A - LEVEL MATHEMATICS 2018/2019

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1 A - LEVEL MATHEMATICS 2018/2019 STRUCTURE OF THE COURSE Yur maths A-Level Maths curse cvers Pure Mathematics, Mechanics and Statistics. Yu will be eamined at the end f the tw-year curse. The assessment will cnsist f three tw-hur eams. WORKLOAD & ORGANISATION Yu will have a heavy wrklad. Class time will be spent giving ntes and eplanatins with questins and reinfrcement set as hmewrk. Fr each hur f taught time yu shuld epect t spend an hur n maths wrk utside f lessns. All wrk will be dne n paper and yu must keep a wellrganised file f ntes and eercises. Each week yu will be set a Cre and Applied hmewrk. In additin yu will be epected t cmplete an independent eercise f yur chice. HOW TO BE A SUCCESSFUL A LEVEL STUDENT The mst successful students at A-level Mathematics: Recgnise that A Level Mathematics is demanding and that they will have t wrk hard t understand the curse fully (even if up until nw they have always fund maths easy) Participate in lessns, answering questins and asking questins abut the ideas studied Keep a well rganised flder f ntes and use their ntes t help them d the wrk set Seek help utside lessns befre the deadline when they get stuck (despite having used their ntes and had a g) Submit wrk n time that is cmplete Revise thrughly with a fcus n aviding previus miscnceptins / mistakes.

2 A-LEVEL MATHEMATICS - PREPARATORY WORK This transitin wrk cvers the techniques that yu need t be 100% cnfident n in rder t be successful at A Level. This wrk cnsists f grade - 8 GCSE algebraic manipulatin. At GCSE such questins are ften wrth marks but at A-level they will be wrth 1 mark and will frm nly part f a mre substantial questin. The prblem slving questins are t shw yu the style f questins that are asked at A-level. These questins have been brken dwn int steps t make them mre accessible but they remain challenging and yu need t spend time wrking thrugh them. Yu need t sure that yur slutins are clear and wellstructured as utlined in the accmpanying eample. Read the instructins belw carefully. INSTRUCTIONS All wrk shuld be cmpleted n file paper nt n the questin sheets! Essential Algebra 1 Read thrugh the eamples n each sectin f the essential algebra befre attempting the questins each eample is designed t help yu avid mistakes. 2 Wrk thrugh the questins shwing yur methds clearly ie shwing each step when rearranging a frmula r slving an equatin. Prblem Slving 1 Read thrugh the infrmatin abut quality f written cmmunicatin carefully. 2 Wrk thrugh the prblems setting yur wrk ut clearly as demnstrated in the eample. We will nt accept wrk that is set ut prly r is nt f a gd standard if yu attempt t hand in wrk like this yu will be made t red the entire sectin f wrk NOTE Yur place n the A-level Mathematics curse is dependent n cmpleting the preparatry wrk t a high standard and being able t answer sme similar questins under test cnditins at the start f the curse. Yu shuld enjy spending the time develping yur algebra skills cntained in this wrk as this is eactly the srt f thing yu are cmmitting yurself t by taking the A level curse.

3 Essential Algebra Sectin A: Epanding Brackets The key t this sectin is knwing hw t add, subtract and multiply negative numbers. A1 2 y 2( y ) Multiply ut each bracket 2 2 = y 2y 8 Nte that yu are multiplying the secnd bracket by -2 nt 2 and that: = Cllect the 2 terms. N further simplificatin is pssible. = 11 2 y 2y A2 2 y 5 2y Multiply ut t achieve terms Nte that + = 2 2 = 10 y 15y 6y Cllect the term in y carefully (nting that = 11) 2 2 = 10 11y 6y N further simplificatin is pssible. A ( + y) 2 ( + y) 2 is nt 2 + y 2 nr is it 2 + 9y 2!! = ( + y )( + y ) Squaring means multiply by itself. = 2 + y + y + 9y 2 Multiply ut t achieve terms = 2 + 6y + 9y 2 Cllect the y terms t simplify. SECTION A QUESTIONS Epand and simplify 1) ( 2y) 2(y ) 2) (2 1)( + 2) ) ( 5)( + 5) 5) 5 ( 2) (hint BIDMAS) 6) 2( + 1) 2 (hint BIDMAS) ) (2 y)( 5y) ) ( 5y) 2

