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**Hint**: We need to solve the two expressions and need to find the values for which the term x will be true in both the cases. And then comparing the two individual sets of values we need to just find the required solution.

**Complete step-by-step answer**:

The two sets A and B are given as below:

A = {x|x ϵ R, $ {x^2} - 3x - 4 = 0 $ }, where x belongs to any real number (R).

B = {x|x ϵ Z, $ {x^2} = x $ }, where x belongs to any integer (Z).

Now, let us solve the first equation using middle term factorization and find the values of x.

A = { x|x ϵ R, $ {x^2} - 3x - 4 = 0 $ }, where x belongs to any real number (R);

$ \Rightarrow {x^2} - 3x - 4 = 0 $

$ \Rightarrow {x^2} - 4x - x - 4 = 0 $

$ \Rightarrow x(x - 4) + 1(x - 4) = 0 $

$ \Rightarrow (x + 1)(x - 4) = 0 $

i.e. $ (x + 1) = 0 $ & $ (x - 4) = 0 $

So, $ x = - 1 $ or $ x = 4 $

Hence, for the given set A = {x|x ϵ R, $ {x^2} - 3x - 4 = 0 $ }

$ \Rightarrow $ A = { -1,4 }

Now, let us solve the second equation and find the values of x.

B = {x|x ϵ Z, $ {x^2} = x $ }, where x belongs to any integer (Z).

$ {x^2} = x $

There are only two values i.e., 0 and 1 for which $ {x^2} = x $ .

$ \Rightarrow $ B = {0,1}

I.Now, considering the above derived two sets we need to find A∩B & A∆B

A∩B means intersection of A and B i.e. the common values in set A and B.

Hence, A∩B = {-1,4}∩{0,1}

i.e. A∩B = ф ; where ф denotes an empty set.

II.And A∆B denotes symmetric difference i.e., the set of elements which are in either of the sets but they are not present in their intersections. In other words, the set of elements of both the sets excluding the common elements present in each set.

So, A∆B = {-1,4,0,1}

Hence, A∩B = ф ; where ф denotes empty set & A∆B = {-1,4,0,1,}.

**So, the correct answer is “ A∆B = {-1,4,0,1,}.AND A∩B = ф ”.**

**Note**: Kindly don’t get confuse in between A∩B & A∆B. The first one resembles the common terms only and the second term resembles the symmetric difference i.e., the set of elements which are in either of the sets but they are not present in their intersections. So, we need to be attentive while finding the answer from the derived sets.

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