1. Additional Mathematics

9. Common errors… PA : PB = 2 : 3 then 2 PA = 3 PB

10. Actually, … PA : PB = 2 : 3 3 PA = 2 PB

11. More mistakes …… 3 2 PA 2 = 2 2 PB 2 9 PA 2 = 4 PB 2

12. Common mistakes … log a x + log a y = 0, then xy = 0 It should be… xy = a 0 = 1

13. Common mistakes … log a (x – 3) = log a x – log a 3 2 x x 2 y = 1 x + y = 1 2 x x 2 y = 2 0 2 x + y = 2 0 x + y = 0

14. Common mistakes … log a x + log a y = 0, then log a xy = 0 So, xy = 0 It should be… xy = a 0 = 1

15. Common mistakes … sin (x + 30 0 ) = ½ , then sin x + sin 30 0 = ½ ………………… gone ! Do NOT use Sin(A+B) = sin A cos B + cos A sin B !

16. Correct way… … sin (x + 30 0 ) = ½ , then x +30 0 = 30 0 , 150 0 So, x = 0 0 , 120 0 If 0 0 is an answer, then 360 0 is also an answer ! ?

17. sin (x + 30 0 ) = ½ , then x +30 0 = 30 0 , 150 0 , 390 0 So, x = 0 0 , 120 0 , 360 0

18. Relationship between Functions and Quadratic Functions ( 1 , 1 ) , ( 2 , 4 ). …. form ordered pairs and can be plotted to obtain a curve. Image Object f(x) = x 2 x O y 1 4 2 1 X Y 1 1 Domain Codomain 2 4

27. SPM 2003 Paper 2 , Question 1 Solve the simultaneous equation 4x + y = - 8 and x 2 + x – y = 2 ( 5 marks)‏ Answer Make x or y the subject P1 N1 x = -2, -3 or y = 0 , 4 Eliminating x or y Solving the quadratic equation : y = 0 , 4 or x = -2, -3 N1 K1 K1

29. SPM 2003 Paper 2 , Question 2 *** Answer 2(a) Writing f(x) in the form (x – p) 2 + q (x – 2k) 2 – 4k 2 + 5k 2 + 1 N1 k = 0 , 4 (b) Equating (his) - (x – p) = 0 N1 r = -1, 3 Equating q ( q* = r 2 + 2k)‏ (k – 1) 2 = r 2 r = k – 1 Eliminating r or k by any valid method N1 K1 K1 N1 K1 K1

30. 1. Functions F4 f(x) = x – 3, g(x) = 3x gf (1) = g [ f(1) ] = g [-2] = -6 f : x x - 3 , g : x 3 x , find gf(1) . 2. Given f : x x 2 - 2 . 1. Given Find the values of x which map onto itself. x 2 - 2 = x x 2 – x – 2 = 0 (x+1)(x-2) = 0 x = -1 , x = 2 f ( x ) = x

31. Functions : Inverse Functions T4 BAB 1 4. Given f (x) = 3 – 2x, find f -1. F4 Let f -1 (x) = y Then x = f (y) x = 3 – 2y Method 1 Method 2 Let f (x) = y Then 3 – 2x = y 3 – y = 2x

32. Functions : Applying the Idea of Inverse functions T4 BAB 1 F4 Method 2 ( No need f -1 )‏ = 8 Let f -1 (a) = 11 Then a = f (11)‏ 5. Given , find the value of a if f -1 (a) = 11 Let f -1 (x) = y Then x = f(y)‏ Method 1 (Find f -1 )‏ x = y = f -1 (a) = = 11 a = 8

33. Functions : Given composite function and one function, find the other function. T4 BAB 1 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 F4 6. Given find fg.

34. **Functions : To skecth the graphs of y = |f(x)| T4 BAB 1 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 !!! Range : 1 ≤ y ≤ 6 F4 x y 0 4 6 5 3 4 2 1

35. 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 2x 2 + 5x – 3 = 0 F4

36. 2. Quadratic Equations x 2 – ( S.O.R ) x + ( P.O.R. ) = 0 a x 2 + b x + c = 0 F4 P.O.R. = S.O.R =

37. 2. The Quadratic Equation : Types of roots The quadratic equation a x 2 + b x + c = 0 has 1. Two distinct roots if 2. Two equal roots if 3. No real roots if b 2 - 4ac b 2 - 4ac b 2 - 4ac > 0 < 0 = 0 ** The straight line y = mx -1 is a tangent to the curve y = x 2 + 2 ……. ??? F4

38. 3 Quadratic Functions : Q uadratic Inequalities SPM 2004, K1, S5 Find the range of values of x for which x(x – 4) ≤ 12 F4 x (x – 4) ≤ 12 x 2 – 4x – 12 ≤ 0 (x + 2)(x – 6) ≤ 0 – 2 ≤ x ≤ 6 6 x -2

