undecidability

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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 number of configurations in the com computation on w is q × 2, where q is the number of states of M; the factor 2 is for the choices re. the direction of heads movement; there is no factor for the written symbol because that is always blank. So the problem is decidable, decided by the following machine: input <M,w>, run M on w for q × 2 steps; if it M ever writes a non blank symbol, stop with yes answer; if M never writes a non blank symbol, stop with no answer.

why is it that q*2 is taken as an upper bound for writing a non blank symbol? there can be some machine which can write non blank on more than q*2 moves.

in fact we can apply rice's theorem here.. as this is a non-trivial property of turing machine and every non trivial property of turing machine is undecidable, so this is also undecidable.

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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.
Use Rice’s theorem, to prove the undecidability of each of the following languages. $INFINITE_{TM} = \{\langle M \rangle \mid \text{M is a TM and L(M) is an infinite language}\}$. $\{\langle M \rangle \mid \text{M is a TM and }\:1011 \in L(M)\}$. $ALL_{TM} = \{\langle M \rangle \mid \text{ M is a TM and}\: L(M) = Σ^{\ast} \}$.
Rice's theorem. Let $P$ be any nontrivial property of the language of a Turing machine. Prove that the problem of determining whether a given Turing machine's language has property $P$ is undecidable. In more formal terms, let $P$ be a language consisting of Turing ... any $TMs$. Prove that $P$ is an undecidable language. Show that both conditions are necessary for proving that $P$ is undecidable.