CBSE Class 12 Mathematics Relations And Functions Worksheet Set A

Read and download free pdf of CBSE Class 12 Mathematics Relations And Functions Worksheet Set A. Students and teachers of Class 12 Mathematics can get free printable Worksheets for Class 12 Mathematics Chapter 1 Relations and Functions in PDF format prepared as per the latest syllabus and examination pattern in your schools. Class 12 students should practice questions and answers given here for Mathematics in Class 12 which will help them to improve your knowledge of all important chapters and its topics. Students should also download free pdf of Class 12 Mathematics Worksheets prepared by school teachers as per the latest NCERT, CBSE, KVS books and syllabus issued this academic year and solve important problems with solutions on daily basis to get more score in school exams and tests

Worksheet for Class 12 Mathematics Chapter 1 Relations and Functions

Class 12 Mathematics students should refer to the following printable worksheet in Pdf for Chapter 1 Relations and Functions in Class 12. This test paper with questions and answers for Class 12 will be very useful for exams and help you to score good marks

Class 12 Mathematics Worksheet for Chapter 1 Relations and Functions

Case Based Questions

1. Consider the mapping f : A → B is defined by f(x) = x - 1/x - 2 such that f is a bijection.

Based on the above information, answer the following questions:

Question. Domain of f is
(a) R – {2}
(b) R
(c) R – {1, 2}
(d) R – {0}
Answer : A

Question. If g : R – {2} → R – {1} is defined by g(x) = 2f(x) – 1, then g(x) in terms of x is
(a) x + 2 / x
(b) x + 1 / x - 2
(c) x - 2 /x
(d) x / x − 2
Answer : D

Question. Range of f is
(a) R
(b) R – {1}
(c) R – {0}
(d) R – {1, 2}
Answer : B

Question. A function f(x) is said to be one-one if
(a) f(x1) = f(x2) ⇒ –x1 = x2
(b) f(–x1) = f(–x2) ⇒ –x1 = x2
(c) f(x1) = f(x2) ⇒ x1 = x2
(d) None of these
Answer : C

Question. The function g defined above, is
(a) One-one
(b) Many-one
(c) into
(d) None of these
Answer : A

2. A relation R on a set A is said to be an equivalence relation on A if it is
• Reflexive i.e., (a, a) ∈ R ∀ a ∈ A.
• Symmetric i.e., (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A.
• Transitive i.e., (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R ∀ a, b, c ∈A.

Based on the above information, answer the following questions:

Question. If the relation R = {(1, 1), (1, 2), (1, 3), (2, 2),
(2, 3), (3, 1), (3, 2), (3, 3)} defined on the set A = {1, 2, 3}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : A

Question. If the relation R on the set N of all natural numbers defined as R = {(x, y) : y = x + 5 and (x < 4), then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : C

Question. If the relation R = {(1, 2), (2, 1), (1, 3), (3, 1)} defined on the set A = {1, 2, 3}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : B

Question. If the relation R on the set A = {1, 2, 3} defined as R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, then R is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) equivalence
Answer : D

Question. If the relation R on the set A = {1, 2, 3, ... 13, 14} defined as R = {(x, y) : 3x – y = 0}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : D

1. Show that the relation R in the set N of Natural numbers given by R = {(a,b): |a-b| is a multiple of 3} is an equivalence relation. 

Determine whether each of the following relations are reflexive, symmetric, and Transitive.

2. Check whether the relation R in R defined by R = {(a,b):a< b3} is reflexive, symmetric, transitive.

3. Prove the relation R on the set N x N defined by (a, b) R (c,d)↔ a+d = b + c, for all (a, b) (c, d) є N x N is an equivalence relation.

