Mark scheme - NLCS Maths Department



Mark schemeQuestionAnswer/Indicative contentMarksPart marks and guidance1i1.2r = 4.2M1or with θ to 3 sf or betterB2 if correct answer unsupportedi3.5 caoA1Examiner's CommentsAlmost all candidates achieved full marks on this question. Some converted to degrees and rounded prematurely, thus losing the accuracy mark for the final answer, and a few used the formula for the area of a sector.iiM1or with θ to 3 sf or betteror correct use of Sine Rule with 0.9708(55.623°)or area = 5.709 = 0.5 × h × 3.952,or 3.52 ? 1.9762 = d2ii2.888.. to 2.9A1Examiner's CommentsThis straightforward question defeated a surprisingly large number of candidates. Many of these misunderstood the question and used the Cosine Rule to calculate the length AB, or simply answered their own question and calculated the area of the sector or the segment. Many of the successful candidates used convoluted methods, such as finding AB and then using Pythagoras — premature rounding sometimes caused a mark to be lost; forgetting to halve AB cost both marks. The Sine Rule was sometimes used successfully — but this was sometimes spoiled by the use of sinπ in conjunction with 3.5. A few candidates found the area of the triangle and then used ? base × height. Surprisingly few were able to use the expected approach: d = 3.5cos0.6.Total42iM1isin θ = cos2θ and completion to given resultA1www Examiner's CommentsMany candidates answered this question well, although there were a number of attempted fudges using . Some adopted a scattergun approach and it was not always possible to follow their method.iisin2θ + sin θ ? 1[= 0]M1allow 1 on RHS if attempt to complete squarecondone y2 + y ? 1 = 0ii?oe may beimplied by correct answerA1may be implied by correct answersmark to benefit of candidateignore any work with negative root & condone omission of negative root with no comment eg M1 for 0.618…ii[θ =] 38.17…, or 38.2 and 141.83…, 141.8 or 142A1Ignore extra values outside range, A0 if extra values in range or in radiansNB 0.6662 and 2.4754 if working in radian mode earns M1A1A0 Examiner's CommentsThis defeated a significant minority of candidates. However, many obtained the correct quadratic equation. Most then went on to attempt factorisation, going wrong and failing to score. A minority successfully completed the square or used the formula. Many of these went on to score full marks, but some candidates missed the last mark because they presented extra values in the range, or because they didn't realise that further work was needed after obtaining the roots of the quadratic.If unsupported, B1 for one of these, B2 for both. If both values correct with extra values in range, then B1.NB 0.6662 and 2.4754 to 3sf or moreTotal532sec2 θ = 5 tanθ6? 2(1 + tan2θ) = 5 tanθM1sec2θ = 1 + tan2 θ used? 2tan2θ ? 5 tanθ + 2 = 0A1correct quadratic oe? (2tanθ ? 1)(tanθ ? 2) = 0M1solving their quadratic for tanθ (follow rules for solving as in Question 1 [*,*]? tanθ = ? or 2A1www? θ = 0.464,A1first correct solution (or better)????????1.107A1second correct solution (or better) and no others in the rangeIgnore solutions outside the range.SC A1 for both 0.46 and 1.11SC A1 for both 26.6° and 63.4° (or better)Do not award SCs if there are extra solutions in range............................................................OR...........................................................2/cos2θ = 5sinθ/cosθM1using both sec = 1/cos and tan = sin/cos? 2cosθ = 5sinθcos2θ, cosθ ≠ 0A1correct one line equation 2 ? 5 sinθ cosθ = 0 or 2 cosθ = 5 sinθ cos2 θ oe (or common denominator). Do not need cosθ ≠ 0 at this stage.? cosθ(2 ? 5sinθ cosθ) 0? cosθ = 0, or sin 2θ = 0.8M1using sin 2θ = 2 sinθ cosθ oe ????eg 2= 5 sinθ√(1 ? sin2θ) and squaring? sin 2θ = 0.8A1sin2θ = 0.8 ???? or, say, 25 sin4θ ? 25sin2θ + 4 = 0? 