Question

Let $f\left(x\right)=(x+1{)}^2-1,(x\ge -1)$. Then the set $S=\{x:f(x)=f^{-1}(x\left)\right\}$. is

• A. $Empty$
• B. $\{0,-1\}$
• C. $\{0,1,-1\}$
• D. $\{0,-1,\frac{-3+i\sqrt{3}}{2},\frac{-3-i\sqrt{3}}{2}\}$

Answer 1

The correct option is D $\{0,-1,\frac{-3+i\sqrt{3}}{2},\frac{-3-i\sqrt{3}}{2}\}$

$Letf\left(x\right)=(x+1{)}^2-1,x\ge -1.Sincef(x)=f^{-1}(x)\therefore (x+1{)}^2-1=\sqrt{1+x}-1(\because f^{-1}(x)=\sqrt{1+x}-1)\Rightarrow (x+1{)}^4=1+x\Rightarrow (x+1\left)\right[(x+1{)}^3-1]=0\Rightarrow x=-1or(x+1{)}^3=1\Rightarrow x+1=1,\omega ,{\omega }^2\Rightarrow x=0,-1,\frac{-3+i\sqrt{3}}{2},\frac{-3-i\sqrt{3}}{2}$