GCE. Mathematics (MEI) Mark Schemes for the Units. June /7895-8/MS/R/08. Advanced GCE A Advanced Subsidiary GCE AS

Transcription

1 GCE Mathematics (MEI) Advanced GCE A Advanced Subsidiary GCE AS 89-8 Mark Schemes for the Units June /789-8/MS/R/08 Oxford Cambridge and RSA Examinations

2 OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the University of Cambridge Local Examinations Syndicate and the RSA Examinations Board in January 998. OCR provides a full range of GCSE, A level, GNVQ, Key Skills and other qualifications for schools and colleges in the United Kingdom, including those previously provided by MEG and OCEAC. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 008 Any enquiries about publications should be addressed to: OCR Publications PO Box 00 Annesley NOTTINGHAM NG 0DL Telephone: Facsimile: publications@ocr.org.uk

3 CONTENTS Advanced GCE Further Mathematics (MEI) (7896) Advanced GCE Further Mathematics (Additional) (MEI) (7897) Advanced GCE Mathematics (MEI) (789) Advanced GCE Pure Mathematics (MEI) (7898) Advanced Subsidiary GCE Further Mathematics (MEI) (896) Advanced Subsidiary GCE Further Mathematics (Additional) (MEI) (897) Advanced Subsidiary GCE Mathematics (MEI) (89) Advanced Subsidiary GCE Pure Mathematics (MEI) (898) MARK SCHEME FOR THE UNITS Unit/Content Page 7 (C) Introduction to Advanced Mathematics 7 (C) Concepts for Advanced Mathematics 6 7 (C) Methods for Advanced Mathematics 8 7 (C) Applications of Advanced Mathematics 7 (FP) Further Concepts for Advanced Mathematics 9 76 (FP) Further Methods for Advanced Mathematics 77 (FP) Further Applications of Advanced Mathematics 78 Differential Equations 8 76 Mechanics 76 Mechanics 7 76 Mechanics 76 Mechanics Statistics Statistics Statistics Statistics Decision Mathematics 8 77 Decision Mathematics Decision Mathematics Computation Numerical Methods Numerical Computation 0 Grade Thresholds 0

4 7 Mark Scheme June (C) Introduction to Advanced Mathematics Section A x > 6/ o.e. isw for x > 6 or for 6/ o.e. found or for their final ans ft their x > k or kx > 6 (i) (0, ) and (6, 0) (ii) /6 o.e. or ft their (i) isw each; allow x = 0, y = etc; condone x = 6, y = isw but 0 for (6, ) with no working for x 6 or / 6 or /6 o.e. or ft (accept 0.67 or better) 0 for just rearranging to y = x+ (i) 0 or / o.e. each (i) T (ii) E (iii) T (iv) F (ii) k < 9/8 o.e. www M for ( )( 8k) < 0 o.e. or 9/8 found or for attempted use of b ac (may be in quadratic formula); SC: allow for 9 8k < 0 and ft for k > 9/8 for all correct, for correct. for correct y (x ) = (x + ) xy y = x + or ft [ft from earlier errors if of comparable difficulty no ft if there are no xy terms] for multiplying by x ; condone missing brackets for expanding bracket and being at stage ready to collect x terms xy x = y + or ft [ ] y + x = y o.e. or ft for collecting x and other terms on opposite sides of eqn for factorising and division alt method: y = + x y = x x = y x = + y for either method: award marks only if fully correct 8

5 7 Mark Scheme June (i) www (ii) 8x 0 y z or x 0 y z allow for ±; for / seen or for / seen or for using / = with another error (ie for coping correctly with fraction and negative index or with square root) mark final answer; B for elements correct, for elements correct; condone multn signs included, but from total earned if addn signs 7 (i) or + ( ) or or a =, b =, c = ; for attempt to multiply numerator and denominator by (ii) 7 7 isw www for 7 and for 7 or for correct terms from or or o.e. eg using 6 or M for or or or o.e.; for but not other equivs www M for 0 ( [x]) o.e. or M for two of these elements or for 0 or ( )/( ) o.e. used [ C is not sufficient] or for 0 0 seen; 9 (y )(y ) [= 0] y = or cao x =± or ± cao B or B for 000; condone x in ans; equivs: M for e.g [ ] 0 x o.e. [ may be outside a bracket for whole expansion of all terms], M for two of these elements etc similarly for factor of taken out at start for factors giving two terms correct or attempt at quadratic formula or completing square or B (both roots needed) for roots correct or ft their y (condone and for ) 8