4 Sectin B: Factrising Epressins Factrising int 1 bracket (cmmn factrs) B1 12 y 18 2 y State the questin. Lk t factrise cmpletely. = 6 2 y ( ) The highest number that each term divides by is 6. The highest pwer f in each term is 2. The highest pwer f y in each term is y. These are the cmmn factrs t be pulled utside f the bracket. = 6 2 y (2 y 2 ) Nw divide each term by the cmmn factrs t get the cntents f the bracket. The terms in the bracket have n cmmn factrs s yu ve factrised cmpletely. Check by epanding that yu ve factrised crrectly. Factrising Quadratics (where the cefficient f 2 = 1) Standard term quadratic B It s a term quadratic and it nly has 2 s if it will factrise it will g int 2 brackets ( )( ). ( + )( ) ( + )( ) The 12 at the end means the numbers in the bracket must multiply t make -12. T get a negative by multiplying the signs must be different. Yu need t try different cmbinatins (-12 & 1, 12 & -1, -6 & 2, 6 & -2, - &, & -) and mentally multiply ut until yu get + 1 as yur middle term. There is nly ne pssible cmbinatin that wrks. Only 2 terms missing cnstant term B It s a 2 term quadratic with a missing cnstant term s if it will factrise int ne bracket by cmmn factr. ( + 5 ) The nly cmmn factr is. There is nly ne pssible cmbinatin that wrks. Only 2 terms missing term B This epressin will nt factrise. The nly quadratics f this srt that will are the difference f tw squares srt see net eample. B5 2 ( + )( ) ( + 2)( 2) 2 is a special kind f quadratic called the difference f tw squares (dts!) because it is ne square number subtract anther square number. Dts factrises int ( 1 st term + 2 nd term) ( 1 st term + 2 nd term) Check by epansin that ( + 2)( 2) gives ( ).

5 Factrising Quadratics (where the cefficient f 2 1) Lk fr cmmn factrs first B = 2( 2 + ) It is a standard term quadratic s will factrise int 2 brackets. There is a cmmn factr f 2. Pull this ut f each term t give an easier quadratic t factrise. = 2( )( ) = 2( )( ) = 2( )( 1) The + at the end shws the signs in the brackets must be the same. The is negative s the signs in the brackets must bth be. This means the nly pssible values are & -1. Check by epansin that ( )( 1) gives ( 2 + ). B = 9( 2 9) = 9( + )( ) State the questin. There is a cmmn factr f 9. Pull this ut f each term t give an easier quadratic t factrise. 2 9 is a difference f tw squares (dts!) because it is ne square number subtract anther square number. = 9( + )( ) Dts factrises int ( 1 st term + 2 nd term) ( 1 st term + 2 nd term) Check by epansin that ( + )( ) gives ( ). Standard term quadratic B = 0 State the questin. There are n cmmn factrs. It s a quadratic s if it will factrise it will g int 2 brackets but as it is 2 then it culd be either ( )( ) r (2 )(2 ). ( + )( ) = 0 (2 + 5)(2 1) = 0 The 5 at the end means the signs in the bracket are different and the end ns are either 5 & -1 r 5 & 1. Yu need t try different cmbinatins and mentally multiply ut until yu get + 8 as yur middle term. There is nly ne pssible cmbinatin that wrks. SECTION B QUESTIONS Factrise cmpletely 1) 6pq 15p 2 2) 2 6 ) 2 6 ) ) ) ) ) ) ) 1 2 2

