Find the range of $f (x) = {(2018)}^{x} + (2018) f (x) = 2018$

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Solution

$(2018)^{x} + (2018) f (x) = 2018$
$f (x) (2018) = 2018 - (2018)^{x}$
$\Rightarrow f (x) = 1 - (2018)^{x - 1} = y (s a y)$
As $x \in R, y \in R =$ Range

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