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

e2 2e2 3e2 2e3

2e2

$\frac{f'(x)}{f(x)}=1 \Rightarrow \log f(x)=x+c$ 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) = e2+log2 = e2 . elog2 = 2e2 Hence (2) is the correct answer.