Edexcel Higher Mathematics Revision Guide

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Edexcel Higher MathematicsRevision GuideFull worked solutionsNumberUsing fractions1Integers, decimals and symbols1a23 8.7 200.14.86 29 140.94b 0.486 2.9 1.40944 0.505 10a0.1240a41 151016545124840403101122 7 133 13 2 15 533Fraction of weekly wage spent 13 15 1420 12 1547 60600.51347Ravi has 1 of his wage left.6060c 3 1Different types of numbera 2 7 51a16 is one more than 15, which is a multiple of 5.b 3 5 8b 5 is one less than 6, which is a factor of 12.cc3 5 15d 12 2 6e 1 7 10 42aIt is best to write the product of prime factors inascending order (i.e. smallest number first).1083 478 1561b 2445 89 513 3047c66.55 3.38 69.93316, 32, 48, 64, 80, 96, 112, 128, 144, 160 .4556 1737 281918, 36, 54, 72, 90, 108, 126, 144, 162 .b 674 387 287cNow look for the lowest number that appears in all threelists: 144.12.935 4.75 8.185d 5.77 0.369 5.401aThis means that all three grandchildren will call onthe same day every 144 days. (The lowest number toappear in only two lists is 36, so the earliest that onlytwo grandchildren will call on the same day is after36 days.)634 47 29 798b 7.7 3.8 29.26c8.32 4.9 40.768a1058 23 46b 617.4 1.8 343c88.5 2.5 35.4Find the multiples of each number.12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156 .d 7.08 4.5 12.343 23.923a16 is not a prime number as it has factors 1, 2, 4,8 and 16.300 30 10 15 2 5 2 5 3 2 5 2 2 2 3 5 5Addition, subtraction, multiplication anddivision440212b 1 59 4339b 1 34 12 74 12 74 21 72 3 12Ascending order means going up in size.28314094 48.6 29031515164 1d 140.94 29 4.8654161282510323 5 2e0.012316d 1 15 1 15 3 15 4 2550.87c5In ascending order, 58c2001d 8723a5b 1 15 65b 2.3 0.87 2.0012.001 2.3c216 3 1a4a756 252 3 126 2 3 63 2 2 3 7 9 2 2 3 7 3 3 2 2 3 2² 3³ 71

Edexcel Higher Mathematics Revision GuideFull worked solutionsb Comparing the two products and looking for thefactors that are common we see that 2² 3² 36Hence 36 is the highest common factor.Surds1a 3 2 6Listing strategiesb ( 5 )2 5c 2 3 3 3 181d (2 5 )2 2 2 5 5 20There are 2 possible outcomes to tossing a coin.There are 6 possible outcomes to rolling a six-sided dice.So 2 6 12 different possible outcomes.2As the first number cannot be zero, it could be anynumber from 1 to 9 (i.e. any of 9 numbers). The secondnumber could be any number from 0 to 9 (i.e. anyof 10 numbers). For the whole 3-digit number to bedivisible by 5 the last number must be 0 or 5 (i.e. 2numbers).So total number of 3-digit numbers that could be picked 9 10 2 180.2 28 4 7 2 7 , hence a 23a 45 9 5 3 5b 72 36 2 6 24a3 24 7a3 2 23 24 74 772 18 718b 5 16 16 28 2162 316 7 8 2 7(1 5 )(1 5 ) 1 5 5 5 4The order of operations in calculationsb (2 3 )2 (2 3 ) (2 3 ) 4 2 3 2 3 3 7 4 31ca10 4 2 10 8 18b 5 3 4 2 15 2 13c2a(7 4)2 (8 2)2 32 42 9 16 252 3 1 1 2 9 1 103Note that the division must be performed firstand that 3 31 3 13 9(1 3 )(2 3 ) 2 3 2 3 3 5 3 3Standard form1a5 10–3 0.005b 5.65 105 565 0002a25 000 2.5 104b 0.00125 1.25 10 3cb 15 (4 6)3 15 ( 2)3 15 ( 8) 15 8 230.05 104 5 102d 14 10–3 1.4 10 2Brackets are worked out first.c3 4 3 ( 7) 4 21 25 5 or 5c45b 3 2 34 3( 2 4) 3 62(54)5 54 5 520a54 56 54 6 3 57c 6 3 6 6 6 3 6 6 ( 3) 6 32 22 54 5 7 210 5 31a43 0.43100b 38 0.375ac11 0.55a8 450.8 10160 13b 100 2 100 10 or 1023c 64 3 ( 64 )2 42 16d 25e 122125 2113 255527 3 36 2 27 36 3 6 18 or 3 6 18(32x)2 81The powers on each side must be the same, so4x 4x 120459 b 0.45 1002011 1 (or 0.2) or 1 (or 0.2) 34x 342Converting between fractions and decimals7 7 11d 74 6 113 4 710 11 111461415 500 1.55 1042.5 1022.5 102843a1d 1.6 107 (1.6 107) 2 4 1035b 613.3 billion 1.33 1010 pounds1.33 1010 1.33 2 107b Number of people 500b 6.55 105 655 0001.25 10512.5 104 5 102c3( 2) 3a 26 600 00077 73 77 3 710c8 104 4 10² 2 102d 8 104 4 10² 102 (8 102 4) 102 804 8.04 104Indicesa(3 10–²)² 9 10 4b 8 104 3 10–² 24 102 2.4 103Note that the whole calculation inside thesquare root sign should be completed first.1a3c58473 0.584 1000125aLet x 0.7 0.777777. and 10x 7.777777.To eliminate the block of recurring digits after thedecimal point we take x from 10x.10x x 7.777777 0.777777 9x 7x 79

