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5 mark questions

5 mark questions. X-STD MATHEMATICS. PREPARED BY: R.RAJENDRAN. M.A., M . Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21. Prove That A – (B  C ) = (A – B)  (A – C). (1)B C. (2) A-(B C). (3) A – B. A. B. B. B. A. A. C. C. C. (4) A – C.

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5 mark questions

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  1. 5 mark questions X-STD MATHEMATICS PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

  2. Prove That A – (B  C) = (A – B)  (A – C) (1)BC (2) A-(BC) (3) A – B A B B B A A C C C (4) A – C (5) (A – B)  (A – C) B A B A C C From the diagrams (2) and (5) A – (B  C) = (A – B)  (A – C)

  3. Prove that A – (B  C) = (A – B)  (A – C) using Venn diagram (2) A – ( BC) (3) A – B (1) BC A B A B B A C C C (4) A – C (5) (A – B)(A – C) A B B A C C From the diagram (2)and(5) A – (BC) = (A – B)(A – C)

  4. Prove that A  (B  C) = (A  B)  (A  C) using Venn diagram. B A A B A B C C C A  (B  C) A B B C A B A B From the figures 2 and 5 A(B  C)=(AB)(AC) C C A C (A B)(A  C)

  5. Prove that A  (B  C) = (A  B)  (A  C) using Venn diagram. B A A B A B C C C A  (B  C) A B B C A B A B From the figures 2 and 5 A(BC)=(AB)(AC) C C A C (AB)  (AC)

  6. Prove that (A  B)’ = A’  B’ using Venn diagram. U U U A B B A B A 2. (AB)’ 3. A’ 1. AB U U B A B A From the diagrams 2 and 5 (A  B)’ = A’  B’ 5. A’  B’ 4. B’

  7. Prove that (A  B)’ = A’  B’ using Venn diagram. U U U A B B A B A 2. (A  B)’ 3. A’ 1. A  B U U B A B A From the diagrams 2 and 5 (AB)’ = A’B’ 5. A’  B’ 4. B’

  8. Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}, show that (i) A (B C) = (A B) C. (ii) Verify (i) using Venn diagram. Solution (i) B C = {3, 4, 5, 6}  {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8} ` A  (B C) = {1, 2, 3, 4, 5}  { 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}………… (1) A B = {1, 2, 3, 4, 5}  {3, 4, 5, 6} = {1,2,3,4,5,6} `(A B) C = {1,2,3,4,5,6}  {5,6,7,8} = {1, 2, 3, 4, 5, 6, 7, 8}……………. (2) From (1) and (2), we have A  (B C) = (A B) C.

  9. A A B Using Venn diagram, we have A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8} B 3 3 4 1 4 5 5 2 6 6 8 8 7 7 C C (1) B  C (2) A  (B  C)

  10. Let A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. (i) Show that A  (B C) = (A B) C. (ii) Verify (i) using Venn diagram. Solution Given A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. B C = {a,c,e}  {a,e} = {a,e} A  (B C) = {a,b,c,d}  {a,e} = {a}…… (1) A B = {a,b,c,d.}  {a,c,e} = {a,c}. (A B) C = {a,c}  {a,e} = {a}…….. (2) From (1) and (2) A  (B C) = (A B) C.

  11. A A B B A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. a a e C C (1) BC (2) A(BC)

  12. A A B B A = {a,b,c,d}, B = {a,c,e} and C = {a,e}. c a a C C (1) AB (2) (AB)C from (2) and (4) , it is verified that A  (B C) =(A B) C

  13. Let A = {0,1,2,3,4}, B = {1, - 2, 3,4,5,6} and C = {2,4,6,7}. (i) Show that A  (B C) = (A B) (A C). (ii) Verify using Venn diagram. Solution (i) B C = {1, - 2, 3, 4, 5, 6}  {2, 4, 6, 7 } = {4, 6}; A  (B C) = {0,1, 2, 3, 4}  {4, 6} = {0,1,2,3,4,6}…….(1) A B = {0,1,2,3,4}  {1, - 2, 3,4,5,6} = {- 2, 0,1, 2, 3, 4, 5, 6}, A C = {0,1,2,3,4}  {2,4,6,7} = {0,1, 2, 3, 4, 6, 7}. (A B)  (A C) = {- 2, 0,1, 2, 3, 4, 5, 6}  {0,1, 2, 3, 4, 6, 7} = {0,1, 2, 3, 4, 6}. …….(2) From (1) and (2) ,we get A  (B C) = (A B)  (A C).

