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 y-intercept nd the - intercept (i vilble) (b) Determine the mimum or minimum vlue. (c) Determine the third point opposite to the y-intercept. Sketch the grph () = () Y-intercept = 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 non-liner 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 = mid-point 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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