Question

$f: X \to \mathbb{R}, X = \{x | 0<x<1\}$ is defined as $f(x) = \frac{2x - 1}{1 - |2x - 1|}$. Then

• A. f is only injective
• B. f is only surjective
• C. f is bijective
• D. f is neither injective nor surjective

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = \frac{2x - 1}{1 - |2x - 1|} \] defined on the domain \( X = \{ x \mid 0 < x < 1 \} \).

Step 1: Understanding the behavior of the function
The function involves the absolute value term \( |2x - 1| \), which affects the function's behavior depending on the value of \( x \). We analyze the function on two intervals: - For \( 0 < x < \frac{1}{2} \), \( 2x - 1 < 0 \), so \( |2x - 1| = -(2x - 1) \). - For \( \frac{1}{2} < x < 1 \), \( 2x - 1 > 0 \), so \( |2x - 1| = 2x - 1 \). 

Step 2: Injectivity of the function
To check if the function is injective, we need to verify that for \( f(x_1) = f(x_2) \), it follows that \( x_1 = x_2 \). Given the form of \( f(x) \), the function is continuous and strictly monotonic in each interval \( (0, \frac{1}{2}) \) and \( (\frac{1}{2}, 1) \). Therefore, it is injective. 

Step 3: Surjectivity of the function
To check if the function is surjective, we need to determine if for every \( y \in \mathbb{R} \), there exists an \( x \in X \) such that \( f(x) = y \). Upon analysis, the function \( f(x) \) takes values in the entire range of \( \mathbb{R} \), making the function surjective. 

Step 4: Bijectivity
Since the function is both injective and surjective, it is bijective.

Answer:

\[ \boxed{\text{Bijective}} \]