The range of the function, `f(x)= (1+sec^-1x) (1 + cos^-1 x)` is

Options

A. `(-oo,oo)`
B. `(-oo,0]uu[4,oo)`
C. `{1,(1+pi)(2)}`
D. `{0,(1+pi^(2))}`

Explanation

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

### Step 1: Understand the components of the function
The function consists of two parts: \( \sec^{-1} x \) and \( \cos^{-1} x \). We need to determine the valid ranges for these inverse trigonometric functions.

### Step 2: Determine the domain of \( \sec^{-1} x \)
The function \( \sec^{-1} x \) is defined for \( x \leq -1 \) or \( x \geq 1 \). Therefore, the domain of \( \sec^{-1} x \) is:
- \( (-\infty, -1] \) and \( [1, \infty) \)

### Step 3: Determine the range of \( \sec^{-1} x \)
The range of \( \sec^{-1} x \) is:
- \( [0, \pi/2) \cup (\pi/2, \pi] \)

### Step 4: Determine the domain of \( \cos^{-1} x \)
The function \( \cos^{-1} x \) is defined for \( x \) in the interval \( [-1, 1] \). Thus, the domain of \( \cos^{-1} x \) is:
- \( [-1, 1] \)

### Step 5: Determine the range of \( \cos^{-1} x \)
The range of \( \cos^{-1} x \) is:
- \( [0, \pi] \)

### Step 6: Combine the domains
Since \( f(x) \) involves both \( \sec^{-1} x \) and \( \cos^{-1} x \), we need to find the intersection of their domains. The combined domain of \( f(x) \) is:
- \( x = 1 \) or \( x = -1 \)

### Step 7: Evaluate \( f(x) \) at the endpoints of the domain
1. **When \( x = 1 \)**:
\[ f(1) = (1 + \sec^{-1}(1))(1 + \cos^{-1}(1)) = (1 + 0)(1 + 0) = 1 \]

2. **When \( x = -1 \)**:
\[ f(-1) = (1 + \sec^{-1}(-1))(1 + \cos^{-1}(-1)) = (1 + \pi)(1 + \pi) = (1 + \pi)^2 \]

### Step 8: Determine the range of \( f(x) \)
From the evaluations:
- At \( x = 1 \), \( f(1) = 1 \)
- At \( x = -1 \), \( f(-1) = (1 + \pi)^2 \)

Thus, the range of \( f(x) \) is:
\[ [1, (1 + \pi)^2] \]

### Final Answer
The range of the function \( f(x) = (1 + \sec^{-1} x)(1 + \cos^{-1} x) \) is:
\[ [1, (1 + \pi)^2] \]