C4 mark schemes - International A level (150 minute papers). First mark scheme is June 2014, second mark scheme is Specimen paper

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1 C4 mark schemes - International A level (0 minute papers). First mark scheme is June 04, second mark scheme is Specimen paper. (a) f (.).7, f () M Sign change (and f ( ) is continuous) therefore there is a root {lies in the interval., } A [] (.) M.69,.69 cao Acao , awrt.6 and awrt.66 A (c) f (.6) , f (.66) Sign change (and as f ( ) is continuous) therefore a root lies in the interval.6, (4 dp) Notes MA (a) M: Attempts to evaluate both f (.) and f () and finds at least one of f (.) awrt. or truncated.7 or f() Must be using this interval or a sub interval e.g.[.,.9] not interval which goes outside the given interval such as [.6,.] A: both f (.) awrt. or truncated.7 and f(), states sign change { or f(.) < 0 < f() or f(.) f() < 0 } or f(.) <0 and f() >0; and conclusion e.g. therefore a root [lies in the interval., ]or so result shown or qed or tick etc M: An attempt to substitute 0. into the iterative formula e.g. see (.). Or can be implied by awrt.6 A:.69 This eact answer to 4 decimal places is required for this mark A: awrt.6 and awrt.66 (so e.g..66 and.6649 would be acceptable here) (c) M: Choose suitable interval for, e.g..6,.66 and at least one attempt to evaluate f(). A minority of candidate may choose a tighter range which should include.66 (alpha to dp), e.g..69,.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) [] [] 7

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

3 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 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 cos M: Applies the Quotient rule, a form of which appears in the formula book, to 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, vsin, u'.., v'... followed by their, then only accept answers of the form v ( sin ) Asin cos ( Bcos ) 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 Bcos. Condone invisible brackets for the M. If double angle formulae are used give marks for correct work. A: Any fully correct (unsimplified) form of If double angle formulae are used give marks for correct work. d sin ( sin ) cos Accept versions of for use of the quotient rule or versions of d ( sin ) d sin cos ( ) sin sin cos for use of the product rule. M: Applies sin cos or sin cos correctly to eliminate squared trig. terms from the numerator to obtain an epression 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

4 ( ) ()() 4. (a) ( ) c or 4 dt ln(4 ) c ln( 4 ) {+k } Notes (a) M: Gives ( ) where λ is a constant or ( ) A: Coefficient does not need to be simplified so is awarded for ( ) ()() ( ) ()() ( ) M c (Ignore + c ) A or for ( ) i.e. M A [] [] ( ) Ignore subsequent errors and condone lack of constant c N.B. If a binomial epansion is attempted, then it needs all thirteen terms to be correctly integrated for MA M: Gives ln(4 ) where µ is a constant or ln( ) or indeed ln( k(4 )) 4 May also be awarded for ln(4 ) or ln( ), where coefficient / is correct and there is a slip writing down the bracket. It may also be given for ln( u) where u is clearly defined as (4 ) or equivalent substitutions such as ln(4u ) where u A: N.B. ln(4 ) Also allow ln 4 ln( 4 ) or o.e. The modulus sign is not needed but allow 0.6ln(4 ) and condone lack of constant c with no bracket can be awarded MA0 ln 4

5 k k!... or B M A ! 9 6 ; ;... A; A 4 [] ()( ) 7 () () (7 ) () (7 )! B or Method Method : Any two of three (un-simplified or simplified) terms correct All three (un-simplified or simplified) terms correct. 9 6 ;... 4 Notes B: or outside brackets then isw or or as candidate s constant term in their binomial epansion. M: Epands... k to give any terms out of terms correct for their k simplified or un-simplified k M A A; A Eg: k or k or... k [Allow for ]!! where k are acceptable for M. Allow omission of brackets. [k will usually be 7, 7/ or 7/ ] A: A correct simplified or un-simplified k k! epansion with consistent k {or (k) for special case only}. Note that k. The bracketing must be correct and now need all three terms correct for their k. 9 A: 4 - allow. or 4 6 A: allow or (Ignore etra terms of higher power) Method : B: 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 Special case (either method) uses instead of throughout to obtain ;... gets BMAA0A0 4 []