4. Prove that the function f: R →R, given by f (x) = |x| + 5, is not bijective.

5. Prove that the function f: R→R, given by f (x) =4x3 -7, is bijective 

Question. Show that the relation 𝑅 in set 𝑍 given by 𝑅{(𝑎 , 𝑏) ∶ 2 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 – 𝑏} is an Equivalence relation.
Answer :
 We have, 𝑅 = {(𝑎 , 𝑏) ∶ 2 𝑑𝑖𝑣𝑖𝑑𝑒 𝑎 – 𝑏}
Symmetric :
let (𝑎 , 𝑏) ε 𝑅
⇒ 𝑎 – 𝑏 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2
⇒ 𝑎 – 𝑏 = 2𝜆 …...{𝜆𝜖𝑍}
⇒ 𝑏 – 𝑎 = −2𝜆 which is also divisible by 2
⇒ (𝑏 , 𝑎) ε
∴ R is Symmetric
Reflexive :
for each 𝑎 ε 𝑍
⇒ 𝑎 – 𝑎 = 0 which is divisible by 2
⇒ (𝑎 , 𝑎) ε 𝑅
∴ R is Reflexive
Transitive :
let (𝑎, 𝑏) ε 𝑅 and (𝑏 , 𝑐) ε 𝑅
⇒ 𝑎 – 𝑏 = 2𝜆 and 𝑏 – 𝑐 = 2 𝑘 …..{𝜆, 𝑘𝜖𝑍}
Now , 𝑎 – 𝑐 = (𝑎 – 𝑏) + (𝑏 – 𝑐)
⇒ 𝑎 – 𝑐 = 2𝜆 + 2𝑘
⇒ 𝑎 – 𝑐 = 2(𝜆 + 𝑘)which is also divisible by 2
⇒ (𝑎 , 𝑐) ε 𝑅
∴ R is transitive
since 𝑅 is Symmetric , Reflexive and transitive
∴ 𝑅 is an Equivalence relation ans.

Question. Show that the relation R in the set A, 𝐴 = {𝑥 ε 𝑧 ∶ 0 ≤ 𝑥 ≤ 12} given by 𝑅 = {(𝑎, 𝑏) ∶ (𝑎 – 𝑏) is multiple of 4} is an equivalence relation. Find the set of all the elements in set A which are related to 1.
Answer :
 We have , 𝑅 = {(𝑎 , 𝑏) ∶ |𝑎 – 𝑏| is multiple of 4}
Symmetric :
let (𝑎, 𝑏) ε 𝑅
⇒ |𝑎 – 𝑏| is multiple of 4
⇒ |𝑎 – 𝑏| = 4𝜆 …...(𝜆𝜖𝑧)
⇒ |𝑏 – 𝑎| = 4𝜆 which is multiple by 4
⇒ (𝑏, 𝑎) ε 𝑅
∴ R is Symmetric
Reflexive :
for each 𝑎 ε 𝐴
we have, |𝑎 – 𝑎| = 0 which is multiple of 4
⇒ (𝑎 , 𝑎) ε 𝑅
∴ R is Reflexive
Transitive :
let (𝑎 , 𝑏) ε 𝑅 and (𝑏 , 𝑐) ε 𝑅
⇒ |𝑎 – 𝑏| = 4𝜆 and |𝑏 – 𝑐| = 4𝑘 …..{𝜆, 𝑘𝜖𝑍}
⇒ (𝑎 – 𝑏) = ±4𝜆 and (𝑏 – 𝑐) = −4𝑘
Now , (𝑎 – 𝑐) = (𝑎 – 𝑏) + (𝑏 – 𝑐)
⇒ (𝑎 – 𝑐) = ±4𝜆 ± 4𝑘
⇒ (𝑎 – 𝑐) = ±4(𝜆 + 𝑘)
⇒ |𝑎 – 𝑐| = |𝜆 + 𝑘| which is multiple of 4
⇒ (𝑎 , 𝑐) ε 𝑅
∴ R is transitive
∴ R is an Equivalence relation
The elements which related to 1 are 1, 5, 9
∴ required set is {1, 5, 9} ans.

Question. Let R be a relation on the set “A” of ordered pairs defined by (𝑥 , 𝑦) 𝑅(𝑢 , 𝑣) if and only if 𝑥𝑣 = 𝑦𝑢.
Show that R is an equivalence relation.