Edexcel Higher Mathematics Revision Guideb Let x 0.04 0.0444444. 2050 100 cmSo 10x 0.444444 and 100x 4.444444 205 000 cmSubtract: 100x 10x 4.444444 0.444444 21 day 24 hours 24 60 minutes 24 60 60 seconds 86400 seconds90x 442x 90c 8.64 104 seconds453Let x 0.954 0.954545454.So 10x 9.54545454. and 1000x 954.545454.Subtract: 1000x 10x 954.545454. 9.54545454.0.34 20 12 81.60Rounding numbers1ac990a1259 1260 (correct to 3 s.f.)b 14.919 14.9 (correct to 3 s.f.)990x 945945x 4Full worked solutionsDividing both parts of fraction by 5 and then by 921gives 220.0003079 0.000308 (correct to 3 s.f.)Remember you don’t start counting the numbersfor significant figures, until the first non-zeronumber.Let x 0.518 0.518518.1000x 518.518518.d 9 084 097 9 080 000 (correct to 3 s.f.)1000x x 518.518518. 0.518518.999x 51851814x 99922714Dividing both parts of fraction by 37 gives 57619 b 0.76 100250.7 73 125% 1417b 85% 2026565 100% 81.25%Maths:81.25 8030.00195 0.002 (correct to 3 decimal places)f4.098 4.10 (correct to 2 decimal places)a1989 2000 (correct to 1 significant figure)c41a7Question3 of 640 3 640 3 160 4804a15 30 4.50b 15% of 30 10029595% of 80 kg 80 76 kg100There are twice as manyhalves as units in 1200.EstimationAnswer 3 3 0.09 0.8Ab12.56 1.87 0.45 10 2 0.5 10Cc120 0.45 100 0.5 200B3.45 2.78 0.09d0.01 0.15 109 0.01 0.2 100 0.2Be0.12 300 0.53 0.1 300 0.5 15APercentage of boys in School B 100 35 65%f6.07 3.67 0.1 6 4 0.1 2BNumber of boys in School B 65% of 70065 700 455100g20.75 6.98 20 7 3Ch0.01 145 35 0.01 100 40 40ANumber of boys in School A 56% of 60056 600 336 100Standard measurement units10.52Fractions and percentages as operatorsc0.5b 4.65 28.9 6 5 30 6 5 5 10 33 42.9% (to 3 s.f.)40.54Write each numberto 1 significantfigure.10a5.9 1896 2001200 2400804 16%b2511989 1990 (correct to 3 significant figures)3.755 10 4 0.0003755 0.0004 (correct to4 decimal places)So Charlie did better at maths.3 30%aceEstimation1768% 25800.02557 0.03 (correct to 2 decimal places)b 1989 2000 (correct to 2 significant figures)Converting between fractions and percentagesc10.565 10.6 (correct to 1 decimal place)d 3.9707 3.971 (correct to 3 decimal places)9aacSo 0.7 29 79 2911.8099 10 4 1.81 10 4 (correct to 3 s.f.)b 123.9765 123.977 (correct to 3 decimal places)27 99ea9.7 kg 9.7 1000 9700 gb 850 cm3 850 ml 850 1000 0.85 litresc2.05 km 2.05 1000 m 2050 m3i6.5 0.3 0.01 7 0.3 0.01 0.02Bj65 1050 70 1000 0.07Aa 36 45 49 so 45 is between 6 and 7. 45 6.7 (accept 6.5 to 6.9)3

