Question

Let A = { x : x ∈ R; x is not a positive integer } Define f : A → R as f(x) = \(\frac{2x}{x-1}\), then f is

• A. injective but not surjective
• B. surjective but not injective
• C. bijective
• D. neither injective nor surjective

Answer 1

The Correct Option is A

The function \( f(x) = \frac{2x}{x-1} \) is defined for all \( x \in \mathbb{R} \) except for \( x = 1 \) (since the denominator cannot be zero). Now, let's analyze whether the function is injective (one-to-one) and surjective (onto).
Injective (One-to-one): A function is injective if, for every pair of distinct elements \( x_1 \) and \( x_2 \), \( f(x_1) \neq f(x_2) \). To check if \( f \) is injective, we solve for \( x_1 \) and \( x_2 \) such that: \[ f(x_1) = f(x_2) \] \[ \frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1} \] Cross-multiply and solve: \[ 2x_1(x_2 - 1) = 2x_2(x_1 - 1) \] Simplifying: \[ x_1x_2 - x_1 = x_2x_1 - x_2 \] \[ -x_1 = -x_2 \] \[ x_1 = x_2 \] Therefore, the function is injective. **Surjective (Onto)**: A function is surjective if for every element \( y \in \mathbb{R} \), there exists an \( x \in A \) such that \( f(x) = y \). Let's examine if the function covers all real numbers: \[ f(x) = \frac{2x}{x-1} \] We can see that there are values of \( x \) for which the function will not reach certain values in \( \mathbb{R} \). For example, the function cannot take the value \( 2 \) since: \[ \frac{2x}{x-1} = 2 \quad \Rightarrow \quad 2x = 2(x-1) \quad \Rightarrow \quad 2x = 2x - 2 \] This results in a contradiction, meaning the function is not surjective.

Therefore, \( f \) is injective but not surjective.

Answer 2

Given the set A = { x : x ∈ R; x is not a positive integer } and the function \(f : A → R\) defined as \(f(x) = \frac{2x}{x-1}\).

We need to determine if the function f is injective (one-to-one) and/or surjective (onto).

Injective (One-to-One):

To check if f is injective, we need to verify that if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\). So, let's assume that \(f(x_1) = f(x_2)\).

\(\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}\)

\(2x_1(x_2-1) = 2x_2(x_1-1)\)

\(x_1x_2 - x_1 = x_2x_1 - x_2\)

\(-x_1 = -x_2\)

\(x_1 = x_2\)

Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function f is injective.

Surjective (Onto):

To check if f is surjective, we need to show that for every y in R, there exists an x in A such that f(x) = y. Let's set \(f(x) = y\) and solve for x.

\(\frac{2x}{x-1} = y\)

\(2x = y(x-1)\)

\(2x = yx - y\)

\(2x - yx = -y\)

\(x(2-y) = -y\)

\(x = \frac{-y}{2-y} = \frac{y}{y-2}\)

Now, we need to check if this x is in A. That means x must not be a positive integer. Also, we have the condition \(y \neq 2\), since the denominator cannot be zero. So we check to see if there are conditions where x is a positive integer.

Assume \(x = \frac{y}{y-2} = n\) where n is a positive integer. Then \(y = n(y-2) = ny-2n\), which gives \(y(n-1) = 2n\), so \(y = \frac{2n}{n-1}\).

If n=2, then y = 4; if n = 3, then y=3; if n=4, then y = 8/3; if n=5, then y = 5/2.

In general \(y = \frac{2n}{n-1} = \frac{2(n-1)+2}{n-1} = 2 + \frac{2}{n-1}\). So if n=3, y=3, x=3, and 3 isn't in A.

Thus, for \(y=3\), there is no \(x \in A\) such that \(f(x)=y\), because that x value is a positive integer.

Since there is a \(y \in R\) for which there's no \(x \in A\) satisfying \(f(x) = y\), the function f is not surjective.

Therefore, f is injective but not surjective.