Since
$ e^{2jt}\rightarrow t e^{-2jt} $
and
$ e^{-2jt}\rightarrow t e^{2jt} $,
we can conclude that the system performs the function
$ x(t)\rightarrow t x(-t) $.
Therefore, we assume the response to our input is
$ cos(2t)\rightarrow t cos(-2t) = t cos(2t) $,
since cosine is an even function.
