Let f`(x) = cosec^-1[1 + sin^2x], where [*]` denotes the greatest integer function, then the range of f

Options

A. Is discrete and has two members
B. Is discrete and has four members
C. Is continuous
D. Is `(pi/2,pi/6)`

Answer

To solve the problem, we need to find the range of the function \( f(x) = \csc^{-1}(1 + \sin^2 x) \) where \( [\cdot] \) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Understanding the function**:
The function we are dealing with is \( f(x) = \csc^{-1}(1 + \sin^2 x) \). The term \( \sin^2 x \) varies between 0 and 1 for all \( x \).

2. **Finding the range of \( 1 + \sin^2 x \)**:
Since \( \sin^2 x \) varies from 0 to 1, we can find the range of \( 1 + \sin^2 x \):
\[ \text{Minimum value: } 1 + 0 = 1 \]
\[ \text{Maximum value: } 1 + 1 = 2 \]
Thus, \( 1 + \sin^2 x \) varies from 1 to 2.

3. **Applying the greatest integer function**:
The greatest integer function \( [1 + \sin^2 x] \) will take the values:
- When \( 1 + \sin^2 x = 1 \) (when \( \sin^2 x = 0 \)), the greatest integer is 1.
- When \( 1 + \sin^2 x \) is in the interval \( (1, 2) \) (when \( 0 < \sin^2 x < 1 \)), the greatest integer is also 1.
- When \( 1 + \sin^2 x = 2 \) (when \( \sin^2 x = 1 \)), the greatest integer is 2.

Therefore, the possible values of \( [1 + \sin^2 x] \) are 1 and 2.

4. **Finding the range of \( f(x) \)**:
Now we evaluate \( f(x) \) for the values obtained from the greatest integer function:
- For \( [1 + \sin^2 x] = 1 \):
\[ f(x) = \csc^{-1}(1) = \frac{\pi}{2} \]
- For \( [1 + \sin^2 x] = 2 \):
\[ f(x) = \csc^{-1}(2) \]

5. **Conclusion**:
The range of \( f(x) \) consists of the values \( \frac{\pi}{2} \) and \( \csc^{-1}(2) \). Thus, the range of \( f \) is:
\[ \left\{ \frac{\pi}{2}, \csc^{-1}(2) \right\} \]

### Final Answer:
The range of the function \( f(x) \) is \( \left\{ \frac{\pi}{2}, \csc^{-1}(2) \right\} \).