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Let \([t]\) denote the greatest integer less than or equal to \(t\). If the function \(f\left(x\right)=\left{\begin{matr…

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Let \([t]\) denote the greatest integer less than or equal to \(t\). If the function

\(f\left(x\right)=\left{\begin{matrix}{b}^{2}\text{sin}\left(\frac{\pi }{2}\left[\frac{\pi }{2}\left(\text{cos}x+\text{sin}x\right)\text{cos}x\right]\right),\text{   }x<0 \\ \text{                           }\frac{\sin x−\frac{1}{2}\sin 2x}{{x}^{3}}\text{            },\text{      }x>0 \\ \text{                                        }a\text{                             },\text{      }x=0\end{matrix}\right.\)

is continuous at \(x = 0\), then \(a^2+b^2\) is equal to

[JEE Main 2026, 24 Jan (Shift 2)]

a

\(\frac{9}{16}\)

b

\(\frac{1}{2}\)

c

\(\frac{5}{8}\)

d

\(\frac{3}{4}\)

✓ Correct answer: d)

\(\frac{3}{4}\)

Explanation

Given \(f(x)\) is continuous at \(x=0\)

\(f(0)=a\)

\(\mathrm{RHL}=\lim _{x\to {0}^{+}}\frac{\sin x(1−\cos x)}{{x}^{3}}\\ =\lim _{x\to {0}^{+}}\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{{x}^{2}}\right)\\ =\frac{1}{2}\)

\(\mathrm{LHL}=\lim _{x\to {0}^{−}}\left({b}^{2}\sin \left(\frac{\pi }{2}\left[\frac{\pi }{2}(\sin x+\cos x)\cos x\right]\right)\right)\\ ={b}^{2}\)

\(∴a=\frac{1}{2}\)

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

So, \(\left({a}^{2}+{b}^{2}\right)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

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