N4/5/MATHL/HP/ENG/TZ0/XX/M MARKSCHEME November 04 MATHEMATICS Higher Level Paper pages
N4/5/MATHL/HP/ENG/TZ0/XX/M This markscheme is the property of the International Baccalaureate and must not be reproduced or distributed to any other person without the authorization of the IB Assessment Centre.
N4/5/MATHL/HP/ENG/TZ0/XX/M!()*)*+,-.(/* Abbreviations M (M) A (A) R N Marks awarded for attempting to use a correct Method; working must be seen. Marks awarded for Method; may be implied by correct subsequent working. Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks. Marks awarded for an Answer or for Accuracy; may be implied by correct subsequent working. Marks awarded for clear Reasoning. Marks awarded for correct answers if no working shown. Answer given in the question and so no marks are awarded. Using the markscheme General Mark according to RM Assessor instructions and the document Mathematics HL: Guidance for e- marking November 04. It is essential that you read this document before you start marking. In particular, please note the following. Marks must be recorded using the annotation stamps. Please check that you are entering marks for the right question. If a part is completely correct, (and gains all the must be seen marks), use the ticks with numbers 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 RM Assessor. Method 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, as A mark(s) depend on the preceding M mark(s), if any. Where M and A marks are noted on the same line, for example, M, this usually means M for an attempt to use an appropriate method (for example, substitution into a formula) and for using the correct values. Where the markscheme specifies (M), N, etc, do not split the marks. Once a correct answer to a question or part-question is seen, ignore further working.
4 N4/5/MATHL/HP/ENG/TZ0/XX/M N marks Award 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 it penalizes candidates for not following the instruction to show their working. 4 Implied marks Implied marks appear in brackets, for example,, and can only be awarded if correct work is seen or if implied 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. 5 Follow through marks Follow through (FT) marks are awarded where an incorrect answer from one part of a question is used correctly in subsequent part(s). To award FT marks, there must be working present and not just 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 FT marks. If the error leads to an inappropriate value (for example, sin! =.5), do not award the mark(s) for the final answer(s). Within a question part, once an error is made, no further dependent A marks can be awarded, but M marks may be awarded if appropriate. Exceptions to this rule will be explicitly noted on the markscheme. 6 Mis-read If a candidate incorrectly copies information from the question, this is a mis-read (MR). A candidate should be penalized only once for a particular mis-read. Use the MR stamp to indicate that this has been 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 fewer marks. If the MR leads to an inappropriate value (for example, sin! =.5 ), do not award the mark(s) for the final answer(s). 7 Discretionary marks (d) An examiner uses discretion to award a mark on the rare occasions when the markscheme does not cover the work seen. In such cases the annotation DM should be used and a brief note written next to the mark explaining this decision.
5 N4/5/MATHL/HP/ENG/TZ0/XX/M 8 Alternative methods Candidates will sometimes use methods other than those in the markscheme. Unless the question specifies 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, METHOD, 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 alternatives start and finish. 9 Alternative forms Unless 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 brackets 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 the bracket or the form in brackets (if it is seen). Example: for differentiating f () x = sin(5x! ), the markscheme gives:!( ) = ( cos(5 ) ) 5 ( 0cos(5x ) ) f x x =! Award for ( cos(5x! ) ) 5, even if 0cos(5x! ) is not seen. 0 Accuracy of Answers Candidates 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 the required accuracy. When this is not specified in the question, all numerical answers should be given exactly or correct to three significant figures. Please check work carefully for FT. Crossed out work If a candidate has drawn a line through work on their examination script, or in some other way crossed out their work, do not award any marks for that work.
6 N4/5/MATHL/HP/ENG/TZ0/XX/M Calculators A GDC is required for paper, but calculators with symbolic manipulation features (for example, TI-89) are not allowed. Calculator notation The Mathematics HL guide says: Students must always use correct mathematical notation, not calculator notation. Do not accept final answers written using calculator notation. However, do not penalize the use of calculator notation in the working. More than one solution Where a candidate offers two or more different answers to the same question, an examiner should only mark the first response unless the candidate indicates otherwise. 4. Candidate work Candidates are meant to write their answers to Section A on the question paper (QP), and Section B on answer booklets. Sometimes, they need more room for Section A, and use the booklet (and often comment to 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 some rough work here which they assume will be ignored. If they have solutions on the answer booklets, there is no need to look at the QP. However, if there are whole questions or whole part solutions missing on answer booklets, please check to make sure that they are not on the QP, and if they are, mark those whole questions or whole part solutions that have not been written on answer booklets.
