`f(x) ={{:(1-sqrt(1-x^(2)),if -1lexle1),(1+ log ""(1)/(x), if xgt1):}` (a) continuous and differentiable at `x=1` (b) continuous but not differentiable at `x=1` (c) neither continuous nor differentiable at `x=1` (d) none of these

Options

A. continuous and differentibable at x=1
B. continuous but not differentiable at x=1
C. neither continuous nor differentiable at =1
D. none of these

Answer

To determine the continuity and differentiability of the function \( f(x) \) at \( x = 1 \), we will follow these steps:

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} 1 - \sqrt{1 - x^2} & \text{if } -1 \leq x \leq 1 \\ 1 + \log\left(\frac{1}{x}\right) & \text{if } x > 1 \end{cases} \]

### Step 2: Check continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to find:
1. The left-hand limit as \( x \) approaches 1.
2. The right-hand limit as \( x \) approaches 1.
3. The value of the function at \( x = 1 \).

**Left-hand limit:**
\[ \lim_{x \to 1^-} f(x) = 1 - \sqrt{1 - 1^2} = 1 - \sqrt{0} = 1 \]

**Right-hand limit:**
\[ \lim_{x \to 1^+} f(x) = 1 + \log\left(\frac{1}{1}\right) = 1 + 0 = 1 \]

**Value of the function at \( x = 1 \):**
\[ f(1) = 1 - \sqrt{1 - 1^2} = 1 - 0 = 1 \]

Since the left-hand limit, right-hand limit, and the value of the function at \( x = 1 \) are all equal to 1, we conclude that:
\[ f(x) \text{ is continuous at } x = 1. \]

### Step 3: Check differentiability at \( x = 1 \)
To check differentiability at \( x = 1 \), we need to find the left-hand derivative and the right-hand derivative.

**Left-hand derivative:**
For \( x \leq 1 \):
\[ f'(x) = \frac{d}{dx}(1 - \sqrt{1 - x^2}) = 0 - \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{x}{\sqrt{1 - x^2}} \]
Now, evaluating at \( x = 1 \):
\[ f'(1^-) = \frac{1}{\sqrt{1 - 1^2}} = \frac{1}{\sqrt{0}} = \infty \]

**Right-hand derivative:**
For \( x > 1 \):
\[ f'(x) = \frac{d}{dx}(1 + \log\left(\frac{1}{x}\right)) = 0 - \frac{1}{x} \cdot (-1) = -\frac{1}{x} \]
Now, evaluating at \( x = 1 \):
\[ f'(1^+) = -\frac{1}{1} = -1 \]

### Step 4: Conclusion
Since the left-hand derivative \( f'(1^-) = \infty \) and the right-hand derivative \( f'(1^+) = -1 \) are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Final Answer
Thus, the function \( f(x) \) is continuous but not differentiable at \( x = 1 \). The correct option is:
(b) continuous but not differentiable at \( x = 1 \).