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- 1. Additional Mathematic s
- 2. How to Solve a Problem Understand Plan your Do - Carry out Check your the Problem Strategy Your Strategy Answers •Which Topic / •Carry out the •Is the answer •Subtopic ? •Choose suitable calculations reasonable? •What info has been strategy •Graph sketching given? •Choose the correct •Any other •What is to be formula •Creating tables ... methods ?? found?
- 3. PAPER 1 FORMAT Objective Test : •Short Questions No. of Questions : •25 questions Total Marks : •80 Duration : •2 hours L.O.D. : •L (15) , M(7-8), H(2-3) Additional Materials : Scientific Calculators, • Mathematical Tables, Geometrical sets.
- 4. PAPER 2 FORMAT •Subjective Questions • No. of Questions : A (6), B (4/5), C (2/4) • Total Marks : 100 • Duration : 2 hours 30 minutes • L.O.D : L (6) , M(4-5), H(4-5) • Additional Materials : Scientific Calculators, • Mathematical Tables, • Geometrical sets.
- 5. Key towards achieving 1A … •Read question carefully •Follow instructions •Start with your favourite question •Show your working clearly •Choose the correct formula to be used +(Gunakannya dengan betul !!!) •Final answer must be in the simplest form •The end answer should be correct to 4 S.F. • (or follow the instruction given in the question) • 3.142
- 6. Kunci Mencapai kecemerlangan •Proper / Correct ways of writing mathematical notations •Check answers! •Proper allocation of time (for each question) • •Paper 1 : 3 - 7 minutes for each question •Paper 2 : • Sec. A: 8 - 10 minutes for each question • Sec. B: 15 minutes for each question • Sec. C: 15 minutes for each question
- 7. Common Mistakes… 1. The Quadratic equation 3x2 - 4x = 0 • y = 3x2 + 4x y = 6x + 4 4. sin x = 300 , x = 300 , 1500 5.
- 8. Kesilapan Biasa Calon … • f ' (x) wrongly interpreted as f – 1(x) and / or conversely • x2= 4 x = 2 • • x2> 4 x > ±2 •
- 9. Common errors… PA : PB = 2 : 3 then 2PA = 3 PB
- 10. Actually, … PA : PB = 2 : 3 3 PA = 2 PB
- 11. More mistakes …… 32 PA2 = 22 PB2 9 PA2 = 4 PB2
- 12. Common mistakes … logax + loga y = 0, then xy = 0 It should be… xy = a0= 1
- 13. Common mistakes … loga(x – 3) = loga x – loga 3 2x x 2y = 1 2x x 2y = 20 x + y = 1 2 x + y = 20 x + y = 0
- 14. Common mistakes … logax + loga y = 0, then loga xy = 0 So, xy = 0 It should be… xy = a0= 1
- 15. Common mistakes … sin (x + 30 0) = ½, then sin x + sin 300= ½ …………………gone ! Do NOT use Sin(A+B) = sin A cos B + cos A sin B !
- 16. Correct way… … sin (x + 300) = ½ , then x +300= 300 , 1500 So, x = 00 , 1200 ? If 00 is an answer, then 3600 is also an answer !
- 17. sin (x + 300) = ½ , then x +300= 300 , 1500 , 3900 So, x = 00 , 1200 , 3600
- 18. Relationship between Functions and Quadratic Functions Domain Codomain y X Y f(x) = x 2 4 1 1 1 2 4 x Image O 1 2 Objec t (1, 1) , (2, 4). …. form ordered pairs and can be plotted to obtain a curve.
