Option A. Its O(n)
Let's take a Language (L) = {a^nb^n}
a |
a |
a |
a |
aa |
a |
a |
a |
a |
------n |
b |
b |
b |
b |
b |
b |
b |
b |
b |
------n |
In Standard Turing Machine we do comparison for 1st 'a' to 1st 'b' via R/W header points to the tape, then we have to cross through n cells,
So for 'n' comparison for (a,b), require n*n cells = O(n^2).
But when we use Multi 2-Tape Turing Machine we require to R/W header points to the tape,
So first we copy a^nb^n from 1-tape to 2-tape through scanning the cells, it takes O(n) time to copy the input .
Now we easily compare with an appropriate header in both Tape 1 and Tape 2. in O(n) time .
So total time takes = Copy time + Comparsion time = O(n) +O(n) = O(n).