Question

The range of the function, $f\left(x\right)=\sqrt{3x^2-4x+5}$ is

• A. $(-\infty ,\sqrt{\frac{11}{3}}]$
• B. $(-\infty ,\sqrt{\frac{11}{3}})$
• C. $[\sqrt{\frac{11}{3}},\infty )$
• D. $(\sqrt{\frac{11}{3}},\infty )$

Answer 1

The correct option is C $[\sqrt{\frac{11}{3}},\infty )$

Given,

$f\left(x\right)=\sqrt{3x^2-4x+5}$

For given function to be defined

$3x^2-4x+5>0$ as $a=3>0$ and $b^2-4ac=16-60<0$

Therefore, $f\left(x\right)=\sqrt{3x^2-4x+5}$ exists for all values of $x$

now, minimum value of $f\left(x\right)=\sqrt{\frac{-D}{4a}}=\sqrt{\frac{-16+60}{12}}=\sqrt{\frac{11}{3}}$

Therefore, range of $f\left(x\right)$ is $[\sqrt{\frac{11}{3}},\infty )$