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

$\left.\frac{2}{3}, \infty\right)$ $\left.\frac{11}{3}, \infty\right)$ $\left[\frac{11}{3}, \infty\right]$ $\left(\frac{2}{3}, \infty\right)$

$\left.\frac{11}{3}, \infty\right)$

Given $f(x)=\sqrt{3 x^2-4 x+5}=\sqrt{3\left(x-\frac{2}{3}\right)^2+\frac{11}{3}}$ clearly domain of f(x) is R. Now f(x) will be minimum $\sqrt{\frac{11}{3}}$ when x = 2/3 Hence range is $\left.\frac{11}{3}, \infty\right)$ Hence (2) is the correct answer.