Edexcel Higher Mathematics Revision GuideFull worked solutionsb 100 104 121 so 104 is between 10 and 11. 104 10.2 (accept 10.1 to 10.3)c(2)3 (2.3)3 (3)3 so (2.3)3 is between 8 and 27.(2.3)3 12 (accept 10 to 14)1 21 21 212 Let x 0.72 0.727272.So 100x 72.727272.100x x 72.727272. 0.727272.99x 72872x Upper and lower boundsHalf of the smallest unit is 0.5 cmLower bound least length 144.5 cm9913 ab Upper bound greatest length 145.5 cmc144.5 l 145.5 cmaUpper bound of b24 647.515a8 1.5 6.5b 1 4 1 4c27 3 81d 1.65 3.6 4 1.65 9 10 6.6 9 10 (66 6.6) 10 59.4 10 5.943a6416 5 11231247 9b1313434 34 12aHalf the smallest unit is 0.005 cm.Least width 4.545 cm and least length 2.225 cmLeast area 4.545 2.225 10.112625 cm² 10.113 cm² (3 d.p.)Greatest width 4.555 cm and greatest length 2.235 cmGreatest area 4.555 2.235 10.180425 cm² 10.180 cm² (3 d.p.)b The upper and lower bounds for area are the samewhen written to the nearest cm² (5 and 2).Area 5 2 10 cm²6a6.02 1023 18 10 3.34 1023 molecules (3 s.f.)b 18 g 0.018 kg0.018 6.02 1023 2.99 10 26 kg (3 s.f.)7(0.45 0.78)² (0.5 0.8)² 0.4² 0.168a70 11b 9 2 9 3 or 3c11 8 2 64821111 or d 64 3 344 649Half the smallest unit 0.1 2 0.05Upper bound 5.6 0.05 5.65Lower bound 5.6 0.05 5.55Error interval is 5.55 y 5.654Error interval is 0.287996 c 0.289272b The value of c is the same when rounded to2 significant figures.So c 0.29 (to 2 s.f.)14 a39350 1 2 225502591335251616 51 80b 25 5161115 a0.00000045 4.5 10–7b 12 million 12 000 000 1.2 107Use the largest integer as the whole numberand the next largest as the numerator.52.345 0.287996b 21.5443469 21.5 (1 d.p.)211a: upper bound 0.67545, lower bound 0.67535b: upper bound 2.345, lower bound 2.335Upper bound of a0.67545Greatest value of c 2.335Lower bound of b 0.289272Lower bound of a0.67535 Least value of c Review it!1Error interval is 111.5 a 112.531.4 5 (accept 4 to 6)a11 Upper bound 112.5 cm and lower bound 111.5 cmd 31 31.4 32 so 31.4 is between 3 and 9.1 2 1 21 23 2 210 1 1 12 22 2 2 2 3 1c5640 5.64 10³16 (8 10 5) (4 10³) 8 4 10 5 10³ 32 10–2 3.2 10–117 aFactors of 64: 1, 2, 4, 8, 16, 32, 64b Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100Highest common factor 4