2θ = 0.9273 or 2.2143? θ = 0.464A1first correct solution (or better)????????1.107A1second correct solution (or better) and no others in rangeIgnore solutions outside the rangeSCs as aboveExaminer's CommentsCandidates seemed equally to choose the two approaches in the mark scheme to solve the trigonometric equation. Both were equally successful and few offered extra unnecessary solutions. The main error was to give insufficient accuracy in the final solutions.Where solving tan θ=2 in degrees leads to θ=63.4° to 3sf, giving θ= 1.11 radians = 63.598° (63.6°) and θ=1.1radians=63.0° were insufficiently accurate so we needed θ=1.107radians to achieve the same accuracy as 63.4°.Total64iAC = cosecθM1or 1/sinθi? AD = cosecθ sec ?A1oe but not if a fraction within a fraction Examiner's CommentsThis question was answered well by the most able but many others could not cope with the fractions in part (i). AC was generally correct but often AD=cosφ/sinθ or sinθ/cosφ was the given answer, whilst others left their answer as a fraction within a fraction.iiDE = AD sin (θ + ?)AD sin(θ + ?) with substitution for their ADcorrect compound angle formula usedDo not award both M marks unless they are part of the same method. (They may appear in either order.)simplifying using tan = sin/cos. A0 if no intermediate step as AG...........................................................from triangle formed by using X on DE where CX is parallel to BE to getDX = CD cosθ and CB = 1 (oe trigonometry)substituting for both CD = AD sin? and their AD oe to reach an expression for DE in terms of θ and ? only(M marks must be part of same method)AG simplifying to required formii= cosecθ sec ? sin (θ + ?)M1AD sin(θ + ?) with substitution for their ADii= cosecθ sec ?(sinθ cos ? + cosθ sin ?)M1correct compound angle formula usediiDo not award both M marks unless they are part of the same method. (They may appear in either order.)iiii= 1 + tan ? / tan θ*...........................................................OR equivalent,A1simplifying using tan = sin/cos. A0 if no intermediate step as AGii...........................................................iieg from DE = CB + CD cosθ????????????= 1 + CD cosθ????????????= 1 + AD sin ? cosθM1from triangle formed by using X on DE where CX is parallel to BE to getDX = CD cosθ and CB = 1 (oe trigonometry)ii= 1 + cosec θ sec ? sin ? cosθM1substituting for both CD = AD sin? and their AD oe to reach an expression for DE in terms of θ and ? only(M marks must be part of same method)ii= 1 + tan? / tanθ*A1AG simplifying to required form Examiner's CommentsGood candidates were able to answer this with ease. Quite a few candidates made no response. Much depended upon their answers in part (i) which were followed through for the method marks. Those who then did not obtain the given answer should have realised that they ought to have reconsidered their answer to part (i).Total55i[y = ] 2sin x oe1Examiner's CommentsThis was done very well. Of those who were unsuccessful, nearly all realised that “2” was relevant, giving the answer as sin2x or ?sinx.ii[y = ] sin (0.5x) oe2M1 for [y = ] sin (2x)Examiner's CommentsAgain, this was very well done. A few unsuccessful candidates gave the answer as sin2x or sin?x. Occasionally ?sinx or 2 sinx were seen.Total36isin (? COD) = 3.5/5M1i? COD = 0.7753(97496…)so COD = 1.55M1COD = cos?1 (1/50) (or = 1.550….)must see 88.85 × ifworking in degrees (88.85 or better)iarea of sector = ? × 112 × 1.55(07‥)[= 93.8m2 to 3 sf]M1*or equivalent in degreesiarea of triangle = ? × 52 × sin 1.55(07‥)or ? × 7 × 3.57…M1*or ? × 7 × 5 × cos (? COD) oe12.497… implies M1M0 for area of triangle = ? × 52itheir 93.8 – their 12.497 soiM1dep*Examiner's CommentsThe Cosine Rule was by far the most popular approach, and most candidates were successful in deriving the required result. Most were successful in finding the correct area of the sector, although a few used a radius of 5 or 6 instead of 11. The majority correctly applied ?absinC to find the area of the triangle. Common errors were ?×5×5 or ?×7×5. Many who found the correct perpendicular height were successful in finding the area, but some made slips such as finding the area of either half or double the required triangle.i81 to 81.4 m2A1ii(A) sector angle = 2π ? 1.55 soiM1*may be embedded in circumference – removed arc= 4.73 to 3sfii7.4 × their 4.73(…) soiM1dep*≈ 35 miitheir arc ÷ 0.8M1may be implied by answer 43.78…arc length must be dimensionally correct, and must be an arc, not a radius.ii43 caoA1Examiner's CommentsThere were many excellent responses to this question. A few candidates rounded up to 44 instead of truncating to 43. A significant minority simply used 7.4 × 1.55 for the arc length, but generally went on to earn the method mark for dividing by 0.8. A few candidates worked with areas, or used rθ with θ in degrees, and didn’t score.ii(B) 11 × 4.73(…) ÷ 0.8M1orii22 caoA1Examiner's CommentsThe majority of those who were successful in part (A) went on to be successful in this part.Total127i9.82 + 6.42 ? 