6 Sectin C: Slving Equatins Slving Quadratics by Factrising C = 0 There are n cmmn factrs. It s a quadratic s if it will factrise it will g int 2 brackets but as it is 2 then it culd be either ( )( ) r (2 )(2 ). ( + )( ) = 0 ( + )( 5) = 0 The 15 at the end means the signs in the bracket are different and the end ns are either 15 & -1, -15 & 1, -5 & r 5 & -. Yu need t try different cmbinatins and mentally multiply ut until yu get -1 as yur middle term. There is nly ne pssible cmbinatin that wrks. Yu must nw slve the equatin (ie find ). Yu have tw things multiplied tgether that equal 0. This means that ne r bth f them must equal 0. Either ( + ) = 0 which means = s = / Or ( 5) = 0 = 5 Slving Linear Simultaneus Equatins C2 Slve simultaneusly 2 5y = 16 (i) 2y = 1 (ii) T eliminate ne variable yu need the same amunt f 6 15y = 8 (i) ne f the variables in each equatin. Multiply bth sides f the first equatin by and bth sides 6 y= 26 2 (ii) f the secnd equatin by 2 t give 6 in each equatin. Yu nw eliminate ne variable by either adding r 0 11y = 22 (i) 2(ii) subtracting the equatins. In this case yu d (i) 2(ii) Nte that 15y ( y) which is 15y + y = 9y y = = -2 Nw divide the equatin thrugh by -11 t find y: in (i): in (ii): Substitute this value f y back int the riginal equatins t 2 5(-2) = 16 2(-2) = 1 find. Bth equatins shuld give the same value! = 16 + = 1 2 = 6 = 9 The slutin t the simultaneus eqns is =, y = -2 = = SECTION C QUESTIONS Factrise and then slve these quadratic equatins. 1) = 0 2) t 2 5t + 6 = 0 ) p 2 p = 0 ) g = 0 5) 2 = 0 6) 6y 2 5y 6 = 0 ) = 0 8) = 0 Slve these simultaneus equatins 9) y = 19 (i) 2y = 1 (ii) 10) 2 + y = 2 (i) 5y = -19 (ii)

7 Sectin D Rearranging Frmulae (I) Frmulae are cnstructed using the rder f peratins, BIDMAS Brackets 1st I ndices (pwers) 2nd Divisin }rd Multiplicatin } Additin }th Subtractin } If yu rearrange a frmula yu are unpicking it s yu must fllw BIDMAS in reverse and d the ppsite peratin t bth sides. D1 Make p the subject f this frmula y p This is the same as y = (p + ) because all f the p + is in the numeratr f the fractin. Think abut hw it was cnstructed p ( ) (+) ( ) = y. T make p the subject yu must fllw these steps backwards and d the ppsite. Slutin: T und the yu bth sides T und the + yu bth sides T und the yu bth sides p y y p y p y y p... hence... p D2 Make p the subject f this frmula y p Nte this appears very similar t eample E1 but is nt the same because it was cnstructed in a different rder. This was cnstructed p ( ) ( ) (+) = y. T make p the subject yu must fllw these steps backwards and d the ppsite t bth sides. Slutin: T und the + yu bth sides T und the yu bth sides T und the yu bth sides p y p y ( y ) p ( y ) p p ( y ) r p y 21

8 D Make p the subject f this frmula y p This is eactly the same questin as eample E2. Hwever there is anther methd t rearrange it Yu can eliminate the denminatr f by multiplying thrugh by that mean multiplying the whle f each side by. In effect yu multiply each term by Slutin: Multiply every term by Simplify (nte: denminatr is nw gne) T und the yu bth sides p y p ( y ) ( ) ( ) y p 21 y 21 p... hence... p y 21 r... p ( y ) Bth methds (frm E2 and E) are equally gd. The methd f multiplying thrugh is very useful but yu must remember t multiply every term. D Make g the subject f this frmula m f 6g 5 The prblem with this frmula is that the subject that yu want, g,is negative. The easiest way t slve this questin withut making any errrs with minus signs is t add the whle f the term invlving g t bth sides. This nw gives a frmula with a psitive term in g. Nw fllw the previus methd - lk at hw the frmula was made g ( 6) ( 5) (+m) = f Reverse the rder f the steps and d the ppsite t bth sides f the frmula. 6g m 5 6g m 5 6g f 5 f 6g f m 5 6g 5( f m) 5( f m) g 6 6g 5 (Nte: the rder f cnstructin as g ( 5) ( 6) (+m) = f wuld als be crrect) SECTION D QUESTIONS Make the letter in brackets the subject f the frmula. 1. p = aq r (q) 2. p = a(q r) (q). 2 p y r (p). 2 p r y (p) qp 5. y r (p) 6. a 2 = b 2 + c 2 (b)

