Question

The solution set of the equation ${(x+1)}^2+{[x-1]}^2={(x-1)}^2+{[x+1]}^2$, where $\left[x\right]$ and $\left(x\right)$ are the greatest integer and least integer function of $x$, is

• A. $x\in R$
• B. $x\in N$
• C. $x=1$
• D. $x\in Q$

Answer 1

Answer: (B)

$x\in N$

${(x+1)}^2+{[x-1]}^2={(x-1)}^2+{[x+1]}^2$

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

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

$\Rightarrow \left(x\right)=\left[x\right]$

Therefore, $x\in N$