Question

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

• A. $\left[- \in fty , 2 \sqrt{\frac{11}{3}}\right]$
• B. $\left(- \in fty , 2 \sqrt{\frac{11}{3}}\right)$
• C. $\left[2 \sqrt{\frac{11}{3}}, \infty\right)$
• D. $\left(2 \sqrt{\frac{11}{3}} , \in fty\right)$

Answer 1

Answer: (C)

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