Mark Scheme (FINAL) Summer Pearson Edexcel GCE In Core Mathematics 4 (6666/01)

Transcription

1 Mark Scheme (FINAL) Summer 07 Pearson Edexcel GCE In Ce Mathematics 4 (/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding bo. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes f employers. F further infmation visit our qualifications websites at Alternatively, you can get in touch with us using the details on our contact us page at Pearson: helping people progress, everywhere Pearson aspires to be the wld s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, f all kinds of people, wherever they are in the wld. We ve been involved in education f over 50 years, and by wking across 70 countries, in 00 languages, we have built an international reputation f our commitment to high standards and raising achievement through innovation in education. Find out me about how we can help you and your students at: Summer 07 Publications Code xxxxxxxx* All the material in this publication is copyright Pearson Education Ltd 07

3 General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded f what they have shown they can do rather than penalised f omissions. Examiners should mark accding to the mark scheme not accding to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not wthy of credit accding to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out wk should be marked UNLESS the candidate has replaced it with an alternative response.

4 PEARSON EDEXCEL GCE MATHEMATICS General Instructions f Marking. The total number of marks f the paper is 75. The Edexcel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded f knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used f crect ft cao crect answer only cso - crect solution only. There must be no errs in this part of the question to obtain this mark isw igne subsequent wking awrt answers which round to SC: special case oe equivalent (and appropriate) d dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper ag- answer given d The second mark is dependent on gaining the first mark 4. All A marks are crect answer only (cao.), unless shown, f example, as A ft to indicate that previous wrong wking is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft. 5. F misreading which does not alter the character of a question materially simplify it, deduct two from any A B marks gained, in that part of the question affected.

5 . If a candidate makes me than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out none are crossed out, mark all the attempts and sce the highest single attempt. 7. Igne wrong wking increct statements following a crect answer.

6 General Principles f Ce Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark f solving 3 term quadratic:. Factisation ( x bx c) ( x p)( x q), where pq c, leading to x = ( ax bx c) ( mx p)( nx q), where pq c and mn a, leading to x =. Fmula Attempt to use the crect fmula (with values f a, b and c). 3. Completing the square b Solving x bx c 0 : x q c 0, q 0, leading to x = Method marks f differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n ) Use of a fmula Where a method involves using a fmula that has been learnt, the advice given in recent examiners repts is that the fmula should be quoted first. Nmal marking procedure is as follows: Method mark f quoting a crect fmula and attempting to use it, even if there are small errs in the substitution of values. Where the fmula is not quoted, the method mark can be gained by implication from crect wking with values, but may be lost if there is any mistake in the wking. Exact answers Examiners repts have emphasised that where, f example, an exact answer is asked f, wking with surds is clearly required, marks will nmally be lost if the candidate rests to using rounded decimals.

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8 Question Number. (a) (a) Way (b) (c) Way (c) Way Scheme s Marks x 3t 4, y 5, t 0 t 3 dt, t dt = t- 3 ì = 3t = t - = ü í ý î t þ their dt divided by their d x dt their d y multiplied by their d t dt t 3 to give to give in terms of t in terms of t, simplified un-simplified, in terms of t. See note. A isw Award Special Case st if both d x and d y are stated crectly and explicitly. dt dt : You can recover the wk f part (a) in part (b). 8 y x 4 d x ( x 4) (3 t) Writes d y d x in the fm ( x 4), and writes d y as a function of t. Crect un-simplified simplified answer, in terms of t. See note. 5 t P, 7 x = - 5, y = -7 5 P, 7 seen implied. B and either Some attempt to substitute t = 0.5 into their d y which contains t in der to find m T and either ( ) y - "- 7" = "8" x - "- 5 " "-7" = ("8")("- 5 ") + c So, y = (their m T )x + "c" T : y = 8x +3 ì ít = x + 4 î 3 : their Þ y = 5-8 x + 4 x P, their ü Þý y = 5 - þ y P and their x ö Þ y = 5(x + 4) -8 x + 4 applies y - (their y P ) = (their m T )(x - their x P ) finds c from (their y P ) = (their m T )(their x P ) + c A isw and uses their numerical c in y = (their m T )x + c y = 8x +3 y = 3 + 8x A cso m T must be numerical values in der to award [3] So, y = 5x + x + 4, { x > -4 } y = 5x + x + 4 An attempt to eliminate t. See notes. Achieves a crect equation in x and y only A o.e. [] [] ( implied equation) A cso [3] ì ít = 5 - y Þ ü ý x = 8 î þ 5 - y - 4 An attempt to eliminate t. See notes. Achieves a crect equation in x and y only A o.e. Þ (x + 4)(5 - y) = 8 Þ 5x - xy + 0-4y = 8 { Þ 5x + = y(x + 4) } So, y = 5x + x + 4, { x > -4 } y = 5x + x + 4 ( implied equation) A cso [3] : Some all of the wk f part (c) can be recovered in part (a) part (b) 8

