# The maximum value of sin x cos x is A. $\frac{1}{4}$ B. $\frac{1}{2}$ C. $\sqrt 2$ D. $2\sqrt 2$

Let f(x)= sin x cos x

sin2x=2sin x cos x

$\Rightarrow f(x)=\frac{1}{2} \sin 2 x$
Apply first derivative and get
$\\ \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\frac{1}{2} \sin 2 \mathrm{x}\right)}{\mathrm{dx}}\\ \Rightarrow f^{\prime}(x)=\frac{1}{2} \frac{d(\sin 2 x)}{d x}$
Apply derivative,
$\Rightarrow f^{\prime}(x)=\frac{1}{2} \cdot \cos 2 x \cdot \frac{d(2 x)}{d x}$

$\\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cdot \cos 2 \mathrm{x} \cdot 2 \\\Rightarrow f(x)=\cos 2 x \ldots \ldots(i)$
Put $f^{\prime}(x)=0$ and get
$\\\cos 2 x=0 \\ \Rightarrow \cos 2 x=\cos \frac{\pi}{2}$
Equate the angles and get
$\\\Rightarrow 2 x=\frac{\pi}{2} \\\Rightarrow x=\frac{\pi}{4}$
Now we find second derivative by deriving equation (i) and get
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\cos 2 \mathrm{x})}{\mathrm{dx}}$

Apply derivative,
$\\\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin 2 \mathrm{x} \cdot \frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}} \\\Rightarrow f^{\prime}(x)=-\sin 2 x .2 \\\Rightarrow f^{\prime}(x)=-2 \sin 2 x$
Now we find the value of $f^{\prime \prime}(x) at x=\frac{\pi}{4},$ we get
$\\\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2 x \\\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2\left(\frac{\pi}{4}\right) \\\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin \left(\frac{\pi}{2}\right)$
But $\sin \left(\frac{\pi}{2}\right)=1$ so above equation becomes
$\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \times 1=-2<0$
Hencee at$\mathrm{x}=\frac{\pi}{4}$, $\mathrm{f}(\mathrm{x})$ is maximum and $\frac{\pi}{4}$ is the point of maxima.

Now we will find the maximum value of $\sin \mathrm{x} \cos \mathrm{x}$ by substituting $\mathrm{x}=\frac{\pi}{4},$ in $\mathrm{f}(\mathrm{x}),$ we get
$\\ f(x)=\sin x \cos x \\f\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) \\\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{2}$
So, maximum value of  $\sin \mathrm{x} \cos \mathrm{x}$ is $\frac{1}{2}$
So, the correct answer is option B