CBSE Class 12 Mathematics Relations And Functions Worksheet Set A

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}

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.