Question

The solution set of $(x{)}^2+(x+1{)}^2=25$, where $(x)$ 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

Answer: (B)

If $x=n\in Z,(x{)}^2+(x+1{)}^2=25$
$\Rightarrow n^2+(n+1{)}^2=25$
$\Rightarrow 2n^2+2n-24=0$
$\Rightarrow n^2+n-12=0$
$\Rightarrow n=3,-4$
$\therefore x=3,-4$
If $x=n+k,n\in Z,0<k<1$, then
$(x{)}^2+(x+1{)}^2=25$
$\Rightarrow (n+1{)}^2+(n+2{)}^2=25$
$\Rightarrow 2n^2+6n-20=0$
$\Rightarrow n^2+3n-10=0$
$\Rightarrow n=2,-5$
$\therefore x=2+k,-5+k$,
where $0<k<1$
$\therefore x>2,x>-5$
$\therefore$ Solution set is $(-5,-4]\cup (2,3]$