JEE Main 2026 — 24 January, Evening Shift — Question 20

Given $f(x)$ is continuous at $x=0$, so $\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$.

For $x>0$: $f(x) = \frac{\sin x - \frac{1}{2} \sin 2x}{x^3} = \frac{\sin x - \frac{1}{2} (2 \sin x \cos x)}{x^3} = \frac{\sin x (1 - \cos x)}{x^3}$.

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$.

For $x<0$: As $x \to 0^-$, $\cos x + \sin x \to 1$, so $\frac{\pi}{2} (\cos x + \sin x) \cos x \to \frac{\pi}{2} \cdot 1 \cdot 1 = \frac{\pi}{2}$.

Since $\frac{\pi}{2} \approx 1.57$, the greatest integer $\left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] = 1$ for $x$ sufficiently close to 0 from the left.

Thus $\lim_{x \to 0^-} f(x) = b^2 \sin\left( \frac{\pi}{2} \cdot 1 \right) = b^2 \sin\left( \frac{\pi}{2} \right) = b^2 \cdot 1 = b^2$.

Continuity gives $a = \frac{1}{2}$ and $b^2 = \frac{1}{2}$.

Therefore $a^2 + b^2 = \left( \frac{1}{2} \right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.