Theorem: For any real number x$,$ if $($x $1)($x $+5)<0$ then $5<$ x $<1$

Suppose that the theorem is proven by contrapositive, so the proof starts by assuming that it is not the case that $5<$ x $<1$

and then shows that $($x $1)($x $+5)>=0.$

What are the two cases that cover the set of x in the assumption?

Group of answer choices

Case $1$: x$>=5,$ Case $2$: x$<=1$

Case $1$: x$>=5,$ Case $2$: x$>=1$

Case $1$: x$<=5,$ Case $2$: x$<=1$

Case $1$: x$<=5,$ Case $2$: x$>=1$