Let \(f(x)=\sin x \cos x\)
But we know \(\sin 2 x=2 \sin x \cos x\)
$$
\Rightarrow \mathrm{f}(\mathrm{x})=\frac{1}{2} \sin 2 \mathrm{x}
$$
Applying the first derivative we get
$$
f^{\prime}(x)=\frac{d\left(\frac{1}{2} \sin 2 x\right)}{d x}
$$
$$
\Rightarrow f^{\prime}(x)=\frac{1}{2} \frac{d(\sin 2 x)}{d x}
$$
Applying the derivative,
$$
\begin{aligned}
&\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cdot \cos 2 \mathrm{x} \cdot \frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}} \\
&\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cdot \cos 2 \mathrm{x} \cdot 2 \\
&\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\cos 2 \mathrm{x} \ldots \ldots \text { (i) }
\end{aligned}
$$
Putting \(f^{\prime}(x)=0\), we get critical points as
$$
\begin{aligned}
&\cos 2 x=0 \\
&\Rightarrow \cos 2 x=\cos \frac{\pi}{2}
\end{aligned}
$$
Now equating the angles, we get
$$
\Rightarrow 2 \mathrm{x}=\frac{\pi}{2}
$$
$$
\Rightarrow x=\frac{\pi}{4}
$$
Now we will find out the second derivative by deriving equation (i), we get
$$
\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\cos 2 \mathrm{x})}{\mathrm{dx}}
$$
Applying the derivative,
$$
\begin{aligned}
&\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin 2 \mathrm{x} \cdot \frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}} \\
&\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin 2 \mathrm{x} .2 \\
&\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-2 \sin 2 \mathrm{x}
\end{aligned}
$$
Now we will find the value of \(f^{\prime \prime}(x)\) at \(x=\pi / 4\), we get
$$
\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2 x
$$
$$
\begin{aligned}
&\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2\left(\frac{\pi}{4}\right) \\
&\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin \left(\frac{\pi}{2}\right)
\end{aligned}
$$
But \(\sin \left(\frac{\pi}{2}\right)=1\), so above equation becomes
$$
\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \times 1=-2<0
$$
Therefore at \(x=\pi / 4, f(x)\) is maximum and \(\pi / 4\) is the point of maxima.
Now we will find the maximum value of \(\sin x \cos x\) by substituting \(x=\pi / 4\), in \(f(x)\), we get
$$
f(x)=\sin x \cos x
$$
$$
\begin{aligned}
&\mathrm{f}\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) \\
&\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\
&\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{2}
\end{aligned}
$$
Hence the maximum value of \(\sin x \cos x\) is \(1 / 2\)
So the correct option is option B.
