Mark Scheme - June 2007 - 6666 - Core Mathematics C4

Mark Scheme (Results) Summer 2007

GCE Mathematics (6666/01)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH June 2007 6666 Core Mathematics C4 Mark Scheme

Question Scheme Marks Number ** represents a constant Takes 3 outside the -3 - 3 -3 -3 骣2x 1 骣 2 x bracket to give any of 1. (a) f(x )= (3 + 2 x ) =( 3) 琪 1 + = 琪 1 + -3 1 B1 桫3 27 桫 3 (3) or 27 . See note below.

Expands (1+ * *x )-3 to give a simplified or an M1; un-simplified 1+ ( - 3)(* * x) ; 1 (- 3)( - 4)2 ( - 3)( - 4)( - 5) 3 =27 睚1 + ( - 3)(* * x); + (* * x) + (* * x) + ... 2! 3! A correct simplified or with * * 1 an un-simplified {...... } expansion A1 with candidate’s followed thro’ (* * x)

(- 3)( - 4) ( - 3)( - 4)( - 5) =1睚1 + ( - 3)( 2x ) + ( 2x )2 + ( 2x ) 3 + ... 27 32! 3 3! 3

禳 2 1 8x 80 3 =27 睚1 - 2x + - x + ... 铪 3 27

Anything that 2 3 1 2x 1 2x 8x 80x cancels to - ; A1; = -; + - + ... 27 27 27 27 81 729 8x2 80x 3 Simplified 81- 729 A1 [5]

5 marks

Note: You would award: B1M1A0 for Special Case: If you see the 1 (- 3)( - 4) ( - 3)( - 4)( - 5) constant 27 in a candidate’s =1睚1 + ( - 3)( 2x ) + (2x)2 + (2x) 3 + ... 27 3 2! 3! final binomial expression, then you can award B1 because * * is not consistent.

6666/01 Core Maths C4 2 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Question Scheme Marks Number Aliter 1. f(x)= (3 + 2x)-3 Way 2 1 -3 27 or(3) (See note ) B1 Expands (3+ 2x)-3 to give an un-simplified or (- 3)( - 4) M1 (3)-3+ ( - 3)(3) - 4 (* * x); + (3) - 5 (* * x) 2 simplified 镲 2! -3 - 4 = 睚 (3)+ ( - 3)(3) (* * x) ; (- 3)( - 4)( - 5) 镲 +(3)-6 (* * x) 3 + ... 镲 3! A correct un-simplified or simplified {...... } with * * 1 A1 expansion with candidate’s followed thro’(* * x)

(- 3)( - 4) (3)-3+ ( - 3)(3) - 4 (2x); + (3) - 5 (2x) 2 镲 2! = 睚 (- 3)( - 4)( - 5) 镲 +(3)-6 (2x) 3 + ... 镲 3!

禳1 1 1 2 镲27+( - 3)( 81 )(2x); + (6)( 243 )(4x ) = 睚 ( 10)(1 )(8x3 ) ... 铪镲 + -729 +

Anything that 2 3 1 2x 1 2x 8x 80x cancels to - ; A1; = -; + - + ... 27 27 27 27 81 729 8x2 80x 3 Simplified 81- 729 A1

5 marks

Attempts using Maclaurin expansions need to be escalated up to your team leader.

If you feel the mark scheme does not apply fairly to a candidate please escalate the response up to your team leader.

