`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 \( f(x) = [\tan^{-1} x] \) is discontinuous, we will analyze the behavior of the function and the greatest integer function (also known as the floor function).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = [\tan^{-1} x] \) applies the greatest integer function to the inverse tangent function. The range of \( \tan^{-1} x \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), which approximately translates to \( (-1.57, 1.57) \).

2. **Identifying the Range of \( f(x) \)**:
The values of \( \tan^{-1} x \) will be between \( -1.57 \) and \( 1.57 \). Therefore, the possible integer values (outputs of the greatest integer function) will be \( -2, -1, 0, \) and \( 1 \).

3. **Finding Points of Discontinuity**:
The greatest integer function \( [y] \) is discontinuous at integer values of \( y \). Thus, we need to find when \( \tan^{-1} x \) equals these integers:
- \( \tan^{-1} x = 0 \)
- \( \tan^{-1} x = 1 \)
- \( \tan^{-1} x = -1 \)
- \( \tan^{-1} x = -2 \) (not applicable since \( \tan^{-1} x \) does not reach -2)

4. **Calculating the Values of \( x \)**:
- For \( \tan^{-1} x = 0 \):
\[ x = \tan(0) = 0 \]
- For \( \tan^{-1} x = 1 \):
\[ x = \tan(1) \approx 1.5574 \]
- For \( \tan^{-1} x = -1 \):
\[ x = \tan(-1) \approx -1.5574 \]

5. **Identifying Discontinuity Points**:
The function \( f(x) \) will be discontinuous at:
- \( x = 0 \)
- \( x \approx 1.5574 \)
- \( x \approx -1.5574 \)

### Conclusion:
The function \( f(x) = [\tan^{-1} x] \) is discontinuous at \( x = 0, \, x \approx 1.5574, \, x \approx -1.5574 \).