6 7 Mark Scheme June 008 Section B 0 i (x ) 7 mark final answer; for a =, for b = 7 or for + ; bod for (x ) 7 ii (, 7) or ft from (i) + iii sketch of quadratic correct way up and through (0, ) G accept (0, ) o.e. seen in this part [eg in table] if not marked as intercept on graph t.p. correct or ft from (ii) G accept and 7 marked on axes level with turning pt., or better; no ft for (0, ) as min iv x 6x + = x o.e. or their (i) = x x 8x + 6 [= 0] dep on first ; condone one error (x ) [= 0] or correct use of formula, giving equal roots; allow (x + ) o.e. ft x + 8x + 6 x =, y = 6 if M0M0M0, allow SC for showing (, 6) is on both graphs (need to go on to show line is tgt to earn more) equal/repeated roots [implies tgt] - must be explicitly stated; condone only one root [so tgt] or line meets curve only once, so tgt or line touches curve only once etc] or for use of calculus to show grad of line and curve are same when x =

7 7 Mark Scheme June 008 i f( ) used [= 0] ii division of f(x) by (x + ) x x (x + )(x ) [f(x) =] (x + ) (x + )(x ) or B for (x + )(x x ) here; or correct division with no remainder as far as x + 8x in working, or two terms of x x obtained by inspection etc (may be earned in (i)), or f( ) = 0 found x x seen implies or B; allow final ft their factors if A0 earned iii sketch of cubic correct way up G ignore any graph of y = f(x ) through shown on y axis G or coords stated near graph roots,,. or ft shown on x axis G or coords stated near graph if no curve drawn, but intercepts marked on axes, can earn max of G0GG iv x (x )([x ] ) o.e. or x (x )(x.) or ft their factors or (x ) + 7(x ) 7(x ) or stating roots are 0, and. or ft; condone one error eg x 7 not x correct expansion of one pair of brackets ft from their factors or for correct expn of (x ) [allow unsimplified]; or for showing g(0) = g() = g(.) = 0 in given ans g(x) correct completion to given answer allow M for working backwards from given answer to x(x )(x ) and for full completion with factors or roots

8 7 Mark Scheme June 008 i 9 grad AB = or y 9 = (x ) or y = (x + ) y = x + o.e. ii mid pt of AB = (, ) grad perp = /grad AB y = ½ (x ) o.e. or ft [no ft for just grad AB used] at least one correct interim step towards given answer y + x =, and correct completion NB ans y + x = given alt method working back from ans: x y = o.e. ft their m, or subst coords of A or B in y = their m x + c or B condone not stated explicitly, but used in eqn soi by use eg in eqn ft their grad and/or midpt, but M0 if their midpt not used; allow for y = ½x + c and then their midpt subst no ft; correct eqn only mark one method or the other, to benefit of cand, not a mixture grad perp = /grad AB and showing/stating same as given line finding intn of their y = x + and y + x = [= (, )] showing midpt of AB is (, ) eg stating ½ = or showing that (, ) is on y + x =, having found (, ) first [for both methods: for M must be fully correct] iii showing ( ) + ( ) = 0 showing B to centre = 0 or verifying that (, 9) fits given circle iv (x ) + = 0 at least one interim step needed for each mark; M0 for just 6 + = 0 with no other evidence such as a first line of working or a diagram; condone marks earned in reverse order for subst y = 0 in circle eqn (x ) = x = ± or 0 ± isw condone slip on rhs; or for rearrangement to zero (condone one error) and attempt at quad. formula [allow M0 for (x ) = 0 or for (x ) + = 0 ] or ±

9 7 Mark Scheme June (C) Concepts for Advanced Mathematics Section A 0 c.a.o. for π rads = 80 soi 7 (i). 0 -, 0.00 or 000 for S =. / ( 0.) (ii) 6 www stretch, parallel to the y axis, sf for stretch plus one other element correct [f (x) =] x their f (x) > 0 or = 0 soi < x < (i) grad of chord = (. )/0. o.e. =.7 c.a.o. (ii) correct use of A and C where for C,.9 < x <. answer in range (.6,.7) 6 [y =] kx / [+ c] k = subst of (9, 0) in their eqn with c or c = condone x or between - and or chord with ends x = ± h, where 0 < h 0. s.c. for consistent use of reciprocal of gradient formula in parts (i) and (ii) may appear at any stage must have c; must have attempted integration 8 7 sector area = 8.8 or [cm ] c.a.o. area of triangle = ½ 6 sin.6 o.e. their sector their triangle s.o.i. 0.8 to 0.8 [cm ] for ½ 6.6 must both be areas leading to a positive answer 8 a + 0d = or =.(a+0d) (a + 9d) = 0 o.e. a = s.o.i. www and d = s.o.i. www th term is 9 log x log = log or x = log.9 0 ( cos θ) = cos θ + cos θ = cos θ s.o.i. valid attempt at solving their quadratic in cos θ cos θ = - ½ www θ = 90, 70, 0, 0 A D or =.(a + ) gets M eg a + 9d = or x = log for. or versions of.9 for - cos θ = sin θ substituted graphic calc method: allow M for intersection of y = sin θ and y = cos θ + and A for all four roots. All four answers correct but unsupported scores B. 0 and 0 only:. 8 6