- 19. SPM 2003 Paper 1, Question 1 P = { 1, 2, 3} Q = {2, 4, 6, 8, 10} • The relationship between P and Q is defined by the set of ordered pairs { (1, 2), (1, 4), (2, 6), (2, 8)}. • State • the image of 1, • The object of 2. [2 marks] Answer (a) 2, 4 1 (a) 1 1
- 20. SPM 2003 Paper 1, Question 2 Answer (a) or 0.4 2 B1: or g(x) = 3 25x2 + 2 2 B1: (5x+1)2 – 2(5x+1) + 3
- 21. SPM 2003 Paper 1, Question 3 (SPM 2005,Q5) Solve the quadratic equation 2x(x – 4) = (1- x)(x+2). Write your answer correct to four significant figures. (3 marks) Answer 2.591, - 0.2573 (both + 4 s.f.) 3 B2: B1: 3x2– 7x – 2 = 0
- 22. SPM 2003 Paper 1, Question 4 The quadratic equation x (x+1) = px – 4 has two distinct roots. Find the range of values of p. (3 marks) Answer p < -3, p > 5 (kedua-duanya) 3 B2: (p + 3) (p – 5) > 0 B1: (1 – p)2– 4(1)(4) > 0
- 23. SPM 2003 Paper 1, Question 5 Given that log 2T - log4 V = 3, express T in terms of V. (4 marks) Answer T= 8V½ 4 B1 B2 B3
- 24. SPM 2003 Paper 1, Question 6 Solve the equation 42x – 1 = 7x (4 marks) Answer x = 1.677 4 (2x – 1) log 4 = x log 7 B1 2xlog 4 – log 4 = x log 7 2xlog 4 – x log 7 = log 4 B2 x(2 log 4 – log 7 ) = log 4 B3
- 25. SPM 2007 (???) Solve the equation 42x – 1 = 8x (3 marks) Answer 22(2x – 1) = 23x 2(2x – 1) = 3x 4x – 1 = 3x x = 1 4x – 2 = 3x No !!! x = 2
- 26. SPM 2003 Paper 1, Question 7 The first three terms of an A.P.are k-3, k+3, 2k+2. Find (a) the value of k, (b) the sum of the first 9 terms of the progression. (3 marks) Answer (a) 7 2 (k + 3) – (k – 3) = (2k + 2) – (k + 3) B1 6 = k–1 (b) 252 1
- 27. SPM 2003 Paper 2, Question 1 Solve the simultaneous equation 4x + y = - 8 and x2+ x – y = 2 (5 marks) Answer Make x or y the subject P1 Eliminating x or y K1 Solving the quadratic equation : K1 x = -2, -3 or y = 0 , 4 N1 y = 0 , 4 or x = -2, -3 N1
- 28. SPM 2003 Paper 2, Question 2 The function f(x) = x2 - 4kx + 5k2 + 1has a minimum value of r2 + 2k , with r and k as constants. • By the method of completing the square, show that r=k–1 (4 marks) • Hence, or otherwise, find the value of k and the value of r if the graph of the function is symmetrical about the line x = r2 -1. (4 marks)
- 29. SPM 2003 Paper 2, Question 2 *** Answer 2(a) Writing f(x) in the form (x – p)2 + q (x – 2k)2 – 4k2 + 5k2 + 1 K1 Equating q ( q* = r2 + 2k) K1 (k – 1)2 = r2 N1 r= k–1 N1 (b) Equating (his) - (x – p) = 0 K1 Eliminating r or k by K1 any valid method k=0,4 N1 r = -1, 3 N1
- 30. F4 1. Functions 1. f:x x2 - 2 . Find the values of x which map onto Given itself. f (x) = x x2 - 2 = x x 2– x – 2 = 0 (x+1)(x-2) = 0 x = -1 , x = 2 2. f:x x - 3 , g:x 3x , find gf(1). Given f(x) = x – 3, g(x) = 3x gf (1) = g [ f(1) ] = g [-2] = -6
- 31. T4 BAB 1 F4 Functions : Inverse Functions 4. Given f (x) = 3 – 2x, find f -1. Method 1 Method 2 Let f (x) = y Let f -1(x) =y Then 3 – 2x = y Then x = f (y) 3 – y = 2x x = 3 – 2y
- 32. T4 BAB 1 F4 Functions : Applying the Idea of Inverse functions 5. Given , find the value of a if f -1(a) = 11 Method 1 (Find f-1 ) Method 2 ( No need f-1 ) Let f -1(x) = y Then x = f(y) Let f -1(a) = 11 x= Then a = f (11) y= = 8 f-1(a) = = 11 a= 8
