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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.

–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