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M15/5/MATME/SP2/ENG/TZ2/XX/MMARKSCHEMEMay 2015MATHEMATICSStandard levelPaper 218 pages

–2–M15/5/MATME/SP2/ENG/TZ2/XX/MThis markscheme is the property of the International Baccalaureateand must not be reproduced or distributed to any other personwithout the authorization of the IB Assessment Centre.

–8–M15/5/MATME/SP2/ENG/TZ2/XX/MSection A1.(a)evidence of choosing sine ruleegACBC ˆˆsin ABCsin BAC()()correct substitutioneg(A1)AC10 sin 80 sin 35 AC 17.1695AC 17.2 (cm)(b)(M1)A1ˆ 65 (seen anywhere)ACB(A1)correct substitution(A1)egN2[3 marks]1 10 17.1695 sin 65 2area 77.8047area 77.8 (cm 2 )A1N2[3 marks]Total [6 marks]2.(a)(i)correct substitution6 2 3 2 6 1egu v 24(ii)correct substitution into magnitude formula for u or veg(iii)(A1)A1N2(A1)62 32 62 , 22 22 12 , correct value for vu 9A1N2v 3A1N1[5 marks](b)correct substitution into angle formulaeg(A1)24, 0.89 30.475882, 27.26604 A1N20.476, 27.3⁰[2 marks]Total [7 marks]

–9–3.(a) (i)M15/5/MATME/SP2/ENG/TZ2/XX/Mevidence of set upeg correct value for a , b or ra 4.8 , b 1.2(ii)(b)r 0.988064r 0.988correct substitution into their regression equationeg4.8 7 1.234.8 (millions of dollars) (accept 35 and 34 800 000)(M1)A1A1N3A1N1[4 marks](A1)A1N2[2 marks]Total [6 marks]4.valid approach to find the required termeg(M1) 8 8 r r876 2th x k , Pascal’s triangle to 8 row, x 8 x k 28 x k . r identifying correct term (may be indicated in expansion)(A1) 8 6 2 x k , 2 8 6 2eg x k , r 2 6 setting up equation in k with their coefficient/term 8 28k 2 x 6 63 x 6 , k 2 63eg 6 k 1.5 (exact)(M1)A1A1N3[5 marks]

– 10 –5.M15/5/MATME/SP2/ENG/TZ2/XX/M(a)A1A1A1N3Note: Curve must be approximately correct exponential shape (increasing andconcave up). Only if the shape is approximately correct, award the following:A1 for right end point in circle,A1 for y-intercept in circle,A1 for asymptotic to y 2 , (must be above y 2 ).[3 marks](b)valid attempt to find geg(M1)f ( x 3) 1 , g ( x) e x 1 3 2 1 , e x 1 3 , 2 1 , sketchg ( x) e x 2 1A2N3[3 marks]Total [6 marks]

– 11 –6.M15/5/MATME/SP2/ENG/TZ2/XX/MMETHOD 1recognize that the distance walked each minute is a geometric sequenceegr 0.9 , valid use of 0.9(M1)recognize that total distance walked is the sum of a geometric sequence(M1) 1 r Sn , a 1 r negcorrect substitution into the sum of a geometric sequence(A1) 1 0.9 80 1 0.9 negany correct equation with sum of a geometric sequence(A1) 0.9 1 66n80 660, 1 0.9 80 0.9 1 negattempt to solve their equation involving the sum of a GPeg graph, algebraic approach(M1)n 16.54290788A1since n 15he will be lateR1AGN0Note: Do not award the R mark without the preceding A mark.continued.

