Solve this following


If $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})$ and $\sum_{\mathrm{x}=1}^{\infty} f(\mathrm{x})=2, \mathrm{x}, \mathrm{y} \in \mathrm{N}$

where $\mathrm{N}$ is the set of all natural numbers, then

the value of $\frac{f(4)}{f(2)}$ is


  1. $\frac{1}{9}$

  2. $\frac{4}{9}$

  3. $\frac{1}{3}$

  4. $\frac{2}{3}$

Correct Option: , 2


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

$\sum_{x=1}^{\infty} f(x)=2$ where $x, y \in N$

$f(1)+f(2)+f(3)+\ldots . \infty=2 \ldots(1)($ Given $)$

Now for $f(2)$ put $\mathrm{x}=\mathrm{y}=1$

$f(2)=f(1+1)=f(1) \cdot f(1)=(f(1))^{2}$

$f(3)=f(2+1)=f(2) \cdot f(1)=(f(1))^{3}$

Now put these values in equation (1)

$f(1)+(f(1))^{2}+\left[f(1)^{2}+\ldots \infty=2\right]$


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

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


then the value of $\frac{f(4)}{f(2)}=\frac{\left(\frac{2}{3}\right)^{4}}{\left(\frac{2}{3}\right)^{2}}=\frac{4}{9}$

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