{"id":170496,"date":"2024-05-25T12:05:56","date_gmt":"2024-05-25T06:35:56","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170496"},"modified":"2024-05-25T12:06:43","modified_gmt":"2024-05-25T06:36:43","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-ex-2-2\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.2"},"content":{"rendered":"

Students can cross-reference their work with ML Aggarwal Class 11 Solutions<\/a> Chapter 2 Relations and Functions Ex 2.2 to ensure accuracy.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.2<\/h2>\n

Question 1.
\nIf A and B are two sets such that n(A) = 2 and n(B) = 3, find the number of relations from
\n(i) A to B
\n(ii) B to A
\n(iii) A to A.
\nSolution:
\n(i) Given n (A) = 2 ;
\nn (B) = 3
\nsome relation R be a subset of A \u00d7 B.
\n\u2234 n (A \u00d7 B) = n (A) \u00d7 n (B)
\n= 2 \u00d7 36
\nThus total number of relations from A to B = 2n (A \u00d7 B)<\/sup>
\n= 26<\/sup>
\n= 64<\/p>\n

(ii) Thus, total number of relations from B to A = 2n (B \u00d7 A)<\/sup>
\n= 2n (B) \u00d7 n (A)<\/sup>
\n= 23 \u00d7 2<\/sup>
\n= 26<\/sup> = 64<\/p>\n

(iii) Thus, total no. of relation from A to A = 2n (A \u00d7 A)<\/sup>
\n= 2n (A) \u00d7 n (A)<\/sup>
\n= 22 \u00d7 2<\/sup>
\n= 24<\/sup> = 16<\/p>\n

\"ML<\/p>\n

Question 2.
\nLet A = {1, 2} and B = {3, 4}. Find
\n(i) A \u00d7 B
\n(ii) number of relations from A to B.
\nSolution:
\nGiven A = {1, 2)
\nand B = {3, 4}
\n(i) \u2234 A \u00d7 B = {(1, 3), (1, 4), (2, 3), (2, 4)}<\/p>\n

(ii) Total no. of relati\u00f8n from A to B = 2n (A) \u00d7 n (B)<\/sup>
\n= 22 \u00d7 2<\/sup>
\n= 24<\/sup> = 16<\/p>\n

Question 3.
\nIf a relation R = {(- 2, 1), (0, 2), (3, 1), (0, – 1), (4, 2), (5, 1)}, then write its domain and range.
\nSolu
\nGiven R = {(- 2, 1), (0, 2), (3, 1), (0, – 1), (4, 2), (5, 1)}
\nThen domain of R is the first element of each ordered pair and range of R is the second element of each ordered pair.
\nThus, domain of R = {- 2, 0, 3, 4, 5}
\nRange (R) = {1, 2, – 1}<\/p>\n

Question 4.
\nIf A = {2, 3, 5}, B = {2, 4, 6} and R is the relation from A to B defined by R = {(x, y) : x \u2208 A, y \u2208 B and x < y},then write R in the roster form.
\nSolution:
\nGiven A = {2, 3, 5}
\nand B = {2, 4, 6}
\nGiven R = {(x, y} : x \u2208 A, y \u2208 B and x < y}
\nsince 2 < 4
\n\u2234 (2, 4) \u2208 R ;
\n5 < 6 \u21d2 (5, 6) \u2208 R
\nsince 2 < 6
\n\u2234 (2, 6) \u2208 R
\n3 < 4
\n\u2234 (3, 4) \u2208 R
\n3 < 6
\n\u2234 (3, 6) \u2208 R
\nThus R = {(2,4),(2, 6) (3, 4), (3, 6), (5, 6)}.<\/p>\n

\"ML<\/p>\n

Question 5.
\nIf A = {1, 3, 5, 7, 8} and B = {2,3, 4, 6, 8, 10} and R be the relation is one less than\u2019 from A to B, then write R in the roster form.
\nSolution:
\nGiven A = {1, 3, 5, 7, 8}
\nand B = {2, 3, 4, 6, 8, 10}
\ngiven R be the relation is one less than from A to B
\nsince 1 < 2 = 1 + 1 ;
\n3 < 3 + 1 = 4 ;
\n5 < 5 + 1 = 6 ;
\n7 < 7 + 1 = 8
\n\u2234 R = {(1, 2), (3, 4), (5, 6), (7, 8)}.<\/p>\n

