REVISION SHEET FP1 (AQA) ALGEBRA. E.g., if 2x

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1 The mi ides re: The reltioships betwee roots d coefficiets i polyomil (qudrtic) equtios Fidig polyomil equtios with roots relted to tht of give oe the Further Mthemtics etwork wwwfmetworkorguk V 7 REVISION SHEET FP (AQA) ALGEBRA Roots d coefficiets i polyomil equtios Before the em you should kow: Use the reltios betwee the symmetric fuctios of the roots of polyomil equtios d the coefficiets The method of substitutio which is vilble for fidig polyomil equtio with roots relted to give oe Eg, if 7+ = hs roots α, β the ( y 3) 7( y 3) + = will hve roots α + 3, β + 3 Qudrtic b c If + b + c = hs roots α d β the: i) α + β = ii) αβ = Emple The qudrtic equtio + 3+ = hs roots α d β i) Write dow the vlues of α + β dαβ ii) Fid qudrtic equtio with iteger coefficiets with roots α,β Solutio 3 i) α + β = d αβ = 3 5 ii) ( α ) + ( β ) = ( α + β) = = 3 7 ( α )( β ) = αβ ( α + β) + = + = Therefore qudrtic with iteger coefficiets with roots α,β is = Emple (Substitutio Method) The qudrtic equtio = hs roots α d β Fid qudrtic equtio with iteger coefficiets with roots α +,β + Solutio w Let w= z+ so tht z = Siceα d β re the roots of + 3+ =, α +,β + re the roots w w of = Multiplyig this out gives: ( w ) + 3( w ) + = 3 3 w w+ + w + = w w+ = Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

2 the Further Mthemtics etwork wwwfmetworkorguk V 7 REVISION SHEET FP (AQA) CALCULUS d TRIGONOMETRY The mi ides re: Fidig the Grdiet of the Tget to Curve t Poit Evlutig Simple Improper Itegrls Solvig Trigoometric Equtios usig Ect Vlues Before the em you should kow: How to fid the grdiet of the tget to curve t poit d i prticulr tht: df f( + h) f( ) f'( ) = = lim d h h How to evlutig simple improper itegrls How to solve trigoometric equtios usig ect vlues Grdiet of Tget Cosiderig chords For equtio of the form y = f(), where f() is simple polyomil, ie 3 3, the poit A could hve coordites (,f( )) d poit B could hve coordites ( + h,f( + h)) f( + h) f( ) The stright lie AB hs grdiet: h By lettig h the grdiet of y = f() t give poit A c be foud, provided the pproprite limit o h c be evluted, d the resultig limit is the derivtive of f with respect to Hece, df f( + h) f( ) f' ( ) = = lim This is lso referred to s differetitig from first priciples d h h Emple A grph hs the equtio y =, fid the grdiet of the chord from the poit where = 3 to the poit where = 3 + h d hece fid the grdiet of the tget whe = 3 Solutio Whe = 3, y = Whe = 3 + h, y = (3 + h) (3 + h) = 9 + 6h + h 3 h = + 5h + h y y Grdiet of chord = ( + 5 h+ h ) = = 5 + h h As h, the grdiet of the chord 5 Hece, the grdiet of the tget t = 3 is 5 Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

3 the Further Mthemtics etwork wwwfmetworkorguk V 7 Improper Itegrls A improper itegrl is defiite itegrl for which the itegrd (the epressio to be itegrted) is udefied either withi or t oe or both of the limits of itegrtio, or t some poit betwee the limits of itegrtio You c decide whether or ot improper itegrl hs fiite vlue, d if so, wht it is, by cosiderig its limits Emples Fid, if possible, the vlues of i) d ii) Solutios 3 d i) d ii) 3 foud d = d = = +, s,, so the vlue of the itegrl is = d = =, s,, so the vlue of the itegrl cot be 3 Trigoometry I order to solve trigoometric fuctio usig ect vlues you will eed to recll the specific vlues (which c be see i the tble below) You my lso prefer to thik bout equilterl trigle (of side legth ) to fid the vlues for 3 o d 6 o d right gles isosceles trigle of legths uit, uit d uits (o the hypoteuse) to clculte the vlues for 5 o θ π π π 3, 5, 6, 9,π 6 3 cosθ 3 siθ tθ 3 3 or 3 or or 3 3 Udefied It is lso importt tht you recll i which qudrt, ie < < 9 o, 9 o < < 8 o etc trigoometric o o fuctios re equivlet, ie si(8 θ ) = si θ,cos(8 θ) = cosθ etc Emple Fid the ect vlue of cos 5 o Solutio 5 o is i the third qudrt, so cos 5 o is egtive cos 5 o = cos (8 o + 5 o ) = cos 5 o = or Emple Fid the geerl solutio of the equtio siθ = 5 Solutio The geerl solutio of the equtio siθ = siα is give by: θ = (36 + α) o or θ = ( α) o You kow tht si 3 o = 5, so θ = (36 + 3) o or θ = (36 + 5) o Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

