Question

A function \( f \) is defined from \( R \to R \) as \( f(x) = ax + b \), such that \( f(1) = 1 \) and \( f(2) = 3 \). Find the function \( f(x) \). Hence, check whether the function \( f(x) \) is one-one and onto.

Hint:

For a function \( f(x) = ax + b \), use the given points to form and solve simultaneous equations for \( a \) and \( b \). Check one-one by verifying \( f'(x) \neq 0 \) and onto by solving \( f(x) = y \) for all \( y \in \mathbb{R} \).

Answer 1

From the given conditions, we have two equations: \[ f(1) = a(1) + b = 1 \quad \Rightarrow \quad a + b = 1, \tag{1} \] \[ f(2) = a(2) + b = 3 \quad \Rightarrow \quad 2a + b = 3. \tag{2} \] Solving equations (1) and (2) simultaneously: \[ b = 1 - a. \] Substituting this into equation (2): \[ 2a + (1 - a) = 3 \quad \Rightarrow \quad a = 2. \] Substituting \( a = 2 \) into equation (1): \[ 2 + b = 1 \quad \Rightarrow \quad b = -1. \] Thus, the function is: \[ f(x) = 2x - 1. \] Checking one-one (Injective): Since \( f(x) = 2x - 1 \) is a linear function with a non-zero slope, it is one-one. Checking onto (Surjective): To check onto, solve \( f(x) = y \) for \( x \): \[ y = 2x - 1 \quad \Rightarrow \quad x = \frac{y + 1}{2}. \] Since \( x \in R \) for all \( y \in R \), the function is onto. Final Answer: The function \( f(x) = 2x - 1 \) is both one-one and onto.