Let `f :[2,oo)to {1,oo)` defined by `f (x)=2^(x ^(4)-4x ^(3))and g : [(pi)/(2), pi] to A ` defined by `g (x) = (sin x+4)/(sin x-2)` be two invertible functions, then
The set "A" equals to

Options

A. `[5,2]`
B. `[-2,5]`
C. `[-5,2]`
D. `[-5,-2]`

Answer

To solve the problem, we need to find the set "A" which is the range of the function \( g(x) = \frac{\sin x + 4}{\sin x - 2} \) for \( x \) in the interval \( \left[\frac{\pi}{2}, \pi\right] \).

### Step 1: Determine the range of \( \sin x \)
The function \( \sin x \) varies between its minimum and maximum values in the interval \( \left[\frac{\pi}{2}, \pi\right] \):
- At \( x = \frac{\pi}{2} \), \( \sin\left(\frac{\pi}{2}\right) = 1 \).
- At \( x = \pi \), \( \sin(\pi) = 0 \).

Thus, the range of \( \sin x \) in this interval is \( [0, 1] \).

### Step 2: Substitute the range of \( \sin x \) into \( g(x) \)
Now we will substitute the values of \( \sin x \) into \( g(x) \):
\[ g(x) = \frac{\sin x + 4}{\sin x - 2} \]

### Step 3: Evaluate \( g(x) \) at the endpoints of the range of \( \sin x \)
1. **When \( \sin x = 0 \)**:
\[ g(x) = \frac{0 + 4}{0 - 2} = \frac{4}{-2} = -2 \]

2. **When \( \sin x = 1 \)**:
\[ g(x) = \frac{1 + 4}{1 - 2} = \frac{5}{-1} = -5 \]

### Step 4: Determine the behavior of \( g(x) \) in the interval
Since \( g(x) \) is a continuous function and \( \sin x \) decreases from 1 to 0 in the interval \( \left[\frac{\pi}{2}, \pi\right] \), we can analyze the behavior of \( g(x) \):
- As \( \sin x \) decreases from 1 to 0, \( g(x) \) will increase from \( -5 \) to \( -2 \).

### Step 5: Conclusion about the range
Thus, the range of \( g(x) \) as \( x \) varies from \( \frac{\pi}{2} \) to \( \pi \) is:
\[ [-5, -2] \]

Therefore, the set \( A \) is:
\[ A = [-5, -2] \]