9 Sectin E Rearranging Frmulae (II) E1 Make g the subject f this frmula The prblem with this frmula is that the subject that yu want, g, is in the denminatr f a fractin. The way t slve this prblem is t multiply bth sides by g. Yu must put brackets arund the f + h t indicate that bth terms are multiplied by g.this nw gives a frmula with g n lnger in the denminatr. f h g Nw yu have g multiplied by a bracket equals s divide bth sides by the cntent f the bracket t get g = f h g g ( f h) g g g( f h) g ( f h) hence... g f h E2 Make the subject f this frmula y m c The prblem with this frmula is that the subject yu want,, appears twice in the frmula. The way t slve this prblem is t cllect all the terms With in n ne side and all the terms withut in them n the ther side. T d that add m t bth sides and t bth sides Yu can nw pull ut as a cmmn factr by factrising Nw yu have multiplied by a bracket equals y - s Divide bth sides by the cntent f the bracket t get = y m c y c m y c m y ( c m) ( y ) ( c m) hence... y c m SECTION E QUESTIONS Make the letter in brackets the subject f the frmula. 1. p y 2 r (q) q 2. a + b = c + d () t r. p 5 (u) u 5. p(q + r) = 2(q p) (q) 6. p(r q)= 2 + (q r) (r). qp y r (). t u (u) p 5 u

10 Sectin F Algebraic Fractins F1: Simplify: The first thing we need when adding r subtracting is a cmmn denminatr: 2( + 1) ( ) 6 6 Nw epress as a single fractin and then simplify the numeratr be careful with signs!! F2: Slve this equatin: = 2( + 1) ( ) = = = 1 First write as a single fractin. ( + 1) 2( 1) ( 1)( + 1) = 1 Then multiply bth sides by the denminatr t get rid f the fractin. ( + 1) 2( 1) = ( 1)( + 1) Epand the brackets and simplify smetimes yu may get a quadratic!! = 2 1 As this is a quadratic, rearrange s it is equal t zer. Factrise and slve: 2 6 = 0 ( )( + 2) = 0 = r 2 Sectin F Questins: Simplify: p p q (+1) Slve: = = = = =

11 Sectin G Rules f Indices a b = a+b E.g. 5 = 12 8 a b = a b E.g = ( a ) b = ab E.g. ( ) = 6 21 a = 1 E.g. a 5 = 1 = n n = E.g. 2 1 = 2 = = 2 m n n = ( ) E.g. 2 = ( 2) = = 81 0 = 1 G1 Write 1 as a single pwer f Remember 1 is the same as = 1 1 Nw, what is 1 as a pwer f? = 1 1 = 1 1 G2 Write as a single pwer f Write as a pwer f = 1 2 Nw hw can yu rewrite this s 1 2 is nt in the denminatr = 1 2 Sectin G Questins Simplify the fllwing: 1. (a ) 5 2a 2. 8( 5 ) a 2 b 5ab (a 2 b) (ab) 2 a 2 bc a 2 bc 2a 2 b c Evaluate: ( 2 )2 8. ( 2 ) 2 9. (( 2 5 ) 1 ) 10. ( ) 6 2 Write the fllwing as a single pwer f

12 Sectin H: The quadratic frmula Yu will be required t knw the frmula used t slve quadratic equatins that cannt factrise. If a 2 + b + c = 0 then = b± b2 ac 2a H1 Slve = 0, crrect t tw decimal places. Take the quadratic frmula: = b± b2 ac 2a Put a = 5, b = - 11 and c = -: = ( 11)± ( 11) Simplify as far as pssible = 11± Type int yur calculatr and rund: = 2.52 r 0.2 = 11± Sectin H Questins: Slve the fllwing equatins using the quadratic frmula. Give yur answers t tw decimal places = = = = = = 0

