Let $g\left(x\right)$  be a polynomial of degree one and $f\left(x\right)$ be defined by  $f\left(x\right)=\left\{\begin{array}{l}g\left(x\right),x\le 0\\ {\left|x\right|}^{\sin x},x>0\end{array}\right.$  If $f\left(x\right)$ is continuous satisfying $f\text{'}\left(1\right)=f\left(-1\right),$  then $g\left(x\right)$ is

$f\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }{\left|x\right|}^{\sin x}=1$

$f\left(0^{-}\right)=g\left(0\right)=1$

Let $g\left(x\right)=ax+b$

For $x>0,f\text{'}\left(x\right)=e^{\sin x\ln \left(\left|x\right|\right)}\left[\cos x\ln \left(\left|x\right|\right)+\frac{\sin x}{x}\right]$

$f\text{'}\left(x\right)=1\left[0+\sin 1\right]=\sin 1$

$f\left(-1\right)=-a+1\Rightarrow a=1-\sin 1$

$g\left(x\right)=\left(1-\sin 1\right)x+1$