QJEEM20Ma5bf7: JEE Main 2020 Mathematics - Class 12 JEEM20Ma5bf7 Solution: If and , , then at = is :

$x = 2\sin \theta - \sin 2\theta$

$\Rightarrow$ ${{dx} \over {d\theta }}$ = $2\cos \theta - 2\cos 2\theta$

$y = 2\cos \theta - \cos 2\theta$

$\Rightarrow$ ${{dy} \over {d\theta }}$ = –2sin $\theta$ + 2sin2 $\theta$

${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$ = ${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$

This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$ , leading to the expression:

$\frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$

This is the rate of change of y with respect to x, as a function of θ.

Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$ , yielding :

$\frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$

Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$ , so the equation becomes :

$\frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$

Finally, we evaluate this expression at $\theta = \pi$ , yielding :

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

Therefore, the correct answer is (A) $\frac{3}{8}$ .

Alternate Method :

First, let's find the derivatives of x and y with respect to θ :

1) $\frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta$

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

We know that $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

So, we substitute 1) and 2) into this equation :

We have

$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$

For simplification, let's denote the numerator as $N = \sin 2\theta - \sin \theta$ and the denominator as $D = \cos \theta - \cos 2\theta$ .

We have to compute $\frac{d}{d\theta}(\frac{dy}{dx})$ which is $\frac{d}{d\theta}(\frac{N}{D})$ .

We can use the quotient rule for differentiation, which states that if we have a function of the form $\frac{u}{v}$ , then its derivative is given by $\frac{vu' - uv'}{v^2}$ .

So here, $N' = 2\cos 2\theta - \cos \theta$ and $D' = -\sin \theta + 2\sin 2\theta$ .

Applying the quotient rule :

$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$

Substituting the expressions for $N'$ , $D'$ , $N$ , and $D$ we get :

$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$

Now $\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}$

= $\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$ $\times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)}$

$\begin{aligned}\left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\& =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8}\end{aligned}$