d\theta }{dx}$
$\frac{d^{2}y}{d{x}^2}=(-\frac{3}{2}{\text{cosec}}^2\frac{3\theta }{2})\frac{1}{(2\cos \theta -2\cos 2\theta )}$
${\begin{array}{c}\frac{d^{2}y}{d{x}^2}\end{array}|}_{\theta =\pi }=\frac{3}{8}$
None of the above option satisfies the answer. This is a BONUS question.