Let `A=R-{3},B=R-{1} " and " f:A rarr B` defined by `f(x)=(x-2)/(x-3)`. Is 'f' bijective? Give reasons.

Answer

To determine if the function \( f: A \to B \) defined by \( f(x) = \frac{x - 2}{x - 3} \) is bijective, we need to check if it is both injective (one-to-one) and surjective (onto).

### Step 1: Check if \( f \) is Injective

A function is injective if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \) for any \( x_1, x_2 \in A \).

1. Assume \( f(x_1) = f(x_2) \):
\[ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \]

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. Simplify:
\[ -3x_1 - 2x_2 = -3x_2 - 2x_1 \]

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

6. Thus, we conclude:
\[ x_1 = x_2 \]

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

### Step 2: Check if \( f \) is Surjective

A function is surjective if for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).

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

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

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

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

5. The expression for \( x \) is valid as long as \( y \neq 1 \) (since \( y - 1 \) cannot be zero).

Since \( y \) can take any real value except 1 (which is not in the codomain \( B \)), for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \). Thus, \( f \) is surjective.

### Conclusion

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