The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values of the variable it cotais The relatioships betwee roots ad coefficiets i polyomial equatios Fidig a polyomial equatio with roots related to that of a give oe If A ad B are costats ad A(x ) + B(x + ) = 4x + is a idetity, the by substitutig x =, B = 6 ad by substitutig x =, A = So B = ad A = The method of substitutio for fidig a polyomial equatio with roots related to a give oe Eg, if x + x 7x+ 4= 0 has roots α, βγthe, ( y ) + ( y ) 7( y ) + 4 = 0 will have roots α +, β +, γ + Idetities I mathematics, a idetity is a statemet which is true for all values of the variables it cotais For example ( x + ) x + 6 is a idetity because o matter what value of x you substitute i, the left had side is always the same as the right had side ( x + ) = is ot a idetity, because the left had side oly equals the right had side whe x = If you kow that a statemet is a idetity the you ca substitute ay values for the variables i it ad kow that the resultat expressio is true The followig questio is typical Example You are give that ad C A( x )( x ) + Bx( x ) + Cx( x ) x + x+ is a idetity Fid the values of A, B Sice the statemet is a idetity, it is true for ay value of x Substitutig x = gives: Substitutig x = gives: Substitutig x = 0 gives: A( ) ( ) + B ( ) + C ( ) + + C = C = A( ) ( ) + B ( ) + C ( ) + + 7 B = 7 B = A(0 ) (0 ) + B 0 ( ) + C 0 ( ) 0 + 0 + A= A=
Roots ad coefficiets i polyomial equatios Quadratic: If ax + bx + c = 0 has roots α ad β the b c α + β = ad αβ = a a Cubic: If ax + bx + cx + d = 0 has roots α, β ad γ the α b c + β + γ =, ad a αβ + βγ + αγ = a αβγ = d a Example The cubic equatio x + 4x + x+ = 0 has roots α, β ad γ i) Write dow the values of α + β + γ, αβ + βγ + γα ad αβγ ii) Fid a cubic equatio with iteger coefficiets with roots α,β,γ + + = ad i) α + β + γ =, αβ βγ γα αβγ = ii) ( α ) + ( β ) + ( γ ) = ( α + β + γ) = 4 = 7 ( )( ) ( )( ) ( )( ) ( ) ( α )( β )( γ ) αβγ αβ βγ αγ α β γ α β + β γ + α γ = 4 αβ + βγ + αγ 4( α + β + γ ) + = 6+ 8+ = 7 = 8 4( + + ) + ( + + ) = 4 6 4 = 5 Therefore a cubic with iteger coefficiet with roots α,β,γ is x + 7x + 7x+ 5= 0 Example (Substitutio Method) The cubic equatio x + 4x + x+ =0 has roots α, β ad γ Fid a cubic equatio with iteger coefficiets with roots α +,β +,γ + Let w= z+ so that w z = Sice α, β ad γ are the roots of x + 4x + x+ = 0, α +,β +,γ + are the roots of w w w + 4 + + = 0 Multiplyig this out ad multiplyig both sides by 4 gives ( w ) + 4( w ) + 6( w ) + 4= 0 w w + w + 4w 8w+ 4+ 6w 6+ 4= 0 w + w + w+ = 0
REVISION SHEET FP (MEI) COMPLEX NUMBERS The mai ideas are: Maipulatig complex umbers Complex cojugates ad roots of equatios The Argad diagram Polar Form Maipulatig Complex Numbers Before the exam you should kow: How to multiply two complex umbers quickly ad i oe step as this will save a lot of time i the exam How to geometrically iterpret z z as the distace betwee the complex umbers z ad z i the Argad diagram The fact that z+ z = z ( z) which equals the distace betwee z ad z i the Argad diagram The exact values of the sie ad cosie agles which π π are multiplies of ad, eg 6 4 π cos = 4 Multiplyig, dividig, addig ad subtractig Multiplyig, addig ad subtractig are all fairly straightforward Dividig is slightly more complicated Wheever you see a complex umber o the deomiator of a fractio you ca get rid of it by multiplyig both top ad bottom of the fractio by its complex cojugate + j j j 5j eg + + + = = j j j + Complex Cojugates ad Roots of Equatios The complex cojugate of z = a+ bj is z = a bj Remember zz is a real umber ad it equals the square of the modulus of z Complex roots of polyomial equatios with real coefficiets occur i cojugate pairs This meas that if you are told oe complex root of a polyomial equatio with real coefficiets you are i fact beig told two roots This is key to aswerig some typical exam questios A example of a algebraic trick that it is very useful to kow is: ( z (+ j))( z ( j)) = (( z ) j)(( z ) + j) = ( z ) 4j = z 6z+ each specificatio exactly
