Let `f : R rarr R` be defined as `f(x) = int_(-1)^(e^(x)) (dt)/(1+t^(2)) + int_(1)^(e^(x))(dt)/(1+t^(2))` then

Options

A. `f(x)` is periodic
B. `f(f(x)) = f(x) AA x in R`
C. `f(1) = f'(1) = pi/2`
D. `f(x)` is unbounded

Answer

To solve the problem, we need to analyze the function defined by the integrals:

Given:
\[ f(x) = \int_{-1}^{e^x} \frac{dt}{1+t^2} + \int_{1}^{e^x} \frac{dt}{1+t^2} \]

### Step 1: Evaluate the integrals

The integral \( \int \frac{dt}{1+t^2} \) is known to be \( \tan^{-1}(t) \). Thus, we can evaluate both integrals:

1. For the first integral:
\[ \int_{-1}^{e^x} \frac{dt}{1+t^2} = \tan^{-1}(e^x) - \tan^{-1}(-1) \]
Since \( \tan^{-1}(-1) = -\frac{\pi}{4} \), we have:
\[ \int_{-1}^{e^x} \frac{dt}{1+t^2} = \tan^{-1}(e^x) + \frac{\pi}{4} \]

2. For the second integral:
\[ \int_{1}^{e^x} \frac{dt}{1+t^2} = \tan^{-1}(e^x) - \tan^{-1}(1) \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\[ \int_{1}^{e^x} \frac{dt}{1+t^2} = \tan^{-1}(e^x) - \frac{\pi}{4} \]

### Step 2: Combine the results

Now, we can combine the results of the two integrals into the function \( f(x) \):
\[ f(x) = \left( \tan^{-1}(e^x) + \frac{\pi}{4} \right) + \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \]
This simplifies to:
\[ f(x) = 2 \tan^{-1}(e^x) \]

### Step 3: Analyze the properties of \( f(x) \)

1. **Check if \( f(x) \) is periodic**:
A function is periodic if \( f(x + T) = f(x) \) for some period \( T \). Since \( \tan^{-1}(e^x) \) is not periodic, \( f(x) \) is also not periodic.

2. **Check if \( f(f(x)) = f(x) \)**:
We can substitute \( f(x) \) into itself:
\[ f(f(x)) = f(2 \tan^{-1}(e^x)) = 2 \tan^{-1}(e^{2 \tan^{-1}(e^x)}) \]
This does not equal \( f(x) \), so \( f(f(x)) \neq f(x) \).

3. **Evaluate \( f(1) \)**:
\[ f(1) = 2 \tan^{-1}(e^1) = 2 \tan^{-1}(e) \]

4. **Find \( f'(x) \)**:
Using the chain rule:
\[ f'(x) = 2 \cdot \frac{1}{1+(e^x)^2} \cdot e^x = \frac{2e^x}{1+e^{2x}} \]

### Conclusion

Thus, we have derived \( f(x) = 2 \tan^{-1}(e^x) \) and analyzed its properties.