If the function given by $f(x)= \begin{cases}-2 \sin x & -\pi \leq x < -(\pi / 2) \\ a \sin x+b & -(\pi / 2)…
If the function given by
$f(x)= \begin{cases}-2 \sin x & -\pi \leq x < -(\pi / 2) \\ a \sin x+b & -(\pi / 2) < x < (\pi / 2) \\ \cos x & (\pi / 2) \leq x \leq \pi\end{cases}$
is continuous in $[-\pi, \pi]$, then the value of $(3 a+2 b)^3$ is
1
8
-1
-8
Solution
$\begin{aligned}
& \lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}}-2 \sin x=-2 \sin \left(-\frac{\pi}{2}\right)=2 \\
& \lim _{x \rightarrow \frac{-\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{-\pi^{+}}{2}} a \sin x+b=a \sin \left(\frac{-\pi}{2}\right)+b=-a
\end{aligned}$
Since $f(x)$ is continuous at $x=\frac{-\pi}{2}$, we write
$\begin{aligned}
& 2=-a+b \\
& \lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x) \lim _{x \rightarrow \frac{\pi^{-}}{2}} a \sin x+b=a \sin \left(\frac{\pi}{2}\right)+b=a+b \\
& \lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \cos x=\cos \left(\frac{\pi}{2}\right)=0
\end{aligned}$
Since $f(x)$ is continuous at $x=\frac{\pi}{2}$, we write
$\mathrm{a}+\mathrm{b}=0$
From (1) and (2), we get $\mathrm{a}=-1, \mathrm{~b}=1$
$\therefore(3 a+2 b)^3=(-3+2)^3=-1$