Mark Scheme (Results) January 2010

Transcription

1 . Mark (Results) January 00 GCE Core Mathematics C (666) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

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3 January 00 Core Mathematics C 666 Mark Question Q Notes 6 = 79, 58x, + 5x ( ) x = + 6 ( x) + ( x) for either the x term or the x term. Requires correct binomial coefficient in any form with the correct power of x condone lack of negative sign and wrong power of. This mark may be given if no working is shown, but one of the terms including x is correct. Allow 6,or 6 (must have a power of, even if only power ) 6 First term must be 79 for B, ( writing just is B0 ) can isw if numbers added to this constant later. Can allow 79( Term must be simplified to 58x for Acao. The x is required for this mark. Final Ais c.a.o and needs to be + 5x (can follow omission of negative sign in working) Descending powers of x would be x + 6 ( x) + ( x) i.e. x 8x + 5 x +.. This is BA0A0 if completely correct or B0A0A0 for correct binomial coefficient in any form with the correct power of x as before B,A, A [] Alternative NB Alternative method: ( ) x x x = ( + 6 ( ) + ( )..) + is B0A0A0 answers must be simplified to 79, 58x, + 5x for full marks (awarded as before) The mistake ( ) x x x x = ( ) = (+ 6 ( ) + ( )..) + may also be awarded B0A0A0 6 Another mistake ( 6x+ 5 x...) = would be BA0A0 GCE Core Mathematics C (666) January 00

4 Question Q ( x) 5sin x = + sin sin x+ 5sin x = 0 (*) ( s )( s+ ) = 0 giving s = [ sin has no solution] x = so sin x = x = 0, 50 for a correct method to change cos x into sin x (must use cos x = sin x ) A need term quadratic printed in any order with =0 included for attempt to solve given quadratic (usual rules for solving quadratics) (can use any variable here, s, y, x, or sinx ) A requires no incorrect work seen and is for sin x = or x = sin y = is A0 (unless followed by x = 0) B for 0 (α ) not dependent on method nd B for 80 - α provided in required range (otherwise 50 - α ) Extra solutions outside required range: Ignore Extra solutions inside required range: Lose final B Answers in radians: Lose final B S.C. Merely writes down two correct answers is M0A0BB Or sin x = x = 0, 50 is ABB Just gives one answer : 0 only is M0A0BB0 or 50 only is M0A0B0B NB Common error is to factorise wrongly giving( x )( x ) [ sin x = gives no solution ] sin x= x= 0, 0 This earns A0 B0 Bft sin + sin = 0 Another common error is to factorise correctly ( sin x )( sin x + ) = 0 with sin x =, sin x = then x = 0 This would be A0 B B o o,50 and follow this Acso () A B, Bft () [6] GCE Core Mathematics C (666) January 00

5 Question Q f( ) = + a + b 6 8 f ( ) = 5 a+ b= or a+ b= f ( ) = 6 + a b 6 f ( ) = 0 a b= Eliminating one variable from linear simultaneous equations in a and b a = 5 and b = - ( )( ) x + 5x x 6= x+ x + x = ( x+ )( x+ )( x ) NB ( x+ )( x+ )( x ) is A0 But ( x )( x )( x ) + + is A st for attempting f( ± ) Treat the omission of the 5 here as a slip and allow the M mark. st A for first correct equation in a and b simplified to three non zero terms (needs 5 used) s.c. If it is not simplified to three terms but is correct and is then used correctly with second equation to give correct answers- this mark can be awarded later. nd for attempting f( m ) nd A for the second correct equation in a and b. simplified to three terms (needs 0 used) s.c. If it is not simplified to three terms but is correct and is then used correctly with first equation to give correct answers - this mark can be awarded later. rd for an attempt to eliminate one variable from linear simultaneous equations in a and b rd A for both a = 5 and b = - (Correct answers here imply previous two A marks) st for attempt to divide by (x+) leading to a TQ beginning with correct term usually x nd for attempt to factorize their quadratic provided no remainder A is cao and needs all three factors Ignore following work (such as a solution to a quadratic equation). Alternative; a+ for dividing by (x ), to get x + ( ) x+ constant with remainder as a function of a and b, and A as before for equations stated in scheme. for dividing by ( x + ), to get x + ( a ) x... (No need to see remainder as it is zero and comparison of coefficients may be used) with A as before Alternative; for finding second factor correctly by factor theorem, usually (x ) for using two known factors to find third factor, usually (x ± ) Then A for correct factorisation written as product ( x+ )( x+ )( x ) A A A (6) A () [9] GCE Core Mathematics C (666) January 00 5

