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X2 T08 03 inequalities & graphs (2011)

N
Nigel Simmons

The document discusses solving inequalities and graphs. It provides an example of solving the inequality x^2 ≤ 1/(x+2). It also discusses properties of the graph of y=x, showing that it is increasing for x≥0. It uses this to prove inequalities relating sums and integrals. Finally, it introduces a proof by mathematical induction to show an inequality relating sums and fractions is true for all integers n≥1.

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Inequalities & Graphs
Inequalities & Graphs x2 e.g. i  Solve 1 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2 Oblique asymptote:
Inequalities & Graphs x2 e.g. i  Solve 1 x2 4 x2 Oblique asymptote:  x2 x2 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2 4 x2 Oblique asymptote:  x2 x2 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2 4 x2 Oblique asymptote:  x2 x2 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2
Inequalities & Graphs x2 e.g. i  Solve 1 x2 x2 1 x2 x2  x  2 x2  x  2  0  x  2 x  1  0 x  2 or x  1
Inequalities & Graphs x2 e.g. i  Solve 1 x2 x2 1 x2 x2  x  2 x2  x  2  0  x  2 x  1  0 x  2 or x  1 x2 1 x2 x  2 or  1  x  2
(ii) (1990) Consider the graph y  x
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x dy 1  dx 2 x
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x dy 1  dx 2 x dy   0 for x  0 dx
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x at x  0, y  0 dy 1  dx 2 x dy   0 for x  0 dx
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x at x  0, y  0 dy 1 when x  0, y  0  dx 2 x dy   0 for x  0 dx
(ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x at x  0, y  0 dy 1 when x  0, y  0  dx 2 x dy   0 for x  0  curve is increasing for x  0 dx
b) Hence show that; n 2 1  2    n   xdx  n n 0 3
b) Hence show that; n 2 1  2    n   xdx  n n 0 3
b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve
b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve n 1  2    n   xdx 0
b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve n 1  2    n   xdx 0 n 2 x x  3 0  2  n n 3
b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve n 1  2    n   xdx 0 n 2 x x  3 0  2  n n 3 n 2  1  2    n   xdx  n n 0 3
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 L.H .S  1 1
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer 4k  3 i.e. 1  2    k  k 6
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer 4k  3 i.e. 1  2    k  k 6 Prove the result is true for n  k  1
c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer 4k  3 i.e. 1  2    k  k 6 Prove the result is true for n  k  1 4k  7 i.e. Prove 1  2    k  1  k 1 6
Proof:
Proof: 1  2  k 1  1  2  k  k 1
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6  k  14k  12  1  6 k  1 6
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6  k  14k  12  1  6 k  1 6  k  14k  1  6 k  1 2 6
Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6  k  14k  12  1  6 k  1 6  k  14k  1  6 k  1 2 6 4k  1 k  1  6 k  1  6  4k  7  k  1 6
Hence the result is true for n = k +1 if it is also true for n =k
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6 666700  1  2    10000  666700
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6 666700  1  2    10000  666700  1  2    10000  666700 to the nearest hundred
Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6 666700  1  2    10000  666700  1  2    10000  666700 to the nearest hundred Exercise 10F

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X2 T08 03 inequalities & graphs (2011)

