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

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

Hint:

For piecewise functions with limits, always compute LHL and RHL separately before applying continuity.

Answer 1

The Correct Option is C

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.
\[ a^2+b^2=\frac{1}{36}+\frac{1}{6}=\frac{5}{8} \]