Question

The set of values of $x,$ which satisfy $x^2-13x+\left[x\right]+36=0$ is

• A. $[5,7)$
• B. $(6,7)$
• C. $[6,7)$
• D. $(5,7)$

Answer 1

The correct option is D $(5,7)$

$\left[x\right]=x-\left\{x\right\}$
$x^2-13x+\left[x\right]+36=0$
$\Rightarrow x^2-13x+x-\left\{x\right\}+36=0$
$\Rightarrow x^2-12x+36=\left\{x\right\}\Rightarrow (x-6{)}^2=\{x\}$
$\Rightarrow (x-6{)}^2\in [0,1)$
$\Rightarrow x\in (5,7)$