9 Question Number. (c) Way 3. (a) Scheme s Marks A full method leading to the value 3at - 4a + b y = = 3at 3t t - 4a - b = a - 4a - b of a being found Þ a = 5 3t 3t 4a b y a and a 5 A 3t 4a b b 4(5) (3) Both a 5 and b A 3 ö Condone = t f A 3 Question s You can igne subsequent wking following on from a crect expression f in terms of t. (b) - Using a changed gradient (i.e. applying - their ( their ) ) is M0. their Final A: A crect solution is required from a crect d y. Final A: You can igne subsequent wking following on from a crect solution. (c) st : A full attempt to eliminate t is defined as either rearranging one of the parametric equations to make t the subject and substituting f t in the other parametric equation (only the RHS of the equation required f M mark) rearranging both parametric equations to make t the subject and putting the results equal to each other. x 4 Award A f 5 y 3 equivalent. [3]

10 Question Number ì ï. ( + k x) -3 = -3 + k x ö í îï (a) { A = } Scheme s Marks = 8 + (-3) k xö 3 + (-3)(-3 -)! k xö ö ü ï +... ý, k > 0 þï 0.5, clearly identified as A as their answer to part (a) B cao Uses the x term of the binomial expansion to give [] (b) (c) ( 3)( 4) k 8! ( 3)( 4) k either! kx ( 3)( 4) (-3)(- 4) k ö either (their A)!, (-3)(- 4) k x ö (their A)! where (their A) ¹, o.e. 3 k 3 kx 5 ( 3)( 4) ( ) ( kx) ( -5 ) (-3)(-4) (k)!! ( 3)( 4) k So, k k 8 8! So, k = 9 k 9 cao A cso : k = ± 9 with no reference to k 9 only is A0 [3] Uses the x term of the binomial expansion to give either " " k ö ö k ö (their A)(-3) 8 (-3) (their A)(-3) k x ö ; where (their A) ¹, 3k () -4 (-3)(k) () -4 (-3)(k x) ì ö So, B = 8 (-3) 9 ö í Þ ü ý B = A cso î þ [] Question s NOTE IN THIS QUESTION IGNORE LABELLING AND MARK ALL PARTS TOGETHER. ( + k x) -3 = 8-3 kx + 3 k ö x +... = 8-3 kx + 3 k x k x k x ( 3)( 3 ) k x Writing down ( 3)...! gets (b) st Writing down 3 k x ( 3)( 3 ) k x kx ( ) ( 3)... 8! gets (b) st nd and (c) Writing down ( + k x) -3 { } = -3 + (-3)( -4 )(kx) + (-3)(-4) gets (b) st nd and (c) k x ö Writing down {( + k x) } -3 = (their A) + (-3) where (their A) ¹, gets (b) st nd and (c) ( -5 )(kx) + (-3)(-3 -)!. k x ö ö +...

11 . (b), (c) (their A) is defined as either their answer to part (a) their stated A =... Question s their " -3 " in their stated -3 + k x ö Give nd M0 in part (b) if (their A) = Give M0 in part (c) if (their A) =. (c) Allow f (their A)(-3). (b), (c) Award A0 f B = - 7 x 7 k 9 leading to B Give A0 f finding both B = - 7 their k from (b) Allow A f B = - 7 x followed by 7 B.875 is A0 ö and B = (without rejecting B = ) as their final answer. The A mark in part (c) is f a crect solution only. Be careful! It is possible to award M0A0 in part (c) f a solution leading to B = - 7. E.g. f(x) = ( + k x) -3 = -3 (+ k x) -3 = 8 + (-3) k x ( ) + (-3)(-4) ( k x) +...! ö = 8-3k 8 x + 3k 4 x +... leading to (a) A = 8, (b) k = 9 7, (c) B = -, gets (a) B, (b) M0A0 (c) M0A0-3 C 0 () C () - 4 (k x)+ -3 C () - 5 (k x) with the C terms not evaluated gets (b) st M0 nd M0 and (c) M0

