Question

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

• A. $(-\infty ,{\log }_e\frac{11}{3}]$
• B. $[{\log }_e\frac{11}{3},\infty )$
• C. $[-{\log }_e\frac{11}{3},{\log }_e\frac{11}{3}]$
• D. None of these

Answer 1

Answer: (B)

$[{\log }_e\frac{11}{3},\infty )$

$f\left(x\right)$ is defined if $3x^2-4x+5>0$
$\Rightarrow 3[x^2-\frac{4}{3}x+\frac{5}{3}]>0\Rightarrow 3[{(x-\frac{2}{3})}^2+\frac{11}{9}]>0$
which is true for all real $x$

$\therefore$ Domain $f\left(x\right)=(-\infty ,\infty )$

Let $y={\log }_e(3x^2-4x+5)\Rightarrow e^y=3x^2-4x+5\Rightarrow 3x^2-4x+(5-e^y)=0$

For $x$ to be real

$16-12(5-e^y)\ge 0\Rightarrow 12e^y\ge 44\Rightarrow e^y\ge \frac{11}{3}\Rightarrow y={\log }_e\frac{11}{3}$

$\therefore$ Range of $f\left(x\right)=[{\log }_e\frac{11}{3},\infty )$