Let `A=R-{3}, B=R-{1}.` Let `f:ArarrB` be defined by `f(x)=((x-2)/(x-3))`, then

Options

A. a. f is bijective
B. b. Surjective
C. c. Injective
D. d. None of above

Answer

To solve the problem, we need to analyze the function \( f: A \to B \) defined by \( f(x) = \frac{x - 2}{x - 3} \), where \( A = \mathbb{R} \setminus \{3\} \) and \( B = \mathbb{R} \setminus \{1\} \).

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

To check if the function is injective, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

1. Start with the equation:
\[ f(x_1) = f(x_2) \]
This implies:
\[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \]

2. Cross-multiply to eliminate the fractions:
\[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \]

3. Expand both sides:
\[ x_1x_2 - 3x_1 - 2x_2 + 6 = x_2x_1 - 3x_2 - 2x_1 + 6 \]

4. Simplify:
\[ -3x_1 - 2x_2 = -3x_2 - 2x_1 \]
Rearranging gives:
\[ -3x_1 + 2x_1 = -3x_2 + 2x_2 \]
Which simplifies to:
\[ -x_1 = -x_2 \quad \text{or} \quad x_1 = x_2 \]

Since we have shown that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is injective.

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

To check if the function is surjective, we need to show that for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ y = \frac{x - 2}{x - 3} \]

2. Rearranging gives:
\[ y(x - 3) = x - 2 \]
This leads to:
\[ yx - 3y = x - 2 \]

3. Rearranging terms:
\[ yx - x = 3y - 2 \]
Factor out \( x \):
\[ x(y - 1) = 3y - 2 \]

4. Solving for \( x \):
\[ x = \frac{3y - 2}{y - 1} \]

5. We need to ensure that \( x \) is defined and belongs to \( A \). The function \( x \) is undefined when \( y = 1 \), which is excluded from set \( B \).

Since for every \( y \in B \) (where \( y \neq 1 \)), we can find an \( x \in A \) such that \( f(x) = y \), the function \( f \) is surjective.

### Conclusion

Since \( f \) is both injective and surjective, we conclude that \( f \) is bijective.

### Final Answer

The function \( f \) is injective, surjective, and therefore bijective.

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