{... expansion. January Core Mathematics C4 Mark Scheme. Question Number ** represents a constant

Transcription

1 January Core Mathematics C Mark Question ** represents a constant. 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) A ( )( ) ( )( )( ) ;... 5x 5x 5x !! 75x 5x 5x; x 75x 5x + ; Anything that cancels to 5x + ; 75x Simplified 5x + A; 6 8 A 5 + x; + x + 5 x [5] 5 marks 9

2 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 A ( )( ) () + ( )() ( 5x); + () ( 5x)! ( )( )( ) 5 + () ( 5x) +...! + ( )( 5x); + ()(5x ) ( )( 5x ) x 75x 5x + ; Anything that cancels to 5x + ; 75x Simplified 5x + A; 6 8 A 5 + x; + x + 5 x [5] 5 marks Attempts using Maclaurin expansions need to be referred to your team leader. 0

3 . (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 (+ x) 9 ( )() Integrating to give ± p( + x) ( x) + A + 9 ( x) 9 () ( ) 9 Use of limits to give exact values of or or or aef 6 (b) From Fig., AB ( ) units A aef [5] As units cm then scale factor k. Hence Volume of paperweight ( ) (their answer to part (a)) 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

4 Aliter. (a) Volume Way x 6x ( + ) ( + ) ( ) ( + ) 6x Use of V y. Can be implied. Ignore limits. Moving their power to the top. (Do not allow power of -.) Can be implied. Ignore limits and B ( + 6x) ( )(6) Integrating to give ± p( + 6x) ( + 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).

5 . (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 A 7cos t 7cos7t 7 sin t + 7 sin7t Candidate s B [] (b) 7 7cos 7cos When t, m(t) sin 7sin ; Substitutes t 6 or 0 o into their 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. When t 6, 7 x 7cos cos 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 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 Aliter. (a) x 7cos t cos7t, y 7 sin t sin7t, Way 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 theform ± Ccost ± Dcos7t Correct and A 7cost 7cos7t 7( sint sint) tant 7sint + 7sin7t 7(cost sint) Candidate s B [] (b) When t, m(t) tan ; 6 6 Substitutes t 6 or 0 o into their ( ) ( ) 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 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 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

7 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. Beware: A candidate finding an m(t) can obtain Aft for m(n) 0, and also obtains if they write y 0(x ) or y. 5

8 . (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 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. 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 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 Using the power law for logarithms (x ) ln y ln + ln0 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. y 0(x ) (x ) y 0(x ) (x ) or aef. isw A aef [] marks 6

9 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 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 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 logarithmic term with/without constant c. ln y A(x ) ln x where c lna or (x ) (x ) ln + c ln x x lny e e e e c y A(x ) (x ) y 0, x gives A 0 A 0 for B award 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. 7

10 Aliter (b) & (c) Way (x ) y (x )(x ) Separates variables as shown Can be implied + Replaces RHS with their partial (x ) (x ) fraction to be integrated. B 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 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 (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. 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. 8

11 5. (a) sin x + cos y 0.5 ( eqn ) cos x sin y 0 ( eqn # ) Differentiates implicitly to include ± sin y. (Ignore ( ).) 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. 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 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 specified range ( x, y, ) and (, ) Only exact coordinates of, and, Do not award this mark if candidate states other coordinates inside the required range. A [5] 7 marks 9

12 6. x x ln y e (a) Way Aliter ln.e ln.e xln xln Hence x (a) ln y ln( ) Way x x ln. ln AG x ln AG A cso leads to ln y x ln Takes logs of both sides, then uses the power law of logarithms ln y and differentiates implicitly to give ln y [] Hence x yln ln AG x ln AG A cso [] (b) (x ) y (x ) x..ln (x ) Ax (x ) x..ln or x.y.ln if y is defined A When x, () ln Substitutes x into their which is of the form ± k x or Ax (x ) 6ln ln or awrt. A [] 6 marks 50

13 Aliter 6. (b) ln y ln( ) x Way leads to x.ln y ln y x ln Ax.ln y x.ln y A When x, () ln Substitutes x into their which is of the form ± k x or Ax (x ) 6ln ln or awrt. A [] 5

14 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 BO 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. A and therefore OA is perpendicular to OB and hence OACB is a rectangle. perpendicular and OACB is a rectangle A cso Area (c) OD d ( i + j k ) 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 ( + ) A i j k B [6] [] 5

15 (d) Way using dot product formula 5 DA ± ( i + j + k ) BA ± i + j + 5k or & DC ± ( i + j k ) & OC ± ( i + j k) 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. A d A D cos Attempts to find the correct angle D rather than 80 D. dd Aliter o D using dot product formula and direction vectors (d) d BA ± ( i + j + 5k) Way & d OC ± ( i + j k) cos D ( ± ) ( ± ) ( ± ) or awrt09 or.9 c 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. A A d A [6] D cos Attempts to find the correct angle D rather than 80 D. dd o D or awrt09 or.9 c A [6] 5

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

17 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 A 8 7 sin( D), 7 cos( D) or 8 tan( D) Uses correct sohcahtoa to find D Correct ft application of sohcahtoa d A eg. 8 D tan Attempts to find the correct angle D by doubling their angle for D. dd o D or awrt09 or.9 c A [6] 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) Attempts to find two out of the three lengths in OAC Any two correct A 8 sin( D), cos( D) or tan( D) Uses correct d sohcahtoa to find D Correct ft application of sohcahtoa A eg. 8 D tan Attempts to find the correct angle D by doubling their angle for D. dd o D or awrt09 or.9 c A [6] 55

18 Aliter 7. (b) (i) Way c OC ± i + j k AB ± 5 ( i j k) A complete method of proving that the diagonals are equal. OC () + () + ( ) () + () + ( 5) AB As OC AB 7 Correct result. A then the diagonals are equal, and OACB is a rectangle. 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 diagonals are equal and OACB is a rectangle A cso [] Aliter ( OA) + ( AC) ( OC) 7. (b) (i) or ( BC) + ( OB) ( OC) or ( OA) + ( OB) ( AB) or equivalent Way or ( BC) + ( AC) ( AB) + () ( 8) 7 A complete method of proving that Pythagoras holds using their values. Correct result 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 56

19 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; (sf) 0.6 A cao [] Beware: In part (b) candidates can add up the individual trapezia: (b) ( + ) + ( + ) + ( + ) + ( + ) + ( + ) I.e e.e e.e e.e e.e e 57

20 (c) t (x + )..(x + ) or t x+ t A(x ) + or t A (x + ) or t A so.(x + ) t t Candidate obtains either or in terms of t (x+ ) I e e t. t t e.. and moves on to substitute this into I to convert an integral wrt x to an integral wrt t. d I t te t te A change limits: when x 0, t & when x 5, t changes limits x t so that 0 and 5 B t Hence I te ; where a, b, k [5] (d) du u t dv t t e v e Let k be any constant for the first three marks of this part. 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. A A d oe (e ) e either e or awrt 09. A [5] 5 marks Note: d denotes a method mark which is dependent upon the award of the previous method mark dd denotes a method mark which is dependent upon the award of the previous two method marks. 58