Question

Let $f$ be a real valued function satisfying $f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$ and $g\left(x\right)=f_x^{x+8}f\left(t\right)dt$ then :

• A. $g\left(1\right)=g\left(5\right)$
• B. $f\left(2\right)=f\left(10\right)$
• C. $g\left(4\right)=g\left(7\right)$
• D. $f\left(1\right)=f\left(6\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $g\left(1\right)=g\left(5\right)$
B $f\left(2\right)=f\left(10\right)$
C $g\left(4\right)=g\left(7\right)$

$f\left(x\right)+f(x+4)=f(x+2)+f(x+6)$

put $x\to x+2$

$f(x+2)+f(x+6)=f(x+4)+f(x+8)$

$\Rightarrow f\left(x\right)=f(x+8)$

Now $g\left(x\right)={\int }_x^{x+8}f\left(t\right)dt$

$\Rightarrow g^{\prime }\left(x\right)=f(x+8)-f\left(x\right)=0$

$\Rightarrow g\left(x\right)$ is constant function