koorio.com

海量文库 文档专家

海量文库 文档专家

- Mathematics HL paper 2 TZ1 May 2010 MS
- Mathematics HL paper 2 TZ1ms
- Mathematics HL paper 1 TZ1
- Mathematics HL paper 2 TZ1
- Mathematics HL paper 1 TZ1 May 2010 MS
- Mathematics HL paper
- Chemistry HL paper 1 TZ1 (2)
- Mathematics HL paper 3 TZ1 series and differential equationsms
- Mathematics HL paper 1 TZ2
- Mathematics HL paper 1 marksheme
- Mathematics HL paper 1 marksheme
- Mathematics HL paper 1 TZ2
- Chinese A1 HL paper 2
- Mathematics 1993 Paper 2
- AL Pure Mathematics 2008 Paper 2 Marking Scheme

M13/5/MATHL/HP2/ENG/TZ1/XX/M

MARKSCHEME

May 2013

MATHEMATICS

Higher Level

Paper 2

15 pages

–2–

M13/5/MATHL/H

P2/ENG/TZ1/XX/M

This markscheme is confidential and for the exclusive use of examiners in this examination session. It 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.

–3–

M13/5/MATHL/HP2/ENG/TZ1/XX/M

Instructions to Examiners Abbreviations M (M) A (A) R N AG 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 1 General Write the marks in red on candidates’ scripts, in the right hand margin. ? Show the breakdown of individual marks awarded using the abbreviations M1, A1, etc. ? Write down the total for each question (at the end of the question) and circle it. 2 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 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 an attempt to use an appropriate method (e.g. 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. 3 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–

4 Implied marks

M13/5/MATHL/HP2/ENG/TZ1/XX/M

Implied marks appear in brackets e.g. (M1), 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 (e.g. sin? ? 1.5 ), 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). Apply a MR penalty of 1 mark to that question. Award the marks as usual and then write –1(MR) next to the total. Subtract 1 mark from the total for the question. A candidate should be penalized only once for a particular mis-read. ? 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 (e.g. sin? ? 1.5 ), 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. The mark should be labelled (d) and a brief note written next to the mark explaining this decision. 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 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 alternatives start and finish.

–5–

9 Alternative forms

M13/5/MATHL/HP2/ENG/TZ1/XX/M

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 ) ? 2sin (5 x ? 3) , the markscheme gives:

f ? ( x) ? ? 2cos(5x ? 3) ? 5

? ? 10cos(5x ? 3) ?

A1

Award A1 for ? 2cos(5x ? 3) ? 5 , even if 10cos(5 x ? 3) is not seen. 10 Accuracy of Answers The method of dealing with accuracy errors on a whole paper basis by means of the Accuracy Penalty (AP) no longer applies. Instructions to examiners about such numerical issues will be provided on a question by question basis within the framework of mathematical correctness, numerical understanding and contextual appropriateness. The rubric on the front page of each question paper is given for the guidance of candidates. The markscheme (MS) may contain instructions to examiners in the form of “Accept answers which round to n significant figures (sf)”. Where candidates state answers, required by the questio n, to fewer than n sf, award A0. Some intermediate numerical answers may be required by the MS but not by the question. In these cases only award the mark(s) if the candidate states the answer exactly or to at least 2sf. 11 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. 12 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.

–6–

SECTION A 1.

5 ? 6 ? 6k ? 7 ? 3 ? 8 ? 1 ? 9 ? 2 ? 10 ? 1 ? 6.5 (or equivalent) 13 ? k

M13/5/MATHL/HP2/ENG/TZ1/XX/M

(M1)(A1)(A1)

Note: Award (M1)(A1) for correct numerator, and (A1) for correct denominator.

0.5k ? 2.5 ? k ? 5

A1 [4 marks]

2.

METHOD 1 determinant ? 0 k ( ?2 ? 16) ? (0 ? 12) ? 2(0 ? 3) ? 0 ?18k ? 18 ? 0 k ?1 METHOD 2 writes in the form M1 (M1)(A1) (A1) A1

?k 1 2 4? ? 0 ?1 4 5 ? ? ? ? 3 4 2 1? ? ?

(or attempts to solve simultaneous equations)

(M1)

Having two 0’s in first column (obtaining two equations in the same two variables)

M1 A1

2 4 ?k 1 ? ? 0 ?1 4 5 ? ? ? (or isolating one variable) ? 0 0 18k ? 18 21k ? 27 ? ? ?

Note: The A1 is to be awarded for the 18k ?18 . The final column may not be seen.

k ?1

(M1)A1 [5 marks]

3.