4 the Further Mthemtics etwork wwwfmetworkorguk V 7 The mi ides re: Sketchig Grphs of Rtiol Fuctios d Prbols, Ellipses d Hyperbols Solvig Iequlities Grph Sketchig Rtiol fuctios To sketch the grph of y REVISION SHEET FP (AQA) GRAPHS AND INEQUALITIES N( ) D( ) = : Fid the itercepts tht is where the grph cuts the es Fid y symptotes the verticl symptotes occur t vlues of which mke the deomitor zero Emie the behviour of the grph er to the verticl symptotes; good wy to do this is to fid out wht the vlue of y is for vlues of very close to the verticl symptote Before the em you should kow: There re three mi cses of horizotl symptotes Oe is curve which is lier polyomil divided by lier + polyomil, for emple y = This hs horizotl 3 coefficiet of o the top symptote t y= Here this would coefficiet of o the bottom be y = 3 The secod is curve give by qudrtic polyomil divided by qudrtic polyomil, for emple, y = + s ±, This hs horizotl symptote t coefficiet of o the top 5 y = Here this would be y = coefficiet of o the bottom Thirdly, whe the curve is give by lier polyomil divided by qudrtic polyomil, it will geerlly hve the -is (y = ) s horizotl symptote Whe you solve iequlity, try substitutig few of the vlues for which you re climig it is true bck ito the origil iequlity s check About grphs of prbols, ellipses d hyperbols Emie the behviour roud y o-verticl symptotes, ie s teds to ± Emple Sketch the curve y = Solutio (Sketch) The curve c be writte s y = ( + )( ) If = the y = -5 So the y itercept is (, -5) Settig y = gives, = - d = So the itercepts re (-, ) d (, ) The deomitor is zero whe = d whe = so these re the verticl symptotes Also ( + ) + y = = = + so y = - is horizotl symptote Rtiol fuctios You re lso required to kow the grphs of prbols, ellipses d hyperbols with equtios: i) y = ii) y + = iii) y = iv) y = c Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

5 the Further Mthemtics etwork wwwfmetworkorguk V 7 Solvig Iequlities Brodly spekig iequlities c be solved i oe of three wys, or sometimes i combitio of more th oe of these wys Method Drw sketch of the iequlity For emple, if you re sked to solve iequlity of the form g( ) f ( ) the sketch both f d g, d idetify poits where the grph of f is lower th the grph of g These poits will lie betwee poits for which g( ) = f( ) d so these usully eed to be clculted Method Use lgebr to fid equivlet iequlity which is esier to solve Whe delig with iequlities remember there re certi rules which eed to be obeyed whe performig lgebric mipultios The mi oe is DON T MULTIPLY BY A NUMBER UNLESS YOU KNOW IT S SIGN, IF IT S NEGATIVE YOU MUST REVERSE THE INEQUALITY SIGN, IF IT S POSITIVE THEN LEAVE THE INEQUALITY SIGN AS IT IS For emple, do t multiply by ( ) becuse tht s positive whe > d egtive whe < O the other hd ( ) is lwys positive so you c sfely multiply by this (with o eed to reverse the iequlity sig) Method 3 Sometimes is esier to rerrge iequlity of the from g( ) f ( ) to g( ) f ( ) (you do t hve to worry bout reversig the iequlity for such rerrgemet) Idetify poits where g() f() = or where g() f() hs verticl symptote Filly test whether the iequlity is true i the vrious regios betwee these poits Emple Solve the iequlity 3 + Solutio (usig Method 3) (3 )( ) ( + ) ( ) Lookig t the epressio, 3 = if =, = if = d = if = This mes tht the truth of the iequlity should be tested i ech of the followig regios < < < < < > It c be see tht the iequlity is TRUE if <, flse if < <, TRUE if < < d FALSE if > The solutio is therefore, < C you see why = d = re icluded s vlues for which the iequlity is true, but = is ot? Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