Special Case: If you see the 1 constant 27 in a candidate’s final binomial expression, then you can award B1

6666/01 Core Maths C4 3 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Question Scheme Marks Number

1 2x 2. dx , with substitution u = 2x (2x + 1)2 0

du x du du x dx 1 = 2 .ln2 or = u.ln2 = 2 .ln2 dx dx � x B1 dx du 2 .ln2 1 du or ( u) d x =ln 2

2x 骣 1 1 1 dx= 琪 d u k2 d u 蝌(2x + 1)2桫ln2 (u + 1) 2 (u + 1) M1* where k is constant

-2 - 1 骣1骣 - 1 (u+ 1)� a ( u 1) M1 =琪琪 + c 桫ln2桫 (u + 1) (u+ 1)-2� + 1.( u 1) - 1 A1

change limits: when x = 0 & x = 1 then u = 1 & u = 2

1 2 2x 1轾 - 1 dx = (2x + 1)2 ln 2犏 (u + 1) 0 臌 1

1轾骣 1 骣 1 Correct use of limits =犏琪 - - 琪 - depM1* ln2臌桫 3 桫 2 u = 1 and u = 2

1 1 1 1 1 1 = 6ln2 or ln4- ln8 or 2ln2- 3n2 A1 aef 6ln2 Exact value only! [6] Alternatively candidate can revert back to x …

1 1 2x 1轾 - 1 dx = (2x+ 1)2 ln2 犏 (2 x + 1) 0 臌 0

1轾骣 1 骣 1 Correct use of limits =犏琪 - - 琪 - depM1* ln2臌桫 3 桫 2 x = 0 and x = 1

1 1 1 1 1 1 = 6ln2 or ln4- ln8 or 2ln2- 3ln2 A1 aef 6ln2 Exact value only! 6 marks

If you see this integration There are other acceptable 1 1 applied anywhere in a answers for A1, eg: 2ln8or ln64 candidate’s working then you can NB: Use your calculator to check award M1, A1 eg. 0.240449…

禳 du 镲u= x � dx 1 3. (a) 睚 dv cos 2x v1 sin2 x 铪镲dx = � 2 (see note below) Use of ‘integration by Int=x cos 2 x d x =1 x sin2 x - 1 sin2 x .1 d x parts’ formula in the M1 蝌 2 2 correct direction. Correct expression. A1

1 sin2x� 2 cos 2 x =1xsin2 x - 1 - 1 cos2 x + c 1 2 2( 2 ) or sinkx� k cos kx dM1 with k�1 , k 0

1 1 =2xsin2 x + 4 cos2 x + c Correct expression with +c A1 [4]

Substitutes correctly (b) xcos2 x d x= xcos2x + 1 d x for 2 in the M1 蝌 ( 2 ) cos x given integral

1 1 =xcos2 x d x + x d x 2蝌 2

1 1骣 1 1 1 their answer to (a) ; =琪 xsin2 x + cos 2 x ; + x d x ( ) A1; 2桫 2 4 2 2 or underlined expression

1 1 1 Completely correct =xsin2 x + cos2 x + x2 ( + c ) A1 4 8 4 expression with/without +c [3]

7 marks

Notes:

Int=x cos 2 x d x = 1 x sin2 x 1 sin2 x .1 d x (b) 蝌 2 2 This is acceptable for M1 M1

禳 du 镲u= x � dx 1 睚 dv cos 2x v sin2 x 铪镲dx = � l

This is also Int=x cos 2 x d x = l x sin2 x l sin2 x .1 d x M1 蝌 acceptable for M1

Aliter Substitutes correctly 3. (b) xcos2 x d x= xcos2x + 1 d x for 2 in the 蝌 ( 2 ) cos x Way 2 given integral … M1 禳 du … or 镲u= x � dx 1 睚 u= x dv =1cos2x + 1 dv 1cos 2x 1 v 1 sin2 x 1 x and dx 2 2 铪镲dx = 2 + 2� + 4 2

1 12 1 1 =4xsin2 x + 2 x -( 4 sin2 x + 2 x) d x

1 1 12 1 1 2 (their answer to (a)) ; =4xsin2 x + 2 x + 8 cos2 x - 4 x + c 2 A1 or underlined expression

1 1 1 2 Completely correct =xsin2 x + cos2 x + x ( + c ) A1 4 8 4 expression with/without +c [3]

Substitutes correctly Aliter 2 for cos 2x in (b) 蝌xcos 2 x d x= x( 2cos x - 1) d x M1 Way 3 xcos 2 x d x

