Let $$f:\left( {4,\,6} \right) \to \left( {6,\,8} \right)$$ be a function defined by $$f\left( x \right) = x + \left[ {\frac{x}{2}} \right]$$ (where [.] denotes the greatest integer function), then $${f^{ - 1}}\left( x \right)$$ is 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 :
$$x - 2$$
Solution :
$$\eqalign{
& {\text{Since }}f:\left( {4,\,6} \right) \to \left( {6,\,8} \right) \Rightarrow f\left( x \right) = x + 2 \cr
& \therefore \,{f^{ - 1}}\left( x \right) = x - 2 \cr} $$
Releted MCQ Question on Calculus >> Sets and Relations
Releted Question 1
If $$X$$ and $$Y$$ are two sets, then $$X \cap {\left( {X \cup Y} \right)^c}$$ equals.