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PEDAGOGY OF
MATHEMATICS – PART II
By
Dr. I. Uma Maheswari
Principal
Peniel Rural College of Education,Vemparali, Dindigul District
iuma_maheswari@yahoo.co.in
STD IX
CHAPTER 1 – SET LANGUAGE
Ex - 1.5
Solution :
(i) A - B = {4, 6, 3}
(ii) B - C = {-1, 5, 7}
(iii) A' U B'
A' = {-3, 0, 1, 2, 5, 7, 8}
B' = {-3, 0, 1, 2, 3, 4, 6}
A' U B' = {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(iv) A' n B'
A' n B' = {-3, 0, 1, 2}
(v) (B U C)'
B U C = {-3, -,2, -1, 0, 3, 5, 7, 8}
(B U C)' = {4, 6, 1, 2}
(vi) A − (B U C)
A = {-2, -1, 3, 4, 6} and B U C = {-3, -,2, -1, 0, 3, 5, 7, 8}
A − (B U C) = {4, 6}
(vii) A − (B n C)
B n C = {-2, 8}
A = {-2, -1, 3, 4, 6}
A − (B n C) = {-1, 3, 4, 6}
Solution :
(i) K U (L n M)
(L n M) = { b, c, d }
K U (L n M) = {a, b, c, d, e, f}
(ii) K n (L U M)
(L U M) = {a, b, c, d, g, h}
K n (L U M) = {a, b, d}
(iii) (K U L) n (K U M)
(K U L) = {a, b, c, d, e, f, g}
(K U M) = {a, b, c , d, e, f, h}
(K U L) n (K U M) = {a, b , c, d , e, f}
(iv) (K n L) U (K n M)
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(K n L) = {b, d}
(K n M) = {a, b, d}
(K n L) U (K n M) = {a, b, d}
Solution :
A = {x : x ∊ Z,−2 < x ≤ 4}
A = {-1, 0, 1, 2, 3, 4}
B = {x : x ∈ W, x ≤ 5}
B = {0, 1, 2, 3, 4, 5}
C = {−4,−1, 0, 2, 3, 4}
A U (B n C) = (A U B) n (A U C)
(B n C) = {0, 2, 3, 4}
A U (B n C) = {-1, 0, 1, 2, 3, 4} ----(1)
(A U B) = {-1, 0, 1, 2, 3, 4, 5}
(A U C) = {-4, -1, 0, 1, 2, 3, 4}
(A U B) n (A U C) = {-1, 0, 1, 2, 3, 4}----(2)
(1) = (2)
Solution :
Hence proved.
Solution :
(B n C) = {a, d, g}
A - (B n C) = {b, c, e, h} ----(1)
(A − B) = {b, e, h}
(A − C) = {b, c}
(A − B) U (A−C) = {b, c, e, h} ----(2)
(1) = (2)
Hence proved
Solution :
A = {x : x = 6n, n ∈ W and n < 6}
A = {0, 6, 12, 18, 24, 30}
B = {x : x = 2n, n ∈ N and 2 < n ≤ 9}
B = {6, 8, 10, 12, 14, 16, 18}
and C = {x : x = 3n, n ∈ N and 4 ≤ n < 10}
C = {12, 15, 18, 21, 24, 27}
(B n C) = {12, 18}
A−(B n C) = {0, 6, 24, 30} ----(1)
(A − B) = {0, 24, 30}
(A − C) = {0, 6, 30}
(A − B) U (A − C) = {0, 6, 24, 30} ----(2)
Solution :
(B U C) = {-1, 0, 2, 5, 6, 7}
A − (B U C) = {-2, 1, 3} ----(1)
(A − B) = {-2, 1, 3}
(A − C) = {-2, 0, 1, 3}
(A − B) n (A − C) = {-2, 1, 3} ----(2)
(1) = (2)
Hence proved.
Solution :
A = {y : y = (a + 1)/2, a ∈ W and a ≤ 5}
A = {1/2, 3/2, 2, 5/2, 3}
B = {y : y = (2n - 1)/2, n ∈ W and n < 5}
B = {-1/2, 1/2, 3/2, 5/2, 7/2}
C = {-1, -1/2, 1, 3/2, 2}
A - (BUC) = (A - B) n (A - C)
B U C = {-1/2, -1, 1/2, 1, 3/2, 2, 5/2, 7/2}
A - (BU C) = {3} -----(1)
(A - B) = {2, 3}
(A - C) = {1/2, 5/2, 3}
(A - B) n (A - C) = {3} -----(2)
(1) = (2)
Hence proved.
Solution :
Verified
Solution :
De morgan's laws for complementation :
(i) (A U B)' = A' n B'
(ii) (A n B)' = A' U B'
A = {7, 8, 11, 12} and B = {4, 8, 12, 15},
U = {4, 7, 8, 10, 11, 12, 15, 16}
(i) (A U B) = {4, 7, 8, 11, 12, 15}
(A U B)' = {10, 16} ----(1)
A' = {4, 10, 15, 16}
B' = {7, 10, 11, 16}
A' n B' = {10, 16} ----(2)
Hence proved.
(ii) (A n B) = {8, 12}
(A n B)' = {4, 7, 10, 11, 15, 16} ----(1)
A' = {4, 10, 15, 16}
B' = {7, 10, 11, 16}
A' U B' = {4, 7, 10, 11, 15, 16} ----(2)
Hence proved.