2 × 9.8 × 6.4 × cos 53.4M1i9.82 + 6.42 ? 74.79… [= 62.2…]M1for evidence of correct order of operations used; may be implied by correct answer6.89 implies M0262.4368 implies M1 (calc in radian mode), (NB √262.436‥ = 16.199…)i7.887… or 7.89 or 7.9A1if M0, B3 for 7.89 or more precise wwwExaminer's CommentsThis was very well done, with most candidates scoring full marks. A small number lost the last mark because they worked in radians or through premature rounding. Others lost the second mark by evaluating (6.42 + 9.82 - 2×6.4×9.8)cos53.4°, and a small number failed to score because they assumed the triangle to be right angled and used Pythagoras or calculated 9.8sin53.4°. Occasionally candidates used the sine ratio instead of cosine in the correct formula; similarly a few candidates attempted to use the Sine Rule.NB 9.8sin53.4 = 7.87ii? × 9.8 × 7.3 × sin(180 ? 53.4) oe seenM1or sin 53.4 used; may be embeddedmay be split into height = 9.8 × sin 53.4 then Area = ? × 7.3 × heightii28.716…or 28.72 or 28.7 or 29 iswA1if M0, B2 for 28.7 or more precise wwwExaminer's CommentsThere were many correct answers to this question, but a significant minority used the correct formula with CD instead of BC, thus failing to score. Similarly, some candidates found the area of ABD instead of ABC. As with part (i), some candidates worked in radians, although some worked in degrees in part (i) and radians in part (ii) and vice versa. Similarly, a significant minority assumed a right angle and calculated ?× base × height. Of those who found the area of ABD and subtracted the area of ACD, only a tiny fraction successfully obtained the correct answer. Some candidates lost the second mark through poor rounding.Total5845 × ? r2 × 1.6 oeM1r2 = 90/1.6 oeM1r = 7.5 or exact equivalent caoA1or B3 wwwallow recovery to 7.5 if working in degrees, but A0 for (e.g.) 7.49(their 7.5) × 1.6M112 implies M127A1or B2 wwwExaminer's CommentsThere were many completely correct answers to this question. The majority were comfortable working in radians, but some of those who converted to degrees did so successfully and managed to convert back without losing accuracy. A few candidates simply stopped when they had found the radius, and some thought they had found the perimeter when all they had done was find the arc length. A few candidates used the wrong formula initially: ?rθ, ?r θ2, r2θ and πr2θ were all seen. Often they were able to go on and earn the final method mark. A small number of candidates thought the angle was 1.6π, and a few converted to degrees and worked with ?×r2×91.7.Total590.775397.. soiM1or 44.427..°0.388, 1.18, 3.53, 4.32in degrees: 22.2, 67.8, 202, 248*A4A1 each valueif A0 then B1 for at least two of 2.366…, 7.058…, 8.649…for 2θ or all of 135.57…, 404.427…, 495.57…Examiner's CommentsMany candidates divided by 2 before using the inverse sine function, and thus failed to score. A few made an initial step of θ = 0.7 ÷ sin2, and others multiplied by 2 having correctly found arcsin0.7. Many worked in degrees and then lost marks either by failing to convert back to radians, or through premature rounding. Only a few were able to obtain all four angles in the correct form to the specified accuracy. Having found 0.388 correctly, many gave the next answer as 1.18 (or surprisingly often) 2.75 and stopped. A few candidates found one or more correct values and then multiplied them all by π.if any of final answers not given to three sf deduct 1 mark from total A marks*if final answers in degrees deduct 1 from total A marksignore extra values outside range if four correct answers in degrees or radians, deduct 1 for extra values in rangeTotal510cot 2θ = 3? tan 2θ = 1/3? 2θ = 18.43°M1tan = 1/ cot used soiθ = 9.22°A1for first correct solution (9.22 or better e.g. 9.217)2θ = 198.43°M1for method for second solution for θ.θ = 99 .22°orA1for second correct solution and no others in range (99.22 or better) or SC ft A1 for 90 + their first solution(2 tan θ)/(1 ? tan2 θ) = 1/3? 6 tan θ = 1 ? tan2 θM1use of correct double angle formula? tan2 θ + 6tan θ ? 