If `f(x) = |1 -x|,` then the points where `sin^-1 (f |x|)` is non-differentiable are

Options

A. {0,1}
B. {0,-1}
C. {0,1,-1}
D. none of these

Answer

To solve the problem, we need to determine the points where the function \( \sin^{-1}(f(|x|)) \) is non-differentiable, given that \( f(x) = |1 - x| \).

### Step-by-Step Solution:

1. **Define the function \( f(x) \)**:
\[ f(x) = |1 - x| \]
This function can be expressed piecewise:
- For \( x < 1 \): \( f(x) = 1 - x \)
- For \( x \geq 1 \): \( f(x) = x - 1 \)

2. **Find \( f(|x|) \)**:
Since we are interested in \( f(|x|) \), we need to consider the absolute value of \( x \):
- For \( |x| < 1 \): \( f(|x|) = 1 - |x| \)
- For \( |x| \geq 1 \): \( f(|x|) = |x| - 1 \)

3. **Determine the points of non-differentiability of \( f(|x|) \)**:
The function \( f(x) \) is non-differentiable at the points where the expression inside the absolute value changes:
- At \( x = 1 \) (where \( f(x) \) changes from \( 1 - x \) to \( x - 1 \))
- At \( x = -1 \) (since \( |x| \) will also be 1)

4. **Evaluate \( \sin^{-1}(f(|x|)) \)**:
The function \( \sin^{-1}(y) \) is non-differentiable at points where \( y \) is not in the interval \([-1, 1]\). However, since \( f(|x|) \) is always between 0 and 1 for \( |x| < 1 \) and takes values outside this range for \( |x| \geq 1 \), we focus on the points where \( f(|x|) \) is non-differentiable:
- \( f(|x|) \) is non-differentiable at \( x = 1 \) and \( x = -1 \).

5. **Check for non-differentiability at \( x = 0 \)**:
The function \( f(|x|) \) is also non-differentiable at \( x = 0 \) since \( f(0) = |1 - 0| = 1 \) and the derivative does not exist at this point due to the absolute value function.

6. **Conclusion**:
The points where \( \sin^{-1}(f(|x|)) \) is non-differentiable are:
\[ x = -1, 0, 1 \]

### Final Answer:
The points where \( \sin^{-1}(f(|x|)) \) is non-differentiable are \( x = -1, 0, 1 \).