Question

The correct option(s) for the function $f\left(x\right)=\sqrt{\left(\right[x]-1)}+\sqrt{(4-[x\left]\right)}$ is/are
(where [.] represents the greatest integer function).

• A. Range of the function is $\{\sqrt{2},1+\sqrt{3}\}$
• B. Domain of the function is $[1,5)$
• C. Domain of the function is $(0,4]$
• D. Range of the function is $\{\sqrt{3},1+\sqrt{2}\}$

Answer 1

Answer: (B), (D)

The correct options are
B Domain of the function is $[1,5)$
D Range of the function is $\{\sqrt{3},1+\sqrt{2}\}$

$f\left(x\right)=\sqrt{\left(\right[x]-1)}+\sqrt{(4-[x\left]\right)}$
For $f\left(x\right)$ to be defined,
$\left[x\right]-1\ge 0$ and $4-\left[x\right]\ge 0$
$\Rightarrow 1\le \left[x\right]\le 4$
$\Rightarrow 1\le x<5$
For,
$1)x\in [1,2)$
$\left[x\right]=1$
$\Rightarrow \sqrt{\left(\right[x]-1)}+\sqrt{(4-[x\left]\right)}=\sqrt{3}$

$2)x\in [2,3)$
$\left[x\right]=2$
$\Rightarrow \sqrt{\left(\right[x]-1)}+\sqrt{(4-[x\left]\right)}=1+\sqrt{2}$

$3)x\in [3,4)$
$\left[x\right]=3$
$\Rightarrow \sqrt{\left(\right[x]-1)}+\sqrt{(4-[x\left]\right)}=1+\sqrt{2}$

$4)x\in [4,5)$
$\left[x\right]=4$
$\Rightarrow \sqrt{\left(\right[x]-1)}+\sqrt{(4-[x\left]\right)}=\sqrt{3}$

Thus, range of the function is $\{\sqrt{3},1+\sqrt{2}\}$