Question

The value of $x$ graphically satisfying $\left[x\right]-1+x^2\ge 0$ where $[.]$ denotes the greatest integer function.

• A. $(\infty ,-\sqrt{3}]\cup [1,\infty )$
• B. $(\infty ,\sqrt{3}]\cup [1,\infty )$
• C. $(\infty ,\sqrt{3}]\cup [-1,-\infty )$
• D. $(-\infty ,-\sqrt{3}]\cup [1,\infty )$

Answer 1

The correct option is D $(-\infty ,-\sqrt{3}]\cup [1,\infty )$

As, $\left[x\right]-1+x^2\ge 0$
$\Rightarrow x^2-1\ge -\left[x\right]$
Thus, to find the points for which $f\left(x\right)=x^2-1$ is greater then or equals to
$g\left(x\right)=-\left[x\right]$
where two functions $f\left(x\right)$ and $g\left(x\right)$ could be plotted as shown below.
From the adjoining figure; the solution set lies when$x\le A$ or $x\ge B$
Thus, to find $A$ and $B$.
It is clear that $f\left(x\right)$ and $g\left(x\right)$ intersects when;
$-\left[x\right]=2$
$\therefore x^2-1=2$
or $x=\pm \sqrt{3}$
$\Rightarrow x=-\sqrt{3}$
(neglecting $x=+\sqrt{3}$ as $A$ lies for $x<0$)
Thus, for $A:x=-\sqrt{3}$ and for $B:x=1$
$\therefore$ Solution set for which $x^2-1\ge -\left[x\right]$ holds.
$\Rightarrow x\in (-\infty ,-\sqrt{3}]\cup [1,\infty )$