(i) If ƒ: R → R and g: R → R defined by ƒ(x) = x2 and g(x) = x

1.	(i) If ƒ: R → R and g: R → R defined by ƒ(x) = x2 and g(x) = x + 1, then goƒ (x) is(a) (x + 1)2(b) x3 + l(c) x2 + l(d) x + l(ii) Consider the function ƒ: N → N, given by ƒ(x) = x3. Show that the function ‘ƒ’ is injective but not surjective.(iii) The given table shows an operation on A = {p,q}pPPPPPPP(a) Is a binary operation?(b) * commutative? Give reason.
Answer» (i) (C) x² + 1 (ii) ƒ : N → N , given by ƒ(x) = x³ for x,y ∈ N ⇒ ƒ(x) = ƒ(y) x³ = y³ ⇒ x = y There fore f is injective. Now 2 ∈ N, but there does not exists any element x in domain N such that ƒ(x) = x³ = 2 their fore f is not surjective. (iii) (a) Yes (b) No, because pq = q; qp = p ⇒ pq ≠ qp

(i) If ƒ: R → R and g: R → R defined by ƒ(x) = x2 and g(x) = x + 1, then goƒ (x) is(a) (x + 1)2(b) x3 + l(c) x2 + l(d) x + l(ii) Consider the function ƒ: N → N, given by ƒ(x) = x3. Show that the function ‘ƒ’ is injective but not surjective.(iii) The given table shows an operation on A = {p,q}pPPPPPPP(a) Is a binary operation?(b) * commutative? Give reason.

Answer»

(i) (C) x² + 1

(ii) ƒ : N → N , given by ƒ(x) = x³

for x,y ∈ N ⇒ ƒ(x) = ƒ(y)

x³ = y³ ⇒ x = y

There fore f is injective.

Now 2 ∈ N, but there does not exists any element x in domain N such that ƒ(x) = x³ = 2 their fore f is not surjective.

(iii) (a) Yes

(b) No, because p*q = q; q*p = p

⇒ p*q ≠ q*p