Mr Stevenson's Maths Class



1.2(1 – sin2x) + 1 = 5sinxM12sin2x + 5sinx – 3 = 0(2sinx – 1)(sinx + 3) = 0M1sinx = M1, A1x = M1, M1, A1cso6 Use of cos2 x = 1 – sin2x.Condone invisible brackets in first line if 2 – 2 sin2 x is present(or implied) in a subsequent line.Must be using cos2 x = 1 – sin2 x. Using cos2 x = 1 + sin2 x is M0.M1 Attempt to solve a 2 or 3 term quadratic in sin x up to sin x = …Usual rules for solving quadratics. Method may be factorising, formulaor completing the squareM1Correct factorising for correct quadratic and sin x = .So, e.g. (sin x + 3) as a factors ? sin x = 3 can be ignored.A1 Method for finding any angle in any range consistent with (either of)their trig. equation(s) in degrees or radians (even if x not exact).[Generous M mark]Generous mark. Solving any trig. equation that comes from minimalworking (however bad).So x = sin–1/cos–1/tan–1 (number) ? answer in degrees or radianscorrect for their equation (in any range)M1 Method for finding second angle consistent with (either of) theirtrig. equation(s) in radians.Must be in range 0 ? x < 2?. Must involve using ?(e.g. ? ± …, 2? – …)but … can be inexact.Must be using the same equation as they used to attempt the 3rd M mark.Use of ? must be consistent with the trig. equation they are using(e.g. if using cos–1 then must be using 2? – …)If finding both angles in degrees: method for finding 2nd angleequivalent to method above in degrees and an attempt to changeboth angles to radians.M1 c.s.o. Recurring decimals are okay (instead of and ).Correct decimal values (corrected or truncated) before the finalanswer of is acceptable.A1 csoIgnore extra solutions outside range; deduct final A mark for extrasolutions in range.Special caseAnswer only M0, M0, A0, M1, M1 A1Answer only M0, M0, A0, M1, M0 A0Finding answers by trying different values (e.g. trying multiples of π) in2cos2x + 1 = 5sinx: as for answer only.[6] 2.(a)5sin x =1+ 2M12sin2x+ 5sinx–3 = 0(*)A1cso2NoteM1 for a correct method to change cos2x into sin2x(must use cos2x =1–sin2x)A1 need 3 term quadratic printed in any order with = 0 included (b)(2s–1)(s+3)=0 giving s =M1[sin x = –3 has no solution] so sin x=A1 x = 30, 150B1 B1ft4NoteM1 for attempt to solve given quadratic (usual rules forsolving quadratics) (can use any variable here, s, y, x, or sinx)A1 requires no incorrect work seen and is for sin x= or x = sin–1 y= is A0 (unless followed by x = 30)B1 for 30 (α) not dependent on method2nd B1 for 180 – αprovided in required range (otherwise 540 – α)Extra solutions outside required range: IgnoreExtra solutions inside required range: Lose final B1Answers in radians: Lose final B1S.C. Merely writes down two correct answers is M0A0B1B1Or x = 30, 150 is M1A1B1B1Just gives one answer : 30 only is M0A0B1B0 or 150 only is M0A0B0B1 NB Common error is to factorise wrongly giving (2sinx + 1)(sinx – 3) = 0[sin x = 3 gives no solution] sin x=–x= 210, 330This earns M1 A0 B0 B1ftAnother common error is to factorise correctly (2sin x –1)(sin x + 3) = 0 and follow this sin x=,sin x = 3 then x = 30°, 150°This would be M1 A0 B1 B1[6] 3.2cos2 ? – cos? – 1 = 1 – cos2 ?M13cos2 ? – cos? – 2 = 0A1(3cos? + 2)(cos? – 1) = 0cos? = or 1M1 A1? = 0,? = 131.8?B1 A1? = (360 ? “131.8”)? = 228.2?M1 A1 ft[8]4.