10 7 Mark Scheme June 008 Section B i (x + )(x )(x + ) for a (x + )(x )(x + ) ii [(x + )](x + x 0) x + x 0x + x + 6x 0 o.e. for correct expansion of one pair of their brackets for clear expansion of correct factors accept given answer from (x + )(x ) as first step iii iv y = x + 0x their x + 0x = 0 s.o.i. x = 0.6 from formula o.e. (.7,.6) (-.8,.6) M + + if one error or for substitution of 0. if trying to obtain 0, and for correct demonstration of sign change accept (.9,.6) or f.t.( ½ their max x, their max y) 6 i Area = (-)0.6 seen [m ] www Volume = 0. [m ] or ft from their area. M for 0./ ( [ ]) M for one slip; for two slips must be positive ii x x 0. x 0.x o.e. value at 0. [ value at 0] = 0.7 area of cross section (of trough) or area between curve and x-axis 0.7 r.o.t. to or more sf [m ] m seen in (i) or (ii) M E U for terms correct dep on integral attempted must have neg sign 7 i log P = log a + b log t www comparison with y = mx + c intercept = log 0 a must be with correct equation condone omission of base ii log t log P plots f.t. ruled line of best fit iii gradient rounding to 0. or 0. a = 0.9 s.o.i. P = t m allow the form P = 0 0.logt +.9 accept to or more dp for y step / x-step accept.7.0 for intercept accept answers that round to 0, their positive m iv answer rounds in range 60 to 6 7

11 7 Mark Scheme June (C) Methods for Advanced Mathematics Section A x x x x or (x ) 9 x x 8 0 ()(x + )(x ) 0 x Let u = x, dv/dx = e x v = e x / x x x x e d x = xe e.. dx x x = xe e + c 9 (i) f( x) = f(x) Symmetrical about Oy. [] [] [] x ( or =) x x ( or =) x squaring and forming quadratic = 0 (or ) factorising or solving to get x =, x x (www) parts with u = x, dv/dx = e x v = xe e 9 +c x x (ii) (A) even (B) neither (C) odd [] Let u = x + du = x dx x 8/ dx = du x + u = [ ln u ] 8 = ½ (ln8 ln ) = ½ ln(8/) = ½ ln 6* y = x ln x dy = x. + x ln x dx x = x + xlnx dy/dx = 0 when x +xln x = 0 x( + ln x) = 0 ln x = ½ x = e ½ = / e * E [] E [6] / du or k ln (x + ) u ½ ln u or ½ ln(x + ) substituting correct limits (u or x) must show working for ln 6 product rule d/dx (ln x) = /x soi oe their deriv = 0 or attempt to verify ln x = ½ x = e ½ or ln (/ e) = ½ 8

12 7 Mark Scheme June 008 6(i) Initial mass = e 0 = 0 grams Long term mass = 0 grams [] (ii) 0 = e 0.t e 0.t = / 0.t = ln (/) =.0986 t =.0 mins [] anti-logging correctly,.0, 0.99, (not more than d.p) (iii) m 0 0 t [] correct shape through (0, 0) ignore negative values of t 0 as t 7 x + xy + y = dy dy x + x y y 0 dx + + dx = dy ( x + y) = x y dx dy x + y = dx ( x + y) [] Implicit differentiation dy x y dx + correct equation collecting terms in dy/dx and factorising oe cao 9

