Mark Scheme (Results) January 2007

Transcription

1 Mark (Results) January 007 GCE GCE Mathematics Core Mathematics C (6666) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 ** represents a constant. January Core Mathematics C Mark 5x 5x f(x) ( 5x) Takes outside the bracket to give any of () - or. B + ( )(* * x); + (* * x) + (* * x) +...!! Expands ( + * * x) to give an unsimplified + ( )(* * x) ; A correct unsimplified {... expansion } with candidate s (**x) M A ( )( ) ( )( )( ) ;... 5x 5x 5x !! 75x 5x 5x; x 75x 5x + ; Anything that cancels to 5x + ; 75x 5x Simplified A; A 5 + x; + x + 5 x [5] 5 marks

3 Aliter. Way f(x) ( 5x) ( )( ) () + () (* * x); + () (* * x)! ( )( )( ) 5 + () (* * x) +...! or () Expands ( 5x) to give an unsimplifed () + ( )() (* * x) ; A correct unsimplified {...} expansion with candidate s (**x) B M A ( )( ) () + ( )() ( 5x); + () ( 5x)! ( )( )( ) 5 + () ( 5x) +...! + ( )( 5x); + ()(5x ) ( )( 5x ) x 75x 5x + ; Anything that cancels to 5x + ; 75x 5x Simplified A; A 5 + x; + x + 5 x [5] 5 marks Attempts using Maclaurin expansions need to be referred to your team leader.

4 . (a) Volume x 9 x ( + ) ( + ) Use of V y. Can be implied. Ignore limits. B 9 ( + ) x Moving their power to the top. (Do not allow power of -.) Can be implied. Ignore limits and 9 M (+ x) 9 ( )() Integrating to give ± p( + x) ( x) M + A + 9 ( x) 9 () ( ) 9 Use of limits to give exact values of or or or aef (b) From Fig., AB ( ) units 6 A aef [5] As units cm then scale factor k. Hence Volume of paperweight ( ) (their answer to part (a)) M V 6 cm cm 6 or awrt or or aef A [] Note: 9 (or implied) is not needed for the middle three marks of question (a). 7 marks

5 Aliter. (a) Volu me Way x 6x ( + ) ( + ) ( ) ( + ) 6x Use of V y d x. Can be implied. Ignore limits. Moving their power to the top. (Do not allow power of -.) Can be implied. Ignore limits and B M ( + 6x) ( )(6) Integrating to give ± p( + 6x) M ( + 6x) 6 A + ( 6 6x) 6(6) 6 ( ) 6 9 Use of limits to give exact values of or or aef or 6 A aef [5] Note: is not needed for the middle three marks of question (a).

6 . (a) x 7cos t cos7t, y 7 sin t sin7t, 7sint + 7sin7t, 7cost 7cos7t Attempt to differentiate x and y with respect to t to give in the form ± A sin t ± B sin7t in the form ± Ccost ± Dcos7t Correct and M A 7cos t 7cos7t 7 sin t + 7 sin7t Candidate s B [] (b) 7 7cos 7cos W hen t, m(t) sin 7sin ; o Substitutes t or 0 into their 6 expression; M awrt to give any of the four underlined expressions oe (must be correct solution only) A cso Hence m(n) or awrt 0.58 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A oe. W hen t, x 7co co 6 6 s s y 7sin sin N : y ( ) x The point (, ) or ( awrt 6.9, ) Finding an equation of a normal with their point and their normal gradient or finds c by using y (their gradient)x + "c ". B M N: y x or y x or y x Correct simplified EXACT equation of normal. This is dependent on candidate using correct (, ) A oe or + c c 0 Hence N: y x or y x or y x [6] 9 marks 5

7 Aliter. (a) x 7cos t cos7t, y 7 sin t sin7t, Way 7sint + 7sin7t, 7cost 7cos t 7 cos7t 7( sin t sint) 7 sin t + 7 sin7t 7(cos t sint) 7cos7t Attempt to differentiate x and y with respect to t to give in the form ± A sin t ± B sin7t in theform ± Ccost ± Dcos7t Correct and tant Candidate s M A B [] (b) When t, m(t) tan ; 6 6 o Substitutes t or 0 into their 6 expression; M ( ) ( ) awrt.7 () to give any of the three underlined expressions oe (must be correct solution only) A cso Hence m(n) or awrt 0.58 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A oe. When t 6, 7 x 7cos cos ( 8 ) 6 6 y 7sin sin N: y ( ) x The point (, ) or ( awrt 6.9, ) Finding an equation of a normal with their point and their normal gradient or finds c by using y (their gradient)x + "c ". B M N: y x or y x or y x Correct simplified EXACT equation of normal. This is dependent on candidate using correct (, ) A oe or + c c 0 Hence N: y x or y x or y x [6] 9 marks 6

8 Beware: A candidate finding an m(t) 0 can obtain Aft for m(n), but obtains M0 if they write y (x ). If they write, however, N: x, then they can score M. B eware: A candidate finding an m(t) can obtain Aft for m(n) 0, and also obtains M if they write y 0(x ) or y. 7

