Let `F:R to R` be such that F for all `x in R` `(2^(1+x)+2^(1-x)), F(x) and (3^(x)+3^(-x))` are in A.P., then the minimum value of F(x) is:

Options

A. 4
B. 0
C. 2
D. 3

Answer

To solve the problem, we need to determine the minimum value of the function \( F(x) \) given that \( 2^{1+x} + 2^{1-x}, F(x), 3^x + 3^{-x} \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression (A.P.):**
For three terms \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
In our case, let:
- \( a = 2^{1+x} + 2^{1-x} \)
- \( b = F(x) \)
- \( c = 3^x + 3^{-x} \)

Therefore, we can write:
\[ 2F(x) = (2^{1+x} + 2^{1-x}) + (3^x + 3^{-x}) \]

2. **Simplifying the Expression:**
We can simplify \( 2^{1+x} + 2^{1-x} \):
\[ 2^{1+x} + 2^{1-x} = 2 \cdot 2^x + 2 \cdot 2^{-x} = 2(2^x + 2^{-x}) \]
Thus, we have:
\[ 2F(x) = 2(2^x + 2^{-x}) + (3^x + 3^{-x}) \]

3. **Dividing by 2:**
Dividing the entire equation by 2 gives:
\[ F(x) = (2^x + 2^{-x}) + \frac{1}{2}(3^x + 3^{-x}) \]

4. **Finding Minimum Values:**
We know that \( 2^x + 2^{-x} \) and \( 3^x + 3^{-x} \) both have minimum values. Using the AM-GM inequality:
- The minimum value of \( 2^x + 2^{-x} \) is 2 (occurs at \( x = 0 \)).
- The minimum value of \( 3^x + 3^{-x} \) is also 2 (occurs at \( x = 0 \)).

5. **Calculating Minimum of \( F(x) \):**
Substituting these minimum values into the equation for \( F(x) \):
\[ F(x) = 2 + \frac{1}{2} \cdot 2 = 2 + 1 = 3 \]

Thus, the minimum value of \( F(x) \) is \( \boxed{3} \).