Specimen Paper Mathematics: Analysis And Approaches

SPEC/5/MATAA/HP1/ENG/TZ0/XX/MMarkschemeSpecimen paperMathematics:analysis and approachesHigher levelPaper 116 pages

–2–SPEC/5/MATAA/HP1/ENG/TZ0/XX/MInstructions to ExaminersAbbreviationsMMarks awarded for attempting to use a correct Method.AMarks awarded for an Answer or for Accuracy; often dependent on preceding M marks.RMarks awarded for clear Reasoning.AGAnswer given in the question and so no marks are awarded.Using the markscheme1GeneralAward marks using the annotations as noted in the markscheme eg M1, A2.2Method and Answer/Accuracy marks Do not automatically award full marks for a correct answer; all working must be checked, andmarks awarded according to the markscheme. It is generally not possible to award M0 followed by A1, as A mark(s) depend on the precedingM mark(s), if any. Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for anattempt to use an appropriate method (e.g. substitution into a formula) and A1 for using thecorrect values. Where there are two or more A marks on the same line, they may be awarded independently;so if the first value is incorrect, but the next two are correct, award A0A1A1. Where the markscheme specifies M2, A2, etc., do not split the marks, unless there is a note. Once a correct answer to a question or part-question is seen, ignore further correct working.However, if further working indicates a lack of mathematical understanding do not award the finalA1. An exception to this may be in numerical answers, where a correct exact value is followed byan incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,and correct working shown, award FT marks as appropriate but do not award the final A1 in thatpart.ExamplesCorrect answer seen1.2.3.8 21sin 4 x4log a log bFurther working seen5.65685.(incorrect decimal value)ActionAward the final A1(ignore the further working)sin xDo not award the final A1log ( a b )Do not award the final A1

–3–3SPEC/5/MATAA/HP1/ENG/TZ0/XX/MImplied marksImplied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or ifimplied in subsequent working. Normally the correct work is seen or implied in the next line. Marks without brackets can only be awarded for work that is seen.4Follow through marks (only applied after an error is made)Follow through (FT) marks are awarded where an incorrect answer from one part of a question isused correctly in subsequent part(s) or subpart(s). Usually, to award FT marks, there must beworking present and not just a final answer based on an incorrect answer to a previous part.However, if the only marks awarded in a subpart are for the answer (i.e. there is no workingexpected), then FT marks should be awarded if appropriate. Within a question part, once an error is made, no further A marks can be awarded for workwhich uses the error, but M marks may be awarded if appropriate. If the question becomes much simpler because of an error then use discretion to award fewerFT marks. If the error leads to an inappropriate value (e.g. probability greater than 1, use of r 1 for thesum of an infinite GP, sin 1.5 , non integer value where integer required), do not award themark(s) for the final answer(s). The markscheme may use the word “their” in a description, to indicate that candidates may beusing an incorrect value. Exceptions to this rule will be explicitly noted on the markscheme. If a candidate makes an error in one part, but gets the correct answer(s) to subsequent part(s),award marks as appropriate, unless the question says hence. It is often possible to use adifferent approach in subsequent parts that does not depend on the answer to previous parts.5Mis-readIf a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MRpenalty of 1 mark to that question If the question becomes much simpler because of the MR, then use discretion to awardfewer marks. If the MR leads to an inappropriate value (e.g. probability greater than 1, sin 1.5 , non-integervalue where integer required), do not award the mark(s) for the final answer(s). Miscopying of candidates’ own work does not constitute a misread, it is an error. The MR penalty can only be applied when work is seen. For calculator questions with noworking and incorrect answers, examiners should not infer that values were read incorrectly.

–4–6SPEC/5/MATAA/HP1/ENG/TZ0/XX/MAlternative methodsCandidates will sometimes use methods other than those in the markscheme. Unless the questionspecifies a method, other correct methods should be marked in line with the markscheme Alternative methods for complete questions are indicated by METHOD 1,METHOD 2, etc. Alternative solutions for part-questions are indicated by EITHER . . . OR.7Alternative formsUnless the question specifies otherwise, accept equivalent forms. As this is an international examination, accept all alternative forms of notation. In the markscheme, equivalent numerical and algebraic forms will generally be written inbrackets immediately following the answer. In the markscheme, simplified answers, (which candidates often do not write in examinations),will generally appear in brackets. Marks should be awarded for either the form preceding thebracket or the form in brackets (if it is seen).8Accuracy of AnswersIf the level of accuracy is specified in the question, a mark will be linked to giving the answer to therequired accuracy. There are two types of accuracy errors, and the final answer mark should not beawarded if these errors occur. Rounding errors: only applies to final answers not to intermediate steps. Level of accuracy: when this is not specified in the question the general rule applies to finalanswers: unless otherwise stated in the question all numerical answers must be given exactly orcorrect to three significant figures.9CalculatorsNo calculator is allowed. The use of any calculator on paper 1 is malpractice, and will result in nograde awarded. If you see work that suggests a candidate has used any calculator, please followthe procedures for malpractice. Examples: finding an angle, given a trig ratio of 0.4235.

