The value of the parameter α , for which the function f(x)=1+αx, α≠0 is the inverse of itself, is
–2
–1
1
2
Let y=f(x) ⇒y=1+αx⇒x=y−1α
⇒ f−1(y) =y−1α
Now, f(x)=f−1(x) ⇒1+αx=x−1α
⇒ 1+αx=1αx−1α
⇒ α=1α and 1=−1α ⇒ α=−1