13 Sectin I: Pythagras and Trignmetry Pythagras Therem: a 2 + b 2 = c 2 a c b Trignmetry: sin = pp hyp cs = adj hyp tan = pp adj pp adj pp hyp A b c C a Csine Rule: Yu can use the csine rule t find the length f a side when tw sides and the included angle are given. B a 2 = b 2 + c 2 2bc cs A Alternatively, yu can use csine rule t find an unknwn angle if the lengths f all three sides are given. cs A = b2 +c 2 a 2 2bc Sine Rule: Yu can use sine rule t find the length f a side when its ppsite angle and anther ppsite side and angle are given. a = b = c sin A sin B sin C Alternatively, yu can use the sine rule t find an unknwn angle if the ppsite side and anther ppsite side and angle are given. sin A a = sin B b = sin C c Area f a Triangle: The area f a triangle is 1 ab sin C 2 Sectin I Questins 1. Find the missing side n the fllwing triangles: cm a 12.cm 6cm 5.2cm

14 2. Find the value f the missing side r angle crrect t significant figures. a b c d e f. Wrk ut the unknwn side r angle labelled with a letter in each triangle. Give yur answers crrect t significant figures. a b c. d. Wrk ut the area if the triangle.

15 Prblem Slving This set f prblems is designed t shw yu the kind f wrk that A level starts with. If yu enjy ding these questins, even when yu find them difficult, then yu are likely t enjy the A level curse. In ding this wrk yu shuld cncentrate as much n cmmunicating yur answers clearly as yu d n slving the prblems. Lk critically at the answers belw which shw tw appraches t slving the same prblem. Which d yu think is better and why? First Answer: 6 πr = 8πr 8πr πr 2 = 8r 2πr 2 + 2πr(8r) = 18πr 2 Secnd answer: Vl f a sphere = πr Vl f 6 spheres = 6 πr = 8πr Vl f 6 spheres = vl f cylinder with same radius: 8πr = πr 2 h Finding h in terms f r: h = 8πr πr 2 = 8r Surface Area f cylinder = 2πr 2 + 2πrh = 2πr 2 + 2πr(8r) IMPORTANT! = 2πr πr 2 = 18πr 2 Quality f written cmmunicatin [QWC] isn t just abut setting yur wrk ut well. It enables yu t structure yur thinking and helps yu t wrk thrugh cmplicated prblems as it makes yu eplain yur methd. It als makes it easy fr ther peple t fllw yur wrk which makes it mre likely that they will be able t help yu if yu get stuck! Features f quality f written cmmunicatin: (i) Yu can understand what the questin was frm lking nly at the slutin (ii) Wrking is clearly labelled eg: Vl f sphere = (iii) Frmulae are stated befre they are used (iv) The algebraic steps flw ie: They wrk dwn the page with each step equivalent t the ne befre and there is n misuse f the equals sign eg: GOOD: A = ½ bh BAD: A = ½ bh = ½ A = = 12 2 = 6 [cm 2 ] = 6 [cm 2 ] In the tw answers at the start f this sectin the secnd answer has all the features f QWC and yu shuld have been able t deduce that the riginal questin was: 6 spheres f radius, r, are melted dwn and refrmed t make a cylinder als with radius r. Find the surface area f the cylinder in terms f r.

16 Prblem Slving Questins Yu need t attempt each questin. These questins are designed t develp yur prblem slving skills and develp yur resilience when tackling challenging prblems. Sme tpics cvered in the questins are new t yu but yu have been taught the skills required t slve the prblems at GCSE. When we mark this transitin wrk we are mre interested in hw yu apprach and tackle the prblems than yu getting the crrect answer. One mark will be awarded in each questin fr the quality f yur written cmmunicatin. (1) An pen tpped tray is made frm a rectangular piece f paper 16cm by 10cm by cutting ut a square f side length frm each crner and flding a. Find an epressin fr the vlume f the b giving yur answer in the frm a +b 2 + c where a, b and c are integers. [] b. Shw that the eternal surface area f the b is [] c. Given that the eternal surface area is 96cm 2, find the dimensins f the b. [] QWC [1] (2) [Yu may find it useful t draw a sketch in this questin hwever yu will nt scre any marks by slving the questin using an eact graphical methd] Pints A and B have crdinates (-1, 5) and (, ) a. Shw that the line that passes thrugh A and B has a gradient f ½. [1] b. M is the midpint f AB. Find the crdinates f M. [1] c. Line L is the perpendicular bisectr f AB. This means that line L is perpendicular t the line thrugh A and B and it als passes thrugh the midpint f AB. Using the fact that the gradients f perpendicular lines multiply t make -1, find the gradient f line L. [1] d. Using yur gradient frm(c) and the pint M frm (b), find the equatin f line L. [] QWC [1] () A cylinder has radius r and height h. a. Write dwn in terms f r and h the frmula fr the vlume f the cylinder. [1] b. Shw that the surface area f the cylinder is given by 2πr(r + h). [2] c. Given that the surface area f the cylinder is fur times its vlume find an epressin fr the height f the cylinder in terms f the radius r. [] QWC [1] r h

