QJEEM22Ma838c: JEE Main 2022 Mathematics - Class 11 JEEM22Ma838c Solution: Let f : N R be a function such that for natural numbers x and…

Given,

$f(x + y) = 2f(x)f(y)$

and $f(1) = 2$

For x = 1 and y = 1,

$f(1 + 1) = 2f(1)f(1)$

$\Rightarrow f(2) = 2{\left( {f(1)} \right)^2} = 2{(2)^2} = {2^3}$

For x = 1, y = 2,

$f(1 + 2) = 2f(1)y(2)$

$\Rightarrow f(3) = 2\,.\,2\,.\,{2^3} = {2^5}$

For x = 1, y = 3,

$f(1 + 3) = 2f(1)f(3)$

$\Rightarrow f(4) = 2\,.\,2\,.\,{2^5} = {2^7}$

For x = 1, y = 4,

$f(1 + 4) = 2f(1)f(4)$

$\Rightarrow f(5) = 2\,.\,2\,.\,{2^7} = {2^9}$ ..... (1)

Also given

$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)}$

$\Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,...\,\, + \,\,f(\alpha + 10) = {{512} \over 3}({2^{20}} - 1)$

$\Rightarrow f(\alpha + 1) + f(\alpha + 2) + f(\alpha + 3)\, + \,\,....\,\, + f(\alpha + 10) = {{{2^9}\left( {{{({2^2})}^{10}} - 1} \right)} \over {{2^2} - 1}}$

This represent a G.P with first term = 2⁹ and common ratio = 2²

$\therefore$ First term $= f(\alpha + 1) = {2^9}$ ..... (2)

From equation (1), $f(5) = {2^9}$

$\therefore$ From (1) and (2), we get

$f(\alpha + 1) = {2^9} = f(5)$

$\Rightarrow f(\alpha + 1) = f(5)$

$\Rightarrow f(\alpha + 1) = f(4 + 1)$

Comparing both sides we get,

$\alpha = 4$