� 2x cos2 x d x= x d x +1 x sin2 x + 1 cos2 x c 蝌 2 4

1 (their answer to (a)) ; 2 1骣 1 1 1 A1; � xcos x d x+琪 x sin2 x + cos2 x ; x d x 2 蝌 2桫 2 4 2 or underlined expression

1 1 1 2 Completely correct =xsin2 x + cos2 x + x ( + c ) A1 4 8 4 expression with/without +c [3]

7 marks

4. (a) A method of long division gives, Way 1 2(4x 2 + 1) 4 � 2 A = 2 B1 (2x+ 1)(2 x - 1) (2 x + 1)(2 x - 1)

4 B C � (2x+ 1)(2 x - 1) (2 x + 1) (2 x - 1)

4�B (2 x+ 1) + C (2 x 1) Forming any one of these two M1 or their remainder,Dx+ E� B (2 x + 1) + C (2 x 1) identities. Can be implied.

1 Let x= - 2 , 4= - 2B� B - 2 See note below either one of B = - 2 or C = 2 A1 Let x= 1 , 4= 2C� C 2 2 both B and C correct A1 [4]

Aliter 2(4x2 + 1) B C 4. (a) � A + (2x+ 1)(2 x - 1) (2 x + 1) (2 x - 1) Way 2 decide to award B1 here!! … See below for the award of B1 B1 … for A = 2

Forming this identity. 2(4x2 + 1)� A (2 x - 1)(2 x + 1) - B (2 x + 1) + C (2 x 1) M1 Can be implied.

Equate x2, 8= 4A� A 2

1 Let x= - 2 , 4= - 2B� B - 2 See note below either one of B = - 2 or C = 2 A1 Let x= 1 , 4= 2C� C 2 2 both B and C correct A1 [4]

If a candidate states one of either B or C correctly then the method mark M1 can be implied.

2(4x 2 + 1) 2 2 4. (b) dx= 2 - + d x 蝌(2x+ 1)(2 x - 1) (2 x + 1) (2 x - 1)

Either pln(2 x + 1) or qln(2 x - 1) M1* or either pln2 x + 1 or qln2 x - 1 A Ax B1 =2x -2 ln(2 x + 1) + 2 ln(2 x - 1) ( + c ) 2 2 -2ln(2x + 1) + 2 ln(2 x - 1) 2 2 A1 -ln(2x + 1) + ln(2 x - 1) or cso & aef See note below.

2 2 2(4x + 1) 2 dx=[ 2 x - ln(2 x + 1) + ln(2 x - 1)] (2x+ 1)(2 x - 1) 1 1

Substitutes limits of 2 and 1 =(4 - ln5 + ln3) -( 2 - ln3 + ln1) and subtracts the correct way depM1* round. (Invisible brackets okay.)

=2 + ln3 + ln3 - ln5

Use of correct product (or power) and/or quotient laws 骣3(3) =2 + ln琪 for logarithms to obtain a single M1 桫 5 logarithmic term for their numerical expression.

骣9 骣9 =2 + ln琪 2+ ln琪 A1 桫5 桫5 5 9 Or 2- ln( 9 ) and k stated as 5 . [6]

10 marks

Some candidates may find rational values for B To award this M1 mark, the candidate and C. They may combine the denominator of must use the appropriate law(s) of their B or C with (2x +1) or (2x – 1). Hence: logarithms for their ln terms to give a a kln( b (2 x 1)) Either b(2 x- 1) � or one single logarithmic term. Any error a b(2 x+ 1) � kln( b (2 x 1)) is okay for M1. in applying the laws of logarithms would then earn M0.

Candidates are not allowed to fluke -ln(2x + 1) + ln(2 x - 1) for A1. Hence cso. If they do fluke this, however, they can gain the final A1 Note: This is not a mark for this part of the question. dependent method mark.

