Mathematics - Functions Question with Solution | TestHub

For the first term, $\sqrt{{\log }_2\left(\frac{x-1}{x-2}\right)}$, we need two conditions: (1) $\frac{x-1}{x-2}>0$, which implies $x\in (-\infty ,1)\cup (2,\infty )$. (2) ${\log }_2\left(\frac{x-1}{x-2}\right)\ge 0$. Since the base is $2>1$, this means $\frac{x-1}{x-2}\ge 2^0=1$. So $\frac{x-1}{x-2}-1\ge 0\Rightarrow \frac{x-1-(x-2)}{x-2}\ge 0\Rightarrow \frac{1}{x-2}\ge 0$. This implies $x-2>0$ (as denominator cannot be zero), so $x>2$. Combining (1) and (2), the domain for the first term is $x\in (2,\infty )$. For the second term, $\frac{1}{\sqrt{3-|x|}}$, we need $3-|x|>0$. This means $|x|<3$, which implies $-3<x<3$. The overall domain of $f(x)$ is the intersection of $(2,\infty )$ and $(-3,3)$. This intersection is $(2,3)$.