If `f(x)=3+cos^-1(cos(pi/2+x)cos(pi/2-x)+sin(pi/2+x)sin(pi/2-x)) , x in [0,pi]` then find the minimum value of f(x) is

Options

A. 4
B. -2
C. -3
D. 3

Answer

To find the minimum value of the function \( f(x) = 3 + \cos^{-1}(\cos(\frac{\pi}{2} + x) \cos(\frac{\pi}{2} - x) + \sin(\frac{\pi}{2} + x) \sin(\frac{\pi}{2} - x)) \) for \( x \in [0, \pi] \), we can follow these steps:

### Step 1: Simplify the expression inside the inverse cosine
Using the angle addition formula, we know that:
\[ \cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A - B) \]
Let \( A = \frac{\pi}{2} + x \) and \( B = \frac{\pi}{2} - x \). Then:
\[ A - B = \left(\frac{\pi}{2} + x\right) - \left(\frac{\pi}{2} - x\right) = 2x \]
Thus, we can rewrite the expression as:
\[ \cos(\frac{\pi}{2} + x) \cos(\frac{\pi}{2} - x) + \sin(\frac{\pi}{2} + x) \sin(\frac{\pi}{2} - x) = \cos(2x) \]

### Step 2: Substitute back into the function
Now substituting back into \( f(x) \):
\[ f(x) = 3 + \cos^{-1}(\cos(2x)) \]

### Step 3: Simplify \( \cos^{-1}(\cos(2x)) \)
The function \( \cos^{-1}(\cos(\theta)) \) returns \( \theta \) if \( \theta \) is in the range \( [0, \pi] \). Since \( 2x \) varies from \( 0 \) to \( 2\pi \) as \( x \) varies from \( 0 \) to \( \pi \), we need to consider the principal value:
- For \( 0 \leq 2x \leq \pi \), \( \cos^{-1}(\cos(2x)) = 2x \)
- For \( \pi < 2x \leq 2\pi \), \( \cos^{-1}(\cos(2x)) = 2\pi - 2x \)

Since \( x \in [0, \pi] \), we have \( 2x \in [0, 2\pi] \). Thus, we can express \( f(x) \) as:
\[ f(x) = \begin{cases} 3 + 2x & \text{if } 0 \leq 2x \leq \pi \\ 3 + (2\pi - 2x) & \text{if } \pi < 2x \leq 2\pi \end{cases} \]

### Step 4: Determine the intervals
1. For \( 0 \leq x \leq \frac{\pi}{2} \): \( f(x) = 3 + 2x \)
2. For \( \frac{\pi}{2} < x \leq \pi \): \( f(x) = 3 + (2\pi - 2x) \)

### Step 5: Find the minimum value
- In the interval \( [0, \frac{\pi}{2}] \):
- \( f(0) = 3 + 2(0) = 3 \)
- \( f(\frac{\pi}{2}) = 3 + 2(\frac{\pi}{2}) = 3 + \pi \)

- In the interval \( [\frac{\pi}{2}, \pi] \):
- \( f(\frac{\pi}{2}) = 3 + (2\pi - 2(\frac{\pi}{2})) = 3 + \pi \)
- \( f(\pi) = 3 + (2\pi - 2\pi) = 3 \)

### Conclusion
The minimum value occurs at \( x = 0 \) and \( x = \pi \):
\[ \text{Minimum value of } f(x) = 3 \]