The Argad Diagram Imagiary axis + j (, ) (, ) Real axis 4j j ( 0, 4) I the Argad diagram the poit (x, y) correspods to the complex umber x + yj You should be aware that the set of complex umbers z with for example z 5+ j = 6is a circle of radius 6 cetred at 5 j (or (5, )) i the Argad plae (Note: Poits o the diagram above do ot correspod to this example) The argumet of a complex umber z, deoted arg( z ) is the agle it makes with the positive real axis i the Argad diagram, measured aticlockwise ad such that π < arg( z) π Whe aswerig exam questios about poits i the Argad diagram be prepared to used geometrical argumets based aroud equilateral triagles, similar triagles, isosceles triagles ad parallel lies to calculate legths ad agles Polar Form Other sets of poits i the complex plae Where a ad b are complex umbers, the set of complex umbers z such that arg(z a) = θ, is a half lie startig from a i the directio θ arg( z a) = arg( z b), is the lie through a ad b with the sectio betwee a ad b (iclusive) removed arg( z a) = arg( z b) + π, is the lie from a to b (ot icludig a ad b themselves) If z = x+ yj has z = r ad arg( z) = θ the z = r(cosθ + jsi θ) This is called the polar or modulus-argumet form Example Write z = j i polar form π π π z = + ( ) = ad arg( z ) = Therefore i polar form z is z = cos jsi 4 + 4 4 π π Example If z = 6 cos + jsi 7 7, what are z ad arg( z )? π Sice z is give i polar form we ca just read off that z =6 ad arg( z) = 7 each specificatio exactly
The mai ideas are: Sketchig Graphs of Ratioal Fuctios Solvig Iequalities Graph Sketchig Ratioal fuctios To sketch the graph of y REVISION SHEET FP (MEI) GRAPHS AND INEQUALITIES N( x) D( x) = : Fid the itercepts that is where y the graph cuts the axes Fid ay asymptotes the vertical asymptotes occur at values of x which make the deomiator zero Examie the behaviour of the graph ear to the vertical asymptotes; a good way to do this is to fid out what the value of y is for values of x very close to the vertical asymptote Before the exam you should kow: There are three mai cases of horizotal asymptotes Oe is a curve which is a liear polyomial divided by a liear 4x + polyomial, for example y = This has a horizotal x coefficiet of x o the top asymptote at y= Here this would coefficiet of x o the bottom 4 be y = The secod is a curve give by a quadratic polyomial divided by 5x + x+ 5 a quadratic polyomial, for example, y = x x + as x ±, This has a horizotal asymptote at coefficiet of x o the top 5 = Here this would be y = coefficiet of x o the bottom Thirdly, whe the curve is give by a liear polyomial divided by a quadratic polyomial, it will geerally have the x-axis (y = 0) as a horizotal asymptote Whe you solve a iequality, try substitutig a few of the values for which you are claimig it is true back ito the origial iequality as a check Examie the behaviour aroud ay o-vertical asymptotes, ie as x teds to ± x Example Sketch the curve y = 4 x (Sketch) x The curve ca be writte as y = ( x+ )( x ) If x = 0 the y = -05 So the y itercept is (0, -05) Settig y = 0 gives, x = - ad x = So the x itercepts are (-, 0) ad (, 0) The deomiator is zero whe x = ad whe x = so these are the vertical asymptotes x ( x + 4) + Also y = = = + 4 x 4 x 4 x so y = - is a horizotal asymptote
