The function f(t) = 4 `sin^(3) t -6 sin^(2) ` t +12 sin t + 100 is strictly :

Options

A. increasing in `(pi , (3pi)/(2))`
B. decreasing in `((pi)/(2) , pi)`
C. decreasing in ` [-(pi)/(2) , (pi)/(2)]`
D. decreaing in `[0, (pi)/(2)]`

Answer

To determine whether the function \( f(t) = 4 \sin^3 t - 6 \sin^2 t + 12 \sin t + 100 \) is strictly increasing or decreasing, we will follow these steps:

### Step 1: Differentiate the function
To find the intervals where the function is increasing or decreasing, we first need to compute the derivative \( f'(t) \).

\[ f(t) = 4 \sin^3 t - 6 \sin^2 t + 12 \sin t + 100 \]

Using the chain rule and the power rule, we differentiate:

\[ f'(t) = \frac{d}{dt}(4 \sin^3 t) - \frac{d}{dt}(6 \sin^2 t) + \frac{d}{dt}(12 \sin t) + \frac{d}{dt}(100) \]

Calculating each term:

1. \( \frac{d}{dt}(4 \sin^3 t) = 4 \cdot 3 \sin^2 t \cdot \cos t = 12 \sin^2 t \cos t \)
2. \( \frac{d}{dt}(-6 \sin^2 t) = -6 \cdot 2 \sin t \cdot \cos t = -12 \sin t \cos t \)
3. \( \frac{d}{dt}(12 \sin t) = 12 \cos t \)
4. \( \frac{d}{dt}(100) = 0 \)

Combining these results, we have:

\[ f'(t) = 12 \sin^2 t \cos t - 12 \sin t \cos t + 12 \cos t \]

### Step 2: Factor out common terms
We can factor out \( 12 \cos t \):

\[ f'(t) = 12 \cos t (\sin^2 t - \sin t + 1) \]

### Step 3: Analyze the sign of \( f'(t) \)
To determine where \( f(t) \) is increasing or decreasing, we need to analyze the sign of \( f'(t) \).

1. The term \( 12 \cos t \) will determine the sign based on the value of \( \cos t \).
2. The quadratic \( \sin^2 t - \sin t + 1 \) can be analyzed. The discriminant of this quadratic is:

\[ D = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]

Since the discriminant is negative, the quadratic \( \sin^2 t - \sin t + 1 \) has no real roots and is always positive.

Thus, \( f'(t) \) will have the same sign as \( \cos t \).

### Step 4: Determine intervals for \( \cos t \)
- \( \cos t > 0 \) in the intervals \( (2n\pi, (2n+1)\pi) \) for \( n \in \mathbb{Z} \).
- \( \cos t < 0 \) in the intervals \( ((2n+1)\pi, (2n+2)\pi) \).

For the specific interval \( [0, 2\pi] \):
- \( \cos t \) is positive in \( [0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi] \).
- \( \cos t \) is negative in \( (\frac{\pi}{2}, \frac{3\pi}{2}) \).

### Step 5: Conclusion
Since \( f'(t) < 0 \) when \( \cos t < 0 \), the function \( f(t) \) is strictly decreasing in the interval:

\[ \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \]

### Final Answer
The function \( f(t) \) is strictly decreasing in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \).

---