Key Concepts and Formulas

• Function Composition: For functions $f$ and $g$, the composition $f(g(x))$ is found by substituting $g(x)$ into $f(x)$. The domain of $f(g(x))$ is the set of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.
• One-one (Injective) Function: A function $h$ is one-one if for any $x_1, x_2$ in its domain, $h(x_1) = h(x_2)$ implies $x_1 = x_2$.
• Onto (Surjective) Function: A function $h: A \to B$ is onto if its range is equal to its codomain $B$.
• Rational Function Simplification: A rational function of the form $h(x) = \frac{ax+b}{cx+d}$ can be analyzed by expressing it as $h(x) = k + \frac{m}{cx+d}$.

Step-by-Step Solution

Step 1: Determine the Composition Function $f(g(x))$

We are given $f(x) = 2x - 1$ and $g(x) = \frac{x - 1/2}{x - 1}$. The domain of $g$ is $\mathbb{R} - \{1\}$. The domain of $f$ is $\mathbb{R}$. Since the range of $g$ is a subset of the domain of $f$, the composition $f(g(x))$ is well-defined.

To find $f(g(x))$, we substitute $g(x)$ into $f(x)$: $f(g(x)) = 2(g(x)) - 1$ $f(g(x)) = 2 \left( \frac{x - 1/2}{x - 1} \right) - 1$

First, simplify the term inside the parenthesis: $x - \frac{1}{2} = \frac{2x - 1}{2}$. $f(g(x)) = 2 \left( \frac{\frac{2x - 1}{2}}{x - 1} \right) - 1$ $f(g(x)) = 2 \left( \frac{2x - 1}{2(x - 1)} \right) - 1$

Cancel out the factor of 2: $f(g(x)) = \frac{2x - 1}{x - 1} - 1$

To combine the terms, find a common denominator: $f(g(x)) = \frac{2x - 1 - (x - 1)}{x - 1}$ $f(g(x)) = \frac{2x - 1 - x + 1}{x - 1}$ $f(g(x)) = \frac{x}{x - 1}$

Let $h(x) = f(g(x))$. The domain of $h(x)$ is the domain of $g(x)$, which is $\mathbb{R} - \{1\}$. The codomain of $h(x)$ is the codomain of $f(x)$, which is $\mathbb{R}$. So, we have the function $h: \mathbb{R} - \{1\} \to \mathbb{R}$ defined by $h(x) = \frac{x}{x - 1}$.

Step 2: Analyze $h(x)$ for One-one (Injectivity)

To check if $h(x)$ is one-one, we assume $h(x_1) = h(x_2)$ for $x_1, x_2 \in \mathbb{R} - \{1\}$ and see if it implies $x_1 = x_2$. $\frac{x_1}{x_1 - 1} = \frac{x_2}{x_2 - 1}$

Cross-multiply: $x_1(x_2 - 1) = x_2(x_1 - 1)$ $x_1x_2 - x_1 = x_1x_2 - x_2$

Subtract $x_1x_2$ from both sides: $-x_1 = -x_2$ $x_1 = x_2$ Since $h(x_1) = h(x_2)$ implies $x_1 = x_2$, the function $h(x)$ is one-one.

Step 3: Analyze $h(x)$ for Onto (Surjectivity)

To check if $h(x)$ is onto, we need to find its range and compare it with its codomain, which is $\mathbb{R}$. Let $y = h(x)$ and solve for $x$ in terms of $y$. $y = \frac{x}{x - 1}$ $y(x - 1) = x$ $yx - y = x$ $yx - x = y$ $x(y - 1) = y$ $x = \frac{y}{y - 1}$

For $x$ to be defined, the denominator cannot be zero, so $y - 1 \neq 0$, which means $y \neq 1$. This implies that the range of $h(x)$ is $\mathbb{R} - \{1\}$. The codomain of $h(x)$ is $\mathbb{R}$. Since the range ($\mathbb{R} - \{1\}$) is not equal to the codomain ($\mathbb{R}$), the function $h(x)$ is not onto. Specifically, the value $y=1$ is not in the range of $h(x)$.

Common Mistakes & Tips

• Domain Issues: Always pay close attention to the domain of the composite function. The domain of $f(g(x))$ is restricted by the domain of $g(x)$ and the condition that $g(x)$ must be in the domain of $f(x)$. In this case, $x \neq 1$ is the primary restriction.
• Simplification Errors: Algebraic mistakes during the simplification of the composite function can lead to incorrect analysis of its properties. Double-check each step.
• Range vs. Codomain: For a function to be onto, its range must be exactly equal to its codomain. If the range is a proper subset of the codomain, the function is not onto.

Summary

We first calculated the composite function $f(g(x))$ by substituting $g(x)$ into $f(x)$, which resulted in $h(x) = \frac{x}{x - 1}$. We then analyzed this function for injectivity by showing that $h(x_1) = h(x_2)$ implies $x_1 = x_2$, confirming it is one-one. Finally, we determined the range of $h(x)$ by solving for $x$ in terms of $y$, finding the range to be $\mathbb{R} - \{1\}$. Since the range is not equal to the codomain ($\mathbb{R}$), the function is not onto. Therefore, $f(g(x))$ is one-one but not onto.

The final answer is \boxed{A}.