Proof that L(M) = {accepts the string 1100 } is undecidable
The usual reduction from the halting problem: for example, the same reduction that shows the zero-input halting problem to be undecidable.


   Suppose you can determine whether a machine $M$ accepts some magic string (such as $1100$).  You can then decide whether an arbitrary machine $M$ halts when given input $x$ as follows. Produce a machine&nbsp;$M_x$ that ignores its input and simulates $M$ running on input $x$. If $M$ halts, $M_x$ accepts whatever input it was given; otherwise, $M$ doesn't halt so nor does the simulation $M_x$. So $M_x$ accepts the magic string if, and only if, $M$ halts on input&nbsp;$x$. Hence, if you could decide whether a machine accepts the magic string, you could decide the halting problem.