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`
Least positive value fo `c` if `c,k,b` are n A.P. is

Options

A. `0`
B. `1`
C. `alpha`
D. `2alpha`

Explanation

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

### 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 \( a \neq 0 \).

Additionally, we have another continuous function \( g(x) \) that satisfies the equation:
\[ g(x + \alpha) + g(x) = 0 \quad \text{for all } x \in \mathbb{R}, \text{ where } \alpha > 0. \]

### Step 2: Analyzing the Function \( g(x) \)
From the equation \( g(x + \alpha) + g(x) = 0 \), we can rearrange it to find:
\[ g(x + \alpha) = -g(x). \]
This implies that \( g(x) \) is periodic with a period of \( 2\alpha \). This is because if we apply this transformation multiple times, we can see that:
\[ g(x + 2\alpha) = -g(x + \alpha) = -(-g(x)) = g(x). \]
Thus, \( g(x) \) is periodic with period \( 2\alpha \).

### Step 3: Independence of the Integral
We are also given that:
\[ \int_{b}^{2k} g(t) dt \]
is independent of \( b \). This means that the integral must evaluate to a constant value regardless of the choice of \( b \).

### Step 4: Finding the Least Positive Value of \( c \)
Now, we need to find the least positive value of \( c \) such that \( c, k, b \) are in Arithmetic Progression (A.P.).

In an A.P., the middle term is the average of the other two terms. Therefore, we have:
\[ k = \frac{c + b}{2}. \]
Rearranging gives:
\[ c + b = 2k \quad \Rightarrow \quad c = 2k - b. \]

### Step 5: Determining the Values of \( c, k, b \)
To find the least positive value of \( c \), we can set \( b \) to its minimum value. Since \( b \) can be any real number, we can choose \( b = 0 \) (the smallest non-negative value).

Substituting \( b = 0 \) into the equation gives:
\[ c = 2k. \]
Since \( k \) must also be a positive integer, the smallest value for \( k \) is 1. Therefore:
\[ c = 2 \times 1 = 2. \]

### Conclusion
Thus, the least positive value of \( c \) such that \( c, k, b \) are in A.P. is:
\[ \boxed{2}. \]