JEE Advanced 2018 — Paper 1 — Question 30

Differentiate $\sin x + \sin y = 1$ implicitly: $\cos x + \cos y \cdot \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{\cos x}{\cos y}$. Differentiate again: $\frac{d^2 y}{dx^2} = -\frac{-\sin x \cos y + \cos x (-\sin y \frac{dy}{dx})}{\cos^2 y}$.

Substitute $\frac{dy}{dx}$ to get $\frac{d^2 y}{dx^2} = \frac{\sin x \cos^2 y + \sin y \cos^2 x}{\cos^3 y}$. In the first quadrant, let $\sin x = t$.

Then $\sin y = 1 - t$, $\cos x = \sqrt{1-t^2}$, $\cos y = \sqrt{t(2-t)}$. Substitute: $\frac{d^2 y}{dx^2} = \frac{t \cdot t(2-t) + (1-t)(1-t^2)}{(t(2-t))^{3/2}} = \frac{1-t+t^2}{t^{3/2}(2-t)^{3/2}} = t^{-3/2}\frac{1-t+t^2}{(2-t)^{3/2}}$. As $x\to 0$, $t = \sin x \to 0$ and $\frac{t}{x}\to 1$, so $t \sim x$.

Then $\frac{d^2 y}{dx^2} \sim x^{-3/2}\frac{1}{2^{3/2}}$. Then $L = \lim_{x\to 0} x^{\alpha} \frac{d^2 y}{dx^2} \sim \frac{1}{2^{3/2}} \lim_{x\to 0} x^{\alpha - 3/2}$.

For L to exist and be non-zero, $\alpha - \frac{3}{2} = 0 \Rightarrow \alpha = \frac{3}{2}$, and $L = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}$.