  14. A A B B A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}. 3 1 0 4 4 6 6 2 C C (1) BC (2) A(BC)

  15. A = {0, 1, 2, 3, 4}, B = {1, –2, 3, 4, 5, 6},C = {2, 4, 6, 7}. A A -2 B B 1 3 3 5 1 0 0 4 6 4 6 2 2 7 C C (3) AB (4) A C from (2) and (4) , it is verified that A  (B C) =(A B) C

  16. A B 3 1 0 4 6 2 C (5) (AB)(AC) from (2) and (5) , it is verified that A  (B C) = (A B)  (A C)

  17. Given that U = {a,b,c,d,e, f,g,h}, A = {a, b, f, g}, and B = {a, b, c}, verify De Morgan’s laws of complementation. U = {a, b, c, d, e, f, g, h} A = {a, b, f, g} B = {a, b, c}. De Morgan’s laws (AB)’ = A’  B’ (AB)’ = A’  B’ VERIFICATION A  B = {a, b, c, f, g} (A  B)’ = {d, e, h}……..(1) A’ = {c, d, e, h} B’ = {d, e, f, g, h} A’B’ = {d, e, h}……….(2) From (1) and (2) (A  B)’ = A’B’ A  B = {a, b} (A  B)’ = {c, d, e, f, g, h}....(1) A’ = {c, d, e, h} B’ = {d, e, f, g, h} A’  B’ = {c, d, e, f, g, h}….(2) From (1) and (2) (A  B)’ = A’  B’

  18. Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}.. A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} C = {3,9,10,12,13}.. De Morgan’s laws (1) A\(BC) = (A\B)(A\C) (2) A\(BC) = A\B)  (A\C) VERIFICATION B  C = {1, 2, 3, 5, 7, 9, 10, 12, 13} A\(BC) = {11, 15}……..(1) A\B = {3, 9, 11, 13, 15} A\C = {1, 5, 7, 11, 15} (A\B)(A\C) = {11, 15}……….(2) From (1) and (2) A\(BC) = (A\B)(A\C)

  19. Verify De Morgan’s laws for set difference using the sets given below: A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} and C = {3,9,10,12,13}.. A = {1, 3, 5, 7, 9,11,13,15}, B = {1, 2, 5, 7} C = {3,9,10,12,13}.. BC = { } A\(B C) = {1, 3, 5, 7, 9,11,13,15}……..(1) A\B = {3, 9, 11, 13, 15} A\C = {1, 5, 7, 11, 15} (A\B)  (A\C) = {1, 3, 5, 7, 9,11,13,15}……….(2) From (1) and (2) A\(B C) = (A\B)  (A\C)

  20. Let A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5,10,15, 20, 30}and C = {7, 8,15,20,35,45, 48}. Verify A\(B C) = (A\B)  (A\C) A = {10,15, 20, 25, 30, 35, 40, 45, 50}, B = {1, 5, 10, 15, 20, 30} C = {7, 8, 15, 20, 35, 45, 48} BC = {15, 20} A\(B C) = {10, 25, 30, 35, 40, 45, 50}……..(1) A\B = {25, 30, 35, 40, 45, 50} A\C = {10, 25, 30, 40, 50} (A\B)  (A\C) = {10, 25, 30, 35, 40, 45, 50}……….(2) From (1) and (2) A\(B C) = (A\B)  (A\C)

  21. Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and f : A " B be defined by f(x) = . Represent f by (i) an arrow diagram (ii) a set of ordered pairs (iii) a table (iv) a graph . ARROW DIAGRAM A B 6 9 15 18 21 1 2 4 5 6 SET OF ORDERED PAIR f = {(6, 1), (9, 2), (15, 4), (18, 5), (21, 6)}

