Let `f(x) and g(x)` be any two continuous function in the interval `[0, b]` and 'a' be any point between 0 and b. Which satisfy the following conditions : `f(x)=f(a-x), g(x)+g(a-x)=3, f(a+b-x)=f(x)`. Also `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx, int_(a)^(b)f(x)dx=int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`
`int_(0)^(a)f(x)dx=p" then " int_(0)^(a)f(x)g(x)dx` is

Options

A. `(3p)/(2)`
B. `(p)/(2)`
C. `(2p)/(3)`
D. none of these

Answer

To solve the problem step by step, we will use the properties of the functions \( f(x) \) and \( g(x) \) as given in the question.

### Step 1: Understand the properties of \( f(x) \) and \( g(x) \)

We know:
1. \( f(x) = f(a - x) \) (which means \( f \) is symmetric about \( x = \frac{a}{2} \)).
2. \( g(x) + g(a - x) = 3 \) (which indicates a relationship between \( g(x) \) and \( g(a - x) \)).
3. \( f(a + b - x) = f(x) \) (which indicates periodicity or symmetry in a larger interval).

### Step 2: Set up the integral

We need to evaluate the integral:
\[ \int_0^a f(x) g(x) \, dx \]

### Step 3: Use the properties of \( g(x) \)

From the property \( g(x) + g(a - x) = 3 \), we can express \( g(a - x) \) as:
\[ g(a - x) = 3 - g(x) \]

### Step 4: Change of variable in the integral

Now, we can change the variable in the integral:
Let \( u = a - x \), then \( du = -dx \). The limits change as follows:
- When \( x = 0 \), \( u = a \)
- When \( x = a \), \( u = 0 \)

Thus, the integral becomes:
\[ \int_0^a f(x) g(x) \, dx = \int_a^0 f(a - u) g(a - u) (-du) = \int_0^a f(a - u) g(a - u) \, du \]

### Step 5: Substitute the properties into the integral

Using the properties of \( f \) and \( g \):
\[ \int_0^a f(a - u) g(a - u) \, du = \int_0^a f(u) (3 - g(u)) \, du \]

### Step 6: Expand the integral

Now, we can expand the integral:
\[ \int_0^a f(a - u) g(a - u) \, du = \int_0^a f(u) \cdot 3 \, du - \int_0^a f(u) g(u) \, du \]

### Step 7: Combine the integrals

Let \( I = \int_0^a f(x) g(x) \, dx \). Then we have:
\[ I = 3 \int_0^a f(x) \, dx - I \]

### Step 8: Solve for \( I \)

Now, we can solve for \( I \):
\[ 2I = 3 \int_0^a f(x) \, dx \]
\[ I = \frac{3}{2} \int_0^a f(x) \, dx \]

### Step 9: Substitute the value of the integral

Given that \( \int_0^a f(x) \, dx = p \):
\[ I = \frac{3}{2} p \]

### Final Answer

Thus, the value of \( \int_0^a f(x) g(x) \, dx \) is:
\[ \frac{3p}{2} \]