# if x=2

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 :

1. (1) $\frac{3}{4}$

2. (2) $\frac{3}{8}$

3. (3) $\frac{3}{2}$

4. (4) $-\frac{3}{4}$

Correct Option: , 2

Solution:

It is given that

$x=2 \sin \theta-\sin 2 \theta$$\ldots (i) y=2 \cos \theta-\cos 2 \theta$$\ldots$ (ii)

Differentiating (i) w.r.t. $\theta$, we get

$\frac{d x}{d \theta}=2 \cos \theta-2 \cos 2 \theta$

Differentiating (ii) w.r.t. $\theta$; we get

$\frac{d y}{d \theta}=-2 \sin \theta+2 \sin 2 \theta$

From (ii) $\div$ (i), we get

$\therefore \quad \frac{d y}{d x}=\frac{\sin 2 \theta-\sin \theta}{\cos \theta-\cos 2 \theta}$

$=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$ ...(iii)

Again, differentiating eqn. (iii), we get

$\frac{d^{2} y}{d x^{2}}=\frac{-3}{2} \operatorname{cosec}^{2} \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$

$\frac{d^{2} y}{d x^{2}}=\frac{\frac{-3}{2} \operatorname{cosec}^{2} \frac{3 \theta}{2}}{2(\cos \theta-\cos 2 \theta)}$

$\frac{d^{2} y}{d x^{2}(\theta=\pi)}=-\frac{3}{4(-1-1)}=\frac{3}{8}$