Question

If $f\left(x\right)=\{\begin{array}{cc}x+1& xϵ[-1,0]\\ x^2+1& xϵ(0,1)\end{array}$, then the value of $\frac{f^{-1}\left(0\right)+f^{-1}\left(1\right)+f^{-1}\left(2\right)}{f(-1)+f\left(0\right)+f\left(1\right)}$ is-

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

$0$

$f\left(x\right)=\{\begin{array}{cc}x+1& x\in (-1,0)\\ x^2+1& x\in (0,1)\end{array}$

$\therefore x=\{\begin{array}{cc}{f}^{-1}(x+1)& x\in [-1,0]\\ f^{-1}(x^2+1)& x\in (0,1)\end{array}$

$f^{-1}\left(0\right)$:- put $x=-1$

$\therefore f^{-1}\left(0\right)=x=-1$

$\therefore f^{-1}\left(1\right)$ put $x=0$

$f^{-1}=0$

$f^{-1}\left(2\right)$:- put $x=1$

$f^{-1}\left(2\right)=1$

$\therefore f\left(0\right)=1$

$\therefore f\left(1\right)=1+1=2$

$\therefore f(-1)=-1+1=0$

$\therefore \frac{f^{-1}\left(0\right)+f^{-1}\left(1\right)+f^{-1}\left(2\right)}{f(-1)+f\left(0\right)+f\left(1\right)}=\frac{-1+0+1}{3}$

$=\frac{0}{3}$

$=0$.