CBSE Class 12 Mathematics Relations And Functions Worksheet Set 01

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.

FOUR MARKS QUESTIONS

Question. 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.
Answer: To show that the relation \( R \) is an equivalence relation, we must prove that it is reflexive, symmetric, and transitive on the set \( \mathbb{N} \).
1. Reflexive:
For any \( a \in \mathbb{N} \), we have:
\( |a - a| = 0 \)
Since \( 0 \) is divisible by \( 3 \) (\( 0 = 3 \times 0 \)), it is a multiple of \( 3 \).
Therefore, \( (a, a) \in R \) for all \( a \in \mathbb{N} \). Thus, \( R \) is reflexive.

2. Symmetric:
Let \( (a, b) \in R \). This implies that \( |a - b| \) is a multiple of \( 3 \).
\( |a - b| = 3k \) for some non-negative integer \( k \).
Since \( |b - a| = |a - b| \), we have:
\( |b - a| = 3k \), which is also a multiple of \( 3 \).
Therefore, \( (b, a) \in R \). Thus, \( R \) is symmetric.

3. Transitive:
Let \( (a, b) \in R \) and \( (b, c) \in R \). This implies that \( |a - b| \) and \( |b - c| \) are multiples of \( 3 \).
Therefore, we can write:
\( a - b = \pm 3k_1 \) and \( b - c = \pm 3k_2 \) for some integers \( k_1, k_2 \).
Adding these two equations gives:
\( (a - b) + (b - c) = \pm 3k_1 \pm 3k_2 \)
\( a - c = 3(\pm k_1 \pm k_2) \)
Since \( \pm k_1 \pm k_2 \) is an integer, \( a - c \) is divisible by \( 3 \). Hence, \( |a - c| \) is a multiple of \( 3 \).
Therefore, \( (a, c) \in R \). Thus, \( R \) is transitive.

Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation on \( \mathbb{N} \).

Question. Check whether the relation R in R defined by R = {(a, b): a ≤ b^3} is reflexive, symmetric, transitive.
Answer: Let \( R \) be the relation defined on the set of real numbers \( \mathbb{R} \).
1. Reflexive:
For \( R \) to be reflexive, \( a \le a^3 \) must hold true for all \( a \in \mathbb{R} \).
Let \( a = \frac{1}{2} \).
\( a^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)
Since \( \frac{1}{2} \not\le \frac{1}{8} \), \( \left(\frac{1}{2}, \frac{1}{2}\right) \notin R \).
Therefore, \( R \) is not reflexive.

2. Symmetric:
Let \( (a, b) \in R \). This means \( a \le b^3 \). For \( R \) to be symmetric, it must imply \( b \le a^3 \).
Let \( a = 1 \) and \( b = 2 \).
\( 1 \le 2^3 = 8 \), so \( (1, 2) \in R \).
But \( 2 \not\le 1^3 = 1 \), so \( (2, 1) \notin R \).
Therefore, \( R \) is not symmetric.

3. Transitive:
Let \( (a, b) \in R \) and \( (b, c) \in R \). This means \( a \le b^3 \) and \( b \le c^3 \).
Let \( a = 3 \), \( b = 1.5 \), and \( c = 1.2 \).
\( 3 \le (1.5)^3 = 3.375 \), so \( (3, 1.5) \in R \).
\( 1.5 \le (1.2)^3 = 1.728 \), so \( (1.5, 1.2) \in R \).
But \( 3 \not\le (1.2)^3 = 1.728 \), so \( (3, 1.2) \notin R \).
Therefore, \( R \) is not transitive.

Hence, the relation \( R \) is neither reflexive, nor symmetric, nor transitive.

Question. 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) ∈ N x N is an equivalence relation.
Answer: Let \( R \) be the relation on \( \mathbb{N} \times \mathbb{N} \) defined by \( (a, b) R (c, d) \Leftrightarrow a + d = b + c \).
1. Reflexive:
For any \( (a, b) \in \mathbb{N} \times \mathbb{N} \), we have:
\( a + b = b + a \) (since addition of natural numbers is commutative).
This implies \( (a, b) R (a, b) \).
Therefore, \( R \) is reflexive.

