JEE Advanced 2018 — Paper 2 — Question 33

Given limit condition: $\lim_{t \to x} \frac{f(x)\sin t - f(t)\sin x}{t - x} = \sin^2 x$. Apply L'Hôpital's rule (0/0 form) by differentiating numerator and denominator w.r.t. $t$: $\lim_{t \to x} \frac{f(x)\cos t - f'(t)\sin x}{1} = f(x)\cos x - f'(x)\sin x = \sin^2 x$. Rewrite as $\frac{d}{dx}\left(\frac{f(x)}{\sin x}\right) = \frac{f'(x)\sin x - f(x)\cos x}{\sin^2 x} = -\frac{\sin^2 x}{\sin^2 x} = -1$. Integrate: $\frac{f(x)}{\sin x} = -x + C$, so $f(x) = -x\sin x + C\sin x$. Use $f(\pi/6) = -\pi/12$: $-\pi/12 = -\pi/6 \cdot \frac12 + C \cdot \frac12 \Rightarrow C = 0$. Thus $f(x) = -x\sin x$. Check options: A: $f(\pi/4) = -\pi\sqrt{2}/8 \neq \pi/(4\sqrt{2})$ (false). B: For $x\in(0,\pi)$, series expansion shows $-x\sin x < x^4/6 - x^2$ (true). C: $f'(x) = -\sin x - x\cos x$.

By intermediate value property, there exists $\alpha\in(0,\pi)$ with $f'(\alpha)=0$ (true). D: $f''(x) = -2\cos x + x\sin x$, so $f''(\pi/2) + f(\pi/2) = \pi/2 + (-\pi/2) = 0$ (true). Correct options: B, C, D.