Let f be a real valued function such that for any real x `f(15+x)=f(15-x)and f(30+x)=-f(30-x)`
Then which of the following statements is true ?

Options

A. f is odd and periodic
B. f is odd but not periodic
C. f is even and periodic
D. f is even but not periodic

Answer

To solve the problem, we need to analyze the given conditions for the function \( f \):

1. **Given Conditions:**
- \( f(15 + x) = f(15 - x) \) (1)
- \( f(30 + x) = -f(30 - x) \) (2)

2. **Analyzing the First Condition:**
From the first condition, we can replace \( x \) with \( x - 15 \):
\[ f(15 + (x - 15)) = f(15 - (x - 15)) \]
This simplifies to:
\[ f(x) = f(30 - x) \quad \text{(Equation 1)} \]

3. **Analyzing the Second Condition:**
Now, let's analyze the second condition. We can replace \( x \) with \( x - 30 \):
\[ f(30 + (x - 30)) = -f(30 - (x - 30)) \]
This simplifies to:
\[ f(x) = -f(60 - x) \quad \text{(Equation 2)} \]

4. **Combining the Results:**
Now we have two equations:
- From Equation 1: \( f(x) = f(30 - x) \)
- From Equation 2: \( f(x) = -f(60 - x) \)

We can substitute \( f(30 - x) \) from Equation 1 into Equation 2:
\[ f(30 - x) = -f(60 - (30 - x)) \]
This simplifies to:
\[ f(30 - x) = -f(30 + x) \]
Thus, we have:
\[ f(x) = -f(30 + x) \]

5. **Finding Periodicity:**
Now, let's replace \( x \) with \( x + 30 \) in the second condition:
\[ f(30 + (x + 30)) = -f(30 - (x + 30)) \]
This simplifies to:
\[ f(x + 60) = -f(-x) \]
From the earlier result, we know \( f(-x) = -f(x) \). Therefore, we can conclude:
\[ f(x + 60) = f(x) \]
This shows that \( f(x) \) is periodic with a period of 60.

6. **Checking for Odd or Even Function:**
We need to check if \( f(x) \) is odd or even. We already have:
\[ f(-x) = -f(x) \]
This is the definition of an odd function.

7. **Conclusion:**
Therefore, the function \( f \) is both periodic with a period of 60 and an odd function.

**Final Answer:**
The function \( f \) is periodic with a period of 60 and is an odd function.

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