Question

The domain of the function $f\left(x\right)=\sqrt{\left(x^2-8x+12\right).{\left(ln\right)}^2\left(x-3\right)}$ is

• A. $[3,\infty )$
• B. $[4,\infty )$
• C. $[6,\infty )\cup \{2,4\}$
• D. $[6,\infty )\cup \{4\}$

Answer 1

Answer: (D)

$\left(x^2-8x+12\right){\left(ln\right)}^2\left(x-3\right)\ge 0$ and $x-3>0$
$\Rightarrow x^2-8x+12\ge 0\Rightarrow x\in (-\infty ,2]\cup [6,\infty )$
Also, $ln\left(x-3\right)=0\Rightarrow x=4$
Hence, the solution is $x\in \{4\}\cup [6,\infty )$