- 33. T4 BAB 1 F4 Functions : Given composite function and one function, find the other function. 6. Given find fg. Remember : you need to find g first ! f(x) =2 - x , gf(x) = 2x-2 Let f(x) = u Then u = 2 – x or x=2-u g(u) = 2(2-u) – 2 = 2-2u g(x) = 2-2x fg(x) = f(2-2x) = 2 - (2-2x) = 2x
- 34. T4 BAB 1 F4 **Functions :To skecth the graphs of y = |f(x)| 7. Skecth the graph of y = |3-2x|+1 for domain 0 ≤ x ≤ 4 and state the corresponding range. Tips : Sketch y = |3-2x| first !!! y 6 5 4 3 Range : 1≤ y ≤ 6 2 1 x 0 4
- 35. F4 2. Quadratic equations: SPM 2004, K1, Q4 Form the quadratic equation which has the roots – 3 and ½ . x = – 3 , x = ½ (x+3) (2x – 1) = 0 2x2 + 5x – 3 = 0
- 36. F4 2. Quadratic Equations ax2 + bx + c = 0 x2 – ( S.O.R) x + (P.O.R.) = 0 S.O.R P.O.R. = =
- 37. F4 2. The Quadratic Equation : Types of roots The quadratic ax2 + bx + c= 0 has equation 1. Two distinct roots if b2 - >0 4ac 2. Two equal roots if b2 - =0 4ac 3. No real roots if b2 - <0 4ac **The straight line y = mx -1 is a tangent to the curve y = x2+ 2 ……. ???
- 38. F4 3 Quadratic Functions : Quadratic Inequalities SPM 2004, K1, S5 Find the range of values of x for which x(x – 4) ≤ 12 x (x – 4) ≤ 12 x2 – 4x – 12 ≤ 0 (x + 2)(x – 6) ≤ 0 x -2 6 –2≤ x ≤ 6
- 39. F4 Back to Solve BASIC x 2 > 4 x> ±2 x2 – 4 > 0 ??? (x + 2)(x – 2) > 0 R.H.S must be O ! –2 2 x < -2 or x > 2
- 40. F4 4. Simultaneous Equations • Solve the simultaneous equations x + y =1 x2 + 3y2 = 7 Factorisation • Solve the simultaneous equations, give your answer correct to three decimal places. x +y=1 x2 +3y2 = 8 *** P = Q = R
- 41. F4 5. INDICES Back to basic… … Solve .. 32(x – 1) . 3 (– 3x) = 1 2x – 2 – 3x = 1 – x = 3 x= –3 Betul ke ???
- 42. F4 5. INDICES Solve 3 2(x – 1) . 3 (– 3x) = 1 32x – 2 +(– 3x) = 30 –x–2 =0 x = –2
- 43. F4 5. INDICES Solve or… 9x-1 = 27x 3 2(x – 1)= 3 3x 32x – 2 = 33x 2x – 2 = 3x x = –2
- 44. F4 5. INDICES Solve 2x + 3 = 2x+2 Can U take 2x + 3 = 2x . 22 logon both 2x + 3 = 4 (2x ) sides ??? 3 = 3(2x ) WHY? 1 = (2x ) In the form x = 0 u + 3 = 4u
- 45. F4 5. INDICES Solve the equation , give your answer correct to 2 decimal places. [ 4 marks] 9 (3x) = 32 + (3x) 8 (3x) = 32 3x = 4 x = 1.26 (Mid-Yr
- 46. F4 5. INDICES Solve 22x . 5x = 0.05 ambm = (ab)m 4x . 5x = You can also take 20x = log on both sides. x = –1
- 47. F4 5. INDICES & LOGARITHMS (Mid-Yr 07) Solve the equation [4 marks] x–2 = 4 (4 – x) x = 3.6
- 48. F4 5. INDICES & LOGARITHMS Back to basic… … Solve the the equation log3(x – 4) + log3 (x + 4) = 2 log3(x-4)(x+4) = 2 x2 – 16 = 9 x = 5
- 49. F4 Back to basic… … SPM 2005, P1, Q8 Solve the equation log34x – log3(2x – 1) = 1 4x = 3(2x – 1) = 6x – 3 2x = 3 x =
- 50. F4 5 Indices and Logaritms : Change of base Given that log3 p = m and log4p = n. Find logp36 in terms of m and n. logp 36 = logp 9 + logp K1 4 = 2log p 3 + logp 4 K1 K1 N1 logaa 1 =
- 51. Coordinate Geometry Some extra vitamins 4u …
- 52. Coordinate Geometry Distance between two points Division of line segments : midpoints + the ratio theorem Areas of polygons Equation of straight lines Parallel and perpendicular lines Loci (involving distance between two points)
- 53. Coordinate Geometry Note to candidates: Solutions to this • question by scale drawing will not be accepted.
