R= {(4, 5), (5, 4), (4, 4)} is symmetric since (4, 5) and (5, 4) are converse of each or thus satisfying condition for a symmetric relation and it is transitive as (4, 5)∈R and (5, 4)∈R implies that (4, 4) ∈R. It is not reflexive as every element in set I is not related to itself.

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)

Given that, a*b=4ab.
Then, (a*b)*a=(4ab)*a
=4(4ab)( a )=16a² b.

Given that, f:R→R, f(x)=cos?x and g(x)=7x³+6
Then, f?g(x) = f(g(x))=cos?(g(x))=cos?(7x³+6).

The relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a,a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy condition that for a relation R in a set A if (a₁, a₂)∈R and (a₂, a₃)∈R implies that (a₁, a₃) ∈ R for every a₁, a₂, a₃ ∈ R.

The function f(x)=x²+9 is bijective.
Therefore, f(x)=x²+9
i.e.y=x²+9
x=\(\sqrt{y-9}\)
⇒f^-1 (x)=\(\sqrt{x-9}\).

The function f={(5,3),(5,4),(6,4),(8,9)} is neir one -one nor onto.
The function is not one ? one 8 does not have an image in codomain N and we know that a function can only be one ? one if every element in set M has an image in codomain N.
A function can be onto only if each element in co-domain has a pre-image in domain X. In function f={(5,3),(5,4),(6,4),(8,9)}, 10 in co-domain N does not have a pre- image in domain X.
f={(5,3),(6,4),(7,9),(8,10)} is both one-one and onto.
f={(5,4),(5,9),(6,3),(7,10),(8,10)} and f={(6,4),(7,3),(7,9),(8,10)} are many ? one onto.

The above is a many -one function.
Consider f(x₁)=f(x₂)
?5x₁³-8=5x₂³-8
5x₁³=5x₂³
⇒x₁ = �x₂. Hence, function is many ? one.

The given statement is true. A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive. Hence, an equivalence relation is always symmetric.

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).