(i) Consider ƒ : R → R given by ƒ(x) = 5x + 2 (a) Show that f is

1.	(i) Consider ƒ : R → R given by ƒ(x) = 5x + 2 (a) Show that f is one-one. (b) Is f invertible? Justify your answer. (ii) Let * be a binary operation on N defined by a * b = HCF of a and b (a) Is * commutative? (b) Is * associative?
Answer» (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) (ab) c =HCF(a,b) * c HCF(a,b,c) a * (b * c) = (a * b) * c Hence associative.

(i) Consider ƒ : R → R given by ƒ(x) = 5x + 2 (a) Show that f is one-one. (b) Is f invertible? Justify your answer. (ii) Let * be a binary operation on N defined by a * b = HCF of a and b (a) Is * commutative? (b) Is * associative?

Answer»

(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.