Let `f(x)=sqrt([sin 2x] -[cos 2x])` (where I I denotes the greatest integer function) then the range of f(x) will be

Answer

To find the range of the function \( f(x) = \sqrt{[\sin(2x)] - [\cos(2x)]} \), where \([\cdot]\) denotes the greatest integer function (also known as the floor function), we will analyze the behavior of \(\sin(2x)\) and \(\cos(2x)\) over their respective intervals.

### Step-by-step Solution:

1. **Understanding the Functions**:
- The sine function, \(\sin(2x)\), oscillates between -1 and 1.
- The cosine function, \(\cos(2x)\), also oscillates between -1 and 1.
- Therefore, both \([\sin(2x)]\) and \([\cos(2x)]\) can take values from -1 to 1.

2. **Finding the Values of the Greatest Integer Function**:
- The greatest integer function \([\sin(2x)]\) can take the values:
- -1 when \(-1 \leq \sin(2x) < 0\)
- 0 when \(0 \leq \sin(2x) < 1\)
- 1 when \(\sin(2x) = 1\)
- The greatest integer function \([\cos(2x)]\) can take the values:
- -1 when \(-1 \leq \cos(2x) < 0\)
- 0 when \(0 \leq \cos(2x) < 1\)
- 1 when \(\cos(2x) = 1\)

3. **Evaluating \(f(x)\)**:
- The function \(f(x) = \sqrt{[\sin(2x)] - [\cos(2x)]}\) will be defined only when \([\sin(2x)] - [\cos(2x)] \geq 0\).
- We will evaluate the possible combinations of \([\sin(2x)]\) and \([\cos(2x)]\):
- If \([\sin(2x)] = 0\) and \([\cos(2x)] = 1\): \(f(x) = \sqrt{0 - 1} \) (not defined)
- If \([\sin(2x)] = 1\) and \([\cos(2x)] = 0\): \(f(x) = \sqrt{1 - 0} = 1\)
- If \([\sin(2x)] = 0\) and \([\cos(2x)] = 0\): \(f(x) = \sqrt{0 - 0} = 0\)
- If \([\sin(2x)] = -1\) and \([\cos(2x)] = 0\): \(f(x) = \sqrt{-1 - 0} \) (not defined)
- If \([\sin(2x)] = 0\) and \([\cos(2x)] = -1\): \(f(x) = \sqrt{0 - (-1)} = \sqrt{1} = 1\)
- If \([\sin(2x)] = -1\) and \([\cos(2x)] = -1\): \(f(x) = \sqrt{-1 - (-1)} = \sqrt{0} = 0\)

4. **Identifying the Range**:
- From the evaluations above, we find that \(f(x)\) can take the values:
- 0 (from the cases where both \([\sin(2x)]\) and \([\cos(2x)]\) are 0 or both are -1)
- 1 (from the cases where \([\sin(2x)] = 1\) and \([\cos(2x)] = 0, or vice versa) - Therefore, the range of \(f(x)\) is \(\{0, 1\}\).

### Conclusion:
The range of the function \( f(x) = \sqrt{[\sin(2x)] - [\cos(2x)]} \) is \(\{0, 1\}\).