## Transcription

N14/5/MATHL/HP1/ENG/TZ0/XX/MMARKSCHEMENovember MATHEMATICSHigher LevelPaper 120 pages

–2–N14/5/MATHL/HP1/ENG/TZ0/XX/MThis markscheme is the property of the International Baccalaureateand must not be reproduced or distributed to any otherperson without the authorization of the IB Assessment Centre.

–3–N14/5/MATHL/HP1/ENG/TZ0/XX/MInstructions to ExaminersAbbreviationsMMarks awarded for attempting to use a correct Method; working must be seen.(M)Marks awarded for Method; may be implied by correct subsequent working.AMarks awarded for an Answer or for Accuracy; often dependent on preceding M marks.(A)Marks awarded for an Answer or for Accuracy; may be implied by correct subsequent working.RMarks awarded for clear Reasoning.NMarks awarded for correct answers if no working shown.AGAnswer given in the question and so no marks are awarded.Using the markscheme1GeneralMark according to RM Assessor instructions and the document “Mathematics HL: Guidance for emarking November”. It is essential that you read this document before you start marking. Inparticular, please note the following: Marks must be recorded using the annotation stamps. Please check that you are entering marks for theright question. If a part is completely correct, (and gains all the “must be seen” marks), use the ticks with numbers tostamp full marks. If a part is completely wrong, stamp A0 by the final answer. If a part gains anything else, it must be recorded using all the annotations. All the marks will be added and recorded by RM Assessor.2Method and Answer/Accuracy marks Do not automatically award full marks for a correct answer; all working must be checked,and marks awarded according to the markscheme. It is not possible to award M0 followed by A1, as A mark(s) depend on the preceding M mark(s),if any. Where M and A marks are noted on the same line, eg M1A1, this usually means M1 for an attempt to usean appropriate method (eg substitution into a formula) and A1 for using the correct values. Where the markscheme specifies (M2), N3, etc., do not split the marks. Once a correct answer to a question or part-question is seen, ignore further working.3N marksAward N marks for correct answers where there is no working. Do not award a mixture of N and other marks. There may be fewer N marks available than the total of M, A and R marks; this is deliberate as itpenalizes candidates for not following the instruction to show their working.

–4–4N14/5/MATHL/HP1/ENG/TZ0/XX/MImplied marksImplied marks appear in brackets eg (M1), and can only be awarded if correct work is seen or if implied insubsequent 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.5Follow through marksFollow through (FT) marks are awarded where an incorrect answer from one part of a question is usedcorrectly in subsequent part(s). To award FT marks, there must be working present and not just a finalanswer based on an incorrect answer to a previous part. If the question becomes much simpler because of an error then use discretion to award fewer FT marks. If the error leads to an inappropriate value (eg sin 1.5 ), do not award the mark(s) for the finalanswer(s). Within a question part, once an error is made, no further dependent A marks can be awarded, but Mmarks may be awarded if appropriate. Exceptions to this rule will be explicitly noted on the markscheme.6Mis-readIf a candidate incorrectly copies information from the question, this is a mis-read (MR). A candidate shouldbe penalized only once for a particular mis-read. Use the MR stamp to indicate that this has been amisread.Then deduct the first of the marks to be awarded, even if this is an M mark,but award all others so that the candidate only loses one mark. If the question becomes much simpler because of the MR, then use discretion to award fewer marks. If the MR leads to an inappropriate value (eg sin 1.5 ), do not award the mark(s) for the finalanswer(s).7Discretionary marks (d)An examiner uses discretion to award a mark on the rare occasions when the markscheme does not cover thework seen. In such cases the annotation DM should be used and a brief note written next to the markexplaining this decision.8Alternative methodsCandidates will sometimes use methods other than those in the markscheme. Unless the question specifies amethod, other correct methods should be marked in line with the markscheme. If in doubt, contact your teamleader for advice. Alternative methods for complete questions are indicated by METHOD 1, METHOD 2, etc. Alternative solutions for part-questions are indicated by EITHER . . . OR. Where possible, alignment will also be used to assist examiners in identifying where these alternativesstart and finish.