39. Solve x 2 > 4 Back to BASIC x 2 – 4 > 0 (x + 2 )(x – 2 ) > 0 x < -2 or x > 2 F4 x> ±2 ??? R.H.S must be O ! – 2 2

41. Back to basic… … 3 2(x – 1) . 3 (– 3x) = 1 2x – 2 – 3x = 1 – x = 3 x = – 3 Betul ke ??? 5. INDICES F4 Solve ..

42. 3 2(x – 1) . 3 (– 3x) = 1 3 2x – 2 +(– 3x) = 3 0 – x – 2 = 0 x = – 2 5. INDICES F4 Solve

43. or… 9 x-1 = 27 x 3 2(x – 1) = 3 3x 3 2x – 2 = 3 3x 2x – 2 = 3x x = – 2 5. INDICES F4 Solve

44. Solve 2 x + 3 = 2 x+2 5. INDICES 2 x + 3 = 2 x . 2 2 x = 0 2 x + 3 = 4 (2 x )‏ 3 = 3(2 x )‏ 1 = (2 x )‏ F4 Can U take log on both sides ??? WHY? In the form u + 3 = 4u

45. Solve the equation , give your answer correct to 2 decimal places . [ 4 marks ] 5. INDICES F4 9 (3 x ) = 32 + (3 x )‏ 8 (3 x ) = 32 3 x = 4 x = 1.26 (Mid-Yr 07)‏

46. Solve 2 2x . 5 x = 0.05 5. INDICES x = – 1 a m b m = (ab) m F4 4 x . 5 x = 20 x = You can also take log on both sides.

47. (Mid-Yr 07)‏ Solve the equation [ 4 marks ] 5. INDICES & LOGARITHMS F4 x – 2 = 4 (4 – x) x = 3.6

48. Back to basic… … Solve the the equation log 3 (x – 4) + log 3 (x + 4) = 2 log 3 (x-4)(x+4) = 2 x 2 – 16 = 9 x = 5 5. INDICES & LOGARITHMS F4

49. Back to basic… … Solve the equation log 3 4x – log 3 (2x – 1) = 1 SPM 2005, P1, Q8 F4 4x = 3(2x – 1)‏ = 6x – 3 2x = 3 x =

50. 5 Indices and Logaritms : Change of base Given that log 3 p = m and log 4 p = n. Find log p 36 in terms of m and n. = 2 log p 3 + log p 4 log p 36 = log p 9 + log p 4 log a a = 1 F4 K1 K1 K1 N1

51. Coordinate Geometry Some extra vitamins 4u …

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 n m P(x 1 , y 1 )‏ R(x 2 , y 2 )‏ Q(x, y)‏ ● n m R(x 2 , y 2 )‏ P(x 1 , y 1 )‏ Q(x, y)‏ 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. P(x, y) = ● 1 2 N(6, 2)‏ M(3, 7)‏ P(x, y)‏ = = P(x, y) =

58. 6. Coordinate Geometry m 1 .m 2 = –1 Perpendicular lines : P Q R 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 CB is y = 2x – 1 . Find the coordinates of B . [3 marks ] m CB = 2 m AB = – ½ Equation of AB is y = – ½ x + 4 At B, 2x – 1 = – ½ x + 4 x = 2, y = 3 So, B is the point (2, 3). x y O A(0, 4)‏ C Diagram 5 B ● ● ● y = 2x – 1

60. 6. Coordinate Geometry Given points P(8,0) and Q(0,-6). Find the equation of the perpendicular bisector of PQ. m PQ = m AB = Midpoint of PQ = (4, -3)‏ The equation : 4x + 3y -7 = 0 K1 K1 N1 or P Q x y O

61. 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) 6 Coordinate Geometry

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)‏ ● 2 1 B(4, 8)‏ A(-2, 3)‏ P(x, y)‏ 3 x 2 + 3y 2 + 24 x – 8 y – 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)‏ ● 5 A(-2, 3)‏ P(x, y)‏ ● A(-2,3)‏ Let P = (x, y)‏ is the equation of locus of P.

