PMT. Mark Scheme (Results) Summer Pearson Edexcel International A Level in Core Mathematics C34 (WMA02/01)

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1 Mark Scheme (Results) Summer 04 Pearson Edexcel International A Level in Core Mathematics C4 (WMA0/0)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at or 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 world s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: Summer 04 Publications Code IA08476 All the material in this publication is copyright Pearson Education Ltd 04

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 for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according 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 worthy of credit according 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. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 EDEXCEL IAL MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 5.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded for 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.. 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 for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) d or dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given or d The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as A ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks.

5 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more 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 or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.

6 General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles). Method mark for solving term quadratic:. Factorisation (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 =. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square Solving x bx c 0 : b (x ) q c, q 0, leading to x = Method marks for differentiation and integration:. Differentiation n n Power of at least one term decreased by. ( x x ). Integration Power of at least one term increased by. ( x x ) n n Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but may be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.

7 Question Number Scheme Marks. (a) f (.5).75, f () 8 M Sign change (and f (x) is continuous) therefore there is a root {lies in the interval.5, } A [] (b) x 5 (.5) M x.698, x.698 cao Acao x , x x awrt.6 and x awrt.66 A (c) f (.655) , f (.665) Sign change (and as f (x) is continuous) therefore a root lies in the interval.655, (4 dp) Notes MA [] [] 7 (a) M: Attempts to evaluate both f (.5) and f () and finds at least one of f (.5) awrt.8 or truncated.7 or f() 8 Must be using this interval or a sub interval e.g.[.55,.95] not interval which goes outside the given interval such as [.6,.] A: both f (.5) awrt.8 or truncated.7 and f() 8, states sign change { or f(.5) < 0 < f() or f(.5) f() < 0 } or f(.5) <0 and f() >0; and conclusion e.g. therefore a root [lies in the interval.5, ]or so result shown or qed or tick etc (b) M: An attempt to substitute x.5 into the iterative formula 0 e.g. see 5 (.5). Or can be implied by x awrt.6 A: x.698 This exact answer to 4 decimal places is required for this mark A: x awrt.6 and x awrt.66 (so e.g..66 and would be acceptable here) (c) M: Choose suitable interval for x, e.g..655,.665 and at least one attempt to evaluate f(x). A minority of candidate may choose a tighter range which should include.66 (alpha to 5dp), e.g..659,.66 This would be acceptable for both marks, provided the conditions for the A mark are met. A: needs (i) both evaluations correct to sf, (either rounded or truncated) e.g and or (ii) sign change stated and (iii)some form of conclusion which may be :.66 or so result shown or qed or tick or equivalent N.B. f(.664)=0.000 (to sf)

8 Question Number. Scheme Marks dy dy x y x y 0 dx dx M A M dy x y dx x y not necessarily required. dy () ( ) 4 At,, mt dx () ( ) M 4 T: y x dm T: 4x y 0 or equivalent A [6] 6 Notes dy dy st M: Differentiates implicitly to include either ky or x. dx dx dy (Ignore at start and omission of = 0 at end.) dx dy st A: x x and x y y (so the - should have gone) and = 0 needed here or implied dx dy by further work. Ignore at start. dx dy dy nd M: An attempt to apply the product rule: xy y x or y x o.e. dx dx rd M: Correct method to collect two (not three) dy/dx terms and to evaluate the gradient at x y = - (This stage may imply the earlier =0 ) 4 th dm: This is dependent on all previous method marks 4 4 Uses line equation with their. May use y x c and attempt to evaluate c by substituting x = and y = -. (May be implied by correct answer) nd A: Any positive or negative whole number multiple of 4 x y 0 is acceptable. Must have = 0. dx dy N.B. If anyone attempts the question using instead of, please send to review dy dx

