Key Concepts and Formulas

• Injectivity (One-to-One Function): A function $f: A \to B$ is injective if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Graphically, a function is injective if no horizontal line intersects its graph more than once. For differentiable functions, if $f'(x)$ is strictly positive or strictly negative over its domain, the function is injective.
• Surjectivity (Onto Function): A function $f: A \to B$ is surjective if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$. In other words, the range of the function is equal to its codomain.

Step-by-Step Solution

Step 1: Define the Domain and Codomain of the Function The problem states that $A = \{x \in \mathbb{R} : x \text{ is not a positive integer}\}$. This means the domain $A$ includes all real numbers except for $1, 2, 3, \ldots$. The codomain of the function $f$ is given as $\mathbb{R}$ (all real numbers).

Step 2: Check for Injectivity To check if $f$ is injective, we assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$ and try to prove that $x_1 = x_2$.

Let $f(x_1) = f(x_2)$. $\frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1}$ Since $x_1, x_2 \in A$, neither $x_1$ nor $x_2$ can be $1$. Thus, $x_1 - 1 \neq 0$ and $x_2 - 1 \neq 0$. Cross-multiply: $2x_1(x_2 - 1) = 2x_2(x_1 - 1)$ Divide both sides by 2: $x_1(x_2 - 1) = x_2(x_1 - 1)$ Expand both sides: $x_1x_2 - x_1 = x_2x_1 - x_2$ Subtract $x_1x_2$ from both sides: $-x_1 = -x_2$ Multiply by -1: $x_1 = x_2$ Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is injective.

Alternatively, we can examine the derivative of $f(x)$. $f'(x) = \frac{d}{dx} \left( \frac{2x}{x - 1} \right)$ Using the quotient rule, $(u/v)' = (u'v - uv')/v^2$: Let $u = 2x$, so $u' = 2$. Let $v = x - 1$, so $v' = 1$. $f'(x) = \frac{2(x - 1) - 2x(1)}{(x - 1)^2}$ $f'(x) = \frac{2x - 2 - 2x}{(x - 1)^2}$ $f'(x) = \frac{-2}{(x - 1)^2}$ For any $x \in A$, $(x-1)^2 > 0$. Therefore, $f'(x) = \frac{-2}{\text{positive}} < 0$ for all $x \in A$. Since $f'(x) < 0$ for all $x$ in the domain $A$, the function $f$ is strictly decreasing on its domain and thus is injective.

Step 3: Check for Surjectivity To check if $f$ is surjective, we need to determine if for every $y \in \mathbb{R}$ (the codomain), there exists an $x \in A$ such that $f(x) = y$. Let $y = f(x)$. We need to solve for $x$ in terms of $y$. $y = \frac{2x}{x - 1}$ We must ensure that any solution for $x$ is in the domain $A$ (i.e., $x$ is a real number and $x$ is not a positive integer). Multiply both sides by $(x-1)$: $y(x - 1) = 2x$ $yx - y = 2x$ Rearrange the terms to group $x$: $yx - 2x = y$ Factor out $x$: $x(y - 2) = y$ If $y \neq 2$, we can divide by $(y-2)$: $x = \frac{y}{y - 2}$ Now we need to check if this value of $x$ is always in the domain $A$.

1. Is $x$ always a real number? The expression for $x$ is well-defined as long as $y \neq 2$. If $y=2$, the equation $x(y-2)=y$ becomes $x(0)=2$, which has no solution for $x$. This means that $y=2$ is not in the range of $f$. Since $2$ is in the codomain $\mathbb{R}$, but not in the range of $f$, the function is not surjective.

