Let `f(x)=sec^(-1)[1+cos^(2)x],` where [.] denotes the greatest integer function. Then the

Options

A. domain of `f` is R
B. domain of `f` is `[1,2]`
C. domain of `f` is `[1,2]`
D. range of `f " is " {sec^(-1) 1, sec^(-1)2}`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \sec^{-1}[1 + \cos^2 x] \), where \([.]\) denotes the greatest integer function. We will find the domain and range of this function step by step.

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

The function \( \sec^{-1}(y) \) is defined for \( y \leq -1 \) or \( y \geq 1 \). Therefore, we need to find the values of \( x \) for which:

\[ 1 + \cos^2 x \leq -1 \quad \text{or} \quad 1 + \cos^2 x \geq 1 \]

#### Analysis of \( 1 + \cos^2 x \)

1. **For \( 1 + \cos^2 x \leq -1 \)**:
- This inequality is impossible because \( \cos^2 x \) is always non-negative (i.e., \( \cos^2 x \geq 0 \)).
- Therefore, \( 1 + \cos^2 x \) is always at least 1.

2. **For \( 1 + \cos^2 x \geq 1 \)**:
- This inequality is always true since \( \cos^2 x \geq 0 \).

Thus, the function \( f(x) \) is defined for all \( x \in \mathbb{R} \).

### Step 2: Determine the Range of \( f(x) \)

Next, we need to find the range of \( f(x) \). Since we have established that \( f(x) = \sec^{-1}[1 + \cos^2 x] \), we need to analyze the expression \( 1 + \cos^2 x \).

1. **Finding the range of \( 1 + \cos^2 x \)**:
- The minimum value of \( \cos^2 x \) is 0 (when \( \cos x = 0 \)), which gives:
\[ 1 + \cos^2 x \geq 1 + 0 = 1 \]
- The maximum value of \( \cos^2 x \) is 1 (when \( \cos x = \pm 1 \)), which gives:
\[ 1 + \cos^2 x \leq 1 + 1 = 2 \]
- Therefore, \( 1 + \cos^2 x \) varies between 1 and 2:
\[ 1 \leq 1 + \cos^2 x \leq 2 \]

2. **Applying the greatest integer function**:
- The greatest integer function \([1 + \cos^2 x]\) can take values:
- When \( 1 + \cos^2 x \) is in the interval \([1, 2)\), \([1 + \cos^2 x] = 1\).
- When \( 1 + \cos^2 x = 2\), \([1 + \cos^2 x] = 2\).

Thus, \([1 + \cos^2 x]\) can take values 1 or 2.

3. **Finding the range of \( f(x) \)**:
- For \([1 + \cos^2 x] = 1\):
\[ f(x) = \sec^{-1}(1) = 0 \]
- For \([1 + \cos^2 x] = 2\):
\[ f(x) = \sec^{-1}(2) \]
- Therefore, the range of \( f(x) \) is \( \{0, \sec^{-1}(2)\} \).

### Conclusion

- **Domain**: \( x \in \mathbb{R} \)
- **Range**: \( \{0, \sec^{-1}(2)\} \)