Question

A real valued function is defined as $f\left(x\right)={(x+1)}^2-1,x\ge -1$ then the set $A=\{x:f\left(x\right)=f^{-1}\left(x\right)\}$ is equal to

• A. $x=0$
• B. $x=-1$
• C. $x=1$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $x=0$
B $x=-1$

$y={(x+1)}^2-1$

$\Rightarrow \sqrt{y+1}-1=x$

$\therefore f^{-1}\left(y\right)=x=\sqrt{y+1}-1$

or $f^{-1}\left(x\right)=\sqrt{x+1}-1$

Now $f\left(x\right)=f^{-1}\left(x\right)$

$\Rightarrow {(x+1)}^2-1=\sqrt{x+1}-1$

or ${(x+1)}^4=(x+1)$

$\therefore x+1=0$ or $x+1=1^{1/3}=1,w,w^2$

$\therefore x=-1,$ or $0$ only as $x$ is real.