Let function f(x) be defined as
`f(x) = |sin^(-1)x| + cos^(-1) (1/x) `. Then which of the following is /are TRUE.

Options

A. f(x) is injective in its domain.
B. f(x) is many - one in its domain.
C. Range of f is singleton set
D. sgn (f(x)) =1 , where sgn x denotes signum function of x.

Answer

To solve the problem, we need to analyze the function defined as:

\[ f(x) = |\sin^{-1}(x)| + \cos^{-1}\left(\frac{1}{x}\right) \]

### Step 1: Determine the Domain of \( f(x) \)

1. **For \( \sin^{-1}(x) \)**:
- The domain of \( \sin^{-1}(x) \) is \( x \in [-1, 1] \).

2. **For \( \cos^{-1}\left(\frac{1}{x}\right) \)**:
- The expression \( \frac{1}{x} \) must lie in the interval \([-1, 1]\).
- This implies \( x \) must be either \( x \geq 1 \) or \( x \leq -1 \).

3. **Finding the Intersection**:
- The domain of \( f(x) \) is the intersection of the domains of both functions:
- From \( \sin^{-1}(x) \): \( x \in [-1, 1] \)
- From \( \cos^{-1}\left(\frac{1}{x}\right) \): \( x \in (-\infty, -1] \cup [1, \infty) \)
- The intersection gives us \( x = -1 \) and \( x = 1 \).

Thus, the domain of \( f(x) \) is \( \{-1, 1\} \).

### Step 2: Calculate \( f(1) \) and \( f(-1) \)

1. **Calculate \( f(1) \)**:
\[ f(1) = |\sin^{-1}(1)| + \cos^{-1}\left(\frac{1}{1}\right) = |\frac{\pi}{2}| + \cos^{-1}(1) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]

2. **Calculate \( f(-1) \)**:
\[ f(-1) = |\sin^{-1}(-1)| + \cos^{-1}\left(\frac{1}{-1}\right) = |-\frac{\pi}{2}| + \cos^{-1}(-1) = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \]

### Step 3: Determine the Range of \( f(x) \)

- From the calculations:
- \( f(1) = \frac{\pi}{2} \)
- \( f(-1) = \frac{3\pi}{2} \)

Thus, the range of \( f(x) \) is \( \left\{ \frac{\pi}{2}, \frac{3\pi}{2} \right\} \).

### Step 4: Analyze Injectivity and Other Properties

1. **Injectivity**:
- A function is injective (one-to-one) if different inputs map to different outputs.
- Here, \( f(1) \neq f(-1) \), so \( f(x) \) is injective in its domain.

2. **Many-One**:
- Since \( f(x) \) is injective, it cannot be many-one.

3. **Singleton Set**:
- The range of \( f(x) \) contains two elements, so it is not a singleton set.

4. **Signum Function**:
- The signum function \( \text{sgn}(f(x)) \) is 1 for both outputs since both \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \) are positive.

### Conclusion

Based on the analysis:
- \( f(x) \) is injective in its domain.
- \( f(x) \) is not many-one.
- The range of \( f(x) \) is not a singleton set.
- The signum function \( \text{sgn}(f(x)) = 1 \).

### Summary of True Statements:
- \( f(x) \) is injective in its domain: **True**
- \( f(x) \) is many-one in its domain: **False**
- The range of \( f(x) \) is a singleton set: **False**
- \( \text{sgn}(f(x)) = 1 \): **True**