Question

The solution 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 and nearest integer to $x$, is

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

Answer 1

The correct option is C

$x\in I$

Explanation for correct option

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

Here, $\left[x\right]$ is the greatest integer and $\left(x\right)$ is the nearest integer

Case 1: When $x\in I$ and $I$ is integer,

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

$=2I^2+2$

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

$=2I^2+2$

So, here, LHS $=$ RHS

Case 2: When $x\in I+ƒ$, $ƒ$ is some positive fraction.

We have ${(I+ƒ+1)}^2+{\left[I+ƒ-1\right]}^2$,

if $ƒ>0.5$, then LHS will be,

$$\begin{gathered} ={(I+2)}^2+{\left[I-1\right]}^2 \\ =I^2+4+4I+I^2+1-2I \\ =2I^2+5+2I \end{gathered}$$

and when ${(I+ƒ-1)}^2+{\left[I+ƒ+1\right]}^2$, then RHS will be,

$$\begin{gathered} =I^2+I^2+1+2I \\ =2I^2+2I+1 \end{gathered}$$

So, here, LHS $\ne$ RHS

So, the given condition where $\left[x\right]$ is the greatest integer and $\left(x\right)$ is the nearest integer for equation ${(x+1)}^2+{[x-1]}^2={(x-1)}^2+{[x+1]}^2$ is only possible for integral values of $x$.

Thus, $x\in I$.

Hence, the correct option is (C).