Question

If a real-valued function \( f : [-1, 2] \to B \) is defined by: \[ f(x) = \begin{cases} 1 - x, & \text{when } -1 \leq x \leq 1 \\ x - 1, & \text{when } 1 < x \leq 2 \end{cases} \] and if \( f \) is a surjection, then \( B = ? \)

• A. \( [-1, 2] \)
• B. \( [-1, 1] \)
• C. \( [0, 2] \)
• D. \( [0, 1] \)

Hint:

To determine the range of a piecewise function, analyze each part separately and take the union of all possible outputs.

Answer 1

The Correct Option is C

Step 1: Analyze the first piece of the function: \[ f(x) = 1 - x \quad \text{for} \quad x \in [-1, 1] \] When \( x = -1 \), \( f(x) = 1 - (-1) = 2 \) When \( x = 1 \), \( f(x) = 1 - 1 = 0 \) So for this part: \[ f(x) \in [0, 2] \] Step 2: Analyze the second piece of the function: \[ f(x) = x - 1 \quad \text{for} \quad x \in (1, 2] \] When \( x = 1^+ \), \( f(x) \to 0^+ \), when \( x = 2 \), \( f(x) = 2 - 1 = 1 \) So this part also gives: \[ f(x) \in (0, 1] \] Step 3: Combine both intervals: From the first part: \( f(x) \in [0, 2] \) From the second part: \( f(x) \in (0, 1] \) The union of these is: \[ B = [0, 2] \] Since the function is a surjection onto \( B \), all values in \( [0, 2] \) must be covered. We’ve shown that’s true.