Let `f(x) = sin^(3)x - 3 sinx + 6, AA x` `in (0, pi)`.The number of local maximum/maxima of the function f(x) is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of local maxima of the function \( f(x) = \sin^3 x - 3 \sin x + 6 \) for \( x \in (0, \pi) \), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of \( f(x) \).

\[ f'(x) = \frac{d}{dx}(\sin^3 x - 3 \sin x + 6) \]

Using the chain rule and the derivative of sine, we get:

\[ f'(x) = 3 \sin^2 x \cdot \cos x - 3 \cos x \]

### Step 2: Factor the derivative
Next, we can factor out \( 3 \cos x \):

\[ f'(x) = 3 \cos x (\sin^2 x - 1) \]

### Step 3: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):

\[ 3 \cos x (\sin^2 x - 1) = 0 \]

This gives us two equations to solve:

1. \( \cos x = 0 \)
2. \( \sin^2 x - 1 = 0 \)

### Step 4: Solve for critical points
**For \( \cos x = 0 \):**
This occurs at \( x = \frac{\pi}{2} \) within the interval \( (0, \pi) \).

**For \( \sin^2 x - 1 = 0 \):**
This simplifies to \( \sin^2 x = 1 \), which gives \( \sin x = \pm 1 \). In the interval \( (0, \pi) \), this occurs at \( x = \frac{\pi}{2} \).

### Step 5: Analyze the critical points
Now we have one critical point \( x = \frac{\pi}{2} \). We will check the behavior of \( f'(x) \) around this point to determine if it is a local maximum or minimum.

- For \( x < \frac{\pi}{2} \) (e.g., \( x = \frac{\pi}{4} \)):
- \( \cos(\frac{\pi}{4}) > 0 \) and \( \sin^2(\frac{\pi}{4}) < 1 \) → \( f'(\frac{\pi}{4}) > 0 \)

- For \( x > \frac{\pi}{2} \) (e.g., \( x = \frac{3\pi}{4} \)):
- \( \cos(\frac{3\pi}{4}) < 0 \) and \( \sin^2(\frac{3\pi}{4}) < 1 \) → \( f'(\frac{3\pi}{4}) < 0 \)

### Step 6: Conclusion
Since \( f'(x) \) changes from positive to negative at \( x = \frac{\pi}{2} \), this indicates that \( x = \frac{\pi}{2} \) is a local maximum.

### Final Answer
Thus, the number of local maxima of the function \( f(x) \) in the interval \( (0, \pi) \) is **1**.

---