2. Symmetric:
Let \( (a, b) R (c, d) \). This implies:
\( a + d = b + c \)
\( \Rightarrow c + b = d + a \) (by commutativity of addition on \( \mathbb{N} \))
\( \Rightarrow (c, d) R (a, b) \).
Therefore, \( R \) is symmetric.

3. Transitive:
Let \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \). This implies:
\( a + d = b + c \) --- (i)
\( c + f = d + e \) --- (ii)
Adding equations (i) and (ii), we get:
\( (a + d) + (c + f) = (b + c) + (d + e) \)
\( \Rightarrow a + d + c + f = b + c + d + e \)
Canceling \( c \) and \( d \) from both sides yields:
\( a + f = b + e \)
\( \Rightarrow (a, b) R (e, f) \).
Therefore, \( R \) is transitive.

Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Question. Prove that the function f: R → R, given by f (x) = |x| + 5, is not bijective.
Answer: A function is bijective if it is both one-one (injective) and onto (surjective).
1. Testing for One-One:
Let \( x_1 = 1 \) and \( x_2 = -1 \).
\( f(1) = |1| + 5 = 6 \)
\( f(-1) = |-1| + 5 = 6 \)
Since \( f(1) = f(-1) = 6 \) but \( 1 \neq -1 \), the function \( f \) is not one-one.

2. Testing for Onto:
Since the absolute value \( |x| \ge 0 \) for all \( x \in \mathbb{R} \), we have:
\( f(x) = |x| + 5 \ge 5 \)
The range of the function \( f \) is \( [5, \infty) \). However, the co-domain of the function is the set of all real numbers \( \mathbb{R} \).
Since the range \( [5, \infty) \neq \mathbb{R} \) (the co-domain), there are elements in the co-domain (such as \( 0, 1, 2, -3 \)) that have no pre-images in the domain.
Therefore, \( f \) is not onto.

Since \( f \) is neither one-one nor onto, it is not bijective.

Question. Prove that the function f: R → R, given by f (x) = 4x^3 -7, is bijective
Answer: To prove that \( f(x) = 4x^3 - 7 \) is bijective, we must show it is both one-one and onto.
1. One-One:
Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \).
\( \Rightarrow 4x_1^3 - 7 = 4x_2^3 - 7 \)
\( \Rightarrow 4x_1^3 = 4x_2^3 \)
\( \Rightarrow x_1^3 = x_2^3 \)
Taking the real cube root on both sides:
\( \Rightarrow x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \), \( f \) is one-one.

2. Onto:
Let \( y \in \mathbb{R} \) be an arbitrary element in the co-domain. We need to find \( x \in \mathbb{R} \) such that \( f(x) = y \).
\( y = 4x^3 - 7 \)
\( \Rightarrow 4x^3 = y + 7 \)
\( \Rightarrow x^3 = \frac{y+7}{4} \)
\( \Rightarrow x = \left(\frac{y+7}{4}\right)^{1/3} \)
For any real number \( y \), \( \left(\frac{y+7}{4}\right)^{1/3} \) is a unique, well-defined real number. Thus, \( x \in \mathbb{R} \).
Now, let's verify \( f(x) \):
\( f\left(\left(\frac{y+7}{4}\right)^{1/3}\right) = 4\left(\left(\frac{y+7}{4}\right)^{1/3}\right)^3 - 7 = 4\left(\frac{y+7}{4}\right) - 7 = y + 7 - 7 = y \).
Therefore, every element \( y \) in the co-domain has a pre-image in the domain, so \( f \) is onto.

Since the function \( f \) is both one-one and onto, it is bijective.

Question. Prove that the Greatest Integer Function f: R → R given by f(x) =[x], is neither one-one nor onto where [x] denotes the greatest intger less than or equal to x .
Answer: 1. Not One-One:
Let \( x_1 = 1.2 \) and \( x_2 = 1.8 \). Both belong to the domain \( \mathbb{R} \).
\( f(1.2) = [1.2] = 1 \)
\( f(1.8) = [1.8] = 1 \)
Since \( f(1.2) = f(1.8) = 1 \) but \( 1.2 \neq 1.8 \), different elements in the domain have the same image. Hence, the function is not one-one.