2. Is $x$ never a positive integer for any $y$ in the range? We found that $y=2$ is not in the range. Let's consider the case when $x$ is a positive integer. If $x$ is a positive integer ($x \in \{1, 2, 3, \ldots\}$), then $x$ is not in the domain $A$. We need to see if there is any $y \in \mathbb{R}$ such that the calculated $x = \frac{y}{y-2}$ results in a positive integer. Let $x = k$, where $k \in \{1, 2, 3, \ldots\}$. $k = \frac{y}{y - 2}$ $k(y - 2) = y$ $ky - 2k = y$ $ky - y = 2k$ $y(k - 1) = 2k$ If $k \neq 1$: $y = \frac{2k}{k - 1}$ We need to check if these values of $y$ are in the codomain $\mathbb{R}$ and if the corresponding $x=k$ is excluded from the domain $A$.

   • If $k=1$, then $x=1$. However, $x=1$ is not in the domain $A$. From the equation $y(k-1)=2k$, if $k=1$, we get $y(0)=2$, which is $0=2$, an impossibility. This confirms that $x=1$ cannot be generated from any $y$.
   • If $k=2$, then $x=2$. This $x$ is not in $A$. The corresponding $y$ is $y = \frac{2(2)}{2-1} = \frac{4}{1} = 4$. So, if $y=4$, then $x=2$. Since $x=2$ is not in $A$, $y=4$ is not in the range of $f$ for the domain $A$.
   • If $k=3$, then $x=3$. This $x$ is not in $A$. The corresponding $y$ is $y = \frac{2(3)}{3-1} = \frac{6}{2} = 3$. So, if $y=3$, then $x=3$. Since $x=3$ is not in $A$, $y=3$ is not in the range of $f$ for the domain $A$.
   • In general, for any positive integer $k \ge 2$, the value $x=k$ is not in the domain $A$. The corresponding $y = \frac{2k}{k-1}$. We can rewrite this as $y = \frac{2(k-1) + 2}{k-1} = 2 + \frac{2}{k-1}$. For $k=2$, $y = 2 + \frac{2}{1} = 4$. For $k=3$, $y = 2 + \frac{2}{2} = 3$. For $k=4$, $y = 2 + \frac{2}{3} = \frac{8}{3}$. The values $y = 3, 4, \frac{8}{3}, \ldots$ are in the codomain $\mathbb{R}$, but they are not in the range of $f$ because the required $x$ values ($3, 4, \frac{8}{3}, \ldots$) are positive integers and thus not in the domain $A$.

   We have shown that $y=2$ is not in the range of $f$. We have also shown that for any positive integer $k \ge 2$, the value $y = \frac{2k}{k-1}$ is not in the range of $f$ because the corresponding $x$ value ($x=k$) is not in the domain $A$. Since there are values in the codomain $\mathbb{R}$ (e.g., $y=2$, $y=3$, $y=4$) that are not attained by $f(x)$ for any $x \in A$, the function $f$ is not surjective.

Step 4: Determine the Nature of the Function From Step 2, we concluded that $f$ is injective. From Step 3, we concluded that $f$ is not surjective.

Therefore, $f$ is injective but not surjective.

Common Mistakes & Tips

• Domain Restrictions: Always pay close attention to the domain of the function. Here, the exclusion of positive integers is crucial for determining surjectivity.
• Solving for x: When checking for surjectivity, be careful about the values of $y$ that make the denominator zero or lead to $x$ values outside the domain.
• Derivative Test: The derivative test for injectivity is a powerful shortcut, but it only works if the function is differentiable on its entire domain (or on each connected component).

Summary

We analyzed the function $f(x) = \frac{2x}{x-1}$ with domain $A = \{x \in \mathbb{R} : x \text{ is not a positive integer}\}$ and codomain $\mathbb{R}$. We proved injectivity by showing that $f(x_1) = f(x_2)$ implies $x_1 = x_2$, or by observing that $f'(x) < 0$ for all $x \in A$. We demonstrated that the function is not surjective by showing that $y=2$ is not in the range, and also that values of $y$ corresponding to positive integer $x$ values are not in the range of $f$ for the given domain. Thus, $f$ is injective but not surjective.

Final Answer The function $f$ is injective but not surjective. This corresponds to option (D).

The final answer is \boxed{A}.