- 54. Coordinate Geometry Note to candidates: A diagram is usually given • (starting from SPM 2004). You SHOULD make full use of the given diagram while answering the question.
- 55. Coordinate Geometry Note to candidates: Sketch a simple diagram • to help you using the required formula correctly.
- 56. 6. Coordinate Geometry 6.2.2 Division of a Line Segment Q divides the line segment PR in the ratio PQ : QR = m : n R(x2, y2) n m n ● P(x1, y1) m Q(x, y) Q(x, y) R(x2, y2) P(x1, y1) Q(x, y) =
- 57. 6. Coordinate Geometry (Ratio Theorem) The point P divides the line segment joining the point M(3,7) and N(6,2) in the ratio 2 : 1. Find the coordinates of point P. 1 N(6, 2) ● P(x, y) P(x, y) = 2 M(3, 7) = = P(x, y) =
- 58. 6. Coordinate Geometry Perpendicular lines : R m1.m2 = –1 P Q S
- 59. 6. Coordinate Geometry (SPM 2006, P1, Q12) Diagram 5 shows the straight line AB which is perpendicular to the straight line CB at the point B. The equation of CBis y = 2x – 1 . Find the coordinates of B. [3 marks] y mCB= 2 y = 2x – 1 mAB = – ½ A(0, 4) ● Equation of AB is y= –½x+4 ●B Diagram 5 At B, 2x – 1 = – ½ x + 4 O x x = 2, y = 3 ●C So, Bis the point (2, 3).
- 60. 6. Coordinate Geometry Given points P(8,0) and Q(0,-6). Find the equation of the perpendicular bisector of PQ. y mPQ= mAB= K1 O P x Midpoint of PQ (4, -3) = Q K1 The equation 4x + 3y -7 = : or 0 N1
- 61. 6 Coordinate Geometry TASK: To find the equation of the locus of the moving point P such that its distances from the points A and B are in the ratio m : n (Note : Sketch a diagram to help you using the distance formula correctly)
- 62. 6. Coordinate Geometry Find the equation of the locus of the moving point P such that its distances from the points A(-2,3) and B(4, 8) are in the ratio 1 : 2. (Note : Sketch a diagram to help you using the distance formula correctly) A(-2,3), B(4,8) and m:n=1:2 Let P = (x, y) B(4, 8) A(-2, 3) 2 1 ● P(x, y) 3x2 + 3y2 + 24x – 8y– 28 = 0
- 63. 6. Coordinate Geometry Find the equation of the locus of the moving point P such that its distance from the point A(-2,3)is always 5 units. (≈ SPM 2005) A(-2,3) Let P = (x, y) A(-2, 3) ● 5 ● P(x, y) is the equation of locus of P.
