Mathematics - Functions Question with Solution | TestHub

For $f(x)$ to be defined, two conditions must be met: 1) The argument of the logarithm must be positive: $x^2-3x+2>0⟹(x-1)(x-2)>0⟹x\in (-\infty ,1)\cup (2,\infty )$. 2) The argument of the square root must be non-negative: ${\log }_{1/2}(x^2-3x+2)\ge 0$. Since the base $1/2<1$, this inequality implies $x^2-3x+2\le (1/2{)}^0=1$. So, $x^2-3x+1\le 0$. The roots of $x^2-3x+1=0$ are $x=\frac{3\pm \sqrt{(-3{)}^2-4(1)(1)}}{2}=\frac{3\pm \sqrt{5}}{2}$. Thus, $x\in [\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}]$. Combining both conditions, we find the intersection: $[\frac{3-\sqrt{5}}{2},1)\cup (2,\frac{3+\sqrt{5}}{2}]$. Therefore, option A is correct.