Question

The maximum value of $sinx\cdot cosx$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

Answer: (B)

Given $f\left(x\right)=sinx\cdot cosx=\frac{sin2x}{2}$
$f^{\prime }\left(x\right)=\frac{1}{2}\left(cos2x\right)\cdot 2=cos2x$
For maxima or minima, $f^{\prime }\left(x\right)=0$
$\Rightarrow cos2x=0$
$\Rightarrow 2x=\frac{\pi }{2}$, $\frac{3\pi }{2}$
$\Rightarrow x=\frac{\pi }{4}$, $\frac{3\pi }{4}$
Now, $f^{\prime \prime }\left(x\right)=-2sin2x$
At $x=\frac{\pi }{4}$
$f^{\prime \prime }\left(x\right)=-2sin2(\frac{\pi }{4})=-2<0$
$\therefore x=\frac{\pi }{4}$ local max point
$\therefore$ max. value is
$f\left(\frac{\pi }{4}\right)=sin\frac{\pi }{4}\cdot cos\frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{2}$