7 N4/5/MATHL/HP/ENG/TZ0/XX/M SECTION A. n = 4!! and n = ()() use of cos! = n in n n 7! cos! = = 9 7 99 ()() Note: Award for a correct numerator and for a correct denominator.!! = 69 Note: Award for. Total [6 marks]. (a) P( X > x) = 0.99 ( = P( X < x) = 0.0)! x = 54.6(cm) (b) P(60.5! X! 60.5) () = 0.066 [ marks] [ marks] Total [5 marks]
8 N4/5/MATHL/HP/ENG/TZ0/XX/M k!. f i x i i= + x + y +0 +7 use of µ = to obtain = 8 n 5 x+ y= EITHER k f i (x i!µ) use of! i= (! 6) + ( x! 8) + ( y! 8) + + 9 = to obtain = 7.6 n 5 ( x! 8) + ( y! 8) = 7 OR k fx i i i= use of! = µ to obtain + x + y +0 +7! 8 = 7.6 n 5 x + y = 65 THEN attempting to solve the two equations x = 4 and y = 7 (only as x< y) N4 Note: Award A0 for x = 7 and y = 4. Note: Award (M0)A0 for x+ y=! x= 4 and y= 7. Total [6 marks]
9 N4/5/MATHL/HP/ENG/TZ0/XX/M 4. METHOD attempt to set up (diagram, vectors) correct distances x = 5 t, y = 0t () () the distance between the two cyclists at time t is s = (5 t) + (0 t) = 5t (km) ds 5 dt ) hence the rate is independent of time METHOD attempting to differentiate dx dy ds x + y = s dt dt dt x + y = s implicitly () the distance between the two cyclists at time t is (5 t) + (0 t) = 5t (km) () ds (5 t)(5) + (0 t)(0) = (5 t) d t M Note: Award M for substitution of correct values into their equation involving d s dt. ds 5 dt ) hence the rate is independent of time METHOD s = x + y dx dy x + y ds = dt dt dt x + y () () Note: Award M for attempting to differentiate the expression for s. d s (5 t)(5) + (0 t)(0) = d t (5 t) + (0 t) M Note: Award M for substitution of correct values into their d s dt. ds d t = 5 (km h ) hence the rate is independent of time Total [5 marks]
0 N4/5/MATHL/HP/ENG/TZ0/XX/M 5. (a) attempting to find a normal to! eg! ( 8 6! ( 8 6 = 7! () r i! = 5 i! M 4 x y z! + = (or equivalent) [4 marks] (b) l : r = 4 0 8! + t!, t (! () attempting to solve 4 + t!t 8 + t i! = 4 for t ie 9t + 6 = 4 for t M 4 t =! 4, 8, 0! [4 marks] Total [8 marks]
N4/5/MATHL/HP/ENG/TZ0/XX/M 6. using pa= ( )! 7 to obtain ( a )(a a 7) 0 a a 5a 7 0 + + + = M +! + = () Note: Award M for a cubic graph with correct shape and for clearly showing that the above cubic crosses the horizontal axis at (!, 0) only. a =! EITHER showing that a a 7 0! + = has no real (two complex) solutions for a R OR showing that a + a + 5a + 7 = 0 has one real (and two complex) solutions for a R Note: Award R for solutions that make specific reference to an appropriate graph. Total [6 marks] u u 7. (a) using r = u = u to form a + d a + 6d = a a + d + = + aa ( 6 d) ( a d) d( d! a) = 0 (or equivalent) a since d! 0 d = [ marks] (b) a substituting d = into a+ 6d = and solving for a and d a = and d = 4 8 () r = n n! 4 + (n ) 8 ( ( ( ) 00 () attempting to solve for n n!.68 so the least value of n is [6 marks] Total [9 marks]
