Consider `f (x)= sin ^(5) x-1, x in [0, (pi)/(2)],` which of the following is/are correct ?

Options

A. f is strictly decreasing in `[0, (pi)/(4)]`
B. f is strictly increasing in `[(pi)/(4),(pi)/(2)]`
C. There exist a numbe 'c' in `(0,(pi)/(2))` such that `f (c ) =0`
D. The equation `f (x)=0` has only two roots in` [0,(pi)/(2)]`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^5 x - 1 \) over the interval \( [0, \frac{\pi}{2}] \).

### Step 1: Analyze the function
The function is defined as:
\[ f(x) = \sin^5 x - 1 \]
We need to determine the behavior of this function in the given interval.

### Step 2: Find the derivative
To analyze whether the function is increasing or decreasing, we find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^5 x - 1) = 5 \sin^4 x \cos x \]
This derivative will help us understand the monotonicity of the function.

### Step 3: Determine the sign of the derivative
1. **At \( x = 0 \)**:
\[ f'(0) = 5 \sin^4(0) \cos(0) = 5 \cdot 0^4 \cdot 1 = 0 \]

2. **At \( x = \frac{\pi}{4} \)**:
\[ f'\left(\frac{\pi}{4}\right) = 5 \left(\frac{1}{\sqrt{2}}\right)^4 \cdot \frac{1}{\sqrt{2}} = 5 \cdot \frac{1}{4} \cdot \frac{1}{\sqrt{2}} = \frac{5}{4\sqrt{2}} > 0 \]
This indicates that the function is increasing at \( x = \frac{\pi}{4} \).

3. **At \( x = \frac{\pi}{2} \)**:
\[ f'\left(\frac{\pi}{2}\right) = 5 \sin^4\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) = 5 \cdot 1^4 \cdot 0 = 0 \]

### Step 4: Analyze the intervals
- From \( x = 0 \) to \( x = \frac{\pi}{4} \), since \( f'(x) \) goes from 0 to positive, \( f(x) \) is **increasing**.
- From \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \), since \( f'(x) \) is positive, \( f(x) \) is also **increasing**.

### Step 5: Find the roots of the function
To find where \( f(x) = 0 \):
\[ \sin^5 x - 1 = 0 \implies \sin^5 x = 1 \implies \sin x = 1 \]
The solution to this equation in the interval \( [0, \frac{\pi}{2}] \) is:
\[ x = \frac{\pi}{2} \]

### Step 6: Conclusion
1. **Monotonicity**: The function is strictly increasing in the entire interval \( [0, \frac{\pi}{2}] \).
2. **Roots**: The function has only one root in the interval \( [0, \frac{\pi}{2}] \) at \( x = \frac{\pi}{2} \).

### Final Answers
- The function is strictly increasing in the interval \( [0, \frac{\pi}{2}] \).
- There exists a number \( c \) in \( (0, \frac{\pi}{2}) \) such that \( f(c) = 0 \) (specifically at \( c = \frac{\pi}{2} \)).
- The equation \( f(x) = 0 \) has only one root in \( [0, \frac{\pi}{2}] \) at \( x = \frac{\pi}{2} \).