Let X represent the length of time a journey takes on a particular day. (a) (b)

P ( X ? 15) ? 0.0912112819? ? 0.0912

(M1)A1 (M1)

Use of correct Binomial distribution N ~ B(5,0.091...) 1 ? 0.0912112819?? 0.9087887181? 1 ? (0.9087887181?)5 ? 0.380109935? ? 0.380

(M1)A1

Note: Allow answers to be given as percentages. [5 marks]

–7–

4. volume ? ? ? x 2dy

x ? arcsin y ? 1

1 0

M13/5/MATHL/HP2/ENG/TZ1/XX/M

(M1) (M1)(A1) A1

volume ? ? ? (arcsin y ? 1)2 dy Note: A1 is for the limits, provided a correct integration of y.

? ?????????? ? 8.20

A2

N5 [6 marks]

5.

1 2 r ?1 ? 7 2

M1

r ? 3.7? ? 14 (or 37… mm)

1? ? ? ?1? ? or 2r sin ? height ? 2r cos ? ? ? 2? ? 2 ? ? 3.59 or anything that rounds to 3.6 so the dimensions are 3.7 by 3.6 (cm or 37 by 36 mm)

?

?

(A1) (M1)(A1) A1 A1 [6 marks]

6.

other root is 2 ? i a quadratic factor is therefore ( x ? 2 ? i )( x ? 2 ? i )

? x2 ? 4 x ? 5 x ? 1 is a factor ( x ? 2)2 is a factor

(A1) (M1) A1 A1 A1 (M1) A1

p ( x ) ? a ( x ? 1)( x ? 2)2 ( x 2 ? 4 x ? 5) 1 p (0) ? 4 ? a ? 5 1 p ( x ) ? ( x ? 1)( x ? 2) 2 ( x 2 ? 4 x ? 5) 5

[7 marks]

–8–

7. (a) let the distance the cable is laid along the seabed be y y 2 ? x 2 ? 2002 ? 2 ? x ? 200cos600 (or equivalent method) y 2 ? x 2 ? 200 x ? 40000 cost ? C ? 80 y ? 20 x

M13/5/MATHL/HP2/ENG/TZ1/XX/M

(M1) (A1) (M1) A1 [4 marks]

C ? 80( x 2 ? 200 x ? 40000) 2 ? 20 x

1

(b)

? x ? 100 ?

x ? 55.2786 … ? 55 (m to the nearest metre)

2000

(A1)A1

?

[2 marks] Total [6 marks] 8. (a) the three girls can sit together in 3! ? 6 ways (A1) this leaves 4 ‘objects’ to arrange so the number of ways this can be done is 4! (M1) so the number of arrangements is 6 ? 4! ? 144 A1 [3 marks] (b) Finding more than one position that the girls can sit Counting exactly four positions number of ways = 4 ? 3!? 3! ? 144 (M1) (A1) M1A1

N2 [4 marks]

Total [7 marks]

? ? b2 ? 4ac ? 4k 2 ? 4 ? 3 ? (k ? 1) ? 4k 2 ? 12k ? 12

9.

(a)

M1A1

Note: Award M1A1 if expression seen within quadratic formula. EITHER

144 ? 4 ? 4 ?12 ? 0 ? always positive, therefore the equation always has two distinct real roots (and cannot be always negative as a ? 0 )

OR sketch of y ? 4k 2 ? 12k ? 12 or y ? k 2 ? 3k ? 3 not crossing the x-axis ? always positive, therefore the equation always has two distinct real roots OR write ? as 4(k ? 1.5)2 ? 3 ? always positive, therefore the equation always has two distinct real roots

M1 R1

M1 R1

M1 R1 [4 marks] continued …

–9–

Question 9 continued (b) closest together when ? is least minimum value occurs when k ? 1.5

M13/5/MATHL/HP2/ENG/TZ1/XX/M

(M1) (M1)A1 [3 marks] Total [7 marks]

10.

(a)

X ? Po(0.25T) Attempt to solve P ( X ? 3) ? 0.6 T ? 12.8453? ? 13 (minutes)

(A1) (M1) A1 [3 marks]

Note: Award A1M1A0 if T found correctly but not stated to the nearest minute.

(b)

let X 1 be the number of cars that arrive during the first interval and X 2 be the number arriving during the second. X 1 and X 2 are Po(2.5) P (all get on) ? P( X 1 ? 3) ? P ( X 2 ? 3) ? P ( X 1 ? 4) ? P ( X 2 ? 2) ? P ( X 1 ? 5) ? P ( X 2 ? 1) ? P ( X 1 ? 6) ? P ( X 2 ? 0) ? 0.573922 ? ? 0.072654 ? ? 0.019192 ? ? 0.002285? ? 0.668 (053?)

