Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be defined as \( f(x) = 2x - 1 \) and \( g : \mathbb{R} \setminus \{1\} \to \mathbb{R} \) be defined as \( g(x) = \dfrac{x - \frac{1}{2}}{x - 1} \). Then the composition function \( f(g(x)) \) is :

• A. one-one but not onto
• B. onto but not one-one
• C. neither one-one nor onto
• D. both one-one and onto

Hint:

For any linear fractional transformation $y = \frac{ax+b}{cx+d}$, the function is always one-one, and its range is $\mathbb{R} \setminus \{a/c\}$.

Answer 1

The Correct Option is A

Step 1: Calculate the composition $f(g(x))$. $f(g(x)) = 2\left(\frac{x - 1/2}{x - 1}\right) - 1 = \frac{2x - 1}{x - 1} - 1 = \frac{2x - 1 - (x - 1)}{x - 1} = \frac{x}{x - 1}$.
Step 2: Check for one-one. Let $f(g(x_1)) = f(g(x_2)) \Rightarrow \frac{x_1}{x_1 - 1} = \frac{x_2}{x_2 - 1}$. $x_1x_2 - x_1 = x_1x_2 - x_2 \Rightarrow x_1 = x_2$. So, it is one-one.
Step 3: Check for onto. Let $y = \frac{x}{x - 1} \Rightarrow yx - y = x \Rightarrow x(y - 1) = y \Rightarrow x = \frac{y}{y - 1}$. For $x$ to be a real number, $y \neq 1$. Since the codomain is $\mathbb{R}$ but the range is $\mathbb{R} \setminus \{1\}$, it is not onto.