Question

The inequality.
$maxf\left(x\right)<0.77[-\pi ,\pi ]$ holds for the function $f\left(x\right)=\left(sinx\right)\left(sin2x\right).$

• A. True
• B. False

Answer 1

The correct option is A True

$f\left(x\right)=\sin x\sin 2x$

$f^{\prime }\left(x\right)=\cos x\sin 2x+\sin x\cos 2x.2$

$=\cos x\left(2\sin x\cos x\right)+2sinx({\cos }^{2}x-{\sin }^{2}x)$

$=2{\cos }^{2}x\sin x+2\sin x{\cos }^{2}x-2{\sin }^{3}x$

$=4\sin x{\cos }^{2}x-2{\sin }^{3}x$

$=2\sin x(2{\cos }^{2}x-{\sin }^{2}x)$

$=2\sin x(3{\cos }^{2}x-1)$

roots : $\sin x=0$ or ${\cos }^{2}x=\frac{1}{3},{\sin }^{2}x=\frac{2}{3}$

For maxima, $\sin x=\sqrt{\frac{2}{3}}\Rightarrow \sin 2x=2\times \sqrt{\frac{2}{3}}\times \sqrt{\frac{1}{3}}$

$\therefore f_{max}=\left(\sqrt{\frac{2}{3}}\right)\left(\frac{\sqrt[2]{22}}{3}\right)=\frac{4}{\sqrt[3]{33}}=0.7698<0.77$