If `f(x)` is a periodic function with period `lambda` and f(ax+b) is perodic with period T/a then period of `f(lambda x + u)` where `mu` is any constant is

Options

A. ` lamda `
B. `1 `
C. `(lamda )/( a)`
D. none of these

Explanation

To solve the problem, we need to analyze the periodicity of the function \( f(\lambda x + \mu) \) given that \( f(x) \) is a periodic function with period \( \lambda \), and \( f(ax + b) \) is periodic with period \( \frac{T}{a} \).

### Step-by-Step Solution:

1. **Understanding the Periodicity of \( f(x) \)**:
- Since \( f(x) \) is periodic with period \( \lambda \), it means that:
\[ f(x + \lambda) = f(x) \quad \text{for all } x. \]

2. **Analyzing \( f(ax + b) \)**:
- We are given that \( f(ax + b) \) is periodic with period \( \frac{T}{a} \). This implies:
\[ f(ax + b + \frac{T}{a}) = f(ax + b) \quad \text{for all } x. \]
- To find the relationship between \( T \) and \( \lambda \), we can substitute \( a = 1 \):
\[ f(x + b + T) = f(x + b). \]
- This means that \( T \) must equal \( \lambda \) when \( a = 1 \).

3. **Finding the Period of \( f(\lambda x + \mu) \)**:
- Now, we need to determine the periodicity of \( f(\lambda x + \mu) \). We can express this as:
\[ f(\lambda x + \mu + P) = f(\lambda x + \mu) \quad \text{for some period } P. \]
- To find \( P \), we need to find a value such that:
\[ \lambda x + \mu + P = \lambda x + \mu + \lambda. \]
- This implies:
\[ P = \lambda. \]

4. **Conclusion**:
- Therefore, the period of \( f(\lambda x + \mu) \) is \( \lambda \).

### Final Answer:
The period of \( f(\lambda x + \mu) \) is \( \lambda \).

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