Let `f(x)={(|x+1|)/(tan^(- 1)(x+1)), x!=-1 ,1, x!=-1` Then `f(x)` is

Options

A. continuous at x =-1
B. Differentiable at x=-1
C. Discontinuous at x =-1
D. Continuous but not derivable at x =1

Answer

To determine the continuity and differentiability of the function \( f(x) \) defined as:

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

we need to analyze the function at the point \( x = -1 \).

### Step 1: Determine the left-hand limit as \( x \) approaches -1

For \( x < -1 \), we have:

\[ f(x) = \frac{|x+1|}{\tan^{-1}(x+1)} = \frac{-(x+1)}{\tan^{-1}(x+1)} \]

Now, we compute the left-hand limit:

\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{-(x+1)}{\tan^{-1}(x+1)} \]

As \( x \to -1 \), \( x + 1 \to 0 \). We can use the fact that:

\[ \lim_{u \to 0} \frac{\tan^{-1}(u)}{u} = 1 \implies \tan^{-1}(x+1) \sim (x+1) \text{ as } x \to -1 \]

Thus, we have:

\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{-(x+1)}{\tan^{-1}(x+1)} = \lim_{x \to -1^-} \frac{-(x+1)}{(x+1)} = -1 \]

### Step 2: Determine the right-hand limit as \( x \) approaches -1

For \( x > -1 \), we have:

\[ f(x) = \frac{x+1}{\tan^{-1}(x+1)} \]

Now, we compute the right-hand limit:

\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{x+1}{\tan^{-1}(x+1)} \]

Using the same limit property as before:

\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{x+1}{\tan^{-1}(x+1)} = \lim_{x \to -1^+} \frac{x+1}{(x+1)} = 1 \]

### Step 3: Check the value of the function at \( x = -1 \)

From the definition of the function, we know:

\[ f(-1) = 1 \]

### Step 4: Analyze continuity at \( x = -1 \)

Now we compare the left-hand limit, right-hand limit, and the function value at \( x = -1 \):

- Left-hand limit: \( -1 \)
- Right-hand limit: \( 1 \)
- Function value: \( f(-1) = 1 \)

Since the left-hand limit \( (-1) \) is not equal to the right-hand limit \( (1) \), we conclude that:

\[ \lim_{x \to -1} f(x) \text{ does not exist} \]

### Conclusion

Thus, the function \( f(x) \) is discontinuous at \( x = -1 \).

### Final Answer

The function \( f(x) \) is discontinuous at \( x = -1 \).

---