Correct Answer - Option 1 : Commutative but not associative

__Concept: __

Associative

A binary operation ∗ on a set S is said to be associative if it satisfies the associative law:

a ∗ (b ∗c) = (a ∗ b) ∗ c for all a, b, c ∈S.

Commutative

A binary operation ∗ on a set S is said to be commutative if it satisfies the condition:

a ∗b=b ∗a for all a, b, ∈S.

__Calculations:__

First started with checking for associative, x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z.

L.H.S: x ⊕ (y ⊕ z) = x ⊕ (y^{2} + z^{2}) = x^{2} + (x^{2} + z^{2})^{2 }.

R.H.S: (x ⊕ y) ⊕ z = (x^{2} + y^{2}) ⊕ z = (x^{2} + y^{2})^{2} + z^{2} .

L.H.S ≠ R.H.S then the operator is not associative.

For commutative, x ⊕ y = y ⊕ x

L.H.S: x ⊕ y = x^{2} + y^{2}

R.H.S: y ⊕ x = y^{2} + x^{2}

L.H.S = R.H.S then the operator is Commutative.