(i) (a) Let x , x , ∈ R
ƒ(x1) = ƒ(x2) ⇒ 5x1 + 2
= 5x2 + 2
⇒ 5x2 = 5x2 ⇒ x1 = x2
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.