CBSE Class 12 Mathematics Relations And Functions Assignment Set 05

I. Multiple Choice Questions (MCQs) Choose the correct answer from the given options.

Question. R is a relation over the set N × N and it is defined by (a, b) R(c, d) ⇒ a + d = b + c, then R is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) equivalence relation

Answer: D

Question. Let R be a relation over the set of straight lines in a plane such that l₁ R l₂ such that l₁ ⊥ l₂, then R is
(a) symmetric
(b) reflexive
(c) transitive
(d) equivalence relation

Answer: A

Question. A relation R is defined on plane A of triangles in a given planes defined as R = {(T₁, T₂) : T₁ @ T₂ and T₁, T₂ ∈ A}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence

Answer: D

Question. Consider the non-empty set consisting of children in a family and a relation R defined as a Rb if a is a brother of b. Then R is
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) neither symmetric nor transitive
(d) both symmetric and transitive.

Answer: B

Question. Let R be the relation on the set R of all real numbers defined by a R b iff |a – b| ≤ 1. Then R is
(a) reflexive and symmetric
(b) symmetric only
(c) transitive only
(d) anti symmetric only

Answer: A

Question. A relation R = {(a, b) : a divides b} on the set N of all natural numbers is
(a) reflexive
(b) symmetric
(c) reflexive and transitive
(d) symmetric and transitive

Answer: C

Question. Let R be a relation on the set N defined by {(x, y) : x, y ∈ N, 2x + y = 41}. Then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these.

Answer: D

Question. Let S be the set of all real numbers. Then the relation R = {(a, b) : 1 + ab > 0} on S is
(a) reflexive and symmetric but not transitive.
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
(d) reflexive, symmetric and transitive.

Answer: A

Question. Let W denote the words in the English dictionary. The relation R is defined by R = {(x, y) ∈ W × W : the words x and y have at least one letter in common}. Then R is
(a) not reflexive, symmetric and transitive
(b) reflexive, symmetric and not transitive
(c) reflexive, symmetric and transitive
(d) reflexive, transitive and not symmetric

Answer: B

Question. On the set N of all natural numbers define the relation R by aRb if and only if the G.C.D. of a and b is 2, then R is
(a) reflexive but not symmetric
(b) symmetric only
(c) reflexive and transitive
(d) reflexive, symmetric and transitive.

Answer: B

II. Very Short Answer Type Questions

Question. State the reason why the relation R = {(a, b) : a ≥ b²} defined in the set of real numbers is not reflexive.
Answer: R is not reflexive because a ≥ b² does not imply a ≥ a² ∀ a.

Question. State the reason of the relation R in the set {1, 2, 3} given by: R = {(1, 2), (2, 1)} not to be transitive.
Answer: Relation R is not transitive because: (1, 2), (2, 1) ∈ R but (1, 1) ∉ R.

Question. If R = {(x, y) : x + 2y = 8} is a relation on N. Write the domain of R.
Answer: Here R = {(2, 3), (4, 2), (6, 1)},
Domain of R = {2, 4, 6}

Question. Let R be the equivalence relation in the set A = {0, 1, 2, 3, 4, 5} given by R = {(a, b) : 2 divides (a – b); a, b ∈ A}. Write the equivalence class of [0].
Answer: Equivalence class of [0] = {0, 2, 4}

Question. Let A = {1, 2, 3, 4} and R be a relation in set A given by R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1), (1, 3)} then show that the relation R is reflexive and symmetric.
Answer: Relation R is reflexive and symmetric because (1, 1),
(2, 2), (3, 3) ∈ R
⇒ R is reflexive and (1, 2), (2, 1), (3, 1), (1, 3) ∈ R
⇒ R is symmetric.

Question. Show that the relation R less than in the set of natural numbers is transitive.
Answer: Reason: Take a, b, c ∈ N such that a < b and b < c
⇒ a < c \ (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R.
Hence, relation R is transitive.

Question. For real numbers x and y, we write xRy ⇒ x – y + √2 is an irrational number, then show that the relation R is reflexive.
Answer: Relation R = x – y + 2 is reflexive because
x R x = x –x + √2 = √2 .

Question. A relation R = {(x, y) : x + y < 10; x, y ∈ A} defined in the set A = {1, 2, 3, 4}. Is relation R symmetric? Give reason.
Answer: Relation R is symmetric because
x + y < 10 ⇒ y + x < 10
⇒ (x, y) ∈ R ⇒ (y, x) ∈ R

Question. Take a relation R = {(a, b) : a = b, a, b ∈ A} defined on a set A of real numbers. Is R a reflexive relation? Give reason.
Answer: Relation R is reflexive because a = a
⇒ (a, a) ∈ R for all a ∈ A.

