Mark Scheme (Results) Summer 2008

Transcription

1 Mark (Results) Summer 008 GCE GCE Mathematics (6666/0) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 June Core Mathematics C4 Mark. (a) y 0 e 0.08 e 0. e 0.7 e.8 e e or y Either e and e.8 or awrt.8 and.60 (or a miture of e s and decimals) B [] (b) Area 0.4 ; e + e + e + e + e + e Way ( ) Outside brackets 0.4 or 0. For structure of trapezium rule[... ] ; B; M = = = 4.9 (4sf) 4.9 A cao [] Aliter (b) Way Area which is equivalent to: e + e e + e e + e e + e.8 e + e Area 0.4 ; e + ( e + e + e + e ) + e and a divisor of on all terms inside brackets. One of first and last ordinates, two of the middle ordinates inside brackets ignoring the. B M = = = 4.9 (4sf) 4.9 A cao [] 4 marks Area 0.4 e e e e e e would score BMA0 Note an epression like ( ) Allow one term missing (slip!) in the ( ) brackets for The M mark for structure is for the material found in the curly brackets ie first yordinate + ( intermediate ft y ordinate) + final y ordinate

3 . (a) du u = = d v = e v = e e = e e. Use of integration by parts formula in the correct direction. M (See note.) Correct epression. (Ignore ) A = e e ( c) = e e + Correct integration with/without + c A [] (b) du u = = d v = e v = e e = e e. Use of integration by parts formula M in the correct direction. Correct epression. (Ignore ) A e = e ( ) = e e e + c = + + = e ( + ) + c e e e c Correct epression including + c. (seen at any stage! in part (b)) You can ignore subsequent working. Ignore subsequent working A ISW [] 6 marks Note integration by parts in the correct direction means that u and d v must be assigned/used as u = and dv = e in part (a) for eample + c is not required in part (a). i i d i t (b)

4 . (a) From question, d da A 0.0 dt = 0.0 dt = seen or implied from working. B da A= = by itself seen or implied from working B da da 0.06 = = ( 0.0 ) ; = dt dt da 0.0 Candidate's ; M; d When = cm, d = 0.06 dt Hence, d dt (cm s- ) awrt A cso [4] (b) V = (5 ) = 5 V = (5 ) or 5 B dv = 5 dv = 5 or ft from candidate s V in one variable B dv dv 0.06 = = = dt dt { } 5. ; 0.4 Candidate s d V d ; M dt When = cm, dv dt = 0.4() = 0.48 (cm s ) 0.48 or awrt 0.48 A cso [4] 8 marks 4

5 4. (a) y y 4 + = ( eqn ) dy dy dy = 6 y + y + = 0 Differentiates implicitly to include either dy dy dy ± ky or. (Ignore ( d ) = ) Correct application ( ) of product rule ( ) d y y 6 y d and ( 4 0) M B A dy 6 y = y or dy 6+ y = y not necessarily required. dy 8 6 y 8 = = y giving 8 y = 8 6y giving y = 6 Substituting d y 8 = into their equation. Attempt to combine either terms in or terms in y together to give either a or by. M dm Hence, y = y = 0 simplifying to give y = 0 AG A cso (b) At P & Q, y =. Substituting into eqn [6] gives ( ) + ( ) = 4 Attempt replacing y by in at least one of the y terms in eqn M Simplifying gives, = 4 = ± Either = or = A y = y = ± 4 Hence coordinates are (,4) and (, 4) Both (,4) and (, 4) A [] 9 marks 5

6 5. (a) ** represents a constant (which must be consistent for first accuracy mark) = (4 ) = ( 4) = (4) or outside brackets B (4 ) 4 4 ( )( ) = + + +! ( )(** ); (** )... with ** Epands ( + ** ) to give a simplified or an un-simplified + ( )(** ) ; A correct simplified or an un-... epansion simplified [ ] with candidate s followed through (**) M; A ( )( ) ( )( 4 ) ( = 4 ) ! Award SC M if you see ( )( ) ( )(** ) + (** )! + ;... 8 A isw 7 7 = SC: K ; ; 8 A isw 7 = ;... Ignore subsequent working (b) ( 8)... Writing ( + 8) multiplied by candidate s part (a) epansion. M [5] = = ;... Multiply out brackets to find a constant term, two terms and two terms. Anything that cancels to 4 + ; M A; A [4] 9 marks 6

