If `f(x)={(1/2(|x|-1),|x| gt 1),(tan^(-1)x,|x| le 1):}` then f(x) is

Options

A. continuous for `x in R-{0}`
B. continuous for `x in R-{0,1,-1}`
C. not continuous for `x in {-1,0,1}`
D. `f(x)` is continuous for `x in R-{1,-1}`

Answer

To determine the nature of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{1}{2}(|x| - 1) & \text{if } |x| > 1 \\ \tan^{-1}(x) & \text{if } |x| \leq 1 \end{cases} \]

we need to analyze the continuity of the function at the points where the definition changes, specifically at \( x = -1 \) and \( x = 1 \).

### Step 1: Check continuity at \( x = 1 \)

1. **Calculate \( f(1) \)**:
\[ f(1) = \tan^{-1}(1) = \frac{\pi}{4} \]

2. **Calculate the left-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \tan^{-1}(x) = \tan^{-1}(1) = \frac{\pi}{4} \]

3. **Calculate the right-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{2}(|x| - 1) = \frac{1}{2}(1 - 1) = 0 \]

4. **Check continuity**:
Since \( \lim_{x \to 1^-} f(x) = \frac{\pi}{4} \) and \( \lim_{x \to 1^+} f(x) = 0 \), we see that:
\[ \lim_{x \to 1} f(x) \text{ does not exist (left and right limits are different)}. \]
Therefore, \( f(x) \) is discontinuous at \( x = 1 \).

### Step 2: Check continuity at \( x = -1 \)

1. **Calculate \( f(-1) \)**:
\[ f(-1) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

2. **Calculate the left-hand limit as \( x \) approaches -1**:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{1}{2}(-x - 1) = \frac{1}{2}(-(-1) - 1) = \frac{1}{2}(1 - 1) = 0 \]

3. **Calculate the right-hand limit as \( x \) approaches -1**:
\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \tan^{-1}(x) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

4. **Check continuity**:
Since \( \lim_{x \to -1^-} f(x) = 0 \) and \( \lim_{x \to -1^+} f(x) = -\frac{\pi}{4} \), we see that:
\[ \lim_{x \to -1} f(x) \text{ does not exist (left and right limits are different)}. \]
Therefore, \( f(x) \) is discontinuous at \( x = -1 \).

### Conclusion

The function \( f(x) \) is discontinuous at both \( x = 1 \) and \( x = -1 \). Thus, the final answer is that \( f(x) \) is discontinuous at these points.