Edexcel Higher Mathematics Revision Guidex 102x 3 (x 7)(2x 3)Algebra(2x 3)(x 7)2x 3 x 10(x 7)(2x 3)Simple algebraic techniques 1x 7 aFormula1 2x 3Expressiond Identity2eEquationa15x2 4x x2 9x x 6x2 10x2 4xb 7a 5b b 4a 5b 3a bcChanging the subject of a formulaA21 a A π r2π rAr πb A 4πr28yx 5x2 2xy 8x2 3x2 10xy (or 10xy 3x2)A r24πx3 3x 5 2x3 4x 3x3 x 523 P I 2R ( 23) 36 49 36 164 v u at 20 ( 8)(2) 20 16 4dRemoving brackets1 a 2x 8b 63x 21c 1 x or x 1d 3x2 xe 3x2 3xf 20x2 8x2 a 2(x 3) 3(x 2) 2x 6 3x 6 5x 12b 6(x 4) 3(x 7) 6x 24 3x 21 3x 45c 3x2 x x2 x 4x2 2xd 3x2 4x 6x 8 3x2 10x 83 a (t 3)(t 5) t 2 5t 3t 15 t 2 8t 15b (x 3)(x 3) x2 3x 3x 9 x2 9c (2y 9)(3y 7) 6y2 14y 27y 63 6y2 41y 63d (2x 1)2 (2x 1)(2x 1) 4x2 2x 2x 1 4x2 4x 14 a (x 7)(x 2)(2x 3) (x² 9x 14)(2x 3) 2x³ 21x² 55x 42b (2x 1)(3x 2)(4x 3) (6x² 7x 2)(4x 3) 24x³ 28x² 8x 18x² 21x 6 24x³ 46x² 29x 6Factorising1 a 24t 18 6(4t 3)b 9a 2ab a(9 2b)c 5xy 15yz 5y(x 3z)d 24x3y2 6xy2 6xy2 (4x2 1)2 a (x 7)(x 3)b (x 5)(x 3)ccx 10x 1011 x 72x2 11x 21x 7(2x 3)(x 7)Factorise the denominatorof the 2nd fraction.Make bothdenominatorsthe same. V 43 πr 33V 4πr 33V r34π3Vr 4π32y mx cc y mxb v u atu v atc v u atv u atv ua tad v 2 2asv2s (c)Subtract mx from both sides.(u)Subtract at from both sides.(a)Subtract u from both sides.Divide both sides by t.(s)Divide both sides by 2a.2aefv u2 2as (u)v2 2as u2u v2 2as2s 12 (u v)t2s (u v)t2st u v(t)Subtract 2as from both sides.Square root both sides.Multiply both sides by 2.Divide both sides by (u v).Solving linear equations1 a x 7 4x 4 7 3b 9x 27x 27 9 3cTo get 6, use factors 2 and 3, and to get 10 use factors2 and 5. This gives 2x 2 4x and 3x 5 15x,total 19x; so solution is (2x 5)(3x 2)Ar 4πx 45x 4 5 202 a 3x 1 163x 15x 5b2x 1232x 36d Difference of two squares. Factorises to(2x 7)(2x 7)3Combine intoone fractionand simplify(x 7)(2x 3)b IdentitycFull worked solutionscx 183x 4 1653x 1253x 60x 205

Edexcel Higher Mathematics Revision Guide3aFull worked solutions1 7.5501 7.550 or(to 4 s.f.)445(1 x ) 155 5x 15 5x 102ax 2x 2.14 or 1.64 (to 3 s.f.)2x 3 3x 1x 22x 3 (x 2)(3x 1)b 2m 4 m 32x 3 3x2 x 6x 2m 4 3m 1c 9(4x 3) 3(2x 3)36x 27 6x 930x 27 930x 3636x 65 , 1 15 or 1.2300 3x2 5x 1or 3x 5x 1 02b Comparing the equation given, with ax2 bx c 0gives a 3, b 5 and c 1Substituting these values into the quadratic equationformula gives: 5 (5)2 4(3)( 1)x 2(3) 5 25 12 5 37 5 37 5 37x or6666Hence x 0.18 or 1.85 (2 d.p.)Always cancel fractions so that they are in their lowestterms. Here both top and bottom can be divided by 6.Solving simultaneous equationsSolving quadratic equations using factorisation1 a (x 3)(x 2) 0 giving x 2 or 3b (x 3)(x 4) 0 giving x 3 or 4c (2x 7)(x 5) 0 giving x 72 or x 52a1Area of triangle 12 base height2x2 11x 12 182x2 11x 6 00 5x 10(2x 1)(x 6) 0So x 1 or x 65x 102x 2Since x represents a height, only the positive valueis valid.x 1Substituting x 2 into equation (1) we obtainy 3(2) 72 6 7x 0.5, so base is 2 0.5 3 4 cm and height is0.5 4 4.5 cm 1By Pythagoras’ theoremChecking by substituting x 2 into equation (2)we obtain(x 1)2 (x 8)2 132y 2x 3x2 2x 1 x2 16x 64 1692x2 18x 104 0 2(2) 3Dividing by 2 gives 1x 9x 52 0Hence solutions are x 2 and y 1.2(x 4)(x 13) 0b y 2x 6so x 4 or x 13y 3x 14(Disregard x 13 as x is a length.)Subtracting (1) (2) we obtainHence, x 4 cm0 5x 20x 4y 2 4 6y 2.(This also means the sides of the triangle are 5, 12 and13 cm.)Substituting these values into the quadratic equationformula gives:1 ( 1) 4(2)( 7)x 21 57 4 62(2)(2)Subtracting (1) (2) we obtainb 2x 11x 6 02Solving quadratic equations using the formula1 Comparing the equation given, with ax2 bx c givesa 2, b 1 and c 7.(1)Notice that the coefficient of y (the numbermultiplying y, i.e. 1) is the same for both equations.We can eliminate y by subtracting equation (2) fromequation (1).23Firstly write the second equation so that in bothequations the x value and the numerical value arealigned.y 3x 7y 2x 31 (2x 3)(x 4) 9ca2Equating expressions for y gives10x2 5x 2 2x 310x2 7x 1 0Factorising this quadratic gives(5x 1)(2x 1) 0Hence x 15 or x 12(1)(2)(1)