III. Short Answer Type Questions-I

Question. If set A = {1, 2, 3, 4}, set B = {a, b, c, d} and relation R = {(1, a), (2, b), (3, c), (4, d)} is a relation defined from set A to set B. Write relation R–1.
Answer: R^–1 = {(a, 1), (b, 2), (c, 3), (d, 4)}.

Question. A relation R defined in set A = {1, 2, 3, ...., 10} as R = {(x, y) : 3x – y = 0, x, y ∈ A}. Show that the relation is not reflexive.
Answer: Given 3x – y = 0. Put y = x in 3x –y = 0
It gives 3x – x = 0            ⇒ 2x = 0.
It is not true.
∴ R is not reflexive.

Question. Let A = {a, b, c} and a relation R is defined in set A as: R = {(a, a), (b, c), (a, b)}. Write the minimum number of ordered pairs be included in relation R to make it reflexive and symmetric.
Answer: To make relation R reflexive, include (b, b) and (c, c). Now, (a, b) ∈ R and (b, c) ∈ R but (b, a) and (c, b) are not included in R. Therefore, include (b, a) and (c, b) to make R symmetric.
Hence, to make R reflexive and symmetric include (b, b), (c, c), (b, a) and (c, b).

Question. Check whether the relation R defined in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y = x + 1, x, y ∈ A} is reflexive or symmetric.
Answer: (i) Relation: y = x + 1.
If y = x makes it true, R will be reflexive
∴ Put y = x in y = x + 1.
It gives x = x + 1
⇒ 0 = 1 (False)
∴ R is not reflexive.
(ii) Relation: y = x + 1.
Put x = y and y = x in y = x + 1.
It gives x = y + 1.
It is not same as y = x + 1.
Hence, R is no symmetric.

Question. Show that the relation: R = {(1, 2), (2, 3), (3, 2), (1, 3), (2, 1), (2, 2), (3, 3), (3, 1), (1, 1)} is reflexive, symmetric and transitive.
Answer: (i) Reflexive: R is reflexive because
{(1, 1), (2, 2), (3, 3)} ∈ R
(ii) Symmetric: R is symmetric because
{(1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)} ∈ R
(iii) Transitive: R is transitive because (1, 2) ∈ R and (2, 3) ∈ R
⇒ (1, 3) ∈ R and so on.

Question. Show that the relation: R ={(1, 1), (1, 2), (2, 1), (2, 2)} is reflexive, symmetric and transitive.
Answer: (i) Reflexive: R is reflexive because
{(1, 1), (2, 2)} ∈ R
(ii) Symmetric: R is symmetric because
{(1, 2), (2, 1)} ∈ R
(iii) Transitive: R is transitive because {1, 2) ∈ R and (2, 1) ∈ R
⇒ (1, 1) ∈ R and so on.

Question. Show that the relation: R = {(1, 1), (3, 3), (5, 5), (1, 3), (3, 1), (3, 5)} is reflexive and is neither symmetric nor transitive.
Answer: (i) Reflexive: R is reflexive because
{(1, 1), (3, 3), (5, 5)} ∈ R
(ii) Symmetric: R is not symmetric because (3, 5) ∈ R but (5, 3) ∉ R.
(iii) Transitive: R is not transitive because (1, 3) ∈ R and (3, 5)
∈ R does not imply (1, 5) ∈ R.

Question. Take a set A = {1, 3, 7} and find a relation R = A × A, then show that it is reflexive, symmetric and transitive.
Answer: A = {1, 3, 7}
R = A × A = {(1, 1), (1, 3), (1, 7), (3, 1), (3, 3), (3, 7), (7, 1), (7, 3), (7, 7)}
(i) R is reflexive because {(1, 1), (3, 3), (7, 7)} ∈ R
(ii) R is symmetric because
{(1, 3), (3, 1) (1, 7), (7, 1), (3, 7), (7, 3)} ∈ R.
(iii) R is transitive because (1, 3) ∈ R and (3, 7) ∈ R
⇒ (1, 7) ∈ R and so on.