N4/5/MATHL/HP/ENG/TZ0/XX/M t 8. (a)! = 0 t = 6(s) Note: Award A0 if either t =! 0.6 or t = 4.4 or both are stated with t = 6. [ marks] (b) let d be the distance travelled before coming to rest 4 6 t d = 5!( t! ) dt+! dt () 0 4 Note: Award M for two correct integrals even if the integration limits are incorrect. The second integral can be specified as the area of a triangle. 47 d = ( = 5.7)(m) () T t attempting to solve! ( dt = 47 6 (or equivalent) for T M T =.9(s) [5 marks] Total [7 marks]
N4/5/MATHL/HP/ENG/TZ0/XX/M 9. (a) each triangle has area sin 8 x! (use of absin C ) n! there are n triangles so A= nx sin 8 n! 4 8 nx sin! n C =!x n! so C = sin! n [ marks] (b) n! attempting to find the least value of n such that sin 0.99! n > n = 6 nsin! attempting to find the least value of n such that n! + cos! > 0.99! n n = (and so a regular polygon with sides) n! Note: Award (M0)A0 if sin > 0.99 is not considered! n nsin! and n! + cos! > 0.99 is correctly considered.! n Award (M0)A0 for n = 6. [4 marks] (c) EITHER for even and odd values of n, the value of C seems to increase towards the limiting value of the circle ( C = ) ie as n increases, the polygonal regions get closer and closer to the enclosing circular region R OR the differences between the odd and even values of n illustrate that this measure of compactness is not a good one. R [ mark] Total [8 marks]
4 N4/5/MATHL/HP/ENG/TZ0/XX/M SECTION B 0. (a) use of A= qr sin! to obtain ( )(5 ) sin 0 o A= x+! x M = ( )(5 0 ) 4 x x x ( 8 A = x! x + 5 x + 50) 4 [ marks] da x 6x 5 (x )( x 5) dx = 4! + = 4!! (b) (i) ( ) (ii) METHOD EITHER da dx =! 4! OR! 6 + 5 = 0 M da dx =! 4!!! 5 = 0 THEN M so d A 0 dx = when x = METHOD solving d A = 0 for x dx M! < x< 5 x= so d A 0 dx x = METHOD a correct graph of d A dx M the graph clearly showing that d A 0 dx x = so d A 0 dx x = [ marks] continued
5 N4/5/MATHL/HP/ENG/TZ0/XX/M Question 0 continued (c) (i) d A ( x 8) dx =! d A for x =,.5( 0) dx R so x = gives the maximum area of triangle PQR (ii) 4 (.7)(cm ) A max = = 7 (iii) PQ = 7 (cm) and PR = 4! (cm) () QR = 7! + 4 4! 7!! 4 cos0! () = 9.70 QR = 9.8(cm) [7 marks] Total [ marks]
! 0.6. (a) (i) P( X 0) 0.549( e ) 6 N4/5/MATHL/HP/ENG/TZ0/XX/M = = = (ii) P( X! ) = P( X ) P( X! ) = 0.0 [ marks] (b) EITHER using Y! Po() OR using 5 (0.549) THEN! ( ) P( Y = 0) = 0.0498 = e [ marks] continued
7 N4/5/MATHL/HP/ENG/TZ0/XX/M Question continued (c) P( X = 0) (most likely number of complaints received is zero) EITHER calculating P( X = 0) = 0.549 and P( X = ) = 0.9 M OR sketching an appropriate (discrete) graph of P( X = x) against x M OR finding P( X 0) e! 0.6 = = and stating that P( 0) 0.5 X = > M OR µ using P( X = x) = P( X = x! ) where µ < M x [ marks] (d) P(X = 0) = 0.8(! e! = 0.8) ()! = 0. = ln 5 4,=!ln 4 5 [ marks] Total [0 marks]
