Let `f(x)= sin^(-1)((2x)/(1+x^2))AAx in R`. The function `f(x)` is continuous everywhere but not differentiable at x is/ are

Options

A. `0,1`
B. `-1,1`
C. `-1,0`
D. `0,2`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) and determine where it is not differentiable.

### Step-by-step Solution:

1. **Substitution**:
Let \( x = \tan(\theta) \), where \( \theta \) is in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). This substitution helps us simplify the expression.

2. **Transform the function**:
Using the identity for tangent, we can rewrite \( f(x) \):
\[ f(x) = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \]
By using the double angle formula for sine, we have:
\[ \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]
Thus, we can express \( f(x) \) as:
\[ f(x) = \sin^{-1}(\sin(2\theta)) \]

3. **Determine the range of \( 2\theta \)**:
The value of \( 2\theta \) will vary depending on the value of \( \theta \):
- If \( 2\theta \) is in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), then \( f(x) = 2\theta \).
- If \( 2\theta \) is in the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), then \( f(x) = \pi - 2\theta \).
- If \( 2\theta \) is in the interval \( \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right) \), then \( f(x) = -\pi - 2\theta \).

4. **Find the piecewise definition**:
We can derive the piecewise function for \( f(x) \):
- For \( x > 1 \): \( f(x) = \pi - 2\tan^{-1}(x) \)
- For \( -1 \leq x \leq 1 \): \( f(x) = 2\tan^{-1}(x) \)
- For \( x < -1 \): \( f(x) = -\pi - 2\tan^{-1}(x) \)

5. **Differentiate the function**:
Now we differentiate \( f(x) \) in each piece:
- For \( x > 1 \):
\[ f'(x) = -\frac{2}{1+x^2} \]
- For \( -1 < x < 1 \):
\[ f'(x) = \frac{2}{1+x^2} \]
- For \( x < -1 \):
\[ f'(x) = -\frac{2}{1+x^2} \]

6. **Check differentiability at critical points**:
We need to check the points \( x = -1 \) and \( x = 1 \):
- As \( x \) approaches \( -1 \):
- \( f'(-1)^+ = \frac{2}{1+(-1)^2} = 1 \)
- \( f'(-1)^- = -\frac{2}{1+(-1)^2} = -1 \)
- As \( x \) approaches \( 1 \):
- \( f'(1)^+ = -\frac{2}{1+1^2} = -1 \)
- \( f'(1)^- = \frac{2}{1+1^2} = 1 \)

Since the left-hand and right-hand derivatives do not match at both \( x = -1 \) and \( x = 1 \), the function is not differentiable at these points.

### Conclusion:
The function \( f(x) \) is not differentiable at \( x = -1 \) and \( x = 1 \).