Solvig Iequalities Broadly speakig iequalities ca be solved i oe of three ways, or sometimes i a combiatio of more tha oe of these ways Method Draw a sketch of the iequality For example, if you are asked to solve a iequality of the form g( x) f ( x) the sketch both f ad g, ad idetify poits where the graph of f is lower tha the graph of g These poits will lie betwee poits x for which g(x ) = f(x ) ad so these usually eed to be calculated Method Use algebra to fid a equivalet iequality which is easier to solve Whe dealig with iequalities remember there are certai rules which eed to be obeyed whe performig algebraic maipulatios The mai 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 example, do t multiply by (x ) because that s positive whe x > ad egative whe x < O the other had (x ) is always positive so you ca safely multiply by this (with o eed to reverse the iequality sig) Method Sometimes is easier to rearrage a iequality of the from g( x) f ( x) to g( x) f ( x) 0 (you do t have to worry about reversig the iequality for such a rearragemet) Idetify poits where g(x) f(x) = 0 or where g(x) f(x) has a vertical asymptote Fially test whether the iequality is true i the various regios betwee these poits Example Solve the iequality x x + x (usig Method ) x + x + x x 0 x x (x )( x ) ( x+ ) 0 x x 6x 0 x xx ( ) 0 x Lookig at the expressio, x = 0 if x = 0, x = 0 if x = ad x = 0 if x = This meas that the truth of the iequality should be tested i each of the followig regios x < 0 0 0< x < < x < x > It ca be see that the iequality is TRUE if x < 0, false if 0 < x <, TRUE if < x < ad FALSE if x > The solutio is therefore x 0, < x Ca you see why x = 0 ad x = are icluded as values for which the iequality is true, but x = is ot?
The mai ideas are: Maipulatig matrices Addig ad subtractig matrices are straightforward Maipulatig matrices Usig matrices to represet trasformatios Matrices ad simultaeous equatios Ivariat Poits REVISION SHEET FP (MEI) MATRICES Before the exam you should kow: How to add, subtract ad multiply matrices How to calculate the determiat of a matrix How to fid the iverse of a matrix That matrix multiplicatio is associative, so ABC ( ) = ( ABC ) but ot commutative, so AB BA The stadard matrices for rotatio, reflectio ad elargemet ad uderstad how matrices ca be combied to represet composite trasformatios A liear trasformatio maps a straight lie oto aother straight lie The origi is ivariat uder a liear trasformatio The determiat of a trasformatio matrix gives the area factor of the trasformatio How matrices ca be used to represet simultaeous equatios The ivariat poits of a trasformatio are ot moved by the trasformatio Ad are able to idetify the ivariat poits of trasformatios represeted by matrices Multiplyig matrices is slightly more difficult Matrices may oly be multiplied if they are coformable, that is if the umber of colums i the multiplyig matrix (the left-had oe) is the same as the umber of rows i the matrix beig multiplied (the right-had oe) Matrix multiplicatio is ot commutative This meas that for two matrices, A ad B, it is ot geerally true that AB = BA (it is vital that you remember this) 4 4 If A = ad B = the 7 8 + + 4 4+ + 4 + + 4 AB = = + 7 + 8 4+ 7 + 8 + 7 + 8 6 45 8 Notice that BA is ot possible as, this way roud, the matrices are ot coformable a c The determiat of a matrix is ad bc If the matrix is represetig a dimesioal b d trasformatio, the determiat gives the area factor of the trasformatio The iverse of matrix A is deoted A It has the property that AA = A A = I, where I is the idetity matrix Oly square matrices have iverses Idetity matrices are always square The idetity matrix is 0 All idetity matrices have s o the left to right dowwards diagoal ad 0 s everywhere else The 0 iverse of a matrix is give by swappig the etries o the left to right dowwards diagoal, chagig the sig of the etries o the other diagoal ad dividig the whole thig by the determiat of the origial matrix a c - d c so, if A =, A = b d ad bc b a (ote, you ca oly do this whe the determiat is t zero!) each specificatio exactly