2. Not Onto:
For any \( x \in \mathbb{R} \), the value of \( f(x) = [x] \) is always an integer.
Therefore, the range of the function \( f \) is the set of integers \( \mathbb{Z} \).
However, the co-domain is the set of real numbers \( \mathbb{R} \).
Since the range \( \mathbb{Z} \neq \mathbb{R} \) (the co-domain), non-integral real values (such as \( 1.5, 2.7, \pi \), etc.) in the co-domain do not have any pre-image in the domain.
Hence, \( f \) is not onto.

Thus, the Greatest Integer Function is neither one-one nor onto.

Question. Let f: N → N be defined by \( f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases} \) for all n ∈ N, State whether the function f is bijective.
Answer: Let us check if the function \( f \) is one-one and onto.
1. One-One:
Consider \( n = 1 \) (odd) and \( n = 2 \) (even). Both are natural numbers.
\( f(1) = \frac{1+1}{2} = 1 \)
\( f(2) = \frac{2}{2} = 1 \)
Here, \( f(1) = f(2) = 1 \) but \( 1 \neq 2 \).
Therefore, \( f \) is not one-one.

2. Onto:
Let \( m \in \mathbb{N} \) be any natural number in the co-domain.
- If \( n \) is odd, let \( n = 2m - 1 \). Since \( m \ge 1 \), \( 2m - 1 \) is a positive odd integer (natural number).
\( f(2m - 1) = \frac{(2m - 1) + 1}{2} = \frac{2m}{2} = m \).
- If \( n \) is even, let \( n = 2m \). Since \( m \ge 1 \), \( 2m \) is a positive even integer.
\( f(2m) = \frac{2m}{2} = m \).
In both cases, we can find a pre-image in \( \mathbb{N} \) for every element in the co-domain. Thus, \( f \) is onto.

Since the function \( f \) is onto but not one-one, it is not bijective.

Question. Consider that f: N → N given by f(x) = x^2 + x + 1 . Show that f is not invertible.
Answer: A function is invertible if and only if it is bijective (both one-one and onto). Let us examine the properties of \( f(x) = x^2 + x + 1 \) defined on the domain of natural numbers \( \mathbb{N} \).
1. Checking for One-One:
Let \( x_1, x_2 \in \mathbb{N} \) such that \( f(x_1) = f(x_2) \).
\( \Rightarrow x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1 \)
\( \Rightarrow x_1^2 - x_2^2 + x_1 - x_2 = 0 \)
\( \Rightarrow (x_1 - x_2)(x_1 + x_2) + (x_1 - x_2) = 0 \)
\( \Rightarrow (x_1 - x_2)(x_1 + x_2 + 1) = 0 \)
Since \( x_1, x_2 \in \mathbb{N} \), their sum \( x_1 + x_2 \ge 2 \), which means \( x_1 + x_2 + 1 \ge 3 \neq 0 \).
Thus, we must have:
\( x_1 - x_2 = 0 \Rightarrow x_1 = x_2 \).
Therefore, \( f \) is one-one.

2. Checking for Onto:
Let \( y \in \mathbb{N} \) be an element in the co-domain. We need to check if there exists a pre-image \( x \in \mathbb{N} \) such that \( f(x) = y \).
Consider \( y = 1 \).
\( x^2 + x + 1 = 1 \)
\( \Rightarrow x^2 + x = 0 \)
\( \Rightarrow x(x + 1) = 0 \)
Since \( x \in \mathbb{N} \), \( x \ge 1 \), so \( x \neq 0 \) and \( x \neq -1 \). Thus, \( y = 1 \) has no pre-image in the domain \( \mathbb{N} \).
(Similarly, for \( y = 2 \), \( x^2 + x + 1 = 2 \Rightarrow x^2 + x - 1 = 0 \Rightarrow x = \frac{-1 + \sqrt{5}}{2} \notin \mathbb{N} \)).
Since there are elements in the co-domain that have no pre-images in the domain, \( f \) is not onto.

