Question

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

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

Answer 1

Answer: (B)

$(-5,-4]\cup (2,3]$

If $x=k$, where $k$ is an integer.
The equation is $k^2+(k+1{)}^2=25$
$\Rightarrow k^2+k-12=0$
$k=3,-4\Rightarrow x=3,-4$.
If $x=k+f$ , where $k$ is an integer $0<f<1$,
The equation is
${(k+1)}^2+{(k+2)}^2=25$
$k^2+3k-10=0$
$k=2,-5$.
$\therefore$ $x=2+f,x=-5+f$
$x=-5+f$ and $x=-4$ $\Rightarrow x\in (-5,-4]$
$x=2+f$ and $x=3$ $\Rightarrow x\in (2,3]$
$\therefore$ solution set is $(-5,-4]\cup (2,3]$.