Question

Let $f\left(x\right)=\cos x\cdot \sin 2x$ and $x\in [0,\frac{\pi }{2}],$ then

• A. $f\left(x\right)$ has global minimum at $x=0$
• B. $f\left(x\right)$ has global maximum at $x=\frac{\pi }{2}$
• C. Global maximum value of $f\left(x\right)$ is $1$
• D. Global maximum value of $f\left(x\right)$ is $\frac{4}{3\sqrt{3}}$

Answer 1

Answer: (A), (D)

Global maximum value of $f\left(x\right)$ is $\frac{4}{3\sqrt{3}}$

Given,$f\left(x\right)=\cos x\cdot \sin 2x$
$\Rightarrow f\left(x\right)=2\sin x-2{\sin }^{3}x$
$\Rightarrow f^{\prime }\left(x\right)=2\cos x-6{\sin }^{2}x\cos x$
Now, $f^{\prime }\left(x\right)=0$
$\Rightarrow \cos x=0,\sin x=\frac{1}{\sqrt{3}}$
$\therefore x=\frac{\pi }{2},x={\sin }^{-1}\frac{1}{\sqrt{3}}$
$\{\because x\in [0,\frac{\pi }{2}]\}$
For global maxima or minima check the values of $f\left(0\right),f\left(\frac{\pi }{2}\right),f\left(si{n}^{-1}\frac{1}{3}\right)$
$\therefore f\left(0\right)=0$
$f\left(\frac{\pi }{2}\right)=0$
$f\left({\sin }^{-1}\frac{1}{\sqrt{3}}\right)=2\left(\frac{1}{\sqrt{3}}\right)-2{\left(\frac{1}{\sqrt{3}}\right)}^3=\frac{4}{3\sqrt{3}}$