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(B) Given that $(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P$.By the definition of $A.P.$,$2f(x) = (2^{1+x}+2^{1-x}) + (3^x+3^{-x})$.$f(x)

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$3$

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(B) Given that $(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P$.By the definition of $A.P.$,$2f(x) = (2^{1+x}+2^{1-x}) + (3^x+3^{-x})$.$f(x) = \frac{2(2^x+2^{-x}) + (3^x+3^{-x})}{2} = (2^x+2^{-x}) + \frac{1}{2}(3^x+3^{-x})$.Using the $A.M. \geq G.M.$ inequality,we know that for any $a > 0$,$a^x + a^{-x} \geq 2\sqrt{a^x \cdot a^{-x}} = 2$.Thus,$2^x+2^{-x} \geq 2$ and $3^x+3^{-x} \geq 2$.Substituting these minimum values into the expression for $f(x)$:$f(x) \geq 2 + \frac{1}{2}(2) = 2 + 1 = 3$.Therefore,the minimum value of $f(x)$ is $3$.

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:

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