Mark Scheme (Results) January 2011

Transcription

1 Mark (Results) January 0 GCE GCE Core Mathematics C3 (6665) Paper Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

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3 General Instructions for Marking. The total number of marks for the paper is 75.. The Edecel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) should not be subdivided. 3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

4 January 0 Core Mathematics C Mark. (a) 7cos 4sin Rcos( + α ) 7 cos 4sin Rcos cosα Rsin sinα Equate cos : 7 Rcosα Equate sin : 4 Rsinα R ; 5 R 5 B tanα α c tanα 7 or tanα 4 awrt.87 A Hence, 7cos 4sin 5cos( +.87) (b) Minimum value 5 5 or R Bft () (c) 7cos 4sin 0 5cos( +.87) ( + ) cos ( their α ) cos.87 0 ± ( their R) c PV or For applying 0 cos their R c c So, +.87 { , , c } either π + or their PV c or or their PV gives, { , } awrt 3.84 OR 6.6 awrt 3.84 AND 6.6 A A (5) [9] GCE Core Mathematics C3 (6665) January 0

5 . (a) 4 3 ( ) ( )( ) (4 )( ) 3 ( )( ) 8 6 { ( )( ) } An attempt to form a single fraction Simplifies to give a correct quadratic numerator over a correct quadratic denominator A aef ( )(4 + ) { ( )( ) } 4 + (b) 4 3 f( ), > ( ) ( )( ) An attempt to factorise a 3 term quadratic numerator A (4) (4 + ) f( ) ( ) (c) (4 + ) ( ) ( ) ( ) 3 ( ) 3 f( ) 3( ) ( ) An attempt to form a single fraction Correct result A () f( ) 3( )( ) () k ± ( ) 6 f() Either or 3 A aef A [9] GCE Core Mathematics C3 (6665) January 0

6 3. cosθ sinθ ( θ ) sin sinθ Substitutes either sin θ or cos θ or cos θ sin θ for cos θ. θ 4sin sin θ θ θ Forms a quadratic in sine 0 (*) 4sin sin 0 sinθ ± 4 4(4)( ) 8 Applies the quadratic formula See notes for alternative methods. PVs: α 54 or α 8 θ { 54, 6, 98, 34} Any one correct answer 80-their pv All four solutions correct. A d(*) A [6] GCE Core Mathematics C3 (6665) January 0 3

7 4. (a) θ 0 + Ae kt (eqn *) (0) { t 0, θ 90 } 90 0 Ae k + Substitutes t 0 and θ 90 into eqn * A A 70 A 70 (b) θ e kt A () (5) { t 5, θ 55 } e k 35 e ln ( 70 ) 5k 5k ln 5k ( ) Substitutes t 5 and θ 55 into eqn * and rearranges eqn * to make e ±5k the subject. Takes lns and proceeds to make ±5k the subject. d 5k ln ln 5k ln k ln 5 Convincing proof that k ln 5 A (c) 5 ln θ e t ± α e kt dθ where k ln t 5 ln.(70)e 5 ln dt 5 5 4ln e t ln A oe When t 0, dθ 4ln e dt ln dθ 7 ln dt Rate of decrease of θ.46 C / min (3 dp.) awrt ±.46 A [8] GCE Core Mathematics C3 (6665) January 0 4

8 5. (a) Crosses -ais f( ) 0 (8 )ln 0 Either (8 ) 0 or ln 0 8, Either one of {} OR {8} B Coordinates are A (, 0) and B (8, 0). Both A (, { 0} ) and B( 8, { 0} ) B () (b) Apply product rule: u (8 ) v ln du dv d d vu + uv f( ) 8 ln + Any one term correct A Both terms correct A (c) f (3.5) f (3.6) Sign change (and as f( ) is continuous) therefore the -coordinate of Q lies between 3.5 and 3.6. Attempts to evaluate both f(3.5) and f(3.6) both values correct to at least sf, sign change and conclusion A () (d) At Q, 8 f( ) 0 ln + 0 Setting f( ) 0. 8 ln ln + 8 (ln + ) Splitting up the numerator and proceeding to 8 ln + (as required) For correct proof. No errors seen in working. A GCE Core Mathematics C3 (6665) January 0 5

9 (e) Iterative formula: n + 8 ln + n 8 ln(3.55) An attempt to substitute into the iterative formula. Can be implied by 3.58(97)... Both awrt 3.59 and awrt A 3.59, 3.538, , to 3 dp.,, 3 all stated correctly to 3 dp A [3] GCE Core Mathematics C3 (6665) January 0 6

10 6. (a) 3 y y( 5) 3 5 Attempt to make (or swapped y) the subject y 5y 3 y y ( y + ) 3+ 5y Collect terms together and factorise y y f ( ) A oe (b) Range of g is -9 g() 4 or -9 y 4 Correct Range B () (c) Deduces that g() is 0. Seen or implied. g g() g (0) 6, from sketch. -6 A () (d) fg(8) f (4) Correct order g followed by f 3 4() A () GCE Core Mathematics C3 (6665) January 0 7

11 (e)(ii) y Correct shape B -6 Graph goes through ({ } ) ( 6, { 0} ) 0,and which are marked. B (4) (f) Domain of g is -9 4 Either correct answer or a follow through from part (b) answer B () [3] GCE Core Mathematics C3 (6665) January 0 8

12 7 (a) y 3 + sin + cos Apply quotient rule: u 3+ sin v + cos du dv cos sin d d dy cos ( + cos ) sin (3 + sin ) d cos ( + ) Applying Any one term correct on the numerator Fully correct (unsimplified). A A 4cos + cos + 6sin + sin ( + cos) 4cos + 6sin + (cos + sin ) ( + cos) ( + cos) 4cos + 6sin + (as required) For correct proof with an understanding that cos + sin. No errors seen in working. A* (4) π (b) When, 3 + sinπ 3 y 3 + cosπ y 3 B At π 6sinπ + 4cosπ + 4+ (,3 ),m( T ) ( + cos π ) π Either T: y 3 ( ) or y + c and π 3 + c c 3 + π ; ( ) m( T) π y y m( ) with their TANGENT gradient and their y ; or uses y m+ c with their TANGENT gradient ; B T: y ( π 3) + + y + π + 3 A (4) [8] GCE Core Mathematics C3 (6665) January 0 9

13 8. (a) y sec (cos ) cos dy d (cos ) ( sin ) Writes sec as (cos ) and gives dy ± ((cos ) (sin ) ) d (cos ) ( sin ) or (cos ) (sin ) A dy sin sin d cos cos cos sec tan Convincing proof. Must see both underlined steps. A AG (b) sec y, y (n+ ), n. π 4 d Ksecytan y sec ytan y dy secytan y A () (c) dy Applies d secytany dy d d ( dy ) dy Substitutes for sec y. d tany A A y y + tan sec tan sec Attempts to use the identity + tan A sec A So y tan dy d ( ) dy d ( ) A (4) [9] GCE Core Mathematics C3 (6665) January 0 0

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15 Further copies of this publication are available from Edecel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fa Order Code US0638 January 0 For more information on Edecel qualifications, please visit Edecel Limited. Registered in England and Wales no Registered Office: One90 High Holborn, London, WCV 7BH