Let `f(x) = 1 + 2 sin x + 2 cos^2 x, 0 le x le pi//2`. Then

Options

A. f (x) is greatest at `pi//6`
B. least at `0, pi//2`
C. increasing in `[0,pi//6]` and decreasing in `(pi//6, pi//2)`
D. f(x) is continuous in `[0,pi//2]`

Answer

To solve the problem, we need to analyze the function \( f(x) = 1 + 2 \sin x + 2 \cos^2 x \) over the interval \( 0 \leq x \leq \frac{\pi}{2} \). We will find the critical points by taking the derivative, setting it to zero, and determining the nature of these points (maximum or minimum).

### Step 1: Find the derivative of \( f(x) \)

The function is given as:
\[ f(x) = 1 + 2 \sin x + 2 \cos^2 x \]

To find the derivative \( f'(x) \), we differentiate each term:
\[ f'(x) = 2 \cos x - 2 \sin x \cos x \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the derivative as:
\[ f'(x) = 2 \cos x (1 - \sin x) \]

### Step 2: Set the derivative to zero to find critical points

To find the critical points, we set \( f'(x) = 0 \):
\[ 2 \cos x (1 - \sin x) = 0 \]

This gives us two cases:
1. \( 2 \cos x = 0 \) → \( \cos x = 0 \) → \( x = \frac{\pi}{2} \)
2. \( 1 - \sin x = 0 \) → \( \sin x = 1 \) → \( x = \frac{\pi}{2} \)

Thus, the only critical point in the interval \( [0, \frac{\pi}{2}] \) is \( x = \frac{\pi}{2} \).

### Step 3: Evaluate the function at the endpoints and critical points

Now, we need to evaluate \( f(x) \) at the endpoints and the critical point:
1. At \( x = 0 \):
\[ f(0) = 1 + 2 \sin(0) + 2 \cos^2(0) = 1 + 0 + 2(1) = 3 \]
2. At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = 1 + 2 \sin\left(\frac{\pi}{2}\right) + 2 \cos^2\left(\frac{\pi}{2}\right) = 1 + 2(1) + 2(0) = 3 \]

### Step 4: Determine the behavior of the function

We will check the sign of \( f'(x) \) in the intervals \( [0, \frac{\pi}{2}] \):
- For \( x \in [0, \frac{\pi}{6}) \), \( \sin x < \frac{1}{2} \) → \( f'(x) > 0 \) (increasing)
- For \( x \in (\frac{\pi}{6}, \frac{\pi}{2}] \), \( \sin x > \frac{1}{2} \) → \( f'(x) < 0 \) (decreasing)

### Conclusion

- The function \( f(x) \) is increasing on \( [0, \frac{\pi}{6}) \).
- The function \( f(x) \) is decreasing on \( (\frac{\pi}{6}, \frac{\pi}{2}] \).
- The maximum value occurs at \( x = 0 \) and \( x = \frac{\pi}{2} \) where \( f(0) = f\left(\frac{\pi}{2}\right) = 3 \).
- The minimum value occurs at \( x = \frac{\pi}{6} \).

### Final Results
- Maximum value: \( 3 \) at \( x = 0 \) and \( x = \frac{\pi}{2} \)
- Minimum value: \( f\left(\frac{\pi}{6}\right) \) (to be calculated)