JEE Main 2020 — Differentiation Question with Solution

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

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

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

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

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

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 $\frac{d^{2}y}{d{x}^2}$, yielding :

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

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

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

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

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

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 $$\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$$.

So here, $$N^{\prime }=2\cos 2\theta -\cos \theta$$ and $$D^{\prime }=-\sin \theta +2\sin 2\theta$$.

Applying the quotient rule :

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

Substituting the expressions for $$N^{\prime }$$, $$D^{\prime }$$, $$N$$, 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 $$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)\times \frac{d\theta }{dx}$$

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

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