(A1) (M1) (M1) A1 [4 marks] Total [7 marks]

– 10 –

SECTION B

M13/5/MATHL/HP2/ENG/TZ1/XX/M

11.

(a)

?8 ? ? PQ ? ? ?6? ?4? ? ?

?

(A1)

? ?3 ? ? 8 ? ? ? ? ? equation of line: r ? ? ?1 ? ? t ? 6 ? (or equivalent) ? 2? ? 4? ? ? ? ?

Note: Award M1A0 if r = is omitted.

M1A1

[3 marks]

(b)

METHOD 1

x : ?4 ? 5s ? ?3 ? 8t 2 s ? ?1 ? 6t y: 4 ? 2 ? 4t z: solving any two simultaneously t ? 0.5, s ? 1 (or equivalent) verification that these values give R when substituted into both equations (or that the three equations are consistent and that one gives R)

METHOD 2

M1 M1 A1 R1

(1, 2, 4) is given by t ? 0.5 for L1 and s ? 1 for L2 because (1, 2, 4) is on both lines it is the point of intersection of the two lines

M1A1A1 R1 [4 marks]

(c)

? 5? ? 4? ? 2 ??? 3 ? ? 26 ? 29 ? 29 cos ? ? ?? ? ? 0? ? 2? ? ?? ? 26 cos ? ? 29 ? ? 0.459 or 26.3?

M1

(A1)

A1

[3 marks]

continued …

– 11 –

Question 11 continued

M13/5/MATHL/HP2/ENG/TZ1/XX/M

(d)

? ?3 ? ? 1 ? ? ?4 ? ? ? ? ? ? ? RP ? ? ? ?1? ? ? 2 ? ? ? ?3 ? , RP ? 29 ? 2 ? ? 4 ? ? ?2 ? ? ? ? ? ? ?

?

(M1)A1

?8? ? ? Note: This could also be obtained from 0.5 6 ? ? ? 4? ? ?

EITHER

? ?4 ? ? 1 ? ? ?5 ? ? ? ? ? ? ? RS ? 29 RS1 ? ? 1 ? 0 ? ? ? 2 ? ? ? ?2 ? , ? 4 ? ? 4? ? 0 ? ? ? ? ? ? ? ? ?4 ? ? 5 ? ? ? ? ? ? ? ? OS2 ? OS1 ? 2S1R ? ? ? 0 ? ? 2?2? ? 4 ? ? 0? ? ? ? ?

?

A1

M1A1

? ? 1? ? 5?? ? ? ? ? ? ? ? ?? ? or OS2 ? OR ? S1R ? ? 2 ? ? ? 2 ? ? ? 4? ? 0?? ? ? ? ? ?? ? ? 6? ? ?? ?4? ?4? ? ?

S2 is (6, 4, 4)

OR

A1

? ?4 ? 5s ? ? 1 ? ? 5s ? 5 ? ? 2s ? ? ? 2 ? ? ? 2s ? 2 ? ? ? ? ? ? ? ? 4 ? ? 4? ? 0 ? ? ? ? ? ? ? 2 2 (5s ? 5) ? (2 s ? 2) ? 29

29 s ? 58s ? 29 ? 29 s ( s ? 2) ? 0 , s ? 0, 2 (6, 4, 4) (and ( ?4, 0, 4) )

2

M1 M1A1

A1

Note: There are several geometrical arguments possible using information obtained in previous parts, depending on what forms the previous answers had been given. [6 marks]

– 12 –

(e) EITHER midpoint of

?

M13/5/MATHL/HP2/ENG/TZ1/XX/M

? PS1 ?

is M ( ?3.5, ? 0.5, 3)

M1A1 A1

? ?4.5 ? ? RM ? ? ? ?2.5 ? ? ?1 ? ? ?

OR

? ?5 ? ? RS1 ? ? ? ?2 ? ? 0? ? ?

?

M1

? ?

the direction of the line is RS1 ? RP

? ?5 ? ? ?4 ? ? ?9 ? ? ?2 ? ? ? ?3 ? ? ? ?5 ? ? ? ? ? ? ? ? 0 ? ? ?2 ? ? ?2 ? ? ? ? ? ? ?