  22. Table Graph – 7 – 6 – 5 – 4 – 3 – 2 – 1  (21, 6)  (18, 5)  (15, 4)  (9, 2)  (6, 1) | | | | | | | | | 3 6 9 12 15 18 21 24 27

  23. Let A= { 0, 1, 2, 3 } and B = { 1, 3, 5, 7, 9 } be two sets. Let f : A B be a function given by f (x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow diagram and (iv) a graph. Given A= {0, 1, 2, 3}, B = {1, 3, 5, 7, 9} f(x) = 2x + 1 f(0) = 2(0) + 1 = 0 + 1 = 1 f(1) = 2(1) + 1 = 2 + 1 = 3 f(2) = 2(2) + 1 = 4 + 1 = 5 f(3) = 2(3) + 1 = 6 + 1 = 7 (i)Set of ordered pairs {(0, 1), (1, 3), (2, 5), (3, 7)} (ii) Table 0 1 2 3 x 1 3 5 7 f(x)

  24. 4 6 8 10 ARROW DIAGRAM B A 3 4 5 6 7 – 7 – 6 – 5 – 4 – 3 – 2 – 1 Graph  (10, 6)  (8, 5)  (6, 4)  (4, 3) | | | | | | | | 0 2 4 6 8 10 12 14 16

  25. A function f: [1, 6)R is defined as follows f(x) = Find the value of (i) f(5), (ii) f(3), (iii) f(1), (iv) f(2) – f(4) (v) 2f(5) – 3f(1) Since 5 lies between 4 and 6 f(x) = 3x2 – 10 f(5) = 3(5)2 – 10 = 75 – 10 = 65 Since 3 lies between 2 and 4 f(x) = 2x – 1 = 2(3) – 1 = 6 – 1 = 5 Since 1 lies in the interval 1  x < 2 f(x) = 1 + x f(1) = 1 + 1 = 2 Since 2 lies in the interval 2  x < 4 f(x) = 2x – 1 = 2(2) – 1 = 4 – 1 = 3 Since 4 lies in the interval 4  x < 6 f(x) = 3x2 – 10 f(4) = 3(4)2 – 10 = 48 – 10 = 38

  26. f(2) – f(4) = 3 – 38 = – 34 2f(5) – 3 f(1) = 2 (65) – 3(2) = 130 – 6 = 124

  27. A function f: [-3, 7)R is defined as follows Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (i) Since 5 lies between 4 and 6 f(x) = 3x – 2 f(5) = 3(5) – 2 = 15 – 2 = 13 Since 6 lies in the interval 4 < x  6 f(x) = 2x – 3 f(6) = 2(6) – 3 = 12 – 3 = 9 f(5) + f(6) = 13 + 9 = 22

  28. A function f: [-3, 7)R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4)(iv) (ii) Since 1 lies in the interval -3  x < 2 f(x) = 4x2 – 1 f(1) = 4(1)2 – 1 = 4 – 1 = 3 Since –3 lies in the interval-3  x < 2 f(x) = 4x2 – 1 = 4(-3)2 – 1 = 36 – 1 = 35 f(1) – f(-3) = 3 – 35 = – 32

  29. A function f: [-3, 7)R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (iii) Since -2 lies in the interval -3  x < 2 f(x) = 4x2 – 1 f(1) = 4(-2)2 – 1 = 16 – 1 = 15 Since 4 lies in the interval 2  x  4 f(x) = 3x – 2 = 3(4) – 2 = 12 – 2 = 10 f(-2) – f(4) = 15 – 10 = 5

  30. A function f: [-3, 7)R is defined as follows f(x) = Find the value of (i) f(5) + f(6) (ii) f(1) – f(– 3), (iii) f(–2) – f(4) (iv) (iv) Since 3 lies in the interval 2  x  4 f(x) = 3x – 2 f(3) = 3(3) – 2 = 9 – 2 = 7 Since -1 lies in the interval -3  x < 2 f(x) = 4x2 – 1 = 4(-1)2 – 1 = 4 – 1 = 3 f(6) = 9, f(1) = 3