5. (a) If l1 and l2 intersect then:

骣1 骣 1 骣 1 骣 2 琪琪琪琪琪0+ l 琪 1 = 琪 3 + m 琪 1 琪琪琪琪桫-1 桫 0 桫 6 桫 - 1

i : 1+ l = 1 + 2 m (1) Writes down any two of these Any two of j :l = 3 + m (2) M1 equations correctly. k :- 1 = 6 - m (3)

(1) & (2) yields l= 6, m = 3 Solves two of the above equations to find … (1) & (3) yields l= 14, m = 7 either one of l or m correct A1 (2) & (3) yields l= 10, m = 7 both l and m correct A1

checking eqn (3), -1 3 Complete method of putting their values of l and m into a third Either checking eqn (2), 14 10 B1 equation to checking eqn (1), 11 15 show a contradiction.

or for example: this type of explanation is also checking eqn (3), LHS = -1 , RHS = 3 allowed for B1 .

Lines l1 and l2 do not intersect [4]

Aliter k :- 1 = 6 - m = 7 5. (a) 辪 Uses the k component to find m Way 2 and substitutes their value of m i : 1+ l = 1 + 2 m� 1 l = + 1 2(7) into either one of the i or j M1 j :l = 3 + m辧 = 3 + (7) component.

i :l = 14 either one of the l ’s correct A1 j :l = 10 both of the l ’s correct A1

Either: These equations are then inconsistent Complete method giving rise to Or: 14 10 any one of these three B1

Or: Lines l1 and l2 do not intersect explanations. [4]

6666/01 Core Maths C4 9 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Question Scheme Marks Number Aliter 5. (a) If l1 and l2 intersect then: Way 3 骣1 骣 1 骣 1 骣 2 琪琪琪琪琪0+ l 琪 1 = 琪 3 + m 琪 1 琪琪琪琪桫-1 桫 0 桫 6 桫 - 1

i : 1+ l = 1 + 2 m (1) Writes down any two of these Any two of j :l = 3 + m (2) M1 equations k :- 1 = 6 - m (3)

(1) & (2) yields m = 3 either one of the m ’s correct A1 (3) yields m = 7 both of the m ’s correct A1

Either: These equations are then inconsistent Complete method giving rise to Or: 3 7 any one of these three B1

Or: Lines l1 and l2 do not intersect explanations. [4]

Aliter i : 1+ l = 1 + 2 m (1) Writes down any two of these 5. (a) Any two of j :l = 3 + m (2) M1 equations Way 4 k :- 1 = 6 - m (3)

(1) & (2) yields m = 3 m = 3 A1 (3) RHS= 6 - 3 = 3 RHS of (3) = 3 A1

Complete method giving rise to (3) yields -1 3 B1 this explanation. [4]

Only one of either 骣2 骣5 骣2 骣 5     琪琪琪琪 OA = 琪1 or OB = 琪5 or 5. (b) l=1�OA = 琪 1 & m � 2 OB 琪 5 琪琪 B1 琪琪桫-1 桫4 桫-1 桫 4 A(2,1,- 1) or B(5,5,4) . (can be implied)

骣5 骣 2 骣 3 骣-3    琪琪琪  琪 Finding the difference between AB= OB - OA =琪5 - 琪 1 = 琪 4 or BA =琪 -4 their OB and OA . M1 琪琪琪琪桫4 桫- 1 桫 5 桫-5 (can be implied)

Applying the dot product formula between “allowable” M1  vectors. See notes below. AB= 3i + 4 j + 5 k , d1 = i + j + 0 k & q is angle

 Applies dot product formula AB d1 骣3+ 4 + 0  M1 cos q = = 琪 between d1 and their AB. AB . d 桫 50 . 2 1 Correct expression. A1

cos q = 7 7 7 10 10 or 0.7 or 100 A1 cao 7 [6] but not 50 2

10 marks

Candidates can score this mark if there is a complete method for finding the dot product between their vectors in the following cases:

 d= i + j + 0 k Case 1: their ft �AB �+ ( 3i 4 j 5 k) Case 2: d1 = i + j + 0 k Case 3: 1 d=2(2 i + j - 1 k ) and d1 = i + j + 0 k and d2 =2 i + j - 1 k and 2 骣3+ 4 + 0 2+ 1 + 0 4+ 2 + 0 cos q � cos q � cos q � 琪 2 . 6 2 . 24 桫 50 . 2

