(1) Functions A relationship between two variables that assigns to each element in the domain exactly one element in the range.

Size: px

Start display at page:

Download "(1) Functions A relationship between two variables that assigns to each element in the domain exactly one element in the range."

Elmer Bertram Adams
5 years ago
Views:

1 -. ALGEBRA () Fuctios A reltioship etwee two vriles tht ssigs to ech elemet i the domi ectly oe elemet i the rge. () Fctorig Aother ottio for fuctio of is f e.g. Domi: The domi of fuctio Rge: The rge of f, stted s f evluted t (or f of ) f is the set of vlues of for which f f is the set of ll possile vlues of f. is defied. Itervl ottio will ssist i descriig the domi d rge of fuctio. Give tht d re umers, with <, the:,,,,,,, is represeted y is represeted y is represeted y is represeted y, is represeted y, is represeted y is represeted y is represeted y is represeted y () () Commo Fctors y y y y y Decompositio for qudrtics c (fid two umers tht multiply to c d dd to ) (c) Pttered Fctorig y y y (differece of squres)... y y y y (sum or differece of cues) y y y y y (d) Fctor Theorem is fctor of fuctio f if d oly if f

2 - (e) Rtiol Zero Theorem p If is rtiol zero of polyomil P with itegrl coefficiets, the p is fctor of the costt q term of the polyomil, d q is fctor of the ledig coefficiet of the term with the lrgest degree. P q r... s p (f) Cuic Equtio c d If the roots re p, q, d r the p q r, pq pr qr c, d pqr d () Qudrtic Formul The roots of the qudrtic equtio c re, 4 c The epressio 4 c is clled the discrimit (D) ) If D> the we hve two distict rel roots ) If D= the we hve two equl rel roots ( distict rel root) c) If D< the we hve two imgiry roots (o rel roots) The sum of roots is (4) Comitorics. The product of the roots is c The symol! (red s fctoril) stds for the product of the first turl umers. ;! ; The umer of comitios of ojects tke r t time is give y., C r r! r! r!

3 - (5) Biomil Theorem () The epsio of p q, where W, is, r r p q p q r r p q p q p q p q!!!! p p q p q q!!!!!!!! p p q p q q Altertely, coefficiets c e otied frompscl s Trigle, ) The rth term i the epsio is, r r tr p q r (6) Solvig (I)Equlities () c d,, or d () c d d or c (7) Divisio of Polyomils 4 4 c

4 -4 (8) Sythetic Divisio Suppose polyomil is divided y r r r r r where r r r (9) Frctios () y y (), y y y (c) y y (d) (e) (f) y y y y y y y () Epoets () () (c) y y

5 -5 (d) (e) (f) y y ; (g) (h) m m m () Logrithms () log y log log y () log l e (c) log log log y y log (d) log y log y y (e) log y log l l y (f) (g) (h) e l log y y l e log (i) c log log c () Sequeces () Arithmetic Geerl term: t d where t d d t t () Geometric: t r where t d t r t

6 -6 () Series () Arithmetic : S t i i t d () Geometric: S t i i if r i r i S r r. TRIGONOMETRY (4) Pythgore Theorem () c r c (5) Primry d Reciprocl Rtios () () (c) (d) (e) (f) si y r r cos y t r y csc sec r y cot

7 -7 (6) Degrees vs. Rdis (7) Trigoometric Idetities () si cos (i) si cos cos cos (ii) cos si si si () t sec (i) t sec sec sec (c) cot csc (i) cot csc csc csc (d) si y si cos y si y cos (e) cos y cos cos y si si y (f) t y t y y t t t (g) si si cos

8 -8 (h) cos cos si si cos (i) Sie Lw c si si si A B C A (j) Cosie Lw (i) c c cos A c (ii) c c cos B (iii) c cos C B C (8) Trigoometric Agle Vlues si cos t udefied 6 si cos t udefied 6

9 -9 y=si() y=cos() y=t(). ANALYTIC GEOMETRY (9) Distce etwee two poits d y y y () Slope of lie through two poits rise y y y m ru () Agle etwee two lies t m m m m () Prllel Lies l l m m () Perpediculr Lies l l m m (4) Lier Equtios ) Stdrd Form: A By C ) Two Poit: y y y y

