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 odd function, then

Options

A. (a) `phi(x)` is also an odd function
B. (b) `phi(x)` is an even function
C. (c) `phi(x)` is neither an even nor an odd function
D. (d) for `phi(x)` to be an even function, it must satisfy `int_(0)^(a)f(x)dx=0`

Explanation

To solve the problem, we start by analyzing the given functions and their properties.

### Step 1: Understanding the Functions
We have two continuous functions \( f(x) \) and \( \phi(x) \) defined as:
\[ \phi(x) = \int_{a}^{x} f(t) \, dt \]
where \( a \neq 0 \). We also know that \( f(x) \) is an odd function, which means:
\[ f(-x) = -f(x) \]

### Step 2: Evaluating \( \phi(-x) \)
We need to find \( \phi(-x) \):
\[ \phi(-x) = \int_{a}^{-x} f(t) \, dt \]

### Step 3: Changing the Limits of Integration
To evaluate this integral, we can change the variable. Let \( t = -y \), then \( dt = -dy \). The limits change as follows:
- When \( t = a \), \( y = -a \)
- When \( t = -x \), \( y = x \)

Thus, we have:
\[ \phi(-x) = \int_{-a}^{x} f(-y)(-dy) = \int_{-a}^{x} -f(y) \, dy = -\int_{-a}^{x} f(y) \, dy \]

### Step 4: Splitting the Integral
Now, we can split the integral:
\[ \phi(-x) = -\left( \int_{-a}^{0} f(y) \, dy + \int_{0}^{x} f(y) \, dy \right) \]
This can be rewritten as:
\[ \phi(-x) = -\int_{-a}^{0} f(y) \, dy - \int_{0}^{x} f(y) \, dy \]

### Step 5: Relating \( \phi(x) \) and \( \phi(-x) \)
Now, we can express \( \phi(x) \):
\[ \phi(x) = \int_{a}^{0} f(t) \, dt + \int_{0}^{x} f(t) \, dt \]
Now we can see that:
\[ \phi(-x) = -\left( \int_{-a}^{0} f(y) \, dy + \int_{0}^{x} f(y) \, dy \right) \]

### Step 6: Conclusion
Since \( f(x) \) is odd, we can conclude that:
\[ \phi(-x) = -\phi(x) \]
This shows that \( \phi(x) \) is an even function.

### Final Answer
Thus, we conclude that \( \phi(x) \) is an even function.

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