Question. Relation R defined in a set of natural numbers N as: R = {(x, y) : x ≥ y ∀ x, y ∈ N}. Show that relation R is reflexive and transitive but not symmetric.
Answer: R = {(x, y) : x ≥ y, x, y ∈ N}
(i) R is reflexive: Since x ≥ x            ⇒ (x, x) ∈ R
⇒ R is reflexive.
(ii) R is not symmetric: x ≥ y does not imply y ≥ x.
∴ R is not symmetric.
(iii) Transitive: Take three natural numbers x, y and z such that
x ≥ y and y ≥ z            ⇒ x ≥ z

Question. Relation R defined in a set of natural numbers N as: R = {(a, b) : (a + b) is an even natural number, where a, b ∈ N}. Show that R is reflexive, symmetric and transitive.
Answer: R = {(a, b) : (a + b) is an even natural number and a, b ∈ N}
(i) R is reflexive: Take a ∈ N. Since (a + a) is an even number ∀ a ∈ N.
(a, a) ∈ R ⇒ R is reflexive.
(ii) Symmetric: Let a, b ∈ N.
If (a + b) is an even number then (b + a) is also an even number.
∴ (a, b) ∈ R ⇒ (b, a) ∈ R
⇒ R is symmetric.
(iii) Transitive: Since sum of two even numbers is even or sum of two odd numbers is even.
∴ Let a, b, c ∈ N are either all even numbers or odd numbers.
∴ (a + b) is even and (b + c) is even
⇒ (a + c) is also even
∴ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
⇒ R is transitive.

Question. Relation R defined in a set of natural numbers N as: R = {(a, b) : (a – b) is an even natural number and a, b ∈ N}. Show that R is neither reflexive nor symmetric but transitive.
Answer: R = {(a, b) : (a – b) is an even natural number and a, b ∈ N}
(i) Reflexive: Since (a – a) = 0 (not a natural no.)
⇒ (a, a) ∉ R            ∴ R is not reflexive.
(ii) Symmetric: If (a – b) is even natural number then (b – a) is not an even natural number.
∴ (a, b) ∈ R whereas (b, a) ∉ R.
Hence, R is not symmetric.
(iii) Transitive: a, b, c ∈ N such that (a – b) and (b – c) are even natural numbers then (a – c) will also be an even natural numbers.
∴ (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R.
Hence, R is transitive.

Question. Relation R defined in a set of real numbers R as: R = {(x, y) : (x + y) is divisible by 5 ∀ x, y ∈ R}. Show that R is neither reflexive nor transitive but symmetric. Note: In a set of real numbers R, if relation R = {(a, b) : (a + b) is a multiple of 3, 4, 5, 6, …, n ∀ a, b ∈ R} then R is neither reflexive nor transitive but symmetric.
Answer: R = {(x, y) : (x + y) is divisible by 5 ∀ x, y ∈ R}
(i) Reflexive: Take x ∈ R.
Since (x + x) is not divisible by 5 for any real number x.
⇒ R is not reflexive.
(ii) Symmetric: Let x, y ∈ R be such that (x + y) is divisible by 5, then (y + x) is also divisible by 5.
∴ (x, y) ∈ R            ⇒ (y, x) ∈ R
Hence, R is symmetric.
(iii) Transitive: Take x, y, z ∈ R such that (x + y) and (y + z) are divisible by 5, then (x + z) may or may not be divisible by 5.
∴ (a, b) ∈ R and (b, c ) ∈ R does not imply (a, c) ∈ R.
Hence, R is not transitive.

Question. Show that the relation R in set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
Answer: in set A = {1, 2, 3} is
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
(i) Reflexive: Since {(1, 1), (2, 2), (3, 3)} ∈ R
⇒ R is reflexive.
(ii) Symmetric: Since (1, 2) ∈ R but (2, 1) ∉ R and (2, 3) ∈ R but (3, 2) ∉ R.
⇒ R is not symmetric.
(iii) Transitive: Since (1, 2) ∈ R and (2, 3) ∈ R does not imply (1, 3) ∈ R.
Hence R is not transitive.

Question. Let L be the set of lines in a plane and R be the relation in L defined as: R = {(L₁, L₂) : L₁ is perpendicular to L₂, L₁ and L₂ ∈ L}. Show that R is symmetric but neither reflexive nor transitive.
Answer: R = {(L₁, L₂) : L₁ ⊥ L₂ and L_1, L₂ ∈ L}
(i) Reflexive: Let L₁ ∈ L. Since line L₁ cannot be perpendicular to itself
∴ (L₁, L₁) ∉ R
Hence, R is not reflexive.
(ii) Symmetric: Let L1, L₂ ∈ L. If line L₁ ⊥ L₂ then L₂ ⊥ L₁
Hence (L₁, L₂) ∈ R and (L₂, L₁) ∈ R
⇒ R is symmetric.
(iii) Transitive: Let L₁, L₂, L3 ∈ L and L₁ ⊥ L₂ and L₂ ⊥ L3 then
L₁ is not perpendicular to L3
∴ (L₁, L₂) ∈ R and (L₂, L3) ∈ R does not imply that (L₁, L3) ∈ R.
Hence, R is not transitive

Short Answer Type Questions

1. If f (x) = |x| and g(x) = [x] . Evaluate – (fog) (-5/3) – (gof) (-5/3)

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