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 find 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 1: Set up the A.P. condition
Since \( a, b, c \) are in A.P., we have the relationship:
\[ 2F(x) = (2^{1+x} + 2^{1-x}) + (3^x + 3^{-x}) \]

### Step 2: Simplify the terms
Let's simplify \( 2^{1+x} + 2^{1-x} \):
\[ 2^{1+x} + 2^{1-x} = 2 \cdot 2^1 \cdot \cosh(x \ln 2) = 2 \cdot (2 \cosh(x \ln 2)) \]
Thus,
\[ 2^{1+x} + 2^{1-x} = 2(2 \cosh(x \ln 2)) = 4 \cosh(x \ln 2) \]

Now simplify \( 3^x + 3^{-x} \):
\[ 3^x + 3^{-x} = 2 \cosh(x \ln 3) \]

### Step 3: Substitute back into the A.P. equation
Substituting back, we have:
\[ 2F(x) = 4 \cosh(x \ln 2) + 2 \cosh(x \ln 3) \]

### Step 4: Factor out 2
Dividing everything by 2 gives:
\[ F(x) = 2 \cosh(x \ln 2) + \cosh(x \ln 3) \]

### Step 5: Find the minimum value of \( F(x) \)
To find the minimum value of \( F(x) \), we can use the property of the hyperbolic cosine function, which has a minimum value of 1 at \( x = 0 \):
\[ F(0) = 2 \cosh(0) + \cosh(0) = 2(1) + 1 = 3 \]

### Conclusion
Thus, the minimum value of \( F(x) \) is:
\[ \boxed{3} \]