Answer : Given : A → set of ordered pairs
(𝑥 , 𝑦) 𝑅(𝑢 , 𝑣) ⇒ 𝑥𝑣 = 𝑦𝑢
Symmetric :
let (𝑥 , 𝑦) 𝑅 (𝑢 , 𝑣)
⇒ 𝑥𝑣 = 𝑦𝑢                                    (Rough work)
⇒ 𝑣𝑥 = 𝑢𝑦                                   ((𝑢 , 𝑣) 𝑅(𝑥 , 𝑦))
⇒ 𝑢𝑦 = 𝑣𝑥 ⇒ (4 , 𝑣) 𝑅(𝑥 , 𝑦)           (𝑢𝑦 = 𝑣𝑥)
∴ R is Symmetric
Reflexive :
for each (x , y) ε A
⇒ 𝑥𝑦 = 𝑦𝑥                          {Rough work}
⇒ (𝑥 , 𝑦) 𝑅(𝑥 , 𝑦)                 {(𝑥 , 𝑦) 𝑅(𝑥𝑦)}
∴ R is Reflexive                 {𝑥𝑦 = 𝑦𝑥}
Transitive :
let (𝑥 , 𝑦) 𝑅(𝑢 , 𝑣) and (𝑢 , 𝑣) 𝑅(𝑎 , 𝑏)
⇒ 𝑥𝑣 = 𝑦𝑢 and 𝑢𝑏 = 𝑣𝑎
⇒ 𝑥𝑣 = 𝑦𝑢 and 𝑣 = 𝑢𝑏/𝑎    … . . {𝑅𝑜𝑢𝑔ℎ (𝑥 , 𝑦) 𝑅(𝑎 , 𝑏) , 𝑥𝑏 = 𝑦𝑎}
⇒ 𝑥 (𝑢𝑏/𝑎)= yu
⇒ 𝑥𝑏 = 𝑦𝑎
⇒ (𝑥 , 𝑦) 𝑅(𝑎 , 𝑏)
∴ R is transitive
since R is Symmetric , Reflexive as well as transitive
∴ R is an Equivalence relation ans.

Question. If 𝑅1 and 𝑅2 are equivalence relations in set A , show that 𝑅1 ∩ 𝑅2 is also on equivalence relation.
Answer : Given :- 𝑅1 and 𝑅2 are equivalence relations
Symmetric :
let (𝑎 , 𝑏) ε 𝑅1 ∩ 𝑅2
⇒ (𝑎 , 𝑏) ε 𝑅1 and (𝑎 , 𝑏) ε 𝑅2
⇒ (𝑏 , 𝑎) ε 𝑅1 and (𝑏 , 𝑎) ε 𝑅2          ….{... R and R are symmetric relations}
⇒ (𝑏 , 𝑎) ε 𝑅1 ∩ 𝑅2
∴ 𝑅1 ∩ 𝑅2 is Symmetric
Reflexive :
for each 𝑎 ε 𝐴
we have, (𝑎 , 𝑎) ε 𝑅1 and (𝑎 , 𝑎) ε 𝑅2 … … … … . { 𝑅1 𝑎𝑛𝑑 𝑅2 𝑎𝑟𝑒 𝑟𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒}
⇒ (𝑎 , 𝑎) ε 𝑅1 ∩ 𝑅2
∴ 𝑅1 ∩ 𝑅2is Reflexive
Transitive :
let (𝑎 , 𝑏) ε 𝑅1 ∩ 𝑅2 and (𝑏 , 𝑐)𝑅1 ∩ 𝑅2
⇒ (𝑎 , 𝑏) ε 𝑅1 𝑎𝑛𝑑 (𝑎 , 𝑏) ε 𝑅2 𝑎𝑛𝑑 (𝑏 , 𝑐) ε 𝑅1 & (𝑏 , 𝑐) ε 𝑅2
⇒ (𝑎 , 𝑏) ε 𝑅1 𝑎𝑛𝑑 (𝑏 , 𝑐) ε 𝑅1       | (𝑎 , 𝑏) 𝑅 𝑎𝑛𝑑 (𝑏 , 𝑐) ε 𝑅2
⇒ (𝑎 , 𝑐) ε 𝑅              | (𝑎 , 𝑐) ε 𝑅2
                        … . {𝑅1 & 𝑅2 𝑎𝑟𝑒 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑣𝑒}
⇒ (𝑎 , 𝑐) ε 𝑅1 ∩ 𝑅2
∴ 𝑅1 ∩ 𝑅2is transitive
since 𝑅1 ∩ 𝑅2 is Symmetric , Reflexive as well as transitive
∴ 𝑅1 ∩ 𝑅2 is an Equivalence relation ans.

