Let $[t]$ denote the greatest integer less than or equal to $t$. If the function $f(x)=\left\{\begin{aligned} b^{2} \sin \left(\frac{\pi}{2}\left[\frac{\pi}{2}(\cos x+\sin x) \cos x\right]\right) & , x 0 \\ a & , x=0\end{aligned}\right.$ is continuous at $\mathrm{x}=0$, then $\mathrm{a}^{2}+\mathrm{b}^{2}$ is equal to

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Continuity and Differentiability

Let

[t]

denote the greatest integer less than or equal to

t

. If the function

f (x) = ⎩ ⎨ ⎧ b^{2} sin (\frac{π}{2} [\frac{π}{2} (cos x + sin x) cos x]) \frac{sin x - \frac{1}{2} sin 2 x}{x ^{3}} a, x < 0, x > 0, x = 0

is continuous at

x = 0

, then

a^{2} + b^{2}

is equal to

Answer: (C)

Continuity at

0

needs both one-sided limits to equal

a

. For

x \to 0^{+}

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

, so

a = \frac{1}{2}

. For

x \to 0^{-}

\frac{π}{2} (cos x + sin x) cos x \to \frac{π}{2}

from below, so its greatest integer is

1

, and the left limit is

b^{2} sin (\frac{π}{2}) = b^{2}

; continuity forces

b^{2} = \frac{1}{2}

. Therefore

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

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option