  Case 4: their ft �AB �+ ( 3i 4 j 5 k) Case 5: their ft OA= 2i + 1 j - 1 k  and d2 =2 i + j - k and their ft OB= 5i + 5 j + 4 k 骣6+ 4 - 5 骣10+ 5 - 4 cos � cos q 琪 � q 琪桫 50 . 6 桫 6 . 66

Note: If candidate use cases 2, 3, 4 and 5 they cannot gain the final three marks for this part. Note: Candidate can only gain some/all of the final three marks if they use case 1. 6666/01 Core Maths C4 11 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Examples of awarding of marks M1M1A1 in 5.(b)

Example Marks

50 . 2 cosq =�( 3 + 4 0) M1M1A1 (Case 1)

2 . 6 cosq = 3 M1M0A0 (Case 2)

2 . 24 cosq = 4 + 2 M1M0A0 (Case 3)

6. (a) x= tan2 t , y= sin t

dx dy dx dy = 2(tant )sec2 t , = cost Correct and B1 dt dt dt dt

cost M1 dy cos t 骣 cos4 t their dx \ = 琪= dt dx 2 tant sec2 t 桫 2sint + cost dx A1 their dt [3]

1, 1 1, awrt 0.71 p x=1 , y = 1 The point ( 2 ) or ( ) (b) When t = 4 , 2 (need values) B1, B1 These coordinates can be implied. p ( y = sin( 4 ) is not sufficient for B1) p dy cos p = 4 When t = , m(T) = p2 p 4 dx 2 tan4 sec 4

1 1 1 2 2 2 1 2 =2 = = = = any of the five underlined 1 2.(1)(2)4 2 8 B1 aef 1 2.(1) 1 2.(1) 1 ( ) expressions or awrt 0.18 2 ( 2 )

Finding an equation of a tangent with their point and their tangent M1 y-1 = 1 ( x -1) T: 2 4 2 gradient or finds c by using aef y=(their gradient) x + " c " .

1 3 2 3 2 Correct simplified A1 aef T: y= x + or y= x + 4 2 4 2 8 8 EXACT equation of tangent cso

1= 11 +c� c - 1 = 1 3 or 2 4 2( ) 2 4 2 4 2

y=1 x + 3 2 3 2 Hence T: 4 2 4 2 or y=8 x + 8 [5]

A candidate who incorrectly differentiates tan2 t to give Note: The x and y coordinates dx = 2sec2 t dx = sec4 t must be the right way round. dt or dt is then able to fluke the correct answer in part (b). Such candidates can potentially get: (a) B0M1A1 (b) B1B1B1M1A0 cso. Note: cso means “correct solution only”. Note: part (a) not fully correct implies candidate can achieve a maximum of 4 out of 5 marks in part (b). 6666/01 Core Maths C4 13 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Question Scheme Marks Number

sin2 t 6. (c) x=tan2 t = y= sin t cos2 t Way 1 sin2 t x = Usescos2t= 1 - sin 2 t M1 1- sin2 t

y 2 Eliminates ‘t’ to write an equation x = M1 1- y 2 involving x and y.

x(1- y2 ) = y 2� x = xy 2 y 2

Rearranging and factorising with x= y2 + xy 2� x + y 2 (1 x ) ddM1 an attempt to make y 2 the subject.

x x y 2 = A1 1+ x 1+ x [4] Aliter 6. (c) 1+ cot2t = co sec 2 t Uses1+ cot2t = co sec 2 t M1 Way 2 1 1 = Uses cosec2 t = M1 implied sin2 t sin2 t

1 1 Eliminates ‘t’ to write an equation Hence, 1+ = ddM1 x y 2 involving x and y.