6 the Further Mthemtics etwork wwwfmetworkorguk V 7 REVISION SHEET FP (AQA) COMPLEX NUMBERS The mi ides re: Mipultig comple umbers Comple cojugtes d roots of equtios Before the em you should kow: How to dd d subtrct comple umbers How to multiply two comple umbers quickly d i oe step s this will sve lot of time i the em Tht o-rel roots of qudrtic equtios with rel coefficiets occur i cojugte pirs Mipultig Comple Numbers Addig, subtrctig d multiplyig Addig d subtrctig both ivolve firly strightforwrd clcultios It is importt to remember to equte rel d imgiry prts Multiplyig is slightly more difficult d cre should be tke to mke sure tht the correct sig is writte whe two imgiry prts re multiplied Emple Multiply the two comple umber 3 + i d 3i Solutio (3 + i)( 3 i) = 6 9i+ i 3i = 6 7i 3( ) = 6 7i + 3 = 9 7i Comple Cojugtes d Roots of Equtios The comple cojugte of z = + bi is z = bi ( z (3 + i))( z (3 i)) = (( z 3) i)(( z 3) + i) = ( z 3) i = z 6z+ 3 Remember zz is rel umber d it equls the squre of the modulus of z Comple roots of polyomil equtios with rel coefficiets occur i cojugte pirs This mes tht if you re told oe comple root of polyomil equtio with rel coefficiets you re i fct beig told two roots This is key to swerig some typicl em questios A emple of lgebric trick tht it is very useful to kow is: You should be comfortble with mipultig equtios primrily ivolvig comprig rel d imgiry prts Eg z+z* = +i, where z* is the cojugte of z Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

7 The mi ides re: Mipultig mtrices the Further Mthemtics etwork wwwfmetworkorguk V 7 Usig mtrices to represet trsformtios REVISION SHEET FP (AQA) MATRICES Before the em you should kow: How to dd, subtrct d multiply mtrices Tht mtri multiplictio is ssocitive, so ABC ( ) = ( ABC ) but ot commuttive, so AB BA The stdrd mtrices for rottio, reflectio d elrgemet d uderstd how mtrices c be combied to represet composite trsformtios Mipultig mtrices Addig d subtrctig mtrices re strightforwrd Multiplyig mtrices is slightly more difficult Mtrices my oly be multiplied if they re coformble, tht is if the umber of colums i the multiplyig mtri (the left-hd oe) is the sme s the umber of rows i the mtri beig multiplied (the right-hd oe) Mtri multiplictio is ot commuttive This mes tht for two mtrices, A d B, it is ot geerlly true tht AB = BA (it is vitl tht you remember this) If A = d the 7 B = AB = 7 = = Notice tht BA is ot possible s, this wy roud, the mtrices re ot coformble I is the idetity mtri The idetity mtri is All idetity mtrices hve s o the left to right dowwrds digol d s everywhere else Usig mtrices to represet trsformtios These re some of the stdrd trsformtio mtrices, d re worth rememberig Remember tht the first colum of mtri is where (, ) moves to d the secod colum is where (, ) moves to This c be useful i ems becuse usig this ide it is possible to derive the mtri of trsformtio described i words cosθ siθ Rottio through gle θ, ticlockwise bout the origi: siθ cosθ Elrgemet, scle fctor k, cetre the origi: k k Stretch, fctor horizotlly, fctor b verticlly: b A composite trsformtio is mde up of two or more stdrd trsformtios, for emple rottio through gle π, followed by reflectio i y= The mtri represetig composite trsformtio is obtied by multiplyig the compoet trsformtio mtrices together The order is importt The mtri for the composite of the trsformtio with mtri M, followed by the trsformtio with mtri N is NM The order is right to left Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

