If f(x) = f'(x) and f(1) = 2, then f(3) =

$\frac{f'(x)}{f(x)}=1 \Rightarrow \log f(x)=x+c$

Since f(1) = 2

∴ log 2 = 1 + c

∴ log f(x) = x + log 2 1

∴ log f(3) = 3 + log 2 1 = 2 + log 2

⇒ f(3) = e^2+log2 = e² . e^log2 = 2e²

Hence (2) is the correct answer.