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Q. If $[x]^2+[x]-2<0$ and $\{x\}=\frac{1}{2}$, then the number of possible real values of $x$, is
[Note: $[ x ]$ and $\{ x \}$ denote the greatest integer less than or equal to $x$ and fractional part of $x$ respectively.]

Relations and Functions - Part 2

Solution:

$[x]^2+[x]<2 \Rightarrow([x]+2)([x]-1)<0 \Rightarrow-2<[x]<1$
$\therefore[ x ]=-1,0 \Rightarrow x \in[-1,1)$
So, $\{x\}=\frac{1}{2} \Rightarrow x=\frac{-1}{2}, \frac{1}{2}$ ]