Edexcel Higher Mathematics Revision GuideSubstituting x 15 into y 2x 3 givesy 2 33ay5Substituting x 12 into y 2x 3 gives1 2x 11x 612345678910{x: x 11}410x2 3x 10x2 3x 10 0(x 5)(x 2) 0x 2 and x 5Problem solving using algebra1 Let the larger number x and the smaller number y.x y 77x y 25Adding gives 2x 102 which gives x 51 so y mustLet the number added x15 x 5631 x6(15 x) 5(31 x)90 6x 155 5x11x 6512131415161715 6580 56Check the answer31 6596183x 5 73192021222324Substitute into equation (2):25262x 4 x 6x 4 6x 10b 4 x 6 4x4 6 5x 2 5x 2 x 5x 0.4 or 25c 2x 9 5(x 3)2x 9 5x 159 3x 1524 3x8 xor x 8(2)x(12 x) 27So, 12x x2 27Hence, x2 12x 27 0Factorising gives (x 3)(x 9) 0So x 3 or x 9{x: x 26}a(1)xy 27From equation (1) y 12 xAdd 5 to both sides.x 2618Perimeter: 2x 2y 24 so x y 12Area:Multiply both sides by 3.x 5 21x 262xb (0, 2), (0,1), (0, 0), (0, 1), (1, 2), (1, 1), (1, 0), (2, 2),(2, 1), (3, 2)x 11175 5Add 7 to both sides.Divide both sides by 2.x 11164 422x 22c3be 26.b 2x 7 1592 3{x: x 6}81 2x 602 5 4 3 2 1 0 1Subtract 1 from both sides.Divide both sides by 2 andreverse sign. 2x 12x y 231Solving inequalitiesa4y 2Points are (3, 20) and ( 2, 10)15x 1y 2Hence x 15 and y 2 35 or x 12 and y 23 Equating the y values givesx2 5x 4 6x 2x2 x 6 0(x 3)(x 2) 0x 3 or 2When x 3, y 6 3 2 20When x 2, y 6 ( 2) 2 10Full worked solutionsSubstituting each of these values into equation (1)we have3 y 12 or 9 y 12, giving y 9 or y 3.Hence, length 9 cm and width 3 cm.Use of functions1a1f(0) 10 111b f( 12 ) 2313 2 1cLet y 1x 1 2y(x 1) 1xy y 1xy y 1y 1x yx 1f 1(x) x7

Edexcel Higher Mathematics Revision Guide2aFull worked solutionsfg(x) ((x 4)2 9)2(x 3)2 17 0 x 8x 16 92(x 3)2 1717(x 3)2 2 x2 8x 72172172b gf(x) (x2 9) 4cx 3 gf(3) (32 9) 4x 4Iterative methods1 x0 1.5x1 1.5182945x2 1.5209353x3 1.5213157x4 1.5213705 1.521 (correct to three decimal places)Check value of x3 x 2 for x 1.5205, 1.5215xf(x)1.5205 0.0052251.52150.0007151Since there is a change of sign, a 1.521 is correct tothree decimal places.Equation of a straight line1 a 2y 4x 5y 2x 52Comparing this to y mx c we have gradient, m 211b Gradient m 21c y x 5 or 2y x 104Roots are x 0.1 and x 5.9 (1 d.p.)c(3,2a3ay (x 1)(x 5) or y x² 4x 5x2 12x 16 (x 6)2 36 16 (x 6)2 52b Turning point is at ( 6, 52)Recognising and sketching graphs offunctions1aBb FcEd A2eDfCay210y yx1 x2y y 2 64b (, 1 2 2) (, 0 (2, 2)22 )2 2i Gradient 2 (i.e invert 12 and change the sign)90º180º270º360º xb Read up from 60 to the graph, then read acrossuntil you hit the graph again.ii y y1 m(x x1)x 240 y 2 2(x 2)y 2 2x 4y 2x 63aAb GcQuadratic graphs1 a 2x2 12x 1 2[x2 6x 12 ] 2[(x 3)2 9 12 ]17 2[(x 3) 2]2 2(x 3)2 17i Turning point is at (3, 17)ii At the roots,17)b y (x 2)(x 7) or y x² 9x 14 1b5.9x4 0214 1Gradient x2 x1 6 ( 2) 8 2ac8y0.12y y1 m(x x1) where m 3 and (x1, y1) (2, 3).y 3 3(x 2)y 3 3x 6y 3x 33 y y1 m(x x1) where m 2 and (x1, y1) ( 1, 0)y 0 2(x ( 1))y 2(x 1)y 2x 2 y 2x 2 0 (or 2x y 2 0)2 3Fd ETranslations and reflections of functions1a(3, 5) (i.e. a movement of one unit to the right)b ( 1, 5) (i.e. a movement of three units to the left)c(2, 5) (i.e. a reflection in the x-axis)d ( 2, 5) (i.e. a reflection in the y-axis)

