Question

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

Answer 1

As $\left[x\right]=x-\left\{x\right\}$

$\Rightarrow [x{]}^2+[x-2]<0\Rightarrow (x-\left\{x\right\}{)}^2+(x-2)+\{x-2\}<0$

$\Rightarrow (x-\frac{1}{2}{)}^2+x-2+\{x-2\})<0$

As $\left\{x\right\}=\frac{1}{2}\Rightarrow x$ is of from $2.5,3.5,--$

$\Rightarrow (x-\frac{1}{2}{)}^2+x-2+\frac{1}{2}<0$

$\Rightarrow x^2+\frac{1}{4}-x+x-2+\frac{1}{2}<0$

$\Rightarrow x^2-2+\frac{3}{4}<0$

$\Rightarrow x^2-\frac{5}{4}<0\Rightarrow x^2<\frac{5}{4}\Rightarrow \frac{-\sqrt{5}}{2}<x<\frac{\sqrt{5}}{2}$

$\Rightarrow -1.12<x<1.12\Rightarrow x=-0.5,0.5\Rightarrow \left(c\right)$