Question

The difference between greatest and least value
of $f(x)=2\sin x+\sin 2x,x∈\left[0,\frac{3π}{2}\right]$ is-

• A. $\frac{3\sqrt{3}}{2}$
• B. $\frac{3\sqrt{3}}{2}−2$
• C. $\frac{3\sqrt{3}}{2}+2$
• D. None of these

Answer 1

Answer: (C)

$f(x)=2\sin x+\sin 2x$
$f^{′}(x)=2\cos x+2\cos 2x=2(\cos x+\cos 2x)$
$∴f^{′}(x)=0⇒2\cos 2x+\cos x−1=0$
$\cos x=\frac{−1Â\pm 3}{4}=−1,\frac{1}{2}$
$∴x=π,\frac{π}{3}$
Now, $f(0)=0,f\left(\frac{3π}{2}\right)=−2$
$f(Ï€)=0,f\left(\frac{Ï€}{3}\right)=2\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$
$∴$ difference between greatest value and least value
$=\frac{3\sqrt{3}}{2}+2$