TOPICS

Miscellaneous

Relations and Functions

**Question-1 :-** Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = IR.

We have given that : f : R → R be defined as f(x) = 10x + 7 Now, gof = g(f(x)) and fog = f(g(x)) = 10g(x) + 7 ⇒ 10g(x) + 7 = IR(X) = x ⇒ g(x) = (x-7)/10

**Question-2 :-** Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1. i.e., For Injective : Let n and m be any two odd real numberrs, then f(n) = f(m) so, n - 1 = m - 1 ⇒ n = m Let n and m be any two odd real numberrs, then f(n) = f(m) so, n + 1 = m + 1 ⇒ n = m and now n is even and m is odd, then n ≠ m. Also, if f(n) odd and f(m) is even, then f(n) ≠ f(m) Hence, n ≠ m ⇒ f(n) ≠ f(m) ∴ f is an injective mapping. For Surjective : Let n be an arbitrary whole number. If n is an odd no., then there exists an even whole no. (n + 1) such that f(n + 1) = n + 1 - 1 = n. If n is an even no., then there exists an even whole no. (n - 1) such that f(n - 1) = n - 1 + 1 = n. Therefore, every n ∈ W has its pre-image in W. So, f : W → W is a surjective. Thus f is invertible and f^{-1}exists. For f^{-1}: y = n - 1 ⇒ n = y + 1 and y = n + 1 ⇒ n = y - 1 Hence, f^{-1}(y) = y.

**Question-3 :-** If f : R → R is defined by f(x) = x² – 3x + 2, find f (f(x)).

Given that f : R → R is defined by f(x) = x² – 3x + 2. ⇒ f(f(x)) = f(x² – 3x + 2) ⇒ (x² – 3x + 2)² - 3(x² – 3x + 2) + 2 ⇒ (x² – 3x + 2)² - 3x² + 9x - 6 + 2 ⇒ x⁴ - 6x³ + 10x² - 3x

**Question-4 :-** Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(1+|x|) x ∈ R is one one and onto function.

Given that f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(1+|x|) x ∈ R For f is one-one : For any x, y ∈ R – {+1}, we have f(x) = f(y) ⇒ x/(1+|x|)= y/(1+|y|) ⇒ xy + x = xy + y ⇒ x = y Hence, f is one-one. For on-to : If f is one-one, let y = R - {1}, then f(x) = y ⇒ x/(x+1) = y ⇒ x = xy + y ⇒ x(1-y) = y ⇒ x = y/(1-y) Now, f(x) = f(y/(1-y)) Therefore, f is on-to function.

**Question-5 :-** Show that the function f : R → R given by f(x) = x³ is injective.

We have f : R → R given by f(x) = x³ For Injective : ⇒ Let f(x₁) = f(x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂ Hence, f is injective function.

**Question-6 :-** Give Questions of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective. (Hint : Consider f(x) = x and g(x) = |x|).

Given that f : N → Z and g : Z → Z Let f(x) = x and g(x) = x Here, gof(x) = f(f(x)) = g(x) Therefore, gof is an injective function but g is not injective.

**Question-7 :-** Give Questions of two functions f : N → N and g : N → N such that g o f is onto but f is not onto.
(Hint : Consider f(x) = x + 1 and

Let f(x) = x + 1 g(x) = x - 1 if x > 1 and 1 if x = 1 These are two Questions in which gof is on-to but f is on-to.

**Question-8 :-** Given a non empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

(i) A ⊂ A Therefore R is Reflexive. (ii) A ⊂ B ≠ B ⊂ A Therefore R is not Commutative. (ii) A ⊂ B, B ⊂ C, then A ⊂ C Therefore R is transitive. Therefore, R is not equivalent relation.

**Question-9 :-** Given a non-empty set X, consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Let S be a non-empty set and P(S) be its power set. Let any two subsets A and B of S. ⇒ A ∪ B ⊂ S ⇒ A ∪ B ∈ P(S) Therefore, '∪' is an binary operation on P(S). Similarly, if A, B ∈ P(S) and A - B ∈ P(S), then the intersection of sets ∩ and difference of sets are also binary operation on P(S) and A ∩ S = A = S ∩ A for every subset A of sets ⇒ A ∩ S = A = S ∩ A for all A ∈ P(S). ⇒ S is the identity element for intersection (∩) on P(S).

**Question-10 :-** Find the number of all onto functions from the set {1, 2, 3, ... , n} to itself.

The number of onto functions that can be defined from a finite set A containing n elements onto a finite set B containing n elements = 2^{n}- n.

**Question-11 :-** Let S = {a, b, c} and T = {1, 2, 3}. Find F^{-1} of the following functions F from S to T, if it exists.

