Question

The domain of the function $f\left(x\right)=\sqrt{({\log }_{0.2}x{)}^3+({\log }_{0.2}{x}^3\left)\right({\log }_{0.2}0.0016x)+36}$ is

• A. $[0,625]$
• B. $(0,625)$
• C. $(0,125]$
• D. $[0,125]$

Answer 1

The correct option is C $(0,125]$

Let ${\log }_{0.2}x=t,x>0\cdots \left(1\right)$
For $f\left(x\right)$ to be defined,
$({\log }_{0.2}x{)}^3+(3{\log }_{0.2}x\left)\right({\log }_{0.2}(0.2{)}^4+{\log }_{0.2}x)+36\ge 0$
$\Rightarrow t^3+3t(t+4)+36\ge 0$
$\Rightarrow t^3+3t^2+12t+36\ge 0\Rightarrow (t+3)(t^2+12)\ge 0$
$t+3\ge 0\Rightarrow t\ge -3\Rightarrow {\log }_{0.2}x\ge -3\Rightarrow x\le (0.2{)}^{(-3)}$
$\Rightarrow x\le 125\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$x\in (0,125]$