Question

Let the absolute maxima/minima value of $f\left(x\right)=\sin x+\frac{1}{2}\cos 2x;x\in [0,\frac{\pi }{2}]$ be max and min.Find $4\left(max\right)+2\left(min\right)$

Answer 1

Given,

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

$f^{\prime }\left(x\right)=\cos x-\sin 2x=\cos x-2\sin x.\cos x=\cos x(1-2\sin x)=0$ for min. or max.

$\Rightarrow x=\frac{\pi }{6},\frac{\pi }{2}$

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

Clearly $f^{\prime \prime }\left(\frac{\pi }{6}\right)=-\frac{1}{2}-2\frac{1}{2}=-\frac{3}{2}<0$

and $f^{\prime \prime }\left(\frac{\pi }{2}\right)=-1-2(-1)=1>0$

$\therefore \text{max}=f\left(\frac{\pi }{6}\right)=\frac{3}{4},\text{min}=f\left(\frac{\pi }{2}\right)=\frac{1}{2}$

$4\left(max\right)+2\left(min\right)=3+1=4$