ADVANCED HIGHER MATHS

1. ADVANCED HIGHER MATHS UNIT 1 REVISION AND FORMULAE St Mungo's Academy

2. Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS r r Expand r r Expand Expand (a − b)5 a5 − 5a4b + 10a3b² − 10a²b3 + 5ab4 − b5. 8x3 − 12x² + 6x − 1 Expand (2x − 1)3 St Joseph’s College

3. Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS St Joseph’s College

4. Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS St Mungo's Academy

5. Unit 1 Outcome 2 DIFFERENTIATION f(x) function c x 2x xn (ax + b)n sin x cos x sin(ax + b) cos(ax + b) f’(x) derivative 0 1 These are your standard derivatives for now! All of these were covered in the Higher course 2 nxn−1 an(ax+b)n−1 cos x −sin x acos(ax + b) -asin(ax + b) St Joseph’s College

6. Unit 1 Outcome 2 DIFFERENTIATION New Trigonometric Functions AND St Joseph’s College

7. Unit 1 Outcome 2 DIFFERENTIATION Product Rule Quotient Rule St Joseph’s College

8. Unit 1 Outcome 2 DIFFERENTIATION The six basic trigonometric derivatives 1 - Derivative of sin x. The derivative of f(x) = sin x is given by f '(x) = cos x 2 - Derivative of cos x. The derivative of f(x) = cos x is given by f '(x) = - sin x 3 - Derivative of tan x. The derivative of f(x) = tan x is given by f '(x) = sec 2 x St Joseph’s College

9. Unit 1 Outcome 2 DIFFERENTIATION The six basic trigonometric derivatives 4 - Derivative of cot x. The derivative of f(x) = cot x is given by f '(x) = - cosec 2 x 5 - Derivative of sec x. The derivative of f(x) = sec x is given by f '(x) = sec x tan x 6 - Derivative of cosec x. The derivative of f(x) = cosec x is given by f '(x) = - cosec x cot x St Joseph’s College

10. Unit 1 Outcome 2 DIFFERENTIATION Higher derivatives St Joseph’s College

11. Unit 1 Outcome 2 DIFFERENTIATION Motion a = dv/dt = d2x/dt2 v = dx/dt These are used in the example over the page St Joseph’s College

12. Unit 1 Outcome 2 DIFFERENTIATION • Find velocity and acceleration after • (a) t secs and (b) 4 secs for • particles travelling along a straight line if: • x = 2t3 – t2 +2 • x = t2 +8/t • x =8t + et 6t2 – 2t 12t – 2 46 m/s2 88 m/s 2t – 8/t2 2+ 16/t3 7.5 m/s 2.25 m/s2 8 + et et e4 m/s2 8+ e4 m/s 63m/s 55 m/s2 St Mungo's Academy

13. Unit 1 Outcome 3 INTEGRATION St Mungo's Academy

14. Unit 1 Outcome 3 INTEGRATION + St Joseph’s College

15. Unit 1 Outcome 3 INTEGRATION St Joseph’s College

16. Unit 1 Outcome 3 INTEGRATION Example of rotating the region about x-axis Example 1 In the picture shown, a solid is formed by revolving the curve y = x about the x-axis, between  x = 0 and x = 3. FIND THE VOLUME St Joseph’s College

17. Unit 1 Outcome 3 INTEGRATION y 8 o x Example of rotating the region about y-axis Example 2 Find the volume of the solid obtained by the region bounded by y=x3, y=8, and x=0 around the y-axis. St Joseph’s College

18. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS x-axis: When Sketch the graph of . [You need not find the coordinates of any stationary points.] Solution y-axis: When , The curve cuts the y-axis at The curve cuts the x-axis at (3, 0). St Joseph’s College

19. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS or x –21 –2 –19 09 1 11 As y Vertical Asymptotes:   , Non-Vertical Asymptote: (since the degree of the denominator is higher than the degree of the numerator). This means that is a non-vertical asymptote. St Joseph’s College

20. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

21. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

22. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

23. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS Non-Vertical Asymptote St Joseph’s College

24. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

25. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

26. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

27. St Joseph’s College

28. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

29. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS f(-x) St Joseph’s College

30. Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS St Joseph’s College

31. Unit 1 Outcome 5 GAUSSIAN ELIMINATION Past Paper 2002 Q1 5 marks Use Gaussian elimination to solve the following system of equations x +y +3z = 2 2x +y + z = 2 3x + 2y + 5z = 5 R2 –2R1 R3 -3R1 R3 –R2 y = -3 z = 1 -y –5z = -2 x - 3 +3 = 2 x = 2 -y = -2 +5 (x,y,z) = (2, -3, 1) St Joseph’s College

32. Unit 1 Outcome 5 GAUSSIAN ELIMINATION Past Paper 2003 Q6 6 marks Use elementary row operations to reduce the following system of Equations to upper triangle form x +y +3z = 1 3x + ay + z = 1 x + y + z = -1 Hence express x, y and z in terms of parameter a. Explain what happens when a = 3 R2 –3R1 R3 -R1 R3/-2 z = 1 y = 6/(a-3) (a-3)y –8 = -2 x + 6/(a –3) + 3 = 1 x = -2 –6/(a-3) (a-3)y = 6 a=3 gives z = ¼ from R2 and z = 1 from R3. Inconsistent equations! St Joseph’s College

33. Unit 1 Outcome 5 GAUSSIAN ELIMINATION Use Gaussian elimination to solve the system of equations below when . Past Paper 2005 Q6 6 marks x +y +2z = 1 2x +ly + z = 0 3x + 3y + 9z = 5 Explain what happens when l = 2 R2 –2R1 R3 -3R1 R3/3 z = 2/3 x = 1 –4/3 (l-2)y –2 = -2 x = -1/3 y = 0 l=2 gives R2 = R3. Infinite number of solutions! St Joseph’s College

34. Unit 1 Outcome 5 GAUSSIAN ELIMINATION Past Paper 2006 Q9 5 marks Use Gaussian elimination to obtain solutions of the equations 2x -y +2z = 1 x +y - 2z = 2 x -2y + 4z = -1 R2 –R1 R3 -2R1 R3-R2 z = t x +2t+1 –2t = 2 -3y +6t = -3 x = 1 y = 2t+1 (x,y,z:x=1, y=2t+1 and z = t) St Joseph’s College