9 . (a) x A B + (x )(x ) (x ) (x ) x A(x ) + B(x ) Let x, B B Forming this identity. NB: A & B are not assigned in this question M Let x, A( ) A either one of A or B. A both correct for their A, B. A giving + (x ) (x ) [] (b) & (c) (x ) y (x )(x ) Separates variables as shown Can be implied B + (x ) (x ) Replaces RHS with their partial fraction to be integrated. M ln y ln(x ) + ln(x ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c M A A [5] y 0, x gives c ln0 c ln0 B ln y ln(x ) + ln(x ) + ln0 ln y ln(x ) + ln(x ) + ln0 (x ) ln y ln + ln0 or (x ) 0(x ) ln y ln (x ) Using the power law for logarithms Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. M M y 0(x ) (x ) y 0(x ) (x ) or aef. isw A aef [] marks 8

10 Aliter. (b) & (c) Way (x ) y (x )(x ) + (x ) (x ) Separates variables as shown Can be implied Replaces RHS with their partial fraction to be integrated. B M ln y ln(x ) + ln(x ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c M A A See below for the award of B decide to award B here!! B ln y ln(x ) + ln(x ) + c (x ) ln y ln + c x Using the power law for logarithms Using the product and/or quotient laws for logarithms to obtain a single RHS logarithm ic term with/without constant c. M M ln y A(x ) ln x where c lna o r (x ) (x ) ln + c ln x x ln y e e e e c y A(x ) (x ) y 0, x gives award A 0 A 0 for B above y 0(x ) (x ) y 0(x ) (x ) or aef & isw A aef [5] & [] Note: The B mark (part (c)) should be awarded in the same place on epen as in the Way approach. 9

11 Aliter (b) & (c) Way (x ) y (x )(x ) Separates variables as shown Can be implied + es RHS with their partial (x ) (x ) Replacfraction to be integrated. B M ln y ln(x ) + ln(x ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c M A A [5] y 0, x gives c ln0 ln ln 0 c ln0 ln( ) or c ln0 B oe ln y ln(x ) + ln(x ) + ln 0 ln ) + y ln(x ) + ln(x ln0 Using the power law for logarithms M (x ) ln y ln + ln 0 or (x ) 0(x ) ln y ln (x ) Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. M y 0(x ) (x ) y 0(x ) (x ) or aef. isw A aef [] Note: Please mark parts (b) and (c) together for any of the three ways. 0

12 5. (a) sin x + cos y 0.5 ( eqn ) cos x sin y 0 ( eqn # ) Differentiates implicitly to include ± sin y. (Ignore ( ).) M cos x sin y cos x sin y A cso [] (b) cos x 0 0 cosx 0 sin y Candidate realises that they need to solve their numerator 0 or candidate sets 0 in their (eqn #) and attempts to solve the resulting equation. M giving x or x both x o, or x ± 90 or awrt x ±.57 required here A When x When x, sin + cos y 0.5, ( ) sin + cos y 0.5 x Substitutes either their or x into eqn M cos y.5 y has no solutions cos y 0.5 y or Only one of y o or or 0 or 0 or awrt -.09 or awrt.09 A In ( ) (, and, ) A specified range ( x, y ) (, ) and (, ) Only exact coordinates of Do not award this mark if candidate states other coordinates inside the required range. [5] 7 marks

13 6. (a) Way Aliter (a) Way x x ln y e ln.e ln.e xln xln Hence x x x ln. ln AG ln AG A cso x ln y ln leads to ln y x ln Takes logs of both sides, then uses the power law of logarithms and differentiates implicitly to ln y give ln y M M [] H ence x x yln ln AG ln AG A cso [] (b) (x ) y (x ) x..ln (x ) Ax (x ) x..ln or x.y.l n if y is define d M A When x, () ln Substitutes x into th eir which is of the form ± k x or Ax (x ) M 6ln ln or awrt. A [] 6 marks

14 Aliter 6. ) ln y ln( ) x (b leads to ln y Way x.ln y x ln Ax.ln y x.ln y M A When x, () ln Substitutes x into their which is of the form ± k x (x ) or Ax M 6ln ln or awrt. A []

15 7. a OA i + j + k OA b OB i + j k OB 8 BC ± ( i + j + k) BC AC ± i + j k AC 8 (a) c OC i + j k i + j k B cao [] (b) OA OB + 0 or B uu ur O BC + 0 or AC BC + 0 or AO AC + 0 An attempt to take the dot product between either OA and OB OA and AC, AC and BC or OB and BC Showing the result is equal to zero. M A and therefore OA is perpendicular to OB and hence OACB is a rectangle. perpendicular and OACB is a rectangle A cso Area Using distance formula to find either the correct height or wih. Multiplying the rectangle s height by its wih. exact value of 8, 9, 6 or aef (c) OD d ( i + j k ) ( + ) M M A i j k B [6] []