Usig matrices to represet trasformatios These are some of the stadard trasformatio matrices, ad are worth rememberig Remember that the first colum of a matrix is where (, 0) moves to ad the secod colum is where (0, ) moves to This ca be useful i exams because usig this idea it is possible to derive the matrix of a trasformatio described i words cosθ siθ Rotatio through agle θ, aticlockwise about the origi: siθ cosθ k 0 Elargemet, scale factor k, cetre the origi: 0 k a 0 Stretch, factor a horizotally, factor b vertically: 0 b A composite trasformatio is made up of two or more stadard trasformatios, for example a rotatio through agle π, followed by a reflectio i y= x The matrix represetig a composite trasformatio is obtaied by multiplyig the compoet trasformatio matrices together The order is importat The matrix for the composite of the trasformatio with matrix M, followed by the trasformatio with matrix N is NM The order is right to left Matrices ad simultaeous equatios Matrices ca be used to represet systems of simultaeous equatios Matrix techiques ca be used to solve them x+ y= 7 x 7 Eg is equivalet to = Usig the iverse matrix we have x+ 7y= 4 7 y 4 x 7 7 = = x = ad y = This ca be exteded to systems of equatios i ukows y 4 Ivariat Poits The ivariat poits of a trasformatio are ot affected by the trasformatio (0, 0) is a ivariat poit of ay matrix trasformatio, but there could be others too: Example Fid the ivariat poits of the matrix 7 y = x x x x+ y = x x+ y = 0 = So ay poit o 7 y y x+ 7y = y x+ 6y = 0 y = x Matrices with zero determiat y = x is a ivariat poit 8 Example Show that the matrix has determiat zero ad that it maps all poits oto the lie 4 y = x The determiat of the matrix is ( 4) ( 8) = 0 Takig a geeral poit (x, y) 8 x x+ 8y = 4 y x+ 4y Sice ( x + 4 y) = ( x+ 8y ) the poit (x + 8y, x + 4y) is o y = x each specificatio exactly
the Further Mathematics etwork wwwfmetworkorguk V 07 REVISION SHEET FP (MEI) SERIES AND INDUCTION The mai ideas are: Summig Series usig stadard formulae Telescopig (or method of differeces) Proof by Iductio Summig Series Before the exam you should kow: The stadard formula: r, r, r Ad be able to spot that a series like ( ) + ( ) + + ( + ) ca be writte i sigma otatio as: rr ( + ) How to do proof by iductio Usig stadard formulae Fluecy is required i maipulatig ad simplify stadard formulae sums like: ( ) r r + = r + r ( ) ( ) + + = + 4 = ( ) ( ) 4 + + + = ( )( ) 4 + + + The Method of Differeces (Telescopig) r + 4 Sice = + rr ( + )( r+ ) r r+ r+ true first) it is possible to demostrate that: r + 4 = + + + + + + rr ( + )( r+ ) 4 4 5 (frequetly i exam questios you are told to show that this is + + + + + + + + + I this kid of expressio may terms cacel with each other For example, the (+) i the first bracket cacels with the ( ) i the secod bracket ad the (+) i the third bracket (subsequet fractios that are cacellig are doig so with terms i the part of the sum)
the Further Mathematics etwork wwwfmetworkorguk V 07 This leaves r + 4 = + rr ( + )( r+ ) + + Proof by Iductio Usig proof by iductio to prove a formula for the summatio of a series, Eg, Prove that (r ) = Other miscellaeous questios These are usually very easy, i fact easier tha the questios which fall ito the categories above, so log as you do t paic, keep a clear head ad apply what you kow 5 8 + 4 8 Eg, show that if M = the M = for all atural umbers 4 Example Prove by iductio that, for all positive itegers, r+ = (+ 5) Whe = the left had side equals ( ) 4 statemet is true whe = + = The right had side is (( ) + 5) = 4 So the k Assume the statemet is true whe = k I other words r+ = k(k + 5) k + We must show the statemet would the be true whe = k +, ie that r+ = ( k+ )(k+ 8) Now, k+ ( ) ( ) r+ = r+ + (( k+ ) + ) k = k ( k+ 5) + ( k+ 4) k 5 k 6 k 8 = + + + k k 8 = + + = ( k+ )( k+ 8) So the statemet is true whe = ad if it s true whe = k, the it s also true whe = k + Hece, by iductio the statemet is true for all positive itegers,