The value of the parameter $\alpha$ , for which the function $f\left(x\right)=1+\alpha x,\alpha \ne 0$ is the inverse of itself, is

Let $y=f\left(x\right)$ $\Rightarrow y=1+\alpha x\Rightarrow x=\frac{y-1}{\alpha }$

$\Rightarrow f^{-1}\left(y\right)$ $=\frac{y-1}{\alpha }$

Now, $f\left(x\right)=f^{-1}\left(x\right)$ $\Rightarrow 1+\alpha x=\frac{x-1}{\alpha }$

$\Rightarrow 1+\alpha x=\frac{1}{\alpha }x-\frac{1}{\alpha }$

$\Rightarrow \alpha =\frac{1}{\alpha }\text{and}1=-\frac{1}{\alpha }\Rightarrow \alpha =-1$