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:

\textbf{Part (b):} For a function \(f(x) = ax + b\), use the given points to form and solve simultaneous equations for \(a\) and \(b\). Check one-to-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: 

From equation (1): \( b = 1 - a \). 
Substitute this into equation (2): \[ 2a + (1 - a) = 3 \quad \Rightarrow \quad 2a + 1 - a = 3 \quad \Rightarrow \quad a = 2. \] 

Substituting \( a = 2 \) into equation (1): \[ 2 + b = 1 \quad \Rightarrow \quad b = -1. \] 
Thus, the function is: \[ f(x) = 2x - 1. \] 
One-one (Injective): A function is one-one if distinct inputs lead to distinct outputs. Since \( f(x) = 2x - 1 \) is a linear function with a non-zero slope, it is one-one. 
Onto (Surjective): A function is onto if for every element \( y \in R \), there exists \( x \in R \) such that \( f(x) = y \). For \( f(x) = 2x - 1 \), for any \( y \in R \), we can solve \( y = 2x - 1 \) for \( x \), which gives \( x = \frac{y + 1}{2} \). Hence, the function is onto. 

Answer: The function \( f(x) = 2x - 1 \) is both one-one and onto. \bigskip