8 N4/5/MATHL/HP/ENG/TZ0/XX/M. (a) P(Ava wins on her first turn) = [ mark]! (b) P(Barry wins on his first turn) = 4 = ( = 0.444) 9 [ marks] (c) P(Ava wins in one of her first three turns) = +!! +!!!! M Note: Award M for adding probabilities, award for a correct second term and award for a correct third term. Accept a correctly labelled tree diagram, awarding marks as above. 0 = ( = 0.44) 4 [4 marks] (d) P(Ava eventually wins) = +!! +!!!! a using S =! r with a = and + () r = () 9 Note: Award for using r =. 9 a S =! r and award () for a = and = ( = 0.49) 7 [4 marks] Total [ marks]
9 N4/5/MATHL/HP/ENG/TZ0/XX/M b. (a) attempting to use V =!! x dy attempting to express a x in terms of y ie x = 4( y+ 6) h for y= h, V = 4!! y+ 6dy 0! V = 4! h + 6h [ marks] (b) (i) METHOD dh dh dv =! dt dv dt dv 4! ( h 6) dh () dh dt = 4!(h + 6)! 50 h!(h + 6) Note: Award M for substitution into d h d h d V =!. dt dv dt M dh 50 h =! dt 4! ( h+ 6) METHOD dv dt = 4!(h +6) dh dt!50 h dh = 4!(h +6)!(h +6) dt dh dt = 4!(h +6)! 50 h!(h +6) (implicit differentiation) (or equivalent) M dh dt =! 50 h 4! (h +6) (ii) dt 4! ( h+ 6) =! dh 50 h 4! ( h + 6) t =! 50 h dh 4! ( h + h+ 56) t =! d h 50 h t =!4! h + h + 56h! 50 ( dh continued
0 N4/5/MATHL/HP/ENG/TZ0/XX/M Question continued (iii) METHOD t =!4! 50 0 h + h + 56h! ( 48 dh t = 688.756 (s) () 45 minutes (correct to the nearest minute) METHOD t =!4! 50 5 5 h + 64 h + 5h + c when t = 0,h = 48! c = 688.756... c = 4! 5 50 5 48 + 64 48 + 5 48 (( when h = 0, t = 688.756... t = 4! 5 50 5! 48 + 64! 48 + 5! 48 () 45 minutes (correct to the nearest minute) [0 marks] (c) EITHER the depth stabilises when dv dt = 0 ie 8.5! 50 h!(h +6) = 0 attempting to solve 8.5! 50 h!(h +6) = 0 for h R OR the depth stabilises when dh dt = 0 ie 50 h 8.5! = 0 4!(h +6)!(h +6) attempting to solve 50 h 8.5! = 0 for h 4!(h +6)!(h +6) R THEN h = 5.06 (cm) [ marks] Total [6 marks]
N4/5/MATHL/HP/ENG/TZ0/XX/M 4. (a) METHOD squaring both equations M 9sin B+ 4sin Bcos C + 6cos C = 6 () 9cos B 4cosBsin C 6sin C + + = () adding the equations and using cos! + sin! = to obtain 9 + 4sin ( B+ C) + 6 = 7 M 4(sin BcosC+ cos Bsin C) = 4sin ( B+ C) = () sin ( B+ C) = METHOD substituting for sin B and cos B to obtain sin( B + C) = 6! 4cosC! 4sinC cosc + sinc M 6cosC + sinc! 4 = (or equivalent) substituting for sin C and cosc to obtain sin( B + C) = sin B 6! sin B! cos B 4 + cos B 4 M cos B + 6sin B! = (or equivalent) 4 sin B+ C : Adding the two equations for ( ) (8sin B+ 4cos C) + (4sin C+ cos B)! 5 sin ( B+ C) = 6 +! 5 sin ( B+ C) = 4 () sin ( B+ C) = METHOD substituting for sin B and sin C to obtain sin( B + C) = 6! 4cosC! cos B cosc + cos B 4 substituting for cos B and cosc to obtain sin( B + C) = sin B 6! sin B! 4sinC 4 + sinc Adding the two equations for sin ( B C) + : 6cosC + sinc! 4 6sin B + cos B! sin( B + C) = + 4 (or equivalent) M M
N4/5/MATHL/HP/ENG/TZ0/XX/M (8sin B + 4cosC) + (4sinC + cos B)! 5 sin( B + C) = 6 +! 5 sin( B + C) = 4 sin( B + C) = (b) sin A= sin ( 80 o!( B+ C) ) so sin A= sin ( B+ C) R sin ( B+ C) =! sin A=! A = 0 o or A = 50 () [6 marks] if A = 50 o, then B < 0 R for example, sin B+ 4cosC < + 4 < 6, ie a contradiction R only one possible value ( A = 0 ) [5 marks] Total [ marks]