Function Domain and Range Matching Solution

The problem requires matching the domain and range of two given functions, $f(x) = \frac{1}{\sqrt{x^2 - 1}}$ and $f(x) = \sqrt{x - 2}$, with the provided intervals.

Analysis of Functions

Part A: Domain of $f(x) = \frac{1}{\sqrt{x^2 - 1}}$

For the function to be defined:

• The expression under the square root must be non-negative: $x^2 - 1 \ge 0$.
• The denominator cannot be zero: $\sqrt{x^2 - 1} \neq 0$, which implies $x^2 - 1 \neq 0$.
• Combining these, we need $x^2 - 1 > 0$.
• Solving $x^2 > 1$, we get $x < -1$ or $x > 1$.
• Therefore, the domain is $(-\infty, -1) \cup (1, \infty)$. This corresponds to List II.

Part B: Range of $f(x) = \frac{1}{\sqrt{x^2 - 1}}$

Let $y = \frac{1}{\sqrt{x^2 - 1}}$.

• Since $\sqrt{x^2 - 1}$ is always positive for the domain $(-\infty, -1) \cup (1, \infty)$, $y$ must be positive.
• As $x$ approaches $1$ or $-1$, $x^2 - 1$ approaches $0^+$, so $\sqrt{x^2 - 1}$ approaches $0^+$, and $y$ approaches $\infty$.
• As $|x|$ approaches $\infty$, $x^2 - 1$ approaches $\infty$, so $\sqrt{x^2 - 1}$ approaches $\infty$, and $y$ approaches $0$.
• Thus, the range is $(0, \infty)$. This corresponds to List III.

Part C: Domain of $f(x) = \sqrt{x - 2}$

For the function to be defined, the expression under the square root must be non-negative:

• $x - 2 \ge 0$.
• Solving for $x$, we get $x \ge 2$.
• Therefore, the domain is $[2, \infty)$. This corresponds to List I.

Part D: Range of $f(x) = \sqrt{x - 2}$

Let $y = \sqrt{x - 2}$.

• The square root function $\sqrt{\cdot}$ outputs non-negative values. So, $y \ge 0$.
• The minimum value occurs at the minimum value of the domain, $x=2$. Here, $y = \sqrt{2 - 2} = 0$.
• As $x$ increases, $y$ increases without bound.
• Thus, the range is $[0, \infty)$. This corresponds to List IV.

Correct Matching

Based on the analysis:

• A matches with II.
• B matches with III.
• C matches with I.
• D matches with IV.

The correct option is the one that lists these matches: A-II, B-III, C-I, D-IV.