17 () Shw that the equatin / = 1 can be rearranged in the frm a 2 + b + c = 0 where a, b and c are integers. Hence slve / = 1 giving yur answers as fractins. [] QWC [1] (5) The equatin f a circle with radius 5 units and its centre at the rigin is 2 + y 2 = 25. a. On a single diagram draw 2 + y 2 = 25 and the line y = 5. [2] b. By substituting y = 5 int 2 + y 2 = 25 shw that yu get the equatin = 0 [2] c. Factrise cmpletely [1] d. Hence slve = 0 [1] e. Hw d the slutins t (d) relate t the diagram frm (a)? [1] QWC [1] MECHANICS (6) In frmulae relating t mtin: u = initial velcity, v = final velcity, a = acceleratin, t = time and s = displacement. Tw frmulae relating t mtin f bjects are: s = ut + ½ at 2 and v = u + at An bject is travelling alng a straight rad at a velcity f m/s when it starts t accelerate at m/s 2. It accelerates until it has travelled 65 m. a. Use s = ut + ½ at 2 t wrk ut hw lng the bject accelerated fr. [] b. Use v = u + at t wrk ut the velcity f the bject nce it had stpped accelerating. [2] QWC [1] () The diagram shws tw bjects cnnected by a light inetensible string ver a smth pulley. Initially the weights are held at rest. The frce called tensin in the string is marked T. The frce called weight f each bject is its mass [in kg] multiplied by gravity [called g]. When the bjects are allwed t mve they will accelerate in the directins shwn at a m/s 2 a T 5kg kg T a The frce acts in the directin f mtin. As the 5kg bject will mve dwnwards the frce that acts n it is 5g T. 5g g Newtn s secnd law states that Frce = Mass Acceleratin. Fr the 5kg particle this gives the equatin: 5g T = 5a. a. Given that the kg bject mves upward, use Newtn s secnd law t give an equatin fr the kg particle. [2] b. Slve the tw equatins simultaneusly t shw that a = g/. [2] c. Find T in terms f g, giving yur answer as at fractin. [2] QWC [1]

18 STATISTICS (8) T find the mean f list f data yu add all f the data values in the list and then divide by hw many data values there are. In statistics the frmula fr the mean is = n. is the mean, Σ represents the sum f, is a data value frm the list and n is the number f values in the list. The standard deviatin is an alternative measure f spread t the range r interquartile range. The smaller the standard deviatin, the mre cnsistent the data is. The standard deviatin f a set f data is calculated as s = ( )2 n 1 a. Daniel recrds his scres in the weekly mental test fr 6 weeks: 8, 5, 9, 11, 12, 15. Find Daniel s mean scre. [1] b. Shw that the standard deviatin in Daniel s scres is.6 given t significant figures. [] c. Paul als recrds his si test scres. Paul s mean scre is 9.8 and his standard deviatin is 1.5. Write tw cmments cmparing the test perfrmances f the tw bys. [2] QWC [1] (9) Bag A and Bag B cntain nly Blue and Yellw cunters Bag A has Blue cunters and 1 Yellw cunter. Bag B als has five cunters. A cunter is taken at randm frm Bag A and put int Bag B. Then a cunter is taken frm bag B and put int Bag A. Bag A nw cntains nly Blue cunters. The prbability f this happening is 1 / 10. a. Hw many Yellw cunters are in Bag B at the start? [] b. Befre the cunters were swapped if ne cunter was taken frm each bag what wuld have been the prbability f getting ne f each clur? [2] QWC [1] TOTAL = 6 MARKS