Range of the function f (x)=x^2+x+1 is equal to - Vidyadip

MCQ

Range of the function $f (x)=\sqrt{x^2+x+1}$ is equal to

A
$[0, \infty]$
✓
$\left[\frac{\sqrt{3}}{2}, \infty\right)$
C
$\left(\frac{\sqrt{3}}{2}, \frac{-\sqrt{3}}{2}\right)$
D
(0,0)

✓

Answer

Correct option: B.

$\left[\frac{\sqrt{3}}{2}, \infty\right)$

(B)
Here, $f (x)=\sqrt{x^2+x+1}$
$\Rightarrow y^2=x^2+x+1$
$\Rightarrow x^2+x+\left(1-y^2\right)=0$
$\Rightarrow x=\frac{-1+\pm{1-4\left(1-y^2\right)}}{2}$
$\Rightarrow x=\frac{-1 \pm \sqrt{4 y^2-3}}{2}$
For $x$ real, $4 y^2-3 \geq 0$
$\therefore y \geq \pm \frac{\sqrt{3}}{2}$
$\therefore R _{ f }=\left[\frac{\sqrt{3}}{2}, \infty\right)$

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