64. 6. Coordinate Geometry Find the equation of the locus of point P which moves such that it is always equidistant from points A(-2, 3) and B(4, 9) . Constraint / Condition : PA = PB PA 2 = PB 2 (x+2) 2 + (y – 3) 2 = (x – 4) 2 + (y – 9) 2 x + y – 7 = 0 is the equation of locus of P. Note : This locus is actually the perpendicular bisector of AB A(-2, 3)‏ ● B(4, 9)‏ ● Locus of P ● P(x, y)‏

66. (SPM 2006, P2, Q9) : ANSWERS 9(a)‏ = 9 x y O A(-3, 4)‏ Diagram 3 C ● ● ● B(6, -2)‏ 3 2 9(b) K1 N1 Use formula correctly N1 K1 Use formula To find area

67. (SPM 2006, P2, Q9) : ANSWERS √ AP = 2PB AP 2 = 4 PB 2 (x+3) 2 + (y – 4 ) 2 = 4 [(x – 6) 2 + (y + 2) 2 x 2 + y 2 – 18x + 8y + 45 = 0 N1 9(c) (i)‏ K1 Use distance formula K1 Use AP = 2PB x y O A(-3, 4)‏ C ● ● ● B(6, -2)‏ 2 1 P(x, y)‏ ● AP =

68. (SPM 2006, P2, Q9) : ANSWERS 9(c) (ii) x = 0, y 2 + 8y + 45 = 0 b 2 – 4ac = 8 2 – 4(1)(45) < 0 So, the locus does not intercept the y-axis. Use b 2 – 4ac = 0 or AOM K1 K1 Subst. x = 0 into his locus N1 √ (his locus & b 2 – 4ac)

69. 6. Coordinate Geometry : the equation of locus Given that A(-1,-2) and B(2,1) are fixed points . Point P moves such that the ratio of AP to PB is 1 : 2. Find the equation of locus for P. 2 AP = PB x 2 + y 2 + 4x + 6y + 5 = 0 4 [ (x+1) 2 + (y+2) 2 ] = (x -2 ) 2 + (y -1) 2 3x 2 + 3y 2 + 12x + 18y + 15 = 0 F4 K1 J1 N1

71. To estimate median from Histogram F5 10 20 30 40 50 60 70 80 0.5 20.5 40.5 60.5 80.5 100.5 Modal age = 33.5 Age Number of people 33.5 Graph For Question 6(b)‏

73. 8. CIRCULAR MEASURE F4 Diagram shows a sector of a circle OABC with centre O and radius 4 cm. Given that AOC = 0.8 radians, find the area of the shaded region. C A B O 0.8 c Area of sector OABC = ½ x 4 2 x 0.8 = 6.4 cm 2 = ½ x 4 2 x sin 0.8 = 5.738 8 cm 2 Area of triangle OAC Area of shaded region = 6.4 – 5.738 8 = 0 . 6612 cm 2 K1 N1 K1 K1 In radians !!!!

74. DIFFERENTIATION : F4 Given that , find

75. 9 Differentiation : The second derivative Given that f(x) = x 3 + x 2 – 4x + 5 , find the value of f ” (1)‏ f’ (x) = 3x 2 + 2x – 4 f” (x) = 6x + 2 f” ( 1 ) = 8 F4

76. 9 Differentiation : The second derivative Given that , find the value of g ” (1) . g’ (x) = 10x (x 2 + 1) 4 F4 g’’ (x) = 40x (x 2 + 1) 3 . 2x Ya ke ??

77. g’ (x) = 10x (x 2 + 1) 4 F4- 9 g’’ (x) = 10x . 4(x 2 + 1) 3 .2x +(x 2 +1) 4 . 10 g’’ ( -1 ) = 10( -1 ) . 4[( -1 ) 2 + 1] 3 +[( -1 ) 2 +1) 4 . 10 = 800 Mid-year, Paper 2 Given that , find the value of g ” (-1) .

78. Differentiation : Small increments F4 Given that y = 2x 3 – x 2 + 4, find the value of at the point (2, 16). Hence, find the small increment in x which causes y to increase from 16 to 16.05. K1 K1 N1 = 6x 2 – 2x = 20 , x = 2

79. Progressions : A.P & G.P A.P. : a, a+ d , a+2 d , a+3 d , …….. Most important is “ d ” F5 G.P. : a, a r , a r 2 , a r 3 , …….. Most important is “ r ” !!

80. Progressions : G.P - Recurring Decimals SPM 2004, P1, Q12 Express the recurring decimal 0.969696 … as a fraction in the simplest form. F5 x = 0. 96 96 96 … (1)‏ 100x = 96. 96 96 ….. (2)‏ (2) – (1) 99x = 96 x = =

81. Back to basic… … Usual Answer : S 10 – S 5 = ……. ??? Correct Answer : S 10 – S 4 Progressions Given that S n = 5n – n 2 , find the sum from the 5 th to the 10 th terms of the progression. Ans :-54 F5

82. Linear Law 1. Table for data X and Y 2. Correct axes and scale used 3. Plot all points correctly 4. Line of best fit 5. Use of Y-intercept to determine value of constant 6. Use of gradient to determine another constant F5 Y X 1 1-2 1 1 2-4