Edexcel Higher Mathematics Revision Guide2y43a2 y f (x 1)(2, 2)b112y43(1, 4)2y f (x) 210 1 21Hence 50 12.5u210 1 2The total distance travelled 50 my43c3x0 1 2x2123xu 4 m/sy f (x 1)10 1 223cxVelocity 4 m/s and time for deceleration 3 sDeceleration gradient 43 1.33 m/s2(2, 2)y43dFull worked solutions(2, 4)Since deceleration is negative acceleration,a positive answer is appropriate.y f (x 1) 2213Generating sequencesx1a17: sequence goes up by 3b 3.0: sequence goes up by 0.2Equation of a circle and tangent to a circlec1ad 432: last term is multiplied by 61: last term is multiplied by 1e2aCentre is (0, 0)b radius 49 748x2 y2 100fb Gradient of radius to (8, 6) 68 34Gradient of tangent 4cy y1 m(x x1)Second term is ( 4)2 1 17 and third term is172 1 2903Reverse the process: to find the preceding term,subtract 1 and halve.11 5.5Second term is (12 1) 2 Second term is 17, third term is 290.y 43 x 16 23 or 3y 4x 5024.5First term is (5.5 1) 2 2.25Real-life graphsa208:00 to 09:00 is 1 hour (h), which is 1 unit on x axis.2.5 5 km/hAverage speed gradient 0.5b 15 mins 0.25 hourscAverage speed gradient between 09:30 and 09:456 24 km/h First term is 2.25, second term is 5.5The nth term1aWhen n 1, 50 3(1) 47When n 2, 50 3(2) 440.25221 : last term is multiplied by 121623y 6 43 (x 8)1 12: sequence goes down by 3When n 3, 50 3(3) 41First three terms are 47, 44, 41NAILIT!b Use the nth term formula to find the value of n whenthe nth term 3450 3n 34When drawing a velocity-time graph, ensure thatthe axes are labelled with quantities and units.3n 16n 16 3Any values and letters for quantities that need tobe found should be labelled on the graph.aThe value of n is not an integer so 34 is not anumber in the sequence.cyUse the nth term formula to find the value of n whenthe nth term is less than zero (i.e. negative).50 3n 0 (subtracting 50 from both sides)uVelocity (m/s) 3n 50 (dividing both sides andreversing the inequality sign)50n 3n 16 23021215As n has to be an integer, its lowest possible value isn 17.Check that you get a negative term when n 17 isput back into the nth term formula.xTime (s)b Total distance travelled Area under the velocitytime graphUse the formula for the area of a trapezium:Distance 1 (15 10) u2 12.5uUse the formula for the areaof a trapezium.17th term 50 3 17 50 51 12aThe first four terms are: 2 31, 2 32, 2 33, 2 34 6, 18, 54, 1629