9 Question Scheme Marks Number Apply quotient rule : Or apply product rule to y cos ( sin ). u cos v sin u cos v ( sin ) du dv du dv sin cos sin cos ( sin ) d d d d dy sin ( sin ) cos sin sin cos sin M A d ( sin ) sin sin cos ( sin ) { sin sin cos ( sin ) sin ( sin ) ( sin ) { sin } M ( sin ) A cso ( sin ) sin [4] 4 Notes M: Applies the Quotient rule, a form of which appears in the formula book, to cos sin If the formula is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted nor implied by their working, meaning that terms are written out vu ' uv ' u cos, v sin, u '.., v '...followed by their, then only accept answers of the form v ( sin ) Asin cos (B cos ) where A and B are constant (could be ) Condone invisible ( sin ) brackets for the M mark. If double angle formulae are used give marks for correct work. Alternatively applies the product rule with u cos, v ( sin ) If the formula is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted nor implied by their working, meaning that terms are written out u cos, v ( sin ), u '.., v '... followed by their vu ' uv ', then only accept answers of the form sin Asin cos sin B cos. Condone invisible brackets for the M. If double angle formulae are used give marks for correct work. dy A: Any fully correct (unsimplified) form of If double angle formulae are used give marks for correct work. d dy sin ( sin ) cos Accept versions of for use of the quotient rule or versions of d ( sin ) dy sin sin cos ( ) sin cos for use of the product rule. d M: Applies sin cos or sin cos correctly to eliminate squared trig. terms from the numerator to obtain an expression of the form k sin where k and are constants (including ) If double angle formulae have been used give marks only if correct work leads to answer in correct form. (If in doubt, send to review) A: Need to see factorisation of numerator then answer, which is cso a so or and a = -, with no previous errors sin sin

10 Question Number 4. (a) (b) Scheme (x ) (x ) (x ) dx c (x ) ()() c (Ignore + c ) ()() 5x 5 5 dt ln(4 x ) c or ln( x ) {+k } 4 4x 8 8 Notes (a) M: Gives (x ) where λ is a constant or (x ) (x ) A: Coefficient does not need to be simplified so is awarded for or for (x ) i.e. ()() M A M A Marks [] [] (x ) Ignore subsequent errors and condone lack of constant c N.B. If a binomial expansion is attempted, then it needs all thirteen terms to be correctly integrated for MA (b) M: Gives ln(4x ) where µ is a constant or ln( x ) or indeed ln( k(4x )) May also be awarded for ln(4 x ) or ln( x ), where coefficient 5/8 is correct and there is a slip 8 8 writing down the bracket. It may also be given for ln( u) where u is clearly defined as (4x ) or equivalent substitutions such as ln(4u ) where u x A: ln(4 x ) or ln( x ) o.e. The modulus sign is not needed but allow ln 4 x Also allow 0.65ln(4x ) and condone lack of constant c N.B. 5 ln 4 x with no bracket can be awarded MA0 8

11 Question Number 5. Method Method : 8 or Scheme Marks 7x 7 x 8 7 x 8 8 or B 8 8 kx kx... M A! 7 x 7 x... 8! x ; x x ; x... A; A 4 [5] 5 ()( ) 8 7 x (8) (8) (7 x ) (8) (7 x )! B 8 or Any two of three (un-simplified or simplified) terms correct M All three (un-simplified or simplified) terms correct. A x ; x... 4 A; A [5] 5 B: outside brackets then isw or Notes 8 or as candidate s constant term in their binomial expansion. M: Expands... kx to give any terms out of terms correct for their k simplified or un-simplified Eg: kx!! where k are acceptable for M. Allow omission of brackets. [k will usually be 7, 7/8 or 7/ ] or kx kx or... kx [Allow for A: A correct simplified or un-simplified kx kx expansion with consistent kx {or (kx) for! special case only}. Note that k. The bracketing must be correct and now need all three terms correct for their k. ] 9 x or x 4 A: x - allow A: x allow.55 x or x (Ignore extra terms of higher power) Method : B: 8 or M: Any two of three (un-simplified or simplified) terms correct condone missing brackets A: All three (un-simplified or simplified) terms correct. The bracketing must be correct but it is acceptable for them to recover this mark following invisible brackets. AA: as above. 9 8 Special case (either method) uses x instead of x throughout to obtain x; x... gets BMAA0A0 4

