Question

(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}

*	p	P
P	P	P
P	P	P

(a) Is * a binary operation?

(b) * commutative? Give reason.

Answer 1

(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