The range of the function `f(x)=cosec^(-1)[sinx] " in " [0,2pi]`, where `[*]` denotes the greatest integer function , is

Options

A. `[0,pi//2) cup (pi,3pi//2]`
B. {-pi/2} and {pi/2}
C. `(0,pi] cup {3pi//2}`
D. `(pi//2,pi) cup (3pi//2,2pi)`

Answer

To find the range of the function \( f(x) = \csc^{-1}[\sin x] \) for \( x \in [0, 2\pi] \), where \([\cdot]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Determine the range of \(\sin x\) in the interval \([0, 2\pi]\)

The sine function oscillates between -1 and 1. Therefore, for \( x \in [0, 2\pi] \):
\[ \sin x \in [-1, 1] \]

### Step 2: Apply the greatest integer function

The greatest integer function \([\sin x]\) takes the value of \(\sin x\) and rounds it down to the nearest integer. Thus, we need to find the integer values that \(\sin x\) can take in the interval \([-1, 1]\).

- When \( \sin x = 1 \): \([\sin x] = 1\)
- When \( \sin x \) is in the interval \((0, 1)\): \([\sin x] = 0\)
- When \( \sin x = 0 \): \([\sin x] = 0\)
- When \( \sin x \) is in the interval \((-1, 0)\): \([\sin x] = -1\)
- When \( \sin x = -1 \): \([\sin x] = -1\)

Thus, the possible values of \([\sin x]\) are \(-1, 0, 1\).

### Step 3: Evaluate \(f(x) = \csc^{-1}([\sin x])\)

Now we evaluate \(f(x)\) for each of the possible values of \([\sin x]\):

1. **If \([\sin x] = 1\)**:
\[ f(x) = \csc^{-1}(1) = 0 \]

2. **If \([\sin x] = 0\)**:
\[ f(x) = \csc^{-1}(0) \text{ is undefined (since \(\csc^{-1}(y)\) is defined for } |y| \geq 1\text{)} \]

3. **If \([\sin x] = -1\)**:
\[ f(x) = \csc^{-1}(-1) = -\frac{\pi}{2} \]

### Step 4: Compile the range of \(f(x)\)

From the evaluations above, the function \(f(x)\) can take the values:
- \(0\) (when \([\sin x] = 1\))
- \(-\frac{\pi}{2}\) (when \([\sin x] = -1\))

Thus, the range of the function \(f(x)\) is:
\[ \text{Range of } f(x) = \{-\frac{\pi}{2}, 0\} \]

### Final Answer
The range of the function \( f(x) = \csc^{-1}[\sin x] \) for \( x \in [0, 2\pi] \) is \(\{-\frac{\pi}{2}, 0\}\).

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