11th Biology

Analysis of Function:

\( f(x) = \begin{cases} x + 5, & -2 < x < 0 \\[6pt] -x + 5, & 0 \le x < 2 \end{cases} \)

1. Symmetric Domain Check:

The domain of this function is \( (-2, 0) \cup [0, 2) = (-2, 2) \).
This domain is symmetric about the origin. So, the first condition is met.

2. Functional Property Check:

Case 1: \( x \in (0, 2) \)

• \( f(x) = -x + 5 \)
• Now consider \( -x \). Since \( x \in (0, 2) \), then \( -x \in (-2, 0) \).
• For \( -x \in (-2, 0) \), \( f(-x) = (-x) + 5 = -x + 5 \).
• Is \( f(-x) = f(x) \)? Is \( -x + 5 = -x + 5 \)? Yes. This suggests it might be even.

Case 2: \( x \in (-2, 0) \)

• \( f(x) = x + 5 \)
• Now consider \( -x \). Since \( x \in (-2, 0) \), then \( -x \in (0, 2) \).
• For \( -x \in (0, 2) \), \( f(-x) = -(-x) + 5 = x + 5 \).
• Is \( f(-x) = f(x) \)? Is \( x + 5 = x + 5 \)? Yes. This also suggests it might be even.

Check at \( x = 0 \):

• \( f(0) = -(0) + 5 = 5 \).
• For an even function, \( f(0) \) can be any value. For an odd function, \( f(0) \) must be 0. Since \( f(0) = 5 \neq 0 \), it cannot be an odd function.

Conclusion: This function is even.