7 6. (a) Lines meet where: λ = + µ Any two of i : 9 + λ = + µ () j: λ = µ () k : 0 λ = µ () Need any two of these correct equations seen anywhere in part (a). M () () gives: 9 = + 5µ µ = Attempts to solve simultaneous equations to find one of either λ or µ dm () gives: λ = = Both λ = & µ = A 9 r = 0 + or r = Substitutes their value of either λ or µ into the line l or l respectively. This mark can be implied by any two correct,, 7. components of ( ) ddm Intersect at r = or r = i + j+ 7k 7 7 or i + j+ 7k or (,, 7) A Either check that λ =, µ = Either check k: in a third equation or check λ = : LHS= 0 λ = 0 = 7 that λ =, B µ = give the same µ = : RHS= 7+ 5µ = 7 0= 7 coordinates on the other line. Conclusion not needed. (As LHS = RHS then the lines intersect.) [6] (b) d = i + j k, d = i j+ 5k As d d = = ( ) + ( ) + ( 5) = 0 5 Then l is perpendicular to l. Dot product calculation between the two direction vectors: ( ) + ( ) + ( 5) or 6 5 M Result =0 and appropriate conclusion A [] 7

8 6. (c) Equating i ; 9 + λ = 5 λ = r = = 7 0 ( = OA uuur. Hence the point A lies on l.) Substitutes candidate s λ = 7 into the line l and finds 5i + 7j+ k. The conclusion on this occasion is not needed. B [] uuur (d) Let OX = i + j+ 7k be point of intersection 5 8 uuur uuur uuur AX = OX OA = 7 = uuur uuur uuur uuur uuur OB = OA + AB = OA + AX Finding the difference between uuur their OX (can be implied) and uuur OA. 5 uuur AX =± 7 7 M ± uuur OB 5 8 = uuur 7 + their AX dm Hence, uuur OB = uuur or OB = i j+ k or i j+ k or (,, ) A [] marks 8

9 7. (a) A B + 4 y ( y)( + y) ( y) ( + y) A( + y) + B( y) y = = B( 4) B = Let, Forming this identity. NB: A & B are not assigned in this question M y = A( 4) A Let, = = Either one of A= or B = A giving + ( y) ( + y) +, aef A cao ( y) ( + y) (If no working seen, but candidate writes down correct partial fraction then award all three marks. If no working is seen but one of A or B is incorrect then M0A0A0.) [] 9

10 7. (b) dy = y 4 cot Separates variables as shown. Can be implied. Ignore the integral signs, and the. B ( y) + ( + y) = dy tan ( ) ln(sec ) or ln(cos ) B Either ± aln( λ y) or ± bln( λ + y) M; ln( y) + ln( + y) = ln(sec ) + c their = LHS correct with ft cot for their A and B and no error with the with or without + c A y = 0, = ln + ln = ln ( cos( )) + c Use of y = 0 and = in an integrated equation containing c ; M* { 0 = ln + c c = ln} ln( y) + ln( + y) = ln(sec ) ln + y sec ln = ln y + y sec ln = ln y + y sec ln = ln y Using either the quotient (or product) or power laws for logarithms CORRECTLY. Using the log laws correctly to obtain a single log term on both sides of the equation. M dm* + y sec = y 4 Hence, 8+ 4y sec = y 8+ 4y sec = y A aef [8] marks 0

11 8. (a) At P (4, ) either 4= 8cost or = 4sint 4= 8cost or = 4sin t M only solution is t = where 0 t t = or awrt.05 (radians) only stated in the range 0 t A [] (b) = 8cost, y = 4sin t Attempt to differentiate both and y wrt t to give ± psint and 8sint dt =, dy 8cost dt = ± qcos trespectively Correct d and d y dt d t M A At P, dy 8cos = 8sin ( ) ( ) Divides in correct way round and attempts to substitute their value of t (in degrees or radians) into their d y epression. M* ( ) 8 = = = ( 8)( ) awrt 0.58 You may need to check candidate s substitutions for M* Note the net two method marks are dependent on M* Hence m(n) = or Uses m(n) = their m( T ). dm* N: y = ( 4) Uses y = ( their m )( 4) or finds c using = 4 and y = and uses y = (their m ) + " c". N N dm* N: y = + 6 AG y = + 6 A cso AG = 4 + c c = + 4 = 6 or ( ) so N: y= + 6 [6]

12 4 8. (c) = = ( ) 0 A y 4sin t. 8sint dt ( ) A = sin t.sin t dt = sintcos t.sin t dt attempt at A= y d t dt correct epression (ignore limits and dt ) Seeing sin t = sin tcost anywhere in PART (c). M A M A = 64.sin tcost dt A 64.sin tcost dt = Correct proof. Appreciation of how the negative sign affects the limits. Note that the answer is given in the question. A AG [4] (d) du {Using substitution u = sint dt = cost} {change limits: when t =, u = & when t =, u = } sin t A = 64 or u A = 64 ksin t or ku with u = sint M Correct integration ignoring limits. A A = Substitutes limits of either t = and t = or ( ) ( u andu ) = = and subtracts the correct way round. dm 64 A = 64 = 8 8 (Note that a=, b = 8) A aef isw Aef in the form a + b, [4] with awrt. and anything 64 that cancels to a = and b = 8. 6 marks