Mark (Final) January 009 GCE GCE Core Mathematics C (6666/0) Edecel Limited. Registered in England and Wales No. 96750 Registered Office: One90 High Holborn, London WCV 7BH
General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly 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. Eaminers 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. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers 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 eemplification may be limited. When eaminers are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
January 009 6666 Core Mathematics C Mark. (a) C: y y + 8 y Differentiates implicitly to include either ky ± or ±. (Ignore d d.) M Correct equation. ( y ) A correct (condoning sign error) attempt to combine or factorise their y. d d Can be implied. M y y oe [] (b) y 9 () + 8 Substitutes y into C. M 8 Only d () d (,) y y 6 from correct working. Also can be ft using their value and y in the correct part (a) of y [] 7 marks (b) final. Note if the candidate inserts their value and y into y, then an answer of their, may indicate a correct follow through.
. (a) Area(R) (+ ) ( + ) 0 0 Integrating ( + ) to give ( + ) ± k( + ).. 0 Correct integration. Ignore limits. M ( + ) 0 ( ) 9 ( () ) Substitutes limits of and 0 into a changed function and subtracts the correct way round. M 9 (units) (Answer of with no working scores M0A0M0A0.) [] (b) Volume π ( + ) 0 Use of V π y. Can be implied. Ignore limits and d. B ( ) π 0 9 + ± kln + M 9 ( π ) ln + 0 9 ln + ( ) ( 9 ln 9) ( 9 ln) π Substitutes limits of and 0 and subtracts the correct way round. dm 9 So Volume π ln 9 9 π ln 9 or 9 π ln or 8 π ln oe isw [5] 9 marks Note the answer must be a one term eact value. Note, also you can ignore subsequent working here. Note that ln can be implied as equal to 0. 9 Note that π ln 9 + c (oe.) would be awarded the final A0.
. (a) 7 + + 6 A(+ )( ) + B( ) + C(+ ) Forming this identity M 6 ( 5 ) 0 ( 5 ), + 6 B B B, 7 + + 6 5C 75 5C C Substitutes either or into their identity or equates terms or substitutes in values to write down three simultaneous equations. Both B and C (Note the is dependent on both method marks in this part.) M Equate : 7 A+ 9C 7 A + 7 0 A A 0 0, 6 A + B + C 6 A + + 0 A A 0 Compares coefficients or substitutes in a third -value or uses simultaneous equations to show A 0. B [] (b) f( ) + ( + ) ( ) ( + ) + ( ) Moving powers to top on any one of the two epressions M ( ) + + ( ) ( ) + + ( ) Either ± ( )( ) or ( )( ) + ( )( ); + ( ) +... ± ( )( ) from either first or dm;! second epansions respectively Ignoring and, any one ( )( ) + ( )( ); ( )... correct {...} epansion. + + +! Both {...} correct. 7 {...} {... } + + + + + + + 0 ; + 9 (0 ); 9 + ; [6]
. (c).08 + 6. + 6 Actual f (0.) (6.76)(0.8).8 95.75976... 5.08 676 Or Actual f (0.) + ((0.) + ) ( 0.) 95 +.75.75976... 6.76 676 Attempt to find the actual value of f(0.) or seeing awrt. and believing it is candidate s actual f(0.). Candidates can also attempt to find the actual value by using A + B + C ( + ) (+ ) ( ) with their A, B and C. M 9 Estimate f (0.) + (0.) + 0.9.9 Attempt to find an estimate for f(0.) using their answer to (b) M.9.75976... %age error 00.75976... their estimate - actual actual 00 M.09508....%(sf ).% cao [] marks
. (a) d i + j k, d qi + j+ k As q d d ( q) + ( ) + ( ) Apply dot product calculation between two direction vectors, ie. ( q) + ( ) + ( ) M d d 0 q + 8 0 Sets d d 0 q 6 q AG and solves to find q cso [] (b) Lines meet where: 5 q + λ + µ 7 p First two of i : λ 5 + qµ () j: + λ + µ () k :7 λ p + µ () Need to see equations () and (). Condone one slip. (Note that q.) M () + () gives: 5 7 + µ µ () gives: + λ λ 5 Attempts to solve () and () to find one of either λ or µ dm Any one of λ 5orµ Both λ 5andµ () 7 (5) p + ( ) Attempt to substitute their λ and µ into their k component to give an equation in p alone. ddm p 7 0 + p p cso [6] (c) 5 r + 5 or r 7 Substitutes their value of λ or µ into the correct line l or l. M Intersect at r 7 or (, 7, ) 7 or (, 7, ) []
(d) Let OX i + 7j k be point of intersection 9 8 Finding vector AX by finding the AX OX OA 7 difference between OX and OA. Can M ± 6 be ft using candidate sox. OB OA + AB OA + AX OB 9 8 + 6 9 + their AX dm Hence, OB 7 9 or OB 7i + j 9k 7 9 or 7i + j 9k or ( 7,, 9) [] marks
5. (a) Similar triangles r 6 h r h Uses similar triangles, ratios or trigonometry to find either one of these two epressions oe. M h π h V π r h π h 7 AG Substitutes r h into the formula for the volume of water V. [] (b) From the question, d V 8 dt dv dt 8 B dv π h π h dh 7 9 d V π h π or h B dh 7 9 dh dv dv 9 8 8 dt dt dh π h π h π h 8 7 Candidate s d V dt or d V ; M; dh 9 8 8 or π h π h oe When dh 8 h, dt π 8π 8 π or 8π oe isw [5] 7 marks Note the answer must be a one term eact value. Note, also you can ignore subsequent working after 8 π.
