The value of parameter k for which the function $f(x)= 1+kx , k ≠0$ is the inverse of itself is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $k=-1$
$f(x)= 1+kx$
$f(f(x))=x$
$1+(1+kx)k=x$
$k+k^x+1=x$
$(k^2-1)x+(k+1)=0$
$1(k+1)((k+1)x+1)=0$
$⇒k=-1$