1 = 0? tan θ = [?6 ± √(36 + 4)]/2 = 0.1623 or ?6.1623M1rearranged to a quadratic = 0 and attempt to solve by formula oe? θ = 9.22°, 99.22°A1first correct solutionA1second correct solution and no others in the range (9.22, 99.22 or better) or SC ft A1 for 90 + their first solution?1 MR if radians used (0.16, 1.73 or better)Examiner's CommentsCandidates used both the approach from invtan 2θ and from the double angle formula, the latter method being more common. Most candidates scored marks here. Common errors included only giving one solution (particularly in the invtan method) and failing to give the required accuracy.Total4113sin x + 2cos x = R sin(x + α) = R sin x cos α + R cos x sin α? R cos α = 3, R sin α = 2M1Correct pairs. Condone omission of R if used correctly. Condone sign error.? R2 = 32 + 22 = 13, R = √13B1or 3.6 or better, not ±√13 unless + √13 chosentan α = 2/3,M1ft from first M1α = 0.588A10.588 or better (accept 0.59), with no errors seen in method for angle (allow 33.7° or better)? 3sin x + 2cos x = √13 sin(x + 0.588)maximum when x + 0.588 = π/2M1any valid method e.g. differentiating? x = π/2 ? 0.588 = 0.98 radsA10.98 only. Do not accept degrees or multiples of π.? y = √13 = 3.61So coords of max point are (0.98, 3.61)B1condone √13, ft their R if, say = √14Examiner's CommentsMost candidates scored the first four marks with the ‘R’ method generally being well understood. The second part was also often fully correct. Common errors included failing to find both coordinates, using an incorrect method for finding the angle such as equating the correct expression to 1 instead of to R, using degrees instead of radians or failing to give the required accuracy.Total712LHS = sec2θ + cosec2θM1Use of secθ = 1/cosθ and cosecθ = 1/sinθ not just statingM1addingA1use of cos2θ + sin2θ = 1 soiA1AGORM1Correct formulae oeM1addingA1Use of PythagorasA1AGOR working with both sidesEg LHS sec2θ + cosec2θ = tan2θ + 1 + cot2θ + 1 = tan2θ + cot2θ + 2M1Correct formulae used on one sideRHS = (1 + tan2θ)(1 + cot2θ) = 1 + tan2θ + cot2θ + tan2θ cot2θM1Use of same formulae on other side= tan2θ + cot2θ + 2 = LHSA1Use of tanθ cotθ = 1 oe,dependent on both method marksA1Showing equalExaminer's CommentsThe most common and most successful method seen was from those that changed sec?θ+cosec?θ to 1/cos?θ+ 1/sin?θ and added these fractions together and then used sin?θ+ cos?θ=1.?There were other successful methods including changing both sides to tan?θ+cot?θ+2 or starting with?sec?θcosec?θ=sec?θ(1+cot?θ)=sec?θ+ sec?θcot?θ=sec?θ+1/cos?θxcos?θ/sin?θ=sec?θ+1/sin?θ=sec?θ+cosec?θ?Some candidates seemed to write down every relevant trig identity they could think of and make multiple starts of attempts without any clear structure to their methods.?Some attempts included taking reciprocals term by term. In general, candidates need to be encouraged to produce more structured responses when proving identities.Total413sin(x + 45°) = sin x cos 45° + cos x sin 45°M1Use of correct compound angle formula= sin x .1/√2 + cos x .1/√2= (1/√2)(sinx + cos x) = 2cos xA1? sin x + cos x = 2 √2cos x *A1Since AG, sin x cos 45° + cos x sin 45° = 2cos xsin x + cos x = 2√2 cos x only gets M1need the second line or statement of cos 45° = sin 45° = 1/√2 oe as an intermediate step to get A1 A1? sin x = (2√2 ? 1) cos x? tan x = 2√2 ? 1M1terms collected and tan x = sin x /cos x used? x = 61.32°,A1for first correct solution241.32°A1for second correct solution and no others in the range2dp but allow overspecificationignore solutions outside the rangeSC A1 for both 61.3° and 241.3°SC A1 for both 1.07 and 4.21 radians (or better)SC A1 for incorrect answers that round to 61.3° and180° + their anseg 61.33° and 241.33°Do not award SC marks if there are extra solutions in the range.Examiner's CommentsMost candidates correctly expanded the double angle formula, substituted the values for sin 45? and cos 45? and gained the first three marks.?Many candidates then proceeded correctly to obtain full marks. Others squared term by term and lost the last three marks.?There were few instances of additional solutions in the range being given although not all gave their final solutions to the required degree of accuracy.Total6 ................
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