CBSE Class 12 Mathematics Relations And Functions Worksheet Set 03

Relation and Functions Case Study Questions

CASE STUDY 1:

A general election of Lok Sabha is a gigantic exercise. About 900 million people were eligible to vote and voter turnout was about 75%, the highest ever. Let I be the set of all citizens of India who were eligible to exercise their voting right in general election held in 2019. A relation ‘R’ is defined on I as follows:
R = {(𝑉1,2)∶ 𝑉1,𝑉2 ∈𝐼 and both use their voting right in general election – 2019}. Based on given information, answer the following questions
Q1) Is R a reflexive relation? Justify your answer. 
Q2) Is R a symmetric relation? Justify your answer. 
Q3) Is R a transitive relation? Justify your answer. 
Q4) Is R an equivalence relation? Justify your answer. 

CASE STUDY 2:

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 i.e. A ={S, D} and B = {1,2,3,4,5,6}, based on given information, answer the following questions.
Q1) Let R : B ⟶ B be defined by R = {(x, y) : y is divisible by x}. Is R an equivalence relation? Justify your answer.
Q2 ) Raji wants to know the number of functions from A to B. How many number of functions are possible?

CASE STUDY 3:

An organization conducted bike race under 2 different categories-boys and girls. Totally there were 250 participants. Among all of them finally three from Category 1 and two from Category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project. Let G={g₁,g₂} and B = {b₁,b₂,b₃} where B represents the set of boys selected and G the set of girls who were selected for the final race. based on given information, answer the following questions.
Q1) Ravi wishes to form all the relations possible from B to G. How many such relations are possible? 
Q2) Ravi wants to know among those relations, how many functions can be formed from B to G? 
Q3) Let R: B→B be defined by R = {(𝑥, 𝑦): 𝑥 and y are students of same sex}, then verify whether R is an equivalence relation? 

CASE STUDY 4:

Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by 𝑦 = 𝑥². Answer the following questions using the above information.
Q1) Let 𝑓: 𝑅→𝑅 be defined by (𝑥)= 𝑥², then verify whether 𝑓 is a bijective function? 
Q2) Let 𝑓: N→R be defined by (𝑥)= 𝑥², then find the range of R? 
Q3) Let 𝑓: Z→R be defined by (𝑥)= 𝑥², Is f Injective function? Justify your answer?

Question. A relation R on set A = {1, 2, 3, 4, 5} is defined as R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)} then R is _______________ relation
a) Reflexive
b) Symmetric
c) Transitive
d) Equivalence.
Answer : D

Question. Let R be the relation on set A = { x ∈ Z : x ≤ 20}, defined by R = {(a, b) : Ia - bI is a multiple of 3}, then [4], the equivalence class of 4, is
a) {0, 4, 8, 12, 16, 20}
b) {1, 4, 7, 10, 13, 16, 19}
c) {0, 1, 4, 7, 10, 13, 16, 19}
d) A
Answer : B

Question. If a relation R on the set {a, b, c, d} is defined as R = {(a, b)}, then R is _______________ relation
a) Reflexive
b) Symmetric
c) Transitive
d) simply a relation
Answer : C

Question. The function 𝑓:R → Z, defined as 𝑓(𝑥) = [𝑥] is (Z is set of integers)
a) neither one – one nor onto
b) one – one but not onto
c) onto but not one – one
d) one – one and onto
Answer : C

Question. The function 𝑓(𝑥) = 5 − |sin(4𝑥)| has maximum value ‘a’ and minimum value ‘b’, then (a, b) =
a) (4, 5)
b) (5, 4)
c) (5, 6)
d) (6, 5)
Answer : B

Question. The function 𝑓: R → R, given by 𝑓(𝑥) = |𝑥| is
a) Surjective
b) Injective
c) Bijective
d) neither surjective nor injective.
Answer : D