12 Question Number 3. (a) x Scheme s Marks x y y x ( e ) At 0., y.854 (5 dp).854 B cao : Look f this value on the given table in their wking. [] (b) (0.) é+.75+ their ë ( ) ù û Outside brackets (0.) 0 B o.e. 5 F structure of éë... ù û (.483) (4 dp) 0 (c) { u = e x x = lnu Þ } (d) Way du = u(u + ) 3 u - 3 ö (u + ) du anything that rounds to.43 A du = ex du = u du = u du = u etc., and = (e x + ) (u + )u du See notes {x = 0} Þ a = e 0 Þ a = a and b e b e {x = } Þ b = e Þ b = e evidence of 0 and e NOTE: st B mark CANNOT be recovered f wk in part (d) NOTE: nd B mark CAN be recovered f wk in part (d) u(u + ) º A u + B Writing, o.e. (u + ) u(u + ) º A u + B (u + ) u(u + ) º P u + Q (u + ), Þ º A(u + ) + Bu o.e., and a complete method f finding the value of at least one of their A their B ( their P their Q) u = 0 Þ A = 3 Both their A = 3 and their B = -3. (Or their P = and their u = - Þ B = -3 Q = - with the fact of in front of the integral sign) Integrates M u ± N u ± k, M, N, k ¹ 0; = 3lnu - 3ln(u + ) = 3lnu - 3ln(u + 4) So éë e { 3lnu - 3ln(u + ) ù û } ( ) - ( 3ln -3ln3) = 3ln(e) -3ln(e + ) [: A proper consideration of the limit of u = is required f this mark] 33ln(e ) 3ln3 3(- ln(e + ) + ln3) e ö 3ln e + - 3ln ö 3 (i.e. a two term partial fraction) to obtain either ±l ln(a u) ± mln(b(u ± k)); l, m, a, b ¹ 0 Integration of both terms is crectly followed through from their M and from their N. dependent on the nd M mark Applies limits of e and ( their b and their a, where b > 0, b ¹, a > 0) in u applies limits of and 0 in x and subtracts the crect way round ln e e + ö 3-3ln 3 3e 3ln e ln 7e 3 ö (e + ) 3 B * B A A ft d see notes A cso : Allow e in place of e f the final A mark. [] : Give final A0 f 3-3lne + + 3ln3 (i.e. bracketing err) unless recovered. : Give final A0 f 3-3ln(e + ) + 3ln3-3ln, where 3ln has not been simplified to 0 : Give final A0 f 3lne - 3ln(e + ) + 3ln3, where 3lne has not been simplified to 3 [3] []

13 Question 3 s 3. (b) : Do not allow an extra y-value a repeated y value in their [ ] Do not allow an omission of a y-dinate in their [ ] f unless they give the crect answer of awrt.43, in which case both and A can be sced. A: Wking must be seen to demonstrate the use of the trapezium rule. (Actual area is ) Full marks can be gained in part (b) f awrt.43 even if B0 is given in part (a) Award BA f 0 (+.75) + (their ) = awrt.43 5 Bracketing mistakes: Unless the final answer implies that the calculation has been done crectly Award BM0A0 f (0.) + + (their ) +.75 (=.583) Award BM0A0 f (0.)( +.75) + (their ) (= ) Award BM0A0 f (0.)() + (their ) +.75 (=4.83) Alternative method: Adding individual trapezia é Area» 0. + ".854" + ".854" ê + ë =.483 B 0. and a divis of on all terms inside brackets First and last dinates once and two of the middle dinates inside brackets igning the A anything that rounds to (c) st B Must start from either y, with integral sign and (e x + ) (e x + ) and state either du = ex du and end at u(u + ) So, just writing du = ex and, with integral sign and du, with integral sign and du du du = u du = u du = u du, with integral sign and du, with no increct wking. = (e x + ) u(u + ) du is sufficient f st B Give nd B0 f b =.78..., without reference to a and b e b e You can also give the st B mark f using a reverse process. i.e. Proceeding from u(u + ) du to (e x + ), with no increct wking, and stating either du = ex du = u du = du = u u 3. (d) Give final A0 f 3-3ln(e + ) + 3ln3 simplifying to - ln(e + ) + ln3 (i.e. dividing their crect final answer by 3) Otherwise, you can igne increct wking (isw) following on from a crect exact value. A decimal answer of (without a crect exact answer) is final A0 e éë -3ln(u + ) + 3lnuù û followed by awrt.4 (without a crect exact answer) is final A0 ù ú û

