Which of the following is wrong ?

Options

A. Every constant function is an even function
B. A constant function may be odd function also.
C. Every constant function is an odd as well as an even function.
D. Every constant function is a periodic function

Answer

To determine which of the statements regarding constant functions is incorrect, let's analyze each option step by step.

### Step 1: Understand Constant Functions
A constant function is defined as a function that always returns the same value, regardless of the input. Mathematically, it can be represented as:
\[ f(x) = k \]
where \( k \) is a constant real number.

### Step 2: Analyze Each Option
1. **Option A: Every constant function is an even function.**
- A function is even if \( f(-x) = f(x) \) for all \( x \).
- For a constant function \( f(x) = k \):
\[ f(-x) = k \quad \text{and} \quad f(x) = k \]
- Since \( f(-x) = f(x) \), this statement is **true**.

2. **Option B: Every constant function may be an odd function also.**
- A function is odd if \( f(-x) = -f(x) \) for all \( x \).
- For a constant function \( f(x) = k \):
\[ f(-x) = k \quad \text{and} \quad -f(x) = -k \]
- For \( f(-x) = -f(x) \) to hold, \( k \) must equal \( 0 \) (i.e., the function must be the zero function). Thus, this statement is **partially true** but not universally applicable.

3. **Option C: Every constant function is an odd as well as an even function.**
- This statement combines the conclusions of options A and B. Since not all constant functions are odd (only the zero function is), this statement is **false**.

4. **Option D: Every constant function is a periodic function.**
- A function is periodic if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \).
- Since a constant function does not change with \( x \), it can be considered periodic with any period \( T \) (including \( T = 1 \)). Thus, this statement is **true**.

### Conclusion
The incorrect statement among the options is **Option C**: "Every constant function is an odd as well as an even function."