Let `f(x)=sin^(-1)2x + cos^(-1)2x + sec^(-1)2x`. Then the sum of the maximum and minimum values of f(x) is
(a)`pi`
(b)`pi/2`
(c)`2pi`
(d)`(3pi)/2`

Options

A. `pi`
B. `pi/2`
C. `2pi`
D. `(3pi)/2`

Answer

To solve the problem, we need to find the sum of the maximum and minimum values of the function \( f(x) = \sin^{-1}(2x) + \cos^{-1}(2x) + \sec^{-1}(2x) \).

### Step 1: Determine the domain of the function
1. **For \( \sin^{-1}(2x) \)**: The argument \( 2x \) must be in the range \([-1, 1]\). Thus, we have:
\[ -1 \leq 2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \]

2. **For \( \cos^{-1}(2x) \)**: The argument \( 2x \) must also be in the range \([-1, 1]\), which gives the same condition:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

3. **For \( \sec^{-1}(2x) \)**: The argument \( 2x \) must satisfy \( |2x| \geq 1 \). Thus, we have:
\[ 2x \leq -1 \quad \text{or} \quad 2x \geq 1 \implies x \leq -\frac{1}{2} \quad \text{or} \quad x \geq \frac{1}{2} \]

4. **Finding the intersection of the domains**:
- From \( \sin^{-1} \) and \( \cos^{-1} \), we have \( -\frac{1}{2} \leq x \leq \frac{1}{2} \).
- From \( \sec^{-1} \), we have \( x \leq -\frac{1}{2} \) or \( x \geq \frac{1}{2} \).
- The only values that satisfy both conditions are \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \).

### Step 2: Evaluate the function at the endpoints of the domain
1. **At \( x = \frac{1}{2} \)**:
\[ f\left(\frac{1}{2}\right) = \sin^{-1}(1) + \cos^{-1}(1) + \sec^{-1}(1) \]
\[ = \frac{\pi}{2} + 0 + 0 = \frac{\pi}{2} \]

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

### Step 3: Find the maximum and minimum values
- The minimum value of \( f(x) \) is \( \frac{\pi}{2} \) (at \( x = \frac{1}{2} \)).
- The maximum value of \( f(x) \) is \( \frac{3\pi}{2} \) (at \( x = -\frac{1}{2} \)).

### Step 4: Calculate the sum of maximum and minimum values
\[ \text{Sum} = \text{Maximum} + \text{Minimum} = \frac{3\pi}{2} + \frac{\pi}{2} = 2\pi \]

### Final Answer
The sum of the maximum and minimum values of \( f(x) \) is \( 2\pi \).