Mathematics - Quadratic Equation Question with Solution | TestHub

$f(x)$ is defined if $3x^2-4x+5\ge 0$
$\Rightarrow 3\left[x^2-\frac{4}{3}x+\frac{5}{3}\right]\ge 0\Rightarrow 3\left[{\left(x-\frac{2}{3}\right)}^2+\frac{11}{9}\right]\ge 0$
Which is true for all real $x\therefore$ Domain of $f(x)=$ $(-\infty ,\infty )$
Let $y=\sqrt{3x^2-4x+5}$
$\Rightarrow y^2=3x^2-4x+5$ i.e. $3x^2-4x+\left(5-y^2\right)=0$
For $x$ to be real , $16-12\left(5-y^2\right)\ge 0\Rightarrow y\ge \sqrt{\frac{11}{3}}$
$\therefore$ Range of $y=\left[\sqrt{\frac{11}{3}},\infty \right)$