1 x 1 x Hence, y 2 =1 - or 1- or A1 (1+x ) 1 + x (1+x ) 1 + x [4]

1 1 is an acceptable response for the final accuracy A1 mark. 1+ x

6666/01 Core Maths C4 14 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Question Scheme Marks Number Aliter 6. (c) x= tan2 t y= sin t Way 3 1+ tan2t = sec 2 t Uses1+ tan2t = sec 2 t M1

1 1 = Uses sec2 t = M1 cos2 t cos2 t

1 = 1- sin2 t

1 Eliminates ‘t’ to write an equation Hence, 1+x = ddM1 1- y 2 involving x and y.

1 x 1 x Hence, y 2 =1 - or 1- or A1 (1+x ) 1 + x (1+x ) 1 + x

[4] Aliter 6. (c) y2=sin 2 t = 1 - cos 2 t Uses sin2t= 1 - cos 2 t M1 Way 4 1 1 =1 - Uses cos2 t = M1 sec2 t sec2 t

1 =1 - then uses sec2t= 1 + tan 2 t ddM1 (1+ tan2 t )

1 x 1 x Hence, y 2 =1 - or 1- or A1 (1+x ) 1 + x (1+x ) 1 + x [4]

6666/01 Core Maths C4 15 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION Question Scheme Marks Number Aliter 6. (c) x= tan2 t y= sin t Way 5 x= tan2 t� tan t x

Draws a right-angled triangle and M1 places both and 1 on the (1+ x ) x x triangle t 1 Uses Pythagoras to deduce the hypotenuse M1

x Eliminates ‘t’ to write an equation Hence, y=sin t = ddM1 1+ x involving x and y.

x x Hence, y 2 = A1 1+ x 1+ x [4]

12 marks

There are so many ways that a candidate can proceed with part (c). If a candidate produces a correct solution then please award all four marks. If they use a method commensurate with the five ways as detailed on the mark scheme then award the marks appropriately. If you are unsure of how to apply the scheme please escalate your response up to your team leader.

x 0 p p 3p p 7. (a) 16 8 16 4 y 0 0.445995927… 0.643594252… 0.817421946… 1

Enter marks into ePEN in the correct order. 0.446 or awrt 0.44600 B1 awrt 0.64359 B1 awrt 0.81742 B1 0 can be [3] implied Outside brackets 1 p p B1 2 16 or 32 For structure of (b) 1 p trapezium rule{...... } Area淮 ;�{ 0 + 2( 0.44600 + 0.64359 + 0.81742) 1} M1 Way 1 2 16 ; Correct expression inside brackets which all A1 h must be multiplied by 2 .

p =�4.81402... = 0.472615308... 0.4726 (4dp) for seeing 0.4726 A1 cao 32 [4]

p 16 and a divisor of 2 on B1 Area 淮p 0+ 0.44600+ 0.44600 + 0.64359 + 0.64359 + 0.81742 + 0.81742 + 1 all terms inside brackets. 16{ 2 2 2 2 } One of first and last Aliter ordinates, two of the (b) middle ordinates inside M1 Way 2 which is equivalent to: brackets ignoring the 2. Correct expression inside 1 p 1 Area淮 ;�{ 0 + 2( 0.44600 + 0.64359 + 0.81742) 1} brackets if 2 was to be 2 16 factorised out. A1

p =�2.40701... = 0.472615308... 0.4726 0.4726 A1 cao 16 [4]

1 p Area =2创 20 {0 + 2(0.44600 + 0.64359 + 0.81742) + 1} = 0.3781, gains B0M1A1A0

In (a) for x = p writing 0.4459959… then 0.45600 gains B1 for awrt 0.44600 even though 0.45600 is incorrect. 16

Question Scheme In (b) you can follow though a candidate’s values from part (a) to award M1 ft, A1 ft Marks Number