14 Question 3 s Continued 3. (d) BE CAREFUL! Candidates will assign their own A and B f this question. A B Writing down in the fm with at least one of A B crect is st ( u ) u ( u ) u 3 3 Writing down as is st st A. ( u ) u ( u ) u 3 Condone u - 3 ö (u + ) du to give 3lnu - 3lnu + (po bracketing) f nd A. Award M0A0Aft f a candidate who writes down e.g. du = u(u + ) u + ö (u + ) du = lnu + ln(u + ) AS EVIDENCE OF WRITING AS PARTIAL FRACTIONS. u(u + ) Award M0A0M0A0 f a candidate who writes down du = lnu + ln(u + ) du = lnu + ln(u + ) u(u + ) u(u + ) WITHOUT ANY EVIDENCE OF WRITING as partial fractions. u(u + ) Award AA f a candidate who writes down du = 3lnu - 3ln(u + ) u(u + ) WITHOUT ANY EVIDENCE OF WRITING as partial fractions. u(u + ) If they lose the and find e du we can allow a maximum of A0AftA0 u(u + )

15 Question 3 s Continued 3. (d) 3(u ) u du du du Way u u u u u u ±a(u + ) d 3(u ) { du du u u u u + u du } ± { u + du },,, 0 Crect expression A ` 3ln( u u) ln( u ) Integrates ± M(u + ) u + u ± N, M, N, k ¹ 0, to obtain u ± k any one of ±l ln(u + u) ± mln(b(u ± k)); l, m, b ¹ 0 Integration of both terms is crectly followed through from their M and from their N A ft dependent on the So, 3ln(u e { ë + u) - ln(u + ) ù } M mark û Applies limits of e and ( their b and their a, where b > 0, b ¹, a > 0) in u = ( 3ln(e + e) - ln(e + ) ) - ( 3ln3- ln3) applies limits of and 0 in x and subtracts the crect way round. d 3ln(e e) ln(e ) 3ln3 3ln(e e) ln(e ) 3ln3 A o.e. [] 3. (d) Applying u = q - Way 3 e e e e du d du 3ln uu ( ) ( )( ) AA = 3ln + e - ö e ln - ö + = 3ln e ö e + - 3ln ö 3 rd M mark is dependent 3 on nd M mark da []

16 Question Number 4. 4x - y 3-4xy + y = 0 Scheme s Marks (a) Way ì í = ü ý î þ 8x - 3y - 4y - 4x + y ln = 0 A B 8(-) - 3(4) - 4(4) - 4(-) + 4 ln = ln = 0 dependent on the first M mark d = ln ln 4-5+ ln 4 exact equivalent A cso -5+ ln4 NOTE: You can recover wk f part (a) in part (b) [] 40 ln 40 ln Applying m (b) e.g. mn N = - to find a numerical m m N 3 3 T Can be implied by later wking (a) Way 40 ln 3 y 4 x 4 = Cuts y-axis x 0 Þ y - 4 = 40 - ln 3 ö - ( ) + c 40 - ln 3 ö ( ) Using a numerical m N (¹ m T ), either y 4 mn ( x ) and sets x 0 in their nmal equation 4 = (their m N )(-) + c ì 40 - ln 04 - ln ü íþ c = 4 +, so y = Þý î þ y ( c) = 3 - ln 04 - ln 3 - ln -ln + 3 A cso isw : Allow exact equivalents in the fm p - ln f the final A mark [3] 9 ì í îï = ü ý þï 8x - 3y - 4y - 4x + y ln = 0 8(-) - 3(4) - 4(4) - 4(-) + 4 ln = 0 dependent on the first M mark d A B = ln ln 4-5+ ln 4 exact equivalent -5+ ln4 A cso : You must be clear that Way is being applied befe you use this scheme [] Question 4 s 4. (a) F the first four marks Writing down from no wking = = 4y - 8x -3y - 4x + y ln 8x - 4y 3y + 4x - y ln 8x - 4y -3y - 4x + y ln 4y - 8x 3y + 4x - y ln sces AB sces A0B Writing 8x - 3y - 4y - 4x + y ln = 0 sces AB

