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# Add maths module form 4 & 5

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### Add maths module form 4 & 5

1. 1. FORMAT&COMPONENT TO EXCELL in You need to…  set TARGET  familiar with FORMAT of PAPER  do EXERCISESExercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercis e Exercis e Exercis e Exercis e Exercis e My TARGET 1
2. 2. PAPER 1 (3472 / 1)No Items Comments1 No. of questions 25 questions (Answer ALL)2 Total Marks ……80 marks………….3 Time …….2 hours4 Level of difficulty Low : Moderate : High 6 : 3 : 1 PAPER 2 (3472 / 2)No Items Comments1 Three sections Section A Section B Section C2 No of questions 6 5 4 Answer Choose Choose (need to answer 12 questions ) …ALL….. …four…. …two…..3 Total marks (100) 40 marks 40 marks 20 marks4 Time ………2 ½ hours…………….. Low : Moderate : High5 Level of difficulty 4 3 3 2
3. 3. FORMAT&COMPONENT 1 mark  1.5 minutes Check answers * Extra Time = …………………………………. 1. Functions 2. Quadratic Equations 3. Quadratic Functions ALGEBRA 4. Indices & Logarithms 5. Simultaneous Equations 6. Progressions 7. Linear Law 1. Coordinate Geometry Geometry 2. Vector 1. Differentiation Calculus 2. Integration 1. Circular Measures Triginometry 2. Trigonometric Functions 1. Statistics 2. Probability STATISTICS 3. Permutation & Combination 4. Distribution Probability 1. Index Numbers Social Science 2. Linear Programming 1. Solutions to TriangleScience & Technology 2. Motion in a Straight Line. 3
4. 4. FORM 4 TOPICS P(X) Learn with your heart and you’ll see the wonders … 4
5. 5. FORM 4 TOPICSTOPIC 1 : FUNCTIONS a b f(a) = b  object = …………………….. image = ……………….. -1 Given f (x) and gf(x). Find g(x) .  Thus, g(x) = gf f 3x  5 f ( x)  , X1 x 1TOPIC 2: QUADRATIC EQUATIONS [ ax2 + bx + c = 0 ]Types of roots - two distinct real roots >0 - intersects at two points 0 Real roots - two equal roots b2 – 4ac =0 - touches / tangent <0 - no root - does not intersect - f(x) is always positiveSum and Product of Roots [ ax2 + bx + c = 0 ] b cSum of roots, ( + ) = Product of roots ( ) = x2 – Sx + P =0 a aTOPIC 3: QUADRATIC FUNCTIONS 1. General form CTS form f (x) = ax2 + bx + c = a( x + p)2 + q Similarity Same value of a Same value of a c = y-intercept. q = max/min value Difference of f(x) Able to find: Able to find Specialty - shape - turning point - y intercept ( - p , q) 5
6. 6. FORM 4 TOPICS2. Sketch Graphs: y = ax2 + bx + c (a) From the graph i value of a : Positive ii value of b2-4ac: <0 iii type of roots : No roots iii y-intercept : C Equation of axis of b iv X=  symmetry : 2a3. Inequalities : Solving 2 inequalities  use Graph (x – a) (x – b) > 0 (x – c) (x – d) < 0 1. Let the right hand side = 0 - factorise 2. Find the roots of the equation 3. Sketch the graph 3. Determine the region  c d positive or negative a b x<a , x > b, c<x<d ………………………. …………….………TOPIC 4: SIMULTANEOUS EQUATIONS Substitution Use ……………………………….. method intersection To find the ………………………………… points between a straight line and a curve. 6
7. 7. FORM 4 TOPICSTOPIC 5: INDICES & LOGARITHM Change the base to x+1 x 1 Use Index rule : (i) 5 . 125 = the same number ………………………. 25 5x +1. 5 3x = 5–2 x + 1 + 3x = – 2 x = –INDEX Insert log on both sides Use log : (ii) 8  3x = 7 ............................ IND n+1 n Steps of solutions Use substitution : (iii) 3 + 3 = 12 or can be factorise 3 . 3 + 3n = 12 n a(3) + a = 12 1. separate the index 4a = 12 a = 3 2. substitute n 3 =3  n=1 2n n (iv) 3 + 5. 3 = 6 a2 + 5a – 6 = 0 (a – 1) (a + 6) = 0 a = 1 , a = –6 3n = 1 3n  –6  n=0LOGARITHM : Use Rules of logarithms to simplify or to solve logarithmic equations log2 (x + 9) = 3 + 2 log2 x log2 (x + 9) – log2 x 2 =3 log2 (x + 9) =3 x2 x + 9 = 23 x2 x + 9 = 8x 2 8x2 – x – 9 = 0 (8x -9) (x + 1) = 0 x –1, x=  7
