Question

If $f\left(x\right)$ be a real valued function $f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$,
$g\left(x\right)={\int }_x^{x+8}f\left(t\right)dt$. Then $g^{\prime }\left(x\right)$ is equal to

• A. $f\left(x\right)$
• B. $f(x+8)$
• C. $8$
• D. $0$

Answer 1

The correct option is D $0$

$f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$
Substitute $x\to x+2$
$\therefore f(x+2)+f(x+6)=f(x+6)+f(x+8)\Rightarrow f\left(x\right)=f(x+8)g\left(x\right)={\int }_x^{x+8}f\left(t\right)dt\Rightarrow g^{\prime }\left(x\right)=f(x+8)-f\left(x\right)=0$