17 Question 4 s Continued 4. (a) st 3 Differentiates implicitly to include either 4x y y y ± m y (Igne ). l, m are constants which can be st A 3 Both 4x y 8x 3y and = 0 = 0 e.g. 8x - 3y - 4y - 4x + y ln - 3y - 4x + y ln = 4y - 8x e.g ln ln = 3 will get st A (implied) as the " = 0" can be implied by the rearrangement of their equation. nd -4xy -4y - 4x 4y - 4x B y y ln y e yln ln If an extra term appears then award st A0 3 rd d dependent on the first M mark - 4y + 4x 4y + 4x F substituting x and y = 4 into an equation involving can be gained by seeing at least one example of substituting x and at least one example of substituting y = 4 unless it is clear that they are instead applying x = 4 and y = - Otherwise, you will NEED to check (with your calculat) that x = -, y = 4 that has been substituted into their equation involving A cso: If the candidate s solution is not completely crect, then do not give this mark. isw: You can, however, igne subsequent wking following on from crect solution. (b) The nd mark can be implied by later wking. y 4 Eg. Award st and nd f their m evaluated at x and y 4 4. (a) Way A: Allow the alternative answer y = { } ln st Differentiates implicitly to include either ± 4y 4x ± lx (Igne = ö ). l is a constant which can be st A Both 4x - y 3 8x - 3y and = 0 = 0 T ö + 3 (ln) which is in the fm p + qln ln nd -4xy -4y - 4x 4y - 4x - 4y + 4x 4y + 4x B y y ln 3 rd d dependent on the first M mark F substituting x and y = 4 into an equation involving

18 Question Number 5. y = e x + e -x, x ³ 0 Way ln4 { V = } p e x + e - x 0 ln 4 Scheme s Marks ( ) x x e 4e 4 0 Expands e x + e - x x x F e e Igne limits and. Can be implied. ( ) ±a e x ±b e -x ± d where,, 0. Igne, integral sign, limits and. This can be implied by later wk. B Integrates at least one of either ±a e x to give ± a ex ±b e -x to give ± b é = { p} ê ln4 ù e- x a, b ¹ 0 ex - e -x + 4xú dependent on the nd M mark ë û0 x x x x e 4e e e, A which can be simplified un-simplified 4 4x 4e 0 x B cao dependent on the previous method mark. Some evidence of ö = { p} e(ln4) - e -(ln4) + 4(ln4) - e0 - e 0 + 4(0) ö ö applying limits of ln 4 o.e. and 0 to a changed function in x and d subtracts the crect way round. : A proper consideration of the limit of 0 is required. 8 4ln 4 8 = p + 4p ln4 8 8 p + 8p ln 75 4ln ln p + ln8p 75 8 p + p ln5 ln 75 8p 8 e p ö p ( ln4), etc 8 Question 5 s 5. is only required f the st B mark and the final A mark. Give st B0 f writing p y followed by p e x + e - x Give st f e x + e - x ( ) ( ) e x + 4e -x + e 0 + e 0 because d = e 0 + e 0 A decimal answer of p( ) (without a crect exact answer) is A0 ln4 A isw é p ù ê ex - e -x + 4xú followed by awrt 4.9 (without a crect exact answer) is final da0 ë û0 Allow exact equivalents which should be in the fm ap + bp lnc p(a + blnc), where a = Do not allow a = 8 9 Give BM0AB0A0 f the common response ln4 p ( e x + e ) - x ln4 é p e ( x + 4e -x ) = p ù ê ex - e -x ú 0 0 ë û0 ln4 = 75 8 p [7] 7

19 Question Number 5. y = e x + e -x, x ³ 0 Way ln4 { V = } p e x + e - x u = e x V = p = p 0 ( ) Scheme s Marks Þ du = ex = u and x = ln4 Þ u = 4, x = 0 Þ u = e 0 = { } u + u 4 ö u du { } u + 4 u u 4 ö du { } u + 4 u + 4 = p 4 ö u du x x F e e Igne limits and. Can be implied. ( e x + e ) - x ±a u ±b u -3 ± du - where u = e x,,, 0. Igne, integral sign, limits and du. This can be implied by later wk. B é = p ê ë { } u - u + 4lnu ù ú û 4 Integrates at least one of either ±a u to give ± a u ±b u -3 to give ± b u- a, b ¹ 0, where u = e x dependent on the nd M mark u + 4u -3 u - u -, A = { p} (4) - (4) + 4ln4 ö - () - () + 4ln öö simplified un-simplified, where u = e x 4u - 4lnu, where u = e x dependent on the previous method mark. Some evidence of applying limits of 4 and to a changed function in u [ ln 4 o.e. and 0 to an integrated function in x] and subtracts the crect way round. 8 4ln 4 8 = p + 4p ln4 8 8 p + 8p ln 75 4ln ln p + ln8p 75 8 p + p ln5 ln 75 8p 8 e p ö p ( ln4), etc 8 B cao d A isw [7]

20 Question Number 4. l : 8 5, 4 (a) (b) Way r l l l and Scheme s Marks 5 3 : r 3 0 ; lies on l So, X (, 3, 9) (, 3, 9) Let q Acute be the acute angle between l and l l = 5+ 3m and 4 + l = - 4m 5 (Can be implied). Puts l = l and solves to find l and/ m and substitutes their value f into l 3 9 their value f into l i 3j 9k condone 3 3 Realisation that the dot product d 5, d is required between d and d 4 4 a multiple of d and d 3 dependent on the 5 0 st M mark. Applies dot product fmula 4 d 7 cos between d and d a ( ) ( 5) (). (3) (0) ( 4) 7. 5 multiple of d and d { q = Þ} Acute ( dp) awrt seen in (b) only A (c) A l =, X l = 5 Þ AX = 3 d, { d = 7} AX = (-3) + (-5) + (3) 3 7 { = 43} = B A cao Full method f finding AX XA 9 3 seen in (c) only A cao : You cannot recover wk f part (c) in either part (d) part (e). [] [3] [3] (d) Way YA, where is "9 3" = tan(" ") their acute obtuse angle between l and l YA ( dp) anything that rounds to 55.7 A (e) A, X 5 So AX AB So at B, ) Way Substitutes either l = (their l found in (a)) + X l b =3 - (their l found in (a)) X into l At least one position vect is crect. (Also allow codinates). Both position vects are crect. (Also allow codinates). ; A A [] [3] 3

21 Question Number Scheme s Marks. (e) Way Applies either where ; At least one position vect is crect (Also allow codinates) Both position vects are crect (Also allow codinates) A A [3]. (e) Way 3. (e) Way 4. (e) Way 5 ; AX = 43 Þ AB = 7( - l) AX = AB Þ AX = 4AB Þ 43 = 4(7)( - l) Þ ( - l) = 9 4 7l - 08l = 0 08l - 43l + 89 = 0 4l - l + 7 = 0 Þ l = 3.5 l = 0.5 Full method of solving f l the equation : AX = 4AB using and and substitutes at least one of their values f l into l At least one position vect is crect (Also allow codinates) Both position vects are crect (Also allow codinates). Hence, l = 3.5 l = 0.5 can be found from solving either x : -3 = ± ( - l) y : -5 = ± (0-5l) z : -3 = ± (- + l) Applies either where At least one position vect is crect (Also allow codinates) Both position vects are crect (Also allow codinates) ; A A ; A A Applies ; [3] [3] At least one position vect is crect (Also allow codinates) Both position vects are crect (Also allow codinates) A A [3]

22 Question Number Scheme s Marks. (e) Way where Applies either ; At least one position vect is crect (Also allow codinates) Both position vects are crect (Also allow codinates) [3] Question s. (a) can be implied by at least two crect follow through codinates from their l from their m (b) Evaluating the dot product (i.e. (-)(3) + (-5)(0) + ()(-4)) is not required f the, d marks. A A F d: Allow one slip in writing down their direction vects, d and d Allow d f ( - ö 3 ö (-) + (-5) + (). (3) + (0) + (-4) ) q = c, (without evidence of awrt 74.37) is A0. (b) Alternative Method: Vect Cross Product Way Only apply this scheme if it is clear that a vect cross product method is being applied. - ö 3 ö ì i j k ü Realisation that the vect ï ï cross product is required d d = -5 0 = í - -5 = 0i - j + 5k ý between d - 4 ï ï and d îï þï a multiple of d and d Applies the vect product (0) + (-) + (5) sin q = fmula betweend and d (-) + (-5) + (). (3) + (0) + (-4) a multiple of d and d sin q = 7. 5 Þ q = = ( dp) awrt seen in (b) only A [3] d. (c) Finds the difference between their and and applies Pythagas to the result to find AX XA ( ) -. (-) + (-5) + () OR applies their l X found in (a) F : Allow one slip in writing down their and Allow A f 3 ö 5 leading to AX = (3) + (5) + (3) = 43 = (e) Imply f no wking leading to any two components of one of the which are crect.