Since \( f \) is not onto, it is not bijective, and therefore, it is not invertible.

Question. Let A=N x N and * be the binary operation on A defined by (a, b) *(c, d) = (ad + bc, bd) .Show that * is commutative and associative . Find the identity element for * on A , if any.
Answer: Let \( (a, b), (c, d), (e, f) \in \mathbb{N} \times \mathbb{N} \).
1. Commutativity:
\( (a, b) * (c, d) = (ad + bc, bd) \)
\( (c, d) * (a, b) = (cb + da, db) = (ad + bc, bd) \) (since addition and multiplication are commutative on natural numbers).
Since \( (a, b) * (c, d) = (c, d) * (a, b) \), the operation \( * \) is commutative.

2. Associativity:
\( [ (a, b) * (c, d) ] * (e, f) = (ad + bc, bd) * (e, f) = ((ad + bc)f + bd(e), (bd)f) = (adf + bcf + bde, bdf) \)
\( (a, b) * [ (c, d) * (e, f) ] = (a, b) * (cf + de, df) = (a(df) + b(cf + de), b(df)) = (adf + bcf + bde, bdf) \)
Since \( [ (a, b) * (c, d) ] * (e, f) = (a, b) * [ (c, d) * (e, f) ] \), the operation \( * \) is associative.

3. Identity Element:
Let \( (x, y) \in \mathbb{N} \times \mathbb{N} \) be the identity element in \( A \) such that for any \( (a, b) \in A \):
\( (a, b) * (x, y) = (a, b) \)
\( \Rightarrow (ay + bx, by) = (a, b) \)
Comparing the coordinates:
\( by = b \Rightarrow y = 1 \) (since \( b \in \mathbb{N} \))
Substituting \( y = 1 \) into \( ay + bx = a \):
\( a(1) + bx = a \)
\( \Rightarrow bx = 0 \)
Since \( b \in \mathbb{N} \), \( b \ge 1 \neq 0 \), we must have \( x = 0 \).
However, \( 0 \notin \mathbb{N} \), which means \( (0, 1) \notin \mathbb{N} \times \mathbb{N} \).
Hence, there is no identity element for \( * \) on \( A \).

Question. Let f: R→R be the function defined by f(x) = 1/(2-cosx) ∀ x ∈ R .Then find the range of f. (Exemplar).
Answer: We know that for all \( x \in \mathbb{R} \):
\( -1 \le \cos x \le 1 \)
Multiplying the inequality by \( -1 \):
\( -1 \le -\cos x \le 1 \)
Adding \( 2 \) to all parts of the inequality:
\( 2 - 1 \le 2 - \cos x \le 2 + 1 \)
\( \Rightarrow 1 \le 2 - \cos x \le 3 \)
Taking the reciprocal of each term (which reverses the inequality signs since all terms are positive):
\( \frac{1}{3} \le \frac{1}{2 - \cos x} \le 1 \)
\( \Rightarrow \frac{1}{3} \le f(x) \le 1 \)

Therefore, the range of the function \( f \) is \( \left[ \frac{1}{3}, 1 \right] \).

FOUR MARKS QUESTIONS

Question. Let f: N→ R be a function defined as f(x) = 4x^2 +12x +15. Show that f:N→ S, where S is the range of f is invertible .Find the inverse of f.
Answer: To show that the function \( f: \mathbb{N} \rightarrow S \) is invertible, we must show that it is both one-one and onto.
1. One-One:
Let \( x_1, x_2 \in \mathbb{N} \) such that \( f(x_1) = f(x_2) \).
\( \Rightarrow 4x_1^2 + 12x_1 + 15 = 4x_2^2 + 12x_2 + 15 \)
\( \Rightarrow 4(x_1^2 - x_2^2) + 12(x_1 - x_2) = 0 \)
\( \Rightarrow 4(x_1 - x_2)(x_1 + x_2) + 12(x_1 - x_2) = 0 \)
\( \Rightarrow 4(x_1 - x_2)(x_1 + x_2 + 3) = 0 \)
Since \( x_1, x_2 \in \mathbb{N} \), we have \( x_1 + x_2 + 3 \ge 5 > 0 \). Thus, \( x_1 + x_2 + 3 \neq 0 \).
Therefore:
\( x_1 - x_2 = 0 \Rightarrow x_1 = x_2 \).
Thus, the function is one-one.

