The document discusses inequalities and graphs. It provides examples of solving inequalities like x^2 ≤ 1 and discusses the graphs of functions like y = x. It shows that y = x is an increasing function for x ≥ 0. It also uses integrals to calculate the area under the graph of y = x and proves results using mathematical induction.

1. Inequalities & Graphs

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
 x2
4
x2 Oblique asymptote:
x2 x2

8. Inequalities & Graphs
x2
e.g. i  Solve 1 x2
 x2
4
x2 Oblique asymptote:
x2 x2

9. Inequalities & Graphs
x2
e.g. i  Solve 1 x2
 x2
4
x2 Oblique asymptote:
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

38. Proof:

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