CBSE Class 12 Mathematics Relations and Functions Case Studies

Read the following and answer any four questions from (i) to (v).

A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, 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: \( R = \{(V_1, V_2) : V_1, V_2 \in I \text{ and both use their voting right in general election – 2019}\} \)

Question. Two neighbours \( X \) and \( Y \in I \). \( X \) exercised his voting right while \( Y \) did not cast her vote in general election – 2019. Which of the following is true?
(a) \( (X, Y) \in R \)
(b) \( (Y, X) \in R \)
(c) \( (X, X) \notin R \)
(d) \( (X, Y) \notin R \)
Answer: (d) Since \( X \) exercised his voting right while \( Y \) did not cast her vote in general election – 2019. Therefore, \( (X, Y) \notin R \).

Question. Mr.‘X’ and his wife ‘W’ both exercised their voting right in general election -2019, Which of the following is true?
(a) both \( (X,W) \) and \( (W,X) \in R \)
(b) \( (X,W) \in R \) but \( (W,X) \notin R \)
(c) both \( (X,W) \) and \( (W,X) \notin R \)
(d) \( (W,X) \in R \) but \( (X,W) \notin R \)
Answer: (a) Since Mr. 'X' and his wife 'W' both exercised their voting right in general election – 2019, both \( (X, W) \) and \( (W, X) \in R \).

Question. Three friends \( F_1, F_2 \) and \( F_3 \) exercised their voting right in general election- 2019, then which of the following is true?
(a) \( (F_1, F_2 ) \in R, (F_2, F_3) \in R \text{ and } (F_1, F_3) \in R \)
(b) \( (F_1, F_2 ) \in R, (F_2, F_3) \in R \text{ and } (F_1, F_3) \notin R \)
(c) \( (F_1, F_2 ) \in R, (F_2, F_2) \in R \text{ but } (F_3, F_3) \notin R \)
(d) \( (F_1, F_2 ) \notin R, (F_2, F_3) \notin R \text{ and } (F_1, F_3) \notin R \)
Answer: (a) Since three friends \( F_1, F_2 \) and \( F_3 \) exercised their voting right in general election – 2019, therefore \( (F_1, F_2) \in R, (F_2, F_3) \in R \text{ and } (F_1, F_3) \in R \).

Question. The above defined relation \( R \) is
(a) Symmetric and transitive but not reflexive
(b) Universal relation
(c) Equivalence relation
(d) Reflexive but not symmetric and transitive
Answer: (c) This relation is an equivalence relation.

Question. Mr. Shyam exercised his voting right in General Election – 2019, then Mr. Shyam is related to which of the following?
(a) All those eligible voters who cast their votes
(b) Family members of Mr. Shyam
(c) All citizens of India
(d) Eligible voters of India
Answer: (a) Mr. Shyam exercised his voting right in General election – 2019, then Mr. Shyam is related to all those eligible votes who cast their votes.

Read the following and answer any four questions from (i) to (v). 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. \( A = \{S, D\}, B = \{1,2,3,4,5,6\} \)

Question. Let \( R : B \rightarrow B \) be defined by \( R = \{(x, y): y \text{ is divisible by } x\} \) 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) Reflexive: Let \( x \in B \), since \( x \) always divide \( x \) itself. \( \therefore (x, x) \in R \). It is reflexive. Symmetric: Let \( x, y \in B \) and let \( (x, y) \in R \). \( \Rightarrow y \) is divisible by \( x \). \( \Rightarrow \frac{y}{x} = k_1 \), where \( k_1 \) is an integer. \( \Rightarrow \frac{x}{y} = \frac{1}{k_1} \neq \text{integer} \). \( \therefore (y, x) \notin R \). It is not symmetric. Transitive: Let \( x, y, z \in B \) and let \( (x, y) \in R \Rightarrow \frac{y}{x} = k_1 \), where \( k_1 \) is an integer. and, \( (y, z) \in R \Rightarrow \frac{z}{y} = k_2 \), where \( k_2 \) is an integer. \( \therefore \frac{y}{x} \times \frac{z}{y} = k_1 \cdot k_2 = k \text{ (integer)} \). \( \Rightarrow \frac{z}{x} = k \Rightarrow (x, z) \in R \). It is transitive. Hence, relation is reflexive and transitive but not symmetric.