  31. A function f: [-7, 6)R is defined as follows Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (i) Since -4 lies in the interval -5  x  2 f(x) = x + 5 f(-4) = – 4 + 5 = 1 Since 2 lies in the interval -5  x  2 f(x) = x + 5 f(2) = 2 + 5 = 7 2f(-4) + 3f(2) = 2(1) + 3(7) = 2 + 21 = 23

  32. A function f: [-7, 6)R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (ii) Since -7 lies in the interval -7  x < -5 f(x) = x2 + 2x + 1 f(-7) = (–7)2 + 2(–7) + 1 = 49 – 14 + 1 = 36 Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(2) = -3 + 5 = 2 f(-7) – f(–3) = 36 – 2 = 34

  33. A function f: [-7, 6)R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (iii) Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(-3) = -3 + 5 = 2 Since -6 lies in the interval -7  x < -5 f(x) = x2 + 2x + 1 f(-6) = (–6)2 + 2(–6) + 1 = 36 – 12 + 1 = 25 4 lies in the interval 2 < x < 6 f(x) = x – 1 f(4) = 4 – 1 = 3 Since 1 lies in the interval -5  x  2 f(x) = x + 5 f(1) = 1 + 5 = 6

  34. A function f: [-7, 6)R is defined as follows f(x) = Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (iii) f(-3) = 2 f(-6) = 25 f(4) = 3 f(1) = 6

  35. If verify that (AB)T= BTAT . LHS = RHS  (AB)T= BTAT .

  36. If verify that (AB)T= BTAT .

  37. If verify that (AB)T= BTAT . LHS = RHS  (AB)T= BTAT .

  38. If then show that A2 – 4A + 5I2 = O A2 – 4A + 5I2 = O

  39. 7. Find X and Y if 2X + 3Y =

  40. 11. An electronic company records each type of entertainment device sold at three of their branch stores so that they can monitor their purchases of supplies. The sales in two weeks are shown in the following spreadsheets. Find the sum of the items sold out in two weeks using matrix addition.

  41. If verify (AB)C = A(BC) . LHS = RHS (AB)C =A(BC)

  42. If then prove that (A + B)2  A2 + 2AB + B2.

  43. If then prove that (A + B)2  A2 + 2AB + B2. LHS RHS (A + B)2  A2 + 2AB + B2.

  44. If find (A + B)C and AC + BC. Is (A + B)C = AC + BC ?

  45. If find (A + B)C and AC + BC. Is (A + B)C = AC + BC ?  (A+B)C = AC + BC

  46. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term. (2) – (1)  8d = 32 d = 32/8 d = 4 Sub d = 4 in (1) a + 9(4) = 41 a + 36 = 41 a = 41 – 36 a = 5 t27 = a + (27 – 1)d = 5 + (26)4 = 5 + 104 = 109 Given t10 = 41 t18 = 73 tn = a + (n – 1)d t10 = 41 a + (10 – 1)d = 41 a + 9d = 41…………(1) t18 = a + (18 – 1)d = 73 a + 17d = 73 …….(2) (2) a + 17d = 73 (1)  a + 9d = 41 (-) (-) (-)

  47. The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers. Let a – d, a, a + d be three consecutive terms in an A.P. Sum = 6 a – d + a + a + d = 6 3a = 6 a = 6/3 = 2 Product = – 120 (a – d)  a  (a + d) = – 120 (2 – d)  2  (2 + d) = – 120 (22 – d2)  2 = – 120 4 – d2 = – 120/2 = – 60 – d2 = – 60 – 4 – d2 = – 64 d2 = 64 d = 8 The given numbers are 2 – 8, 2, 2 + 8 – 6, 2, 10 (or) 10, 2, –6

  48. A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the 15th year. Given t7 = 1000 t10 = 1450 tn = a + (n – 1)d t7 = 1000 a + (7 – 1)d = 1000 a + 6d = 1000…………(1) t10 = a + (10 – 1)d = 1450 a + 9d = 1450 …….(2) (2) a + 9d = 1450 (1) a + 6d = 1000 (2) – (1)  3d = 450 d = 450/3 d = 150 Sub d = 150 in (1) a + 6(150) = 1000 a + 900 = 1000 a = 1000 – 900 a = 100 t15 = a + (15 – 1)d = 100 + (15)150 = 100 + 2250 = 2350

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