6. (a) tan NB : sec A + tan A gives tan A sec A The correct underlined identity. M oe sec tan ( + c) Correct integration with/without + c [] (b) ln du u ln v dv ln. Use of integration by parts formula in the correct direction. Correct epression. M ln d + ln + + ( c) An attempt to multiply through k n, n, n by and an attempt to... integrate (process the result); M correct solution with/without + c oe [] Correct direction means that u ln.
(c) e + e du u + e e,, du e du u Differentiating to find any one of the three underlined B e.e ( u ).e d. du + e u e ( u ) or. du u ( u ) Attempt to substitute for e f( u), their d and u + du e e or e f( u), their d and du u u + e. M* ( u ) u du ( u ) u du u u+ d u u u + du u u u + lnu + c ( ) An attempt to multiply out their numerator to give at least three terms and divide through each term by u Correct integration with/without +c dm* ( + e ) (+ e ) + ln(+ e ) + c Substitutes u + e back into their integrated epression with at least two terms. dm* + + + + + c e e e ln( e ) + + + + + c e e e ln( e ) e e + ln(+ e ) + c + + + k AG e e ln( e ) must use a + + + k e e ln( e ) + c and " " combined. cso [7] marks
7. (a) At A, + y A A(7,) 8 7 & ( ) (7,) B [] (b) t t y t 8,, d t y d 8 t, d t t t t 8 Their d y divided by their d dt dt Correct d y M ( ) At A, m( T ) Substitutes for t to give any of the ( ) 8 8 5 5 four underlined oe: ( ) T ( ( )) T : y their m their 7 7 + c c 9 or ( ) 5 5 5 9 Hence T : y 5 5 Finding an equation of a tangent with their point and their tangent gradient or finds c and uses y (their gradient) + " c". dm gives T : 5y 9 0 AG 5y 9 0 cso [5] (c) ( t 8 t) 5t 9 0 Substitution of both t 8t and y t into T M t 5t 6t 9 0 { } ( t+ ) (t 7t 9) 0 { } ( t+ ) ( t+ )(t 9) 0 9 { } t (at A) t atb A realisation that t + is a factor. ( ) 9 t dm Candidate uses their value of t to 9 9 79 ( ) 8( ) 8 6 8 55.5 or awrt 55. find either the or y coordinate ddm 9 8 y ( ) 0.5 or awrt 0. One of either or y correct. Both and y correct. B 8, awrt [6] Hence ( ) 8 marks
Note: dm denotes a method mark which is dependent upon the award of the previous method mark. ddm denotes a method mark which is dependent upon the award of the previous two method marks. Oe or equivalent.