13 7 Mark Scheme June 008 Section B 8(i) y = /(+cosπ/) = /. [] or 0.67 or better (ii) f'( x) = (+ cos x). sinx sin x = (+ cos x) When x = π/, sin( π / ) f'( x) = (+ cos( π / )) / = = = ( ) 9 9 [] chain rule or quotient rule d/dx (cos x) = sin x soi correct expression substituting x = π/ oe or 0.8 or better. (0.8, 0.89) (iii) deriv = ( + cos x) cos x sin x.( sin x) (+ cos x) cos x + cos x+ sin x = (+ cos x) cos x + = (+ cos x) = * + cosx π / Area = dx 0 + cosx π / = sin x + cosx 0 = sin π / ( 0) + cos π / = = dep E cao [7] Quotient or product rule condone uv u v for correct expression cos x + sin x = used dep www substituting limits or / - must be exact (iv) y = /( + cos x) x y x = /( + cos y) + cos y = /x cos y = /x y = arccos(/x ) * Domain is ½ x E attempt to invert equation www reasonable reflection in y = x [] 0

14 7 Mark Scheme June (i) y= x y = x x + y = which is equation of a circle centre O radius Square root does not give negative values, so this is only a semi-circle. [] squaring x + y = + comment (correct) oe, e.g. f is a function and therefore single valued (ii) (A) Grad of OP = b/a grad of tangent = a b (B) / f( x) = ( x ).( x) = x x a f'( a) = a (C) b = ( a ) so a f'( a ) = as before b E [6] chain rule or implicit differentiation oe substituting a into their f (x) (iii) Translation through followed by 0 stretch scale factor in y-direction 6 [6] Translation in x-direction through or to right ( shift, move A0) 0 alone is SC 0 stretch in y-direction (condone y axis ) (scale) factor elliptical (or circular) shape through (0, 0) and (, 0) and (, 6) (soi) if whole ellipse shown (iv) y = f(x ) = ( (x ) ) = ( x + x ) = (x x ) y = 9(x x ) 9x + y = 6x * E [] or substituting ( (x ) ) oe for y in 9x + y x x www

15 7 Mark Scheme June (C) Applications of Advanced Mathematics Section A x x + = + x x+ ( x )( x+ ) x+ x+ ( x ) = ( x+ )( x ) x = ( x+ )( x ) [] combining fractions correctly factorising and cancelling (may be x²+x-8) V = π ydx= 0 = x π x+ e 0 = π ( + e ) = ( π + e ) * cos θ = sin θ sin θ = sin θ sin θ sin θ = 0 x π ( + e ) dx 0 ( sin θ)( + sin θ) = 0 sin θ = ½ or θ = π/6, π/6, π/ sec θ = x/, tan θ = y/ sec θ = + tan θ x / = + y /9 x / y /9 = * OR x²/-y²/9= sec²θ/-9tan²θ/9 =sec²θ-tan²θ = E [] A,,0 [7] E [] must be π x their y² in terms of x x + e x only substituting both x limits in a function of x www cosθ = sin θ oe substituted forming quadratic( in one variable) =0 correct quadratic www factorising or solving quadratic ½, oe www cao penalise extra solutions in the range sec θ = + tan θ used (oe, e.g. converting to sines and cosines and using cos θ + sin θ = ) eliminating θ (or x and y) www (i) dx/du = u, dy/du = 6u dy / 6 = dy du = u dx dx / du u = u OR y=(x-) /, dy/dx=(x-) / =u [] both u and 6u (y=f(x)), differentiation, (ii) At (, 6), u = dy/dx = 6 [] cao

16 7 Mark Scheme June 008 ( ).( ) 6(i) (+ x ) =. x + ( x ) +...! = x + 6 x +... Valid for < x < - ½ < x < ½ [] binomial expansion with p= -/ x + 6x (ii) x = x = x + 6x x + x +... = x + 8x + ( x )( x 6 x...) [] substitituting their x + 6 x +... and expanding ft their expansion (of three terms) cao 7 sin x cos x = Rsin(x α) = R(sin xcos α cos x sin α) = Rcos α, = R sin α R = + = R = tan α = / α = π/6 y = sin(x π/6) Max when x π/6 = π/ x = π/6 + π/ = π/ max value y = So maximum is (π/, ) [6] correct pairs soi R = ft cao www cao ft their R SC (, π/) no working