6 6. (a) 4 A B ()( ) () ( ) so 4 A( ) B( ) B (i), (ii) Method for (ii) Let, 9 B B Let A = and B = - or, A A 4 ()( ) () ( ) 4 y ()( ) B = C ln() Dln( ) () ( ) M "" ln( ) ""ln( ) ln y ln() ln( ) c ln 4 ln(() ) ln( ) c c ln 6 6( ) ln y ln() ln( ) ln 6 so ln y ln ( ) 6( ) So y ( ) M A Aft A M M A [] [7] Method for (ii) Solution as Method up to ln y ln() ln( ) c so first four marks as before BMAA A() Writes y as general solution which would earn the rd M mark. ( ) Then may substitute to find their constant A, which would earn the nd M mark. M M 6( ) Then A for y ( ) as before. A [7] 0

7 7. (a) Method Method y y y ( ) y y y y y ( y) y y Hence f ( ) (, ff ( ) ( ) ( ) ( ) ( ) y y so ) Hence M y f ( ) ff ( ) ( ) ff ( ) 4 4 M (, ) A oe M A (note that a. ) A [4] (c) fg() f (4 6) f ( ) ( ) "" ( ) ; or substitute into fg( ) ;= M; A [] (d) g( ) (.).. Hence gmin. M Either gmin. or g( ). or g() 0 B. g( ) 0 or. y 0 A [] M []

8 . B: dv 0 dt 4 dv V r 4 r dr V r r dr dr dv 0 dt dv dt 4 r When r 9000, dr So, dt dv dr dr 0 dt cms awrt A Notes 4 r. This may be stated or used and need not be simplified 4 Applies 000 r and rearranges to find r using division then cube root with accurate algebra May state V r then substitute V = 000 later which is equivalent. r does not need to be evaluated. 4 M: Uses chain rule correctly so 0 dv their d r dm: Substitutes their r correctly into their equation for d r 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 approimation usually results in all marks being earned prior to this one. B B M dm []

9 9. (a) y e e 6 e 7 e e e M B oe e e 6 7 e e e e ( ) M ( dp) A Special case (s.c.) Uses h = /4 with ordinates giving answer 6.76 award M0B0MA(s.c.) [4] See note below du u or u du B u M A e e u du u u ue e du M u u ue e u u u A e e e e e e ddm 4e e or e (e ) etc. A [7]

10 0. (a) A B sin A sin A A sin Acos Acos Asin A or sin Acos A sin Acos A M Hence, sin A sin Acos A (as required) * A * [] Way A: sec sec Way B y ln tan tan M A tan tan cos sin tan cos sin cos dm. tan sin cos cos cosec * cosec * A * sin cos sin sin cos sin [4] Way : y ln sin ln cos cos sin ; cosec sin cos sin cos sin sin cos M A M;A [4] Way: quotes cosec = ln(tan ) d tan cosec (As differentiation is reverse of integration) M A M A [4] (c) Way 0 (c) B y ln tan sin cosec cos 0 cosec cos 0 cos 0 M sin sincos (sincos ) so sin k, where k and k 0 M So sin A So , , Method (Squaring Method) y ln tan sin cosec cos B 0 cosec cos 0 cos 0 sin M 4 4 9cos so 9cos 9cos 0 or 9sin 9sin 0 cos M So cos 0.7 or 0.7 or sin 0.7 or 0.7 A A A So , A A [6] [6]

11 Way 0c) t method y ln tan sin cosec cos B 0 cosec cos 0 cos 0 sin M t t 4 0 so t 6t t 6t 0 t t M t = 0.4 or 0.6 A , A A So Notes [6] (a) M: This mark is for the underlined equation in either form sin Acos Acos 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 Acos Asin A or sin Acos A sin Acos A = sinacosa at the end (b )M: For epression of the form ksec tan, where k is constant ( could even be ) sec A: Correct differentiation so tan Way A: sin dm: Use both tan and sec in their differentiated epression. This may be implied. cos cos ( ) This depends on the previous Method mark. A*: Simplify the fraction, use double angle formula, see and obtain correct answer with completely sin correct work and no errors seen (NB Answer is given) Way B dm: Use both sin sec tan ( ) and tan cos A*: Simplify the fraction, use double angle formula, see correct work and no errors seen (NB Answer is given) Way : M:Split into y ln sin ln cos sin then differentiate to give cos sin A: Correct answer sin cos M: Obtain cos sin sin cos A*: As before Way : Alternative method: This is rare, but is acceptable. Must be completely correct. Quotes cosec = ln(tan ) and follows this by cosec gets 4/4 and obtain correct answer with completely kcos sin csin cos