8 There mi ides re: the Further Mthemtics etwork wwwfmetworkorguk V 7 Usig the method of itervl bisectio, lier iterpoltio d the Newto Rphso Method to pproimte the solutio of equtio The step by step method for solvig differetil equtios Reducig Reltio to Lier Lw REVISION SHEET FP (AQA) NUMERICAL METHODS Before the em you should kow The formul ssocited with lier iterpoltio f( b) bf( ) c = f( b) f( ) The formul ssocited with the Newto Rphso f( r ) Method: = r + r f( r ) Tht i the step by step method for solvig dy f( ) d = : + h y = y + hf ( ) + =, + The Bisectio Method + b If root of equtio f() = lies betwee d b, the c = gives pproimtio to the root You c the determie whether the root lies betwee d c or betwee b d c d repet this ide, obtiig better d better itervl estimtes to the root Evetully you c give pproimtio of the root to the desired degree of ccurcy Lier Iterpoltio If root of equtio f() = lies betwee d b, the f( b) bf( ) c = gives pproimtio of the root If eeded you f( b) f( ) c determie whether the root lies betwee d c or betwee b d c d repet this process to obti sequece of pproimtios to the root Emple Show tht the equtio f( ) = 7= hs root, α, betwee d 3 By usig the rule of flse positio, strtig with these two vlues, fid estimte to this root f() f(b) f() c α b Solutio Sice f( ) = 5< d f( 3) = > the fuctio must cross the -is betwee = d = 3 d so there will be root of f() = there With = d b = 3, the lier iterpoltio gives first f( b) bf( ) pproimtio to the solutio of c = I this cse this is f( b) f( ) ( ) ( ) ( ) ( ) f(3) 3 f() = = = 6 f(3) f() ( 3) 5 Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

9 the Further Mthemtics etwork wwwfmetworkorguk V 7 Newto Rphso Method f( r ) The sequece of vlues geerted by r+ = r with f( ) pproprite estimte, usully coverges to root of f( ) = er to r Figure f() Emple Use the Newto Rphso method to fid the root of the equtio + 3= er = 5 With = 5, use the method three times (i other words clculte s fr s root d stte its ccurcy Solutio ) Hece, give pproimtio to the The itertive formul i the cse of f( ) = + 3 is s follows: f( ) r r + r 3 r+ = r = r 3 f( r) r Thus = = = 3 + ( 5 ) Similrly = =, 3 = 6, d = From this evidece it looks s though 6 my be estimte to the root which is correct to three deciml plces You c check this by verifyig tht the fuctio chges sig betwee 635 d 65 I fct y α f (635) = 39 < d f (65) = 3399 > Step by Step Method for solvig Differetil Equtios + h y = y + hf ( + =, + ) Emple dy A solutio of the differetil equtio d = + stisfies y = 5 whe = Approimte the vlue of y whe = usig step size h = Solutio y dy/d=f() I ech cse, this vlue is simply dy/d t the curret vlue of I ech cse, this is the previous y vlue plus (h the vlue of dy/d t the previous vlue), h the vlue of dy/d t the previous vlue is good pproimtio to how much y chges by whe chges by h Reducig Reltio to Lier Lw Suppose you hve some pirs of dt (, y) tht re supposed to be i the reltioship y = You wish to determie vlues of d which give best fit to your dt Tkig logrithms gives log( y) = log( ) + log( ) Therefore by plottig log y gist log you should see stright lie whose grdiet is d whose itercept with the y-is is t log() Thik y = m + c Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly

10 the Further Mthemtics etwork wwwfmetworkorguk V 7 REVISION SHEET FP (AQA) SERIES The mi ides re: Summig Series usig stdrd formule Fidig Polyomil Epressios Summig Series Before the em you should kow: The stdrd formul: 3 r, r, r Ad be ble to see tht series like ( 3) + ( ) + + ( + ) c be writte i sigm ottio s: rr ( + ) Usig stdrd formule It is importt to kow the stdrd results for summtios, mely: r = ( + r = ( + )( + ), ), 3 r = ( + ) Also required is fluecy i mipultig d simplify the stdrd formule sums Emple ( ) 3 r r + = r + r = ( + ) + ( + ) 6 = ( ) ( ) = ( )( ) It is lso useful to uderstd tht series such s ( ) + ( 3) + + ( + ) c be writte i sigm ottio For emple, the sequece just described c be writte s emple rr ( + ) d the solved s i the previous It is usul tht the th term of the sequece gives wht the sigm ottio should be, so if the th term is ot give i sequece it is importt to be ble to estblish wht it is Disclimer: Every effort hs goe ito esurig the ccurcy of this documet However, the FM Network c ccept o resposibility for its cotet mtchig ech specifictio ectly