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A function $f: \mathbf{R} \rightarrow \mathbf{R}$ is such that $f(\mathrm{l})=2$ and $f(x+y)=f(x) \cdot f(y) \forall x, y$. The area (in square units) enclosed by the lines $2|x|+5|y| \leq 4$ expressed interms of $f(1), f(2)$ and $f(4)$ is
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The correct answer is:
$\frac{f(4)}{1+f(2)}$
Given, $f(x+y)=f(x) \cdot f(y)$
$\therefore \quad f(x)=a^x$
$f(\mathrm{1})=a^{\mathrm{l}}=2 \Rightarrow a=2$
$\therefore \quad f(x)=2^x$
Area enclosed by the lines $2|x|+5|y| \leq 4$
$4\left(\frac{1}{2} \times 2 \times \frac{4}{5}\right)=\frac{16}{5}$
$=\frac{16}{1+(2)^2}=\frac{(2)^4}{1+(2)^2}=\frac{f(4)}{1+f(2)}$
$\therefore \quad f(x)=a^x$
$f(\mathrm{1})=a^{\mathrm{l}}=2 \Rightarrow a=2$
$\therefore \quad f(x)=2^x$
Area enclosed by the lines $2|x|+5|y| \leq 4$
$4\left(\frac{1}{2} \times 2 \times \frac{4}{5}\right)=\frac{16}{5}$
$=\frac{16}{1+(2)^2}=\frac{(2)^4}{1+(2)^2}=\frac{f(4)}{1+f(2)}$
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