Question

The solution set of $(x{)}^2+(x+1{)}^2=25$, where $\left(x\right)$ is the least integer greater than or equal to $x$ is

• A. $(2,4)$
• B. $(-5,-4]\cup (2,3]$
• C. $[-4,-3)\cup [3,4)$
• D. none of these

Answer 1

The correct option is B $(-5,-4]\cup (2,3]$

For an integer $x$, $\left(x\right)=x$
The equation $x^2+(x+1{)}^2=25$ then will be satisfied by the pythagorean triplet $3,4,5$
This gives $x=3$ and $-4$
Least integer would make $-4$ of every value of $x$ in $(-5,-4]$
Similarly, if $x=(2,3]$ $\left(x\right)=3$
Thus, the solution set is $x\in (5,4]\cup (2,3]$.