- 64. 6. Coordinate Geometry Find the equation of the locus of point Pwhich moves such that it is always equidistant from points A(-2, 3) and B(4, 9). Constraint / Condition : B(4, 9) ● PA = PB A(-2, 3) PA2 = PB2 ● (x+2)2 + (y – 3)2 = (x – 4)2 + (y – 9)2 ● P(x, y) x + y – 7 = 0 is the equation of Locus of P locus of P. Note : This locus is actually the perpendicular bisector of AB
- 65. Solutions to this question by scale drawing will not be accepted. (SPM 2006, P2, Q9) Diagram 3 shows the triangle AOB where Ois the origin. Point C lies on the straight line AB. A(-3, 4) y ● Diagram 3 C ● x O ● B(6, -2) (a) Calculate the area, in units2, of triangle AOB. [2 marks] (b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks] • A point P moves such that its distance from point A is always twice its distance from point B. (i) Find the equation of locus of P, (ii) Hence, determine whether or not this locus intercepts the y-axis. [6 marks]
- 66. (SPM 2006, P2, Q9): ANSWERS A(-3, 4) y ● 3 Diagram 3 C ● 2 x O 9(a) K1 ● B(6, -2) N1 Use formula = 9 To find area 9(b) K1 Use formula correctly N1
- 67. (SPM 2006, P2, Q9): ANSWERS A(-3, 4) y ● 2 ● P(x, y) 9(c) (i) C ● 1 O x AP = ● B(6, -2) K1 Use distance formula AP = 2PB K1 AP2 = 4 PB2 Use AP = 2PB (x+3)2 + (y – 4 )2 = 4 [(x – 6)2+ (y + 2)2 x2 + y2– 18x + 8y + 45 = 0 N1 √
- 68. (SPM 2006, P2, Q9): ANSWERS 9(c) (ii) x = 0, y2+ 8y + 45 = 0 K1 Subst. x = 0 into his locus b2 – 4ac = 82 – 4(1)(45) < 0 K1 Use b2 – 4ac = 0 or AOM So, the locus does not interceptthe y-axis. N1 √ (his locus & b2 – 4ac)
- 69. F4 6. Coordinate Geometry : the equation of locus Given that A(-1,-2) and B(2,1) are fixed points . Point P movessuch that the ratio of AP to PB is 1 : 2. Find the equation of locus for P. 2AP = PB K1 4[ (x+1)2 + (y+2)2 ] = (x -2 )2 + (y -1)2 J1 3x2 + 3y2 + 12x + 18y + 15 = N1 0 x2 + y2 + 4x + 6y + 5 = 0
- 70. F4 Statistics From a given set of data, Marks f (e.g. The frequency distribution of marks of a group of students) 6-10 12 11-15 20 Students should be able to find …. 16-20 27 • the mean, mode & median 21-25 16 • Q1, Q3 and IQR • the variance & S.Deviations 26-30 13 •Construct a CFT and draw an ogive 31-35 10 •Use the ogive to solve related problems 36-40 2 Total 100
- 71. F5 To estimate median from Graph For Question 6(b) Histogram Number of people 80 70 60 50 40 30 20 10 33.5 0.5 20.5 40.5 60.5 80.5 100.5 Age Modal age = 33.5
- 72. F4 CHAPTER 8 8. CIRCULAR MEASURE ‘Radian’ ‘Degrees’ S = rθ (θ must be in θ RADIANS) A = ½ r2 θ Always refer to diagram when answering this question.
- 73. F4 8. CIRCULAR MEASURE Diagram shows a sector of a circle A OABC with centre O and radius 4 B 0.8c O cm.Given that AOC = 0.8 radians, find the area of the shaded region. C K1 Area of sector = ½ x 42 x 0.8 OABC = 6.4 cm 2 K1 Area of triangle = ½ x 42 x sin 0.8 OAC In radians !!!! = 5.7388 cm2 K1 Area of shaded = 6.4 – region 5.7388 N1 = 0.6612 cm2
- 74. F4 DIFFERENTIATION : Given that , find
- 75. F4 9 Differentiation : The second derivative Given that f(x) = x3 + x2 – 4x + 5, find the value of f ” (1) f’ (x) = 3x2 + 2x – f” (x) = 6x + 2 f” (1) = 8
- 76. F4 9 Differentiation : The second derivative Given that , find the value of g ” (1) . g’ (x) = 10x (x2 + 1)4 g’’ (x) = 40x (x2 + 1) 3 . 2x Ya ke ??