Question 6.
\nIf A = {2, 3, 4}, B = {4, 6, 9, 10} and R = {(x, y) : (x, y) \u2208 A \u00d7 B such that x is a factor of y}, then write R in roster form.
\nSolution:
\nGiven A = {2, 3, 4}
\nand B = {4, 6, 9, 10}
\nand R = {(x, y) : (x, y) \u2208 A \u00d7 B s.t. x is a factor of y}
\nsince 2\/4, 2\/6, 2\/10, 3\/6, 3\/9, 4\/4
\nThus R = {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}.<\/p>\n

Short Answer questions (7 to 11) :<\/p>\n

Question 7.
\nIf A = (1, 2, 3, ………………, 17} and R is a relation on A defined by R ={(x, y) : 3x – y = 0, x, y \u2208 A), then write R \u00a1n the roster form.
\nSolution:
\nGiven A = {1, 2, 3, …………….., 17}
\nand R be a relation defined by
\nR = {(x, y) : 3x – y = 0, x, y \u2208 A}
\nsince 3x – y = 0
\n\u21d2 y = 3x
\nWhen x = 1
\n\u21d2 y = 3 \u00d7 1 = 3 ;
\nWhen x = 2
\n\u21d2 y = 3 \u00d7 2 = 6 ;
\nWhen x = 3
\n\u21d2 y = 3 \u00d7 3 = 9 ;
\nWhen x = 4
\n\u21d2 y = 3 \u00d7 4 = 12 ;
\nWhen x = 5
\n\u21d2 y = 3 \u00d7 5 = 15 ;
\nWhen x = 6
\n\u21d2 y = 3 \u00d7 6 = 18 \u2209 A
\nand for all other values of x \u2208 A gives no value of y that in A.
\n\u2234 R = {(1, 3), (2, 6), (3, 9), (4, 12), (5, 15)}.<\/p>\n

Question 8.
\nWrite the following relations in the roster form :
\n(i) R = {(x, x3<\/sup>) : x is a prime number less than 10}
\n(ii) R={(x – 2, x2<\/sup>) : x is a prime number less than 10}.
\nSolution:
\n(i) Given R = {(x, x3<\/sup>) : x is a prime number less than 10}
\nNow prime number less than 10 are 2, 3, 5, 7
\n\u2234 (2, 8), (3, 27), (5, 125), (7, 343) \u2208 R
\nThus R = {(2, 8), (3, 27), (5, 125), (7, 343)}.<\/p>\n

(ii) Given R = {(x – 2, x2<\/sup>) : x is a prime number less than 10}
\nNow prime numbers less than 10 are 2, 3, 5 and 7.
\nR = {(0, 22<\/sup>), (3 – 2, 32<\/sup>), (5 – 2, 52<\/sup>), (7 – 2, 72<\/sup>)}
\n\u21d2 R = {(0, 4), (1, 9), (3, 25), (5, 49)}<\/p>\n

\"ML<\/p>\n

Question 9.
\nLet a relation R = {(1, – 1), (2, 0), (3, 1), (4, 2), (5, 3)}, then
\n(i) write the domain and the range of R.
\n(ii) write R in the builder form.
\nSolution:
\nGiven relation R = {(1, – 1), (2, 0), (3, 1), (4, 2), (5, 3)}
\n(i) domain of R = {1, 2, 3, 4, 5}
\nand Range of R = {- 1, 0, 1, 2, 3}<\/p>\n

(ii) Clearly x \u2208 N, 1 \u2264 x \u2264 5
\nWhen x = 1, y = – 1
\n\u21d2 y = x – 2
\nWhen x = 2, y = 0
\n\u21d2 y = x – 2
\nsimilarly the result y = x – 2 holds for other pairs of (x, y).
\nThus R = {(x, y) : x \u2208 N, 1 \u2264 x \u2264 5, y = x – 2}
\nbe the relation \u00a1n set builder form.<\/p>\n

Question 10.
\nLet A = {2, 4, 6}, B = {4, 6, 18} and R be the relation \u2018is a factor of’ from A to B. Find R as a set of ordered pairs and represent it by an arrow diagram.
\nSolution:
\nGiven A = {2, 4, 6}
\nand B = {4, 6, 18}
\nand R be the relation from A to B defined by \u2018is a factor of’.
\nsince, 2 be a factor of 4
\ni.e. 2\/4 \u21d2 (2, 4) \u2208 R
\n2\/6 \u21d2 (2, 6) \u2208 R
\n2\/18 \u21d2 (2, 18) \u2208 R
\n4\/4 \u21d2 (4, 4) \u2208 R
\n6\/6 \u21d2 (6, 6) \u2208 R
\nand 6\/18 \u21d2 (6, 18) \u2208 R
\nThus R = {(2, 4), (2, 6), (2, 18), (4, 4), (6, 6), (6, 18)}.<\/p>\n

\"ML<\/p>\n

Question 11.
\nIf R1<\/sub> = {(x, | x |) : x is a real number}, then find the domain and the range of the relation R1<\/sub>. (Exemplar)
\nSolution:
\nSince x can be any real number
\n\u2234 DR1<\/sub><\/sub> = R
\nRange of R1<\/sub> = {|x|, x \u2208 R}
\nSince x \u2265 0 \u2200 x \u2208 R
\n\u2234 Range of R1<\/sub> = [0, \u221e).<\/p>\n

\"ML<\/p>\n

Long answer questions (12 to 21) :<\/p>\n

Question 12.
\nDetermine the domain and the range of the relation R defined by R = {(x, x + 5) : x \u2208 (0, 1, 2, 3, 4, 5)}.
\nSolution:
\nGiven R = {(x, x + 5) : x \u2208 (0, 1, 2, 3, 4, 5)}
\ngiven y = x + 5
\nWhen x = 0
\n\u21d2 y = 0 + 5 = 5
\n\u21d2 (0, 5) \u2208 R
\nWhen x = 1
\n\u21d2 y = 1 + 5 = 6
\n\u21d2(1, 6) \u2208 R
\nWhen x = 2
\n\u21d2 y = 2 + 5 = 7
\n\u21d2 (2, 7) \u2208 R
\nWhen x = 3
\n\u21d2 y = 3 + 5 = 8
\n\u21d2 (3 8) \u2208 R
\nWhen x = 4
\n\u21d2 y = 4 + 5 = 9
\n\u21d2 (4, 9) \u2208 R
\nWhen x = 5
\n\u21d2 y = 5 + 5 = 10
\n\u21d2 (5, 10) \u2208 R
\nThus, R = {(0, 5),(1,6),(2,7), (3, 8), (4, 9), (5, 10)}.
\nHence domain of R = (0, 1, 2, 3, 4, 5)
\nand Range of R = {5, 6, 7, 8, 9, 10}.<\/p>\n

Question 13.
\nThe adjoining diagram shows a relationship between the sets P and Q. Write this relation in
\n(i) roster form
\n(ii) set builder form.
\nWhat is its domain and range?<\/p>\n

\"ML<\/p>\n

Solution:
\n(i) \u2234 R = {(5, 3), (6, 4), (7, 5)}<\/p>\n

(ii) In set builder form,
\nR = {(x, y) ; 5 \u2264 x \u2264 7 ; y = x – 2}
\ngiven y = x – 2
\nWhen x = 5
\n\u21d2 y = 5 – 2 = 3
\n\u21d2 (5, 3) \u2208 R
\nWhen x = 6
\n\u21d2 y = 6 – 2 = 4
\n\u21d2 (6, 4) \u2208 R
\nWhen x = 7
\n\u21d2 y = 7 – 2 = 5
\n\u21d2 (7, 5) \u2208 R
\nDomain of R = {5, 6, 7}
\nand Range of R= {3, 4, 5}<\/p>\n

\"ML<\/p>\n

Question 14.
\nLet R be the relation on N defined by R = {(a, b) : a \u2208 N, b \u2208 N and a +3b = 12}, then
\n(i) List the elements of R
\n(ii) find the domain of R
\n(iii) find the range of R.
\nSolution:
\n(i) Given R be a relation on N defined by
\nR = {(a, b) : a \u2208 N, b \u2208 N and a + 3b = 12}
\nGiven a + 3b = 12
\n\u21d2 b = \\(\\frac{12-a}{3}\\)
\nsince a \u2208 N
\nWhen a = 1
\n\u21d2 b = \\(\\frac{12-1}{3}=\\frac{11}{3}\\) \u2209 N
\nWhen a = 2
\n\u21d2 b = \\(\\frac{10}{3}\\) \u2209 N
\nWhen a = 3
\n\u21d2 b = \\(\\frac{9}{3}\\) = 3 \u2208 N
\n\u21d2 (3, 3) \u2208 N
\nWhen a = 4
\n\u21d2 b = \\(\\frac{12-4}{3}=\\frac{8}{3}\\) \u2209 N
\nWhen a = 5
\n\u21d2 b = \\(\\frac{12-5}{3}=\\frac{7}{3}\\) \u2209 N
\nWhen a = 6
\n\u21d2 b = \\(\\frac{12-6}{3}=\\frac{6}{3}\\) = 2 \u2208 N
\n\u21d2 (6, 2) \u2208 N
\nWhen a = 7
\n\u21d2 b = \\(\\frac{12-7}{3}=\\frac{5}{3}\\) \u2209 N
\nWhen a = 8
\n\u21d2 b = \\(\\frac{12-8}{3}=\\frac{4}{3}\\) \u2209 N
\nWben a = 9
\n\u21d2 b = \\(\\frac{12-9}{3}=\\frac{3}{3}\\) = 1 \u2209 N
\n\u21d2 (9, 1) \u2208 N
\nFor all other values of a, we do not get b \u2208 N
\nThus, R = {(3, 3), (6, 2), (9, 1)}
\n(ii) \u2234 domain of R = {3, 6, 9}
\n(iii) Range of R = {3, 2, 1}<\/p>\n

Question 15.
\nIf R = {(x, y) : x, y \u2208 W, x2<\/sup> + y2<\/sup> = 100}, then find the domain and the range of R. Also write R in roster form.
\nSolution:
\nGiven R = {(x, y) : x, y \u2208 W, x2<\/sup> + y2<\/sup> = 100}
\nsince x, y \u2208 W
\nand x2<\/sup> + y2<\/sup> = 100
\n\u21d2 y2<\/sup> = 100 – x2<\/sup>
\nWhen x = 0
\n\u21d2 y2<\/sup> = 100
\n\u21d2 y = \u00b1 10,
\nsince y \u2208 W
\n\u2234 y = 10
\n\u2234 (0, 10) \u2208 R
\nWhen x = 1
\n\u21d2 y2<\/sup> = 100 – 1 = 99
\n\u21d2 y = \u00b1 \\(\\sqrt{99}\\) \u2209 W
\nWhen x = 2
\n\u21d2 y2<\/sup> = 100 – 4 = 96
\n\u21d2 y = \u00b1 \\(\\sqrt{96}\\) \u2209 W
\nWhen x = 3
\n\u21d2 y2<\/sup> = 100 – 9 = 91
\n\u21d2 y = \u00b1 \\(\\sqrt{91}\\) \u2209 W
\nWhen x = 4
\n\u21d2 y2<\/sup> = 100 – 16 = 84
\n\u21d2 y = \u00b1 \\(\\sqrt{84}\\) \u2209 W
\nWhen x = 5
\n\u21d2 y2<\/sup> = 100 – 25 = 75
\n\u21d2 y = \u00b1 \\(\\sqrt{75}\\) \u2209 W
\nWhen x = 6
\n\u21d2 y2<\/sup> = 100 – 36 = 64
\n\u21d2 y = \u00b1 8
\nSince y \u2208 W
\n\u2234 y = 8 \u2208 W
\n\u21d2 (6, 8) \u2208 R
\nWhen x = 7
\n\u21d2 y2<\/sup> = 100 – 49 = 51
\n\u21d2 y = \u00b1 \\(\\sqrt{51}\\) \u2209 W
\nWhen x = 8
\n\u21d2 y2<\/sup> = 100 – 64 = 36
\n\u21d2 y = \u00b1 6,
\nsince y \u2208 W
\n\u2234 y = 6
\n\u21d2 (8, 6) \u2208 R
\nWhen x = 9
\n\u21d2 y2<\/sup> = 100 – 81 = 19
\ny = \u00b1 \\(\\sqrt{19}\\) \u2209 W
\nWhen x = 10
\n\u21d2 y2<\/sup> = 100 – 100 = 0
\n\u21d2 y = 0 \u2208 W
\n\u21d2 (10, 0) \u2208 R
\nFor all other values of x, we donot get y \u2208 W
\nThus, R = {(0, 10), (6, 8), (8, 6), (10, 0)}
\n\u2234 domain of R = {0, 6, 8, 10}
\nand Range of R = {10 8, 6, 0}.<\/p>\n

\"ML<\/p>\n

Question 16.
\nIf R is the relation on N defined by R = {(x, y) : y = x + \\(\\frac{12}{x}\\), x, y \u2208 N), then find
\n(i) R in roster form
\n(ii) domain of R
\n(iii) range of R
\nSolu:
\n(i) Given relation R is defined by
\nR = {(x, y) : y = x + \\(\\frac{12}{x}\\), x, y \u2208 N}
\nWhen x = 1
\n\u21d2 y = 1 + \\(\\frac{12}{1}\\)
\n= 13 \u2208 N
\n\u21d2 (1, 13) \u2208 R
\nWhen x = 2
\n\u21d2 y = 2 + \\(\\frac{12}{2}\\) = 8
\n\u21d2 8 \u2208 N
\n\u21d2 (2, 8) \u2208 R
\nWhen x = 3
\n\u21d2 y = 3 + 4 = 7 \u2208 N
\n\u21d2 (3, 7) \u2208 R
\nWhen x = 4
\n\u21d2 y = 4 + 3 = 7 \u2208 N
\n\u21d2 (4, 7) \u2208 R
\nWhen x = 6
\n\u21d2 y = 6 + 2 = 8 \u2208 N
\n\u21d2 (6, 8) \u2208 R
\nWhen x = 12
\n\u21d2 y = 12 + 1 = 13 \u2208 N
\n\u21d2 (12, 13) \u2208 R
\nWhile at other values of x, we donot get y \u2208 N.
\nHence R = {(1, 13), (2, 8), (3, 7), (4, 7), (6, 8), (12, 13)}
\n(ii) domain of R = {1, 2, 3, 4, 6, 12}
\nrange of R = {13, 8, 7}<\/p>\n

Question 17.
\nWrite down the domain and the range of the relation (x, y) : x = 3y and x and y are natural numbers less than lo.
\nSolution:
\nGiven relation R = {(x, y) : x = 3y ; 1 \u2264 x, y < 10, x, y \u2208 N}
\nHere 1 \u2264 x, y < 10, x, y \u2208 N} since x = 3y
\n\u21d2 y = \\(\\frac{x}{3}\\) When x = 3
\n\u21d2 y = 1 \u2208 N \u21d2 (3, 1) \u2208 R When x = 6
\n\u21d2 y = 2 \u2208 N \u21d2 (6, 2) \u2208 R When x = 9
\n\u21d2 y = 3 \u2208 N \u21d2 (9, 3) \u2208 R
\nWhile at all other values of x, we donot get y \u2208 N
\nThus, R = {(3, 1), (6, 2), (9, 3)}
\n\u2234 domain of R = {3, 6, 9}
\nand Range of R = {1, 2, 3}.<\/p>\n

Question 18.
\nLet A = {- 2, – 1, 0, 1, 2}, list the ordered pairs satisfying each of the following relations on A :
\n(i) \u2018is greater than\u2019
\n(ii) \u2018is the square of’
\n(iii) \u2018is the negative of’.
\nSolution:
\nGiven A = {- 2. – 1, 0, 1, 2}
\nand given relation be \u2018is greater than\u2019 (1)
\nsince – 1 > – 2
\n\u21d2 (- 1, – 2) \u2208 R ;
\n1 > – 2
\n\u21d2 (1 – 2) \u2208 R
\no > – 2
\n\u21d2 (0, – 2) \u2208 R ;
\n1 > – 1
\n\u21d2 (1, – 1) \u2208 R
\n0 > – 1
\n\u21d2 (0, – 1) \u2208R;
\n1 > 0
\n\u21d2 (1, 0) \u2208 R
\n2 > – 2
\n\u21d2 (2, – 2) \u2208 R ;
\n2 > – 1
\n\u21d2 (2, – 1) \u2208 R
\n2 > 0
\n\u21d2 (2, 0) \u2208 R ;
\n2 > 1
\n\u21d2 (2, 1) \u2208 R
\n\u2234 R = {(- 1, – 2), (1, – 2), (0, – 2), (1, – 1), (0, – 1), (1. 0), (2, -2), (2, – 1), (2, 0), (2, 1)}.<\/p>\n

(ii) Here relation be \u2018is the square of\u2019
\nSince 0 = 02<\/sup>
\n\u21d2 (0, 0) \u2208 R
\n1 = (- 1)2<\/sup>
\n\u21d2 (1, – 1) \u2208 R
\nAlso, 1 = 12<\/sup>
\n\u21d2 (1, 1) \u2208 R
\n\u2234 R = {(0, 0), (1, – 1), (1, 1)}<\/p>\n

(iii) Here, relation be \u2018is the negative of\u2019
\nHere, – 1 be the negative of 1
\n\u2234 (- 1, 1) \u2208 R
\n1 be the negative of – 1
\n\u2234 (1, – 1) \u2208 R
\n0 be the negative of 0
\n\u2234 (0, 0) \u2208 R
\n2 be the negative of – 2
\n\u2234 (2, – 2) \u2208 R
\nand – 2 be the negative of 2
\n\u2234 (- 2, 2) \u2208 R
\nHence, R = {(- 1, 1), (1, – 1), (0, 0), (2, – 2), (- 2, 2)}.<\/p>\n

\"ML<\/p>\n

Question 19.
\nIf A = {1, 3, 5, 6) and B = {3, 4, 5}, write the relation R as a set of ordered pairs if
\n(i) R = {(x, y) : (x, y) \u2208 A \u00d7 B : x + y is even)
\n(ii) R = {(x, y) : (x, y) \u2208 A \u00d7 B : x y is odd}.
\nSolution:
\nGiven A = { 1, 3, 5, 6}
\nand B = {3, 4, 5}
\n(i) and Relation R = {(x, y) : (x, y) \u2208 A \u00d7 B : x + y is even}
\nHere A \u00d7 B = {(1, 3), (1, 4),(1, 5), (3, 3), (3, 4), (3, 5), (5, 3), (5, 4), (5, 5), (6, 3), (6, 4), (6, 5)}
\nsince 1 + 3 = 4 (even)
\n\u21d2 (1, 3) \u2208 R
\nsince 1 + 5 = 6 (even)
\n\u21d2 (1, 5) \u2208 R
\nsince 3 + 3 = 6 (even)
\n\u21d2 (3, 3) \u2208 R
\nsince 3 + 5 = 8 (even)
\n\u21d2 (3, 5) \u2208 R
\nsince 5 + 3 = 8 (even)
\n\u21d2 (5, 3) \u2208 R
\nsince 5 + 5 = 10 (even)
\n\u21d2 (5, 5) \u2208 R
\nsince 6 + 4 = 10 (even)
\n\u21d2 (6, 4) \u2208 R
\nR = {(1, 3), (1, 5), (3, 3), (5, 3), (3, 5), (5, 5), (6, 4)}<\/p>\n

(ii) Given R = {(x, y) : (x, y) \u2208 A \u00d7 B : xy is odd}
\nsince 1 . 3 = 3 (odd)
\n\u21d2 (1, 3) \u2208 R
\n1 . 5 = 5 (odd)
\n\u21d2 (1, 5) \u2208 R
\n3 . 3 = 9 (odd)
\n\u21d2 (3, 3) \u2208 R
\n3 . 5 = 15 (odd)
\n\u21d2 (3, 5) \u2208 R
\n5 . 3 = 15 (odd)
\n\u21d2 (5, 3) \u2208 R
\n5 . 5 = 25 (odd)
\n\u21d2 (5, 5) \u2208 R
\nThus, R = {(1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5)}<\/p>\n

Question 20.
\nLet R = {(x, y) : x, y \u2208 Z, y = 2x – 4}. If (a, – 2) and (4, b2<\/sup>) belong to R, find the values of a and b.
\nSolution:
\nGiven R = {(x, y) : x, y \u2208 Z, y = 2x – 4}
\nsince (a, – 2) \u2208 R
\n\u21d2 – 2 = 2 \u00d7 a – 4
\n\u21d2 – 2 = 2a – 4
\n\u21d2 2a = 2
\n\u21d2 a = 1
\nand (4, b2<\/sup>) \u2208 R
\n\u21d2 b2<\/sup> = 2 \u00d7 4 – 4 = 4
\n\u21d2 b = \u00b1 2.<\/p>\n

\"ML<\/p>\n

Question 21.
\nFind the linear relation between the components of the ordered pairs of the relation R where
\n(i) R = {(- 1,- 1), (0, 2), (1, 5), ………………}.
\n(ii) R = {(0, 2), (- 1, 5), (2, – 4), ……………..}.
\nSolution:
\n(i) Given R = ((- 1,- 1), (0, 2), (1, 5), ……………..}
\nLet the linear relation be
\ny = ax + b ………….(1)
\nsince (- 1, – 1) \u2208 R
\n\u2234 When x = – 1, y = – 1
\n\u2234 from (1) ;
\n– 1 = – a + b ………………..(2)
\nAlso (0, 2) \u2208 R
\ni.e. When x = 0, y = 2
\n\u2234 from (1) ;
\n2 = a \u00d7 0 + b
\n\u21d2 b = 2
\n\u2234 from (2) ;
\n– 1 = – a + 2
\n\u21d2 – a = – 3
\n\u21d2 a = 3
\nHence the required relation be, y = 3x + 2
\nClearly (1, 5) \u2208 R.
\nSince 5 = 3 \u2208 1 + 2<\/p>\n

(ii) Given R = {(0, 2), (- 1, 5), (2, – 4), …………….}
\nLet the required linear relation be
\ny = ax + b ………………….(1)
\nsince (0, 2) \u2208 R
\ni.e. when x = 0, y = 2
\n\u2234 from (1) ;
\n2 = 0 + b
\n\u21d2 b = 2
\nNow (- 1, 5) \u2208 R
\ni.e. When x = – 1, y = 5
\n\u2234 from (1) ;
\n5 = – a + b
\n\u21d2 a = b – 5
\n= 2 – 5 = – 3
\nHence the required relation be
\ny= – 3x + 2 …………………(2)
\nAlso, (2, – 4) satisfies eqn. (2).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can cross-reference their work with ML Aggarwal Class 11 Solutions Chapter 2 Relations and Functions Ex 2.2 to ensure accuracy. ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.2 Question 1. If A and B are two sets such that n(A) = 2 and n(B) = 3, find …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170496"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170496"}],"version-history":[{"count":4,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170496\/revisions"}],"predecessor-version":[{"id":170506,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170496\/revisions\/170506"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170496"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170496"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170496"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}