(a)tan ? = Use of tan ? = M1? = 56.3°caoA1= 236.3° ft 180° + their principle valueA1ft3Maximum of one mark is lost if answers not to 1 decimal place (b)2 – cos θ = 2(1 – cos2 θ) Use of sin2 θ + cos2 θ = 1M12cos2 θ – cos θA1Allow this A1 if both cos θ = 0 and cos θ = are givencos θ = 0?θ = 90°, 270°M1 one solutionM1 A1cos θ = ?θ = 60°, 300°M1 one solutionM1 A16[9] 5.Using sin2? + cos2? = 1 to give a quadratic in cos?.M1Attempt to solvecos2? + cos? = 0M1(cos ?= 0) ? ? = ,B1, B1(cos ? = –1) ? ? = ?B15(Candidate who writes down 3 answers only with no working scoresa maximum of 3)[5] 6.(a)sin(θ + 30) = ( on RHS)B1θ + 30 = 36.9(α = AWRT 37)B1or = 143.1(180 – α)M1θ = 6.9, 113.1A1cao4(b)tan θ = ±2) or sin θ = ? or cos θ = ± B1(tan θ = 2 ?) θ = 63.4(β = AWRT 63.4)M1or 243.4(180 + β)M1(tan θ = –2 ?) θ = 116.6(180 – β)M1or 296.6(180 + their 116.6)M15[9] (a)M1 for 180 – their first solution. Must be at the correct stage i.e. for θ + 30(b) ALL M marks in (b) must be for θ = ...1st M1 for 180 + their first solution2nd M1 for 180 – their first solution3rd M1 for 180+ their 116.6 or 360 – their first solution Answers Only can score full marks in both partsNot 1 d.p.: loses A1 in part (a). In (b) all answers are AWRT.Ignore extra solutions outside rangeRadiansAllow M marks for consistent work with radians only, but allA and B marks for angles must be in degrees. Mixing degreesand radians is M0. MISREADS5x2 misread as 5x3M1 A0f(x) = 3x + (+C)M1 A1ft6 = 3 + + 4 + CM1C = –, f(x) = 3x + A0, A17.(a)arctan = 56.3° (= ?) (seen anywhere)B1? – 20°, (? – 20)° ÷ 3M1M1? + 180° (= 236.3°), ? – 180° (= – 123.7°) (One of these)M1x = ?47.9?, 12.1?, 72.1?A1A16First M1: Subtracting (allow adding) 20° from ?Second M1: Dividing that result by 3 (order vital !)[So 12.1° gains B1M1M1]Third M1: Giving a third quadrant resultFirst A1 is for 2 correct solutions,Second A1 for third correct solution.B1: Allow 0.983 (rads) or 62.6 (grad), and possible Ms but A0A0]EXTRAUsing expansion of tan(3x + 20°) = Getting as far as tan 3x = number (0.7348..)First M1tan 3x = 36.3°, 216.3°, – 143.7°36.3° B1x = 12.1°, 72.1°, – 47.9°Third quad result Third M1Divide by 3 Second M1Answers as scheme A1A1 (b)2sin2 x + (1 – sin2 x) = or 2(1 – cos2x) + cos2 x = M1sin2 x = or cos2x = or tan2x = or sec2x = or cos 2x = A1x = 19.5°, –19.5°A1A1ft4M1 for use of sin2 x + cos2 x = 1 or sin2 x and cos2 x interms of cos2xNote: Max. deduction of 1 for not correcting to 1 dec. place.Record as 0 first time occurs but then treat as f.t. Answers outside given interval, ignoreExtra answers in range, max. deduction of 1 in each part[Final mark](i.e. 4 or more answers within interval in (a), –1 from any gained A marks;3 or more answers within interval in (b), –1 from anygained A marks[10]8.(a)(x + 10 =)60?B1120M1(M: 180 – ? or ? – ?)x = 50 x = 110 (or 50.0 and 110.0)M1 A14(M: subtract 10)First M: Must be subtracting from 180 beforesubtracting 10. (b)(2x =) 154.2 ?B1Allow a.w.r.t. 154 or a.w.r.t. 2.69 (radians)205.8M1(M: 360 – ? or 2? – ?)x = 77.1 x = 102.9M1 A14M: Divide by 2First M: Must be subtracting from 360 before dividingby 2, or dividing by 2 then subtracting from 180.[8] In each part:Extra solutions outside 0 to 180: Ignore.extra solutions between 0 and 180 : A0. Alternative for (b): (Double angle formula)1 – 2sin2 x = –0.92sin2 x = 1.9B1sin x = M1x = 77.1x = 180 – 77.1 = 102.9M1 A1 9.