`f: (0,oo) to (-pi/2,pi/2)" be defined as, "f(x)=tan^(-1) (log_(e)x)`.
If `x_(1),x_(2) and x_(3)` are the points at which g(x) =[f (x)] is discontinuous (where, [.] denotes the greatest integer function), then `(x_(1)+x_(2)+x_(3))" is :"`

Options

A. 2
B. 3
C. greater than 3
D. belong to

Answer

To solve the problem, we need to analyze the function \( f(x) = \tan^{-1}(\log_e x) \) and determine the points where the greatest integer function \( g(x) = [f(x)] \) is discontinuous.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = \tan^{-1}(\log_e x) \) is defined for \( x > 0 \). The range of \( \log_e x \) is \( (-\infty, \infty) \), and the range of \( \tan^{-1}(y) \) for \( y \in (-\infty, \infty) \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

2. **Finding the Discontinuities**:
The greatest integer function \( g(x) = [f(x)] \) is discontinuous at points where \( f(x) \) takes integer values. We need to find \( x \) such that:
\[ f(x) = n \quad \text{for integers } n \]
This translates to:
\[ \tan^{-1}(\log_e x) = n \]
which implies:
\[ \log_e x = \tan(n) \]
Hence:
\[ x = e^{\tan(n)} \]

3. **Identifying Relevant Integers**:
Since \( f(x) \) ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), the integers \( n \) that fall within this range are \( -1, 0, 1 \).

4. **Calculating Points**:
- For \( n = -1 \):
\[ x_1 = e^{\tan(-1)} = e^{-\tan(1)} \]
- For \( n = 0 \):
\[ x_2 = e^{\tan(0)} = e^0 = 1 \]
- For \( n = 1 \):
\[ x_3 = e^{\tan(1)} \]

5. **Summing the Points**:
We need to find \( x_1 + x_2 + x_3 \):
\[ x_1 + x_2 + x_3 = e^{-\tan(1)} + 1 + e^{\tan(1)} \]

6. **Simplifying the Expression**:
Using the property of exponentials:
\[ e^{-\tan(1)} + e^{\tan(1)} = 2 \cosh(\tan(1)) \]
Therefore:
\[ x_1 + x_2 + x_3 = 2 \cosh(\tan(1)) + 1 \]

### Final Answer:
The sum \( (x_1 + x_2 + x_3) \) is \( 2 \cosh(\tan(1)) + 1 \).