The value of parameter k for which the function $f(x)= 1+kx , k ≠0$ is the inverse of itself is :

$k=-2$ $k=-1$ $k=1$ $k=2$

$k=-1$

The correct answer is Option (2) → $k=-1$ $f(x)= 1+kx$ $f(f(x))=x$ $1+(1+kx)k=x$ $k+k^x+1=x$ $(k^2-1)x+(k+1)=0$ $1(k+1)((k+1)x+1)=0$ $⇒k=-1$