Question

Let $f:(2,3)\to (0,1)$ be defined by $f(x)=x-[x]$. Then, $f^{-1}(x)$ equals to

• A. $x-2$
• B. $x+1$
• C. $x-1$
• D. $x+2$

Answer 1

Answer: (D)

The given function is
$f:(2,3)\to (0,1)$ defined by
$f(x)=x-[x]$
Let $y\in (0,1)$ such that $y=f(x)$
$\therefore y=x-2$
$\{\because 2<x<3\Rightarrow [x]=2\}$
$\Rightarrow x=y+2$
$\therefore f^{-1}(y)=y+2$
$\left[\because x=f^{-1}(y)\right]$
$\Rightarrow f^{-1}(x)=x+2$