Let `A={x in R:x" is not a positive integer "}`define a function `f:A to R" such that "f(x)=(2x)/(x-1)`. Then f is

Options

A. injective but not surjective
B. not injective
C. surjective but not injective
D. neither inhective nor surjective

Answer

To determine the nature of the function \( f: A \to \mathbb{R} \) defined by \( f(x) = \frac{2x}{x-1} \), where \( A = \{ x \in \mathbb{R} : x \text{ is not a positive integer} \} \), we need to check if the function is injective (one-to-one) and surjective (onto).

### Step 1: Check for Injectivity
To check if \( f \) is injective, we assume that \( f(x_1) = f(x_2) \) and show that this implies \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \):
\[ \frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1} \]

2. Cross-multiply:
\[ 2x_1(x_2 - 1) = 2x_2(x_1 - 1) \]

3. Expand both sides:
\[ 2x_1x_2 - 2x_1 = 2x_2x_1 - 2x_2 \]

4. Rearranging gives:
\[ -2x_1 = -2x_2 \implies x_1 = x_2 \]

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

### Step 2: Check for Surjectivity
To check if \( f \) is surjective, we need to see if for every \( y \in \mathbb{R} \), there exists an \( x \in A \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ y = \frac{2x}{x - 1} \]

2. Rearranging gives:
\[ y(x - 1) = 2x \]
\[ yx - y = 2x \]
\[ yx - 2x = y \]
\[ x(y - 2) = y \]
\[ x = \frac{y}{y - 2} \]

3. The function \( f \) is surjective if \( x \) is in the domain \( A \). We need to ensure that \( \frac{y}{y - 2} \) is not a positive integer.

4. The expression \( \frac{y}{y - 2} \) is undefined when \( y = 2 \). For \( y \neq 2 \):
- If \( y > 2 \), then \( y - 2 > 0 \) and \( x \) will be positive.
- If \( y < 2 \), then \( y - 2 < 0 \) and \( x \) will be negative.

However, there are certain values of \( y \) that make \( x \) a positive integer. For instance, if \( y = 4 \):
\[ x = \frac{4}{4 - 2} = 2 \quad (\text{which is a positive integer}) \]

Thus, not all values of \( y \) yield \( x \in A \). Therefore, \( f \) is not surjective.

### Conclusion
The function \( f(x) = \frac{2x}{x-1} \) is injective but not surjective.