- 77. F4- 9 Given that , find the value of g ” (-1) . g’ (x) = 10x (x2 + 1)4 g’’ (x) = 10x . 4(x2 + 1) 3.2x +(x2+1)4. 10 g’’ (-1) = 10(-1) . 4[(-1)2 + 1] 3 +[(-1)2+1)4. 10 = 800 Mid-year, Paper 2
- 78. F4 Differentiation : Small increments Given that y = 2x3 – x2 + 4, find the value of at the point (2, 16). Hence, find the small increment in xwhich causes y to increase from 16 to 16.05. = 6x2 – 2x = 20 , x= K1 2 K1 N1
- 79. F5 Progressions : A.P & G.P A.P. : a, a+d, a+2d, a+3d ,… ….. Most important is “d” G.P. : a, ar, ar2, ar3,…….. Most important is “ r ” !!
- 80. F5 Progressions : G.P - Recurring Decimals SPM 2004, P1, Q12 Express the recurring decimal 0.969696 … as a fraction in the simplest form. x = 0. 96 96 96 … (1) 100x = 96. 96 96 ….. (2) (2) – (1) 99x = 96 x= =
- 81. F5 Back to basic… … Progressions Given that Sn = 5n – n2 , find the sum from the 5th to the 10th terms of the progression. Usual Answer : S10 – S5 = ……. ??? Correct Answer : S10 – S4 Ans :- 54
- 82. F5 Linear Law 1. Table for data X and Y 1-2 Y 2. Correct axes and scale used 1 3. Plot all points correctly 1 4. Line of best fit 1 5. Use of Y-intercept to determine X value of constant 2-4 6. Use of gradient to determine another constant
- 83. F5 Linear Law Bear in mind that …...... 1. Scale must be uniform Y 2. Scale of both axes may defer : FOLLOW given instructions ! 3. Horizontal axis should start from 0 ! 4. Plot ……… against ………. X Vertical Axis Horizontal Axis
- 84. Linear law F5 Y 4.5 x 3.5 x 3.0 x 2.5 2.5 x 1.5 x 1.0 x 0.5 Read this value !!!!! 0 2 4 6 8 10 12 x
- 85. F5 INTEGRATION = = = =
- 86. F5 INTEGRATION SPM 2003, P2, Q3(a) 3 marks Given that = 2x + 2 and y= 6 when x= – 1, find y in terms of x. Answer: = 2x + 2 y = = x2+ 2x + c x = -1, y = 6: 6 = 1 +2 + c c = 3 Hence y = x2 + 2x + 3
- 87. F5 INTEGRATION SPM 2004, K2, S3(a) 3 marks The gradient function of a curve which passes through A(1, -12) is 3x2 – 6. Find the equation of the curve. Answer: = 3x2 – 6 y = Gradient = x3– 6x + c Function x = 1, y = – 12 : – 12 = 1 – 6 + c c = –7 Hence y = x3 – 6 x – 7
- 88. F5 Vectors : Unit Vectors Given that OA = 2i + j and OB = 6i + 4j, find B the unit vectorin the direction of AB AB = OB - OA = ( 6i + 4j ) – ( 2i + j ) A = 4i + 3j K1 l AB l = = 5 Unit vector in the direction of AB = K1 N1
- 89. F5 Parallel vectors Given that a and bare parallel vectors, with a = (m-4)i +2 j and b= -2i + mj. Find the the value of m. a=kb a= b (m-4) i + 2 j = k (-2i + mj) K1 m- 4 = -2k 1 mk = 2 2 K1 m=2 N1
- 90. F5 5 TRIGONOMETRIC FUNCTIONS Prove that tan2 x – sin2 x = tan2 x sin2 x tan2 x – sin2 x = sin 2x K1 K1 N1
- 91. F5 5 TRIGONOMETRIC FUNCTIONS Solve the equation 2 cos 2x + 3 sin x - 2 =0 2( 1 - 2sin2 x) + 3 sin x - 2 = 0 K1 -4 sin2 x + 3 sin x = 0 sin x( -4 sin x + 3 ) = 0 K1 sin x= 0 sin x = , x = 00, 1800, 3600 N1 x = 48.590, 131.410 N1
- 92. F5 5 TRIGONOMETRIC FUNCTIONS (Graphs) (Usually Paper 2, Question 4 or 5) - WAJIB 1. Sketch given graph : (4 marks) (2003) y = 2 cos x, (2004) y = cos 2x for (2005) y = cos 2x , (2006) y = – 2 cos x ,
- 93. F5 PERMUTATIONS AND COMBINATIONS Find the number of four digit numbers exceeding 3000 which can be formed from the numbers 2, 3, 6, 8, 9 if each number is allowed to be used once only. No. of ways = 4. 4. 3. 2 = 96 3, 6, 8, 9