(a)45 (?) (This mark can be implied by an answer 65)B1180 – ?, Add 20 (for at least one angle)M1, M165 155A14Extra solution(s) in range: Loses the A mark.Extra solutions outside range: Ignore (whether correct or not).Common solutions:65 (only correct solution) will score B1 M0 M1 A0 (2 marks)65 and 115 will score B1 M0 M1 A0 (2 marks)44.99 (or similar) for a is B0, and 64.99, 155.01 (or similar) is A0.(b)120 or 240 (?): (This mark can be implied by an answer 40 or 80)B1(Could be achieved by working with 60, using 180 – 60 and/or 180 + 60)360 – ?, 360 + ?(or 120 + an angle that has been divided by 3)M1, M1Dividing by 3 (for at least one angle)M140 80 160 200 280 320First A1: at least 3 correctA1A16 Extra solution(s) in range: Loses the final A mark.Extra solutions outside range: Ignore (whether correct or not).Common solutions:40 (only correct solution) will score B1 M0 M0 M1 A0 A0 (2 marks)40 and 80 (only correct solutions) B1 M1 M0 M1 A0 A0 (3 marks)40 and 320 (only correct solutions) B1 M0 M0 M1 A0 A0 (2 marks) Answers without working:Full marks can be given (in both parts), B and M marks by implication.Answers given in radians:Deduct a maximum of 2 marks (misread) from B and A marks.(Deduct these at first and second occurrence.)Answers that begin with statements such as sin(x – 20) = sin x – sin 20 orcos x = , then go on to find a value of ‘?’ or ‘?’, however badly, cancontinue to earn the first M mark in either part, but will score no further marks.Possible misread: cos3x = , giving 20, 100, 140, 220, 260, 340Could score up to 4 marks B0 M1 M1 M1 A0 A1 for the above answers.[10] 10.sin2 x = 4 cos xM11 – cos2 x = 4 cos xM1cos2 x + 4 cos x – 1 = 0A1cos x= M1= , second root has no real solution for xA1, B1x = 76.3? or 283.7?A1 A1 ft8[8]11.(a)(or 0.4)(i.s.w. if a value of ? is subsequently found)B11Requires the correct value with no incorrect working seen. (b)awrt B1(Also allow awrt 68.2, ft from in (a), but no other ft)(This value must be seen in part (b). It may be implied by a correctsolution, e.g. 10.9),or (if division by 2 hasalready occurred)M1( found from or or or ), or M1( found from or or )OR, or ( found from )Dividing at least one of the angles by 2M1( found from or or )(Allow awrt)A15 NoteExtra solution(s) in range: Loses the final A mark.Extra solutions outside range: Ignore (whether correct or not).Common answers:10.9 and 100.9 would score B1 M1 M0 M1 A0 (Ensure that theseM marks are awarded)10.9 and 190.9 would score B1 M0 M1 M1 A0 (Ensure that theseM marks are awarded)Alternatives:(i) 2cos 2x – 5sin 2x = 0R cos(2x + λ) = 0λ = 68.2 2x + 68.2 = 90B12x + λ = 270M12x + λ = 450 or 2x + λ = 630M1Subtracting λ and dividing by 2 (at least once)M1(ii) 25sin2 2x = 4cos2 2x = 4(1 – sin2 2x)29sin2 2x = 42x = 21.8B1The M marks are scored as in the main scheme, but extra solutionswill be likely, losing the A mark.Using radians:B1: Can be given for awrt 0.38 (β)M1: For π + β or 180 + βM1: For 2π + β or 3π + β (Must now be consistently radians)M1: For dividing at least one of the angles by 2A1: For this mark, the answers must be in degrees.(Correct) answers only (or by graphical methods):B and M marks can be awarded by implication, e.g.10.9 scores B1 M0 M0 M1 A010.9, 100.9 scores B1 M1 M0 M1 A010.9, 100.9, 190.9, 280.9 scores full marks.Using 11, etc. instead of 10.9 can still score the M marks by implication.[6] ................
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