Question

The range of the function $f\left(x\right)=\sqrt{x^2-3x+5}$ is

• A. $(-\infty ,-\frac{\sqrt{11}}{2}]\cup [\frac{\sqrt{11}}{2},\infty )$
• B. $[-\frac{\sqrt{11}}{2},\frac{\sqrt{11}}{2})$
• C. $[\frac{\sqrt{11}}{2},\infty )$
• D. $[\sqrt{5},\infty )$

Answer 1

The correct option is C $[\frac{\sqrt{11}}{2},\infty )$

$f\left(x\right)=\sqrt{x^2-3x+5}$
$x^2-3x+5>0$ for all real values of $x$
So, domain of $f$ is $R$.

Now, $x^2-3x+5={(x-\frac{3}{2})}^2+5-\frac{9}{4}$
$={(x-\frac{3}{2})}^2+\frac{11}{4}$
$0\le {(x-\frac{3}{2})}^2<\infty$
$\Rightarrow \frac{11}{4}\le {(x-\frac{3}{2})}^2+\frac{11}{4}<\infty$
$\Rightarrow \sqrt{\frac{11}{4}}\le \sqrt{{(x-\frac{3}{2})}^2+\frac{11}{4}}<\infty$
$\Rightarrow \frac{\sqrt{11}}{2}\le f\left(x\right)<\infty$