Question. 𝑅 is a relation on set 𝑁 given by 𝑎𝑅𝑏 ↔ 𝑏 is divisible by 𝑎; 𝑎. 𝑏 ε 𝑁check whether R is Symmetric , reflexive and transitive.
Answer :
 We have,𝑎𝑅𝑏 ↔ 𝑏 is divisible by 𝑎
Symmetric :
2𝑅6 ⇒ 6 is divisible by 2 ….{6/2 = 3}
but 6𝑅2 ⇒ 2is not div by 6 .....{2/6 = 1/2}
∴ R is not symmetric
Reflexive : for each 𝑎 ε 𝑁
a is always divisible by a
⇒ 𝑎𝑅𝑎
∴ R is Reflexive
Transitive :
let 𝑎𝑅𝑏 and 𝑏𝑅𝑐
⇒ b is divisible by a and c is div by b
⇒ 𝑏 = 𝑎𝜆 and 𝑐 = 𝑏𝑘         ….{𝜆, 𝑘𝜖𝑁}
⇒ 𝑐 = (𝑎𝜆)𝑘                     ….{.. . 𝑏 = 𝑎𝜆}
⇒ 𝑐/𝑎 = 𝜆𝑘
clearly c is div by a
⇒ 𝑎𝑅𝑐
∴ R is transitive ans.

Question. R be relation in P(x) , where x is a non-empty set , given by
ARB if only if ACB , where A & B are subsets in 𝑃(𝑥). Is R is an equivalence relation on 𝑃(𝑥) ? Justify your answer.
Answer : Let ARB
⇒ 𝐴 ⊂ 𝐵
then it is not necessary that B is a subset of A
i.e. 𝐵 ⊄ 𝐴
⇒ B R A
∴ R is not symmetric and hence R is not an equivalence relation
eg. 𝑥 = {1,2,3}
𝑃(𝑥) = {{1}{2}{3}{1 , 2}{2 , 3}{1 , 3}{1 , 2 , 3}}
clearly {2} ⊂ {1,2}
between{1,2} ⊂ {2}
∴ R is not symmetric ans.

Question. Show that the relation R defined in the set A of all triangles as R -{(T1 , T2) : T1 is similar to T2} is equivalence relation.
Consider three right angle 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 : A → set of are triangles
𝑅 = {(𝑇1 , 𝑇2) ∶ 𝑇1 ∼ 𝑇2}
Symmetric :
        let (𝑇1 , 𝑇2) ε 𝑅
⇒ 𝑇1 ∼ 𝑇2
⇒ 𝑇2 ∼ 𝑇1
⇒ (𝑇2 , 𝑇1) ε 𝑅
∴ R is symmetric
Reflexive : for each triangle 𝑇 ε 𝐴
              (𝑇, 𝑇) ε 𝑅
since every triangle is similar to itself
∴ R is reflexive
Transitive :
let (𝑇1 , 𝑇2) ε 𝑅 and (𝑇2 ∼ 𝑇3) ε 𝑅
⇒ 𝑇1 ∼ 𝑇2 𝑎𝑛𝑑 𝑇2 ∼ 𝑇3
⇒ 𝑇1 ∼ 𝑇3
⇒ 𝑇1 , 𝑇3) ε 𝑅
∴ R is transitive
and hence R is an equivalence relation
𝑇1 ∶ 3 , 4 , 5
𝑇2 ∶ 5 , 12 , 13
𝑇3 ∶ 6 , 8 , 10
clearly sides of triangles T1 and T3 are in equal proportion i.e 3/6 = 4/8 = 5/10
∴ T1 ∼ T3
⇒ T1 and T3 are related to each other ans.

Question. Check whether the relation R in R (real no's) define by 𝑅 = (𝑎, 𝑏): 𝑎 ≤ 𝑏3is reflexive, symmetric or transitive.
Answer :
 Symmetric :
(1,2) ε 𝑅
as1 ≤ 23
but (2,1) ∉ 𝑅
since2 ≰ 13
... R is not symmetric
Reflexive : 1/2 ∈ 𝑅
but (1/2, 1/2) ∉ 𝑅
as 1/2 ≰ (1/2)3
∴ R is not reflexive
Transitive :
(9,4) ∈ 𝑅and(4,2) ∈ 𝑅
as 9 ≤ 43 and 4 ≤ 23
but(9,2) ∉ 𝑅
since 9 ≰ 23
∴ R is not transitive ans.

