Question

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

• A. $(-\infty ,\infty )$
• B. $[{\log }_{e}5,\infty )$
• C. $[{\log }_e\frac{11}{3},\infty )$
• D. $[{\log }_e\frac{11}{3},{\text{log}}_{e}5]$

Answer 1

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

$D\left(f\right)=R$ as $3x^2-4x+5>0$ for any real value of $x$

$f\left(x\right)={\log }_e(3x^2-4x+5)$
$\Rightarrow f\left(x\right)={\log }_e(3(x^2-\frac{4x}{3})+5)$
$={\log }_e(3{(x-\frac{2}{3})}^2+\frac{11}{3})$

Now, $\frac{11}{3}\le 3{(x-\frac{2}{3})}^2+\frac{11}{3}<\infty$
As ${\log }_{e}x$ is strictly increasing function,
${\log }_e\frac{11}{3}\le {\log }_e(3{(x-\frac{2}{3})}^2+\frac{11}{3})<\infty$

Hence, $R\left(f\right)=[{\log }_e\frac{11}{3},\infty )$