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12 October, 2014 St Mungo's Academy 1 ADVANCED HIGHER MATHS REVISION AND FORMULAE UNIT 1.

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Presentation on theme: "12 October, 2014 St Mungo's Academy 1 ADVANCED HIGHER MATHS REVISION AND FORMULAE UNIT 1."— Presentation transcript:

1 12 October, 2014 St Mungo's Academy 1 ADVANCED HIGHER MATHS REVISION AND FORMULAE UNIT 1

2 12 October, 2014 St Joseph’s College 2 Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS Expand (a − b) 5 a 5 − 5a 4 b + 10a 3 b² − 10a²b3 + 5ab 4 − b 5. Expand (2x − 1) 3 8x 3 − 12x² + 6x − 1 r r rr Expand

3 12 October, 2014 St Joseph’s College 3 Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS

4 12 October, 2014 St Mungo's Academy 4 Unit 1 Outcome 1 BINOMIAL and PARTIAL FRACTIONS

5 12 October, 2014 St Joseph’s College 5 f(x) function c x 2x x n (ax + b) n sin x cos x sin(ax + b) cos(ax + b) 0 f’(x) derivative 1 2 nx n−1 an(ax+b) n−1 cos x −sin x acos(ax + b) -asin(ax + b) These are your standard derivatives for now! All of these were covered in the Higher course Unit 1 Outcome 2 DIFFERENTIATION

6 12 October, 2014 St Joseph’s College 6 New Trigonometric Functions AND Unit 1 Outcome 2 DIFFERENTIATION

7 12 October, 2014 St Joseph’s College 7 Unit 1 Outcome 2 DIFFERENTIATION Product Rule Quotient Rule

8 12 October, 2014 St Joseph’s College 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 The six basic trigonometric derivatives Unit 1 Outcome 2 DIFFERENTIATION

9 12 October, 2014 St Joseph’s College 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 The six basic trigonometric derivatives Unit 1 Outcome 2 DIFFERENTIATION

10 12 October, 2014 St Joseph’s College 10 Unit 1 Outcome 2 DIFFERENTIATION Higher derivatives

11 12 October, 2014 St Joseph’s College 11 Unit 1 Outcome 2 DIFFERENTIATION Motion v = dx / dt a = dv / dt = d 2 x / dt 2 These are used in the example over the page

12 12 October, 2014 St Mungo's Academy 12 Find velocity and acceleration after (a) t secs and (b) 4 secs for particles travelling along a straight line if: (i) x = 2t 3 – t 2 +2 (ii) x = t / t (iii) x = 8t + e t 88 m/s 6t 2 – 2t 12t – 2 46 m/s 2 2t – 8 /t m/s /t m/s 2 e 4 m/s 2 etet 8 + e t 8+ e 4 m/s 63 m/s 55 m/s 2 Unit 1 Outcome 2 DIFFERENTIATION

13 12 October, 2014 St Mungo's Academy 13 Unit 1 Outcome 3 INTEGRATION

14 12 October, 2014 St Joseph’s College 14 Unit 1 Outcome 3 INTEGRATION +

15 12 October, 2014 St Joseph’s College 15 Unit 1 Outcome 3 INTEGRATION

16 12 October, 2014 St Joseph’s College 16 Example of rotating the region about x-axis 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 Example 1 Unit 1 Outcome 3 INTEGRATION

17 12 October, 2014 St Joseph’s College 17 Find the volume of the solid obtained by the region bounded by y=x 3, y=8, and x=0 around the y-axis. o 8 x y Example of rotating the region about y-axis Example 2 Unit 1 Outcome 3 INTEGRATION

18 12 October, 2014 St Joseph’s College 18 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 x-axis:When, The curve cuts the x-axis at (3, 0). Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

19 12 October, 2014 St Joseph’s College 19 This means that is a non-vertical asymptote. Vertical Asymptotes:  or  x–2  1 –2–1  9 0  9 11  1 y Non-Vertical Asymptote: As, (since the degree of the denominator is higher than the degree of the numerator). Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

20 12 October, 2014 St Joseph’s College 20 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

21 12 October, 2014 St Joseph’s College 21 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

22 12 October, 2014 St Joseph’s College 22 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

23 12 October, 2014 St Joseph’s College 23 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS Non-Vertical Asymptote

24 12 October, 2014 St Joseph’s College 24 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

25 12 October, 2014 St Joseph’s College 25 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

26 12 October, 2014 St Joseph’s College 26 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

27 12 October, 2014 St Joseph’s College 27

28 12 October, 2014 St Joseph’s College 28 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

29 12 October, 2014 St Joseph’s College 29 f(-x) Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

30 12 October, 2014 St Joseph’s College 30 Unit 1 Outcome 4 PROPERTIES OF FUNCTIONS

31 12 October, 2014 St Joseph’s College 31 Past Paper 2002Q1 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 z = 1 -y –5z = -2 -y = y = -3 x = 2 x = 2 (x,y,z) = (2, -3, 1) Unit 1 Outcome 5 GAUSSIAN ELIMINATION

32 12 October, 2014 St Joseph’s College 32 Past Paper 2003Q6 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 R2 –3R1 R3 -R1 R3/-2 z = 1 (a-3)y –8 = -2 (a-3)y = 6 y = 6/(a-3) x + 6/( a –3) + 3 = 1 x = -2 –6/(a-3) a=3 gives z = ¼ from R2 and z = 1 from R3. Inconsistent equations! Hence express x, y and z in terms of parameter a. Explain what happens when a = 3 Unit 1 Outcome 5 GAUSSIAN ELIMINATION

33 12 October, 2014 St Joseph’s College 33 Past Paper 2005Q6 6 marks x + y + 2z = 1 2x + y + z = 0 3x + 3y + 9z = 5 R2 –2R1 R3 -3R1 R3/3 z = 2/3 ( -2)y –2 = -2 y = 0 x = 1 –4/3 x = -1/3 =2 gives R2 = R3. Infinite number of solutions! Explain what happens when = 2 Use Gaussian elimination to solve the system of equations below when. Unit 1 Outcome 5 GAUSSIAN ELIMINATION

34 12 October, 2014 St Joseph’s College 34 Past Paper 2006Q9 5 marks 2x - y + 2z = 1 x +  y - 2z = 2 x - 2y + 4z = -1 R2 –R1 R3 -2R1 R3-R2 z = t -3y +6t = -3 y = 2t+1 x +2t+1 –2t = 2 x = 1  x,y,z:x=1, y=2t+1 and z = t) Use Gaussian elimination to obtain solutions of the equations Unit 1 Outcome 5 GAUSSIAN ELIMINATION


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