12 . (a) e e a M A a a e e 0 e e a M = a A So, y P a e a e a ; e a ddm; A [6] Mark parts and (c) together. Method Method Method a a "" e "" 0 e a e e a a "" 0e e e 0 e a e e e M a "" ln "" a ln ln "" a dm a ln or equivalent e.g. a e ln or ln a e etc A [] Method 4 a 0e e a e e and so aln M "" a ln dm a ln o.e. e.g. a e ln or ln a e etc A [] (c) y (0, e a ) y e e a Shape Cusp and behaviour for large (0, e a ). B B B O []

13 (a) change limits: 0 t 0 and t B Uses V ( ) y - in terms of the parameter t M y y t t t t A dt 4tan tsin tdt or 4sin tsin t dt A cos t 4tan t( cos t)dt or 4sin t(sec t) dt dm ( ) d ( ) d ( ) (sin ) sec d d 4 (tan sin ) d * Correct proof. A * [6] 0 0 V y t t t cost (tan t sin t) dt sec t dt Uses tan t sec t (may be implied) Uses cos t sin t(may be implied) M M sec cos d tan sin t t 4 M A tan sin (0) 4 Applies limit of ddm V 4 or oe Two term eact answer A [6] *See back page for methods using integration by parts Notes (a) B: See both 0 t 0 and t ; Allow if just stated as in scheme- must be in part (a) M: attempt at V ( ) y - ignore limits and but need to replace both y and by epressions in terms of the parameter t. Methods using Cartesian approach are M0 unless parameters are reintroduced. 4 A: (sin t 4 4sin t 4 ) sec tdt ignoring limits and 4sin tsec tdt dt 4tan tcos tdt cos t A: Obtain 4tan t sin tdt at some point or 4sin tsin t dt cos t dm: Applies sin t cos t or tan t sec t after reaching 4tan t sin tdt or 4sin tsin t dt cos t A*: Obtains given answer with no errors seen (To obtain this mark must have been included in V y ) This answer must include limits, but can follow B0 scored earlier. Any use of where dt should be used is M0 M: Uses tan t sec t M: Uses cos t sin t M: At least two terms of Atan t Bt Csin 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 ) Substitutes into their integrand (can be implied by answer or by 4.7) A: Two term eact answer for V

14 . (a) R (must be given in part (a)) B tan or sin or cos ( see notes for other values which gain M) M (must be given in part (a)) A [] 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 sin 6.7 to obtain Way cos 4cossin 4sin 4 See notes for variations sin "6.7" cossin cos 0 " " cos (4sin cos ) 0 so tan 4 M arcsin "6.7" arctan their " " 4 or equivalent M Hence, A [] [] [] (d) Way : " " Way : y = 4 B tan"6.9" si n 4 h 4 h 4 h y h M tan"6.9" sin"6.9" A cao h or or. (sf ) h or.(sf) tan 6.9 sin 6.9 []

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

16 WMA0/0: Core Mathematics C4 SPECIMEN PAPER MARK SCHEME. (a) R = B tanα = α = 67. M A 0 cos(θ ) = 0 cos(θ ) = M () θ = 9.7,(0., 99.7 ) A θ = 6. A θ = 66. M A () ( marks). (a) π π π 4 4 π y π π 4 or awrt.44 B awrt 4.04 or B π Area ; { 0 + ( ) + 0} M 4.94 A B () () (c) Uses vu ' + uv ' = e (sin ) (cos ) + e (sin ) M A A = 0 e (sin ) (cos ) + e (sin ) = 0 cos = sin M tan = =.6 M A (6) ( marks) Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 0

17 WMA0/0: Core Mathematics C4. du = sin B cos + u e sin = e du M A u =-e ( + c) ft on sign error A (cos + ) =-e ( + c) π (cos + ) e = ( e) ( e ) = e (e ) 0 cso M A* (6 marks) 4. (a) ( ) = ( )( ) ( )( )( 4) = + ( ) B M A 7 7 = = M A ( ) f( ) = ( a + b) Coefficient of : Coefficient of : a b + = 0 (a+ b= 0) M 4 4 7a b 9 + = (9a+ 4b= ) M A (either correct) A Leading to a =, b = dm A () 7a 7b 7 7 (c) Coefficient of : + = + M Aft = = cao A 6 6 () ( marks) () 0 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics

18 WMA0/0: Core Mathematics C4. (a) ln fg( ) = e +, M ln = e + = + = + M A (+ ) (+ ) e + = 6 e = 4 M A ( + ) = ln 4 ln 4 = M A (c) Let y e = + y = e ln( y ) = = ln( y ) M f ( ) = ln( ), >. A B () (4) () (d) Shape for f () B (0, ) B Shape for f () B (, 0) B (4) (4 marks) Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 0

19 WMA0/0: Core Mathematics C4 6. (a) Differentiating implicitly to obtain ± ay and/or ± b M 4 y A 9 y 9 + y or equivalent B 4y y 4 = 0 M ( ) 4 y 6y = = 4y + 9 6y + A () 6y = 0 M Using = or y y = Leading to 6y + 9 y 4 = 0 y y 4 6y + 6 = 0 or or = M + = M 4 y = or 6 4 = 6 y =, or =, A, A Subs either of their values into y = to obtain a value of other variable. M,,, both A (7) ( marks) 04 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics

20 WMA0/0: Core Mathematics C4 7. (a) cos cos cot cot sin sin B sin cos cos sin sin sin M sin( ) sin sin M sin cosec sin sin sin M A* π π = θ + =.θ + 6 B π π cot.θ + = tan.θ + = 6 6 M () π π 4π. θ + =, 6 π 7π θ =, A, A 9 9 () (0 marks) M Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 0

21 WMA0/0: Core Mathematics C4. (a) 4 ( + ) + 4( + ) + = + + ( + )( + ) ( + )( + ) = = ( + ) ( + )( + ) + ( ) ( + ) h '( ) = ( + ) M A M A* M A (4) (c) 0 h '( ) = ( + ) Maimum occurs when h '( ) = 0 0 = 0 =.. M = A A () When = h( ) = M A Range of h() is 0 h ( ) Aft () ( marks) 06 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics

22 WMA0/0: Core Mathematics C4 9. (a) Equate j components + λ = 9 λ = M A Leading to C = (,9, ) A Choosing correct directions or finding AC and BC. 0 = + = 6 9 cos ABC Use of scalar product. M A ACB = 7.9 A M () (4) (c) A = (,, 4) B = (, 9, ) AC = 6 AND 0 BC = 0 4 AC = = ( 6) 0 4 ( 9 ) BC = + = M A A Area triangle ABC = AC BC sin ACB = ' 6 ' ' 9 ' sin 7.9 M =. A () ( marks) Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 07

23 WMA0/0: Core Mathematics C4 0. (a) tanθ = or sinθ = M π θ = awrt.0 A d sec θ y dθ = θ =, d cos θ () cosθ = ( = cos θ ) M A sec θ At P, π m = cos = Can be implied. Using mm =, m = M For normal y ( ) At Q, y = 0 = ( ) leading to 7 6 = dm 7 = ( ) 6 A k =.06 A (6) (c) dθ y = y dθ = sin θsec θ dθ M A = tan θ dθ A ( sec θ ) = dθ dm ( C) = tanθ θ + A π π [ ] ( ) ( ) π d tan dm V = π y = θ θ = π π π p=, q= A = ( ) (7) ( marks) 0 Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics

24 WMA0/0: Core Mathematics C4. (a) dp = dt B P( P) Hence giving = A( P) + BP M A=, B = A dp = + dp P( P) P ( P) A dp = dt P( P) ln P ln( P) = t ( + c) M Aft { t = 0, P = } ln ln(4) = 0 + c c = ln 4 dm eg: ln P = t ln 4 P 4P ln = t P Using any of the subtraction (or addition) laws for logarithms CORRECTLY. dm eg: 4P e P = t or eg: P 4P t = e Eliminate ln s correctly. dm gives t t t t 4P = e Pe P(4 + e ) = e P t t e ( e ) = t t (4 + e ) ( e ) Make P the subject. dm P = t ( + 4 e ) etc. ( + 0e ) or P = t Note that the dm marks are dependent upon the first two M marks. () t + > P <. So population cannot eceed 000 B 4e A () ( marks) Pearson Edecel International Pearson Education Limited 0 Sample Assessment Materials Advanced Level in Mathematics 09

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

PMT. Mark Scheme (Results) Summer Pearson Edexcel International A Level in Core Mathematics C34 (WMA02/01) Mark Scheme (Results) Summer 04 Pearson Edexcel International A Level in Core Mathematics C4 (WMA0/0) Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest