Let [t] denote the greatest integer less than or equal to t . If the function f(x)=\ array cl b^ 2 ((π)/(2)[(π)/(2)( x+ x) x]), & x 0 a &, x=0 array . is continuous at x=0 , then a^ 2 +b^ 2 is equal to

Correct answer: D. (3)/(4) For x > 0 : x - (1)/(2) 2x = x(1 - x) . _ x → 0^+ ( x(1- x))/(x^3) = _ x → 0^+ ( x)/(x) · (1- x)/(x^2) = 1 · (1)/(2) = (1)/(2) . So a = (1)/(2) . For x f(x) = b^2 ((π)/(2) · 1) = b^2 . For continuity: b^2 = a = (1)/(2) . a^2 + b^2 = (1)/(4) + (1)/(2) = (3)/(4) .

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Continuity and Differentiability

Let [t] denote the greatest integer less than or equal to t . If...

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

Options

A $\frac{5}{8}$
B $\frac{1}{2}$
C $\frac{9}{16}$
D $\frac{3}{4}$

Correct answer

D. $\frac{3}{4}$

Step-by-step 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

.<br>

f(x) = b^2 \sin\left(\frac{\pi}{2} \cdot 1\right) = b^2

.<br>For continuity:

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

.<br>

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

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