Question. A relation R on set A = {a, b, c} is defined as R = {(a, b), (b, b)} then R will be _______________ relation when (b, a) will be added
a) Reflexive
b) Symmetric
c) Transitive
d) Equivalence
Answer : B

Question. The function 𝑓: R → R, given by 𝑓(𝑥) = 3𝑥 + 2 is
a) Surjective
b) Injective
c) Bijective
d) neither surjective nor injective.
Answer : C

Question. The maximum number of equivalence relation on the set A = {a, b, c} are
a) 2
b) 3
c) 5
d) 6
Answer : D

Question. A relation R on set A = {a, b, c} is defined as R = {(a, b), (b, b), (c,c), (a, a)} then R will be _______________ relation when (b, a) will be added
a) Reflexive
b) Symmetric
c) Transitive
d) Equivalence.
Answer : D

Question. The function 𝑓:R → R, defined as 𝑓(𝑥) = [𝑥] + x is
a) neither one – one nor onto
b) one – one but not onto
c) onto but not one – one
d) one – one and onto
Answer : D

Question. Show that the number of equivalence relation in the set {1,2,3} containing (1 , 2) and (2 , 1) is two.
Answer: \( A = \{1, 2, 3\} \)
The maximum possible relation (i.e. universal relation) is
\( R = \{(1,1), (2,2), (3,3), (1,2), (1,3), (2,1), (2,3), (3,1), (3,2)\} \)
The smallest equivalence relation \( R_1 \) containing \( (1 , 2) \) and \( (2 , 1) \) is
\( R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\} \)
we are left with four pairs (from universal relation) i.e. \( (2,3), (3,2), (1,3) \) and \( (3,1) \)
If we add \( (2,3) \) to \( R_1 \), then for symmetric by we must add \( (3,2) \) and now for transitivity we are forced to add \( (1,3) \) and \( (3,1) \)
Thus the only relation bigger than \( R_1 \) is universal relation i.e \( R \)
\( \therefore \) The no. of equivalence relations containing \( (1,2) \) and \( (2,1) \) is two.

Question. If \( R = \{(x , y) : x^2 + y^2 \leq 4 ; x , y \in Z\} \) is a relation on \( Z \). Write the domain of R.
Answer: \( R = \{(0,1), (0,-1), (0,2), (0,-2), (1,1), (1,-1), (-1,0), (-1,1), (-1,-1), (2,0), (-2,0)\} \)
\( \therefore \) Domain of \( R = \{0, 1, -1, 2, -2\} \)
(i.e the first domain of each ordered pairs)

Question. Let R =\(\{ (x, y): |x^2 - y^2| < 1 \}\) be a relation on set A = {1,2,3,4,5}. Write R as a set of ordered pairs.
Answer: \( A = \{1,2,3,4,5\} \)
for \( |x^2 - y^2| < 1 \): \( x \) should be equal to \( y \)
\( \therefore R = \{(1,1), (2,2), (3,3), (4,4), (5,5)\} \)

Question. R is a relation in Z defined as \( (a , b) \in R \iff a^2 + b^2 = 25 \). Find the range.
Answer: We have, \( a^2 + b^2 = 25 \) and \( a, b \in Z \)
\( \therefore R = \{(0,5), (0,-5), (3,4), (3,-4), (-3,4), (-3,-4), (4,3), (4,-3), (-4,3), (-4,-3), (5,0), (-5,0)\} \)
\( \therefore \) Range = \( \{-5, 5, 4, -4, 3, -3, 0\} \)
(i.e. second elements of each order pairs)

Question. * : \( R \times R \rightarrow R \)
(1) a * b = a + b (2) a * b = a - b (3) a * b = ab
Find identity element and inverse in both cases.
Answer: (1) a * b = a + b
Identity element
\( a * e = a \quad e * a = a \)
\( \implies a + e = a \quad e + a = a \)
\( \implies e = 0 \in R \quad e = 0 \in R \)
\( \therefore e = 0 \) is the identity element
Inverse
\( a * b = e \)
\( \implies a + b = 0 \)
\( \implies b = -a \in R \quad \{ a \in R \implies -a \text{ also } \in R \} \)
\( \therefore -a \) is the inverse of a i.e \( a^{-1} = -a \)