23 Question Number Scheme s Marks. (d) Way. (d) Way 3 "9 3" YA = tan(90 - " "), where is the acute obtuse angle between l and l YA ( dp) anything that rounds to 55.7 A YA sin(" ") = "9 3" sin(90 - " ") YA = 9 3sin( ) sin( ) o.e., where is the acute obtuse angle between l and l = = 55.7 ( dp) anything that rounds to 55.7 A [] []. (d) Way 4 d = - ö -5, (Allow a sign slip in copying d ) Þ 3+ 3m m = 0 Þ m = 7 7 YA = ö ö 7 ( ) ö ö Applies to find m and applies Pythagas to find a numerical expression f AY f the distance AY So, YA = - ö 7 + ( 5) ö 7 = = 55.7 ( dp) anything that rounds to 55.7 A : []

24 Question Number Scheme s Marks 7. dh = k (h - 9), 9 < h 00; h = 30, dh = -. dt dt Substitutes h 30 and either d h dh.. (a) -. = k (30-9) Þ k =... dt dt into the printed equation and rearranges to give k =... (b) Way (b) Way so, k 0. 0 dh kdt ( h 9) ( h 9) dh k dt ( h 9) k t c k 0. A 0 Separates the variables crectly. dh and dt should not be in the wrong positions, although this mark can be implied by later wking. Igne the integral signs. Integrates ( h 9) kt ( h 9) (h - 9) ( ) B to give ± m (h - 9) ; l, m ¹ 0 = (their k )t, with/without + c, equivalent, which can be un-simplified simplified. Some evidence of applying both { t = 0, h = 00 Þ} (00-9) = k(0) + c t = 0 and h = 00 to changed equation containing a constant of integration, e.g. c A Þ c = 9 Þ ( h 9) 0.t 9 dependent on the previous M mark Applies h 50 and their value of c to h 50 (50 9) 0.t 9 their changed equation and rearranges t... to find the value of t... t t (minutes) (nearest minute) 50 T dh 00 ( h 9) 0 50 T 00 0 k dt ( h 9) dh k dt 50 ( h 9) T kt kt kt t t isw awrt 48 Separates the variables crectly. dh and dt should not be in the wrong positions, although this mark can be implied by later wking. Integral signs and limits not necessary. ( h 9) Integrates kt t t (minutes) (nearest minute) ( h 9) (h - 9) ( ) A d [] A cso B to give ± m (h - 9) ; l, m ¹ 0 = (their k )t, with/without limits, A equivalent, which can be un-simplified simplified. Attempts to apply limits of h = 00, h = 50 and (can be implied) t = 0to their changed equation dependent on the previous M mark Then rearranges to find the value of t... t awrt 48 hours and awrt 8 minutes d [] A cso [] 8

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26 7. (b) Allow first B f writing dt dh = k (h - 9) Question 7 s dt dh = (their k) (h - 9) equivalent dt dh = k (h - 9) leading to t = (h - 9) (+ c) with/without + c is BA k After finding k = 0. in part (a), it is only possible to gain full marks in part (b) by initially writing dh dh = - k (h - 9) = - k dt dt (h - 9) dh dh = - 0. (h - 9) = -0.dt dt (h - 9) Otherwise, those candidates who find k = 0. in part (a), should lose at least the final A mark in part (b).

27 Question Number Scheme s Marks 8. x = 3q sinq, y = sec 3 q, 0 q < p sec cos cos 8 3 Sets 8 k ( x) 3 sin 3 3 (a) When y 8, (b) (c) Way y to find and attempts to substitute their into x = 3q sinq so k ( x) 3 3p 3p 3 A : Obtaining two value f k without accepting the crect value is final A0 [] 3sin 3 cos d 3 sin 3sin 3 cos Can be implied by later wking B y ì í { dq dq ü Applies } ý = (sec 3 q )(3sinq + 3q cosq ){ dq} ( ± K sec î þ q ) their ö dq Igne integral sign and dq ; K ¹ 0 = 3 q sec q + tanq sec q dq Achieves the crect result no errs in their wking, e.g. bracketing manipulation errs. A * Must have integral sign and d in their final answer. evidence of 0 0 and k B 3 3 x 0 and x k 0 and 0 and 3 : The wk f the final B mark must be seen in part (b) only. [4] q sec q Aqg(q) - B g(q ), A > 0, B > 0, where g(q)is a trigonometric function in q and { q sec q dq } = q tanq - tanq { dq g(q) = their sec } q dq. [: g(q) ¹ sec q ] dependent on the previous M mark Either lq sec q Aq tanq - B tanq, A > 0, B > 0 d = q tanq - ln(secq) = q tanq + ln(cosq) { } q sec q q tanq - tanq q sec q q tanq - ln(secq) q tanq + ln(cosq) lq sec q lq tanq - l ln(secq) lq tanq + l ln(cosq) A : Condone q sec q q tanq - ln(sec x) q tanq + ln(cos x) f A tanq sec q dq = tan q sec q where u = cosq u u where u = tanq é { Area( R) } = 3q tanq - 3ln(secq) + 3 ù ê tan q ú ë û tan sec l tanq sec q ± C tan q ± Csec q ± Cu -, where u = cosq tanq sec q tan q sec q tan q - cos q sec q 0.5u -, where u = cosq 0.5u, where u = tanq p ln (3) tan sec tan sec cos 0.5lu -, where u = cosq 0.5lu, where u = tanq é 3q tanq - 3ln(secq) + 3 ù ê sec q ú ë û p ln (4) 3 = 9 + 3p - 3ln 9 + 3p + 3ln ö 9 3 ln8 ln 8 e 9 + 3p ö 0 A A o.e. []