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13 Notes (a) B: Forming the linear identity (this may be implied). Note: A & B are not assigned in this question so other letters may be used M: A valid method to find the value of one of either their A or their B. A: A and B (This is sufficient without rewriting answer provided it is clear what A and B are ) 5 4x Note: In part (a),, from no working, is BMA (cover-up rule). (x )(x ) (x ) (x ) (b) You can mark parts (b)(i) and (b)(ii) together. (i) B: Separates variables as shown. (Can be implied.) Need both sides correct, but condone missing integral signs. M: Uses partial fractions on RHS and obtains two log terms after integration. The coefficients may be wrong e.g. ln (x ) or may follow their wrong partial fractions. Ignore LHS for this mark. Aft: RHS correct integration for their partial fractions do not need LHS nor +c for this mark A : All three terms correct (LHS and RHS) including c. (ii) M: Substitutes y 4 and x into their general solution with a constant of integration to obtain c =. M: A fully correct method of removing the logs must have a constant of integration which must be treated Correctly. Must have had ln y =.earlier 6(x ) A: y isw. (x ) NB If Method is used the third method mark is earned at the end of part (i), then the second method mark is earned when the values are substituted. Special case: A common error using method : (x ) () y A, then 4 A so A would earn M (substitution); M0 (not fully correct removing logs); A0 (x ) () Special case: A possible error using method or : y (x ) (x ) A, then 4 9 A so A would earn M0 (too bad an error); M0 (not fully correct removing logs); A0 i.e. M0M0A0 If there is no constant of integration they are likely to lose the last four marks.

14 Question Scheme Number 7. (a) Method Method x 5 8 y y x x yx ( ) x 5 xy y x y so x x y y 5 x xy y 5 x( y) 8 y 5 x x y y (b) (c) Hence f ( x) ( x, x ) Hence f ( x) ( x, x ) x x x 5 8 x ff ( x ) x 8 ff ( x ) x x 5 x ( x 5) 5( x ) x 8(x ) ff ( x ) (x 5) (x ) 4x 4 x 9x 5 5x 5 4x 0 x 5 x 4x 4 x 5 x 5 (note that a 5. ) x x ( ) "5" (x x) 5 fg() f (4 6) f ( ) ; or substitute into fg( x ) ;= x x (d) g( x ) x x (x.5).5. Hence g min.5 M Either g min.5 or g( x ).5 or g(5) B.5 g( x) 0 or.5 y 0 A Marks M M A oe M A M A M; A [] [4] [] []

15 Notes Method is less likely and the notes apply to Method. (a) M: Brings (x ) to the LHS and multiplies out by y or if x and y swapped first ( y ) to the LHS and multiplies out by x M: A full method to make x (or swapped y) the subject by collecting terms and factorising. x 5 x 5 x 5 8 A: or equivalent e.g. or or etc Ignore LHS. x x x x Does not need to include domain i.e does not need statement that x, x Should now be in x, not y, for this mark. N.B. Use of quotient rule to differentiate and to find f is M0M0A0. This is NOT a misread. f(x) 5 (b) M: An attempt to substitute f into itself. e.g. ff (x). Squaring f( x ) is M0. f(x) f(x) 5 x 5 Allow ff (x) or ff (x) for MA0 x f(x) A: Correct expression. This mark implies the previous method mark. M: An attempt to combine each of the numerator and the denominator into single rational fraction with same common denominator x 5 A: See Does not need to include domain or statement that x, x, x x NB If they use a mixture of methods and then mark accordingly attempt M, correct A, combined into single rational function M then answer is A 8 5 x 8 so may see or 8 x 5 x x (x x) 5 (c) M: Full method of inserting g() ( i.e. - ) into f( x ). Or substitutes into fg(x) x x A: cao (d) M: Full method to establish the minimum of g. (Or correct answer with no method) e.g.: x leading to g min. Or finding derivative, setting to zero, finding x ( =.5) and then finding g(.5) in order to find the minimum. Or obtaining roots of x 0, and using symmetry to obtain g min g(.5). Or listing values leading to g min g(.5). This mark may also be implied by -.5. B: For finding either the correct minimum value of g (can be implied by g(x).5 or g(x).5 ) or for stating that g(5) 0 or finding the value 0 as a maximum A:.5 g(x) 0 or.5 y 0 or.5 g 0. Note that:.5 x 0 (wrong variable)is A0;.5 y 0 (wrong inequality) is A0;.5 f 0 (wrong function) is A0; Accept [ -.5, 0] (correct notation) for A but not (-.5, 0) (strict inequality) which is A0 A correct answer with no working gains M B A i.e. /

16 Question Number Scheme Marks 8. dv 50 dt 4 dv V r 4 r dr V r r dr dr dv 50 dt dv dt 4 r 9000 dr 50 When r, dt dr So, cms awrt dt Notes dv B: 4 r. This may be stated or used and need not be simplified dr 4 Applies 000 r and rearranges to find r using division then cube root with accurate algebra V May state r then substitute V = 000 later which is equivalent. r does not need to be evaluated. 4 M: Uses chain rule correctly so 50 dv their dr dr dm: Substitutes their r correctly into their equation for This depends on the previous method mark dt A: awrt (Units may be ignored) If this answer is seen, then award A and isw. Premature approximation usually results in all marks being earned prior to this one. B B M dm A [5] 5

17 Question Number 9. (a) Scheme Marks x y e 5 e 6 e 7 e 8 e e M B oe e e e e e e ( ) M ( dp) A Special case (s.c.) Uses h = 5/4 with 5 ordinates giving answer award M0B0MA(s.c.) [4] See note below d u d x u x x or u d x d u B x u M A e d x e u du (b) u u u e e du M u u ue e u u u e e e e A e e ddm 4e e or e (e ) etc. A [7]

18 Notes (a) M: Finds y for x = 4, 5, 6, 7, 8 and 9. Need six y values for this mark. May leave as on middle row of table give mark if correct unsimplified answers given, then isw if errors appear later. If given as decimals only, without prior expressions, need to be accurate to significant figures.(allow one slip) May not appear as table, but only in trapezium rule. B: Outside brackets or or h = stated. This is independent of the method marks M: For structure of... ft their y values and allow for 5 or 6 y values so may follow wrong h or table which has x from 5 to 9 or from 4 to 8 NB {4+9+( )} is M A: N.B. Wrong brackets e.g. (e e ) e e e e is M0 unless followed by correct answer which implies MA Special case: uses five ordinates (i.e. four strips) x y e e 6 5 e 7 75 e Then Giving e e e e e = This complete method for special case earns M0 B0 M A i.e. /4 du dx (b) B: States or uses x or u dx du u M: Obtains ue du for a constant value λ A: Obtains e u d u u M: An attempt at integration by parts in the right direction on λ ue u. This mark is implied by the correct answer. There is no need for limits. If the rule is quoted it must be correct. A version of the rule appears in the formula booklet. Accept for this mark expressions of the form e u ue du u e u u u u A: u e u e e. (Candidates just quoting this answer earn MA) ddm: Substitutes limits of and in u (or 9 and 4 in x) in their integrand and subtracts the correct way round. (Allow one slip) This mark depends on both previous method marks having been earned A: Obtains 4e e or e (e ) with terms collected. If then given as a decimal isw. edu u

19 Question Number Scheme Marks 0. (a) A B sin A sin A A sin Acos A cos Asin A or sin Acos A sin Acos A M Hence, sin A sin Acos A (as required) * A * [] (b) Way A: dy sec x dy sec x Way B y ln tan x dx tan x M A dx tan x tan x cos x sin x tan x cos x sin x cos x. tan x sin x cos x cos x dm cosec x * cosec x * A * x cos x sin x sin x cos x sin x [4] sin Way : y ln sin cos x sin x sin dy cos x sin x x cos x ln cos x dx sin x ; cosec x x cos x sin x M A M;A [4] Way: quotes cosecxdx = ln(tan x ) d x cosec x dx (As differentiation is reverse of integration) tan M A M A [4] (c) Way 0 (c) dy B y ln tan x sin x cosec x cos x dx dy 0 cosec x cos x 0 cos x 0 M dx sin x sin x cos x (sin xcos x) so sin x k, where k and k 0 M So sin x A x , So x , Method (Squaring Method) y ln tan x sin x dy 0 cosec x cos x 0 cos x 0 dx sin x dy cosec x cos x dx A A B M [6] 4 4 9cos x so 9cos x 9cos x 0 or 9sin x 9sin x 0 cos x M So cos x 0.87 or 0.7 or sin x 0.87 or 0.7 A So x , A A [6]

20 Way 0c) t method y ln tan dy x sin x cosec x cos x B dx dy 0 cosec x cos x 0 cos x 0 dx sin x M t t 4 0 so t 6t t 6t 0 t t M t = or A So x , A A Notes [6] (a) M: This mark is for the underlined equation in either form sin Acos A cos Asin A or sin Acos A sin Acos A A: For this mark need to see : sina at the start of the proof, or as part of a conclusion sin(a + A) = at the start = sin Acos A cos Asin A or sin Acos A sin Acos A = sinacosa at the end k sec x (b )M: For expression of the form, where k is constant ( could even be ) tan x dy sec x A: Correct differentiation so dx tan x Way A: sin x dm: Use both tan x and sec x in their differentiated expression. This may be implied. cos x cos ( x) This depends on the previous Method mark. A*: Simplify the fraction, use double angle formula, see and obtain correct answer with completely sin x correct work and no errors seen (NB Answer is given) Way B dm: Use both sec sin x x tan ( x) and tan x cos A*: Simplify the fraction, use double angle formula, see correct work and no errors seen (NB Answer is given) Way : x sin x and obtain correct answer with completely dy k cos x c sin x M:Split into y ln sin x ln cos x then differentiate to give dx sin x cos x dy cos x x sin A: Correct answer dx sin x cos x cos x sin x M: Obtain A*: As before sin x cos x Way : Alternative method: This is rare, but is acceptable. Must be completely correct. Quotes cosecxdx = ln(tan x ) and follows this by d gets 4/4 tan x cosecx dx