(i) F = {(a, 3), (b, 2), (c, 1)}

(ii) F = {(a, 2), (b, 1), (c, 1)}

Let S = {a, b, c} and T = {1, 2, 3}. (i) F = {(a, 3), (b, 2), (c, 1)} ⇒ F(a) = 3, F(b) = 2, F(c) = 1 ⇒ F^{-1}(3) = a, F^{-1}(2) = b, F^{-1}(1) = c ⇒ F^{-1}= {(3, a), (2, b), (1, c)} (ii) F = {(a, 2), (b, 1), (c, 1)} F is not one-one function, since element b and c have the same image 1. Therefore F is not one-one function.

**Question-12 :-** Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗ b = |a – b| and a o b = a, ∀a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that ∀a, b, c ∈ R, a ∗ (b o c) = (a ∗ b) o (a ∗ c). [If it is so, we say that the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer.

(i) For Commutative : a * b = |a-b| also b * a = |b-a| = (a - b) Hence, operation * is commutative. For Associative : Now, a * (b * c) = a * |b-c| = |a-(b-c)| = |a-b+c| And (a * b) * c = |a-b| * c = |a-b-c| Here, a * (b * c) ≠ a * (b * c) Operation is * not associative.

(ii) For Coomutative : aob = a∀a, b ∈ R boa = b ⇒ aob ≠ boa Here, operation o is not commutative. For Associative : Now, boa = aob = a and (aob)oc = aoc = a Here, ao(boc) = (aob)oc Here, operation o is associative.

(iii) L.H.S a * (boc) = a * b = |a-b| R.H.S (a * b)o(a * b) = (a - b)o(a - c) = |a-b| = L.H.S (Hence Proved) Now, another distribution law : ao(b * c) = (aob)*(aob) L.H.S ao(b * c) = ao(|b-c|) = a R.H.S (aob) * (aob) = a * a = |a-a| = 0 As L.H.S ≠ R.H.S Therefore, the operation o is does not distribute over.

**Question-13 :-** Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

For every A ∈ P(X), we have φ * A = (φ - A) ∪ (A - φ) = φ ∪ A = A and A * φ = (A - φ) ∪ (φ - A) = A ∪ φ = A φ is the identity element for the operation * on P(X). Also, A * A = (A - A) ∪ (A - A) = φ ∪ φ = ∪ Every Element A of P(X) is invertible with A^{-1}= A.

**Question-14 :-** Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.

A binary operation (or composition) * on a (non-empty) set is a function * : A x A → A. We denote *(a, b) by a * b for every ordered pair (a, b) ∈ A x A. binary operation on a no-empty set A is a rule that associates with every ordered pair of elements a, b (distinct or equal) of A some unique element a * b of A. For all a ∉ A, we have 0 * a(mod 6) = 0 And a * 1 = a(mod 6) = a and a * 1 = a(mod 6) 0 is the identity element for the operation. Also on 0 = 0 - 0 = 0* 2 * 1 = 3 = 1 * 2 0^{-1}= 0 0^{-1}= 5

**Question-15 :-** Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined
by f(x) = x² – x, x ∈ A and g(x) = 2|x - 1/2|-1 x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) ∀a ∈ A, are called equal functions).

When x = -1 then f(x) = 1² + 1 = 2 and g(x) = 2|-1 - 1/2|-1 = 2 At x = 0, f(0) = 0 and g(0) = 2|-1/2|-1 = 0 At x = 1, f(1) = 1² - 1 = 0 and g(1) = 2|1 - 1/2|-1 = 0 At x = 2, f(2) = 2² - 2 = 2 and g(2) = 2|2 - 1/2|-1 = 2 Thus for each a ∈ A, f(a) = g(a) Therefore, f and g are equal function.

**Question-16 :-** Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1 (B) 2 (C) 3 (D) 4

Its clear that 1 is reflexive and symmetric but not transitive. Therefore, option (A) is correct.

**Question-17 :-** Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1 (B) 2 (C) 3 (D) 4

Its clear that 2 is an equivalence relation. Therefore, option (B) is correct.

**Question-18 :-** Let f : R → R be the Signum Function defined as
and g : R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Given that f : R → R. Now, gof : R → R and fog : R → R Consider x = 1/2 which lie on (0, 1) Now, gof(1/2) = g(f(1/2)) = g(1) = [1] = 1 And fog(1/2) = f(g(1/2)) = f(1/2) = f(0) = 0 gof fog in (0, 1]. Therefore, option (B) is correct.

**Question-19 :-** Number of binary operations on the set {a, b} are

(A) 10 (B) 16 (C) 20 (D) 8

A = {a, b} A x A = {(a, a), (a, b), (b, b), (b, a)} n(A x A) = 4 Number of subsets = 2^{4}= 16 Hence number of binary operation is 16. Therefore, option (B) is Correct.

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