Range of `f(x)=sin^(-1)x+tan^(-1)x+sec^(-1)x` is

Options

A. `((pi)/(4), (3pi)/(4))`
B. `[(pi)/(4), (3pi)/(4)]`
C. `{(pi)/(4), (3pi)/(4)}`
D. none of these

Explanation

To find the range of the function \( f(x) = \sin^{-1} x + \tan^{-1} x + \sec^{-1} x \), we will follow these steps:

### Step 1: Determine the Domain of Each Function
1. **Domain of \( \sin^{-1} x \)**: The domain is \( [-1, 1] \).
2. **Domain of \( \tan^{-1} x \)**: The domain is \( (-\infty, \infty) \).
3. **Domain of \( \sec^{-1} x \)**: The domain is \( (-\infty, -1] \cup [1, \infty) \).

### Step 2: Find the Common Domain
The common domain of the function \( f(x) \) is the intersection of the domains of the individual functions:
- The intersection of \( [-1, 1] \) (from \( \sin^{-1} x \)) and \( (-\infty, -1] \cup [1, \infty) \) (from \( \sec^{-1} x \)) is just the endpoints \( -1 \) and \( 1 \).

Thus, the common domain is \( \{-1, 1\} \).

### Step 3: Evaluate \( f(x) \) at the Endpoints
1. **Evaluate at \( x = -1 \)**:
\[ f(-1) = \sin^{-1}(-1) + \tan^{-1}(-1) + \sec^{-1}(-1) \]
- \( \sin^{-1}(-1) = -\frac{\pi}{2} \)
- \( \tan^{-1}(-1) = -\frac{\pi}{4} \)
- \( \sec^{-1}(-1) = \pi \)

Therefore,
\[ f(-1) = -\frac{\pi}{2} - \frac{\pi}{4} + \pi = -\frac{\pi}{2} - \frac{\pi}{4} + \frac{4\pi}{4} = \frac{\pi}{4} \]

2. **Evaluate at \( x = 1 \)**:
\[ f(1) = \sin^{-1}(1) + \tan^{-1}(1) + \sec^{-1}(1) \]
- \( \sin^{-1}(1) = \frac{\pi}{2} \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)
- \( \sec^{-1}(1) = 0 \)

Therefore,
\[ f(1) = \frac{\pi}{2} + \frac{\pi}{4} + 0 = \frac{\pi}{2} + \frac{\pi}{4} = \frac{2\pi}{4} + \frac{\pi}{4} = \frac{3\pi}{4} \]

### Step 4: Conclusion on the Range
The values of \( f(x) \) at the endpoints are:
- \( f(-1) = \frac{\pi}{4} \)
- \( f(1) = \frac{3\pi}{4} \)

Thus, the range of \( f(x) \) is:
\[ \left[ \frac{\pi}{4}, \frac{3\pi}{4} \right] \]

### Final Answer
The range of \( f(x) = \sin^{-1} x + \tan^{-1} x + \sec^{-1} x \) is \( \left[ \frac{\pi}{4}, \frac{3\pi}{4} \right] \).