–5–SPEC/5/MATAA/HP1/ENG/TZ0/XX/MSection A1.attempt to substitute into P A B P A P B P A B (M1)Note: Accept use of Venn diagram or other valid method.0.6 0.5 0.4 P( A B)P A B 0.3 (seen anywhere)attempt to substitute into P A B (A1)A1P A B P B (M1)0.30.4 P A B 0.75 3 4 A1Total [5 marks]2.(a)attempting to expand the LHSLHS 4n 4n 1 4n 4n 1 22 8n 2 2 RHS (M1)A1AG[2 marks](b)METHOD 1recognition that 2n 1 and 2n 1 represent two consecutive oddintegers (for all odd integers n )8n 2 2 4n 1 22valid reason egdivisible by 2 (2 is a factor)so the sum of the squares of any two consecutive odd integers is evenR1A1R1AG[3 marks]METHOD 2recognition, eg(for n )that n and n 2 represent two consecutive odd integersR1n 2 n 2 2 n 2 2n 2 2valid reason egdivisible by 2 (2 is a factor)so the sum of the squares of any two consecutive odd integersis evenA1R1AG[3 marks]Total [5 marks]

–6–3.SPEC/5/MATAA/HP1/ENG/TZ0/XX/Mattempt to integratedu 4xdx8x2dx du2u2x 1(M1)u 2x2 1 (A1)EITHER 4 u C A1OR 4 2 x 2 1 C A1THENcorrect substitution into their integrated function (must have C )5 4 C C 1f x 4 2 x2 1 1(M1)A1Total [5 marks]

–7–SPEC/5/MATAA/HP1/ENG/TZ0/XX/M4.no y values below 1horizontal asymptote at y 2 with curve approaching from below as x A1A1 1,1 local minima 0,5 local maximumA1smooth curve and smooth stationary points5.(a)attempt to form composition x 3 x 3 correct substitution g 8 5 4 4 g f x 2 x 11A1A1Total [5 marks]M1A1AG[2 marks](b)attempt to substitute 4 (seen anywhere)correct equation a 2 4 11a 19(M1)(A1)A1[3 marks]Total [5 marks]

g to use the change of base ruleM1log 9 (cos 2 x 2) A1log 3 (cos 2 x 2)log 3 91 log 3 (cos 2 x 2)2 log 3 cos 2 x 2A1AG[3 marks](b)log 3 (2sin x) log 3 cos 2 x 22sin x cos 2x 2M14sin x cos 2 x 2 (or equivalent)use of cos 2 x 1 2sin 2 x6sin 2 x 31sin x 2πx 42Note: Award A0 if solutions other than x A1(M1)A1A1πare included.4[5 marks]Total [8 marks]

–9–7.SPEC/5/MATAA/HP1/ENG/TZ0/XX/Mattempting integration by parts, egπxπ6 πx πx , du dx, dv sin dx, v cos 3636π 6 6 33π 6x πx 6 πx P 0 X 3 cos cos dx (or equivalent)36 π 6 0 π 0 6 u Note: Award A1 for a correct uv and A1 for a correctattempting to substitute limits(M1)A1A1 v du .M13π 6x πx cos 0 36 π 6 0(A1)31 πx so P 0 X 3 sin (or equivalent)π 6 0 1πA1A1Total [7 marks]8.recognition that the angle between the normal and the line is 60 (seen anywhere) R1attempt to use the formula for the scalar productM1cos 60 2 1 1 2 2 p 9 1 4 p22p1 2 3 5 p2A1A13 5 p2 4 pattempt to square both sides9 5 pp 32 16 p2M1 7 p 4525(or equivalent)7A1A1Total [7 marks]

– 10 –9.(a)SPEC/5/MATAA/HP1/ENG/TZ0/XX/Mattempt to differentiate and set equal to zerof ( x) 2e 6e 2e (e 3) 0minimum at x ln 3a ln 32xxxxM1A1A1[3 marks](b)Note: Interchanging x and y can be done at any stage.y e x 3 4(M1)ex 3 y 4A12 as x ln 3 , x ln 3 y 4so f 1 x ln 3 x 4domain of f 1 is x , 4 x 5R1A1A1[5 marks]Total [8 marks]