83. Linear Law Bear in mind that …...... 1. Scale must be uniform 2. Scale of both axes may defer : FOLLOW given instructions ! 3. Horizontal axis should start from 0 ! 4. Plot ……… against ………. F5 Y X Vertical Axis Horizontal Axis

84. Linear law 0 2 4 6 8 10 12 x 0.5 1.0 1.5 2.5 2.5 3.0 3.5 4.5 Y x x x x x x F5 Read this value !!!!!

85. INTEGRATION F5 = = = =

86. INTEGRATION SPM 2003, P2, Q3(a) 3 marks Given that = 2 x + 2 and y = 6 when x = – 1 , find y in terms of x. F5 Answer: = 2x + 2 y = = x 2 + 2x + c x = -1, y = 6: 6 = 1 + 2 + c c = 3 Hence y = x 2 + 2x + 3

87. INTEGRATION SPM 2004, K2, S3(a) 3 marks The gradient function of a curve which passes through A(1, -12) is 3x 2 – 6 . Find the equation of the curve. F5 Answer: = 3x 2 – 6 y = = x 3 – 6x + c x = 1, y = – 12 : – 12 = 1 – 6 + c c = – 7 Hence y = x 3 – 6 x – 7 Gradient Function

88. Vectors : Unit Vectors Given that OA = 2 i + j and OB = 6 i + 4 j , find the unit vector in the direction of AB AB = OB - OA = ( 6 i + 4 j ) – ( 2 i + j ) = 4 i + 3 j l AB l = = 5 Unit vector in the direction of AB = F5 A B K1 N1 K1

89. Parallel vectors Given that a and b are parallel vectors, with a = (m-4) i +2 j and b = -2 i + m j . Find the the value of m. a = k b (m-4) i + 2 j = k (-2i + mj) m- 4 = -2 k m k = 2 1 2 a = b m = 2 F5 K1 N1 K1

90. Prove that tan 2 x – sin 2 x = tan 2 x sin 2 x sin 2 x tan 2 x – sin 2 x = 5 TRIGONOMETRIC FUNCTIONS F5 K1 N1 K1

91. Solve the equation 2 cos 2x + 3 sin x - 2 = 0 5 TRIGONOMETRIC FUNCTIONS F5 sin x ( -4 sin x + 3 ) = 0 sin x = 0 , 2( 1 - 2sin 2 x ) + 3 sin x - 2 = 0 -4 sin 2 x + 3 sin x = 0 sin x = x = 0 0 , 180 0 , 360 0 x = 48.59 0 , 131.41 0 K1 N1 K1 N1

92. 5 TRIGONOMETRIC FUNCTIONS (Graphs)‏ (Usually Paper 2, Question 4 or 5) - WAJIB ! F5 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. 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 F5 PERMUTATIONS AND COMBINATIONS

94. Vowels : E, A, I C onsonants : B, S, T, R Arrangements : C V C V C V C No. of ways = 4 ! 3 ! = 144 Find the number of ways the word BESTARI can be arranged so that the vowels and consonants alternate with each other [ 3 marks ] F5

95. Two unbiased dice are tossed. Find the probability that the sum of the two numbers obtained is more than 4. n(S) = 6 x 6 = 36 Constraint : x + y > 4 Draw the line x + y = 4 We need : x + y > 4 F5 Dice B, y 4 1 5 6 2 3 Dice A, x 2 3 4 5 1 6 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X P( x + y > 4) = 1 – =

96. The Binomial Distribution r = 0, 1, 2, 3, …..n n = Total number of trials q = probability of ‘failure’ p = Probability of ‘success’ r = No. of ‘successes’ F5 PROBABILITY DISTRIBUTIONS Mean = np Variance = npq p + q = 1

98. T5 z f(z) 0 1.5 z z f(z) 0 -1.5 1 = – 1 f(z) 0 1 –

104. Linear Programming y ≤ 2x 12. The ratio of the quantity of Q ( y ) to the quantity of P ( x ) should not exceed 2 : 1 x ≥ y + 10 11. x must exceed y by at least 10 y - 2x >10 13. The number of units of model B ( y ) exceeds twice the number of units of model A ( x ) by 10 or more. x + y > 40 10. The sum of x and y must exceed 40 x + y ≥ 50 9. The sum of x and y is not less than 50 3x - 2y ≥ 18 8. The minimum value of 3x – 2y is 18 x + 2y ≤ 60 7. The maximum value of x+ 2y is 60 y ≥ 35 6. The minimum value of y is 35 x ≤ 100 5. The maximum value of x is 100 y ≥ 2x 4. The value of y is at least twice the value of x x ≤ y 3. x is not more than y x ≤ 80 2. x is not more than 80 x ≥ 10 1. x is at least 10 Ketaksamaan Maklumat

105. Selamat maju jaya !