(i) (a) Let x , x , ∈ R

ƒ(x_{1}) = ƒ(x_{2}) ⇒ 5x_{1} + 2

= 5x_{2} + 2

⇒ 5x_{2} = 5x_{2} ⇒ x_{1} = x_{2}

Therefore s one-one.

(b) Yes.

Let y e range of ƒ

⇒ ƒ(x) = y ⇒ 5x + 2 = y

⇒ x = \(\frac{y-2}{5}\) ∈ R

Therefore corresponding to every y ∈ R there existsa real number \(\frac{y-2}{5}\) Therefore f is onto.

Hence bijective, so invertible.

(ii) (a) Yes.

a * b = HCF (a,b) = HCF (b,a) = b * a

Hence commutative.

(b) Yes.

a * (b * c) = a* HF(b,c) = HCF(a,b,c)

(a*b) * c =HCF(a,b) * c HCF(a,b,c)

a * (b * c) = (a * b) * c

Hence associative.