Explanation:
(C) : Given,
\(f(x)=\sin x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
\(\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}\)
For maxima or minima,
\(\mathrm{f}^{\prime}(\mathrm{x})=0\)
\(\Rightarrow \quad \cos x=0\)
\(\Rightarrow \quad \mathrm{x}=-\frac{\pi}{2}\) or \(\mathrm{x}=\frac{\pi}{2}\), for all \(\mathrm{x} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Then, \(\quad \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin \mathrm{x}\)
At, \(\quad x=\frac{\pi}{2}\), we get-
\(\mathrm{f}^{\prime \prime}\left(\frac{\pi}{2}\right)=-\sin \frac{\pi}{2}=-1\lt 0\)
\(\therefore \mathrm{x}=\frac{\pi}{2}\) is the point of local maximum of \(\mathrm{f}(\mathrm{x})\)
At, \(\quad x=-\frac{\pi}{2}\), we have
\(\mathrm{f}^{\prime \prime}\left(-\frac{\pi}{2}\right)=-\sin \left(-\frac{\pi}{2}\right)=\sin \frac{\pi}{2}=1>0\)
\({[\because \sin (-\theta)=-\sin \theta]}\)
\(\therefore \mathrm{x}=-\frac{\pi}{2}\) is the point of local minimum of \(\mathrm{f}(\mathrm{x})\)
Then, maximum value of \(\mathrm{f}(\mathrm{x})=\mathrm{f}\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}=1\)
So, the minimum value of \(f(x) \sin x\) in
\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { is } 1\)