- 94. F5 Find the number of ways the word BESTARIcan be arranged so that the vowels and consonants alternate with each other [ 3 marks ] Vowels : E, A, I Consonants : B, S, T, R Arrangements : C V C V C V C No. of ways = 4! 3 ! = 144
- 95. F5 Two unbiased dice are tossed. Find the probability that the sum of the two numbers obtained is more than 4. Dice B, y n(S) = 6 x 6 = 36 6 X X X X X X Constraint : x + y > 4 5 X X X X X X Draw the line x + y = 4 4 X X X X X X 3 X X X X X X We need : x + y > 4 2 X X X X X X P( x + y > 4) = 1 – 1 X X X X X X Dice A, x = 1 2 3 4 5 6
- 96. F5 PROBABILITY DISTRIBUTIONS The Binomial Distribution r = 0, 1, 2, 3, …..n p+q=1 n = Total number of trials r = No. of ‘successes’ p = Probability of ‘success’ q = probability of ‘failure’ Mean = np Variance = npq
- 97. F5 PROBABILITY DISTRIBUTIONS The NORMAL Distribution Candidates must be able to … f(z) determine the Z-score Z = z 00 0.5 use the SNDT to find the values (probabilities)
- 98. T5 f(z) f(z) f(z) = 1 – – z z z -1.5 0 1 0 1 0 1.5
- 99. F4 Index Numbers • • Index Number = • • Composite Index = • • Problems of index numbers involving two or more basic years.
- 100. Solution of Triangles The Sine Rule • The Cosine Rule • Area of Triangles • Problems in 3-Dimensions. • Ambiguity cases (More than ONE • answer)
- 101. Motion in a Straight Line •Initial displacement, velocity, acceleration... •Particle returns to starting point O... •Particle has maximum / minimum velocity.. •Particle achieves maximum displacement... •Particle returns to O / changes direction... •Particle moves with constant velocity...
- 102. Motion in a Straight Line • Question involving motion of TWO particles. • ... When both of them collide / meet ??? • … how do we khow both particles are of the same direction at time t ??? •The distance travelled in the nth second. •The range of time at which the particle returns …. •The range of time when the particle moves with negative displacement •Speed which is increasing •Negative velocity •Deceleration / retardation
- 103. Linear Programming To answer this question, CANDIDATES must be able to ..... form inequalities from given mathematical information draw the related straight lines using suitable scales on both axes recognise and shade the region representing the inequalities solve maximising or minimising problems from the objective function (minimum cost, maximum profit ....)
- 104. Linear Programming Maklumat Ketaksamaan 1. x is at least 10 x ≥ 10 2. x is not more than 80 x ≤ 80 3. x is not more than y x ≤ y 4. The value of y is at least twice the value of x y ≥ 2x 5. The maximum value of x is 100 x ≤ 100 6. The minimum value of y is 35 y ≥ 35 7. The maximum value of x+ 2y is 60 x + 2y ≤ 60 8. The minimum value of 3x – 2y is 18 3x - 2y ≥ 18 9. The sum of x and y is not less than 50 x + y ≥ 50 10. The sum of x and y must exceed 40 x + y > 40 11. x must exceed y by at least 10 x ≥ y + 10 12. The ratio of the quantity of Q (y) to the quantity of P (x) y ≤ 2x should not exceed 2 : 1 13. The number of units of model B (y) exceeds twice the y - 2x >10 number of units of model A (x) by 10 or more.
- 105. Good Luck ! www.banksoalanspm.blogspot.com

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