2. Onto:
Since the co-domain is given as the range of the function \( S \), every element in \( S \) has at least one pre-image in the domain \( \mathbb{N} \).
Therefore, the function is onto.

Since \( f \) is both one-one and onto, it is invertible.

3. Inverse of f:
Let \( y \in S \). Then there exists \( x \in \mathbb{N} \) such that:
\( y = 4x^2 + 12x + 15 \)
By completing the square on the right side:
\( y = (2x + 3)^2 + 6 \)
\( \Rightarrow (2x + 3)^2 = y - 6 \)
\( \Rightarrow 2x + 3 = \sqrt{y - 6} \) (taking the positive square root since \( x \in \mathbb{N} \))
\( \Rightarrow 2x = \sqrt{y - 6} - 3 \)
\( \Rightarrow x = \frac{\sqrt{y - 6} - 3}{2} \)
Therefore, the inverse function \( f^{-1}: S \rightarrow \mathbb{N} \) is defined by:
\( f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2} \).

Question. Consider f: R+ → [-5, ∞) given by f(x) = 9x^2 + 6x - 5 . Show that f is invertible. Find the inverse of f.
Answer: Let \( \mathbb{R}_+ = [0, \infty) \). To prove that \( f: \mathbb{R}_+ \rightarrow [-5, \infty) \) is invertible, we must show it is one-one and onto.
1. One-One:
Let \( x_1, x_2 \in \mathbb{R}_+ \) such that \( f(x_1) = f(x_2) \).
\( \Rightarrow 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5 \)
\( \Rightarrow 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \)
\( \Rightarrow 3(x_1 - x_2)[3(x_1 + x_2) + 2] = 0 \)
Since \( x_1, x_2 \in \mathbb{R}_+ \), we have \( x_1, x_2 \ge 0 \), which implies \( 3(x_1 + x_2) + 2 \ge 2 > 0 \).
Thus, we must have:
\( x_1 - x_2 = 0 \Rightarrow x_1 = x_2 \).
Therefore, \( f \) is one-one.

2. Onto:
Let \( y \in [-5, \infty) \). We need to find \( x \in \mathbb{R}_+ \) such that \( f(x) = y \).
\( y = 9x^2 + 6x - 5 \)
\( y = (3x + 1)^2 - 6 \)
\( \Rightarrow (3x + 1)^2 = y + 6 \)
\( \Rightarrow 3x + 1 = \sqrt{y + 6} \) (since \( y \ge -5 \Rightarrow y + 6 \ge 1 \), the square root is real; we take the positive root because \( 3x + 1 \ge 1 \))
\( \Rightarrow 3x = \sqrt{y + 6} - 1 \)
\( \Rightarrow x = \frac{\sqrt{y + 6} - 1}{3} \)
Since \( y \ge -5 \), we have \( \sqrt{y + 6} \ge 1 \), which implies \( x = \frac{\sqrt{y + 6} - 1}{3} \ge 0 \). Thus, \( x \in \mathbb{R}_+ \).
Therefore, for every \( y \in [-5, \infty) \), there exists an \( x \in \mathbb{R}_+ \) such that \( f(x) = y \). Hence, \( f \) is onto.

Since \( f \) is both one-one and onto, it is invertible, and the inverse function \( f^{-1}: [-5, \infty) \rightarrow \mathbb{R}_+ \) is given by:
\( f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3} \).

Question. Let * be the binary operation on Z given by a*b = a + b-15. 1) Is * commutative? 2) Is * associative 3) Does the identity for *exist? If yes find the identity. 4) Are the elements of Z invertible? If so find the inverse.
Answer: Let \( * \) be the binary operation on the set of integers \( \mathbb{Z} \) defined by \( a * b = a + b - 15 \).
1) Commutativity:
For any \( a, b \in \mathbb{Z} \):
\( a * b = a + b - 15 \)
\( b * a = b + a - 15 = a + b - 15 \) (since addition on \( \mathbb{Z} \) is commutative).
Since \( a * b = b * a \) for all \( a, b \in \mathbb{Z} \), \( * \) is commutative.

