Question

Let $x$ be a real number $\left[x\right]$ denotes the greatest integer function, and $\left\{x\right\}$ denotes the fractional part and $\left(x\right)$ denotes the least integer function,then solve the following.

$\left[2x\right]-2x=[x+1]$

• A. $\{-1,-\frac{1}{2}\}$
• B. $\{-1,\frac{1}{2}\}$
• C. $\{1,-\frac{1}{2}\}$
• D. $\{1,\frac{1}{2}\}$

Answer 1

The correct option is A $\{-1,-\frac{1}{2}\}$

$\left[2x\right]-2x=[x+1]$

$\Rightarrow \left\{2x\right\}=\left[x\right]+1$ (i)

But range of fractional part is $[0,1)$

$\Rightarrow 0\le \left[x\right]+1<1$

$\Rightarrow -1\le \left[x\right]<0$

$\Rightarrow -1\le x<0$

$\Rightarrow \left[x\right]=-1$

Thus (i) becomes $\left\{2x\right\}=0$ it means that fraction part is zero for that,

$\therefore x=-\frac{1}{2},-1$ in $[-1,0)$