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Mathematics Question on Functions

If f(x)={1+xif 0≤x≤2 3−xif 2<x≤3f(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)]f[f(x)] is

A

f(f(x))={2+xif 0≤x≤1 2−xif 1<x≤2 4−xif 2<x≤3f(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}

B

f(f(x))={2+xif −1≤x≤1 2−xif 1<x≤2 4−xif 2<x≤3f(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}

C

f[f(x)]={2+xif 0≤x≤1 2−xif 1≤x≤2 4−xif 1≤x≤3 xif 0≤x≤1f[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}

D

f(f(x))={2+xif −1≤x≤1 2−xif 1<x≤2 4−xif 2≤x<3f(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}

Answer

f(f(x))={2+xif 0≤x≤1 2−xif 1<x≤2 4−xif 2<x≤3f(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}

Explanation

Solution

The correct option is (A): f(f(x))={2+xif 0≤x≤1 2−xif 1<x≤2 4−xif 2<x≤3f(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}