8. 8. FORM 4 TOPICSTOPIC 6: COORDINATE GEOMETRY Distance   Ratio theorem  Mid point   Equation of straight line y = mx + c Arrange Area (positive) : general form ax + by + c = 0 anticlockwise Gradient : gradient form y = mx + c - parallel m1 = m2 x y : intercept form  1 - perpendicular m1 m2 = –1 a bEquation of locus : …use distance formula………………….Rhombus : ……its diagonal are perpendicular to each other ……….Parallelogram, square, rectangle, rhombus. its diagonal share the same mid pointTOPIC 7: STATISTCSEFFECT ON CHANGES TO DATA The change in Interquartile values when Mean Mode Median Range range  Variance each data is Added with k +k +k +k unchange unchange unchange unchangeMultiplied by m m m m m m m m 2TOPIC 8: CIRCULAR MEASURES  …… radian = ………  180 For s = r and A = ½ r2 , the value of  is in ……radian……. Area of segment = ½ r2 (  - sin )  Shaded angle =  –  rad 8
9. 9. FORM 4 TOPICSTOPIC 9: DIFFERENTIATION gradient of normal mN mT = –1 gradient of tangent dy equation of normal Tangent mT = dx equation of tangent y – y1 = m(x – x1) Rate of change dy = dy  dx dt dx dtApplications  y = dy   x Small Changes approximate value dx y ORIGINAL + y d2y minimum  0 dx 2 Turning points dy maximum d2y dx = 0  0 dx 2TOPIC 10: SOLUTION OF TRIANGLES Sine Rule - Ambiguous Case  two possible angles  acute and obtuse angle Cosine Rule Area =  ab sin CTOPIC 11: INDEX NUMBERS  I A, B  I B, C = I A, C  Given that the price index of an item is 120. If the price index increases at the same rate in the next year, what will be the new price index of the item? 120 ………………………………120  …………………………………………… 100 ………………………………………………………………………………………. 9
10. 10. :10
11. 11. FORM 5 TOPICSTOPIC 1: PROGRESSIONS Janjang Aritmetik Janjang Geometri Examples : 20, 15, 10, …..., …. 4, 3, 2.25, ……., …. T Uniqueness : d = T2 – T1 r= 2 T1 n( a  l ) a Others : Sn  S = 2 1 r Given Sn find Tn Example: Given Sn = n( 3 + 2n), find T8. Thus, T8 = S8 – S7 S 3 to 7 = S7 – S2 Find the sum from the 3rd term until the 7th term. + + + + + + T1 T2 T3 T4 T5 T6 T7TOPIC 2: LINEAR LAWConvert to linear form Y = m X + c b 1 b ay = x + xy = x2 + x a a p y h x y x = h x + p x T 1 T + 1 = a 2 + k = a  + k  b y = ax log y = b log x + log a x y=kp log y = log p x + log k 11
12. 12. TOPIC 3: INTEGRATIONS To find THE EQUATION OF A CURVE given dy/dx dy = ……gradient function …………………… dxEquation of CURVE, y  {gradient function} dx  the integrated function must have c AREA under a curve: Show how you would find the area of the shaded region. str. line: y = –4x + 12 str.line: y = –x + 4 y = x (x–1) (x–2) y = x2 x = y2 – y y = 4x - x2 1 0 1 2 2 3 1 4 0 1. Find the intersection intersection, x = 1 1. Expand y 1. Formula point. thus, y = 4(1) – 12 = 3 y = x3 – 3x2 + 2x when x = 2,  y = 4 1 Shaded Area: A 0 x dy Shaded Area: 1 0 y dx 2. Find the area Area under curve + area  Area under curve – area  Area above = 4  2 2  2 = x dx +  (1)(4) = (4xx ) dx –  (3)(3) 1  2 y dx 0 1 Area below = Total area VOLUME : Show the strategy to find the generated volume.str. line: y = –4x + 12 str. line: y = –x + 4 y = x2– 1 x = y2 – 1 revolved about x-axis revolved about x-axis revolved about y-axis revolved about y-axis 2 y=x 2 y = 4x - x 1 3– 4– 1 2 2 3 1 4 –1 y y x x 2 dx 2 dx V  2 dy 2 dy V  V  V  where y2 = (x2)2 where y2 : (4x – x2)2 where x2 = y + 1 where x2 = (y2 – 1)2 2 2 1    4  0 (x 2 ) 2 dx  1 (4 x  x 2 ) 2 dx I=  1 ( y  1) dy 1 ( y 2  1) 2 dy + Volume cone – volume cone 12
13. 13. FORM 5 TOPICSTOPIC 4: VECTORS If vectors a and b are parallel, thus, ………a = k b …………………….……….. If OA = a and OB = b , thus, AB = … OB - OA……= b – a ………………..…. If T is the mid point of AB thus, OT = ……AB………………………………………... Given m = 2i + 3j and n = i – 4j find, i) m+n = ……2i + 3j + ( i – 4j ) = 3i – j ……………. ii) m + n = …… 32  ( 1)2  10 ………………………………………….. 1 iii) unit vector in the direction of m + n = …… (3i  j ) ……………………… 10 1    2   2  1    3 If A(1, 3) and B(–2, 5) find AB : …OA =   , OB =  3          5   AB =  5  – 3  =  2            can also be written in the form of i and j. Given CD = 2h x + 5y and CD = 8x – 2hky , find the value of h and of k. 2h x + 5y = 8x – 2hky ……………………… (comparison method) compare coef. of x , compare coef of y.TOPIC 5 : TRIGONOMETRIC FUNCTIONS Solving equation : SIMPLE Solve: 2 cos 2x = 3 for 0  x  360 1. Separate coefficient of trigo 3 cos 2x = positive values 2 2. Determine the quadrant  1st and 4th quadrant 3. Find the reference angle 2x = 30 4. Find new range (if necessary). 0  x  720 2x = 30, 330, 390, 690 5. Find all the angles x = 15, 165, 195, 345 13
