Domain of the function $f$, given by $f(x) = \dfrac{1}{\sqrt{(x-2)(x-5)}}$ is
(A) $(-\infty, 2] \cup [5, \infty)$
(B) $(-\infty, 2) \cup (5, \infty)$
(C) $(-\infty, 3) \cup [5, \infty)$
(D) $(-\infty, 3] \cup (5, \infty)$

Question

Accepted Answer

> [!TLDR] > A square root in a denominator forces its argument to be *strictly* positive, so solve the strict inequality $(x-2)(x-5)>0$. **Step 1 — Identify the two constraints.** $$f(x)=\frac{1}{\sqrt{(x-2)(x-5)}}$$ - For $\sqrt{\;\cdot\;}$ to be **real**: $(x-2)(x-5)\ge 0$. - For the fraction to be **defined** (no division by zero): $\sqrt{(x-2)(x-5)} e 0$, i.e. $(x-2)(x-5) e 0$. Combining, the single condition is the **strict** inequality: $$(x-2)(x-5)>0$$ This is the whole point of the question — the endpoints $x=2$ and $x=5$ are *excluded* precisely because the root sits downstairs. **Step 2 — Solve by sign analysis.** Critical points: $x=2$ and $x=5$. Test each interval: | Interval | $(x-2)$ | $(x-5)$ | Product | |---|---|---|---| | $x<2$ (say $x=0$) | $-$ | $-$ | $+$ $\checkmark$ | | $25$ (say $x=6$) | $+$ | $+$ | $+$ $\checkmark$ | (The product of a downward-free upward parabola is positive **outside** its roots — the standard "outside the roots" rule for a positive leading coefficient.) **Step 3 — Write the domain.** $$ ext{Domain}(f)=(-\infty,2)\cup(5,\infty)$$ **Step 4 — Reject the distractors.** - **(A)** $(-\infty,2]\cup[5,\infty)$ — includes $x=2,5$, where $f$ is $ frac{1}{0}$: undefined. - **(C), (D)** — use the number 3, which is not a root of $(x-2)(x-5)$ at all; and $x=3$ actually makes the product **negative** $(1)(-2)=-2$. > [!ANSWER] > The correct option is **(B) $(-\infty, 2) \cup (5, \infty)$**. ANSWER: B

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