Let $$f : R \rightarrow R$$ be such that 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

We are given that for all $$x \in \mathbb{R}$$, the three quantities $$(2^{1+x} + 2^{1-x})$$, $$f(x)$$, and $$(3^x + 3^{-x})$$ are in arithmetic progression (A.P.).

Step 1: Express $$f(x)$$ using the A.P. condition.

For three terms in A.P., the middle term equals the average of the other two:

$$f(x) = \frac{(2^{1+x} + 2^{1-x}) + (3^x + 3^{-x})}{2}$$

Step 2: Simplify the first pair of terms.

$$2^{1+x} + 2^{1-x} = 2 \cdot 2^x + 2 \cdot 2^{-x} = 2(2^x + 2^{-x})$$

So:

$$f(x) = \frac{2(2^x + 2^{-x}) + (3^x + 3^{-x})}{2}$$

Step 3: Apply the AM-GM inequality.

For any positive real number $$a$$ and any real $$t$$, by the AM-GM inequality:

$$a^t + a^{-t} \geq 2\sqrt{a^t \cdot a^{-t}} = 2$$

Applying this to each pair:

$$2^x + 2^{-x} \geq 2 \implies 2(2^x + 2^{-x}) \geq 4$$

$$3^x + 3^{-x} \geq 2$$

Therefore:

$$f(x) \geq \frac{4 + 2}{2} = 3$$

Step 4: Verify that the minimum is attained.

At $$x = 0$$:

$$2^{1+0} + 2^{1-0} = 2 + 2 = 4$$

$$3^0 + 3^{0} = 1 + 1 = 2$$

$$f(0) = \frac{4 + 2}{2} = 3$$

Since $$f(x) \geq 3$$ for all $$x$$ and $$f(0) = 3$$, the minimum value of $$f(x)$$ is $$3$$.

The correct answer is Option B: $$3$$.