The range of f(x)=√(3 x^2-4 x+5) is

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 [√(11/3), ∞)

Hence (2) is the correct answer.