(2) a * b = a - b
Identity element
\( a * e = a \quad e * a = a \)
\( a - e = a \quad e - a = a \)
\( -e = 0 \quad e = 2a \)
\( e = 0 \in R \) but 'e' cannot be a variable as when a changes e also change but should be same for all \( a \in R \)
\( \therefore \) Identity element does not exist
hence inverse does not exist

(3) a * b = ab
Identity element
\( a * e = a \quad e * a = a \)
\( ae = a \quad ea = a \)
\( e = 1 \in R \quad e = 1 \in R \)
\( \therefore 1 \) is the identity element
Inverse
\( a * b = e \)
\( ab = 1 \)
\( b = \frac{1}{a} \in R; a \neq 0 \)
\( \therefore \) all elements of R are invertible except '0' and \( a^{-1} = \frac{1}{a}; a \neq 0 \)

Question. Let * be a binary operation on R (real no's)
* : \( R \times R \rightarrow R \)
\( a * b = a + b + ab \)
(.) Check whether * is binary operation or not
(.) Check the commutativity and Associativity
(.) Find identity element and inverse.
Answer: We have,
\( a * b = a + b + ab \)
since * carries each pair (a, b) in \( R \times R \) to a unique element \( a + b + ab \) in R
\( \therefore * \) is a binary operation on R

Alternate : since \( (a, b) \in R \times R \) and addition and multiplication of real no.s is also a real no.
\( a + b + ab \in R \)
\( \therefore * \) is a binary operation on R

Commutative :
let \( a, b \in R \)
\( a * b = a + b + ab \)
\( b * a = b + a + ba \)
\( = a + b + ab \) ...{\( \therefore \) addition and multiplication are itself commutative}
\( = a * b \)
\( \therefore b * a = a * b \) for all \( a, b \in R \)
\( \therefore * \) is commutative on R

Associative:
let \( a, b, c \in R \)
\( (a * b) * c = (a + b + ab) * c \)
\( = a + b + ab + c + (a + b + ab)c \)
\( = a + b + ab + c + ac + bc + abc \)
\( = a + b + c + ab + bc + ac + abc \)
Now \( a * (b * c) = a * (b + c + bc) \)
\( = a + b + c + bc + a(b + c + bc) \)
\( = a + b + c + bc + ab + ac + abc \)
\( = a + b + c + ab + bc + ac + abc \)
clearly \( (a * b) * c = a * (b * c) \) for all \( a, b, c \in R \)
\( \therefore * \) is Associative on R.

Identity element :
let e be the identity element in R
\( a * e = a \) and \( e * a = a \) for all \( a \in R \)
\( \implies a + e + ae = a \quad e + a + ea = a \)
\( \implies e(1 + a) = 0 \quad e(1 + a) = 0 \)
\( \implies e = 0 \in R \quad e = 0 \in R \)
\( \therefore 0 \) is the identity element

Inverse :
\( a * b = e \)
\( a + b + ab = 0 \)
\( b(1 + a) = -a \)
\( b = \frac{-a}{1+a} \in R \) {except a = -1}
\( \therefore \) -1 is not the invertible element
(.) all elements of R are invertible except -1
(.) and \( a^{-1} = \frac{-a}{1+a}; a \neq -1 \)
(.) e.g. inverse of 2 = \( \frac{-2}{1+2} = \frac{-2}{3} \)

Question. Let * be a binary operation on Z (integers) \( a * b = a + ab \)
Check the commutative, Associativity, identify element and inverse (if it exists).
Answer: We have
\( a * b = a + ab \) where \( a, b \in Z \)
Commutative :
let \( a, b \in Z \), then
\( a * b = a + ab \)
\( b * a = b + ba \)
\( = b + ab \)
\( b * a \neq a * b \)
e.g. \( (1 * 2) = 1 + (1)(2) = 1 + 2 = 3 \)
\( (2 * 1) = 2 + 2(1) = 2 + 2 = 4 \)
clearly \( 1 * 2 \neq 2 * 1 \)
\( \therefore * \) is not commutative on Z

Associative :
let \( a, b, c \in Z \) then
\( (a * b) * c = (a + ab) * c \)
\( = a + ab + (a + ab)c \)
\( = a + ab + ac + abc \)
\( a * (b * c) = a * (b + bc) \)
\( = a + a(b + bc) \)
\( = a + ab + abc \)
\( \neq (a * b) * c \)
e.g. \( (1 * 2) * 3 = (1 + 2) * 3 \)
\( = 3 * 3 \)
\( = 3 + 3(3) = 12 \)
\( 1 * (2 * 3) = 1 * (2 + 6) = 1 * 8 \)
\( = 1 + 1(8) = 9 \)
Clearly * is not Associative on Z

Identity element:
let e be the identity element \( \in Z \), then
\( a * e = a \quad e * a = a \)
\( a + ae = a \quad e + ea = a \)
\( ae = 0 \quad e(1 + a) = a \)
\( e = 0 \in Z \quad e = \frac{a}{1+a} \)
as a changes e also changes, but e must be constant for all value of a
\( \therefore \) identity element does not exist and hence inverse not possible.

Question. Let * be a binary operation on N given by a * b = LCM of a & b
(1) Find 5 * 7 , 20 * 16
(2) Is * commutative ?
(3) If * Associative ?
(4) Find the identity element in N.
(5) which elements of N are invertible ?
Answer: We have
\( a * b = \text{LCM of } a \text{ & } b; a, b \in N \)
(1) \( 5 * 7 = \text{LCM of } 5 \text{ and } 7 = 35 \)
\( 20 * 16 = \text{LCM of } 20 \text{ and } 16 = 80 \)
(2) Commutative :
let \( a, b \in N \)
\( a * b = \text{LCM of } a \text{ and } b \)
\( b * a = \text{LCM of } b \text{ and } a \)
\( = \text{LCM of } a \text{ and } b \)
\( = a * b \)
\( \therefore b * a = a * b \) for all \( a, b \in N \)
\( \therefore * \) is commutative on N
(3) Associative :
let \( a, b, c \in N \)
\( (a * b) * c = (\text{LCM of } a \text{ and } b) * c \)
\( = \text{LCM of } [(\text{LCM of } a \text{ and } b) \text{ and } c] \)
\( = \text{LCM of } a, b \text{ and } c \)
\( a * (b * c) = a * (\text{LCM of } b \text{ and } c) \)
\( = \text{LCM of } [a \text{ and } (\text{LCM of } b \text{ and } c)] \)
\( = \text{LCM of } a, b \text{ and } c \)
clearly \( (a * b) * c = a * (b * c) \) for all \( a, b, c \in N \)
\( \therefore * \) is Associative on N
(4) Identity element
let e be an identity element \( \in N \)
\( a * e = a \)
\( \implies \text{LCM of } a \text{ and } e = a \quad e * a = a \)
\( \implies \text{LCM of } a \text{ & } 1 = a \quad \text{LCM of } e \text{ and } a = a \)
\( \implies e = 1 \in N \quad \text{LCM of } 1 \text{ and } a = a \)
\( \implies e = 1 \in N \)
\( \dots 1 \) is the identity element for all \( a \in N \)
(5) Inverse
\( a * b = e \)
\( \implies \text{LCM of } a \text{ and } b = 1 \)
this is possible only when \( a = 1 \) & \( b = 1 \)
\( \therefore 1 \) is the only invertible element and 1 is its inverse