10 - c) Poit Slope: y y m d) Slope Itercept: y m y e) Two Itercept: (5) Qudrtic Equtios,where,,, re the itercepts ) Stdrd Form: c ) Verte Form: y h k, where verte t h, k (6) Circle h y k r ) cetre (h,k), rdius r: y f k r h ) Equtios of upper d lower semicircles: (7) Shpes d Solids ) Circle i) Circumferece: C r d ii) Are: A r ) Three Dimesiol Solids SHAPE SURFACE AREA ( uits ) VOLUME ( uits ) Sphere 4 r 4 r Cylider rh r Coe r r h r h r h Rectgulr Solid hw wd d hwd

11 - (8) Circle Geometry The Circumcircle of trigle is the circle pssig through the vertices of the trigle. [The Circumcetre is the poit of itersectio of the perpediculr isectors of the sides of the trigle.] The Icircle of trigle is the circle tht touches the sides of the trigle. (sides re tgets to the circle). [The Icetre is the poit of itersectio of the isectors of the gles of the trigle.] A Escried circle of trigle is tget to oe side of the trigle d to the lies formed whe the other two sides re eteded. [The cetre of the escried circle is the poit of itersectio of the isectors of the eterior gles t two of the vertices. Note: trigle hs escried circles] Theorems Tget Theorem for Circles (TTC) ) A lie is tget to circle iff the lie is perpediculr to the rdius t the poit of cotct. ) A lie is tget to circle iff the distce from the cetre of the circle to the lie equls the rdius

12 - c) The two tget segmets drw from the eterl poit: (i) re cogruet (ii) suted cogruet gles t the cetre (iii) mke cogruet gles with the lie segmet joiig the eterl poit to the cetre. Chord Cetre Theorem (CCT) ) The perpediculr isector of y chord of circle psses through the cetre of the circle ) The perpediculr from the cetre of circle to chord isects the chord. c) The lie segmet from the cetre of circle to the midpoit of chord is perpediculr to the chord. Tget Theorem for Circles (TTC) ) A lie is tget to circle iff the lie is perpediculr to the rdius t the poit of cotct. ) A lie is tget to circle iff the distce from the cetre of the circle to the lie equls the rdius. c) the two tget segmets drw from eterl poit: i) Are cogruet ii) Suted cogruet gles t the cetre iii) Mke cogruet gles with the lie segmet joiig the eterl poit to the cetre. Agle Theorem for Circles (ATC) ) A gle iscried i circle equls sme rc. the sector gle suteded y the

13 - ) Agles iscried i circle tht re suteded y the sme rc re cogruet. c) A gle iscried i semicircle is right gle. A qudrilterl with its four vertices lyig o the circumferece of circle is clled cyclic qudrilterl. d) Opposite gles of iscried qudrilterl re supplemetry. e) A eterior gle of iscried qudrilterl equls the iterior opposite gle. Tget Chord Theorem (TCT) If chord d tget re drw through poit of circle, ech of the gles determied y the tget d chord is cogruet to the gle iscried i the circle o the opposite side of the chord. Tget Sect Theorem (TST) ) If the sect from eterl poit P itersects circle t A d B, d the tget from P cotcts the circle t T, the PT PA PB

14 -4 ) If two sects from eterl poit P itersect circle t A, B d C, D respectively, the PA PB PC PD Chord Product Theorem (CPT) AB d CD re two chords tht itersect t M. The, MA MB MC MD (9) Summtios k k k k k k

15 -5 Fuctios Emple : Worked Emples Solve Solutio: (See Rtiol Zero Theorem) Let P 4 7 9, the vi the Rtiol Zero Theorem the we hve q p if p P where p is divisor 9 d q is divisor of. q The divisors of 9 re:,, 9 The divisors of re:,,, 4, 6, Therefore the possile vlues of p q re: ,,,,,,,,, 9,,,,,,,,,,,, 9,, By Tril P therefore is fctor. Now use Polyomil Divisio to fid the other fctor This gives us P 6 Now fctorig 6 we oti P Therefore, the zeroes re:,,.

16 -6 Emple Solve 4 5 Solutio () if d 4 the ( ) if d 4 the 4 5 This solutio is cotrdictio () if d 4 the (4) if d 4 the 4 5 This solutio is cotrdictio. Therefore the solutios re or A grphicl solutio c e foud y grphig the left hd side d right hd side idepedetly d loctio the poits of itersectio.

17 -7 Emple : Solve Solutio: Hit: Isolte the rdicl term, the squre oth sides (rememer to check solutios whe you squre oth sides) 4 4 Therefore = or =4 re the cdidtes. By checkig =4 is the oly solutio Emple 4: For f k k f for ll., determie ll vlues of k such tht Solutio: (See sum of roots d product of roots i Qudrtic Formul) We will fid the vlues of where f d the verte is o the -is. So pplyig the sum d product of zeros formul. k k () () We ote tht if the verte i o the -is, the. Thus k () k () Now solvig for i () d sustitutig ito () d solve for k. k 4 k k 4 k 6 k k k 8 Therefore k= d k=8, sice we wt f, k 8 is the solutio set.

