BRIEF NOTES ADDITIONAL MATHEMATICS FORM


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1 BRIEF NOTES ADDITIONAL MATHEMATICS FORM CHAPTER : FUNCTION. : + is the object, + is the imge : + cn be written s () = +. To ind the imge or mens () = + = Imge or is. Find the object or 8 mens () = 8 wht is the vlue o? + = 8 ; = The object is. I the unction is written in the orm o ordered pirs (, y), is the object nd y is the imge. E.g. (, ), (, 6), (, 7) Imge or is nd 6 wheres object or 7 is. In the rrow digrm, the set o object is {,, } nd the set o imge is {, }. For :, 0, i.e. becuse is undeined. 0 Thus, i :, k then k is.. Function which mps into itsel mens () = I :, ind the vlue o which is mpped into itsel. = = ( ) = Thus, = 0 ( )( + ) = 0 = or. Inverse Function Symbol :  To ind the inverse unction, chnge () to y nd ind in tems o y.  Given :, ind Let () = y y = y( ) = y y = y = + y = ( + y) y = y, thus  () =. Composite Function Given : nd g :, ind () g() (b) g() (c) () (d) g  () () g() = ( ) = ( )  = 6 9 = 9 (b) g() = g( ) = ( ) = 9 + = 9 (c) () = () = (9 ) = () = =. y (d) Let y =, = Thus  () = 8 g  () = g( 8 ) = 8 = 6. To ind () or g() given the composite unction. Given () = + 8 nd g() = 6 +, ind g(). () = + 8 [g()] = g() + 8 = 6 + g() = = 6 + g() = + Given () = nd g() = , ind g()
2 g() = g( ) = y Let y =, = y g (y) = 9 0( y ) + 0 = y + 0y + 0y = y + Thus, g() = + CHAPATER : QUADRATIC EQUATION. Find the roots o qudrtic eqution () Fctoristion b b c (b) ormul = () Solve 6 7 = 0 ( )( + ) = 0 = 0, = + = 0, = (b) I it cnnot be ctorised, use the ormul. Solve = 0 =, b = nd c = ( ) ( ) ( ) = = = 6 =.87 = 6 = Form eqution orm roots. Use the reverse o ctoriston Find the qudrtic eqution with roots =,, =, ( ) = 0 =, ( ) = 0 The eqution is ( )( ) = = 0 nd. Using SOR nd POR nd the ormul (SOR) + POR = 0 Cri persmn kudrtik dengn punc dn SOR = + = 7 POR = = Eqution is 7 + = 0, 7 + = 0. I + b + c = 0 is the generl orm o the qudrtic eqution, b SOR = α + β = POR = αβ = c Given tht one root is twice the other root or the qudrtic eqution + m + 8 = 0, ind the postive vlue o m. The roots re α nd α m SOR = α + α = α = = m POR = α α = α = 8 α = 9 α = 9 When α =, α = 9 = m, m = 9 (not ccepted) When α =, α = 9 = m, thus m = 9. Types o roots () rel nd distinct roots. b c > 0 (b) rel nd equl roots b c = 0 (c) No rel root b c < 0 (d) Rel root (distinct or sme) b c 0 Find the rnge o vlues o k in which the eqution + k = 0 hs two rel nd distinct roots. For two rel nd distinct roots b c > 0
3 () ()k > 0 9 8k > 0 8k < 9 k < 9 8 CHAPTER : QUADRATIC FUNCTIONS. To ind the mimum/minimum vlue by completing the squre. Given () = 6 + 8, ind the mimum or minimum vlue nd stte the corresponding vlue o. () = = [ ] + 8 = [ + ] + 8. Qudrtic Inequlity () Fctorise (b) Find the roots (c) Sketch the grph nd determine the rnge o rom the grph. Find the rnge o vlue o or which 7 8 < < 0 ( 8)( + ) < 0 = 8, = Sketch the grph Note: I the coeicient o is negtive, the shpe o the grph is n =[( ) 9 ] + 8 = ( ) = ( ) + 7 The minimum vlue (the coeicient o is positive nd the grph is u shped) is 7 when = 0, or =.. To sketch qudrtic unction () Determine the yintercept nd the  intercept (i vilble) (b) Determine the mimum or minimum vlue. (c) Determine the third point opposite to the yintercept. Sketch the grph () = () Yintercept = 6 (b) () = = ( ) = ( ) 0 Min vlue = 0 when = 0, =. Min point (, 0) (c) when = 8, (8) = 8 8(8) + 6 = 6 From the sketch, ( 8)( + ) < 0 < < 8. Types o Roots () I the grph intersects the is t two dierent points rel nd distinct roots b c > 0 (b) I the grph touches the is, equl roots b c = 0 (c) I the grph does not intersect the  is,(or the grph is lwys positiv or lwys negtive.) no rel root b c < 0 The grph y = n + + n does not intersect the is or n < nd n > b, ind the vlue o nd b. y = n + + n does not intersect the is no rel root b c < 0 n(n ) < 0 6 n + n < 0 0 < n n 6 n n > 0 (n )(n + ) > 0 n =, n = From the grph, or (n )(n + ) > 0, n < nd n >
4 = nd b = CHAPTER : SIMULTANEOUS EQUATIONS To solve between one liner nd one nonliner eqution. Method : Substitution Emple : Solve + y = () y () y rom (), y + y = y () rom (), = y substitute in () ( y) + y = ( y)y (6 6y + y ) + y = 0y 0y 8y + 0y + y y 0y + = 0 0y y + = 0 y y + 8 = 0 (y 8)(y ) = 0 y = 8 or y = 8, = ( 8 ) = 6 = y =, = = Thus, =, y = nd =, y = 8.!Note Be creul not to mke the mistke ( y) =6 + y wrong I the equtions re joined, you hve to seprte them. Solve + y = + y = + y = nd + y = CHAPTER : INDEX AND LOGARTHM Inde orm: b = Logrithm orm log b = Logrithm Lw :. log + log y = log y. log n = nlog. log b =. log = 6. log = 0 logc log b c Emple: Find the vlue o log 8 log + log 8 log 8 log + log = log 8 = log = log 6 = log 9 = log = = To solve inde equtions, chnge to the sme bse i possible. I not possible to chnge to the sme bse tke logrithm on both sides o the eqution. Emple: Solve.7  = = 9 () = () + = 6 + = 6 = = Emple: Solve + 7 = = 0 + = 7 log + = log 7 ( + )log = log 7 + = log 7 log =.09 =.09 =.79 Emple: Solve log 8 log log 6 =. log log y = log y log 8 6 =
5 log 6 = 6 = = = CHAPTER 6: COORDINATE GEOMETRY. Distnce between A(, y ) nd B(, y ) AB = ( ) ( y y ) Emple: I M(k, k) nd N(k +, k ) re two points equidistnt rom the origin O. Find the vlue o k. MO = ON ( k) k (k ) ( k ) Squre, k + k = k + k + + k 6k = k + 9 k = 9 k = 9. Point which divides line segment in the rtio m : n n m ny my, n m n m Emple: Given Q(, k) divides the line which joins P(, ) nd R(, 9) in the rtio m : n. Find () the rtio m : n (b) the vlue o k () (b) n m = n m n + m = n + m m m = n n m = n m n thus, m : n = : 9 = k = k. Eqution o stright line Grdient orm: y = m + c y Intercept orm: b y int ercept Grident = m = = int ercept Generl orm: + by + c = 0 b The eqution o stright line given the grdient, m, nd psses through the point (, y ) : y y = m( ) Eqution o stright line pssing throug two points (, y ) nd (, y ) is y y y y Emple: Find the equtioon o the stright line () with grdient nd psses through (, ) (b) psses through (, ) nd (, 8) () (b) Eqution o stright line y () = ( ) y + = y = Eqution o stright line y 8 y (y ) = ( ) y 0 = 6 y = +. Prllel nd Perpendiculr Line Prllel lines, m = m Perpendiculr lines, m m = Emple: Find the eqution o the stright line which is prllel to the line y = nd psses through (, ) y =, y =  m =, psses through (, ) Persmn gris lurus ilh