Edexcel Higher Mathematics Revision GuideFull worked solutionsb As the nth term formula is 2 3n both 2 and 3 arefactors, so 6 must also be a factor.3aCommon difference between terms 2 so formulawill start with 2n.As 3 is a factor of this expression, the sum of threeconsecutive integers must be a multiple of 3.4When n 1, you need to subtract 3 from 2n to getan answer of 1.aThe numerator is larger than the denominator so thefraction will always be greater than 1. The statementis false.b As a is larger than b, squaring a will result in alarger number than squaring b. Hence a² b² so thestatement is false.Therefore nth term 2n 3b 59 2x 3c2x 62The square root of a number can have two values,one positive and the other negative so, thisstatement is false.x 3144,First differences17,13Second differences38,21867298As a second difference is needed before a constantdifference is found, there is an n2 term in the nth term.The number in front of this n2 will be 82 4.So first part of the nth term will be 4n2.n1440Term4n²Term 4n2217161338362467643Use this set of information to work out the linear part ofthe sequence (the part with an n term and a number).Difference11Review it!1 a 3(3x 4) 9x 12b 4x 3(x 2) (x 2) 4x 3x 6 x 2 6x 4c (x 3)(2x 1)(3x 5) (2x2 5x 3)(3x 5) 6x3 25x2 16x 152 a 2x2 7x 4 (2x 1)(x 4)b 2x2 7x 4 0x 12 or x 43b 2x 3 3x4 6x15a3bc 51This means that the linear sequence will start with n.When n 2, ‘Term 4n2’ is 1, not 2, so if n is in theterm you also need to subtract 1.(2x2y )3 8x6y3a3a3b25x y 11 (2)(2) 2: 10x 2y 22 (3)This makes the linear part of the sequence n 1.(3) (1): 7x 14Check it with a different value of n. When n 3, n 1equals 2. This is the correct value for ‘Term 4n2’.x 2Substitute into (2) to find yCombining the terms gives nth term 4n n 125 2 y 11Arguments and proofs1 a 2n is always even as it has 2 as a factor. Adding 1 toan even number always gives an odd number. Thestatement is true.5x 73x 3 2x 14 x2 7xn could be a decimal such as 4.25 so squaring itwould not give an integer.3x 3 x2 5x 14x2 8x 11 0b x2 8x 11 0The statement is false.d If n was 1, or a fraction smaller than 1, this wouldnot be true. 8 (8)2 4 1 ( 11)x 2 1The statement is false. 8 108x 2Let the consecutive integers be n, n 1, n 2 andn 3, where n is an integer that can be either odd oreven. 8 6 3x 2 4n 6 2(2n 3)3Let the consecutive integers be x, x 1 and x 2,where x is an integer that can be either odd or even.Sum of the integers x x 1 x 2 3x 3 3(x 1)10x 4 3 3 or x 4 3 3So x 1.20 or x 9.20 (to 2 d.p.)Sum of the integers n n 1 n 2 n 3As 2 is a factor of this expression, the sum of fourconsecutive integers must be a multiple of 2, andtherefore even.x 13(x 1) (2 x)(x 7)The statement is false.2y 11 10y 12 x3 ab x2 9 0 so x2 9 and x 9 3cb3x 2y 8 (1)463y x ax 2z(x)3y x z(ax 2)3y x axz 2z3y 2z axz x3y 2z x(az 1)3 y 2zx az 1

Edexcel Higher Mathematics Revision Guide7y 3x 53y x 15x 3y 15acFull worked solutionsyNow replace x with f 1(x) and y with x.f 1(x) 3x 15 or f 1(x) 3(x 5) 3.9(2(x)2 k)fg(x) 5b3(8 k)So fg(2) 5( 2, 7)3We know that fg(2) 1012 Perimeter of ABCD 2 (4x (2x 3)) 12x 6Perimeter of EFG 2x 1 x 9 5x 2 8x 6(8 k)So 5 103Equate the perimeters to find x(8 k) 15 3012x 6 8x 68 k 154x 12k 78x 3Let n 1: 30 4 1 26aThe height of the triangle, EF 2 3 1 5 cmLet n 2: 30 4 2 22The base of the triangle, EG 3 9 12 cmArea of the triangle 1 12 5 30 cm²Let n 3: 30 4 3 18First three terms are 26, 22, 18.213 Side AB is parallel to side CD, so k 5.5 ( 2)Gradient of BD 7 7b 30 4n 0 4n 30 30n 1 ( 2) 4n must be an integer, so the lowest possible value ofn is n 8Therefore the first negative term of the sequence is:30 4 8 2x 4, y 3(4)² (3)² 16 9 25So x² y² 21Hence the point (4, 3) lies outside the circle.10 ( x 9y )( x 3 y )Simplify terms inside the brackets if possible( x 3 y )( x 3 y ) x 3 xy 3 xy 9y x 9y11 a2x2 8x 1 2(x2 4x) 1 2(x 2)2 8 1 2(x 2)2 7b1Using point B ( 2, 2)n 7.59x 0.1y ( 2) 7(x ( 2))y 2 7x 14Equation of BD is y 7x 1214 x2 y2 42y x 2Rearrange (2) for y: y 12 x 1(1)(2)(3)Substitute (3) into (1):2x2 1 x 1 4(2)xx2 x 1 445x2 4x 4 165x2 4x 12 0(5x 6)(x 2) 0x 65 1.2 or x 22Substitute into (2) to find ySo x 6 , y 8 or x 2, y 055i Turning point is ( 2, 7).ii For 2x2 8x 1 0 8 8 4 2 1x 2 22 8 56x 4 8 2 14x 4 4 14x 2So roots are at x 3.9 and x 0.1 (1 d.p.)11

