Question

Let $R=$ the set of real numbers, $Z=$ the set of integers, $N=$ the set of natural numbers. If $S$ be the solution set of the equation $(x{)}^2+[x{]}^2=(x-1{)}^2+[x+1{]}^2,$ where $\left(x\right)=$ the least integer greater than or equal to $x$ and $\left[x\right]=$ the greatest integer less than or equal to $x$, then

• A. $S=R$
• B. $S=R-Z$
• C. $S=R-N$
• D. $S=\varphi$

Answer 1

Answer: (B)

$S=R-Z$

Here $\left(x\right)$ = least integer $\ge x$

and $\left[x\right]$ = greatest integer $\le x$

For example $\left(2.3\right)=3,[2,3]=2$

$\therefore \left(x\right)-\left[x\right]=1$ if $x$ is not integer and $\left[x\right]=\left(x\right)$ if $x\in Z$

Then $(x-1)=\left(x\right)-1,[x+1]=\left[x\right]+1$

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

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

$\Rightarrow \left[x\right]-\left(x\right)+1=0$

$\Rightarrow -1+1=0$ and $x\notin Z$

and $0+1\ne 0$ if $x\in Z$

Therefore solution set $S=R-Z$