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

Options

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

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(2x) + \cos^{-1}(2x) + \sec^{-1}(2x) \).

### Step 1: Simplify the function
We know that:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
Thus, we can apply this identity to the first two terms of \( f(x) \):
\[ f(x) = \sin^{-1}(2x) + \cos^{-1}(2x) + \sec^{-1}(2x) = \frac{\pi}{2} + \sec^{-1}(2x) \]

### Step 2: Determine the range of \( 2x \)
The function \( \sec^{-1}(x) \) is defined for \( |x| \geq 1 \). Therefore, for \( \sec^{-1}(2x) \) to be defined, we need:
\[ |2x| \geq 1 \implies |x| \geq \frac{1}{2} \]
This means \( x \) must be in the intervals \( (-\infty, -\frac{1}{2}] \) or \( [\frac{1}{2}, \infty) \).

### Step 3: Analyze the range of \( \sec^{-1}(2x) \)
The principal range of \( \sec^{-1}(x) \) is \( [0, \pi/2) \cup (\pi/2, \pi] \). Therefore, we need to find the minimum and maximum values of \( \sec^{-1}(2x) \) within the defined intervals.

- For \( x = \frac{1}{2} \):
\[ \sec^{-1}(2 \cdot \frac{1}{2}) = \sec^{-1}(1) = 0 \]

- For \( x = -\frac{1}{2} \):
\[ \sec^{-1}(2 \cdot -\frac{1}{2}) = \sec^{-1}(-1) = \pi \]

### Step 4: Find the minimum and maximum values of \( f(x) \)
From the previous steps:
- The minimum value of \( \sec^{-1}(2x) \) occurs at \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]

- The maximum value of \( \sec^{-1}(2x) \) occurs at \( x = -\frac{1}{2} \):
\[ f\left(-\frac{1}{2}\right) = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \]

### Step 5: Calculate the sum of the maximum and minimum values
Now we can find the sum of the maximum and minimum values of \( f(x) \):
\[ \text{Sum} = \frac{\pi}{2} + \frac{3\pi}{2} = \frac{4\pi}{2} = 2\pi \]

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