Browse Questions

The value of $\alpha$ for which the function $f(x)=1+\alpha x,\:\:\alpha\neq 0$ is the inverse of itself is:

$\begin{array}{1 1} -2 \\ -1 \\ 1 \\ 2 \end{array}$

Toolbox:
• If f(g(x))=x then g is inverse of x and f is inverse of g
Given f is inverse of itself.
$\Rightarrow\:f(f(x))=x$
$\Rightarrow 1+\alpha(1+\alpha x)=x$
$\Rightarrow\:1+\alpha+(\alpha)^2x=x$
Equating the like terms on either side, we get
$1+\alpha =0\:\:and\:\:(\alpha)^2=1$
$\Rightarrow\:\alpha=-1$