–5–9N14/5/MATHL/HP1/ENG/TZ0/XX/MAlternative 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 in bracketsimmediately following the answer. In the markscheme, simplified answers, (which candidates often do not write in examinations), willgenerally appear in brackets. Marks should be awarded for either the form preceding the bracket or theform in brackets (if it is seen).Example: for differentiating f ( x) 2sin (5 x 3) , the markscheme gives:f ( x) 2cos (5 x 3) 5 10cos (5 x 3) A1Award A1 for 2cos (5 x 3) 5 , even if 10cos (5 x 3) is not seen.10Accuracy of AnswersCandidates should NO LONGER be penalized for an accuracy error (AP).If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to therequired accuracy. When this is not specified in the question, all numerical answers should be given exactlyor correct to three significant figures. Please check work carefully for FT.11Crossed out workIf a candidate has drawn a line through work on their examination script, or in some other way crossed outtheir work, do not award any marks for that work.12CalculatorsNo calculator is allowed. The use of any calculator on paper 1 is malpractice, and will result in no gradeawarded. If you see work that suggests a candidate has used any calculator, please follow the proceduresfor malpractice. Examples: finding an angle, given a trig ratio of 0.4235.13More than one solutionWhere a candidate offers two or more different answers to the same question, an examiner should only markthe first response unless the candidate indicates otherwise.14.Candidate workCandidates are meant to write their answers to Section A on the question paper (QP), and Section B onanswer booklets. Sometimes, they need more room for Section A, and use the booklet (and often commentto this effect on the QP), or write outside the box. This work should be marked.The instructions tell candidates not to write on Section B of the QP. Thus they may well have done somerough work here which they assume will be ignored. If they have solutions on the answer booklets, there isno need to look at the QP. However, if there are whole questions or whole part solutions missing on answerbooklets, please check to make sure that they are not on the QP, and if they are, mark those whole questionsor whole part solutions that have not been written on answer booklets.

–6–N14/5/MATHL/HP1/ENG/TZ0/XX/MSECTION A1.(a)g ( x) 1 1x 3A1A1Note: Award A1 for x 3 in the denominator and A1 for the “ 1 ”.[2 marks](b)x 3y 1A1A1[2 marks]Total [4 marks]2.(a)using the formulae for the sum and product of roots:(i) 4(ii) 12A1A1Note: Award A0A0 if the above results are obtained by solving the originalequation (except for the purpose of checking).[2 marks](b)METHOD 1 2 2 2 2 required quadratic is of the form x 2 x 4q (M1) q 8A1 2 2 p 2( ) M12 412p 16A1 Note: Accept the use of exact rootscontinued

–7–N14/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 2 continuedMETHOD 2replacing x with2xM12 2 2 2 8 1 0 x x 8 16 1 0x2 xx 2 16 x 8 0p 16 and q 8(A1)A1A1Note: Award A1A0 for x 2 16 x 8 0 ie, if p 16 and q 8are not explicitly stated.[4 marks]Total [6 marks]

–8–3.N14/5/MATHL/HP1/ENG/TZ0/XX/MMETHOD 1 OP (1 s ) 2 (3 2 s ) 2 (1 s ) 2 ( 6s 2 12s 11 )A1Note: Award A1 if the square of the distance is found.EITHERattempt to differentiate:attempting to solved OPdsd OPds2 12 s 12 M12 0 for s(M1)s 1(A1)ORattempt to differentiate:attempting to solve d OPds 6 s 2 12 s 11 6s 6d OP 0 for sdsM1(M1)s 1(A1)OR 2attempt at completing the square: OP 6 ( s 1) 2 minimum valueoccurs at s 1 5 M1(M1)(A1)THEN the minimum length of OP is5A1METHOD 2 1 the length of OP is a minimum when OP is perpendicular to 2 1 1 s 1 3 2s 2 0 1 s 1 attempting to solve 1 s 6 4 s 1 s 0 (6s 6 0) for ss 1(R1)A1(M1)(A1) OP 5A1Total [5 marks]

–9–4.(a)(i)(ii)N14/5/MATHL/HP1/ENG/TZ0/XX/Muse of P ( A B ) P ( A) P ( B )P ( A B ) 0.2 0.5 0.7(M1)use of P ( A B ) P ( A) P ( B ) P ( A)P ( B )P ( A B ) 0.2 0.5 0.1 0.6(M1)A1A1[4 marks](b)P ( A B)P ( B)P ( A B ) is a maximum when P ( A B ) P ( A)P ( A B ) is a minimum when P ( A B ) 00 P ( A B ) 0.4P ( A B) A1A1A1Note: A1 for each endpoint and A1 for the correct inequalities.[3 marks]Total [7 marks]5.use of the quotient rule or the product rule3 t 2 2 2t 2t 6 2t 2 24t 2 or C (t ) 2 3 t 2 2 3 t 2 3 t 2 2 3 t 2 M1A1A1Note: Award A1 for a correct numerator and A1 for a correct denominatorin the quotient rule, and A1 for each correct term in the product rule.attempting to solve C (t ) 0 for tt 3 (minutes)C 3 33 mg l or equivalent. 1(M1)A1A1Total [6 marks]

– 10 –6.N14/5/MATHL/HP1/ENG/TZ0/XX/Mdu1 dx 2 xA1dx 2 (u 1) duNote: Award the A1 for any correct relationship between dx and du. 1 xxdx 2 (u 1) 2duu(M1)A1Note: Award the M1 for an attempt at substitution resulting in an integral onlyinvolving u.1 2 u 2 duu2 u 4u 2 ln u ( C ) (A1) x 2 x 3 2 ln 1 x ( C )A1A1Note: Award the A1 for a correct expression in x, but not necessarily fullyexpanded/simplified.Total [6 marks]7.(a)p (3) f (3) g (3) g (3) f (3)(M1)Note: Award M1 if the derivative is in terms of x or 3. 2 4 3 1 11A1[2 marks](b)h ( x) g f ( x) f ( x)h (2) g (1) f (2) 4 4 16(M1)(A1)A1A1[4 marks]Total [6 marks]

– 11 –8.N14/5/MATHL/HP1/ENG/TZ0/XX/Mlet P ( n) be the proposition that (2n)! 2n (n !) 2 , n consider P (1) :2! 2 and 21 1! 2 so P (1) is trueR1assume P ( k ) is true ie (2k )! 2k (k !) 2 , k M12Note: Do not award M1 for statements such as “let n k ”.consider P (k 1) : 2(k 1) ! (2k 2) (2k 1) (2k )! 2(k 1) ! (2k 2) (2k 1) (k !)2 2kM1A1 2(k 1) (2k 1) (k !) 2 2k 2k 1 (k 1) (k 1) (k !) 2 since 2k 1 k 1R1 2k 1 (k 1)! A12P (k 1) is true whenever P (k ) is true and P (1) is true, so P ( n) is true for n R1Note: To obtain the final R1, four of the previous marks must have been awarded.Total [7 marks]

– 12 –9.N14/5/MATHL/HP1/ENG/TZ0/XX/M(a)2 t correct for [1, 2]A12 t correct for [2, 3]A1[2 marks](b)EITHERlet q1 be the lower quartile and let q3 be the upper quartilelet d 2 q1 q3 2 and so IQR 2d by symmetryuse of area formulae to obtain1 2 1d 24(or equivalent)1d or the value of at least one q.2ORM1A1A1let q1 be the lower quartileq1consider1 (2 t ) dt 4M1A11obtain q1 2 12A1THENIQR 2A1Note: Only accept this final answer for the A1.[4 marks]Total [6 marks]

– 13 –10.(a)N14/5/MATHL/HP1/ENG/TZ0/XX/Muse of the addition principle with 3 termsto obtain 4C3 5C3 6C3 ( 4 10 20)number of possible selections is 34(M1)A1A1[3 marks](b)EITHERrecognition of three cases: (2 odd and 2 even or 1 odd and 3 even or 0odd and 4 even)(M1)545454(M1)A1 C2 C2 C1 C3 C0 C4 ( 60 20 1)ORrecognition to subtract the sum of 4 odd and 3 odd and 1 even fromthe totalC4 5C4 5C3 4C1 ( 126 5 40)9(M1)(M1)A1THENnumber of possible selections is 81A1[4 marks]Total [7 marks]

– 14 –N14/5/MATHL/HP1/ENG/TZ0/XX/MSECTION B11.(a)x e3 y 1(i)M1Note: The M1 is for switching variables and can be awarded at any stage.Further marks do not rely on this mark being awarded.taking the natural logarithm of both sides and attempting to transpose f 1 ( x) 13 (ln x 1)x or equivalent, for example x 0 .(ii)M1A1A1[4 marks](b)111ln x (ln x 1) ln x ln x (or equivalent)3331ln x (or equivalent)2x e 12M1A1A1A1 1 coordinates of P are e , 2 1 2A1[5 marks](c)coordinates of Q are (1, 0) seen anywheredy 1 dx xdyat Q , 1dxy x –1A1M1A1AG[3 marks]continued

– 15 –N14/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 11 continued(d)let the required area be Aee11A x 1dx ln xdxM1Note: The M1 is for a difference of integrals. Condone absence of limits here.attempting to use integration by parts to find ln xdx(M1)e x2 x [ x ln x x]1e 2 1A1A1x2Note: Award A1 for x and A1 for x ln x x .2Note: The second M1 and second A1 are independent of the first M1 and the first A1.e21 e2 2e 1 e 222 A1[5 marks](e)(i)METHOD 1consider for example h ( x) x 1 ln x1h (1) 0 and h ( x) 1 xas h ( x) 0 for x 1 , then h ( x) 0 for x 1as h ( x) 0 for 0 x 1 , then h ( x) 0 for 0 x 1so g ( x) x 1 , x (A1)R1R1AGMETHOD 21x2g ( x) 0 (concave down) for x the graph of y g ( x) is below its tangent ( y x 1 at x 1 )so g ( x) x 1 , x g ( x) A1R1R1AGNote: The reasoning may be supported by drawn graphical arguments.continued

– 16 –N14/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 11