# (i) If ƒ: R → R and g: R → R defined by ƒ(x) = x^2 and g(x) = x + 1, then goƒ (x) is (a) (x + 1)^2 (b) x^3 + l

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(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.

## 1 Answer

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(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

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