The binary operation ?*? is both commutative and associative for a*b=a+b.
The operation is commutative on a*b=a+b because a+b=b+a.
The operation is associative on a*b=a+b because (a+b)+c=a+(b+c).

The given statement is true. A function is invertible if it is bijective.
For one ? one: Consider f(x₁)=f(x₂)
&re4; 5x₁+9=5x₂+9
⇒x₁=x₂. Hence, function is one ? one.
For onto: For any real number y in co-domain R, re exists an element x=\(\frac{y-9}{5}\) such that f(x)=\(f(\frac{y-9}{5})=5(\frac{y-9}{5})\)+9=y.
Therefore, function is onto.

The above is a condition for a symmetric relation.
For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}
1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.
Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each or and is a part of relation. Hence, y are symmetric.

The given statement is true. If re is a binary operation *:M�M → M with an identity element a∈ M is said to be invertible with respect to binary operation * if re exists an element b ∈ M such that a*b = e = b*a, b is called inverse of a.

The above function is onto or surjective. A function f:X→Y is said to be surjective or onto if, every element of Y is image of some elements in X.
The condition for a surjective function is for every y∈Y, re is an element in X such that f(x)=y.

For above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3,4)∈R and (4,6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy condition (a,a)∈R, for every a∈A for a relation R in set A.

The binary operation is defined by a*b=a-b+ab².
&re4;4*5=4-5+4(5²)=-1+100=99.

f={(1,4),(2,5),(3,6)} is a bijective function.
One-one: It is a one-one function because every element in set A={1,2,3} has a distinct image in set B={4,5,6}.
Onto: It is an onto function as every element in set B={4,5,6} is image of some element in set A={1,2,3}.
f={(2,4),(2,5),(2,6)} and f={(1,4),(1,5),(1,6)} are many-one onto.
f={(1,5),(2,4),(3,4)} is neir one ? one nor onto.

The given function is bijective i.e. both one-one and onto.
one ? one : Every element in domain X has a distinct image in codomain Y. Thus, given function is one- one.
onto: Every element in co- domain Y has a pre- image in domain X. Thus, given function is onto.

Given that, f(x)=3x-5, g(y)=6y² and h(z)=tan?z,
Then, ho(gof)=h?(g(f(x))=h(6(3x-5)²)=tan?(6(3x-5)²)
&re4; ho(gof)=tan?(6(3x-5)²)