Question

The function $f\left(x\right)=\frac{x+1}{x^2+1}$ can be written as the sum of an even function $g\left(x\right)$ and an odd function $h\left(x\right)$. Then the value of $\left|g\right(0\left)\right|$ is ____.

Answer 1

$f\left(x\right)=\frac{(x+1)}{(x^2+1)}=\frac{x}{(x^2+1)}+=\frac{1}{(x^2+1)}$

since, $g\left(x\right)=\frac{1}{(x^2+1)}$ is even.

$\frac{x}{(x^2+1)}=h\left(x\right)$ is odd

$g\left(0\right)=\frac{x}{(0^2+1)}=1$

so, $|g\left(0\right)|=1$