Let `f(x)` be a continuous function such that `f(a-x)+f(x)=0` for all `x in [0,a]`. Then `int_0^a dx/(1+e^(f(x)))=` (A) `a` (B) `a/2` (C) `1/2f(a)` (D) none of these

Options

A. a
B. `(a)/(2)`
C. f (a)
D. `(1)/(2) f(a)`

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_0^a \frac{dx}{1 + e^{f(x)}} \]

given the condition that \( f(a-x) + f(x) = 0 \) for all \( x \in [0, a] \).

### Step 1: Use the property of the function

From the given condition, we can express \( f(a-x) \) as:

\[ f(a-x) = -f(x) \]

### Step 2: Rewrite the integral

Now, we can rewrite the integral \( I \) by substituting \( x \) with \( a - x \):

\[ I = \int_0^a \frac{dx}{1 + e^{f(a-x)}} \]

### Step 3: Substitute \( f(a-x) \)

Using the property we derived, we substitute \( f(a-x) \):

\[ I = \int_0^a \frac{dx}{1 + e^{-f(x)}} \]

### Step 4: Combine the two integrals

Now we have two expressions for \( I \):

1. \( I = \int_0^a \frac{dx}{1 + e^{f(x)}} \)
2. \( I = \int_0^a \frac{dx}{1 + e^{-f(x)}} \)

### Step 5: Add the two integrals

Adding these two equations, we get:

\[ 2I = \int_0^a \left( \frac{dx}{1 + e^{f(x)}} + \frac{dx}{1 + e^{-f(x)}} \right) \]

### Step 6: Simplify the expression

Now, we simplify the right-hand side:

\[ \frac{1}{1 + e^{f(x)}} + \frac{1}{1 + e^{-f(x)}} = \frac{(1 + e^{-f(x)}) + (1 + e^{f(x)})}{(1 + e^{f(x)})(1 + e^{-f(x)})} \]

This simplifies to:

\[ \frac{2}{1 + e^{f(x)} + e^{-f(x)} + 1} = \frac{2}{2 + e^{f(x)} + e^{-f(x)}} \]

Using the identity \( e^{f(x)} + e^{-f(x)} = 2 \cosh(f(x)) \):

\[ = \frac{2}{2 + 2 \cosh(f(x))} = \frac{1}{1 + \cosh(f(x))} \]

### Step 7: Integrate

Thus, we have:

\[ 2I = \int_0^a \frac{2}{2 + 2 \cosh(f(x))} dx = \int_0^a dx = a \]

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

Dividing both sides by 2 gives:

\[ I = \frac{a}{2} \]

### Conclusion

Thus, the value of the integral is:

\[ \int_0^a \frac{dx}{1 + e^{f(x)}} = \frac{a}{2} \]

The correct answer is (B) \( \frac{a}{2} \).

---