THEN the equation of the line is: ?1 ? ? 9 ? ? ? ? r ?? ? 2 ? ? t ? 5 ? or equivalent ? 4? ? 2? ? ? ? ? Note: Marks cannot be awarded for methods involving halving the angle, unless it is clear that the candidate considers also the equation of the plane of L1 and L2 to reduce the number of parameters involved to one (to obtain the vector equation of the required line).

M1A1

A1

[4 marks] Total [20 marks]

– 13 –

12. (a)

M13/5/MATHL/HP2/ENG/TZ1/XX/M

A1A1A1 Note: Award A1 for general shape, A1 for correct maximum and minimum, A1 for intercepts. Note: Follow through applies to (b) and (c). [3 marks] (b)

? 5? 7 ? 0 ? t ? 0.785, ? or 0 ? t ? ? 3 ? ? (allow t ? 0.785 ) ? 5? 7 ? and t ? 2.55 ? or t ? ? 3 ? ?

A1

A1 [2 marks]

(c)

? 5? 7 ? 0 ? t ? 0.785, ? or 0 ? t ? ? 3 ? ? (allow t ? 0.785 ) ? 5? 7 ? 2 ? t ? 2.55, ? or 2 ? t ? ? 3 ? ? t?3

A1

A1 A1 [3 marks]

(d)

position of A:

x A ? ? t 3 ? 5t 2 ? 6t dt

(M1) A1 R1 [3 marks] continued …

1 5 x A ? t 4 ? t 3 ? 3t 2 ( ? c ) 4 3 when t ? 0 , x A ? 0 so c ? 0

– 14 –

Question 12 continued (e)

M13/5/MATHL/HP2/ENG/TZ1/XX/M

dvB 1 ? ?2vB ? ? dvB ? ? ?2dt dt vB

(M1) (A1) (M1)

?2 t

ln vB ? ?2t ? c

vB ? Ae ?2 t

vB ? ?20 when t ? 0 so vB ? ?20e

A1 [4 marks]

(f)

xB ? 10e ?2 t ( ? c )

(M1)(A1)

?2 t

xB ? 20 when t ? 0 so xB ? 10e ? 10 1 5 meet when t 4 ? t 3 ? 3t 2 ? 10e ?2 t ? 10 4 3 t ? 4.41(290?)

(M1)A1 (M1) A1 [6 marks] Total: [21 marks]

13.

(a)

f (2) ? 9

(A1)

1 3

? 2 3

f ?1 ( x ) ? ( x ? 1)

A1 (M1) A1 (M1) A1

1 ( f ?1 )? ( x ) ? ( x ? 1) 3 1 ( f ?1 )? (9) ? 12 2 ? f ( x ) ? 3x 1 1 1 ? ? f ?(2) 3 ? 4 12

Note: The last M1 and A1 are independent of previous marks. [6 marks] (b)

g ?( x ) ? e x ? 2 x 2e x g ?( x ) ? 0 as each part is positive

2 2

M1A1 R1 [3 marks]

continued …

– 15 –

Question 13 continued (c) to find the x-coordinate on y ? g ( x ) solve

2 ? xe x x ? 0.89605022078…

2

M13/5/MATHL/HP2/ENG/TZ1/XX/M

(M1) (A1)

1 g ?(0.896?)

2

gradient = ( g ?1 )?(2) ?

(M1) A1

?

e

(0.896?)2

?1 ? 2 ? (0.896?) ?

1

? 0.172 to 3sf

(using the

dy function on gdc g ?(0.896?) ? 5.7716028? dx

1 ? 0.173 ) g ?(0.896?)

[4 marks] (d) (i)

( x 3 ? 1)e( x ?1) ? 2 x ? ?0.470191?

3 2

A1 A1

(ii)

METHOD 1

( g f )?( x ) ? 3x 2e( x

3

?1)2

? 2( x

3

? 1) 2 ? 1?

(M1)(A1) (A1) A1

( g f )'( ?0.470191?) ? 3.85755? 1 h?(2) ? ? 0.259 (232 ?) 3.85755?

Note: The solution can be found without the student obtaining the explicit form of the composite function. METHOD 2

h ( x ) ? ( f ?1 g ?1 )( x )

A1 M1 M1

h?( x ) ? ( f ?1 )? ? g ?1 ( x) ? ? ( g ?1 )? ( x)

2 ? 1 ?1 g ( x ) ? 1? 3 ? ( g ?1 )? ( x) ? 3 2 ? 1 h?(2) ? ? g ?1 (2) ? 1? 3 ? ( g ?1 )? (2) 3 2 ? 1 ? (0.89605?? 1) 3 ? 0.171933? 3 ? 0.259 (232 ?)

?

A1

N4 [6 marks]

Total [19 marks]