Question

A minimum value of $\sin x\cos 2x$ is-

• A. $1$
• B. $-1$
• C. $-\frac{2}{3\sqrt{6}}$
• D. None of these

Answer 1

Answer: (B)

$-1$

$f\left(x\right)=\sin x.\cos 2x$
$=\sin x(1-2{\sin }^{2}x)$
$=\sin x-2{\sin }^{3}x$
Hence
$f^{\prime }\left(x\right)=\cos x-6{\sin }^{2}x.\cos x$
$=0$ ... for critical point.
Or
$\cos x(1-6{\sin }^{2}x)=0$
$\cos x=0$ or $\sin x=\frac{1}{\sqrt{6}}$
Now
$f"\left(x\right)$
$=-\sin x-12\sin x.\cos x+6{\sin }^{3}x$
At $x={\sin }^{-1}\frac{1}{\sqrt{6}}$ we get
$f"\left(x\right)<0$.
Hence
$f(x={\sin }^{-1}\frac{1}{\sqrt{6}})$

$=\frac{1}{\sqrt{6}}-\frac{2}{6\sqrt{6}}$

$=\frac{3-1}{3\sqrt{6}}$

$=\frac{2}{3\sqrt{6}}$
This is the maximum value.
for minimum value $f"\left(x\right)>0$ at $x={\cos }^{-1}\left(0\right)$
Substituting, we get
$f\left(\frac{\pi }{2}\right)$
$=1-2$
$=-1$.