Question

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

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

Answer 1

Answer: (D)

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