Question. Show that the relation R in the set {1,2,3} given by R = {(1 , 1), (2 , 2), (3 , 3), (1 , 2), (2 , 3)} is reflexive neither symmetric nor transitive.
Answer :
 We have,
𝐴 = {1,2,3}
𝑅 = {(1,1), (2,2), (3,3), (1,2), (2,3)}
since (1,2) ε 𝑅
but (2,1) ∉ 𝑅
∴ R is not Symmetric
(1,2) ∈ 𝑅and(2,3) ∈ 𝑅
but (1,3) ∉ 𝑅
∴ R is not transitive
for each 𝑎 ε 𝐴
(𝑎 , 𝑎) ε 𝑅 i.e. (1,1), (2,2), (3,3) ε 𝑅
∴ R is reflexive      ans.

Question. Determine whether each of the following relations are reflexive, symmetric and transitive
(i) Relation in set A = {1,2,3,.......... 13,14} defined by 𝑅 = (𝑥, 𝑦): 3x– 𝑦 = 0.
(ii) Relation in N defined as 𝑅 = (𝑥, 𝑦): 𝑦 = 𝑥 + 5; 𝑥 < 4.
(iii) Relation in set A = {1,2,3,4,5,6} defined as 𝑅 = (𝑥, 𝑦): 𝑦is divisible by 𝑥.
(iv) Relation in Z defined as 𝑅 = (𝑥, 𝑦): 𝑥– 𝑦is an integer.
(v) Relation in R (real nos) defined as 𝑅 = (𝑎, 𝑏): 𝑎 ≤ 𝑏2.
Answer :
 (i) 𝑅 = {(1,3), (2,6), (3,9), (4,12)}            …...(𝑦 = 3x)
clearly (1,3) ∈ 𝑅but(3,1) ∉ 𝑅
∴ not symmetric
1 ∈ 𝐴 but (1,1) ∉ 𝑅
∴ not reflexive
(1,3) ∈ 𝑅and (3,9) ∈ 𝑅but(1,9) ∉ 𝑅
∴ not transitive
(ii) 𝑅 = {(1,6), (2,7), (3,8)} … . . {. . . 𝑦 = 𝑥 + 5 𝑎𝑛𝑑 𝑥 < 4}
Do yourself
(iii) 𝑅 = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (5,5), (6,6)}. . . {. . . 𝑦 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑥}
clearly for each 𝑎 ε 𝐴
(𝑎 , 𝑎) ε 𝑅 i.e. (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) ε 𝑅
∴ R is reflexive
(1 , 2) ε 𝑅
but(2,1) ∉ 𝑅
since 1 in not divisible by 2
∴ R is not transitive
for each (𝑎 , 𝑏) and (𝑏 , 𝑐) ε 𝑅
clearly (𝑎 , 𝑐) ε 𝑅
∴ R is transitive
(iv) Symmetric let (𝑥, 𝑦) ε 𝑅
⇒ 𝑥 – 𝑦 = 𝜆 ….. where 𝜆→ integer
⇒ 𝑦 – 𝑥 = −𝜆 which is also an integer
⇒ (𝑦 , 𝑥) ε 𝑅
∴ R is Symmetric
Reflexive and transitive (Do yourself)
(v) give same examples as in case of 𝑎 ≤ 𝑏3
It is neither symmetric, nor reflexive, nor transitive.

Question. Let f: 𝑅 → 𝑅 be defined as f(x) =3x - 2. Choose the correct answer.
a) f is one-one onto
b) f is many one onto
c) f is one-one but not onto
d) f is neither one-one nor onto
Answer : A

Question. Let us define a relation R in R as aRb if a ≥ b. Then R is
(a) an equivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric
Answer : B

Question. let R be the relation in the set N given by R={(a,b):a=b-2,b>6}.Choose the correct answer.
(a) (2,4)€R
(b) (3,8) € R
(c) (6,8)€ R
(d) (8,10) € R
Answer : D

Question. Let R be a relation on set of lines as L1 R L2 if L1 is perpendicular to L2. Then
a) R is Reflexive
b) R is transitive
c) R is symmetric
d) R is an equivalence relation
Answer : C

Question. Let R be a relation on the set N of natural numbers denoted by nRm⇔ n is a factor of m (i.e. n | m). Then, R is
(a) Reflexive and symmetric
(b) Transitive and symmetric
(c) Equivalence
(d) Reflexive, transitive but not symmetric
Answer : D

