x 2 + n(n 1)(n 2) x 3 +

Transcription

1 Core 4 Module Revision Sheet The C4 exm is hour 30 minutes long nd is in two sections Section A 36 mrks 5 7 short questions worth t most 8 mrks ech Section B 36 mrks questions worth bout 8 mrks ech You re llowed grphics clcultor Before you go into the exm mke sureyou re fully wre of the contents of theformul booklet you receive Also be sure not to pnic; it is not uncommon to get stuck on question I ve been there! Just continue with wht you cn do nd return t the end to the questions you hve found hrd If you hve time check ll your work, especilly the first question you ttemptedlwys n re prone to error J MS I cnnot stress enough tht with A modules such s C3 nd C4 success is entirely through prctice The topics nd questions re hrder nd longer nd only diligent prctice will let you gin top grdes This revision sheet will therefore be of only limited use You hve been wrned!!! Algebr Review binomil expnsion from C for xy n for integer n Notice tht it is vlid for ny x nd y nd tht the expnsion hs n terms The generl binomil expnsion is given by x n = nx nn! x nn n x 3 3! nd is vlid for ny n frctionl or negtive but < x < ie x < Notice lso it must strt with in the brckets For exmple expnd 4 x / 4 x / = 4 x / 4 = x [ 4 = = [ / x 3 4! x8 3x 8 5x3 307 It is only vlid for x/4 < x < 4 ] x 3 5 x 3 ] 4 3! 4 Prtil frctions is effectively the reverse of combining together two lgebric frctions For exmple x x Algebric Frctions Prtil Frctions x3 xx To review lgebric frctions quickly; you must be ble to mnipulte lgebric frctions fluently The rules for frctions re: wwwmthshelpercouk JMStone

2 Adding/Subtrcting Multiplying b ± c d = d±bc, bd b c d = c bd, So for exmple simplify the following: Dividing b c d = b d c = d bc = = The generl principle for prtil frctions is tht we need to split n lgebric frction into two or more frctions So we crete n identity tht must be true for ll x nd then find certin missing constnts For exmple we expnd x x 4x to A x 4 B x x x 4x A x 4 B x We need to discover A nd B Multiplying through by x 4x nd cncelling we find x AxBx 4 We cn either multiply out nd equte coefficients or choose vlues of x to work out the constnts A nd B I prefer the ltter Here we would let x = to discover B = /5 nd let x = 4 to discover A = 3/5 Therefore The overll methods you need to know re: Trigonometry By definition nd x x 4x 3 5x 4 5x pxq xbcxd A xb B cxd, px qxr xbcx d A xb BxC cx d, px qxr xbcxd A xb B cxd C cxd secx cosx, cosecx sinx, cotx tnx I remember these by the third letter underlined You must be ble to produce grphs of these in both rdins nd degrees P84 By dividing sin x cos x by sin x nd cos x we cn derive cot x cosec x nd tn x sec x respectively These llow us to solve whole fmily of equtions tht we couldn t solve before With ny eqution with combintion of trigonometric functions nd if one of them is squred we should be using these reltionships to help us It is importnt to note tht is you must be creful when you see things like tnxsinx = tnx It is so tempting to divide by tnx to yield sinx = But you must bring everything to one side nd fctorise; tnxsinx = 0 The full solutions cn then found by solving sinx = 0 nd tnx = 0 [It is completely nlogous to x = x If we divide by x we find x =, but we know this hs missed the solution x = 0 However when we fctorise we find xx = 0 nd both solutions re found] Here we would find x AxABx 4B ABxA 4B Therefore AB = nd A 4B = These solve to A = 3/5 nd B = /5 wwwmthshelpercouk JMStone

3 3 Prmetric Equtions Curves re usully given in the form y = fx However we hve seen in C tht this is not lwys the best form; for exmple circle of centre,b nd rdius r is usully given in the form x y b = r A curve cn lso be expressed by prmeter usully t or θ In this system y = ft nd x = ft Therefore insted of y being linked directly with x, they re linked indirectly through t Usully t cn vry over ll the rel numbers, but it cn lso be restricted eg t 0 or 0 θ π Grphs of more exotic reltionships cn be obtined in prmetrics see P5 To grph the prmetric you just vry the prmeter nd plot the resulting x, y points You must be ble to convert ny resonble prmetric system to norml xy reltionship by eliminting the prmeter If it contins merely polynomil reltionship then it should be esy enough For exmple eliminte t from y = t, x = t From the x reltionship t = x, so y = x = x 4x4 In hrder exmples you must use lgebric trickery; for exmple y = t t, x = t t From the second we discover so by substitution x = t t y = xxt = t t = x x, x x x x = x x x x x x = x x If the reltionship involves sin s nd cos s then you will need to be more cunning Ber in mind tht most of these will be circles or ellipses so we will probbly be deling with x s nd y s Alwys think bout the wys we know to get rid of trig functions like sin θ cos θ = nd the other trig identities from the previous section For exmple eliminte t from y = 3sinθ, x = 43cosθ We find y = 9sin θ nd x 4 = 9cos θ Adding nd using tht 9sin θ 9cos θ = 9 we find x 4 y = 9 So it is circle This is specific exmple of generl type you must lern: x = rcosθ y = brsinθ x x b = r The similr reltionship for the circle centre 0,0 is trivil from the bove if we let = b = 0 Clculus is possible with prmetric reltionship We know from the chin rule tht A simple extension of this shows tht dx = dt dt dx dx = /dt dx/dt For exmple find /dx for x = e t, y = 3e 4t ; dx = /dt dx/dt = e4t e t = 6et e t = 6e t wwwmthshelpercouk 3 JMStone

