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.$\left[x\right]+|x-2|\le 0$ and $-1\le x\le 3$, Number of elements in set is

Answer 1

case 1. $-1\le x<0\Rightarrow \left[x\right]=-1$ and $x-2<0$

Thus given equation becomes,

$\Rightarrow -1-(x-2)\le 0\Rightarrow x\ge 1\Rightarrow$ No solution,

since $-1\le x<0$

In all other cases $\left[x\right]+|x-2|>0$

Hence for the given equation, there does not exist any solution.