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Let f:\; R\rightarrow R be such that for all x\epsilon 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 :
 
Option: 1 0
Option: 2 4
Option: 3 3
Option: 4 2
 

Answers (1)

best_answer

Important Properties of an AP

If a, b, c are in AP, then 2b = a + c

and 

AM ≥ GM

 

Applying this to given question,

\\2 f(x)=\left(2 \frac{1}{2^{x}}+2.2^{x}\right)+\left(3^{x}+\frac{1}{3^{x}}\right)\\ \\Now,\,\,AM \geq GM\\ \\\frac{a+b}{2}\geq\sqrt{ab}\\\frac{3^{x}+\frac{1}{3^{x}}}{2} \geq \sqrt{1}\\\left(3^{x}+\frac{1}{3^{x}}\right) \geqslant 2\\Value\,\,equals\,\,2 \,\,at\;\;3^x=\frac{1}{3^x}\Rightarrow x=0

\\Similarly, \frac{1}{2^{x}}+2^{x} \geq 2, and\,\,this\,\,holds\,\,at\,\, x=0\\ \\So,2 (minimum\,\,of\,\,f(x))=2(2)+(2)\\\min\;f(x)=3

Correct Option (3)

Posted by

Kuldeep Maurya

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