Mathematics - Functions Question with Solution | TestHub

For the function $f(x)$ to be defined, two conditions must be met: 1. The denominator $[x]-1\ne 0$, which implies $[x]\ne 1$. Thus, $x\notin [1,2)$. 2. The expression under the square root must be non-negative: $\frac{1-|x-1|}{[x]-1}\ge 0$. Case 1: $[x]-1>0⟹[x]\ge 2⟹x\ge 2$. In this case, we need $1-|x-1|\ge 0⟹|x-1|\le 1⟹-1\le x-1\le 1⟹0\le x\le 2$. The intersection of $x\ge 2$ and $0\le x\le 2$ is $x=2$. Case 2: $[x]-1<0⟹[x]\le 0⟹x<1$. In this case, we need $1-|x-1|\le 0⟹|x-1|\ge 1$. This implies $x-1\ge 1$ or $x-1\le -1$, so $x\ge 2$ or $x\le 0$. The intersection of $x<1$ and ($x\ge 2$ or $x\le 0$) is $x\le 0$. Combining both cases, the domain is $(-\infty ,0]\cup \{2\}$.