Q40 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function $f(x)=\left\{\begin{array}{cl}b^{2} \sin \left(\frac{\pi}{2}\left[\frac{\pi}{2}(\cos x+\sin x) \cos x\right]\right), & x<0 \\ \frac{\sin x-\frac{1}{2} \sin 2 x}{x^{3}} &, x>0 \\ a &, x=0\end{array}\right.$ is continuous at $x=0$, then $a^{2}+b^{2}$ is equal to
- A$\frac{5}{8}$
- B$\frac{1}{2}$
- C$\frac{9}{16}$
- Correct answerD$\frac{3}{4}$
Solution
For $x > 0$: $\sin x - \frac{1}{2}\sin 2x = \sin x(1 - \cos x)$.
$\lim_{x \to 0^+} \frac{\sin x(1-\cos x)}{x^3} = \lim_{x \to 0^+} \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$.
So $a = \frac{1}{2}$.
For $x < 0$ near $0$: $\frac{\pi}{2}(\cos x + \sin x)\cos x \to \frac{\pi}{2}$ from below, so $\left[\frac{\pi}{2}(\cos x + \sin x)\cos x\right] = 1$.
$f(x) = b^2 \sin\left(\frac{\pi}{2} \cdot 1\right) = b^2$.
For continuity: $b^2 = a = \frac{1}{2}$.
$a^2 + b^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.
$\lim_{x \to 0^+} \frac{\sin x(1-\cos x)}{x^3} = \lim_{x \to 0^+} \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$.
So $a = \frac{1}{2}$.
For $x < 0$ near $0$: $\frac{\pi}{2}(\cos x + \sin x)\cos x \to \frac{\pi}{2}$ from below, so $\left[\frac{\pi}{2}(\cos x + \sin x)\cos x\right] = 1$.
$f(x) = b^2 \sin\left(\frac{\pi}{2} \cdot 1\right) = b^2$.
For continuity: $b^2 = a = \frac{1}{2}$.
$a^2 + b^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.
Hence, the correct answer is option: 'D'.