Question. Raji wants to know the number of functions from \( A \) to \( B \). How many number of functions are possible?
(a) \( 6^2 \)
(b) \( 2^6 \)
(c) \( 6! \)
(d) \( 2^{12} \)
Answer: (a) We have, \( A = \{ S, D\} \Rightarrow n(A) = 2 \) and, \( B = \{1, 2, 3, 4, 5, 6\} \Rightarrow n(B) = 6 \). Number of functions from \( A \) to \( B \) is \( 6^2 \).

Question. 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) \( R \) is not reflexive since \( (1, 1), (3, 3), (4, 4) \notin R \). \( R \) is not symmetric as \( (1, 2) \in R \) but \( (2, 1) \notin R \). and, \( R \) is not transitive as \( (1, 3) \in R \) and \( (3, 1) \in R \) but \( (1, 1) \notin R \). \( \therefore R \) is neither reflexive nor symmetric nor transitive.

Question. Raji wants to know the number of relations possible from \( A \) to \( B \). How many numbers of relations are possible?
(a) \( 6^2 \)
(b) \( 2^6 \)
(c) \( 6! \)
(d) \( 2^{12} \)
Answer: (d) Total number of possible relations from \( A \) to \( B = 2^{12} \).

Question. Let \( R : B \rightarrow B \) 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) \( R \) is reflexive as each elements of \( B \) is related to itself and \( R \) is also transitive as \( (1, 2) \in R \) and \( (2, 2) \in R \Rightarrow (1, 2) \in R \). \( \therefore R \) is reflexive and transitive.

Read the following and answer any four questions from (i) to (v). An organization conducted bike race under 2 different categories-boys and girls. In all, 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 \( B = \{b_1, b_2, b_3\} \), \( G = \{g_1, g_2\} \) where \( B \) represents the set of boys selected and \( G \) the set of girls who were selected for the final race.

Question. Ravi wishes to form all the relations possible from \( B \) to \( G \). How many such relations are possible?
(a) \( 2^6 \)
(b) \( 2^5 \)
(c) 0
(d) \( 2^3 \)
Answer: (a) We have sets \( B = \{b_1, b_2, b_3\}, G = \{g_1, g_2\} \Rightarrow n(B) = 3 \text{ and } n(G) = 2 \). Number of all possible relations from \( B \) to \( G = 2^{3 \times 2} = 2^6 \).

Question. Let \( R : B \rightarrow B \) be defined by \( R = \{(x, y) : x \text{ and } y \text{ are students of same sex}\} \), Then this relation \( R \) is
(a) Equivalence
(b) Reflexive only
(c) Reflexive and symmetric but not transitive
(d) Reflexive and transitive but not symmetric
Answer: (a) On the set \( B \). Since the set is \( B = \{b_1, b_2, b_3\} = \text{all boys} \), \( \therefore \) It is an equivalence relation.

Question. Ravi wants to know among those relations, how many functions can be formed from \( B \) to \( G \)?
(a) \( 2^2 \)
(b) \( 2^{12} \)
(c) \( 3^2 \)
(d) \( 2^3 \)
Answer: (d) We have, \( B = \{b_1, b_2, b_3\} \Rightarrow n(B) = 3 \) and \( G = \{g_1, g_2\} \Rightarrow n(G) = 2 \). \( \therefore \) Total no. of possible functions from \( B \) to \( G = 2^3 \).

Question. Let \( R : B \rightarrow G \) be defined by \( R = \{ (b_1, g_1), (b_2, g_2), (b_3, g_1)\} \), then \( R \) is
(a) Injective
(b) Surjective
(c) Neither Surjective nor Injective
(d) Surjective and Injective
Answer: (b) We have, \( R : B \rightarrow G \) be defined by \( R = \{(b_1, g_1), (b_2, g_2), (b_3, g_1)\} \). It is not injective because \( (b_1, g_1) \in R \) and \( (b_3, g_1) \in R \). So \( b_1 \neq b_3 \Rightarrow \text{same image } g_1 \). It is surjective because its Co-domain = Range. \( \therefore R \) is surjective.

Question. Ravi wants to find the number of injective functions from \( B \) to \( G \). How many numbers of injective functions are possible?
(a) 0
(b) \( 2! \)
(c) \( 3! \)
(d) \( 0! \)
Answer: (a) Since \( R \) is not injective therefore number of injective functions = 0.

