Q. Let \([t]\) denote the greatest integer less than or equal to \(t\). If the function
\[
f(x)=
\begin{cases}
b^2\sin\!\left(\dfrac{\pi}{2}\left[\dfrac{\pi}{2}(\cos x+\sin x)\cos x\right]\right), & x<0,\\[6pt]
\dfrac{\sin x-\dfrac{1}{2}\sin 2x}{x^3}, & x>0,\\[8pt]
a, & x=0,
\end{cases}
\]
is continuous at \(x=0\), then \(a^2+b^2\) is equal to
Explanation
For continuity at \(x=0\),
\[
\lim_{x\to0^-} f(x)=\lim_{x\to0^+} f(x)=f(0)=a
\]
First find the left hand limit.
As \(x\to0^{-}\),
\[
\cos x \to 1,\quad \sin x \to 0
\]
\[
(\cos x+\sin x)\cos x \to 1
\]
\[
\left[\frac{\pi}{2}\cdot 1\right]=\left[\frac{\pi}{2}\right]=1
\]
\[
\lim_{x\to0^-}f(x)=b^2\sin\!\left(\frac{\pi}{2}\right)=b^2
\]
Now find the right hand limit.
\[
\sin x = x-\frac{x^3}{6}+o(x^3),\quad
\sin 2x = 2x-\frac{8x^3}{6}+o(x^3)
\]
\[
\sin x-\frac{1}{2}\sin 2x
= \left(x-\frac{x^3}{6}\right)-\frac{1}{2}\left(2x-\frac{8x^3}{6}\right)
= \frac{x^3}{2}
\]
\[
\lim_{x\to0^+}\frac{\sin x-\frac{1}{2}\sin 2x}{x^3}
= \frac{1}{2}
\]
Thus,
\[
a=\frac{1}{2},\quad b^2=\frac{1}{2}
\]
\[
a^2+b^2=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}
\]
Therefore, the correct answer is \( \dfrac{3}{4} \).