Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b`
If `f(x)` is an even function, then

Options

A. `phi(x)` is also an even function
B. `phi(x)` is an odd function
C. `phi(x)` is an neither even nor odd function
D. None of these

Explanation

To solve the problem, we need to analyze the given functions and their properties step by step.

### Step 1: Understand the given functions
We have two continuous functions \( f(x) \) and \( \phi(x) \) such that:
\[ \phi(x) = \int_a^x f(t) \, dt \]
where \( a \neq 0 \). We also have a continuous function \( g(x) \) that satisfies:
\[ g(x + \alpha) + g(x) = 0 \quad \text{for all } x \in \mathbb{R}, \text{ where } \alpha > 0 \]
Additionally, the integral \( \int_b^{2k} g(t) \, dt \) is independent of \( b \).

### Step 2: Analyze the properties of \( f(x) \)
Given that \( f(x) \) is an even function, we have:
\[ f(-x) = f(x) \quad \text{for all } x \]

### Step 3: Evaluate \( \phi(-x) \)
Now, we need to evaluate \( \phi(-x) \):
\[ \phi(-x) = \int_a^{-x} f(t) \, dt \]
To change the limits of integration, we can use the substitution \( t = -y \) which gives \( dt = -dy \):
\[ \phi(-x) = \int_a^{-x} f(t) \, dt = \int_{-(-x)}^{-a} f(-y)(-dy) = \int_{x}^{-a} f(-y) \, dy \]
Since \( f(y) \) is even, we have \( f(-y) = f(y) \):
\[ \phi(-x) = \int_{x}^{-a} f(y) \, dy = -\int_{-a}^{x} f(y) \, dy \]

### Step 4: Change the limits of integration
Now we can rewrite the integral:
\[ \phi(-x) = -\left( \int_{-a}^{a} f(y) \, dy + \int_{a}^{x} f(y) \, dy \right) \]
This can be simplified to:
\[ \phi(-x) = -\int_{-a}^{x} f(y) \, dy \]

### Step 5: Relate \( \phi(x) \) and \( \phi(-x) \)
Now we have:
\[ \phi(-x) = -\left( \int_{-a}^{a} f(y) \, dy + \int_{a}^{x} f(y) \, dy \right) \]
This shows that \( \phi(-x) \) is related to \( \phi(x) \) and can be expressed as:
\[ \phi(-x) = -\phi(x) \]

### Step 6: Conclusion
Since \( \phi(-x) = -\phi(x) \), we conclude that \( \phi(x) \) is an odd function.

### Final Result
Thus, if \( f(x) \) is an even function, then \( \phi(x) \) is an odd function.

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