If `int_(a)^(b)(f(a+b-x))/(f(x)+f(a+b-x))dx=10`, then (a, b) can have the values

Options

A. `a=20, b=40`
B. `a=-5, b=5`
C. `a=10, b=20`
D. None of these

Answer

To solve the problem, we need to evaluate the given integral and find the values of \(a\) and \(b\) such that the integral equals 10.

Given:
\[ \int_{a}^{b} \frac{f(a+b-x)}{f(x) + f(a+b-x)} \, dx = 10 \]

### Step 1: Use the property of definite integrals
We can use the property of definite integrals that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]
This means we can rewrite our integral as:
\[ \int_{a}^{b} \frac{f(a+b-x)}{f(x) + f(a+b-x)} \, dx = \int_{a}^{b} \frac{f(x)}{f(a+b-x) + f(x)} \, dx \]

### Step 2: Combine the two integrals
Let:
\[ I = \int_{a}^{b} \frac{f(a+b-x)}{f(x) + f(a+b-x)} \, dx \]
Then, from the property, we have:
\[ I = \int_{a}^{b} \frac{f(x)}{f(a+b-x) + f(x)} \, dx \]
Adding these two equations gives:
\[ 2I = \int_{a}^{b} \left( \frac{f(a+b-x)}{f(x) + f(a+b-x)} + \frac{f(x)}{f(a+b-x) + f(x)} \right) \, dx \]
The sum of the fractions simplifies to 1:
\[ 2I = \int_{a}^{b} 1 \, dx \]

### Step 3: Evaluate the integral
The integral of 1 from \(a\) to \(b\) is simply:
\[ \int_{a}^{b} 1 \, dx = b - a \]
Thus, we have:
\[ 2I = b - a \]
Since we know \(I = 10\), we can substitute this into the equation:
\[ 2 \times 10 = b - a \]
This simplifies to:
\[ b - a = 20 \]

### Step 4: Find possible values of \(a\) and \(b\)
From the equation \(b - a = 20\), we can express \(b\) in terms of \(a\):
\[ b = a + 20 \]
Now we can check the provided options:

1. **Option 1:** \(a = 20\), \(b = 40\)
\(b - a = 40 - 20 = 20\) (Correct)

2. **Option 2:** \(a = 5\), \(b = 5\)
\(b - a = 5 - 5 = 0\) (Incorrect)

3. **Option 3:** \(a = 10\), \(b = 20\)
\(b - a = 20 - 10 = 10\) (Incorrect)

Thus, the only valid option is:
\[ (a, b) = (20, 40) \]

### Final Answer:
The values of \(a\) and \(b\) can be \( (20, 40) \).