Question

If $g\left(x\right)=f\left(x\right)-1$ and $f\left(x\right)+f(1-x)=2,\forall x\in R$, then $g\left(x\right)$ is symmetrical about

• A. the origin
• B. the line $x=\frac{1}{2}$
• C. the point $(1,0)$
• D. the point $(\frac{1}{2},0)$

Answer 1

The correct option is D the point $(\frac{1}{2},0)$

$f\left(x\right)+f(1-x)=2\Rightarrow g\left(x\right)+g(1-x)=0[\because g(x)=f(x)-1]$

Replace $x$ by $x+\frac{1}{2}$
$\Rightarrow g(x+\frac{1}{2})+g(\frac{1}{2}-x)=0\Rightarrow g(\frac{1}{2}+x)=-g(\frac{1}{2}-x)$

Therefore, $g\left(x\right)$ is symmetric about the point $(\frac{1}{2},0)$.