4 The questions sked on this cn be rther lgebric! Insted of giving you specific point on the curve they get you to consider generl point where the prmeter is some vlue sy p They then get you to consider the grdient of the tngent or norml t tht point You my then need to substitute this into y y = mx x There re two detiled exmples on P38/9 Plese try to do these by yourself; they re not esy Obviously turning points re found by putting /dx = 0 when ren t they?! Once the vlue of the prmeter is found, put it bck into the originl reltionships to find the crtesin coordintes To discover their nture you must go bck to bsics nd consider points either side; don t try to do double differentition 4 Further Integrtion b πy dx is the volume of revolution of the curve y rotted bout the x-xis between x = nd x = b All tht is needed for you to do is clculte y in terms of x from y For volumes of revolution round the the y-xis switch the x nd the y nd use q p πx between y = p nd y = q You must be on the lookout for integrls of the form to split the terms nd then integrte x x dx = x x dx = x dx x x dx Use prtil frctions dx = ln x You must be very fluent with using prtil frctions to simplify terms x c x The re under ny curve cn be pproximted by the Trpezium Rule The governing formul is given by nd contined in the formul booklet you will hve in the exm b ydx h[y 0 y n y y y n ], where h is the width of ech trpezium, y 0 nd y n re the end heights nd y y y n re the internl heights 5 Vectors Two Dimensionl Vectors The vector 3 4 cn be written 3i4j nd represents vector going 3 right nd 4 up By Pythgors Theorem it cn be shown tht the mgnitude of this vector is 3 4 = 5 nd by trigonometry the direction is tn 4 3 You must be fluent in converting between nd r,θ x y Two vectors re equl if their mgnitudes nd directions re the sme Two vectors re prllel if one is sclr multiple of the other For exmple show tht 3 is prllel to 3 45 ; so show tht 5 3 = 3 45 When multiplying vector by positive sclr it chnges the length of the vector but not the direction If the sclr is negtive then it lso reverses the direction of the vector For exmple 3 3 = 6 9 When dding vectors, you just dd the x components nd dd the y components For exmple 3 5 = 4 It cn be done, but it s fiddly! wwwmthshelpercouk 4 JMStone

5 You must be good t the type of questions similr to those from GCSE on pge 86 Know the geometric interprettion of b nd b Also know tht in generl if you hve position vectors OA = nd OB = b then AB = b A unit vector cn be constructed by dividing vector by its mgnitude For exmple the unit vector from 3 is 3 3 = 3 3 A line cn be written in vector form If you know line goes through point,b nd hs the grdient m then its vector form is r = b λ m where λ is sclr tht tkes different vlues on different points on the line The vector m cn be re-written to mke the components nicer For exmple /3 3 These vectors re not equl, but they hve the sme direction The most generl form is r = λu where is the point it psses through nd u is the direction We cn therefore show tht the eqution of the line through nd b is given by r = λb = λλb see pge 94 These two ides cn be combined to convert to y = mxc form nd vice-vers To find the ngle between two vectors we use the sclr product result b = b cosθ where u represents the mgnitude of vector u From this we cn see tht two vectors re perpendiculr if their sclr product is zero The sclr product is most esily clculted s follows; x y bx by = x b x y b y It is just number, not vector! Three Dimensionl Vectors The following tble sums up the 3D equivlents of the D results we hve lre found: = D 3D i,j i,j,k x y = x y z b = x b x y b y b = x b x y b y z b z Most of the results from the D section bove still hold true for 3D vectors 6 Differentil Equtions If you re told tht y is proportionl to x then we write y x This implies tht y = kx for some constnt k k cn then be determined by putting in one pir of vlues x,y into the eqution The nottion /dx lets us believe the it is norml frction Although this is not the cse we cn mnipulte it like frction in differentil eqution You must move the vribles to different sides of the eqution nd integrte seprtion of vribles Only dd the ever-present c to one side For exmple solve dx = y cosx y = cosxdx y = sinxc wwwmthshelpercouk 5 JMStone