6 Question Q sin( ACB ˆ ) sin 0.6 Either = 5 ACB ˆ = arcsin( ) = [ or.58] Use angles of triangle ABC ˆ = π 0.6 ACB ˆ (But as AC is the longest side so) ABC ˆ =.76 (*)(sf) [Allow00.7 o.76] In degrees o o 0.6 =.77, AĈB =.9 or = b + 5 b 5cos0.6 0 cos 0.6 (00 cos 0.6 6) ± b = = [6.96 or.9] Use sine / cosine rule with value for b sin 0.6 sin B = b or b cos B = 0 (But as AC is the longest side so) ˆ.76 ABC = (*)(sf), A () [ BD ˆ = π.76 =.8].76 =.0 ~. 79. is M0 C Sector area = ( π ) [ ] = 5 sin.76 = 9.8 or 5 sin0 Area of ABC ( ) [ ] Required area = awrt 0.8 or 0.9 or.0 or gives (sf) after correct work. A () [7] st for correct use of sine rule to find ACB or cosine rule to find b (M0 for ABC here or for use of sin x where x could be ABC) nd for a correct expression for angle ACB (This mark may be implied by.785 or by arcsin (.7058)) and needs accuracy. In second method this is for correct expression for b may be implied by [Note 0 cos ] (do not need two answers) rd for a correct method to get angle ABC in method (i) or sinabc or cosabc, in method (ii) (If sin B >, can have A0) Acso for correct work leading to.76 sf. Do not need to see angle 0.85 considered and rejected. st for a correct expression for sector area or a value in the range.0. nd for a correct expression for the area of the triangle or a value of 9.8 Ignore 0. (working in degrees) as subsequent work. A for answers which round to 0.8 or 0.9 or.0. No need to see units. Special case If answer.76 is assumed then usual mark is M0 M0 M0 A0. A Fully checked method may be worth M0 A0. A maximum of marks. The mark is either or 0. Either for AC ˆ B is found to be 0,786 (angles of triangle) then for checking sin( ACB ˆ ) sin 0.6 = with conclusion giving numerical answers 5 This gives a maximum mark of / OR for b is found to be 6.97 (cosine rule) for checking sin( ABC) sin 0.6 b = with conclusion giving numerical answers This gives a maximum mark of / Candidates making this assumption need a complete method. They cannot earn M0. So the score will be 0 or for part. Circular arguments earn 0/.. GCE Core Mathematics C (666) January 00 6

7 Question Q5 log 6 = 6 = x x So x = 8 A () ( x) ( x ) log 6 = log + 6x log = ( x ) 6x = ( x ) { 6 x= 8 x x+ } and so 0= 8x 0x ( ) ( x )( x ) 0= + x =... x =, for getting out of logs A Do not need to see x = -8 appear and get rejected. Ignore x = -8 as extra solution. x= 8 with no working is A st for using the nlogx rule nd for using the logx - logy rule or the logx + logy rule as appropriate rd for using to the power need to see or 8 (May see = log 8 used) If all three M marks have been earned and logs are still present in equation 6x do not give final. So solution stopping at log = log 8 would earn ( x ) M0 st A for a correct TQ th dependent for attempt to solve or factorize their TQ to obtain x = (mark depends on three previous M marks) nd A for.5 (ignore -0.5) s.c.5 only no working is 0 marks A d A (6) [8] Alternatives log 6 Change base : (i) log x =, so log x = and log0 6 (ii) log x =, log x = log 6 so x = 0 x =, is or 6 is then x = 8 is A BUT log x = 0.90 so x = 8 is A0 (loses accuracy mark) (iii) log6 x = so x = 6 is then x = 8 is A GCE Core Mathematics C (666) January 00 7

8 Question Q6 (d) (d) (d) 8000 ( 0.8 ) = 96 [may see or 80% or equivalent]. 5 ( ) < 000 S 5 n n log( 0.8) < log ( ) 8 ( ) log 8 n > = so n =. log(0.8) u ( ) 5 = 00., =.70 or.7 5 ( ) 00. =. or 5 ( ) 00.., = awrt 760 B NB Answer is printed so need working. May see as above or 0.8 in three steps giving 00, 50, 96. Do not need to see sign but should see 96. st for an attempt to use nth term and 000. Allow n or n and allow > or = nd for use of logs to find n Allow n or n and allow > or = A Need n = This is an accuracy mark and must follow award of both M marks but should not follow incorrect work using n for example. Condone slips in inequality signs here. for use of their a and r in formula for 5 th term of GP A cao need one of these answers answer can imply method here NB.7 A0 for use of sum to 5 terms of GP using their a and their r ( allow if formula stated correctly and one error in substitution, but must use n not n - ) st A for a fully correct expression ( not evaluated) Alternative Methods Trial and Improvement See ( or 989 or 990) identified with, or years for first See 6.95 ( or 6 or 7) identified with, or years for second Then n = is A (needs both Ms) n Special case 8000 ( 0.8) < 000 so n = as <000 is M0A0 (not discounted n = ) May see the terms, 50.88, 80.99,.7 with a small slip for A0, or done accurately for A Adds 5 terms (977.) Seeing 977 is A Obtains answer A or awrt 760 NOT 750 Bcso () A cso (), A () A, A () [9] GCE Core Mathematics C (666) January 00 8

