{"id":170493,"date":"2024-05-24T17:08:43","date_gmt":"2024-05-24T11:38:43","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170493"},"modified":"2024-05-24T17:09:26","modified_gmt":"2024-05-24T11:39:26","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-ex-2-1","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-ex-2-1\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.1"},"content":{"rendered":"

Access to comprehensive Class 11 ISC Maths Solutions<\/a> Chapter 2 Relations and Functions Ex 2.1 encourages independent learning.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.1<\/h2>\n

Question 1.
\nFind a and b if
\n(i) (a + 1, b – 2) = (3, 1)
\n(ii) (\\(\\frac{a}{3}\\) + 1, b – \\(\\frac{2}{3}\\)) = (\\(\\frac{5}{3}, \\frac{1}{3}\\))
\n(iii) (2a, a + b) = (6, 2)
\n(iv) (a + b, 3b – 2) = (7, – 5)
\nSolution:
\n(i) Given (a + 1, b – 2) = (3, 1)
\n\u2234 a + 1 = 3
\n\u21d2 a = 2
\nand b – 2 = 1
\n\u21d2 b = 3<\/p>\n

(ii) Given (\\(\\frac{a}{3}\\) + 1, b – \\(\\frac{2}{3}\\)) = (\\(\\frac{5}{3}, \\frac{1}{3}\\))
\ni.e. \\(\\frac{a}{3}\\) + 1 = \\(\\frac{5}{3}\\)
\n\u21d2 \\(\\frac{a}{3}=\\frac{2}{3}\\)
\n\u21d2 a = 2
\nand b – \\(\\frac{2}{3}\\) = \\(\\frac{1}{3}\\)
\n\u21d2 b = \\(\\frac{2}{3}\\) + \\(\\frac{1}{3}\\) = 1<\/p>\n

(iii) Given (2a, a + b) = (6, 2)
\n\u2234 2a = 6
\n\u21d2 a = 3
\nand a + b = 2
\n\u21d2 b = 2 – 3 = – 1<\/p>\n

(iv) Given (a + b, 3b – 2) = (7, – 5)
\n\u2234 a + b = 7
\nand 3b – 2 = – 5
\n\u2234 3b = – 3
\n\u2234 b = – 1
\n\u2234 a = 7 – b
\n= 7 + 1 = 8.<\/p>\n

\"ML<\/p>\n

Question 2.
\nFind x and y if
\n(ij) (4x + 3, y) = (3x + 5, – 2)
\n(ii) (x – y, x + y) = (6, 10)
\nSolution:
\n(i) Given (4x + 3, y) = (3x + 5, – 2)
\n\u2234 4x + 3 = 3x + 5
\n\u21d2 x = 2 and y = – 2<\/p>\n

(ii) (x – y, x + y) = (6, 10)
\n\u2234 x – y = 6 ………………(1)
\nand x + y = 10 …………………(2)
\nOn adding (1) and (2), we have
\n2x = 16
\n\u21d2 x = 8
\n\u2234 from (1) ;
\ny = 8 – 6 = 2<\/p>\n

(ii) Given (2x + y, x – y) = (8, 3)
\n\u2234 2x + y = 8 ……………..(1)
\nx – y = 3 ………………(2)
\nOn adding ( ) and (2) ; we have
\n3x = 11
\n\u21d2 x = \\(\\frac{11}{3}\\)
\n\u2234 from (2) ; we have
\ny = x – 3
\n= \\(\\frac{11}{3}\\) – 3
\n= \\(\\frac{11-9}{3}\\)
\n= \\(\\frac{2}{3}\\)<\/p>\n

Question 3.
\nIf the ordered pars (a, – 1) and (5, b) belong to {(x, y) : y = 2.x – 3}, find the values of a and b.
\nSolution:
\nSince (a, – 1) \u2208 {(x, y) : y = 2x – 3}
\n\u2234 – 1 = 2a – 3
\n\u21d2 2a = 2
\n\u21d2 a = 1
\n(5, b) \u2208 {(x, y) : y = 2x – 3}
\n\u2234 b = 2 \u00d7 5 – 3 = 7<\/p>\n

Question 4.
\nIf A = {- 1, 0, 1} and B = {3, 5), write the following:
\n(i) A \u00d7 B
\n(ii) B \u00d7 A
\n(iii) B \u00d7 B.
\nSolution:
\nGiven A = {- 1, 0, 1}
\nand B = {3, 5}
\n(i) A \u00d7 B = {(- 1, 3), ( 1, 5), (0, 3), (0, 5), (1, 3), (1, 5)}
\n(ii) B \u00d7 A = {(3, – 1), (3, 0), (3, 1), (5, – 1), (5, 0), (5, 1)}
\n(iii) B \u00d7 B = {(3, 3), (3, 5), (5, 3), (5, 5)}<\/p>\n

Question 5.
\nIf A is a set such that n (A) = 3 and B = {3, 4, 5) then what is the number of elements in A \u00d7 B ?
\nSolution:
\nGiven n (A) = 3
\nand B = {3, 4, 5}
\n\u2234 n(B) = 3
\nThus, n (A \u00d7 B) = n (A) \u00d7 n (B)
\n= 3 \u00d7 3 = 9<\/p>\n

\"ML<\/p>\n

Question 6.
\nIf A = {- 3, – 1, 0, 4} and B = {- 1, 0, 1, 2, 3}, then write the number of elements in each of the following cartesian products:
\n(i) A \u00d7 B
\n(ii) B \u00d7 A
\n(iii) A \u00d7 A
\n(iv) B \u00d7 B.
\nSolution:
\nGiven A = {- 3, – 1, 0, 4}
\n\u2234 n(A) = 4
\nand B = {- 1, 0, 1, 2, 3}
\nn(B) = 5
\n(i) \u2234 no. of elements in A \u00d7 B = n (A \u00d7 B)
\n= n (A) \u00d7 n (B)
\n= 4 \u00d7 5 = 20<\/p>\n

(ii) \u2234 no. of elements in B \u00d7 A = n (B \u00d7 A)
\n= n (B) \u00d7 n (A)
\n= 5 \u00d7 4 = 20<\/p>\n

(iii) \u2234 no. of elements in A \u00d7 A = n (A \u00d7 A)
\n= n (A) \u00d7 n (A)
\n= 4 \u00d7 4 = 16<\/p>\n

(iv) \u2234 no. of elements in B \u00d7 B = n (B \u00d7 B)
\n= n (B) \u00d7 n (B)
\n= 5 \u00d7 5 = 25<\/p>\n

Question 7.
\nIf A = {1, 2} and B = {3, 4}, then how many ubets will A \u00d7 B have?
\nSolution:
\nGiven A = {1, 2}
\nand B = {3, 4}
\n\u2234 n (A) = 2 ;
\nn (B) = 2
\nThus number of subsets of A \u00d7 B = 2n (A \u00d7 B)<\/sup>
\n= 2n (A) \u00d7 n (B)<\/sup>
\n= 22 \u00d7 2<\/sup>
\n= 24<\/sup> = 16<\/p>\n

\"ML<\/p>\n

Question 8.
\nIf x \u2208 {2, 3, 5} and y \u2208 {2, 4, 6}, form the set of all ordered pairs (x, y) such that x < y.
\nSolution:
\nGiven x \u2208 {2, 3, 5} and y \u2208 {2, 4, 6}
\nWe want to find all ordered pair (x, y) for which x < y.
\nsince 2 < 4 ;
\n2 < 6 ;
\n3 < 4 ;
\n3 < 6 ;
\n5 < 6 ;
\nHence, required set of all such ordered pairs (x, y) s.t. x < y be {(2, 4) (2, 6), (3,4) (3, 6) (5, 6)}.<\/p>\n

Question 9.
\nIf x \u2208 {- 1, 2, 3, 4, 5} and y \u2208 {0, 3, 6}, form the set of all ordered pairs (x, y) such that x + y = 5.
\nSolution:
\nGiven x \u2208 {- 1, 2, 3, 4, 5}
\nand y \u2208 {0, 3, 6}
\nHence the set of all ordered pair (x, y) for which x + y = 5 be given by {(- 1, 6), (2, 3), (5, 0)}.<\/p>\n

Question 10.
\nIf x \u2208 (2, 3, 4) and y \u2208 {4, 6, 9, 10}, form the set of all ordered pairs (x, y) such that x is a factor of y.
\nSolution:
\nGiven x \u2208 {2, 3, 4}
\nand y \u2208 {4, 6, 9, 10}
\nsince 2\/4, 2\/6, 2\/10, 3\/6, 3\/9, 4\/4
\nHence the set of all ordered pair (x, y) for which x be a factor of y be given by
\n{(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}<\/p>\n

\"ML<\/p>\n

Question 11.
\nState whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.
\n(i) If P = {m, n} and Q = {n, m}, then P \u00d7 Q = {(m, n), (n, m)}.
\n(ii) If A and B are non-empty sets, then A \u00d7 B is a non-empty set of ordered pairs (x, y) such that x \u2208 B and y \u2208 A.
\n(iii) If A = (1, 2), B = {3, 4}, then A \u00d7 (B + \u03a6) = \u03a6.
\nSolution:
\n(i) false, true statement is
\nP \u00d7 Q = {(m, n), (m, m), (n, n), (n, m)}<\/p>\n

(ii) false, If A and B are non empty sets then A \u00d7 B is a non-empty set of ordered pairs (x, y) s.t x \u2208 A and y \u2208 B.<\/p>\n

(iii) True, A \u00d7 (B \u2229 \u03a6) = A \u00d7 \u03a6 = \u03a6<\/p>\n

Question 12.
\nIf A = {- 1, 0, 1, 2, 3), write the subset S of A \u00d7 A such that the second component of the elements of S is 0.
\nSolution:
\nGiven A = {- 1, 0, 1, 2, 3}
\n\u2234 A \u00d7 A = {(- 1,- 1), (- 1, 0), (- 1, 1), (-1, 2), (- 1, 3), (0, – 1), (0, 0), (0, 1), (0, 2), (0, 3), (1, – 1), (1, 0) (1, 1), (1, 2), (1, 3), (2, – 1), (2, 0), (2, 1), (2, 2), (2, 3), (3, – 1), (3, 0), (3, 1), (3, 2), (3, 3)}
\nThus, A \u00d7 A has 25 different ordered pairs
\n\u2234 S = {(- 1, 0), (0, 0), (1, 0), (2, 0). (3,0)} \u2286 A \u00d7 A.<\/p>\n

Question 13.
\nIf A \u00d7 B = {(p, q), (p, r), (m, q), (m, r)}, find A and B.
\nSolution:
\nGiven A \u00d7 B = {(p, q), (p, r), (m, q), (m, r)}
\nsince A \u00d7 B = {(x, y), x \u2208 A, y \u2208 B}
\n\u2234 A = {p, m}
\nand B = {q, r}<\/p>\n

Question 14.
\nIf A = {x, y, z} and some elements of A \u00d7 B are (x, 1), (y, 2), (z, 1) then write the set B such that n (A \u00d7 B) = 6.
\nSolution:
\nGiven A = {x, y, z}
\nsince A \u00d7 B = {(x, y) : x \u2208 A and y \u2208 B} ……………(1)
\nSome elements of A \u00d7 B are (x, 1), (y, 2) and (z, 1)
\nalso, n (A \u00d7 B) = 6
\n\u2234 A \u00d7 B = {x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
\nThus B = {1, 2}.<\/p>\n

Question 15.
\nIf A \u00d7 B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}, then find B \u00d7 A.
\nSolution:
\nGiven A \u00d7 B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}
\n\u2234 A = {x, y}
\nand B = {1, 2, 3}
\nSince A \u00d7 B = {(x, y) : x \u2208 A, y \u2208 B}
\nThus, B \u00d7 A = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)}.<\/p>\n

\"ML<\/p>\n

Long answer questions (16 to 22) :<\/p>\n

Question 16.
\nIf A = {1, 2, 3, 4} and B = {5, 7, 9}, find
\n(i) A \u00d7 B
\n(ii) B \u00d7 A
\n(iii) Is A \u00d7 B = B \u00d7 A ?
\n(iv) Is n (A \u00d7 B) = n (B \u00d7 A) ?
\nSolution:
\nGiven A = {1, 2, 3, 4}
\nand B = {5, 7, 9}<\/p>\n

(i) A \u00d7 B = {(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)}<\/p>\n

(ii) B \u00d7 A = {(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)}<\/p>\n

(iii) Clearly A \u00d7 B \u2260 B \u00d7 A<\/p>\n

(iv) n (A \u00d7 B) = number of distinct elements in A \u00d7 B
\n= n (A) \u00d7 n (B)
\n= 4 \u00d7 3 = 12
\nn (B \u00d7 A) = n(B) \u00d7 n(A)
\n= 3 \u00d7 4 = 12
\n\u2234 n (A \u00d7 B) = n (B \u00d7 A<\/p>\n

Question 17.
\nLet A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
\n(i) A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)
\n(ii) A \u00d7 C is a subset of B \u00d7 D.
\nSolution:
\nGiven A = {1, 2},
\nB = {1, 2, 3, 4}
\nC = {5, 6}
\nand D = {5, 6, 7, 8}<\/p>\n

(i) (B \u2229 C) = {1, 2, 3, 4} \u2229 {5, 6} = \u03a6
\nA \u00d7 (B \u2229 C) = \u03a6
\nNow, A \u00d7 B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
\nA \u00d7 C = {(1, 5), (1, 6), (2, 5). (2, 6)}
\n\u2234 (A \u00d7 B) \u2229 (A \u00d7 C) = \u03a6 …………………….(2)
\n\u2234 from (1) and (2) ; we have
\nA \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n

(ii) Now A \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6)}
\nand B \u00d7 D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
\nClearly all members of set A \u00d7 C are the members of B \u00d7 D.
\n\u2234 A \u00d7 C be a subset of B \u00d7 D.<\/p>\n

\"ML<\/p>\n

Question 18.
\nIf A = {x : x \u2208 W, x < 2), B = {x : x \u2208 N, 1 < x < 5} and C = {3, 5}, find
\n(i) A \u00d7 (B \u2229 C)
\n(ii) A \u00d7 (B \u222a C)
\nSolution:
\n(i) Given A = {x : x \u2208 W, x < 2) = {0, 1}
\nB = {x : x \u2208 N, 1 < x < 5}
\n= {2, 3, 4}
\nand C = {3, 5}
\nB \u2229 C = {2, 3, 4) \u2229 {3, 5) = {3}
\n\u2234 A \u00d7 (B \u2229 C) = {0, 1} \u00d7 {3}
\n= {(0, 3), (1, 3)}<\/p>\n

(ii) B \u222a C = {2, 3, 4, 5}
\n\u2234 A \u00d7 (B \u222a C) = {0, 1} \u00d7 {2, 3, 4, 5}
\n= {(0, 2), (0, 3), (0, 4), (0, 5), (1,2), (1, 3), (1, 4), (1, 5)}<\/p>\n

Question 19.
\nIf A = {x : x \u2208 N and x \u2264 3}, B = {x : x \u2208 I, – 1 \u2264 x \u2264 1} and C = {1, 2}, verify that
\n(i) A \u00d7 (B \u222aC) = (A \u00d7 B) \u222a(A \u00d7 C)
\n(ii) A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)
\n(iii) (A – B) \u00d7 C = A \u00d7 C – B \u00d7 C.
\nSolution:
\nGiven A = {x : x \u2208 N and x \u2264 3} = {1, 2, 3}
\nB = {x : x \u2208 I and – 1 \u2264 x \u2264 1} = {- 1, 0, 1}
\nand C = {1, 2}
\n(i) B \u222a C = {- 10, 1, 2}
\nL.H.S. A \u00d7 (B \u222a C) = {1, 2, 3} \u222a {- 1, 0, 1, 2}
\n= {(1, – 1), (1, 0), (1, 1), (1, 2), (2, – 1), (2, 0), (2, 1), (2, 2), (3, – 1), (3, 0), (3, 1), (3, 2)}
\nNow A \u00d7 B = {(1, – 1), (1, 0), (1, 1), (2, – 1), (2, 0), (2, 1), (3, – 1), (3, 0), (3, 1)}
\nA \u00d7 C = {1, 2, 3) \u00d7 {1, 2}
\n= {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
\n\u2234 R.H.S. = (A \u00d7 B) \u222a (A \u00d7 C)
\n= ((1, – 1), (1, 0), (1, 1), (1, 2), (2, – 1), (2, 0), (2, 1), (2, 2), (3, 1), (3, 2)}
\n\u2234 R.H.S. = (A \u00d7 B) \u2229 (A \u00d7 C)
\n= {(1, 1), (2, 1), (3, 1)}
\nThus L.H.S = R.H.S.<\/p>\n

(iii) A – B = {2, 3}
\nL.H.S. = (A – B) \u00d7 C
\n= {(2, 1), (2, 2), (3, 1), (3, 2)}
\nA \u00d7 C = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
\nand B \u00d7 C = {(- 1, 1), (- 1, 2), (0, 1), (0, 2), (1, 1), (1, 2)}
\n\u2234 R.H.S. = A \u00d7 C – B \u00d7 C
\n= {(2, 1), (2, 2), (3, 1), (3, 2)}
\nThus L.H.S. = R.H.S.<\/p>\n

Question 20.
\nIf P = (1, 2), form the set P \u00d7 P \u00d7 P.
\nSolution:
\nGiven P = {1, 2}
\n\u2234 P \u00d7 P = {(1, 1), (1, 2), (2, 1), (2, 2)}
\n\u2234 P \u00d7 P \u00d7 P = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
\n[using definition]<\/p>\n

\"ML<\/p>\n

Question 21.
\nIf A = [a, b, c] and some elements of A \u00d7 B are (a, p), (b, q), (c, p), write the set B and find the remaining ordered pairs of A \u00d7 B such that n (A \u00d7 B) = 6.
\nSolution:
\nGiven A = {a, b, e} ;
\nn(A) = 3
\nAlso, n(A \u00d7 B) = 6
\nn(A). n(B) = 6
\n\u21d2 n(B) = 2 so B must contain., exactly two distinct elements.
\nAlso, (a, p), (b, q), (c, p) \u2208 A \u00d7 B
\na, b, c \u2208 A and p, q \u2208 B
\nThus B = {p, q)
\nA \u00d7 B = {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)}<\/p>\n

Question 22.
\nGiven B = (2, 3, 5) and some elements of A \u00d7 B are (a, 2), (b, 3), (c, 5). Find the set A and the remaining ordered pairs of A \u00d7 B such that A \u00d7 B is least.
\nSolution:
\nGiven B = {2, 3, 5)
\nand Some elements of A \u00d7 B are (a, 2), (b, 3), (c, 5)
\nand a, b, c are distinct elements
\n\u2234 a, b, c \u2208 A.
\nThus, A = {a, b, e)
\n\u2234 A \u00d7 B = {(a, 2), (a, 3), (a, 5), (b, 2), (b, 3), (b, 5), (c, 2), (e, 3), (c, 5)}
\nso that A \u00d7 B is least.
\nThus, remaining elements of A \u00d7 B are (a, 3), (a, 5), (b, 2), (b, 5), (c, 2), (c, 3).<\/p>\n","protected":false},"excerpt":{"rendered":"

Access to comprehensive Class 11 ISC Maths Solutions Chapter 2 Relations and Functions Ex 2.1 encourages independent learning. ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.1 Question 1. Find a and b if (i) (a + 1, b – 2) = (3, 1) (ii) ( + 1, b …<\/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\/170493"}],"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=170493"}],"version-history":[{"count":2,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170493\/revisions"}],"predecessor-version":[{"id":170495,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170493\/revisions\/170495"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170493"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170493"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170493"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}