6666/01 Core Maths C4 17 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION 2 p p tan x dx tan x dx 4 4 ( ) or 2 7. (c) Volume =(p) 蝌( tanx) d x = ( p ) tan x d x Can be implied. M1 0 0 Ignore limits and (p )

p p tanx lnsec x = (p )[ln sec x] 4 or =(p )[ -lncos x] 4 A1 0 0 or tanx� lncos x

The correct use of limits on a function other than =p 轾lnsecp - lnsec 0 ( ) 臌( 4 ) ( ) tan x; ie or p dM1 x = 4 ‘minus’ x = 0 . =p 轾 -lncosp - lncos0 ( ) 臌( 4 ) ( ) ln(sec 0)= 0 may be implied. Ignore (p )

轾 1 1 轾 =pln1 - ln( 1) = p ln 2 - ln1 臌犏 ( 2 ) 臌 or =p 轾 -ln1 - ln( 1) 臌 ( 2 )

p ln 2 p ln 2 or 2 p ln 2 1 p ln2 -p ln 1 = p ln 2 or 2 or 2 or ( 2 ) or 1 p ln2 -p ln 1 or 2 or ( 2 ) A1 aef p 1 2ln( 2 ) p 1 or 2ln( 2 ) must be exact. [4]

11 marks

If a candidate gives the correct exact answer and then writes 1.088779…, then such a candidate can be awarded A1 (aef). The subsequent working would then be ignored. (isw)

Beware: In part (c) the factor of p is not needed for the first three marks.

Beware: In part (b) a candidate can also add up individual trapezia in this way:

1p 1 p 1 p 1 p Area�2 . 16( 0+ 0.44600) + 2 . 16( 0.44600 + 0.64359) + 2 . 16( 0.64359 + 0.81742) + 2 . 16 ( 0.81742 1)

dP 8. (a) = kP and t=0, P = P (1) dt 0

Separates the variables dP dP with and kd t on = kd t P M1 蝌P either side with integral signs not necessary.

Must see lnP and kt ; lnP= kt ;( + c ) Correct equation A1 with/without + c.

When t=0, P = P0� ln P 0 c Use of boundary condition (1) to attempt to find the M1 or P= Aekt � P A ( 0 ) constant of integration.

lnPkt+ ln P0 kt ln P 0 lnP= kt + ln P0 � e= e e. e

kt kt Hence, P= P0 e P= P0 e A1 [4]

2.5t SubstitutesP= 2 P0 into an (b) P=2 P0 & k = 2.5 � 2P0 P 0 e M1 expression involving P

2.5t 2.5 t e= 2� ln e ln2 or 2.5t = ln2 Eliminates P and takes 0 M1 …or ekt= 2� ln e kt ln2 or kt = ln2 ln of both sides

1 � t = 2.5 ln2 0.277258872... days

t =0.277258872...创 24 60 = 399.252776... minutes

awrt t = 399 or t = 399min or t = 6 hr 39 mins (to nearest minute) A1 6 hr 39 mins [3]

kt P= P0 e written down without the first M1 mark given scores all four marks in part (a).

dP 8. (c) = lPcos l t and t=0, P = P (1) dt 0

Separates the variables dP with and dP P = lcos lt d t M1 蝌P lcos lt d t on either side with integral signs not necessary.

Must see lnP and sinlt ; lnP= sinl t ;( + c) Correct equation A1 with/without + c.

