CBSE Class 12 Mathematics Relations And Functions Assignment Set F

Short Answer Type Questions

Question. Show that the relation R in the set A of all the books in a library of a college given by R = {(x, y) : x and y have the same number of pages and x, y ∈ A} is an equivalence relation.
Answer: A = Set of all the books in a library
R = {(x, y) : x and y have the same number of pages and x, y ∈ A}
(i) Reflexive: Since number of pages of any book x is equal to number of pages of itself.
∴ (x, x) ∈ R              ⇒ R is reflexive.
(ii) Symmetric: Take any two books x, y from the library such that number of pages of book x is equal to number of pages of book y then the number of pages of book y is equal to number of pages of book x.
∴ (x, y) ∈ R              ⇒ (y, x) ∈ R
⇒ R is symmetric.
(iii) Transitive: Take three books x, y, z such that x and y are of equal pages and books y and z are of equal pages. Then x and z will also have equal number of pages.
∴ (x, y) ∈ R and (y, z) ∈ R
⇒ (x, z) ∈ R
Hence, R is transitive.

Question. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by
(A) R = {(a, b) : |a – b| is a multiple of 4 and a, b ∈ A}
(B) R = {(a, b) : a = b and a, b ∈ A} are equivalence relations.
Find the set of all elements related to 1 in each case.
Answer: (A) A = {x ∈ Z : 0 ≤ x ≤ 12}
R = {(a, b) : |a – b| is a multiple of 4 and a, b ∈ A}
(i) Reflexive: Take a ∈ A
Since |a – a| = 0 is divisible by 4
∴ (a, a) ∈ R              ⇒ R is reflexive.
(ii) Symmetric: Take a, b ∈ A such that |a – b| is divisible by
4 then |b – a| will also be divisible by 4.
\ (a, b) ∈ R              ⇒ (b, a) ∈ R
Hence R is symmetric.
(iii) Transitive: Take a, b, c ∈ A such that |a – b| and |b – c| are
divisible by 4 then |a – c| will also be divisible by 4.
∴ (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R              ∴ R is transitive.
Set of elements related to 1 = {1, 5, 9}
(B) A = {x ∈ Z : 0 ≤ x ≤ 12}
R = {(a, b) : a = b and a, b ∈ A}
Prove in the same way as we have proved in Q.No. 3(A).
Set of elements related to 1 = {1}

Question. Show that the relation R in the set A of points in a plane given by: R = {(p, q) : distance of the point p from the origin is same as the distance of the point q from the origin} is an equivalence relation. Further, show that the set of all points related to a point p ≠ (0, 0) is the circle passing through p with origin as centre.
Answer: A = a set of points in a given plane.
R = {(p, q) : distance of p and q from origin is same, p, q ∈ A}
(i) Reflexive: Take p ∈ A.
Since (p, p) ∈ R              ∴ R is reflexive.
(ii) Symmetric: Take p, q ∈ A such that distances of p and q from origin is same.
∴ (p, q) ∈ R                     ⇒ (q, p) ∈ R
⇒ R is symmetric.
(iii) Transitive: Take p, q, r ∈ A such that p and q are at equal distances from O and q and r are also at equal distances from O.
Then (p, q) ∈ R and (q, r) ∈ R
⇒ (p, r) ∈ R
Hence, R is transitive.
Since R is reflexive, symmetric and transitive.
Hence R is equivalence relation,

Question. Show that the relation R defined in the set A of all triangles as: R = {(T₁, T₂) : T₁ is similar to T₂ and T₁, T₂ ∈ T} is an equivalence relation. Consider three right angled triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?
Answer: T₁ is related to T₃ because sides of T₁ are 3, 4, 5 and sides of T₃ are 6, 8, 10 and hence are proportional.

Question. Show that the relation R defined in the set A of all polygons as: R = {(P₁, P₂) : P₁ and P₂ have the same number of sides} is an equivalence relation. What is the set of all the elements in A related to the right angled triangle T with sides 3, 4 and 5?
Answer: Similar solution as given in solution of Q.No. 4 (Short answers-II). Set of elements in set A related to the right angled triangle T with sides 3, 4 and 5 is a set of all possible triangle in set A.

Question. Prove that the relation R in the set A ={1, 2, 3, 4, 5} given by R ={(a, b) : |a – b| and a, b ∈ A is an even number} is an equivalence relation.
Answer: A = {1, 2, 3, 4, 5} and
R = {(a, b) : |a – b| is an even number, a, b ∈ A}
(i) Reflexive: Take a ∈ A since |a – a| = 0 (an even number)
⇒ (a, a) ∈ R              ⇒ R is reflexive
(ii) Symmetric: Take a, b ∈ A such that |a – b| is an even number, then |b – a| also will be an even number.
∴ (a, b) ∈ R              ⇒ (b, a) ∈ R
⇒ R is symmetric.
(iii) Transitive: Take a, b, c ∈ A such that |a – b| and |b – c| are even numbers. Then |a – c| will also be an even number.
∴ (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R
Hence R is transitive.
Since R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.

Question. Let R be a relation defined on the set of natural numbers N as follows:
R = {(x, y) : x, y ∈ N and 2x + y = 24}
Find the domain and range of the relation R. Also find R is an equivalence relation or not.
Answer: R = {(x, y) : x, y ∈ N and 2x + y = 24}
R = {(1, 22), (2, 20), (3, 18), (4, 16), (5, 14),
(6, 12), (7, 10), (8, 8), (9, 6), (10, 4), (11, 2)}
⇒ Domain of R = {1, 2, 3, ....., 11} and
Range of R = {2, 4, 6, ......, 20, 22}
Relation R is neither reflexive, nor symmetric, nor transitive.

Long Answer Type Questions

Question. Let n be a fixed positive integer. Define a relation R in Z as follows ∀ a, b ∈ Z, aRb if and only if (a – b) is divisible by n. Show that R is an equivalence relation.
Answer: R = {(a, b) : (a – b) is divisible by a fixed positive integer, a, b ∈ Z}
(i) Reflexive: Let a ∈ Z.
Since (a – a) = 0 is divisible by n.
∴ (a, a) ∈ R              ⇒ R is reflexive.
(ii) Symmetric: Let a, b ∈ Z such that (a – b) is divisible by n.
If (a – b) is divisible by n then (b – a) is also divisible by n.
∴ (a, b) ∈ R              ⇒ (b, a) ∈ R
⇒ R is symmetric.
(iii) Transitive: Let a, b, c ∈ Z such that (a – b) and (b – c) are divisible by n.
Then (a – c) also will be divisible by n.
                                                    [ ∴ a – b + b – c = a – c]
Hence (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R              ⇒ R is transitive

Question. Let R be a relation defined as
R = {(x, y) : x, y ∈ N and 2x + y = 41}
Find the domain and range of R also verify that R is neither reflexive, nor symmetric nor transitive.
Answer: Domain of R = {1, 2, 3, ...., 20}
Range of R = {1, 2, 3, ...., 37, 39}

Question. Given the relation R = {(1, 2), (2, 3)} in set A = {1, 2, 3}.
Find the minimum number of ordered pairs which when added to R makes it an equivalence relation.
Answer: In a set A = {1, 2, 3}
Relation R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
will be an equivalence relation.
⇒ 7 ordered pairs will be included.

Question. Give an example to show that union of two equivalence relation on set A is not necessarily equivalence on A.
Answer: Hint: A = {1, 2, 3}
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
S = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
Here R and S each is an equivalence relation on set A, but their
union R ∪ S is not an equivalence relation.

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