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 one-one nor onto
D. f is both one-one and onto

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{2x}{x-1} \) defined on the domain \( \mathbb{R} - \{1\} \) and the range \( \mathbb{R} - \{2\} \). We will determine whether the function is one-to-one (1-1) or many-to-one, and whether it is onto (onto) or not onto.

### Step-by-Step Solution:

**Step 1: Check if the function is one-to-one (1-1)**

To check if the function is one-to-one, we need to see if \( f(a) = f(b) \) implies \( a = b \).

Assume \( f(a) = f(b) \):
\[ \frac{2a}{a-1} = \frac{2b}{b-1} \]

Cross-multiplying gives:
\[ 2a(b-1) = 2b(a-1) \]

Expanding both sides:
\[ 2ab - 2a = 2ab - 2b \]

Subtracting \( 2ab \) from both sides:
\[ -2a = -2b \]

Dividing by -2:
\[ a = b \]

Since \( f(a) = f(b) \) implies \( a = b \), the function is one-to-one.

**Step 2: Check if the function is onto (onto)**

To check if the function is onto, we need to determine if every element \( y \) in the range \( \mathbb{R} - \{2\} \) has a corresponding \( x \) in the domain \( \mathbb{R} - \{1\} \) such that \( f(x) = y \).

Starting from the equation:
\[ y = \frac{2x}{x-1} \]

We can rearrange this to find \( x \):
\[ y(x - 1) = 2x \]
\[ yx - y = 2x \]
\[ yx - 2x = y \]
\[ x(y - 2) = y \]
\[ x = \frac{y}{y - 2} \]

Now, we need to check if \( x \) is defined for all \( y \neq 2 \):
- The expression \( \frac{y}{y - 2} \) is defined for all \( y \neq 2 \).
- Since \( y = 2 \) is excluded from the range, we can conclude that for every \( y \neq 2 \), there exists an \( x \) such that \( f(x) = y \).

Thus, the function is onto.

### Conclusion:

The function \( f(x) = \frac{2x}{x-1} \) is both one-to-one and onto. Therefore, it is a bijection.