16 (d) Way using dot product formula 5 DA ± ( i + j + k ) DC ± i + j k BA ± i + j + 5k & OC ± i + j k or & cos D ( ± ) ( ± ) ( ± ) Identifies a set of two relevant vectors Correct vectors ± Applies dot product formula on multiples of these vectors. Correct ft. application of dot product formula M A D cos Attempts to find the correct angle D rather than 80 D. ddm o D or awr c t09 or.9 A Aliter using dot product formula and direction vectors [6] (d) Identifies a set of two M d BA ± ( i + j + 5k) & d OC ± ( i + j k) direction vectors Correct vectors ± A Way Applies dot product formula on multiples dm of these vectors Correct ft. cos D ( ± ) ( ± ) ( ± ) application of dot A product formula. D cos Attempts to find the correct angle D rather than 80 D. ddm o D or awrt09 or.9 c A [6] 5

17 Aliter using dot product formula a nd similar triangles doa i + j + k OC i + j k (d) Way cos ( D) & d D cos Identifies a set of two direction vectors Correct vectors Applies dot product formula on multiples of these vectors. Correct ft. application of dot product formula. Attempts to find the correct angle D by doubling their angle for. D M A dm A ddm o D or A awrt09 or.9 c [6] Aliter using cosine rule (d) D uu ur 5 A i + j + k Way DA 7,, DC + DC 7 i j k, AC 8 ( 8 ) cos D 7 7, AC i + j k Attempts to find all the lengths of all three edges of ADC All Correct Using the cosine rule formula with correct subtraction. Correct ft application of the cosine rule formula M A dm A D cos Attempts to find the correct angle D rather than 80 D. ddm o 09.5 or D A awrt09 or.9 c [6] 6

18 Aliter using trigonometry on a right angled triangle 5 (d) DA i + j + k OA i + j + k AC i + j k Way 5 Let X be the midpoint of AC 7 DA, DX OA, AX AC 8 (hypotenuse), (adjacent), (opposite) Attempts to find two out of the three lengths in ADX Any two correct M A 8 7 sin( D), 7 cos( D) or 8 tan( D) Uses correct sohcahtoa to find D Correct ft application of sohcahtoa dm A eg. 8 D tan Attempts to find the correct angle D by doubling their a ngle for. D ddm D o Aliter using trigonometry on a right angled similar triangle OAC (d) OC i + j k OA i + j + k AC i + j k Way 6 OC 7, OA, AC 8 (hypotenuse), (adjacent), (opposite) 09.5 or awrt09 or.9 c A Attempts to find two out of the three lengths in OAC M Any two correct A [6] 8 sin( D), cos( D) or tan( D) Uses correct sohcahtoa to find dm Correct ft application of sohcahtoa A eg. 8 D tan Attempts to find the correct angle D by doubling their angle for. D ddm o 09.5 or D A awrt09 or.9 c [6] 7

19 Aliter 7. (b) (i) Way c OC ± i + j k AB ± i j 5k A complete method OC () + () + ( ) () + () + ( 5) AB of proving that the diagonals are equal. A s OC AB 7 Correct result. M A then the diagonals are equal, and OACB is a rectangle. diagonals are equal and OACB is a rectangle A cso [] a OA i + j + k OA b OB i + j k OB 8 BC ± ( i + j + k) BC AC ± ( i + j k) AC 8 c OC ± ( i + j k) OC 7 AB ± i j 5k AB 7 Aliter 7. (b) (i) ( OA) + ( AC) ( OC) or ( BC) + ( OB) ( OC) or ( OA) + ( OB) ( AB) Way or ( BC) + ( AC) ( AB) + () ( 8) 7 or equivalent A complete method of proving that Pythagoras holds using their values. Correct result M A and therefore OA is perpendicular to OB or AC is perpendicular to BC and hence OACB is a rectangle. perpendicular and OACB is a rectangle A cso [] marks 8

20 Sch eme 8. (a) x 0 5 y e e e 7 0 e e e or y Either e, e and e or awrt.,.6 and 6.8 or e to the power awrt.65,.6,.6 (or mixture of decimals and e s) At least two correct All three correct B B [] (b) { 7 0 } I ; e + e + e + e + e + e Outside brackets For structure of trapezium rule{...} ; B; M A 0.6 (sf) 0.6 cao [] Beware: In part (b) candidates can add up the individual trapezia: (b) ( + ) + ( + ) + ( + ) + ( + ) + ( + ) I.e e.e e.e e.e e.e e 9

21 (c) A(x ) t (x + )..(x + ) + or t A or t x + t (x + ) or t M A so.(x + ) t t Candidate obt ains either or in terms of t (x+ ) I e t e. t t e.. and moves on to dm substitute this into I to convert an integral wrt x to an integral wrt t. t I te change limits: when x 0, t & when x 5, t t te changes limits x t so that 0 and 5 A B t ence I H te ; where a, b, k (d) du u t dv t t e v e Let k be any constant for the first three marks of this part. [5] k te t k te t e t. t t k te e + c { ( } t te e e e e ) Use of integration by parts formula in the correct direction. Correct expression with a constant factor k. Correct integration with/without a constant factor k Substitutes their changed limits into the integrand and subtracts oe. M A A dm oe either e or awrt 09. A (e ) e 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. [5] 5 marks 0