`f (x) = sin ^(-1) ((1 + x ^(2))/( 2x ))` is

Options

A. differentiable at `x =1`
B. continous `AA x in R`
C. neither continuous nor differentiable at `x =1`
D. continous but not differentiable at `x =1`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1} \left( \frac{1 + x^2}{2x} \right) \) for continuity and differentiability at \( x = 1 \) and \( x = 0 \).

### Step 1: Determine the domain of the function
The function \( f(x) \) involves \( \sin^{-1}(x) \), which is defined for \( x \) in the interval \([-1, 1]\). Therefore, we need to check when \( \frac{1 + x^2}{2x} \) lies within this interval.

1. **Finding when \( \frac{1 + x^2}{2x} \) is defined**:
- The expression is undefined at \( x = 0 \) (division by zero).
- For \( x > 0 \), \( \frac{1 + x^2}{2x} \) is defined.
- For \( x < 0 \), we need to check if \( \frac{1 + x^2}{2x} \) can be negative or not.

2. **Finding the range**:
- For \( x > 0 \):
\[ \frac{1 + x^2}{2x} = \frac{1}{2x} + \frac{x}{2} \]
As \( x \to 0^+ \), \( \frac{1 + x^2}{2x} \to \infty \), and as \( x \to \infty \), \( \frac{1 + x^2}{2x} \to \frac{x}{2} \to \infty \).
- For \( x < 0 \):
\[ \frac{1 + x^2}{2x} \to \text{negative values} \]
- Thus, the function is not defined for \( x = 0 \) and does not yield values in \([-1, 1]\) for \( x < 0\).

### Step 2: Check continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to evaluate the left-hand limit (LHL), right-hand limit (RHL), and the function value at that point.

1. **Calculate \( f(1) \)**:
\[ f(1) = \sin^{-1} \left( \frac{1 + 1^2}{2 \cdot 1} \right) = \sin^{-1}(1) = \frac{\pi}{2} \]

2. **Calculate LHL as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \sin^{-1} \left( \frac{1 + x^2}{2x} \right) = \sin^{-1}(1) = \frac{\pi}{2} \]

3. **Calculate RHL as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \sin^{-1} \left( \frac{1 + x^2}{2x} \right) = \sin^{-1}(1) = \frac{\pi}{2} \]

Since LHL = RHL = \( f(1) = \frac{\pi}{2} \), the function is continuous at \( x = 1 \).

### Step 3: Check differentiability at \( x = 1 \)
To check differentiability, we need to compute the derivative \( f'(x) \) and evaluate it at \( x = 1 \).

1. **Differentiate using the chain rule**:
\[ f'(x) = \frac{1}{\sqrt{1 - \left( \frac{1 + x^2}{2x} \right)^2}} \cdot \frac{d}{dx} \left( \frac{1 + x^2}{2x} \right) \]

2. **Find \( \frac{d}{dx} \left( \frac{1 + x^2}{2x} \right) \)**:
Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{1 + x^2}{2x} \right) = \frac{(2x)(2x) - (1 + x^2)(2)}{(2x)^2} = \frac{2x^2 - 2 - 2x^2}{4x^2} = \frac{-2}{4x^2} = -\frac{1}{2x^2} \]

3. **Evaluate \( f'(1) \)**:
\[ f'(1) = \frac{1}{\sqrt{1 - 1}} \cdot \left(-\frac{1}{2}\right) \text{ (undefined since denominator is zero)} \]

Since \( f'(1) \) does not exist, the function is not differentiable at \( x = 1 \).

### Conclusion
- The function \( f(x) \) is continuous at \( x = 1 \).
- The function \( f(x) \) is not differentiable at \( x = 1 \).
- The function is not defined at \( x = 0 \), hence not continuous or differentiable there.

### Final Answer
- The correct option is: **Continuous but not differentiable at \( x = 1 \)**.