Question:
If \(f(x) = \begin{cases} 1 + x & \text{if } 0 \leq x \leq 2 \\ 3 - x & \text{if } 2 < x \leq 3 \end{cases}\) , then \(f[f(x)]\) is
If \(f(x) = \begin{cases} 1 + x & \text{if } 0 \leq x \leq 2 \\ 3 - x & \text{if } 2 < x \leq 3 \end{cases}\) , then \(f[f(x)]\) is
Updated On: Oct 4, 2024
- \(f(f(x)) = \begin{cases} 2 + x & \text{if } 0 \leq x \leq 1 \\ 2 - x & \text{if } 1 < x \leq 2 \\ 4 - x & \text{if } 2 < x \leq 3 \end{cases}\)
- \(f(f(x)) = \begin{cases} 2 + x & \text{if } -1 \leq x \leq 1 \\ 2 - x & \text{if } 1 < x \leq 2 \\ 4 - x & \text{if } 2 < x \leq 3 \end{cases}\)
- \(f[f(x)] = \begin{cases} 2 + x & \text{if } 0 \leq x \leq 1 \\ 2 - x & \text{if } 1 \leq x \leq 2 \\ 4 - x & \text{if } 1 \leq x \leq 3 \\ x & \text{if } 0 \leq x \leq 1 \end{cases}\)
- \(f(f(x)) = \begin{cases} 2 + x & \text{if } -1 \leq x \leq 1 \\ 2 - x & \text{if } 1 < x \leq 2 \\ 4 - x & \text{if } 2 \leq x < 3 \end{cases}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): \(f(f(x)) = \begin{cases} 2 + x & \text{if } 0 \leq x \leq 1 \\ 2 - x & \text{if } 1 < x \leq 2 \\ 4 - x & \text{if } 2 < x \leq 3 \end{cases}\)
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