Question

The value of the parameter $\alpha$, for which the function $f\left(x\right)=1+\alpha$ where $\alpha \ne x$ is the inverse of itself is

• A. -2
• B. -1
• C. 1
• D. 0

Answer 1

Answer: (A), (D)

-2
0

$f\left(x\right)=1+\alpha =\frac{1}{1+\alpha }$

$(1+\alpha {)}^2=1$

$(\alpha +1{)}^2-1^2=0$

$(\alpha +1+1)(\alpha +1-1)=0$

$(\alpha +2)\left(\alpha \right)=0$

$\alpha =-2,\alpha =0$