2) Associativity:
For any \( a, b, c \in \mathbb{Z} \):
\( (a * b) * c = (a + b - 15) * c = (a + b - 15) + c - 15 = a + b + c - 30 \)
\( a * (b * c) = a * (b + c - 15) = a + (b + c - 15) - 15 = a + b + c - 30 \)
Since \( (a * b) * c = a * (b * c) \), the operation \( * \) is associative.

3) Existence of Identity:
Let \( e \in \mathbb{Z} \) be the identity element. Then for all \( a \in \mathbb{Z} \):
\( a * e = a \)
\( \Rightarrow a + e - 15 = a \)
\( \Rightarrow e = 15 \)
Since \( 15 \in \mathbb{Z} \), the identity element exists and is \( e = 15 \).

4) Invertibility of Elements:
Let \( a \in \mathbb{Z} \). If \( b \) is the inverse of \( a \), then:
\( a * b = e \)
\( \Rightarrow a + b - 15 = 15 \)
\( \Rightarrow b = 30 - a \)
Since \( a \in \mathbb{Z} \), \( 30 - a \) is also an integer. Hence, every element \( a \in \mathbb{Z} \) is invertible, and its inverse is \( 30 - a \).

Question. Let A = R - {3}, B = R - {1}. Let f: A → B defined by f(x) = (x-2)/(x-3) ∀ x ∈ A. Then show that f is bijective.(Exemplar).
Answer: To show that the function \( f: A \rightarrow B \) is bijective, we must prove that it is both one-one and onto.
1. One-One:
Let \( x_1, x_2 \in A \) such that \( f(x_1) = f(x_2) \).
\( \Rightarrow \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
\( \Rightarrow (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \)
\( \Rightarrow x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \)
\( \Rightarrow -3x_1 - 2x_2 = -3x_2 - 2x_1 \)
\( \Rightarrow -3x_1 + 2x_1 = -3x_2 + 2x_2 \)
\( \Rightarrow -x_1 = -x_2 \)
\( \Rightarrow x_1 = x_2 \).
Therefore, \( f \) is one-one.

2. Onto:
Let \( y \in B \) (co-domain, i.e., \( y \in \mathbb{R} \) and \( y \neq 1 \)). We need to find \( x \in A \) such that \( f(x) = y \).
\( y = \frac{x - 2}{x - 3} \)
\( \Rightarrow y(x - 3) = x - 2 \)
\( \Rightarrow xy - 3y = x - 2 \)
\( \Rightarrow xy - x = 3y - 2 \)
\( \Rightarrow x(y - 1) = 3y - 2 \)
Since \( y \neq 1 \), we can divide by \( y - 1 \):
\( x = \frac{3y - 2}{y - 1} \)
Let us check if \( x \in A \) (i.e., \( x \) is a real number and \( x \neq 3 \)).
Since \( y \neq 1 \), \( x \) is a well-defined real number. Let us assume \( x = 3 \):
\( \frac{3y - 2}{y - 1} = 3 \)
\( \Rightarrow 3y - 2 = 3y - 3 \)
\( \Rightarrow -2 = -3 \) (which is impossible).
Thus, \( x \neq 3 \), meaning \( x \in A \).
Therefore, for any \( y \in B \), there exists an \( x = \frac{3y - 2}{y - 1} \in A \) such that:
\( f(x) = f\left(\frac{3y - 2}{y - 1}\right) = \frac{\frac{3y - 2}{y - 1} - 2}{\frac{3y - 2}{y - 1} - 3} = \frac{(3y - 2) - 2(y - 1)}{(3y - 2) - 3(y - 1)} = \frac{3y - 2 - 2y + 2}{3y - 2 - 3y + 3} = \frac{y}{1} = y \).
Hence, \( f \) is onto.

Since the function \( f \) is both one-one and onto, it is bijective.