Question

If $[x{]}^2-5[x]+6=0,$ then $x\in$
Where $[.]$ denotes GIF.

• A. $[2,4)$
• B. $[2,3)$
• C. $[3,4)$

Answer 1

The correct option is A $[2,4)$

Given: $[x{]}^2-5[x]+6=0$
Let $\left[x\right]=t$
$\Rightarrow t^2-5t+6=0$
$\Rightarrow t^2-(3+2)t+6=0$
$\Rightarrow t^2-3t-2t+6=0$
$\Rightarrow t(t-3)-2(t-2)=0$
Either $t=2$ or $t=3$

For $t=2\Rightarrow \left[x\right]=2$
$\Rightarrow \left[x\right]\le x<\left[x\right]+1$
$\Rightarrow 2\le x<2+1$
$\Rightarrow 2\le x<3$
$\Rightarrow x\in [2,3)$

For $t=3\Rightarrow \left[x\right]=3$
$\Rightarrow \left[x\right]\le x<\left[x\right]+1$
$\Rightarrow 3\le x<3+1$
$\Rightarrow 3\le x<4$
$\Rightarrow x\in [3,4)$

$\therefore x\in [2,3)\cup [3,4)$
$\Rightarrow x\in [2,4)$