Question

$f:\left(4,6\right)\to \left(6,8\right)$ be a function defined by $f\left(x\right)=x+\left[\frac{x}{2}\right]$ where $[.]$ denotes the greatest integer function then $f^{-1}\left(x\right)$in equal to.

• A. $x-\left[\frac{x}{2}\right]$
• B. $-x-2$
• C. $x-2$
• D. $\frac{1}{x+\left[\frac{x}{2}\right]}$

Answer 1

The correct option is C $x-2$

$f:(4,6)\to (6,8)$

$f\left(x\right)=x+\left[\frac{x}{2}\right]$

$2<\frac{x}{2}<3\Rightarrow \left[\frac{x}{2}\right]=2$

So; $f\left(x\right)=x+2$

function is one - one and on to So inverse

of this function is defend

$f\left(x\right)=y$ $f\left(x\right)=x+2$

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

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

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