`f(x) = 1 + 2 sin x + 3 cos^2 x, ( 0 le x lt (2pi)/3)` is

Options

A. Min. at `x = 90^@`
B. Max. at `x = sin^(-1) 1/sqrt3`
C. Min. at `x = 30^@`
D. Max. at `x = sin^(-1) (1/3)`

Answer

To find the maximum and minimum points of the function \( f(x) = 1 + 2 \sin x + 3 \cos^2 x \) in the interval \( 0 \leq x < \frac{2\pi}{3} \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(1 + 2\sin x + 3\cos^2 x) \]
Using the chain and product rules, we get:
\[ f'(x) = 2\cos x + 3 \cdot 2\cos x(-\sin x) = 2\cos x - 6\cos x\sin x \]
Thus,
\[ f'(x) = 2\cos x(1 - 3\sin x) \]

### Step 2: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 2\cos x(1 - 3\sin x) = 0 \]
This gives us two conditions:
1. \( \cos x = 0 \)
2. \( 1 - 3\sin x = 0 \)

### Step 3: Solve the conditions
**Condition 1:** \( \cos x = 0 \) gives:
\[ x = \frac{\pi}{2} \]

**Condition 2:** \( 1 - 3\sin x = 0 \) gives:
\[ \sin x = \frac{1}{3} \]
Thus,
\[ x = \sin^{-1}\left(\frac{1}{3}\right) \]

### Step 4: Find the second derivative
Next, we find the second derivative \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(2\cos x(1 - 3\sin x)) \]
Using the product rule:
\[ f''(x) = -2\sin x(1 - 3\sin x) + 2\cos x(-3\cos x) = -2\sin x(1 - 3\sin x) - 6\cos^2 x \]

### Step 5: Evaluate the second derivative at critical points
**At \( x = \frac{\pi}{2} \):**
\[ f''\left(\frac{\pi}{2}\right) = -2\sin\left(\frac{\pi}{2}\right)(1 - 3\sin\left(\frac{\pi}{2}\right)) - 6\cos^2\left(\frac{\pi}{2}\right) \]
\[ = -2(1)(1 - 3) - 6(0) = -2(-2) = 4 > 0 \]
This indicates a local minimum at \( x = \frac{\pi}{2} \).

**At \( x = \sin^{-1}\left(\frac{1}{3}\right) \):**
We need to evaluate:
\[ f''\left(\sin^{-1}\left(\frac{1}{3}\right)\right) \]
Calculating this is more complex, but we can substitute \( \sin x = \frac{1}{3} \) and use \( \cos^2 x = 1 - \left(\frac{1}{3}\right)^2 = \frac{8}{9} \):
\[ f''\left(\sin^{-1}\left(\frac{1}{3}\right)\right) = -2\left(\frac{1}{3}\right)\left(1 - 1\right) - 6\left(\frac{8}{9}\right) = -\frac{48}{9} < 0 \]
This indicates a local maximum at \( x = \sin^{-1}\left(\frac{1}{3}\right) \).

### Step 6: Conclusion
Thus, we have:
- Minimum at \( x = \frac{\pi}{2} \)
- Maximum at \( x = \sin^{-1}\left(\frac{1}{3}\right) \)