The domain function `f(x) = 1/ (sqrt((|sin^-1x|-cos^-1|x|)))` is given by

Options

A. `x in[1/(sqrt2),1]`
B. `x in[-1,-1/(sqrt2))`
C. `x in[-1,-1/(sqrt2))cup(1/(sqrt2),1]`
D. `x in[-1,1]-{1/(sqrt2)}`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{|\sin^{-1} x| - \cos^{-1} |x|}} \), we need to ensure that the expression inside the square root is positive, as the square root of a negative number is not defined in the real number system. Therefore, we need to solve the inequality:

\[ |\sin^{-1} x| - \cos^{-1} |x| > 0 \]

### Step 1: Understanding the Functions
1. **Range of \( \sin^{-1} x \)**: The function \( \sin^{-1} x \) is defined for \( x \in [-1, 1] \) and its range is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
2. **Range of \( \cos^{-1} |x| \)**: The function \( \cos^{-1} |x| \) is also defined for \( |x| \in [0, 1] \) and its range is \( [0, \pi] \).

### Step 2: Analyzing the Inequality
We need to analyze the inequality \( |\sin^{-1} x| > \cos^{-1} |x| \).

- For \( x \in [0, 1] \):
- \( |\sin^{-1} x| = \sin^{-1} x \)
- \( \cos^{-1} |x| = \cos^{-1} x \)

Thus, the inequality becomes:
\[ \sin^{-1} x > \cos^{-1} x \]

Using the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), we can rewrite the inequality:
\[ \sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x \]
\[ 2\sin^{-1} x > \frac{\pi}{2} \]
\[ \sin^{-1} x > \frac{\pi}{4} \]

Now, taking the sine of both sides:
\[ x > \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

- For \( x \in [-1, 0] \):
- \( |\sin^{-1} x| = -\sin^{-1} x \)
- \( \cos^{-1} |x| = \cos^{-1} (-x) = \cos^{-1} |x| \)

Thus, the inequality becomes:
\[ -\sin^{-1} x > \cos^{-1} (-x) \]

Using the same identity:
\[ -\sin^{-1} x > \frac{\pi}{2} - \sin^{-1} (-x) \]
\[ -\sin^{-1} x > \frac{\pi}{2} + \sin^{-1} x \]
This leads to:
\[ -2\sin^{-1} x > \frac{\pi}{2} \]
This inequality does not hold for any \( x \in [-1, 0] \).

### Step 3: Conclusion
From the analysis, we find that \( x \) must be in the intervals where \( x > \frac{1}{\sqrt{2}} \) or \( x < -\frac{1}{\sqrt{2}} \).

Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain: } (-1, -\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1) \]