– 11 –SPEC/5/MATAA/HP1/ENG/TZ0/XX/MSection B10.(a)attempt to use quotient rulecorrect substitution into quotient rule(M1) 1 5kx k ln 5 x 5x (or equivalent)f x 2 kx k k ln 5 x, k k 2 x21 ln 5x kx 2 A1A1AG[3 marks](b)f x 0M11 ln 5 x 0kx 2ln 5x 1ex 5(A1)A1[3 marks](c)f x 0M12 ln 5 x 3 0kx33ln 5 x 2A135x e 2so the point of inflexion occurs at x A11e532AG[3 marks]continued

– 12 –SPEC/5/MATAA/HP1/ENG/TZ0/XX/MQuestion 10 continued(d)attempt to integrate(M1)du 1 u ln 5 x dx xln 5x1dx u du kxk (A1)EITHER u22kA1323 u2 21so u du k1 2 k 1A1OR ln 5x 2A12k331 2e5so e51 2e5 ln 5 x 2 ln 5 xdx kx 2k eA15THEN1 9 1 2k 4 5 8k setting their expression for area equal to 35 38k5k 24A1M1A1[7 marks]Total [16 marks]

– 13 –11.(a)SPEC/5/MATAA/HP1/ENG/TZ0/XX/Mattempt to find modulus r 2 3 12(M1) A1attempt to find argument in the correct quadrant (M1)3 π arctan 3 A15π6A15πi5πi 3 3i 12e 6 2 3e 6 [5 marks](b)attempt to find a root using de Moivre’s theorem1612 eattempt to find further two roots by adding and subtractingthe argument167πi 181617πi1812 e12 eM15πi18A12πto3M1A1A1Note: Ignore labels for u , v and w at this stage.[5 marks]continued

– 14 –SPEC/5/MATAA/HP1/ENG/TZ0/XX/MQuestion 11 continued(c)METHOD 1attempting to find the total area of (congruent) triangles UOV, VOWand UOW 2π 1 Area 3 12 12 sin3 2 1616 1 1 Note:Award A1 for 12 6 12 6 and A1 for sin M1A1A12π.3 3 3 12 (or equivalent)4 13A1[4 marks]METHOD 222 1 1 1 1 2π(or equivalent)UV 12 6 12 6 2 12 6 12 6 cos3 16 UV 3 12 (or equivalent) 1attempting to find the area of UVW using Area UV VW sin 22for exampleA1A1M1 1 πArea 3 12 3 12 sin2 3 3 3 13 12 (or equivalent)4 1616A1[4 marks](d)u v w 01 5π5π17π17π 7π 7π 12 6 cos i sin cos i sin cos i sin 018181818 18 18 R1consideration of real partsM1A116 5π17π 7π 12 cos cos cos 01818 18 7π 7π cos cosexplicitly stated18 18 5π7π17πcos cos cos 0181818A1AG[4 marks]Total [18 marks]

– 15 –12.(a)SPEC/5/MATAA/HP1/ENG/TZ0/XX/Mattempting to use the chain rule to find the first derivativef x cos x eM1A1sin xattempting to use the product rule to find the second derivativef x esin x cos 2 x sin x (or equivalent)M1A1attempting to find f 0 , f 0 and f 0 M1 f 0 1 ; f 0 cos 0 esin 0 1 ; f 0 esin 0 cos 2 0 sin 0 1substitution into the Maclaurin formula f ( x) f (0) xf (0) x2f (0) .2!A1M1x2so the Maclaurin series for f x up to and including the x term is 1 x A122[8 marks](b)METHOD 1attempting to differentiate f ( x)f x cos x esin x cos2x sin x cos x eM1sin x 2sin x 1 (or equivalent)substituting x 0 into their f x A2M1f 0 1 1 0 1 0 1 0so the coefficient of x3 in the Maclaurin series for f x is zeroAGMETHOD 2substituting sin x into the Maclaurin series for e x2(M1)3sin x sin x .2!3!substituting Maclaurin series for sin xesin x 1 sin x esin x2M13 xx x . x . 3!3! x . 1 x . 3!2!3! 3331 1 03! 3!so the coefficient of x3 in the Maclaurin series for f x is zerocoefficient of x3 is A1A1AG[4 marks]continued