Question. A Relation from A to B is an arbitrary subset of:
a) AxB
b) BxB
c) AxA
d) BxB
Answer : A

Question. Let R be a relation defined on Z as R= {(a,b) ; a2+b2=25 } , the domain of R is;
(a) {3,4,5}
(b) {0,3,4,5}
(c) {0,3,4,5,-3,-4,-5}
(d) none
Answer : C

Question. Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is
(a) reflexive but not transitive
(b) transitive but not symmetric
(c) equivalence
(d) None of these
Answer : C

Question. The maximum number of equivalence relations on the set A = {1, 2, 3} are
(a) 1
(b) 2
(c) 3
(d) 5
Answer : D

Question. Let S = {1, 2, 3, 4, 5} and let A = S × S. Define the relation R on A as follows: (a, b) R (c, d) iff ad = cb. Then, R is
(a) reflexive only
(b) Symmetric only
(c) Transitive only
(d) Equivalence relation
Answer : D

Question. Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is
(a) reflexive but not symmetric
(b) reflexive but not transitive
(c) symmetric and transitive
(d) neither symmetric, nor transitive
Answer : A

Question. Let A = {x : -1 ≤ x ≤ 1} and f : A → A is a function defined by f(x) = x |x| then f is
(a) a bijection
(b) injection but not surjection
(c) surjection but not injection
(d) neither injection nor surjection
Answer : A

Question. Let X = {-1, 0, 1}, Y = {0, 2} and a function f : X → Y defined by y = 2x4, is
(a) one-one onto
(b) one-one into
(c) many-one onto
(d) many-one into
Answer : C

Question. Let f : [0, ∞) → [0, 2] be defined by f(x) = 2x/1 + x, then f is
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
Answer : B

Question. Given set A = {a, b, c). An identity relation in set A is
(a) R = {(a, b), (a, c)}
(b) R = {(a, a), (b, b), (c, c)}
(c) R = {(a, a), (b, b), (c, c), (a, c)}
(d) R= {(c, a), (b, a), (a, a)}
Answer : B
 

CASE STUDY
A relation R on a set A is said to be an equivalence relation on A if it is
• Reflexive i.e., (a, a) ∈ R ∀ a ∈ A.
• Symmetric i.e., (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A.
• Transitive i.e., (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R ∀ a, b, c ∈A. Based on the above information, answer the following questions:

Question. If the relation R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} defined on the set A = {1, 2, 3}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : A

Question. If the relation R = {(1, 2), (2, 1), (1, 3), (3, 1)} defined on the set A = {1, 2, 3}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : B

Question. If the relation R on the set N of all natural numbers defined as R = {(x, y) : y = x + 5 and (x <4), then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence
Answer : C
 

CASE STUDY

Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible outcomes of the throw every time belongs to set {1,2,3,4,5,6}. Let A be the set of players while B be the set of all possible outcomes.A = {S, D}, B = {1,2,3,4,5,6}

""CBSE-Class-12-Mathematics-Relations-And-Functions-Worksheet-Set-F

1. Let 𝑅∶ 𝐵→𝐵 be defined by R = {(𝑥,): 𝑦 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏 } is
a. Reflexive and transitive but not symmetric
b. Reflexive and symmetric and not transitive
c. Not reflexive but symmetric and transitive
d. Equivalence
Answer : A

2. Raji wants to know the number of functions from A to B. How many number of functions are possible?
a. 62
b. 26
c. 6!
d. 212
Answer : A

3. Let R be a relation on B defined by R = {(1,2), (2,2), (1,3), (3,4), (3,1), (4,3), (5,5)}. Then R is
a. Symmetric
b. Reflexive
c. Transitive
d. None of these three
Answer : D

4. Raji wants to know the number of relations possible from A to B. How many numbers of relations are possible?
a. 62
b. 26
c. 6!
d. 212
Answer : D

5. Let 𝑅:𝐵→𝐵 be defined by R={(1,1),(1,2), (2,2), (3,3), (4,4), (5,5),(6,6)}, then R is
a. Symmetric
b. Reflexive and Transitive
c. Transitive and symmetric
d. Equivalence
Answer : B
 

ASSERTION AND REASON

Read Assertion and reason carefully and write correct option for each question
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.