18 -8 Epoets d Logrithms Emple : Solve 5 5 Solutio: Fctor whe you hve terms with sme se seprted y dditio d/or sutrctio Emple : Evlute log 5log 7 Solutio: log 5log 7 log 5 5 Emple : Solve Solutio: Let u 5 5u u 5 u u u u u Now 5

19 -9 Emple 4: Fid log log log...log 4 Solutio: 9 4 log log log log log... log log log log log... log 9 log Emple 5: Solve e 5 99 Solutio: 5 e 99 e 5 l 99 l 99 5 e l l 99 5 l l 99 Emple 6: 5.99 Determie ll vlues of t where t t Solutio: t t t t log log log t t t t 4 4 log4 t t t t 4 t 4 t 6 t, t Check for vlidity log log 4 4 Whe t we oti logrithms of egtive umers which is ot llowed, therefore t=.

20 - Trigoometry Emple : cos 7 cos, Solve Solutio: Let u cos u 7 u u or is ot vlid, therefore cos or 5.9 Emple : Prove 4 4 Solutio: Left Side cos si cos cos si cos si cos 4 4 cos si cos cos si cos si cos Right Side Sice Left Side equls Right Side the idetity is true.

21 - Emple : Wht is the period of y si 6 cos 5 Solutio: The period of si 6 is 6 The period of cos 5 is. 5 We eed to fid d such tht Therefore is = or =5 the period is. This is the smllest distce tht oth d divide evely ito. 5 Emple 4: A eperimetl Ferris wheel with m rdius rottes couter clockwise from the oservers side. Strtig from iitil groud positio with uiform rte of revolutio per secod. Determie fuctio tht gives the height of the chir t y time t. Solutio: We will look t this s sie fuctio si = h k t p d with the followig ttriutes Sice period is the, or k k The phse shift is p to the right, d the verticl trsltio is d= 4 The equtio is h si t 4.

22 - Biomil Theorem Emple : Write out the iomil epsio of Solutio: Emple : I the iomil epsio of 6, fid the term idepedet of. Solutio: We wt the term which hs the The geerl term of the epsio is tk. tk k 6 6 k k k Sice we wt the term with, we must simplify k t to oti power of t k k 6 6k k k k 6k k 6 k k 6 k 6k k For to hve zero epoet, the k or k=4 t

23 - Sequeces d Series The geerl term of rithmetic sequece is give y: t d umer of the term, d d is the commo differece. Emple : Give the rithmetic sequece: 8, 4,, 6,... ) fid the th term g where is the first term, the Solutio: t d t Therefore the th term is ) which term is 6 Solutio: t d g g Therefore 6 is the 9 th term of the sequece The geerl term of geometric sequece is give y; t of the term, d r is the commo rtio. r, where is the first term, the umer Emple: Give the geometric sequece:, 6,, 4,... ) Fid the 4 th term Solutio: t t 4 r Therefore the 4 th term is 4576

24 -4 ) which term is 84 Solutio: t r Therefore 84 is the 8 th term of the sequece. I y series: The geerl term is the differece etwee the sum of the first terms d the sum of the first (-) terms t S S The first term is the sme s S : t S For the geerl rithmetic series: d d... d The sum of the first terms is : S d Emple: Fid the sum of the rithmetic series Solutio: First we eed to determie which term is 5 t d g g Net we eed to determie the sum. S d g Therefore the sum of the series is 6. g

25 -5 For the geometric series: r r r... the sum of the first terms is : S r, r r Emple: Fid the sum of the first 7 terms of Solutio: S r r Therefore, the sum of the first seve terms is Alytic Geometry Give the Tget Sect Theorem If the sect from eterl poit P itersects circle t A d B d the tget from P cotcts the sme circle t T, the ccompyig digrm. PT PA PB.,Determie, y i the Solutio: PT PA PB PC PD B D 5 C 4 y A P 6 y 6 9 y y 7 y 9 T

Important Facts You Need To Know/Review:

Important Facts You Need To Know/Review: Importt Fcts You Need To Kow/Review: Clculus: If fuctio is cotiuous o itervl I, the its grph is coected o I If f is cotiuous, d lim g Emple: lim eists, the lim lim f g f g d lim cos cos lim 3 si lim, t