Question

The absolute minimum and maximum values of $f\left(x\right)=\sin x+\frac{1}{2}\cos 2x,xϵ[0,\frac{\pi }{2}]$ are respectively

• A. $\frac{3}{4},\frac{1}{2}$
• B. $0,\frac{1}{2}$
• C. $\frac{1}{2},-\frac{3}{4}$
• D. $-\frac{1}{2},\frac{3}{4}$

Answer 1

Answer: (A)

$\frac{3}{4},\frac{1}{2}$

Differentiate the given function $f\left(x\right)=\sin x+\frac{1}{2}\cos 2x$.

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

Put $f^{\prime }\left(x\right)=0$,

$\cos x-\sin 2x=0$

$\cos x=2\sin x\cos x$

$\sin x=\frac{1}{2}$

$x={\sin }^{-1}\left(\frac{1}{2}\right)$

$x=\frac{\pi }{6}$

Put the value of $x$ and the end points of the given interval in the given function.

$f\left(\frac{\pi }{6}\right)=\sin \left(\frac{\pi }{6}\right)+\frac{1}{2}\cos 2\left(\frac{\pi }{6}\right)$

$=\frac{3}{4}$

$f\left(0\right)=\sin \left(0\right)+\frac{1}{2}\cos 2\left(0\right)$

$=\frac{1}{2}$

$f\left(\frac{\pi }{2}\right)=\sin \left(\frac{\pi }{2}\right)+\frac{1}{2}\cos 2\left(\frac{\pi }{2}\right)$

$=\frac{1}{2}$

The maximum value of the given function is $\frac{3}{4}$ and the minimum value is $\frac{1}{2}$.