N12/5/MATHL/HP1/ENG/TZ0/XX/MMARKSCHEMENovember 2012MATHEMATICSHigher LevelPaper 117 pages
–2–N12/5/MATHL/HP1/ENG/TZ0/XX/MThis markscheme is confidential and for the exclusive use ofexaminers in this examination session.It is the property of the International Baccalaureate andmust not be reproduced or distributed to any other personwithout the authorization of the IB Assessment Centre.
–3–N12/5/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 scoris instructions and the document “Mathematics HL: Guidance for e-markingNov 2012”. It is essential that you read this document before you start marking. In particular, pleasenote the following: Marks must be recorded using the annotation stamps. Please check that you are entering marks forthe right question. If a part is completely correct, (and gains all the “must be seen” marks), use the ticks withnumbers to stamp 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 scoris.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, 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 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.
–4–4N12/5/MATHL/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.5Follow through marksFollow through (FT) marks are awarded where an incorrect answer from one part of a question isused correctly in subsequent part(s). To award FT marks, there must be working present and notjust a final answer 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 FTmarks. If the error leads to an inappropriate value (e.g. sin θ 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, butM marks 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 candidateshould be penalized only once for a particular mis-read. Use the MR stamp to indicate that this hasbeen a misread. 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 fewermarks. If the MR leads to an inappropriate value (e.g. sin θ 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 notcover the work seen. In such cases the annotation DM should be used and a brief note written next tothe mark explaining this decision.8Alternative 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. If in doubt,contact your team leader 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 thesealternatives start and finish.
–5–9N12/5/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 inbrackets immediately 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 orthe form in brackets (if it is seen).Example: for differentiating f ( x) 2sin(5x 3) , the markscheme gives:f ʹ′ ( x ) ( 2cos(5x 3) ) 5( 10cos(5x 3) )A1Award A1 for ( 2cos(5x 3) ) 5 , even if 10cos(5x 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 tothe required accuracy. When this is not specified in the question, all numerical answers should begiven exactly or 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 waycrossed out their 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 nograde awarded. If you see work that suggests a candidate has used any calculator, please follow theprocedures for 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 onlymark the first response unless the candidate indicates otherwise.
–6–N12/5/MATHL/HP1/ENG/TZ0/XX/MSECTION A21.7 3 sin α 1 4 4 attempt to use double angle formula7 3 3 7sin 2α 2 4 4 8(M1)A1M1A17seen would normally be awarded M1A1.4Note:[4 marks]42.4 x y x x 4 y x y y 223 x y y y x y x 6 y x 4 y x x 4(M1)(A1)Award M1 for attempt to expand and A1 for correct unsimplified expansion.Note: 3x4x2y2 y4464 y4y2x2 x4 x8 4 x 6 y 2 6 x 4 y 4 4 x 2 y 6 y 8 x4 y4 Note:Award A1 for powers, A1 for coefficients and signs.Note:Final two A marks are independent of first A mark.A1A1[4 marks]3.(a)METHOD 1f ( x) ( x 1)( x 1)( x 2) x3 2 x2 x 2M1A1A1A1a 2 , b 1 and c 2METHOD 2from the graph or using f (0) 2c 2setting up linear equations using f (1) 0 and f ( 1) 0 (or f (2) 0 )obtain a 2 , b 1A1M1A1A1[4 marks](b)(i)(1, 0) , (3, 0) and (4, 0)(ii)g (0) occurs at 3 f ( 2) 36A1(M1)A1[3 marks]Total [7 marks]
–7–4.N12/5/MATHL/HP1/ENG/TZ0/XX/M2 x ln x (ln x ) 2 2ln x ln x (ln x 2) )(xf '( x) 0 ( x 1, x e 2 )f '( x ) (ln x ) 2 (a)M1A1M1Award M1 for an attempt to solve f ʹ′ ( x) 0 .Note:A(e 2 , 4e 2 ) and B (1, 0)A1A1The final A1 is independent of prior working.Note:[5 marks]2(ln x 1)xf ''( x ) 0 x e 1f ''( x ) (b)(A1)(M1)inflexion point (e 1 , e 1 )A1M1 for attempt to solve f ʹ′ʹ′ ( x) 0 .Note:[3 marks]Total [8 marks]5.(a)0A1[1 mark](b) 10(M1)f ( x)dx 1 a 11 e xdx01 a A11 e x 0e(or equivalent) a e 1Note:A1Award first A1 for correct integration of e xdx .This A1 is independent of previous M mark.[3 marks](c)(11E( X ) xf ( x )dx a xe x dx00attempt to integrate by parts x 1 a xe x e 0 e 2 a e e 2(or equivalent) e 1)M1M1(A1)A1[4 marks]Total [8 marks]
–8–6.N12/5/MATHL/HP1/ENG/TZ0/XX/MMETHOD 1(a) 1 3 a 1 det 2 2 a 2 3 1 a 3 1( 2(a 3) (a 2) ) 3 ( 2(a 3) 3(a 2) ) ( a 1)(2 6)(or equivalent) 0 (therefore there is no unique solution)M1A1A1[3 marks](b) 1 3 a 1 1 1 3 a 1 2 2 a 2 1 : 0 4 a 3 1 a 3 b 0 8 2a 1 3 a 11 : 0 4 a 1 0 0b 1 0 b 1Note:1 1 b 3 M1A1A1A1N2Award M1 for an attempt to use row operations.[4 marks]METHOD 2(a) 1 3 a 1 1 1 2 2 a 2 1 : 0 3 1 a 3 b 0 1 3 a 11 : 0 4 a 1 0 0b 1 0 3 a 1 4 a 8 2a1 1 b 3 (and 3 zeros imply no unique solution)M1A1A1[3 marks](b)b 1Note:A4Award A4 only if “ b 1 ” seen in (a).[4 marks]Total [7 marks]
–9–7.(a)N12/5/MATHL/HP1/ENG/TZ0/XX/Mattempt to apply cosine rule22M12o4 6 QR 2 QR 6cos30(or QR2 6 3 QR 20 0)QR 3 3 7 or QR 3 3 7A1A1A1[4 marks](b)METHOD 1k 6A1M1A1ok 6sin30 3Note:The M1 in (b) is for recognizing the right-angled triangle case.METHOD 2k 6use of discriminant: 108 4 36 k 2 0M1k 3A1(Note:)A1k 3 is M1A0.[3 marks]Total [7 marks]8.(a)attempt to differentiate implicitly2 x cos yNote:dydy y x 0dxdxA1A1A1 for differentiating x 2 and sin y ; A1 for differentiating xy.substitute x and y by πdydydyπ2π π π 0 dxdxdx 1 πNote:M1M1M1A1M1 for attempt to make dy/dx the subject. This could be seen earlier.[6 marks](b)ππ(or seen the other way) arctan41 ππ1 π π π1tan θ tan arctan π1 π 1 41 πθ tan θ 11 2πM1M1A1AG[3 marks]Total [9 marks]
– 10 –9.N12/5/MATHL/HP1/ENG/TZ0/XX/MMETHOD 1(a)9t A 7 4tB and3 6t A 6 7tBM1A1solve simultaneously1t A , tB 13Note:A1Only need to see one time for the A1.therefore meet at (3, 1)A1[4 marks](b)1 boats do not collide because the two times t A , t B 1 3 are different(A1)R1[2 marks]Total [6 marks]METHOD 2(a)path of A is a straight line: y Note:2x 33M1A1Award M1 for an attempt at simultaneous equations.path of B is a straight line: y 725x 442725 x 3 x ( x 3)344so the common point is (3, 1)A1A1[4 marks](b)for boat A, 9t 3 t 1and for boat B, 7 4t 3 t 13times are different so boats do not collideA1R1AG[2 marks]Total [6 marks]
– 11 –N12/5/MATHL/HP1/ENG/TZ0/XX/MSECTION B10.(a)(i)(ii)3π z1 2 3i2z1 z2 2 3i 1 3i 1 3iz1 2 3 cisA1A1*( z1 z2 ) 1 3iA1[3 marks](b)z2 2(i)tanθ 3(M1)z2 lies on the second quadrant2πθ arg z2 32πz2 2cis3(ii)A1A1attempt to use De Moivre’s theoremM12π 2kπ3z 2 cis, k 0, 1 and 233z 3 2 cis2π 38π14π, 2 cis , 3 2 cis999 3 4π 2 cis 9 Note:Award A1 for modulus, A1 for arguments.Note:Allow equivalent forms for z .A1A1[6 marks](c)(i)METHOD 1(z 2 1 1 3iz 3 cis2)( 3 z 3iπ3πor z1 3 cis22so r 3 and θ Note:) π 3 cis 2 M1A1A13π π πor θ 2 2 2Accept r cis(θ ) form.METHOD 2(z 2 1 1 3i2) 3 z 2 3cis ((2n 1) π )r2 3 r 32θ (2n 1) π θ Note:M1A1π3πor θ (as 0 θ 2π)22A1Accept r cis(θ ) form.continued
– 12 –N12/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 10 continued(ii)METHOD 11z 2π2cis31π z cis23so r z cisπ2π2cis3M11πand θ 32A1A1METHOD 2z1 1 3i1 z1 1 3i 1 3i 1 3i()()1 3i1 π z cis4231πso r and θ 32M1z A1A1[6 marks]z15π 3 cisz26(d)(A1)nn z1 5nπ 3 cis6 z2 equating imaginary part to zero and attempting to solveobtain n 12Note:A1M1A1Working which only includes the argument is valid.[4 marks]Total [19 marks]
– 13 –11.(a) 2 1 5 3 A A2 1 1 3 2 2 1 6A 3 A 1 1 32 1 0 22( A 3 A) 0 1 I3 3 N12/5/MATHL/HP1/ENG/TZ0/XX/MA1(A1)A1AG[3 marks](b)(i)(ii) a a 1 1 0 a 3a 3 0 b b 3 2 b 3b 231a and b 42M1(A1)A1METHOD 1 1 0 0 1 A 1 A 3 2 2 3 a a 1 0 1 2a 2 1 b b 2 3 2b 313a and b 22M1A1A1METHOD 2serious attempt at finding A 1M1 1 1 b 1 a A a b b obtain b 3 – 3a 0 and – b 3a 2b or equivalent linear equations A113A1a and b 22[6 marks]continued
– 14 –N12/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 11 continued(c)(i)METHOD 1" a a 1 %" x % " 0 %' '' ''b & # y '& # 1 b M1A 1 x a a 1 0 b 1 y b1 b a 1 A 1 a b b a 1 x b y a b (A1)A1A1METHOD 2ax (a 1) y 0 and bx by 1A1attempt to solveM1 1 a a obtain , b b (ii) aand gradient of l2 is 1a 1 a1the lines are perpendicular 1 a a 12 1 1 so they intersect at , 2b 2b gradient of l1 isA1A1A1A1M1A1A1[9 marks]Total [18 marks]
– 15 –12.(a)N12/5/MATHL/HP1/ENG/TZ0/XX/Mx x 2 x( f o f )( x ) f 2 x 2 x2 xx( f o f )( x ) 4 3xM1A1A1[3 marks](b) f f ) (x) Fn (x)P (n) : ( f n timesP (1) : f ( x ) F1 ( x )xxxand RHS F1 ( x ) 1LHS f ( x ) 12 x2 (2 1) x 2 x P (1) trueassume that