Let `f : [0, pi//2] to R` such that `f(0) = 3 and f'(x) = 1/(1 + cos x)` , then
`3 + pi/4 e le f(pi/2) le 3 + pi/2` True or False ?

Answer

To solve the problem step by step, we will start from the given function and its derivative.

### Step 1: Understand the given information
We are given that \( f(0) = 3 \) and \( f'(x) = \frac{1}{1 + \cos x} \).

### Step 2: Integrate \( f'(x) \)
To find \( f(x) \), we need to integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int \frac{1}{1 + \cos x} \, dx \]

### Step 3: Simplify the integrand
We can simplify \( \frac{1}{1 + \cos x} \) using the identity \( 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \):
\[ f'(x) = \frac{1}{2 \cos^2 \left(\frac{x}{2}\right)} = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right) \]

### Step 4: Integrate using the secant function
Now we integrate:
\[ f(x) = \int \frac{1}{2} \sec^2 \left(\frac{x}{2}\right) \, dx \]
The integral of \( \sec^2 u \) is \( \tan u \), so we have:
\[ f(x) = \frac{1}{2} \cdot 2 \tan \left(\frac{x}{2}\right) + C = \tan \left(\frac{x}{2}\right) + C \]

### Step 5: Use the initial condition to find \( C \)
We know \( f(0) = 3 \):
\[ f(0) = \tan(0) + C = 0 + C = 3 \implies C = 3 \]
Thus, we have:
\[ f(x) = \tan \left(\frac{x}{2}\right) + 3 \]

### Step 6: Find \( f\left(\frac{\pi}{2}\right) \)
Now we calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \tan\left(\frac{\pi}{4}\right) + 3 = 1 + 3 = 4 \]

### Step 7: Verify the inequalities
We need to check if:
\[ 3 + \frac{\pi}{4} \leq f\left(\frac{\pi}{2}\right) \leq 3 + \frac{\pi}{2} \]
Calculating the bounds:
- \( 3 + \frac{\pi}{4} \approx 3 + 0.785 = 3.785 \)
- \( 3 + \frac{\pi}{2} \approx 3 + 1.57 = 4.57 \)

Since \( f\left(\frac{\pi}{2}\right) = 4 \):
- \( 3.785 \leq 4 \leq 4.57 \)

### Conclusion
Both inequalities are satisfied, thus the statement is **True**.

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