Let `f (x)= sin ^(-1) x-cos ^(-1)` x, then the set of values of k for which of `|f (x)|=k` has exactly two distinct solutions is :

Options

A. `(0,pi]`
B. `(0,(pi)/(2)]`
C. `[(pi)/(2), (3pi)/(2))`
D. `[pi, (3pi)/(2)]`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1} x - \cos^{-1} x \) and determine the values of \( k \) for which \( |f(x)| = k \) has exactly two distinct solutions.

### Step-by-Step Solution:

1. **Understand the Function**:
We start with the function:
\[ f(x) = \sin^{-1} x - \cos^{-1} x \]
We know from trigonometric identities that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Therefore, we can rewrite \( f(x) \):
\[ f(x) = \sin^{-1} x - \left(\frac{\pi}{2} - \sin^{-1} x\right) = 2\sin^{-1} x - \frac{\pi}{2} \]

2. **Determine the Domain**:
The domain of \( \sin^{-1} x \) is \( x \in [-1, 1] \). Thus, \( f(x) \) is defined for \( x \in [-1, 1] \).

3. **Find the Range of \( f(x) \)**:
We need to find the maximum and minimum values of \( f(x) \) within the interval \( [-1, 1] \).
- At \( x = -1 \):
\[ f(-1) = 2\sin^{-1}(-1) - \frac{\pi}{2} = 2\left(-\frac{\pi}{2}\right) - \frac{\pi}{2} = -\frac{3\pi}{2} \]
- At \( x = 1 \):
\[ f(1) = 2\sin^{-1}(1) - \frac{\pi}{2} = 2\left(\frac{\pi}{2}\right) - \frac{\pi}{2} = \frac{\pi}{2} \]

Thus, the range of \( f(x) \) is:
\[ \left[-\frac{3\pi}{2}, \frac{\pi}{2}\right] \]

4. **Graph of \( f(x) \)**:
The function \( f(x) \) is continuous and linear between \( -1 \) and \( 1 \). The graph will start at \( -\frac{3\pi}{2} \) and end at \( \frac{\pi}{2} \).

5. **Consider \( |f(x)| = k \)**:
The equation \( |f(x)| = k \) will have two distinct solutions when \( k \) lies within the range of \( f(x) \) but not at the endpoints. Specifically, we need:
\[ k \in (0, \frac{\pi}{2}) \]
Here, \( k = 0 \) gives only one solution (where \( f(x) = 0 \)), and \( k = \frac{\pi}{2} \) also gives only one solution (where \( f(x) = \frac{\pi}{2} \) or \( f(x) = -\frac{\pi}{2} \)).

6. **Final Set of Values for \( k \)**:
Therefore, the set of values of \( k \) for which \( |f(x)| = k \) has exactly two distinct solutions is:
\[ k \in (0, \frac{\pi}{2}) \]