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Question

# Let $f:R\to R$ be such that for all $x\in R$,$\left({2}^{1+x}+{2}^{1-x}\right)$, $f\left(x\right)$and $\left({3}^{x}+{3}^{-x}\right)$ are in A.P, then the minimum value of $f\left(x\right)$is:

A

$0$

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B

$4$

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C

$3$

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D

$2$

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Solution

## The correct option is C $3$Explanation for the correct option: Determine the value of $f\left(x\right)$Step 1: Apply the condition of A.M and find a relation.Given, $\left({2}^{1+x}+{2}^{1-x}\right)$, $f\left(x\right)$and $\left({3}^{x}+{3}^{-x}\right)$ are in A.P.If ${a}_{1},{a}_{2},{a}_{3},...$ are in A.P, then there is a common difference between any two consecutive terms.Also we know that, ${a}_{2}=\frac{{a}_{1}+{a}_{3}}{2}$,Then$f\left(x\right)=\frac{\left({3}^{x}+{3}^{-x}\right)+\left({2}^{1+x}+{2}^{1-x}\right)}{2}\phantom{\rule{0ex}{0ex}}2f\left(x\right)=\left({3}^{x}+{3}^{-x}\right)+\left({2}^{1+x}+{2}^{1-x}\right)\to \left(i\right)$Step 2: Apply the condition $A.M\ge G.M$Let ${3}^{x}$ and ${3}^{-x}$ be two numbers , then A.M=$\frac{\left({3}^{x}+{3}^{-x}\right)}{2}$G.M=$\sqrt{{3}^{x}{3}^{-x}}$We know that $A.M\ge G.M$ applying this inequality we get:$⇒\frac{\left({3}^{x}+{3}^{-x}\right)}{2}\ge \sqrt{{3}^{x}{3}^{-x}}\phantom{\rule{0ex}{0ex}}⇒{3}^{x}+{3}^{-x}\ge 2\sqrt{1}\phantom{\rule{0ex}{0ex}}⇒{3}^{x}+{3}^{-x}\ge 2\to \left(ii\right)$Similarly,Let ${2}^{1+x}$ and ${2}^{1-x}$ be two numbers , then A.M=$\frac{{2}^{1+x}+{2}^{1-x}}{2}$G.M=$\sqrt{{2}^{1+x}{2}^{1-x}}$We know that $A.M\ge G.M$ applying this inequality we get:$⇒\frac{{2}^{1+x}+{2}^{1-x}}{2}\ge \sqrt{{2}^{1+x}{2}^{1-x}}\phantom{\rule{0ex}{0ex}}⇒{2}^{1+x}+{2}^{1-x}\ge 2\sqrt{{2}^{2}}\phantom{\rule{0ex}{0ex}}⇒{2}^{1+x}+{2}^{1-x}\ge 2×2\to \left(iii\right)$Step 3: Find the required resultAdding the equations $\left(ii\right)\text{and}\left(iii\right)$ we get,$\left({3}^{x}+{3}^{-x}\right)+\left({2}^{1+x}+{2}^{1-x}\right)\ge 2+\left(2×2\right)\phantom{\rule{0ex}{0ex}}⇒\left({3}^{x}+{3}^{-x}\right)+\left({2}^{1+x}+{2}^{1-x}\right)\ge 6$Substitute this value in equation $\left(i\right)$$2f\left(x\right)=\left({3}^{x}+{3}^{-x}\right)+\left({2}^{1+x}+{2}^{1-x}\right)\ge 6\phantom{\rule{0ex}{0ex}}⇒2f\left(x\right)\ge 6\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)\ge \frac{6}{2}\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)\ge 3$Hence, the minimum value of $f\left(x\right)$ is $3$.Hence, option (C) is the correct answer.

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