Edexcel Higher Mathematics Revision GuideFull worked solutionsRatio, proportion and ratesof changeScale diagrams and maps1150 km 150 000 m 15 000 000 cm500 000 cm is equivalent to 1 cm on the map.15 000 000 30 cmDistance in cm on the map Introduction to ratios1a2 : 6 1: 3 (divide both sides by 2)b 25 : 60 5 : 12 (divide both sides by 5)23c1.6 : 3.6 4 : 9 (divide both sides by 0.4, or multiplyby ten then divide by four)a250 g : 2 kg 250 g : 2000 g 250 : 2000 1 : 8500 0002ab 25 m : 250 mm 25 000 mm : 250 mm 25 000 : 250 100 : 1We now need to get the units the same.cThe scale is 2 : 1 000 00010 km 10 000 m 1 000 000 cm2 cl : 1 l 2 cl : 100 cl 2 : 100 1 : 50Ratio 3.5 : 2.1 35 : 21 5 : 3Dividing both sides of the ratio by 2 gives1: 500 000b Measuring the actual distance between the twoships gives 1.2 cm85 shares 5 50 2503 shares 3 50 150Actual distance 1.2 500 000 600 000 cm4 parts 180Divide this by 100 and then 1000 to give the actualdistance in km.1 part 45(Dividing both sides by 4)600 000 cm 6 km3 parts 3 45 135Hence there are 180 135 315 members of the gym.5Distance between the ship and the port 2 cm.This is measured using a ruler on the map.Total shares 5 3 8400 501 share 4First convert 150 km to cm.The ratio is 21 : 25 : 29Percentage problems18 100 2.6 2.67% to 2 d.p.2Increase Final earnings Initial earnings 1 100 000 600 000 500 000Increase% increase 100Total shares 21 25 29 75150 000 2000One share 300original value75Youngest daughter receives 21 2000 42 0006500 000 100 600 000Total number of parts in the ratio 5 2 7Now pick a number that is divisible by 7. We will choose 70.Dividing this into the ratio 5 : 2 gives 50 male guestsand 20 female guests. 83.3% to 1 d.p.*3Add 1 to create a multiplier for the originalnumber: 1.03560% of male guests are under 40 and 70% of femaleguests are under 40.60% of 50 30 (males under 40)4So if there were 70 guests, 44 of them would be under40 years old. Use this information to work out thecorrect percentage, whatever the number of guests:44 100 62.9%82% of original price 291.92291.92 3.561% of original price 82100% of original price 3.56 100 356There are 2 more parts of the ratio for 20p coins, and6 more 20p coins.Original price 3565Hence there are 5 3 15 10p coins, and 7 3 2120p coinsTotal amount ( ) in the money box 15 0.1 21 0.2 1.5 4.2 5.70Let x be the number of yellow marbles.So 5x number of red marbles, and 2 5x 10x number of blue marbles.Hence the ratio of blue to red to yellow marbles 10x : 5x : x 10 : 5 : 1Amount of interest in one year 3.5% of 12 0003.5 12000 420 100Total interest paid over 6 years 6 420 25201 part 3 coins.121001 0.18 0.82So 2 parts 6 coins8New salary 1.035 38 000 39 33018 0.1818% 70% of 20 14 (females under 40)7073.5 0.0353.5% 100Direct and inverse proportion1Inverse proportion means that if one quantity doublesthe other quantity halves.2ay kxb 8 k 3 giving k 83y 83 x32When x 4, 83 4 10.7 (1 d.p.)3

Edexcel Higher Mathematics Revision Guide3a5Find the equivalent price in 120 94.491.27The sunglasses are cheaper in the UK.b 94.49 89 5.49V r3 so V kr3When V 33.5, r 2 so 33.5 k 2333.5k 4.18758Substituting this value of k back into the equation givesV

Revision Guide. b Comparing the two products and looking for the factors that are common we see that 2² 3² 36 Hence 36 is the highest common factor. . So Charlie did better at maths. 3 a _3 10 30% b _4 25 16% c 3_ 7 42.

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