L1:{<M> | there exist a Turing machine M' such that <M>$\neq$<M'> and L(M) = L(M')} How this problem becomes trivial? and if it non-trivial then please explain why is that so. According to my understanding, non-trivial properties are the one where a language ... and if it is then is it okay to say M1=M2 because they are kind of same machine but other one is just with some non deterministic nature.

Writes Non Blank: Given a turing machine T, does it ever writes a non-blank symbol on its tape, when started with a blank tape. how the above problem is solvable? somewhere i got this explanation: Let the machine only writes blank symbol. Then its ... this is a non-trivial property of turing machine and every non trivial property of turing machine is undecidable, so this is also undecidable.

Here is my analysis. P1: When we bound the number of steps a turing machine can tape, the total number of input possible that can be taken by such turing machine becomes finite and by running TM in an interleaved mode I can decide whether TM M halts on x within k steps. So, ... hence I can decide P3. Hence, P3 is decidable.->REC. So, I think here answer must be 1. Please let me know what's right.

Use Rice’s theorem, to prove the undecidability of each of the following languages. $INFINITE_{TM} = \{\langle M \rangle \mid \text{M is a TM and L(M) is an infinite language}\}$. $\{\langle M \rangle \mid \text{M is a TM and }\:1011 \in L(M)\}$. $ ALL_{TM} = \{\langle M \rangle \mid \text{ M is a TM and}\: L(M) = Σ^{\ast} \}$.