14. 14.  Solving Equation : Using Identity WHEN? sin 2x cos x = sin x cos 2x cos x = 0 ………Angles are not the same……… sin2 x + 3cos x = 3 2 sec2 x + tan2 x = 5 ………have different functions ……….  Proofing: Use Identity sinA 1 . 1 1 Remember : tanA   , tan  , sec 2x = , cosec A = cosA cot cos 2x sin A  Use of Trigo Ratios: Examples: From the question given, If sin A =  , A is not acute, 1. Determine the quadrant involved. ……second………………. 2. Determine the values of the other trig. fxn cos and tan = negative in the quadrant. cos A =  3 , tan A = –  5 find sin 2A 3. Do you need to use identity? sin 2A = 2 sin A cos A 4. Substitute values = 2 ( ) (  3 ) 5 = 24  25  Sketch Graphs y = a sin b x + c a = max / minimum point a cos b x + c a tan b x + c b = number of basic shape between 0 and 2 c = increase / decrease translation of the Basic Graphs y = sin x y = cos x y = tan x 1- 1- 2 2 2–1- –1- 14
15. 15. FORM 5 TOPICSTOPIC 6: PERMUTATIONS & COMBINATIONS Permutations = …order of arrangement is important Combinations =…order not important…. Three committee members of a society are to be Three committee members of a society are to be chosen from 6 students for the position of chosen from 6 students. Find the number of president, vice president and secretary. Find ways the committee can be chosen. the number of ways the committee can be chosen. Permutations: 6 3 P Combination: 6 3Cwith condition: Find the number of different ways the letters Find the number of ways 11 main players of a H O N E S T can be arranged if it must football team can be chosen from 15 local begin with a vowel. players and 3 imported players on the 2 5 4 3 2 1 condition that not more than 2 imported players are allowed. condition  2 Import. condition Case : (2 Import, 9 local) or 3C2. 15C9 vowels = 2 choices 3 15 (1 import 10 locals) or + C1. C10 3 15 ( 0 import 11 locals) + C0. C11TOPIC 7: PROBABILITY n ( A) P(A) = n ( S) - Probability event A or B occurs = P(A) + P(B) - Probability event C and D occurs = P(A) . P(B) Considering several cases: Probability getting the same colors = Example: (Red and Red) or (Blue and blue) Probability of at least one win in two matches = (win and lose) or (lose and win) or (win and win) Or using compliment event = 1 – (lose and lose) 15
16. 16. TOPIC 8: PROBABILITY DISTRIBUTION P(X)BINOMIAL DISTRIBUTION _ 0.3 0.25 _- The table shows the binomial probability distribution of 0.2 _ an event with n = 4 . 0.15 _ 0.1 _ X=r r=0 r=1 r=2 r=3 r=4 0.05 _ P(X) 0.2 0.15 0.3 0.25 0.1 0 1 2 3 4 X=r Graph of Binomial Prob Distribution total = 1- formula: P(X = r) = n C r p r q n – r- mean,  = np standard deviation = npq variance = npqNORMAL DISTRIBUTION X - Formula : Z  - Type 1 : Given value of X  find the value of Z  find the probability [use formula] [use calculator]- Type 2 : Given the probability Find the value of Z  Find its value of X . [use log book] [use formula]TOPIC 9: MOTION IN STRAIGHT LINES Displacement, s Velocity, v Acceleration, a  ds dv d 2s s= v dt v a = dt dt dt 2 dv Maximum velocity - 0 a=0 dt return to O s=0 - - stops momentarily v=0 da max. acceleration 0 dt 16
17. 17. FORM 5 TOPICSTOPIC 10: LINEAR PROGRAMMINGGiven:(i) y >x–2 (ii) x+y  5 (iii) 4x  y(a) Draw and shade the region, R, that satisfy the three inequalities on the graph paper provided using 2 cm to 2 units on both axes.(b) Hence, find, in the region R, the maximum value of 2x + y where x and y are integers. 2 possible maximum points (x, y intergers) (1, 4) and ( 3, 2) . Point (3, 1) cannot be taken because it is not in R (it’s on dotted line) 2x + y = 2(1) + 4 = 6 = 2(3) + 2 = 8  the max value 6 4x = y 5 4 3 y=x–2 2 1 R –3 –2 –1 0 1 2 3 4 5 6 –1 y+x=5 –2 –3 –4 17