Add maths complete f4 & f5 Notes

Add maths complete f4 & f5 Notes

• 2.
  2 PAPER 1 (3472/ 1) No Items Comments 1 No. of questions 25 questions (Answer ALL) 2 Total Marks ……80 marks…………. 3 Time …….2 hours 4 Level of difficulty Low : Moderate : High 6 : 3 : 1 PAPER 2 (3472 / 2) No Items Comments 1 Three sections Section A Section B Section C 2 No of questions 6 5 4 (need to answer 12 questions ) Answer …ALL….. Choose …four…. Choose …two….. 3 Total marks (100) 40 marks 40 marks 20 marks 4 Time ………2 ½ hours…………….. 5 Level of difficulty Low : 4 Moderate : 3 High 3
• 3.
  FORMAT&COMPONENT 3 * Extra Time= …………………………………. ALGEBRA 1. Functions 2. Quadratic Equations 3. Quadratic Functions 4. Indices & Logarithms 5. Simultaneous Equations 6. Progressions 7. Linear Law Geometry 1. Coordinate Geometry 2. Vector Calculus 1. Differentiation 2. Integration Triginometry 1. Circular Measures 2. Trigonometric Functions STATISTICS 1. Statistics 2. Probability 3. Permutation & Combination 4. Distribution Probability Social Science 1. Index Numbers 2. Linear Programming Science & Technology 1. Solutions to Triangle 2. Motion in a Straight Line. 11 mmaarrkk  mmiinnuutteess1.5 Check answers
• 4.
  FORM 4 TOPICS 4 Learnwith your heart and you’ll see the wonders … P(X)
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  FORM 4 TOPICS 5 ab X  1 TOPIC 1 : FUNCTIONS  f(a) = b  object = …………………….. image = ………………..  Given f (x) and gf(x). Find g(x) .  Thus, g(x) =  1 53 )(    x x xf , TOPIC 2: QUADRATIC EQUATIONS [ ax2 + bx + c = 0 ] Types of roots Sum and Product of Roots [ ax2 + bx + c = 0 ] Sum of roots, ( + ) = a b Product of roots ( ) = a c x2 – S x + P = 0 TOPIC 3: QUADRATIC FUNCTIONS General form CTS form f (x) = ax2 + bx + c = a( x + p)2 + q Similarity Same value of a Same value of a Difference c = y-intercept. q = max/min value of f(x) Specialty Able to find: - shape - y intercept Able to find - turning point ( - p , q) - two distinct real roots - intersects at two points - two equal roots - touches / tangent - no root - does not intersect - f(x) is always positive 1. gf f -1 b2 – 4ac > 0 =0 < 0 0 Real roots
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  FORM 4 TOPICS 6 2.Sketch Graphs: y = ax2 + bx + c (a) From the graph 3. 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  positive or negative a b ………………………. …………….……… TOPIC 4: SIMULTANEOUS EQUATIONS  Use ……………………………….. method  To find the ………………………………… points between a straight line and a curve. i value of a : Positive ii value of b2 -4ac: < 0 iii type of roots : No roots iii y-intercept : C iv Equation of axis of symmetry : X = a2 b  c d x < a , x > b, c < x < d Substitution intersection
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  FORM 4 TOPICS 7 TOPIC5: INDICES & LOGARITHM Use Index rule : (i) 5x+1 . 125x = 25 1 ………………………. INDEX Use log : (ii) 8  3x = 7 ............................ IND Use substitution : (iii) 3 n+1 + 3 n = 12 or can be factorise 3n . 3 + 3n = 12 a(3) + a = 12 4a = 12 a = 3 3n = 3  n = 1 (iv) 3 2n + 5. 3 n = 6 a2 + 5a – 6 = 0 (a – 1) (a + 6) = 0 a = 1 , a = –6 3n = 1 3n  –6  n = 0 LOGARITHM : 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 x 2 x + 9 = 23 x 2 x + 9 = 8x 2 8x2 – x – 9 = 0 (8x -9) (x + 1) = 0 x –1, x =  Steps of solutions 1. separate the index 2. substitute Change the base to the same number Insert log on both sides 5x +1 . 5 3x = 5–2 x + 1 + 3x = – 2 x = –