Solution :
…. (1)
(1) = (2)
Verified
PEDAGOGY OF MATHEMATICS – PART II CHAPTER 1

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PEDAGOGY OF MATHEMATICS – PART II CHAPTER 1

  • 1. PEDAGOGY OF MATHEMATICS – PART II By Dr. I. Uma Maheswari Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
  • 2. STD IX CHAPTER 1 – SET LANGUAGE Ex - 1.5
  • 3.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19.
  • 20. Solution : (i) A - B = {4, 6, 3} (ii) B - C = {-1, 5, 7} (iii) A' U B' A' = {-3, 0, 1, 2, 5, 7, 8} B' = {-3, 0, 1, 2, 3, 4, 6} A' U B' = {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
  • 21. (iv) A' n B' A' n B' = {-3, 0, 1, 2} (v) (B U C)' B U C = {-3, -,2, -1, 0, 3, 5, 7, 8} (B U C)' = {4, 6, 1, 2}
  • 22. (vi) A − (B U C) A = {-2, -1, 3, 4, 6} and B U C = {-3, -,2, -1, 0, 3, 5, 7, 8} A − (B U C) = {4, 6} (vii) A − (B n C) B n C = {-2, 8} A = {-2, -1, 3, 4, 6} A − (B n C) = {-1, 3, 4, 6}
  • 23.
  • 24. Solution : (i) K U (L n M) (L n M) = { b, c, d } K U (L n M) = {a, b, c, d, e, f} (ii) K n (L U M) (L U M) = {a, b, c, d, g, h} K n (L U M) = {a, b, d}
  • 25. (iii) (K U L) n (K U M) (K U L) = {a, b, c, d, e, f, g} (K U M) = {a, b, c , d, e, f, h} (K U L) n (K U M) = {a, b , c, d , e, f} (iv) (K n L) U (K n M) K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} (K n L) = {b, d} (K n M) = {a, b, d} (K n L) U (K n M) = {a, b, d}
  • 26.
  • 27. Solution : A = {x : x ∊ Z,−2 < x ≤ 4} A = {-1, 0, 1, 2, 3, 4} B = {x : x ∈ W, x ≤ 5} B = {0, 1, 2, 3, 4, 5} C = {−4,−1, 0, 2, 3, 4} A U (B n C) = (A U B) n (A U C)
  • 28. (B n C) = {0, 2, 3, 4} A U (B n C) = {-1, 0, 1, 2, 3, 4} ----(1) (A U B) = {-1, 0, 1, 2, 3, 4, 5} (A U C) = {-4, -1, 0, 1, 2, 3, 4} (A U B) n (A U C) = {-1, 0, 1, 2, 3, 4}----(2) (1) = (2)
  • 29.
  • 31.
  • 32. Solution : (B n C) = {a, d, g} A - (B n C) = {b, c, e, h} ----(1) (A − B) = {b, e, h} (A − C) = {b, c} (A − B) U (A−C) = {b, c, e, h} ----(2) (1) = (2) Hence proved
  • 33.
  • 34. Solution : A = {x : x = 6n, n ∈ W and n < 6} A = {0, 6, 12, 18, 24, 30} B = {x : x = 2n, n ∈ N and 2 < n ≤ 9} B = {6, 8, 10, 12, 14, 16, 18} and C = {x : x = 3n, n ∈ N and 4 ≤ n < 10} C = {12, 15, 18, 21, 24, 27}
  • 35. (B n C) = {12, 18} A−(B n C) = {0, 6, 24, 30} ----(1) (A − B) = {0, 24, 30} (A − C) = {0, 6, 30} (A − B) U (A − C) = {0, 6, 24, 30} ----(2)
  • 36.
  • 37. Solution : (B U C) = {-1, 0, 2, 5, 6, 7} A − (B U C) = {-2, 1, 3} ----(1) (A − B) = {-2, 1, 3} (A − C) = {-2, 0, 1, 3} (A − B) n (A − C) = {-2, 1, 3} ----(2) (1) = (2) Hence proved.
  • 38.
  • 39. Solution : A = {y : y = (a + 1)/2, a ∈ W and a ≤ 5} A = {1/2, 3/2, 2, 5/2, 3} B = {y : y = (2n - 1)/2, n ∈ W and n < 5} B = {-1/2, 1/2, 3/2, 5/2, 7/2} C = {-1, -1/2, 1, 3/2, 2} A - (BUC) = (A - B) n (A - C)
  • 40. B U C = {-1/2, -1, 1/2, 1, 3/2, 2, 5/2, 7/2} A - (BU C) = {3} -----(1) (A - B) = {2, 3} (A - C) = {1/2, 5/2, 3} (A - B) n (A - C) = {3} -----(2) (1) = (2) Hence proved.
  • 41.
  • 43.
  • 44. Solution : De morgan's laws for complementation : (i) (A U B)' = A' n B' (ii) (A n B)' = A' U B' A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, U = {4, 7, 8, 10, 11, 12, 15, 16}
  • 45. (i) (A U B) = {4, 7, 8, 11, 12, 15} (A U B)' = {10, 16} ----(1) A' = {4, 10, 15, 16} B' = {7, 10, 11, 16} A' n B' = {10, 16} ----(2) Hence proved.
  • 46. (ii) (A n B) = {8, 12} (A n B)' = {4, 7, 10, 11, 15, 16} ----(1) A' = {4, 10, 15, 16} B' = {7, 10, 11, 16} A' U B' = {4, 7, 10, 11, 15, 16} ----(2) Hence proved.
  • 47.