CUET Preparation Today
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 |
-1 |
Let f(x)=y, then 1+αx=y $\Rightarrow x = \frac{y-1}{\alpha}$ $f^{-1}(y) = \frac{y-1}{\alpha}$ $\Rightarrow f^{-1}(x)=\frac{x-1}{\alpha}$ Now, $f (x)= f^{-1} (x) $ $ 1+\alpha x = \frac{x-1}{\alpha}$ ⇒ $\alpha = –1$ |