21 dy (c) B: Correct differentiation so see cosec x cos x dx dy M: Sets their 0 and uses cosec x dx sin x Way : M: Rearranges and uses double angle formula to obtain sin x k, where k and k 0 (This may be implied by a + b sinx = 0 followed by correct answer) A: sin x (This may be implied by correct answer) A: Either awrt 0.65 or awrt.06 ( answers in degrees lose both final marks) A: Both awrt 0.65 and awrt.06 Ignore y values. Ignore extra answers outside range. Lose the last A mark for extra answers in the range. Way : M: Obtain quadratic in sinx or in cosx. Condone cos ec x 9cos x 0 as part of the working A A A: See scheme Way : x This method is unlikely and uses t tan( ). See scheme for detail

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23 Notes (a) M: At least one term differentiated correctly A: Correct differentiation of both terms dy M: Sets to 0 and applies a correct method for eliminating the exponentials e x to reach x = dx (At this stage the RHS may include ln(e a ) term but should include no x terms ) A: x P a after correct work dy ddm: (Needs both previous M marks) Substitutes their x-coordinate into y ( not into ) dx A: y P e a given as one term (b) Parts (b) and (c) may be marked together. Methods, and : a f( x) a x x e M: Put y = 0 and attempt to obtain e k e.g. e (Method ) or e (Method ) or e "" x e a (Method ) Must have all x terms on one side of the equation for any of these methods dm: This depends on previous M mark. Take logs correctly. e.g. a x ln (Method ) or x aln (Method ) or ln ""x a (Method ) a ln A: cao x Q (must be exact) Method 4: a x x M: Puts e e then takes lns correctly (see scheme) a x ln x dm: Collects x terms on one side a ln A: x Q cao (must be exact to answer requirements of (c)) (c) B: Correct overall shape, so y 0 for all x, curve crossing positive y axis and small portion seen to left of y axis, meets x axis once, one maximum turning point B: Cusp at x x Q (not zero gradient) and no appearance of curve clearly increasing as x becomes large a a a B: Either writes full coordinates (0, e ) in the text or (0, e ) or e marked on the y-axis or even (e a, 0) if marked on the y axis (must be exact) allow e a i.e. allow modulus sign, Can be earned without the graph. a ln No requirement for x Q to be repeated for this mark. It has been credited in part (b)

24 Question Number Scheme Marks (a) change limits: x 0 t 0 and x t B Uses V ( ) ( ) y dx - in terms of the parameter t M y dx d x ( ) y dt ( ) (sin t ) sec t dt dt 4sin t sin t dt A cos t 4sin t(sec t ) d t dm 4tan t sin t dt or 4 tan t( cos t)d t or V y dx 4 (tan 0 0 t sin t) d t * Correct proof. A A * [6] cos t (tan t sin t) d t sec t (b) dt Uses tan t (may be implied) M Uses cos t sin t (may be implied) M sec t cos t dt tan t t t sin t 4 M A tan sin (0) Applies limit of 4 ddm V 4 or oe 8 Two term exact answer A [6] *See back page for methods using integration by parts Notes (a) B: See both x 0 t 0 and x t ; Allow if just stated as in scheme- must be in part (a) M: attempt at V ( ) y dx - ignore limits and but need to replace both y and dx by expressions in terms of the parameter t. Methods using Cartesian approach are M0 unless parameters are reintroduced sin t 4 A: (sin t ) sec t dt ignoring limits and 4sin t sec t dt dt 4 tan t cos t dt cos t cos t A: Obtain 4tan t sin t dt at some point or dm: Applies sin t 4sin t sin t dt cos t cos t or tan t sec t after reaching 4tan t sin t dt or 4sin t sin t dt A*: Obtains given answer with no errors seen (To obtain this mark must have been included in V y dx ) This answer must include limits, but can follow B0 scored earlier. Any use of dx where dt should be used is M0 (b) M: Uses tan t sec t M: Uses cos t sin t M: At least two terms of Atan t Bt C sin t A: Correct integration of tan t sin t with all signs correct ddm: (depends upon the first two M marks being awarded in part (b)) Substitutes into their integrand (can be implied by answer or by 4.75) A: Two term exact answer for V

25 Question Number Scheme Marks. (a) R (must be given in part (a)) B tan or sin or cos 5 5 ( see notes for other values which gain M) M (must be given in part (a)) A [] (b) Way : Uses distance between two lines is 4 (or half distance is ) with correct trigonometry M may state 4sin cos 4 or show sketch Need sketch and 4sin cos 4 and deduction that A * sin cos or cos sin * Way : Alternative method: Uses diagonal of rectangle as hypotenuse of right angle triangle and M obtains 0 sin( ) 4 So from (a) sin cos or cos sin A Way : They may state and verify the result provided the work is correct and accurate See notes below. Substitution of 6.9 (obtained in (c) is a circular argument and is M0A0) [] [] (c) Way: Uses 5 sin 6.57 to obtain Way cos 4cos sin 4sin 4 See notes for variations sin "6.57" cos sin cos 0 M " 5 " cos (4sin cos ) 0 so tan 4 arcsin "6.57" arctan or equivalent 4 their " 5 " Hence, M A [] (d) 4 Way : " x" Way : y = tan"6.9" si n 4 h x 4 h 4 h y 8 h 8 tan"6.9" sin"6.9" h or or. (sf ) h 8 or. (sf ) tan 6.9 sin 6.9 B M A cao []

26 Notes (a) B: R 5 or awrt.4 no working needed must be in part (a) M: tan or tan or sin or sin or cos or cos find alpha. Method mark may be implied by correct alpha. A: accept awrt 6.57; also accept 5sin must be in part (a) Answers in radians (0.46) are A0 and attempt to (b) Way : M: Uses distance between two lines is 4 (or half distance is ) states 4sin cos 4 or shows sketch (may be on Figure 4 on question paper) with some trigonometry A*: Shows sketch with implication of two right angled triangles (may be on Figure 4 on question paper) and follows 4sin cos 4 by stating printed answer or equivalent (given in the mark scheme) and no errors seen. Way : on scheme (not a common method) Way : They may state and verify the result provided the work is correct and accurate. x 4 x M: Verification with correct accurate work e.g., with x shown on figure 4 A: Needs conclusion that sin cos Substitution of 6.9 (obtained in (c) is a circular argument and is M0A0) (c) Way : M: sin their (Uses part (a) to solve equation) their R M: arcsin their (operations undone in the correct order with subtraction) their R A: awrt 6.9 (answer in radians is and is A0) Way : M: Squares both sides, uses appropriate trig identities and reaches 4 tan or sin or cos or sin 5 {One example is shown in the scheme. Another popular one is 4 sin cos 4( cos ) 4 4cos cos 5cos 4cos 0 and so cos for M} 5 M: arctan 4 4 or other correct inverse trig value e.g. arcsin ( ) or arcos ( ) 5 5 A: awrt 6.9 (answer in radians is and is A0) (d) Way : (Most popular) B : States x, where x (not defined in the question) is the non-overlapping length of rectangle tan M: Writes equation h 4 - must be this expression or equivalent e.g. tan gets B M tan 4 h 4 A: accept decimal which round to. or the exact answer i.e. (may follow slight inaccuracies in earlier angle being rounded wrongly) cos sin sin(90 ) N.B. There is a variation which states sin or for B M then A as before 4 h 4 h

27 Way : (Less common) 4 B : States y, where y (not defined in question) is the non-overlapping length of two rectangles si n 4 4 M: Writes equation h 8 - must be this expression or equivalent e.g. sin gets B M si n 8 h A: as in Way There are other longer trig methods possibly using Pythagoras for showing that h =. to sf. If the method is clear award BMA otherwise send to review.

28 Question Number Scheme 4. A (, a, 5), B ( b,, ), l : r i 4j 6k ( i j k) (a) Either at point A : or at point B : M leading to either a or b 5 A leading to both a and b 5 A (b) Attempts ('5 i ' j k ) (i ' j' 5 k) subtraction either way round M AB 4 i j k o.e. subtraction correct way round A (c) Way Way AB 6, AC, BC 6 (AC ) "0" or (CA ) "0" M Marks [] [] ˆ cos CAB ˆ 6 cos CAB Or right angled triangle and (4) () ( ). () (0) ( ) cos CAB ˆ o.e. ˆ 0 6 ˆ cos CAB (o.e.) CAB 0 * ˆ so CAB dm A * cso [] (e) (d) Area CAB 4 8 sin 0 (or k ) 4 b 4 OD 4 5 or a or = ; = or M A M; oe A [] 4 b 4 M; oe OD 4 or a or = ; A 7 9 [4] See notes for a common approach to part (e) using the length of AD 4

29 Notes Throughout allow vectors to be written as a row, with commas, as this is another convention. (a) M: Finds, or implies, correct value of for at least one of the two given points A: At least one of a or b correct A: Both a and b correct (b) M: Subtracts the position vector of A from that of B or the position vector of B from that of A. Allow any notation. Even allow coordinates to be subtracted. Follow through their a and b for this method mark. 4 A: Need correct answer : so AB 4i j k or AB or (4,, -) This is not ft. (c) Way : M: Subtracts the position vector of A from that of C or the position vector of C from that of A. Allow any notation. Even allow coordinates to be subtracted. Follow through their a for this method mark. dm: Applies dot product formula between their AB or BA and their AC or CA. ˆ A*: Correctly proves that CAB 0. This is a printed answer. Must have used AB with AC or BA with CA for this mark and must not have changed a negative to a positive to falsely give the answer, that would result in MMA0 Do not need to see but should see equivalent value. Allow as final answer. 6 Way : M: Finds lengths of AB, AC and BC dm: Uses cosine rule or trig of right angled triangle, either sin, cos or tan A: Correct proof that angle = 0 degrees (d) M: Applies AB AC sin 0 - must try to use their vectors (b a) and (c a) or state formula and try to use it. Could use vector product. Must not be using OB OC sin0 A: cao must be exact and in this form (see question) (e) M: Realises that AD is twice the length of AB and uses complete method to find one of the points. Then uses one of the three possible starting points on the line (A, B, or the point with position vector i 4j 6k ) to reach D. See one of the equations in the mark scheme and ft their a or b. b So accept OD 4 5 or a 4 or = A: Accept (9,, ) or 9i + j + k or cao M: Realises that AD is twice the length of AB but is now in the opposite direction so uses one of the three possible starting points to reach D. See one of the equations in the mark scheme and ft their a or b.

30 4 b 4 So accept OD 4 or a or = A: Accept (-7, -7, 9) or -7i -7j + 9k or 7 cao 9 NB Many long methods still contain unknown variables x, y and z or λ. These are not complete methods so usually earn M0A0M0A0 on part (e) PTO. Mark scheme for a common approach to part (e) using the length of AD is given below: (e) ( ) ( ) ("" ) "96" then obtain 5 0so, then substitute value of to find coordinates. May make a slip in algebra expanding brackets or collecting terms (even if results in two term quadratic) This may be simplified to 6( ) 4 6 or to 6( ) 4 6 NB 6( ) 4 6 is M0 as one side has dimension (length) and the other is length 9 (from =5) Substitute other value of. May make a slip in algebra 7 7 (from =-) 9 Special case uses AD is half AB instead of double AB ( ) ( ) ("" ) "6" then obtain 0 so, then substitute value of to find coordinates 4 (from =0) 6 M A M A [4] Substitute other value of (from =) 4 For this solution score MA0MA0 i.e. /4

31 Qu (b) using integration by parts Qu (b) Some return to V 4tan t sin t dt.there are two ways to proceed and both use integration by parts (b) t t (sec t ) sin t d t Uses cos t Uses tan t sec t Way : (tan t sin ) d sin t cos t sin t tant sin t cos t tan tdt dt sin t tan t t sin t 4 tan sin (0) Applies limit of V 4 or oe Two term exact answer 8 Way : Try to use parts on (sec t ) sin d t t using u sin t and v = tant t Award first two M marks as before Uses tan t sec t and Uses cos t sin t t This needs parts twice and to get down to sin t( tan t t) t sin t cos t sin t Then limits as before to give V 4 or oe 8 M M M A ddm A M M MA ddma [6]

32 Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE

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