Question

If $f\left(x\right)$ is a continuous function in $[0,\pi ]$ such that $f\left(0\right)=f\left(\pi \right)=0$ then the value of ${\int }_0^{\pi /2}\left(f\right(2x)+f^{{}^{\prime \prime }}(2x\left)\right)\sin x.\cos xdx$ is equal to

• A. $\pi$
• B. $2\pi$
• C. $3\pi$
• D. 0

Answer 1

The correct option is D 0

Given : $f\left(x\right)$ is a continuous function , $(0,\pi )f\left(0\right)=f\left(\pi \right)=0$

According to the conditions

$f\left(x\right)$ can be $\sin xor\tan x$

But since $f\left(x\right)$ is continuous $(0,\pi )$

$f\left(x\right)=\sin xf"\left(x\right)=-\sin x$

$I={\int }_0^{\frac{\pi }{2}}\left[f\right(2x)+f"(x\left)\right]\sin x\cos xdxI={\int }_0^{\frac{\pi }{2}}[\sin 2x-\sin 2x]\sin x\cos xdxI=0$

Hence the correct answer is $0$.