9 Question Q7 (d) Puts y = 0 and attempts to solve quadratic e.g.( x )( x ) = 0 Points are (,0) and (, 0) x = 5 gives y = and so (5, ) lies on the curve ( ) 5 x 5 x + d x = x x + x ( + c) Area of triangle = 8 Area under the curve = 5 x d x = x x with limits and 5 to give = 6 = or ( ) 8 + = 5 = 5 8 = or equivalent (allow.8 or.8 here) 6 6 Area of R = 7 r 8 = 6 or or 6.6 (not 6.7) for attempt to find L and M A Accept x = and x =, then isw or accept L =(,0), M = (,0) Do not accept L =, M = nor (0, ), (0, ) (unless subsequent work) Do not need to distinguish L and M. Answers imply A. See substitution, working should be shown, need conclusion which could be just y = or a tick. Allow y = = But not =. ( y = may appear at start) Usually 0 = 0 or = is B0 for attempt to integrate x kx, x kx or x A for correct integration of all three terms (do not need constant) isw. 5 Mark correct work when seen. So e.g. x x + xis A then x 5x + x would be ignored as subsequent work. A () Bcso () A () B A cao A cao (5) [0] (d) B for this triangle only (not triangle LMN) st for substituting 5 into their changed function nd for substituting into their changed function 5 5 (d) Alternative method: ( x ) ( x 5 x + )d x + x 5 x + d x can lead to correct answer Constructs ( x ) ( x 5 x+ )d x is B for substituting 5 and and subtracting in first integral for substituting and and subtracting in second integral A for answer to first integral i.e. (allow 0.7) and A for final answer as before.. GCE Core Mathematics C (666) January 00 9

10 (d) Another alternative 5 ( x ) ( x 5x + )dx + area of triangle LMP 5 Constructs ( x ) ( x 5 x+ )d x is B for substituting 5 and and subtracting in first integral for complete method to find area of triangle (.5) A for answer to first integral i.e. 5 and A for final answer as before. (d) Could also use 5 (x 6) ( x 5x + )dx + area of triangle LMN Similar scheme to previous one. Triangle has area 6 A for finding Integral has value 6 and A for final answer as before. GCE Core Mathematics C (666) January 00 0

11 Question Q8 N (, -) 69 r = = = 6.5 B, B () B () (d) (e) (d) (e) x+ x Complete Method to find x coordinates, x x = and = then solve To obtain x =, x = 8 Complete Method to find y coordinates, using equation of circle or Pythagoras i.e. let d be the distance below N of A then d = d =.5 y =.. So y = y =.5 Let ANB ˆ = θ 6 sin θ = "6.5" θ = (67.8)... So angle ANB is.8 AP is perpendicular to AN so using triangle ANP tanθ = Therefore AP = 5.6 B for (α), B for B for 6.5 o.e. AP "6.5" st for finding x coordinates may be awarded if either x co-ord is correct Aft,Aft are for α 6andα + 6 if x coordinate of N is α nd for a method to find y coordinates may be given if y co-ordinate is correct A marks is for.5 only. for a full method to find θ or angle ANB (eg sine rule or cosine rule directly or finding another angle and using angles of triangle.) ft their 6.5 from radius or wrong y. "6.5" + "6.5" (cos ANB = =-0.70) "6.5" "6.5" A is a printed answer and must be.8 do not accept.76. for a full method to find AP Alternative Methods N.B. May use triangle AXP where X is the mid point of AB. Or may use triangle ABP. From circle theorems may use angle BAP = 67.8 or some variation. AP 6 6 Eg =, AP = or AP = are each worth sin 67. sin 5. sin.6 cos 67. Aft Aft A (5) A () Acao () [] GCE Core Mathematics C (666) January 00

12 Question Q9 y x x = 0 [ y = ] 6x x Puts their 6 0 x = x So x =, = (If x = 0 appears also as solution then lose A) x=, y = 0, so y = 6 y = x x A, A d,a (7) A () [Since x >0] It is a maximum B () [0] st for an attempt to differentiate a fractional power x A a.e.f can be unsimplified nd for forming a suitable equation using their y =0 rd for correct processing of fractional powers leading to x = (Can be implied by x = ) A is for x = only. If x = 0 also seen and not discarded they lose this mark only. th for substituting their value of x back into y to find y value. Dependent on three previous M marks. Must see evidence of the substitution with attempt at fractional powers to give A0, but y = 6 can imply A for differentiating their y again A should be simplified B. Clear conclusion needed and must follow correct y It is dependent on previous A mark (Do not need to have found x earlier). n n x (Treat parts, and together for award of marks) GCE Core Mathematics C (666) January 00

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16 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 FN Telephone Fax Order Code US0707 January 00 For more information on Edexcel qualifications, please visit Edexcel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH