Question:
Let $[t]$ denote the greatest integer less than or equal to $t$.
If the function
\[
f(x)=
\begin{cases}
b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x < 0 \\
\dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\
a, & x = 0
\end{cases}
\]
is continuous at $x=0$, then $a^2+b^2$ is equal to
Let $[t]$ denote the greatest integer less than or equal to $t$.
If the function
\[ f(x)= \begin{cases} b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x < 0 \\ \dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases} \] is continuous at $x=0$, then $a^2+b^2$ is equal to
If the function
\[ f(x)= \begin{cases} b^2 \sin\!\left[\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right], & x < 0 \\ \dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases} \] is continuous at $x=0$, then $a^2+b^2$ is equal to
Show Hint
For piecewise functions with limits, always compute LHL and RHL separately before applying continuity.
Updated On: Mar 25, 2026
- $\dfrac{3}{4}$
- $\dfrac{1}{2}$
- $\dfrac{5}{8}$
- $\dfrac{9}{16}$
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Right-hand limit at $x=0$.
\[ \lim_{x\to0^+}\frac{\sin x-\frac{1}{2}\sin2x}{x^3} \] Using expansions, \[ \sin x-\frac{1}{2}(2x)=\frac{x^3}{6} \Rightarrow \text{RHL}=\frac{1}{6} \] Thus, \[ a=\frac{1}{6} \] Step 2: Left-hand limit at $x=0$.
As $x\to0^-$, \[ \cos x+\sin x\to1 \Rightarrow \left[\frac{\pi}{2}(\cos x+\sin x)\cos x\right]=1 \] \[ \sin\!\left(\frac{\pi}{2}\cdot1\right)=1 \] So, \[ \text{LHL}=b^2 \] Step 3: Apply continuity condition.
\[ b^2=\frac{1}{6} \] Step 4: Compute required sum.
\[ a^2+b^2=\frac{1}{36}+\frac{1}{6}=\frac{5}{8} \]
\[ \lim_{x\to0^+}\frac{\sin x-\frac{1}{2}\sin2x}{x^3} \] Using expansions, \[ \sin x-\frac{1}{2}(2x)=\frac{x^3}{6} \Rightarrow \text{RHL}=\frac{1}{6} \] Thus, \[ a=\frac{1}{6} \] Step 2: Left-hand limit at $x=0$.
As $x\to0^-$, \[ \cos x+\sin x\to1 \Rightarrow \left[\frac{\pi}{2}(\cos x+\sin x)\cos x\right]=1 \] \[ \sin\!\left(\frac{\pi}{2}\cdot1\right)=1 \] So, \[ \text{LHL}=b^2 \] Step 3: Apply continuity condition.
\[ b^2=\frac{1}{6} \] Step 4: Compute required sum.
\[ a^2+b^2=\frac{1}{36}+\frac{1}{6}=\frac{5}{8} \]
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