Consider the function `f:R -{1} to R -{2}` given by `f {x} =(2x)/(x-1).` Then

Options

A. f is one-one but not onto
B. f is onto but not one-one
C. f is neither one-one nor onto
D. f is one-one and onto

Answer

To determine the properties of the function \( f: \mathbb{R} - \{1\} \to \mathbb{R} - \{2\} \) given by \( f(x) = \frac{2x}{x-1} \), we need to check if the function is one-one (injective) and onto (surjective).

### Step 1: Check if the function is one-one (injective)

A function is one-one if for any two different inputs \( x_1 \) and \( x_2 \), the outputs are different.

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

Cross-multiplying gives:
\[ 2x_1(x_2 - 1) = 2x_2(x_1 - 1) \]

Expanding both sides:
\[ 2x_1x_2 - 2x_1 = 2x_2x_1 - 2x_2 \]

Rearranging terms:
\[ -2x_1 = -2x_2 \]

Dividing by -2:
\[ x_1 = x_2 \]

Since we assumed \( x_1 \) and \( x_2 \) were different but derived \( x_1 = x_2 \), this is a contradiction. Therefore, the function is one-one.

### Step 2: Check if the function is onto (surjective)

A function is onto if every element in the codomain has a pre-image in the domain. We need to show that for any \( y \in \mathbb{R} - \{2\} \), there exists an \( x \in \mathbb{R} - \{1\} \) such that \( f(x) = y \).

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

Cross-multiplying gives:
\[ 2x = y(x - 1) \]

Expanding:
\[ 2x = yx - y \]

Rearranging:
\[ yx - 2x = y \]

Factoring out \( x \):
\[ x(y - 2) = y \]

Solving for \( x \):
\[ x = \frac{y}{y - 2} \]

This expression for \( x \) is valid as long as \( y \neq 2 \). Since \( y \) is in the codomain \( \mathbb{R} - \{2\} \), we can find an \( x \) for every \( y \) in the codomain. Thus, the function is onto.

### Conclusion

Since the function \( f(x) = \frac{2x}{x-1} \) is both one-one and onto, we conclude that it is a bijection.