Question

The domain of $f\left(x\right)=\sqrt{(x^2-3x-10)\left[\ln \right(x-3){]}^2}$ is

• A. $[5,\infty )$
• B. $[5,\infty )\cup \left\{4\right\}$
• C. $[4,\infty )$
• D. $[4,\infty )-\left\{5\right\}$

Answer 1

Answer: (B)

Given: $f\left(x\right)=\sqrt{(x^2-3x-10)\left[\ln \right(x-3){]}^2}$
For domain,
$x-3>0\Rightarrow x>3\cdots \left(i\right)$
And
$(x^2-3x-10)\left[\ln \right(x-3){]}^2\ge 0$
$\Rightarrow (x-5)(x+2)\left[\ln \right(x-3){]}^2\ge 0$
$\Rightarrow x\in (-\infty ,-2]\cup [5,\infty )\cup \left\{4\right\}\cdots \left(ii\right)$

(because when $x=4\Rightarrow ln(x-3)=ln1=0$ )
So, taking $\left(i\right)\cap \left(ii\right)$
Domain of $f\left(x\right)$ is $[5,\infty )\cup \left\{4\right\}$