Question

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

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is C $-1$

Let $f\left(x\right)=y,$ then $1+\alpha x=y$

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

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

$\therefore f^{-1}\left(x\right)=\frac{x-1}{\alpha }$

Now $f\left(x\right)=f^{-1}\left(x\right)\Rightarrow 1+\alpha x=\frac{x-1}{\alpha }\Rightarrow 1+\alpha x=\frac{1}{\alpha }x-\frac{1}{\alpha }$ ........ As $f\left(x\right)$ is invertible function itself

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

$\Rightarrow \alpha =-1$