Question

Find the values of $x$ satisfying $-1\le \left[x\right]-x^2+4\le 2,$ where $[.]$ denotes the greatest integer function.

• A. $x\in [-\sqrt{3},-1]\cup [\sqrt{3},\sqrt{7}]$
• B. $x\in [-\sqrt{5},-1]\cup [\sqrt{5},\sqrt{7}]$
• C. $x\in [-\sqrt{5},-2]\cup [\sqrt{5},\sqrt{9}]$
• D. None of these

Answer 1

The correct option is A $x\in [-\sqrt{3},-1]\cup [\sqrt{3},\sqrt{7}]$

As, $-1\le \left[x\right]-x^2+4\le 2$ $\Rightarrow x^2-5\le \left[x\right]\le x^2-2$.
Thus, to find the points for which $f\left(x\right)=x^2-5$ is less than or equal to $g\left(x\right)=\left[x\right]$ and $g\left(x\right)=\left[x\right]$ is less than or equal to $h\left(x\right)=x^2-2$.
Where the three functions $f\left(x\right),g\left(x\right)$ and $h\left(x\right)$ could be plotted as shown in the graph.
Thus, from the graph,
$x^2-5\le \left[x\right]\le x^2-2$ when $x\in [A,B]\cup [C,D]$
where $A$ and $D$ is point of intersection of $x^2-5=\pm 2\Rightarrow x=-\sqrt{3},\sqrt{7}$ and
$C$ is point of intersection of $x^2-2=1\Rightarrow x=\sqrt{3}$
$\therefore -1\le \left[x\right]-x^2+4\le 2$ is satisfied, $x\in [-\sqrt{3},-1]\cup [\sqrt{3},\sqrt{7}]$