Question

$f\left(\pi /2\right)$ equals

• A. $(\sqrt{2}+1)(\sqrt{2}-\pi /2)$
• B. $(\sqrt{2}+1)(\sqrt{2}+\pi /2)$
• C. $(\sqrt{2}-1)(\sqrt{2}+\pi /2)$
• D. $(\sqrt{2}-1)(\sqrt{2}-\pi /2)$

Answer 1

The correct option is C $(\sqrt{2}-1)(\sqrt{2}+\pi /2)$

We have

$f\left(x\right)={\int }_0^x\{2\sqrt{2}{\sin }^{2}t+(2-\sqrt{2})\sin t-1\}dt,0<x<2\pi$

$\Rightarrow f^{\prime }\left(x\right)=2\sqrt{2}=2\sqrt{2}{\sin }^{2}x+(2-\sqrt{2})\sin x-1$

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


$\Rightarrow 2\sin x(\sqrt{2}\sin x+1)-1(\sqrt{2}\sin x+1)=0$

$\Rightarrow (\sqrt{2}\sin x+1)(2\sin x-1)=0$

$\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6},\frac{5\pi }{4},\frac{7\pi }{4}$

$f(\frac{\pi }{2},)$ can be calculated by substituting $x=\frac{\pi }{2},$