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  FORM 4 TOPICS 8  TOPIC6: COORDINATE GEOMETRY  Distance   Ratio theorem   Mid point   Equation of straight line y = mx + c  Area (positive) Arrange anticlockwise : general form ax + by + c = 0  Gradient : gradient form y = mx + c - parallel m1 = m2 : intercept form 1 b y a x  - perpendicular m1 m2 = –1 Equation of locus : …use distance formula…………………. Rhombus : ……its diagonal are perpendicular to each other ………. Parallelogram, square, rectangle, rhombus. its diagonal share the same mid point TOPIC 7: STATISTCS EFFECT ON CHANGES TO DATA The change in values when each data is Mean Mode Median Range Interquartile range  Variance Added with k + k +k +k unchange unchange unchange unchange Multiplied by m  m  m  m  m  m  m  m 2 TOPIC 8: CIRCULAR MEASURES  …… radian = ………  For s = r and A = ½ r2 , the value of  is in ……radian…….   Shaded angle =  –  Area of segment = ½ r2 (  - sin )   rad 180 
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  FORM 4 TOPICS 9 TOPIC9: DIFFERENTIATION gradient of normal gradient of tangent equation of normal Tangent equation of tangent Rate of change Applications Small Changes approximate value y ORIGINAL + y minimum Turning points maximum TOPIC 10: SOLUTION OF TRIANGLES  Sine Rule - Ambiguous Case  two possible angles  acute and obtuse angle  Cosine Rule  Area =  ab sin C TOPIC 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? ……………………………… 100 120 120  …………………………………………… ………………………………………………………………………………………. mT = dx dy y – y1 = m(x – x1) mN mT = –1 dy = dy  dx dt dx dt  y = dy   x dx dx dy = 0 02 2  dx yd 02 2  dx yd
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  10 :
• 11.
  FORM 5 TOPICS 11 T1T2 T3 T4 T5 T6 T7 TOPIC 1: PROGRESSIONS Janjang Aritmetik Janjang Geometri Examples : 20, 15, 10, …..., …. 4, 3, 2.25, ……., …. Uniqueness : d = T2 – T1 r = 1 2 T T Others : 2 )( lan Sn   S = r a 1 Given Sn find Tn Example: Given Sn = n( 3 + 2n), find T8. Thus, T8 = S8 – S7 Find the sum from the 3rd term until the 7th term. S 3 to 7 = S7 – S2 TOPIC 2: LINEAR LAW Convert to linear form Y = m X + c ay = x + x b xy = a 1 x2 + a b xh x p y  xy = h x + p T + 1 = a2 + k  1T = a  + k y = ax b log y = b log x + log a y = k p x log y = log p x + log k + + + + + +
• 12.
  12 TOPIC 3: INTEGRATIONS To find THE EQUATION OF A CURVE given dy/dx dx dy = ……gradient function …………………… Equation of CURVE,  dxfunctiongradienty }{  the integrated function must have c  AREA under a curve: Show how you would find the area of the shaded region. x = y2 – y 1. Find the intersection point. when x = 2,  y = 4 2. Find the area Area under curve + area  =  2 0 2 x dx +  (1)(4) intersection, x = 1 thus, y = 4(1) – 12 = 3 Shaded Area: Area under curve – area  =   4 1 2)4( xx dx –  (3)(3) 1. Expand y y = x3 – 3x2 + 2x Shaded Area: Area above =  1 0 dxy Area below =  1 2 dxy Total area 1. Formula  1 0 dyxA  VOLUME : Show the strategy to find the generated volume. revolved about x-axis revolved about x-axis revolved about y-axis revolved about y-axis  dxyV 2  dxyV 2  dyxV 2  dyxV 2 where y2 = (x2 )2 where y2 : (4x – x2 )2 where x2 = y + 1 where x2 = (y2 – 1)2  2 0 22 dx)x( + Volume cone   4 1 22)4( dxxx – volume cone I=   2 1 dy)1y(   1 1 22 dy)1y( y = x2 str. line: y = –4x + 12 y = 4x - x2 str. line: y = –x + 4 y = x2 – 1 x = y2 – 1 1 –1412 3 4 – 3– y = x2 2 3 1 4 0 1 2 1 0 str. line: y = –4x + 12 y = 4x - x2 str.line: y = –x + 4 y = x (x–1) (x–2) 21
• 13.
  FORM 5 TOPICS 13 TOPIC4: 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 = …… 10)1(3 22  ………………………………………….. iii) unit vector in the direction of m + n = …… )ji3( 10 1  ………………………  If A(1, 3) and B(–2, 5) find AB : …OA =         3 1 , OB =         5 2  AB =         5 2 –         3 1 =         2 3 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 cos 2x = 2 3 positive values  1st and 4th quadrant 2. Determine the quadrant 3. Find the reference angle 2x = 30 4. Find new range (if necessary). 0  x  720 5. Find all the angles 2x = 30, 330, 390, 690 x = 15, 165, 195, 345
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  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 Remember :   cot .1 tan, Acos Asin Atan  , sec 2x = x2cos 1 , cosec 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 = 5 3 , tan A = –  find sin 2A 3. Do you need to use identity? sin 2A = 2 sin A cos A 4. Substitute values = 2 ( ) ( 5 3 ) = 25 24   Sketch Graphs y = a sin b x + c a = a cos b x + c a tan b x + c b = c = Basic Graphs max / minimum point number of basic shape between 0 and 2 increase / decrease translation of the 2 2 2 1- –1- 1- –1- Asin 1 y = sin x y = cos x y = tan x
• 15.
  FORM 5 TOPICS 15 TOPIC6: PERMUTATIONS & COMBINATIONS Permutations = …order of arrangement is important Combinations =…order not important…. Three committee members of a society are to be chosen from 6 students for the position of president, vice president and secretary. Find the number of ways the committee can be chosen. Permutations: 6P3 Three committee members of a society are to be chosen from 6 students. Find the number of ways the committee can be chosen. Combination: 6C3 with condition: Find the number of different ways the letters H O N E S T can be arranged if it must begin with a vowel. 2 5 4 3 2 1 condition vowels = 2 choices Find the number of ways 11 main players of a football team can be chosen from 15 local players and 3 imported players on the condition that not more than 2 imported players are allowed. condition  2 Import. Case : (2 Import, 9 local) or 3 C2. 15 C9 (1 import 10 locals) or + 3 C1. 15 C10 ( 0 import 11 locals) + 3 C0.15 C11 TOPIC 7: PROBABILITY  P(A) = )S(n )A(n - 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)
• 16.
  16 TOPIC 8: PROBABILITYDISTRIBUTION BINOMIAL DISTRIBUTION - The table shows the binomial probability distribution of an event with n = 4 . 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 = npq NORMAL DISTRIBUTION - Formula :    X 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 s =  dtv dt ds v  dt dv a  = 2 2 dt sd Maximum velocity - 0 dt dv  a = 0 return to O s = 0 - - stops momentarily v = 0 max. acceleration 0 dt da  X = r r = 0 r = 1 r = 2 r = 3 r = 4 P(X) 0.2 0.15 0.3 0.25 0.1 _ _ _ _ _ _ 0.3 0.25 0.2 0.15 0.1 0.05 P(X)
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  FORM 5 TOPICS 17 TOPIC10: LINEAR PROGRAMMING Given: (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. –3 –2 –1 0 1 2 3 4 5 6 1 2 3 4 5 6 –1 –2 –3 –4 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 y = x – 2 y + x = 5 4x = y R