Let `A = R -{3} " and " B =R -{1}. " Then " f : A to B : f (x) = ((x-2))/((x-3))` is

Options

A. one-one and into
B. one -one and onto
C. many-one and into
D. many -one and onto

Explanation

To solve the given problem, we need to analyze the function \( f(x) = \frac{x - 2}{x - 3} \) defined from the set \( A = \mathbb{R} - \{3\} \) to the set \( B = \mathbb{R} - \{1\} \). We want to determine if the function is one-to-one (injective) and onto (surjective).

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

To check if the function is one-to-one, 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 means:
\[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \]

2. Cross-multiply:
\[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \]

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

4. Simplifying gives:
\[ -3x_1 - 2x_2 = -3x_2 - 2x_1 \]

5. Rearranging terms:
\[ -3x_1 + 2x_1 = -3x_2 + 2x_2 \]
\[ -x_1 = -x_2 \]

6. Thus, we conclude:
\[ x_1 = x_2 \]
This proves that \( f \) is one-to-one.

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

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

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

2. Cross-multiply:
\[ x - 2 = y(x - 3) \]

3. Expand and rearrange:
\[ x - 2 = yx - 3y \]
\[ x - yx = -3y + 2 \]
\[ x(1 - y) = -3y + 2 \]

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

5. We need to ensure that \( x \) is in \( A \) (i.e., \( x \neq 3 \)):
\[ \frac{-3y + 2}{1 - y} \neq 3 \]

6. Solve the inequality:
\[ -3y + 2 \neq 3(1 - y) \]
\[ -3y + 2 \neq 3 - 3y \]
This is always true since both sides will simplify to the same expression.

Thus, for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \). Therefore, \( f \) is onto.

### Conclusion

Since we have shown that \( f \) is both one-to-one and onto, we conclude that the function \( f: A \to B \) is bijective.

### Final Answer

The function \( f(x) = \frac{x - 2}{x - 3} \) is both one-to-one and onto.

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