Question

Consider a real value continuous function $f\left(x\right)$ such that $f\left(x\right)=\sin x+{\int }_{-\pi /2}^{\pi /2}(\cos x+tf\left(t\right))dt$

then which of the following is correct?

• A. $f_{max}=3(\pi +1)$
• B. $f_{max}=2(\pi +1)$
• C. $f_{min}=(\pi +1)$
• D. $f_{min}=\frac{1}{2}(\pi +1)$

Answer 1

Answer: (A), (C)

$f_{max}=3(\pi +1)$
$f_{min}=(\pi +1)$

$f\left(x\right)=\sin x+{\int }_{-\pi /2}^{\pi /2}(\cos x+tf\left(t\right))dt$

$\Rightarrow f\left(x\right)=\sin x+\sin x(\frac{\pi }{2}+\frac{\pi }{2})+{\int }_{-\pi /2}^{\pi /2}tf\left(t\right)dt$

$\Rightarrow f\left(x\right)=\sin x+\pi \sin x+M=(\pi +1)\sin x+M$ ...(1)

Where $M={\int }_{-\pi /2}^{\pi /2}tf\left(t\right)dt={\int }_{-\pi /2}^{\pi /2}t((\pi +1)\sin t+M)dt$

$=(\pi +1){\int }_{-\pi /2}^{\pi /2}t\sin tdt+M{\int }_{-\pi /2}^{\pi /2}tdt=2(\pi +1){\int }_0^{\pi /2}t\sin tdt+M\left(0\right)$

$\Rightarrow M=2(\pi +1)[t{[-\cos t]}_0^{\pi /2}-{\int }_0^{\pi /2}1.(-\cos t)dt]$

$\Rightarrow M=2(\pi +1)\left[{\left[\sin t\right]}_0^{\pi /2}\right]=2(\pi +1)$ ...(2)

From (1) and (2), we get

$f\left(x\right)=(\pi +1)\sin x+2(\pi +1)$

$\Rightarrow f_{max}=(\pi +1)+2(\pi +1)=3(\pi +1)$

and $f_{min}=-(\pi +1)+2(\pi +1)=(\pi +1)$