When t=0, P = P0� ln P 0 c Use of boundary condition (1) to attempt to find the M1 or P= Aesinlt � P A ( 0 ) constant of integration.

lnPsinlt+ ln P0 sinl t ln P 0 lnP= sinl t + ln P0 � e= e e. e

sinlt sinlt Hence, P= P0 e P= P0 e A1 [4]

sin2.5t (d) P=2 P0 &l = 2.5 � 2P0 P 0 e

sin2.5t e= 2� sin2.5 t ln 2 Eliminates P0 and makes lt sinlt or sin2.5t the M1 e= 2� sinl t ln2 …or … subject by taking ln’s

t = 1 sin-1 ln2 Then rearranges 2.5 ( ) dM1 to make t the subject. (must use sin-1) t = 0.306338477...

t =0.306338477...创 24 60 = 441.1274082... minutes

awrt t = 441 or t = 441min or t = 7 hr 21 mins (to nearest minute) A1 7 hr 21 mins [3]

14 marks

sinlt P= P0 e written down without the first M1 mark given scores all four marks in part (c).

dP = kP and t=0, P = P (1) dt 0

Separates the variables Aliter dP dP with and dt on 8. (a) = 1 dt kP M1 蝌kP Way 2 either side with integral signs not necessary.

1 Must see k lnP and t ; 1 k lnP= t ;( + c) Correct equation A1 with/without + c.

1 When t=0, P = P0� k ln P 0 c Use of boundary condition (1) to attempt to find the M1 or P= Aekt � P A ( 0 ) constant of integration.

1 1 klnP= t + k ln P0� ln P + kt ln P 0 � elnP= ekt+ ln P0 e kt . e ln P 0

Separates the variables Aliter dP dP with and dt on 8. (a) = 1 dt kP M1 蝌kP Way 3 either side with integral signs not necessary.

1 Must see k ln(kP ) and t ; 1 k ln(kP) = t ;( + c) Correct equation A1 with/without + c.

1 Use of boundary condition When t=0, P = P0� k ln( kP 0 ) c (1) to attempt to find the M1 or kP= Aekt kP A ( � 0 ) constant of integration.

1 1 kln(kP) = t + k ln( kP0) � ln( kP) + kt ln( kP 0 ) � eln(kP) = e kt+ ln( kP0) ekt . e ln( kP 0 ) kt kt �kP � e.( kP0) kP kP 0 e kt (or kP= kP0 e )

6666/01 Core Maths C4 22 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION dP = lPcos l t and t=0, P = P (1) dt 0

Separates the variables Aliter dP dP with and coslt d t 8. (c) = coslt d t lP M1 蝌lP Way 2 on either side with integral signs not necessary.

1 Must see l lnP and 1 1 1 l sinlt ; llnP= l sinl t ;( + c) A1 Correct equation with/without + c.

1 When t=0, P = P0� l ln P 0 c Use of boundary condition (1) to attempt to find the M1 or P= Aesinlt � P A ( 0 ) constant of integration.

1 1 1 llnP= l sinl t + l ln P0� ln P + sin l t ln P 0

� elnP= esinlt+ ln P0 e sinl t . e ln P 0

kt P= P0 e written down without the first M1 mark given scores all four marks in part (a).

sinlt P= P0 e written down without the first M1 mark given scores all four marks in part (c).

dP = lPcos l t and t=0, P = P (1) dt 0

Separates the variables Aliter dP dP with and coslt d t 8. (c) = coslt d t lP M1 蝌lP Way 3 on either side with integral signs not necessary.

1 Must see l ln(lP ) and 1 1 1 l sinlt ; lln(lP) = l sin l t ;( + c) A1 Correct equation with/without + c.

1 Use of boundary condition When t=0, P = P0� l ln(l P 0 ) c (1) to attempt to find the M1 or lP= Aesinlt l P A ( � 0 ) constant of integration.

1 1 1 lln(lP) = l sin l t + l ln( l P0 )

� ln(lP) + sin l t ln( l P0 )

� eln(lP) = e sin l t+ ln( l P0) esinlt . e ln( l P 0 )

sinlt � lP e.( l P0 ) sinlt (or lP= l P0 e )

 Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. depM1* denotes a method mark which is dependent upon the award of M1* . ft denotes “follow through” cao denotes “correct answer only” aef denotes “any equivalent form”

6666/01 Core Maths C4 24 25th June 2007 June 2007 Advanced Subsidiary/Advanced Level in GCE Mathematics Version 8: THE FINAL VERSION