A function f: R rightarrow R is such that f(1 | vedclass

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(B) Given, $f(x+y)=f(x) \cdot f(y)$. This is a functional equation whose solution is of the form $f(x)=a^x$.Given $f(1)=2$, so $a^1=2 \Rightarrow a=2$

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$\frac{f(4)}{1+f(2)}$

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(B) Given, $f(x+y)=f(x) \cdot f(y)$. This is a functional equation whose solution is of the form $f(x)=a^x$.Given $f(1)=2$, so $a^1=2 \Rightarrow a=2$. Thus, $f(x)=2^x$.Now, consider the region enclosed by $2|x|+5|y| \leq 4$. This represents a rhombus with vertices at $(\pm 2, 0)$ and $(0, \pm 4/5)$.The area of the rhombus is given by $4 	imes 	ext{Area of one triangle in the first quadrant}$.Area $= 4 	imes (\frac{1}{2} 	imes 2 	imes \frac{4}{5}) = 4 	imes \frac{4}{5} = \frac{16}{5}$.We need to express $\frac{16}{5}$ in terms of $f(1)=2^1=2$, $f(2)=2^2=4$, and $f(4)=2^4=16$.Note that $1+f(2) = 1+4 = 5$.Thus, the area is $\frac{16}{5} = \frac{f(4)}{1+f(2)}$.

$A$ function $f: R \rightarrow R$ is such that $f(1)=2$ and $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. The area (in square units) enclosed by the lines $2|x|+5|y| \leq 4$ expressed in terms of $f(1)$, $f(2)$, and $f(4)$ is

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