Read the following and answer any four questions from (i) to (v). Students of Grade 9, planned to plant saplings along straight lines, parallel to each other to one side of the playground ensuring that they had enough play area. Let us assume that they planted one of the rows of the saplings along the line \( y = x - 4 \). Let \( L \) be the set of all lines which are parallel on the ground and \( R \) be a relation on \( L \).

Question. Let relation \( R \) be defined by \( R = \{(L_1, L_2): L_1 \parallel L_2 \text{ where } L_1, L_2 \in L\} \) then \( R \) is______ relation
(a) Equivalence
(b) Only reflexive
(c) Not reflexive
(d) Symmetric but not transitive
Answer: (a) Reflexive: Let \( L_1 \in L \Rightarrow L_1 \parallel L_1 \Rightarrow (L_1, L_1) \in R \). It is reflexive. Symmetric: Let \( L_1, L_2 \in L \) and let \( (L_1, L_2) \in R \Rightarrow L_1 \parallel L_2 \Rightarrow L_2 \parallel L_1 \Rightarrow (L_2, L_1 ) \in R \). It is symmetric. Transitive: Let \( L_1, L_2, L_3 \in L \) and let \( (L_1, L_2) \in R \) and \( (L_2, L_3) \in R \Rightarrow L_1 \parallel L_2 \text{ and } L_2 \parallel L_3 \Rightarrow L_1 \parallel L_3 \Rightarrow (L_1, L_3) \in R \). \( \therefore \) It is transitive. Hence \( R \) is an equivalence relation.

Question. Let \( R = \{(L_1, L_2): L_1 \perp L_2 \text{ where } L_1, L_2 \in L\} \) which of the following is true?
(a) \( R \) is Symmetric but neither reflexive nor transitive.
(b) \( R \) is Reflexive and transitive but not symmetric.
(c) \( R \) is Reflexive but neither Symmetric nor transitive.
(d) \( R \) is an Equivalence relation.
Answer: (a) Reflexive: Since every line is not perpendicular to itself. \( \therefore (L_1, L_1) \notin R \). It is not reflexive. Symmetric: Let \( L_1, L_2 \in L \) and let \( (L_1, L_2) \in R \Rightarrow L_1 \perp L_2 \Rightarrow L_2 \perp L_1 \Rightarrow (L_2, L_1 ) \in R \). It is symmetric. Transitive: Let \( L_1, L_2, L_3 \in L \) and, let \( (L_1, L_2) \in R \) and \( (L_2, L_3) \in R \Rightarrow L_1 \perp L_2 \text{ and } L_2 \perp L_3 \Rightarrow L_1 \parallel L_3 \Rightarrow (L_1, L_3) \notin R \). \( \therefore \) It is not transitive. Hence relation \( R \) is symmetric but neither reflexive nor transitive.

Question. The function \( f : R \rightarrow R \) defined by \( f(x) = x − 4 \) is
(a) Bijective
(b) Surjective but not injective
(c) Injective but not Surjective
(d) Neither Surjective nor Injective
Answer: (a) Injective: Let \( x_1, x_2 \in R \) such that \( x_1 \neq x_2 \Rightarrow x_1 - 4 \neq x_2 - 4 \Rightarrow f(x_1) \neq f(x_2) \). It is injective. Surjective: Let \( y = x - 4 \Rightarrow x = y + 4 \), for every \( y \in R \) there exists \( x \in R \). i.e, Co-domain = Range. \( \therefore \) It is surjective. Hence given function is bijective.

Question. Let \( f : R \rightarrow R \) be defined by \( f(x) = x − 4 \). Then the range of \( f(x) \) is
(a) \( R \)
(b) \( Z \)
(c) \( W \)
(d) \( Q \)
Answer: (a) Let \( y = f(x) \Rightarrow y = x - 4 \Rightarrow x = y + 4 \Rightarrow x \in R \Rightarrow y \in R \). \( \therefore \) Range of \( f(x) \) is \( R \) (Set of real numbers).

Question. Let \( R = \{(L_1 , L_2 ) : L_1 \text{ is parallel to } L_2 \text{ and } L_1 : y = x – 4\} \) then which of the following can be taken as \( L_2 \)?
(a) \( 2x – 2y + 5 = 0 \)
(b) \( 2x + y = 5 \)
(c) \( 2x + 2y + 7 = 0 \)
(d) \( x + y = 7 \)
Answer: (a) Option (a) is correct choice, because the equation of line \( 2x – 2y + 5 = 0 \) is parallel to \( y = x – 4 \).