January 009 6666 Core Mathematics C Appendi. (a) Way C: y y + 8 y Differentiates implicitly to include either ± k. (Ignore.) Correct equation. M y Applies ( ) dm ( ) y y oe []. (a) Way C: y y + 8 gives y y 8 ( ) 8 y y ( y y 8) ( y ) Differentiates in the form ( y ) ( ) y f( ) f ( ). M Correct differentiation. y y y 8 ( ) ( ) y y 8 y Applies dm ( ) ( ) y y ( ) y or y oe []
. (a) Way. (a) Way Area(R) (+ ) 0 ( + ) 0 du {Using substitution u + } {change limits: When 0, u & when, u 9 } So, Area(R) 9 u du u ( ) 9 u 0 Integrating λ ± u to give ± ku. M Correct integration. Ignore limits. Substitutes limits of either ( u 9andu ) or ( ) 9 ( () ) in, ( an 0) into a changed M function and subtracts the correct way round. 9 (units) Area(R) (+ ) 0 ( + ) 0 {Using substitution d u + u u udu } {change limits: When 0, u & when, u } So, Area(R) d u u u u du Integrating ± λ to give ± ku. M Correct integration. Ignore limits. [] ( ()) ( () ) Substitutes limits of either ( u andu ) or in, ( an 0) into a changed function and subtracts the correct way round. M 9 (units) []
. (a) Way 7 + + 6 A(+ )( ) + B( ) + C(+ ) Forming this identity M terms : 7 A + 9 C () terms : A B + C () constants: 6 A + B + C () () + () gives 8 A + 6C () () + () gives 75 5C C () gives 7 A + 7 0 A A 0 equates terms. M () gives B + 6 B 6 Both B and C Decide to award B for A 0 B []. (a) If the candidate assumes A 0 and writes the identity 7 + + 6 B( ) + C(+ ) and goes on to find B and C then the candidate is awarded M0MA0B0.. (a) If the candidate has the incorrect identity 7 + + 6 A(+ ) + B( ) + C(+ ) and goes on to find B, C and A 0 then the candidate is awarded M0MA0B.. (a) If the candidate has the incorrect identity 7 + + 6 A(+ ) ( ) + B( ) + C(+ ) and goes on to find B, C and A 0 then the candidate is awarded M0MA0B.
. (b) Way f( ) + + ( ) ( ) ( + ) + ( ) Moving powers to top on any one of the two epressions M ( + ) + ( ) Either () ± ( )() ( ) or ( )( ) () ( )() ( ); () ( ) ± ( )( ) from either first or dm;! second epansions respectively Ignoring and, any one ( )( ) + ( )( ); ( )... correct {...} epansion. + + +! Both {...} correct. + + + 7 { } { 6 } + +... + + + +... + 0 ; + 9 9 + ; (0 ); [6]
. (c) Way.08 + 6. + 6 Actual f (0.) (6.76)(0.8).8 95.75976... 5.08 676 Attempt to find the actual value of f(0.) M 9 Estimate f (0.) + (0.) + 0.9.9 Attempt to find an estimate for f(0.) using their answer to (b) M.9 %age error 00 00.75976... their estimate 00 00 actual M 00 0.095.09508....%(sf ).% cao []. (c) Note that:.9.75976... %age error 00.9.099868....%(sf ). (c) Also note that:.75976... %age error 00 00.9.099868....%(sf ) Should be awarded the final marks of M0A0 Should be awarded the final marks of M0A0 so be wary of.099868
. (a) q + 8 is sufficient for M.. (b) Way Only apply Way if candidate does not find both λ and µ. Lines meet where: 5 q + λ + µ 7 p First two of i : λ 5 + qµ () j: + λ + µ () k :7 λ p + µ () Need to see equations () and (). Condone one slip. (Note that q.) M () gives λ 9 + µ () gives (9 + µ ) 5 µ 8 µ 5 µ Attempts to solve () and () to find one of either λ or µ dm gives: 8 + 5 µ µ Any one of λ 5orµ () gives 7 (9 + µ ) p + µ () 7 (9 + ( )) p + ( ) Candidate writes down a correct equation containing p and one of either λ or µ which has alrea been found. Attempt to substitute their value for λ ( 9 + µ ) and µ into their k component to give an equation in p alone. ddm. (c) 7 0 p p p If no working is shown then any two out of the three coordinates can imply the first M mark. cso M [6] Intersect at r 7 or (, 7, ) 7 or (, 7, ) []
