FORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81


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1 FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first terms is. Find the first term nd the common difference () the sum of the first 0 terms (c) the sum from the th to the 0th term. T = 86 S n = n [ + Tn ] = [ 86] 74 = 86 = + 4d = d = 86 4d = 98 d = 7 () S 0 = 0 [() + 9(7)] = 0[4 ] = 090 (c) Sum from th to 0th term = S 0 S = 090 [4 + 0(7)] = 090 () = 87 Geometric Progression T n = r n n n ( r ) r ( ) S n = = r r For < r <, sum to infinity S r Emple: Given the 4th term nd the 6th term of G.P. is 4 nd 0 respectively. Find the 8th term given tht ll the terms re positive. r = () r = 0 = () () () r 4 r r = 4 9 r = Since ll the terms re positive, r = = 4 = = 8 T 8 = 8 7 = Emple: Find the lest numer of terms of the G.P 8, 6,,,... such tht the lst term is less thn Find the lst term. = 8, r = 6 8 T n < n n < < (n ) log < log log n > 8 log [Rememer to chnge the sign s log is negtive] n > 0.0
2 n >.0 n =. T = 8 = Emple: Epress ech recurring decimls elow s single frction in its lowest term () = = 0.7 r = = S r () 0... = = 0., r = = S CHAPTER : LINEAR LAW Chrcteristic of The Line of Best Fit. Psses through s mny points s possile.. All the other points re s ner to the line of est fit s possile.. The points which re ove nd elow the line of est fit re equl in numer. To Convert from Non Liner to Liner Form To convert to the form Y = mx + c Emple: Non Liner m c Liner y = y = + y = + log y = log () + log log log y y = + Emple: The digrm shows the line of est fit y plotting y ginst. Find the reltion etween y nd. Solution: m =, pssing through (6, ) 6 = (6) The eqution is y = +. + c c = y, or Emple: The tle elow shows vlues of two vriles nd y, otined from n eperiment. It is known tht y is relted to y the eqution y =. 4 6 y Eplin how stright line cn e otined from the eqution ove. () Plot the grph of log y ginst y using scle of cm to unit on the is nd cm to 0. unit on the yis. (c) From your grph, find the vlue of nd. Solution: y = log y = log () + log By plotting log y ginst, stright line is otined. () 4 6 log y
3 () 4 6 d = = d 4 = c 4 = 4 + c 4 d 6 6 (c) c = log = 0.48 = log = = 0.7 =. CHAPTER : INTEGRATION n n d c n The re etween the grph nd the is A = y d The re etween the grph nd the yis A = dy The volume generted when shded region is rotted through 60 o out the is V = y d Volume generted when shded region is rotted through 60 o out the yis V = dy Emple: Find () d 4 6 d d = c = c The Rule of Integrtion. kf ( ) d k f ( ) d g. f ( ) g( ) d f ( ) d g( ) d c c. f ( ) d f ( ) d f ( ) d 4. f ( ) d f ( ) d Emple: Given () () f ( ) d 9 4 f ( ) d [ f ( )] d 4 f ( ) d = 4 = 4 9 = 6, find the vlue of f ( ) d [ f ( )] d = d = + 9 = [ ] + 9 = 9 f ( ) d Are Below Grph. The re elow grph nd ounded y the line =, = nd the is is
4 Are under curve = Are of tringle + d = = 9 + [ ] 9] = 6 unit A = y d. Are etween the grph nd the line y = c, y = d nd the yis is Volume of Revolution A = d c dy Emple: Given A is the point of intersection etween the curve y = nd the line y =, find the re of the shded region in the digrm elow. Volume generted when shded region is revolved through 60 o out the is is V = y d Volume generted when shded region is rotted through 60 o out the yis is y = y = = = 0 ( ) = 0 = 0 or. A(, 6) = 0 ( ) = 0, = 0 or = V = d c dy CHAPTER 4: VECTORS Addition of Two Vectors Tringle Lw AB BC AC () Prllelogrm Lw 4
5 AB AD AC Prllel Vectors AB is prllel to PQ if AB k PQ constnt. If AB k BC nd C re colliner., where k is, since B is common point, A, B Vector on Crtesin Plne OAi yj OA y = mgnitude of vector OA Unit vector in the direction of i yj OA y Emple : Given OA = nd OB y. P is point on AB such tht AP : PB = : nd Q is the midpoint of OB. The line OP intersects AQ t the point E. Given OE = k OP nd AE haq, where h nd k re constnts, find OQ nd OP in terms of nd/or y. () Epress OE in terms of (i) k, nd y, (ii) h, nd y. (c) Hence, find the vlue of h nd k. OQ = OB y OP OA AP = = AB ( y ) = ( y) () (i) OE = k OP = k ( y) = k k y (ii) OE OA AE = OA + h AQ = h( y) (c) = ( h) + h y Compre the coefficient of nd y h = k () h k nd, h = k () Sustitute in () k k = = 4 k k = 4 h = k = 4 =. Emple : Given OP nd OQ. 4 () (c) Find OP Find the unit vector in the direction of OP. Given OP = m OA n OQ nd A is the point (, 7). Find the vlue of m nd n. OP = ( 4) () Unit vector in the direction of OP = i 4j (c) OP = m OA n OQ m n 4 7
6 m n = () 7m n = () () 0m n = () 7m n = (4) 7m = 9 m = 9 7 sustitute in (), 8 7 n = n = 8 7 = 7 CHAPTER : TRIGONOMETRIC FUNCTION Angles In The Four Qudrnts The Three Trigonometric Functions Secnt = sec = cos Cosecnt = cosec = sin Cotngent = cot = tn The Reltion Between Trigonometric Functions Sin = cos (90 o ) Cos = sin (90 o ) Tn = cot (90 o ) Tn = sin cos Grphs of Trigonometric Functions y = sin Amplitude = Numer of periods = The Addition Formule Sin (A B) = sin A cos B cos A sin B Cos (A B) = cos A cos B sin A sin B tn A tn B Tn (A B) = tn Atn B The Doule Angle Formule Sin A = sin A cos A Cos A = cos A = sin A tn A Tn A = tn A CHAPTER 6: PERMUTATION AND COMBINATION. Arrngement of n different ojects without repetition. n P = n! n. Arrngement of r ojects from n ojects n n! Pr ( n r)! Emple : Given the word TABLES. Find the numer of wys to rrnge ll the letters in the word. () The numer of wys of rrnging the 6 letters such tht the first letter is vowel. (c) The numer of wys of rrnging 4 letters out of the 6 letters such tht the lst letter is S. Numer of rrngement = 6 P 6 = 6! = 70 () Numer of wys of rrnging vowel out of = P Numer or wys of rrnging the remining letters = P. (c) Totl rrngement = P P = 40. OR: 4 Totl numer of wys =! = 40 If the lst letter is S, the numer of wys of rrnging letters out of the remining letters = P = 60. OR: 4 Numer of wys = 4 = 60 6
7 . Comintion of r ojects from n ojects n n! is Cr r!( n r)! Emple: The PTA committee of school consists of 8 memers. The memers re elected from 7 prents, 6 techers nd the principl of the school. Find the numer of different committees tht cn e formed if the principl is one of the memer of the committee. () the committee consists of the principl, techers nd 4 prents. (c) the committee consists of t lest techers. numer of committees = C C7 = 76 (elect principl, nd 7 committee memers from = 7 prents nd 6 techers) () numer of committees = 6 7 C C C 700 (c) 4 numer of committees = totl numer of committees numer of committees with no techer numer of committees with techer. = 4 C 6 C 8 C 6 C 8 C = CHAPTER 8: PROBABILITY DISTRIBUTION. Binomil Distriution The proility of getting r success in n trils where p = proility of success nd q = proility of filure P(X = r) = n C r n r r p q. Men, = np. Stndrd devition = npq Emple: In survey of district, it is found tht one in every four fmilies possesses computer. If 6 fmilies re chosen t rndom, find the proility tht t lest 4 fmilies possess computers. () If there re 800 fmilies in the district, clculte the men nd stndrd devition for the numer of fmilies which possess computer. P(X 4) = P(X = 4) + P(X = ) + P(X = 6) = 6 C (0.) 4 (0.7) 6 C (0.) (0.7) (0.) = () Men, = np = = 0. Stndrd devition = npq = =.6. Norml Distriution Z = X where Z = stndrd norml score X = norml score = men = stndrd devition P( Z < ) = P(Z >) P(Z < ) = P(Z > ) P(Z > ) = P(Z > ) P( < Z < ) = P(Z > ) P(Z > ) P( < Z < ) = P(Z > ) P(Z > ) Emple: The volume of pcket drink produced y fctory is norml distriuted with men of 00 ml nd stndrd devition of 8 ml. Determine the proility tht pcket drink chosen t rndom hs volume of more thn 0 ml () etween 490 ml nd 0 ml P(X > 0) = P(Z > ) 8 = P(Z >.) = 0.06 () P(490 < X < 0) = P( < Z < 8 8 = P(. < Z <.) = P(Z >.) P(Z >.) = (0.06) = CHAPTER 9: MOTION ALONG A STRAIGHT LINE. Displcement (S) Positive displcement prticle t the right of O ) 7
8 Negtive displcement prticle t the left of O. Return to O gin s = 0 Mimum/minimum displcement ds = 0 dt Distnce trvelled in the nth second = S n S n Emple: Distnce trvelled in the third second = S S Emple: A prticle moves long stright line nd its displcement, s meter, from fied point O, t seconds fter leving O is given y s = t t. Find the displcement of the prticle fter seconds, () the time t which it returns to O gin. (c) the distnce trvelled in the fourth second. s = t t t =, s = 0 = m () Return to O gin s = 0 t t = 0 t( t) = 0 t = 0 or t = second the prticle returns to O gin when t = s. (c) Distnce trvelled in the 4th second = S 4 S = (8 6) (6 9) = 8 + = m. the distnce trvelled in the 4th second is m.. Velocity (v) Velocity is the rte of chnge of displcement with respect to time. v = ds dt Initil velocity the vlue of v when t = 0 Instntneously t rest/chnge direction of movement v = 0 Moves towrds the right v > 0 Moves towrds the left v < 0 Mimum/minimum velocity dv = 0 dt s = v dt Distnce trvelled in the time intervl t = until t = If the prticle does not stop in the time intervl Distnce = v dt () If the prticle stops in the time t = c seconds where c is in the intervl Distnce = c v dt v dt Emple: A prticle moves long stright line pssing through fied point O. Its velocity, v m s , t seconds fter pssing through O is given y v = t +. Find the displcement t the time of 4 second. s = v dt = t dt = t + t + c If the prticle psses through O when t = 0,, s = 0 when t = 0. c = 0 s = t + t When t = 4 s, s = 6 + = 8 m Emple: A prticle moves long stright line pssing through fied point O. Its velocity, v m s , t seconds fter pssing through O is given y v = t + t 6. Find () the initil velocity of the prticle, the time when the prticle is momentrily t rest. initil velocity t = 0 v = 6 m s  () momentrily t rest v = 0 t + t 6 = 0 (t + )(t ) = 0 t = or t = s Since the time cnnot e negtive, t = s.. Accelertion Accelertion is the rte of chnge of velocity with respect to time. = dv dt = d s dt Initil ccelertion when t = 0 Decelertion < 0 Positive ccelertion velocity incresing Negtive ccelertion velocity decresing c 8
9 Zero ccelertion uniform velocity. Mimum/minimum velocity = 0 v = dt Emple: A prticle moves long stright line pssing through fied point O. Its velocity, v m s , t seconds fter leving O is given y v = t 6t 7. Find the initil ccelertion of the prticle () the time when the velocity of the prticle is mimum. = dv = t 6 dt t = 0, = 6 m s  () Mimum velocity, = 0 t 6 = 0 t = s. Emple : A prticle moves long stright line pssing through fied point O with velocity of m s . Its ccelertion, ms  the time t second fter leving O is given y = t 4. Find the mimum velocity of the prticle. = t 4 v = t 4dt = t 4t + c t = 0, v =, c = v = t 4t + For mimum velocity, = 0 t 4 = 0, t = s. v m = = m s  CHAPTER 0: LINEAR PROGRAMMING Steps. Form the liner inequlities.. Construct the region which stisfies the constrints.. Form the optimum eqution + y = k 4. Find the point in the region which gives the mimum or minimum vlue.. Sustitute the vlue of nd y to otin the optimum vlue of k. The tle ove shows the time tken y tilor to prepre shirt nd shorts of school uniform. In week, the tilor sells shirts nd y shorts. Given tht in week, the numer of shirts nd shorts sold must e t lest 0. The time for preprtion is t most 800 minutes. The rtio of the numer of shorts to the shirts must e t lest :. Write three inequlities other thn 0 nd y 0 which stisfy the constrints ove. () By using scle of cm to 0 units on the  nd yes, drw the grph of ll the inequlities ove. Hence, shde the region R which stisfies the constrints ove. (c) The tilor mkes profit of RM nd RM in selling shirt nd shorts respectively. Find the mimum profit mde y the tilor in week. the inequlities re: (i) + y 0 (ii) 0 + 0y y 80 y (iii) y () To drw + y = 0, = 0, y = 0 nd y = 0, = 0 To drw + y = 80 = 0, y = 80, y = 40 y = 0, = 80 To drw y = = 0, y = 0 = 40, y = 0. (c) Profit, k = + y Let k = 0 + y =0 = 0, y = 0, y = 0 y = 0, = 0, = 0 Emple: School uniform Time of preprtion (minutes) Shirt 0 Shorts 0 9
10 From the grph, mimum profit is chieved when = 40 nd y = 0. mimum profit = = RM 60. ALL THE BEST FOR YOUR SPM EXAM. 0
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