JEE Main 2026 Jan 24 Shift 2, Mathematics Q20: Continuity of Function

Right-hand limit ($x\to 0^{+}$):

$\frac{\sin x-\frac{1}{2}\sin 2x}{x^3}=\frac{\sin x-\sin x\cos x}{x^3}=\frac{\sin x(1-\cos x)}{x^3}=\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\to 1\cdot \frac{1}{2}=\frac{1}{2}$.

Left-hand limit ($x\to 0^{-}$): As $x\to 0^{-}$, $\cos x\to 1$ and $\sin x+\cos x\to 1$ from below (since $\sin x<0$ slightly). So $\frac{\pi }{2}(\sin x+\cos x)\cos x$ approaches $\frac{\pi }{2}$ from below, i.e., a value slightly less than $\frac{\pi }{2}\approx 1.57$. The greatest integer $\left[\cdot \right]=1$.

So LHL $=b^2\sin \left(\frac{\pi }{2}\cdot 1\right)=b^2$.

For continuity: $b^2=\frac{1}{2}$ and $a=\frac{1}{2}$.

$a^2+b^2=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$.