Let f:R→R be such that for all x∈R2^(1+x)+2^(1-x), fx and 3^x+3^(-x) are in A.P., then the minimum value of fx is

Correct answer: (B) 3. fx=(2^(1-x)+2^(1+x)+3^x+3^(-x))/2 ⇒fx=(23^(-x)+3^x+3^x+3^(-x))/2 ⇒fx=3^(-x)+3^x+(3^x+3^(-x))/2 Using AM≥GM fx≥3

Q50 / 75 · JEE Main 2020 (08 Jan Shift 1) · Mathematics

Let $f : R \to R$ 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

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

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

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

Using $A M \geq G M$
$f (x) \geq 3$

Hence, the correct answer is option: 'B'.