28 Question Number Scheme s Marks 8. (c) Way f the first 5 marks: Applying integration by parts on (q + tanq)sec q dq Way ì u = q + tanq Þ du dq = + ü sec q ï ï í ý ï dv dq = sec q Þ v = tanq = g(q) ï îï þï h(q) and g(q)are trigonometric functions in q and g(q) = their sec q dq. [: g(q) ¹ sec q ] (q sec q + tanq sec q )dq = (q + tanq)sec q dq, = (q + tanq)tanq - ( + sec q)tanq { dq } = (q + tanq)tanq - (tanq + tanq sec q) { dq } = (q + tanq)tanq - ln(secq) - tanq sec q { dq } = (q + tanq)tanq -ln(secq) - tan q = (q + tanq)tanq -ln(secq) - sec q etc. A(q + tanq)g(q) - B ( + h(q))g(q ), A > 0, B > 0 dependent on the previous M mark Either l é ë (q + tanq)sec q ù û A(q + tanq)tanq - B ( + h(q))tanq, A ¹ 0, B > 0 (q + tanq)tanq - ( + h(q))tanq d (q + tanq)tanq -ln(secq) o.e. A l éë (q + tanq)tanq -ln(secq) ù û o.e. tanq sec q ± C tan q ± Csec q (q + tanq)tanq - tan q (q + tanq)tanq - sec q Allow the first two marks in part (c) f q tanq - tanq embedded in their wking Allow the first three marks in part (c) f q tanq - ln(secq) embedded in their wking A Allow 3 rd nd A marks f either tan q - tan q tan q - sec q embedded in their wking Question 8 s 8. (a) Allow f an answer of k = awrt.7 without reference to 3p Allow f an answer of k = 3( arccos( ) )sin( arccos( ) ) without reference to E.g. allow f q = 0, leading to k = 3(0)sin(0) k = p 3 3p 3p 3

29 Question 8 s Continued 8. (b) To gain A, dq does not need to appear until they obtain 3 (q sec q + tanq sec q )dq F, their, where their ¹ 3q sinq, needs to be a trigonometric function in q dq dq Writing (sec 3 q )(3sinq + 3q cosq ) = 3 (q sec q + tanq sec q )dq is sufficient f BA Writing d x 3sin 3 cos followed by writing y d dq dq = 3 (q sec q + tanq sec q )dq is sufficient f BA The final A mark would be lost f cos 3 q 3sinq + 3q cosq = 3 (q sec q + tanq sec q )dq [lack of brackets in this particular case]. Give nd B0 f a = 0 and b = 0, without reference to b = p 3 (c) A decimal answer of (without a crect exact answer) is A0. First three marks are f integrating sec with respect to Fourth and fifth marks are f integrating tansec with respect to Candidates are not penalised f writing ln secq as either ln(secq) lnsecq q sec q q tanq + ln(secq) WITH NO INTERMEDIATE WORKING is M0M0A0 q sec q q tanq - ln(cosq) WITH NO INTERMEDIATE WORKING is M0M0A0 q sec q q tanq - ln(secq) WITH NO INTERMEDIATE WORKING is A q sec q q tanq + ln(cosq) WITH NO INTERMEDIATE WORKING is A Writing a crect uv - v du with u = q, dv du = tanq, = and v = their g(q) and making dq dq one err in the direct application of this fmula is st only. 8. (c) Alternative method f finding tanq sec q dq ì u = tanq Þ du dq = ü sec q ï ï í ý ï dv dq = sec q Þ v = tanq ï îï þï tanq sec q dq = tan q - tanq sec q dq Þ tanq sec q dq = tan q tanq sec q dq = tan q tan sec ± C tan q tanq sec q tan q A Þ ì u = secq ï í ï dv = secq tanq îï dq Þ Þ du ü = secq tanq dq ï ý v = secq ï þï tanq sec q dq = sec q - sec q tanq dq Þ tanq sec q dq = sec q tanq sec q dq = sec q tan sec ± Csec q tanq sec q sec q A

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