QJEEM20Mab912: JEE Main 2020 Mathematics - Class 11 JEEM20Mab912 Solution: If f(x + y) = f(x)f(y) and ∑ x = 1 ∞ f ( x ) = 2 , x, y ∈ N,…

f(x + y) = f(x)f(y)

$$\sum _{x=1}^{\mathrm{\infty }}f\left(x\right)=2$$

$$\Rightarrow$$ f(1) + f(2) + f(3) + ........$$\mathrm{\infty }$$ = 2 ....(1)

On f(x + y) = f(x) f(y)
* Put x = 1, y = 1
f(2) = (f(1))²
* Put x = 2, y = 1
f(3) = f(2). f(1) = f((1))³
* Put x = 2, y = 2
f(4) = f((2))² = f((1))⁴

Now put these values in equation (1)

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

$$\Rightarrow$$ $$\frac{f\left(1\right)}{1-f\left(1\right)}$$ = 2

$$\Rightarrow$$ f(1) = $$\frac{2}{3}$$

Now f(2) = $${\left(\frac{2}{3}\right)}^2$$

and f(4) = $${\left(\frac{2}{3}\right)}^4$$

$$\therefore$$ $$\frac{f\left(4\right)}{f\left(2\right)}$$

= $$\frac{{\left(\frac{2}{3}\right)}^4}{{\left(\frac{2}{3}\right)}^2}$$ = $$\frac{4}{9}$$