Limits, Continuity and Differentiability – Continuity – JEE Main 24 Jan 2026 Shift 2

Solution:

For continuity at $x=0$, $LHL = RHL = f(0) = a$.

**Right Hand Limit (RHL):**
$$RHL = \lim_{x \to 0^+} \frac{\sin x – \frac{1}{2}(2\sin x \cos x)}{x^3} = \lim_{x \to 0^+} \frac{\sin x (1 – \cos x)}{x^3}$$
$$RHL = \lim_{x \to 0^+} \frac{\sin x}{x} \cdot \frac{2\sin^2(x/2)}{x^2} = 1 \cdot 2 \cdot \lim_{x \to 0^+} \left(\frac{\sin(x/2)}{x}\right)^2$$
$$RHL = 2 \cdot \left(\frac{1}{2}\right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}$$
So, $a = \frac{1}{2}$.

**Left Hand Limit (LHL):**
$$LHL = \lim_{x \to 0^-} b^2 \sin\left(\frac{\pi}{2} \left[\frac{\pi}{2}(\cos x + \sin x)\cos x\right]\right)$$
Let $g(x) = \frac{\pi}{2}(\cos x + \sin x)\cos x = \frac{\pi}{2}(\cos^2 x + \sin x \cos x)$.
As $x \to 0^-$, $\cos x \to 1^-$ and $\sin x \to 0^-$.
$g(x) \approx \frac{\pi}{2}(1 + (-\epsilon)) < \frac{\pi}{2}$. Since $g(x)$ is slightly less than $\frac{\pi}{2} \approx 1.57$, $[g(x)] = 1$. $$LHL = b^2 \sin\left(\frac{\pi}{2} \cdot 1\right) = b^2(1) = b^2$$
Equating limits: $b^2 = \frac{1}{2}$.

Value required:
$$a^2 + b^2 = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$$

Ans. (3)