Question

The domain of the function $f\left(x\right)=\frac{3}{4-x^2}+lo{g}_{10}\left(x^3-x\right)$, is

• A. $\left(-1,0\right)\cup \left(1,2\right)$
• B. $\left(1,2\right)\cup \left(2,\infty \right)$
• C. $\left(-1,0\right)\cup \left(1,2\right)\cup \left(2,\infty \right)$
• D. $(1,2)$

Answer 1

Answer: (C)

Let $g\left(x\right)=\frac{3}{4-x^2}$
$\therefore x\ne \pm 2$
$\therefore D\left(g\left(x\right)\right)=R-\left\{-2,2\right\}$
$h\left(x\right)=lo{g}_{10}\left(x^3-x\right)$
$\therefore x^3-x>0$
$\Rightarrow x\left(x+1\right)\left(x-1\right)>0$


$\therefore x\in \left(-1,0\right)\cup \left(1,\infty \right)$
$\therefore$ Domain of $f\left(x\right)$ is $\left(-1,0\right)\cup \left(1,2\right)\cup \left(2,\infty \right)$.