Question

Let \( A = \mathbb{R} - \{3\} \) and \( B = \mathbb{R} - \{a\} \). Find the value of \( a \) such that the function \( f: A \to B \) defined by \( f(x) = \frac{x-2}{x-3} \) is onto. Also, check whether the given function is one-one or not.

Hint:

For verifying if a function is one-one or onto: Use the definition of onto by solving \( f(x) = y \) and ensuring all \( y \) values in the codomain have preimages in the domain. Use the definition of one-one by equating \( f(x_1) = f(x_2) \) and verifying \( x_1 = x_2 \).

Answer 1

Step 1: Condition for \( f(x) \) to be onto
For \( f(x) \) to be onto, every element \( y \in B \) must have a preimage \( x \in A \). Let \( f(x) = y \): \[ f(x) = \frac{x-2}{x-3} \implies \frac{x-2}{x-3} = y. \] Rewriting this equation: \[ x - 2 = y(x - 3) \implies x - 2 = xy - 3y \implies x(1-y) = 2 - 3y. \] If \( y \neq 1 \), we have: \[ x = \frac{2 - 3y}{1-y}. \] For \( x \in A \), \( x \neq 3 \). Substituting \( x = 3 \) into \( x = \frac{2 - 3y}{1-y} \): \[ 3 = \frac{2 - 3y}{1-y}. \] Solving this: \[ 3(1-y) = 2 - 3y \implies 3 - 3y = 2 - 3y \implies 3 = 2, \] which is not valid. Hence, \( y = 1 \) is excluded from the range of \( f(x) \). Therefore: \[ B = \mathbb{R} - \{1\}, \quad {and } a = 1. \] Step 2: {Check if \( f(x) \) is one-one}
For \( f(x) \) to be one-one, \( f(x_1) = f(x_2) \) should imply \( x_1 = x_2 \). Let: \[ f(x_1) = f(x_2) \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}. \] Cross-multiply: \[ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3). \] Expanding both sides: \[ x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 2x_1 - 3x_2 + 6. \] Simplify: \[ -3x_1 - 2x_2 = -2x_1 - 3x_2. \] Rearrange terms: \[ x_1 = x_2. \] Thus, \( f(x) \) is one-one. 

Conclusion:
The function \( f(x) = \frac{x-2}{x-3} \) is: Onto if \( B = \mathbb{R} - \{1\} \) and \( a = 1 \). One-one since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).