`f(x)=[tan^(-1)x]` where [.] denotes the greatest integer function, is discontinous at

Options

A. `x=(pi)/4,(pi)/4` and 0
B. `x=(pi)/3,-(pi)/3` and 0
C. `x=tan1, -tan1, -1tan1 and 0`
D. None of these

Answer

To determine where the function \( f(x) = [\tan^{-1} x] \) (where \([.]\) denotes the greatest integer function) is discontinuous, we need to analyze the behavior of the function.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = [\tan^{-1} x] \) takes the value of the greatest integer less than or equal to \( \tan^{-1} x \). The function \( \tan^{-1} x \) (or arctan) is continuous and maps \( x \) from \( -\infty \) to \( +\infty \) into the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

**Hint**: Recall that the greatest integer function \([x]\) is discontinuous at integer values of \( x \).

2. **Identifying Points of Discontinuity**:
The greatest integer function \([x]\) is discontinuous at integer values. Therefore, we need to find when \( \tan^{-1} x \) equals an integer. The possible integer values of \( \tan^{-1} x \) within the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) are \( -1, 0, \) and \( 1 \).

**Hint**: Determine the values of \( x \) for which \( \tan^{-1} x = n \) where \( n \) is an integer.

3. **Finding \( x \) Values**:
- For \( \tan^{-1} x = -1 \):
\[ x = \tan(-1) = -\tan(1) \]
- For \( \tan^{-1} x = 0 \):
\[ x = \tan(0) = 0 \]
- For \( \tan^{-1} x = 1 \):
\[ x = \tan(1) \quad (\text{where } 1 \text{ is in radians}) \]

**Hint**: Use the tangent function to find the corresponding \( x \) values for each integer.

4. **Conclusion**:
The function \( f(x) = [\tan^{-1} x] \) is discontinuous at:
- \( x = -\tan(1) \)
- \( x = 0 \)
- \( x = \tan(1) \)

Therefore, the points of discontinuity are \( x = -\tan(1), 0, \tan(1) \).

### Final Answer:
The function \( f(x) = [\tan^{-1} x] \) is discontinuous at \( x = -\tan(1), 0, \tan(1) \).