Question. Let R be a of real no.s and \( A = R \times R \) is a binary operation on A given by \( (a , b) * (c , d) = (ac , bd) \) for all \( (a , b), (c , d) \in A \)
(1) Show that * is Commutative
(2) Show that * is Associative
(3) Find the identity element
(4) Find invertible elements and their inverse.
Answer: We have,
\( (a , b) * (c , d) = (ac , bd) \)
Commutative :
let \( (a, b) \) & \( (c, d) \in A \), then
\( (a, b) * (c, d) = (ac, bd) \)
\( (c, d) * (a, b) = (ca, db) \)
\( = (ac, bd) \)
\( = (a, b) * (c, d) \)
\( \therefore * \) is commutative on A

Associative :
let \( (a, b), (c, d) \) & \( (e, f) \in A \)
\( [(a, b) * (c, d)] * (e, f) \)
\( = (ac, bd) * (e, f) \)
\( = (ace, bdf) \)
\( (a, b) * [(c, d) * (e, f)] \)
\( = (a, b) * (ce, df) \)
\( = (ace, bdf) \)
clearly \( ((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d) * (e, f)) \)
\( \therefore * \) is Associative on A

Identity element
let \( (x, y) \) be the identity element
\( (a, b) * (x, y) = (a, b) \quad (x, y) * (a, b) = (a, b) \)
\( \implies (ax, by) = (a, b) \quad \implies (xa, yb) = (a, b) \)
\( \implies ax = a \) & \( by = b \quad \implies xa = a \) & \( yb = b \)
\( \implies x = 1 \) and \( y = 1 \quad \implies x = 1 \) & \( y = 1 \)
\( \therefore (1, 1) \) is the identity element

Inverse
\( (a, b) * (c, d) = (x, y) \)
\( \implies (ac, bd) = (1, 1) \)
\( \implies ac = 1 \) and \( bd = 1 \)
\( \implies c = \frac{1}{a} \) and \( d = \frac{1}{b} \)
\( \therefore (c, d) = \left(\frac{1}{a}, \frac{1}{b}\right) \in R \) except \( (0, b) = (0, 0) \)
(.) all elements of A are invertible except \( (0, 0) \)
(.) inverse of \( (a, b) \) is \( \left(\frac{1}{a}, \frac{1}{b}\right); (a, b) \neq (0, 0) \)

Question. X is a non-empty set and * is a binary operation \( * : P(X) \times P(X) \rightarrow P(X) \) given by \( A * B = A \cap B \)
(.) Show * is Commutative
(.) Show * is Associative
(.) Find the Identity element
(.) Find the Invertible elements in \( P(X) \) and their inverse.
Answer: We have, \( A * B = A \cap B \)
Commutative :
let \( A, B \in P(X) \)
\( A * B = A \cap B \)
\( B * A = B \cap A \)
\( = A \cap B \) ...{\( \cap \) is commutative}
clearly \( A * B = B * A \) for all \( A, B \in P(X) \)
\( \therefore * \) is commutative on P(x)

Associative :
let \( A, B, C \in P(X) \)
\( (A * B) * C = (A \cap B) * C \)
\( = (A \cap B) \cap C \)
\( A * (B * C) = A * (B \cap C) \)
\( = A \cap (B \cap C) \) ...{\( \cap \) is itself Associative}
clearly \( (A * B) * C = A * (B * C) \) for all \( A, B, C \in P(X) \)
\( \therefore * \) is Associative on P(x)

Identity element
let E is an identity element then
\( A * E = A \quad E * A = A \)
\( \implies A \cap E = A \quad \implies E \cap A = A \)
\( \implies E = X \in P(X) \quad \implies E = X \in P(X) \) {reason : X is the largest subset in P(x)}
\( \therefore X \) is the identity element

Inverse :
\( A * B = E \)
\( \implies A \cap B = X \)
this is possible only when \( A = B = X \)
since \( X \cap X = X \)
\( \therefore X \) is only the invertible element in \( P(X) \) and X is its inverse

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