  • 2. Inequalities & Graphs x2 e.g. i  Solve 1 x2
  • 3. Inequalities & Graphs x2 e.g. i  Solve 1 x2
  • 4. Inequalities & Graphs x2 e.g. i  Solve 1 x2
  • 5. Inequalities & Graphs x2 e.g. i  Solve 1 x2
  • 6. Inequalities & Graphs x2 e.g. i  Solve 1 x2 Oblique asymptote:
  • 7. Inequalities & Graphs x2 e.g. i  Solve 1 x2 4 x2 Oblique asymptote:  x2 x2 x2
  • 8. Inequalities & Graphs x2 e.g. i  Solve 1 x2 4 x2 Oblique asymptote:  x2 x2 x2
  • 9. Inequalities & Graphs x2 e.g. i  Solve 1 x2 4 x2 Oblique asymptote:  x2 x2 x2
  • 10. Inequalities & Graphs x2 e.g. i  Solve 1 x2
  • 11. Inequalities & Graphs x2 e.g. i  Solve 1 x2
  • 12. Inequalities & Graphs x2 e.g. i  Solve 1 x2 x2 1 x2 x2  x  2 x2  x  2  0  x  2 x  1  0 x  2 or x  1
  • 13. Inequalities & Graphs x2 e.g. i  Solve 1 x2 x2 1 x2 x2  x  2 x2  x  2  0  x  2 x  1  0 x  2 or x  1 x2 1 x2 x  2 or  1  x  2
  • 14. (ii) (1990) Consider the graph y  x
  • 15. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0
  • 16. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx
  • 17. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x dy 1  dx 2 x
  • 18. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x dy 1  dx 2 x dy   0 for x  0 dx
  • 19. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x at x  0, y  0 dy 1  dx 2 x dy   0 for x  0 dx
  • 20. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x at x  0, y  0 dy 1 when x  0, y  0  dx 2 x dy   0 for x  0 dx
  • 21. (ii) (1990) Consider the graph y  x a) Show that the graph is increasing for all x  0 dy Curve is increasing when 0 dx y x at x  0, y  0 dy 1 when x  0, y  0  dx 2 x dy   0 for x  0  curve is increasing for x  0 dx
  • 22. b) Hence show that; n 2 1  2    n   xdx  n n 0 3
  • 23. b) Hence show that; n 2 1  2    n   xdx  n n 0 3
  • 24. b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve
  • 25. b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve n 1  2    n   xdx 0
  • 26. b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve n 1  2    n   xdx 0 n 2 x x  3 0  2  n n 3
  • 27. b) Hence show that; n 2 1  2    n   xdx  n n 0 3 As x is increasing; Area outer rectangles  Area under curve n 1  2    n   xdx 0 n 2 x x  3 0  2  n n 3 n 2  1  2    n   xdx  n n 0 3
  • 28. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6
  • 29. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1
  • 30. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 L.H .S  1 1
  • 31. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6
  • 32. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S
  • 33. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1
  • 34. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer
  • 35. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer 4k  3 i.e. 1  2    k  k 6
  • 36. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer 4k  3 i.e. 1  2    k  k 6 Prove the result is true for n  k  1
  • 37. c) Use mathematical induction to show that; 4n  3 1  2  n  n for all integers n  1 6 Test: n = 1 41  3 L.H .S  1 R.H .S  1 6 1 7  6  L.H .S  R.H .S Hence the result is true for n = 1 Assume the result is true for n  k where k is a positive integer 4k  3 i.e. 1  2    k  k 6 Prove the result is true for n  k  1 4k  7 i.e. Prove 1  2    k  1  k 1 6
  • 39. Proof: 1  2  k 1  1  2  k  k 1
  • 40. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6
  • 41. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6
  • 42. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6
  • 43. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6
  • 44. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6  k  14k  12  1  6 k  1 6
  • 45. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6  k  14k  12  1  6 k  1 6  k  14k  1  6 k  1 2 6
  • 46. Proof: 1  2  k 1  1  2  k  k 1 4k  3  k  k 1 6  4k  3 k  6 k  1 2 6 16k 3  24k 2  9k  6 k  1  6 k  116k 2  8k  1  1  6 k  1  6  k  14k  12  1  6 k  1 6  k  14k  1  6 k  1 2 6 4k  1 k  1  6 k  1  6  4k  7  k  1 6
  • 47. Hence the result is true for n = k +1 if it is also true for n =k
  • 48. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n
  • 49. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred
  • 50. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6
  • 51. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6
  • 52. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6 666700  1  2    10000  666700
  • 53. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6 666700  1  2    10000  666700  1  2    10000  666700 to the nearest hundred
  • 54. Hence the result is true for n = k +1 if it is also true for n =k Since the result is true for n = 1 then it is also true for n =1+1 i.e. n=2, and since the result is true for n = 2 then it is also true for n =2+1 i.e. n=3, and so on for all positive integral values of n d) Use b) and c) to estimate; 1  2    10000 to the nearest hundred 2 4n  3 n n  1  2  n  n 3 6 2 410000  3 10000 10000  1  2    10000  10000 3 6 666700  1  2    10000  666700  1  2    10000  666700 to the nearest hundred Exercise 10F