6 y = ( ) y 8 = y = + Emple: Find the eqution o the stright line y which is perpendiculr to the line nd psses through (, ) y, m = m = ( ) = m =, psses through (, ) The eqution o the stright line is y = ( ) y = + 6 y + = 8. Eqution o Locus Emple: Find the eqution o the locus or P which moves such tht its distnce rom Q(, ) nd R(, ) is in the rtio : Let P(, y), Q(, ), R(, ) PQ : PR = : PQ PR PR = PQ ( ) ( y ) ( ) ( y ) Squre, y 6y + 9 = ( + + y y + ) = + y 8 6y = + y y 0y y 0y + 7 = 0 CHAPTER 7: STATISTICS. Ungrouped Dt Men, N Vrince, = ( ) N N = Stndrd devition = vrince Emple: For the dt,,, 6, 7, 8 ind the () men (b) vrince (c) stndrd devition () = N 6 =. 667 (b) vrince, = = =.6 (c) stndrd devition = =.6 =.99. Grouped Dt i Men, = requency Medin, Emple: M = L + N F cu m i = midpoint c L = lower boundry o the medin clss N = totl requency F cu = cumultive requency beore the medin clss m = requency o medin clss c = clss intervl size Mode is obtined rom histogrm Stndrd devition, = requency i Mode ( ) clss 6
7 The tble shows the mrks obtined in test. Mrks Frequency Find, () men mrk (b) medin (c) mode (d) stndrd devition Mrk i i i () Men = (b) (c) i = 0 0 = 8.8 N 0 = Medin clss = 9 M =. + = 8.67 Frequency C.F (d) = i ( ) 760 = = 6.76 = 8.09 CHAPTER 8: DIFFERENTIATION represents the grdient o curve t point. = () = irst derivtive = grdient unction. d n n ( ) n Dierentition o Polynomils. Dierentite with respect to : () y = + (b) y = (c) y = () y = + = + 6 (b) y = = = (c) y = =  =  = Dierentition o Product d ( uv) u dv v du. Dierentite with respect to : y = ( + )( ) From the grph, mode = 8 mrk = ( + ) + ( ) = 0 + = 0 Dierentition o Quotient 7
8 du dv d u v u v v. Dierentite with respect to y = ( ) ( ) ( ) = = ( ) ( ) Dierentition o Composite Function d ( b ) n = n( + b) n. Dierentite with respect to : () ( + ) 8 (b) ( ) ( + ) () y = ( + ) 8 = 8( + )7 = ( + ) 7 (b) y = ( ) ( + ) = ( ) ( + ) + ( + ) ( ) = ( ) ( + ) + 8( ) ( + ) = ( ) ( + ) [( ) + 8( + )] = ( ) ( + ) [ ] = ( ) ( + ) ( + ) Eqution o Tngent nd Norml Grdient o tngent = grdient o curve = Emple: Find the eqution o the tngent to the curve y = + t the point =. y = + = 6 =, y = + = 0 = 6 = Eqution o tngent : Note: you must dierentite the unction in the brckets. y 0 = ( ) y =. Mimum nd Minimum Vlue Given y = 8 +. Find the coordintes o the turning point. Hence, determine i the turning point is mimum or minimum. y = 8 + = 8 For turning point = 0 8 = 0 = =, y = () 6 + = d y = > 0, thus the point (, ) is minimum point. Rte o Chnge o Relted Quntities Emple: The rdius o circle increses which rte o 0. cm s , ind the rte o chnge o the re o the circle when the rdius is cm. A = r da dr = r dr = 0. cm s  dt da da dr dt dr dt = r 0. = 0. r When r = cm, da dt = 0. = cm s  Smll Chnges nd Approimtion y Emple: Given y = +, ind the smll chnge in y when increses rom to.0 y = + = =.0 = 0.0 8
9 y = = ( ) 0.0 Substitute the originl vlue, =, y = (8 ) 0.0 = 0.0 Thus the smll increment in y is 0.0. CHAPTER 9: INDEX NUMBER p. Price Inde, I = p 00 0 p = price t certin time p 0 = price in the bse yer (c) p = p = = RM7. Composite inde I Iw w I = price inde w = weightge Emple: Item Price inde Weightge Book 00 6 Beg Shirt y Shoes 0 The tble bove shows the price indices nd the weightge or our items in the yer 00 bsed in the yer 000 s bse yer. I the price o beg in the yer 000 nd 00 re RM0 nd RM respectively. The composite inde or 00 is 6. Find () the vlue o (b) the vlue o y (c) the price o shirt in 00, i the price in 000 ws RM60. () = 00 = 0 0 (b) y 0 = 6 6 y y 0 = 6 y 0 + y = 6( + y) 0 + y = y y 6y = y = 6 y = 9
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