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
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)
$f(1)+(f(1))^{2}+\left[f(1)^{2}+\ldots \infty=2\right]$
$\frac{f(1)}{1-f(1)}=2$
$\Rightarrow f(1)=\frac{2}{3}$
Now $f(2)=\left(\frac{2}{3}\right)^{2}$
$f(4)=\left(\frac{2}{3}\right)^{4}$
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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