Question. Assertion= {(T1, T2) : T1 is congruent to T2}. Then R is an equivalence relation.
Reason(R) Any relation R is an equivalence relation, if it is reflexive, symmetric and transitive
Answer : A

Question. Assertion (A) 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 same number of pages} is not equivalence relation.
Reason (R) Since R is reflexive, symmetric and transitive.
Answer : C

Question. Assertion (A) Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. R is an equivalence relation
Reason (R) Since R is reflexive, symmetric but R is not transitive.
Answer : C

Question. Assertion (A) A one-one function f : {1, 2, 3} →{1, 2, 3} must be onto.
Reason (R) Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co- domain {1, 2, 3} under f.
Answer : A

Question. Assertion (A) The relation R in R defined as R = {(a, b) :a≤𝑏2} is not equivalence relation.
Reason (R) Since R is not reflexive but it is symmetric and transitive.
Answer : A

Question. Assertion (A)The relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2),(3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
Reason (R) R is not symmetric, as (1, 2) ∈R but (2, 1) ∉R. Similarly, R is not transitive, as (1, 2) ∈R and (2, 3) ∈R but (1, 3) ∉ R.

Answer : A

Question. Assertion (A) The Modulus Function f :R→R, given by f (x) = | x | is not one one and onto function
Reason (R) The Modulus Function f :R→R, given by f (x) = | x | is bijective function
Answer : C

Question. Assertion (A) The function f :N→N, given by f (x) = 2x, is one-one
Reason (R) The function f is one-one, for f (x) = f (y) ⇒2x = 2y⇒x = y.

Answer : A

Question. Assertion (A) Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}. R is not equivalence relation.
Reason (R) R is not Reflexive relation but it is symmetric and transitive
Answer : C

Question. Assertion (A) The relation R in R defined as R = {(a, b) :a≤b} is not equivalence relation.
Reason (R) Since R is not reflexive but it is symmetric and transitive.
Answer : C

 

Q1 Let n be a fixed positive integer. Define a relation R on Z as follows (a, b) Є R ⇔ a-b is divisible by n.
Show that R is an equivalence relation on z.

Q2. Let z be the set of integers show that the relation
R = [a, b) : a, b Є z and a + b is even] is an equivalence relation on z.

Q3. Let S be a relation on the set R of real numbers defined by
S = [(a, b) Є R x R : a2 + b2 = 1} prove that S is not an equivalence relation R.

Q4. Show that the relation R on the set R of real numbers defined as
R = [(a, b) : : a < b2] is neither reflexive nor symmetric nor transitive.

Q5. Show that the relation R on R defined as R = [(a, b) : a < ] is reflexive and transitive but not symmetric.

Q6. Show that f : R → R, defined as f(x) = x3, is a bijection.

Q7. Show that the modulus function f R → R, given by f(x) = [x] is neither one-one nor on-to.

Q8. Show that the function of F : R→R given by f(x) = x3 + x is a bijection.

Q9. Let A = R –[2] and B = B-[1]. If f : A → B is a mapping defined by f(x) x-1/x-2, show that f is bijective.

Q10. Show that f: R→R, given f(x) = x – [x], is neither one-one or onto.

Q11. Ret f(x) = [x] and g(x) = [x], Find
(i) (gof) (-5/3)- (fog) (-5/3) (ii) (gof) (5/3)- (fog) (5/3)
(iii) (f+2g) (-1)

Q12. If f(x) = 3x-2/2x-3, prove that f (f(x) = x for all x - R (3/2)

Q13. Find fog and gof, if (i) f(x) = ex, g(x) = logex (ii) f(x) = x + 1, g(x) = 2x + 3

Q14. Prove that the function f : R→R defined by f(x) = 2x-3 is invertible find f.

Q15. Let F : N → R be a function defined as f(x0 = 4x2 + 12 x + 15. Show that f: N → Range (f) is invertible.
Find the inverse of f.

Q16. Show that f: [-1, 1] →R, given by f(x) = x/x+2 is one-one, find the inverse of the function f: (-1, 1) → Range (f).

Q17. Let ‘x’ be a binary operation on set 2 – [1] defined by a x b = a + b – ab ; a, b, - Q – [1]. Find the identity element with respect to on Q. Also, prove that every element of Q – [1] is invertible.

Q18. Consider the binary operation on the set 3 = {1, 2, 3, 4, 5} defined by A B = Minimum of a and B.
Write the composition table of a and b.

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