. (d) Let OX i + 7j k be point of intersection Way 9 8 AX OX OA 7 6 OB OX + XB OX + AX Finding the difference between their OX (can be implied) and OA. 9 M ± AX ± 7 OB 8 7 + 6 their OX + their AX dm Hence, OB 7 9 or OB 7i + j 9k 7 9 or 7i + j 9k or ( 7,, 9) [] At A, λ. At X, λ 5.. (d) Way Hence at B, λ 5 + (5 ) 9 ( their ) ( their their ) λ ( theirλ ) ( their λ ) λ λ + λ λ B X X A B X A M OB + 9 7 Substitutes their value of λ into the line l. dm Hence, OB 7 9 or OB 7i + j 9k 7 9 or 7i + j 9k or ( 7,, 9) []
. (d) Way OA 9i + j+ k and the point of intersection OX i + 7j k 9 Minus 8 Plus 7 Minus 6 Finding the difference between their OX (can be implied) and OA. 9 AX ± 7 ( ) M ± Minus 8 7 7 Plus Minus 6 9 their OX + their AX dm Hence, OB 7 9 or OB 7i + j 9k 7 9 or 7i + j 9k or ( 7,, 9) []. (d) Way 5 OA 9i + j+ k and OB ai + bj+ ck and the point of intersection OX i + 7j k As X is the midpoint of AB, then 9+ a + b + c, 7,,, ( ) a () 9 7 b (7) c ( ) 9 Writing down any two of these equations correctly. An attempt to find at least two of a, b or c. M dm Hence, OB 7 9 or OB 7i + j 9k 7 9 or 7i + j 9k or ( 7,, 9) or a 7, b, c 9 []
. (d) Let OX i + 7j k be point of intersection Way 6 9 8 AX OX OA 7 6 and AX 6 + 6 + 56 6 λ 0 + λ BX OX OB 7 + λ 5 λ 7 λ 0 + λ Hence BX AX 6 gives ( λ) ( λ) ( λ) 0 + + 5 + 0 + 6 Finding the difference between their OX (can be implied) and OA. 9 AX ± 7 Note AX 6 would imply M. Writes distance equation of BX 6 where BX OX OB and λ OB + λ 7 λ M ± dm 00 0λ + λ + 5 0λ + λ + 00 60λ + 6λ 6 λ 0λ + 55 6 λ 0λ + 89 0 λ 0λ + 9 0 ( λ )( λ 9) 0 At A, λ and at B λ 9, so, OB (9) + 9 7 (9) 7 Hence, OB or OB 7i + j 9k 9 7 9 or 7i + j 9k or ( 7,, 9) []
5 5. (a) Similar shapes either π π (6) V V h or h π (6) r () or V h V h π r () Uses similar shapes to find either one of these two epressions oe. M V h π h 08π 7 AG Substitutes their equation to give the correct formula for the volume of water V. [] 5. (a) Candidates simply writing: V 9 π or h V 6 π h would be awarded M0A0. dv dt (b) From question, 8 V 8t ( + c) dv dt 8 or V 8t B 7 7(8 ) 5 h V h t t t π π π π 7(8 t) 5t t or or π π π B dh t dt π dh ± kt ; M; dt dh t dt π oe When π h, t π So when dh h, dt π π 0π 8π 8π oe [5]
7 7. (a) It is acceptable for a candidate to write 7, y, to gain B. A(7,) B [] (c) Way 8 ( 8) ( 8) t t t t t y So, t ( y 8) y( y 8) 5y 9 0 5y + 9 (5y + 9) Hence, y( y 8) (5y + 9) Forming an equation in terms of y only. M yy ( 6y+ 6) 5y + 90y+ 8 y 6y + 56y 5y + 90y + 8 y 89y + 66y 8 0 ( y )( y )(y 8) 0 A realisation that y is a factor. ( ) Correct factorisation dm 8 y 0.5 (or awrt 0.) Correct y-coordinate (see below!) 8 8 Candidate uses their y-coordinate ( 8) to find their -coordinate. ddm Decide to award here for correct y-coordinate. 8 55.5 (or awrt 55.) Correct -coordinate B 8, Hence ( ) 8 [6]
7. (c) Way t y So ( y) 8( y) 5y 9 0 yields ( ) ( ) y 6 y 5y 9 0 ( y) y ( y) 5 6 9 0 Forming an equation in terms of y only. M { } { } ( y ) ( y y ) + 7 9 0 ( y ) ( y )( y ) A realisation that ( y + ) is a factor. + + 9 0 Correct factorisation. 8 y 0.5 (or awrt 0.) Correct y-coordinate (see below!) dm ( ) ( ) 8 8 8 Candidate uses their y-coordinate to find their -coordinate. ddm Decide to award here for correct y-coordinate. 55.5 (or awrt 55.) 8 Correct -coordinate B 8, Hence ( ) 8 [6]