– 12 –M15/5/MATME/SP2/ENG/TZ2/XX/MQuestion 6 continuedMETHOD 2recognize that the distance walked each minute is a geometric sequenceegr 0.9 , valid use of 0.9(M1)recognize that total distance walked is the sum of a geometric sequence(M1) 1 r Sn , a 1 r negcorrect substitution into the sum of a geometric sequence(A1) 1 0.9 80 1 0.9 negattempt to substitute n 15 into sum of a geometric sequenceegS15(M1)correct substitution(A1) 0.9 1 80 0.9 1 15egS15 635.287A1since S 660he will not be there on timeR1AGN0Note: Do not award the R mark without the preceding A mark.METHOD 3recognize that the distance walked each minute is a geometric sequencer 0.9 , valid use of 0.9eg(M1)recognize that total distance walked is the sum of a geometric sequence(M1) 1 r Sn , a 1 r neglisting at least 5 correct terms of the GP15 correct terms(M1)A180, 72, 64.8, 58.32, 52.488, 47.2392, 42.5152, 38.2637, 34.4373, 30.9936, 27.8942,25.1048, 22.59436, 20.3349, 18.3014attempt to find the sum of the terms(M1)eg S15 , 80 72 64.8 58.32 52.488 . 18.301433S15 635.287A1since S 660he will not be there on timeR1AGN0Note: Do not award the R mark without the preceding A mark.[7 marks]

– 13 –7.M15/5/MATME/SP2/ENG/TZ2/XX/Mattempt to set up equationf g , kx 2 kx x 0.8egrearranging their equation to equal zeroegkx 2 kx x 0.8 0, kx 2 x (k 1) 0.8 0(M1)M1evidence of discriminant (if seen explicitly, not just in quadratic formula)b 2 4ac , (k 1) 2 4k 0.8, D 0eg(M1)correct discriminanteg(k 1) 2 4k 0.8, k 2 5.2k 1(A1)evidence of correct discriminant greater than zeroegk 2 5.2k 1 0 , (k 1) 2 4k 0.8 0 , correct answerboth correct valueseg0.2, 5correct answeregk 0.2, k 0, k 5R1(A1)A2N3[8 marks]

– 14 –M15/5/MATME/SP2/ENG/TZ2/XX/MSection B8.Note: The values of p and q found in (a) are used throughout the question. Please check FTcarefully on their values.(a)attempt to find intersectionegf g(M1) p 1, q 3(b)(c)f ′( p ) 1(i)correct approach to find the gradient of the normalm1m2 1 , egA1A1N3[3 marks]A2N2[2 marks](A1)1,correct value of 1f ′( p )EITHERattempt to substitute coordinates (in any order) and correctnormal gradient to find c(M1)13 1 c , 1 1 3 cf ′( p )egc 2(A1)y x 2A1N2ORattempt to substitute coordinates (in any order) and correctnormal gradient into equation of a straight lineegcorrect workingy ( x 1) 3eg(ii)(d)(A1)y x 2A1N2(0, 2)A1N1[5 marks]appropriate approach involving subtractioneg ba( L g ) dx , ( 3x2 ( x 2) )substitution of their limits or functioneg(M1)1( x 1) , y 1 1 ( x 3)y 3 f ′( p ) p0( L g ) dx ,area 1.5 ( ( x 2) 3x )(M1)(A1)2A1N2[3 marks]Total [13 marks]

– 15 –9.M15/5/MATME/SP2/ENG/TZ2/XX/MNote: There may be slight differences in answers, depending on which values candidatescarry through in subsequent parts. In particular there are a number of ways of doing (d).Accept answers that are consistent with their working.(a)valid approachegL µσ(M1), using a value for σ , using 68% and 95%correct workingP ( 1 Z 2) , correct probabilities ( 0.6826 0.1359 )P (50 σ L 50 2σ ) 0.818594P (50 σ L 50 2σ ) 0.819(b)A1z 1.9599653.92 50σ 1.95996 , σ 2.00004σ 2.00(c)N2[3 marks](A1)A1correct equation

Instructions to Examiners (red changed since M13) Abbreviations . M Marks awarded for attempting to use a valid Method; working must be seen. (M) Marks awarded for a valid Method; may be implied by correct subsequent working. A . Marks awarded for an . Answer . or for . Accuracy; often dependent on preceding