17 7 Mark Scheme June 008 Section B 8(i) At A: ( ) + 00 = 0 At B: ( 0) + 00 = 0 At C: ( ) + 00 = 0 So ABC has equation x + y + 0 z + 00 = 0 uuur 00 uuur 0 (ii) DE = DF 0 = = = = = = = 0 0 Equation of plane is x y + 0z = c At D (say) c = 0 0 = 800 So equation is x y + 0z = 0 A,,0 [] [6] substituting co-ords into equation of plane for ABC OR using two vectors in the plane form vector product then x +y +0z = c = -00 OR using vector equation of plane,elim both parameters, need evaluation need evaluation (iii) Angle is θ, where cosθ = = + ( ) θ = 8.9 cao [] formula with correct vectors top bottom (or 0.6 radians) (iv) RS: r = + λ λ = + λ 0λ (+λ) + (+λ) +0.0λ+00 = 0 + 9λ λ + 00 λ + 00 = 0 + λ = 0 λ = so S is (,, 0) [] λ 0 solving with plane λ = cao

18 7 Mark Scheme June 008 9(i) v = 0e t t dt = 0e + c when t = 0, v = 0 0 = 0 + c c = 0 so v= 0 0 t e [] separate variables and intend to integrate t e 0 finding c cao (ii) As t e / t 0 v 0 So long term speed is 0 m s [] ft (for their c>0, found) (iii) A B = + ( w )( w+ ) w w+ Aw ( + ) + Bw ( ) = ( w )( w+ ) A(w + ) + B(w ) w = : = 9A A = /9 w = : = 9B B = /9 /9 /9 = ( w )( w+ ) w w+ = 9( w ) 9( w+ ) [] cover up, substitution or equating coeffs /9 /9 (iv) dw ( w )( w ) dt = + dw = dt ( w )( w+ ) [ ] dw = dt 9( w ) 9( w+ ) ln( w ) ln( w+ ) = t+ c 9 9 w ln = t + c 9 w + When t = 0, w = 0 6 c = ln = ln 9 9 w 9 ln = t + ln w ln w t+ t =.t e = e = 0.e * w + ft E [6] separating variables substituting their partial fractions integrating correctly (condone absence of c) correctly evaluating c (at any stage) combining lns (at any stage) www (v) As t e. t 0 w 0 So long term speed is m s. []

19 7 Mark Scheme June 008 Comprehension. (i) cao (ii) cao. Dividing the grid up into four x blocks gives Lines drawn on diagram or reference to x blocks. One (or more) block does not contain all of the symbols,, and. oe. E. Many possible answers Row correct Rest correct 6

20 7 Mark Scheme June 008. Either Or B. In the top row there are 9 ways of allocating a symbol to the left cell, then 8 for the next, 7 for the next and so on down to for the right cell, giving = 9! ways. So there must be 9! the number of ways of completing the rest of the puzzle. E 6. (i) Block side length, b Sudoku, s s M , and 6 (ii) M = b b - 7

21 7 Mark Scheme June (i) (ii) There are neither s nor s among the givens. So they are interchangeable and therefore there is no unique solution The missing symbols form a embedded Latin square. There is not a unique arrangement of the numbers, and in this square. E E [8] 8

22 7 Mark Scheme June (FP) Further Concepts for Advanced Mathematics Qu Answer Section A 0 (i) 0 Mark Comment (ii) 0 0 (iii) = [] Multiplication, or other valid method (may be implied) c.a.o. B B Circle, ; centre + j, ; radius =, Line parallel to real axis, ; through (0, ), ; correct half line, - x x = y y x y= x, x+ y = y y= x [7] [] Points + j and + j indicated c.a.o. x x For = y y x x + A( x ) + ( x + Bx + Cx+ D) Ax Ax Ax A x Bx Cx D ( A ) x ( B A) x ( A C) x ( D A) A=, B=, C = 6, D= B [] Attempt to compare coefficients One for each correct value 9

23 7 Mark Scheme June 008 (i) AB = B [] Minus each error to minimum of 0 (ii) A 0 7 = 7 7 [] Use of B c.a.o. 6 w w= x x = w w w + + = 0 Substitution. For substitution x = w give B0 but then follow through for a maximum of marks Substitute into cubic Correct substitution 6 OR w + w w+ = α + β + γ = 6 0 αβ + αγ + βγ = αβγ = A [] Minus for each error (including = 0 missing), to a minimum of 0 Give full credit for integer multiple of equation All three Let new roots be k, l, m then B k + l+ m = ( α + β + γ) = = A ( αβ αγ βγ ) kl + km + lm = + + = 6= D klm = 8αβγ = = A ω + ω 6ω+ = 0 C A A [] Attempt to use sums and products of roots of original equation to find sums and products of roots in related equation Sums and products all correct ft their coefficients; minus one for each error (including = 0 missing), to minimum of 0 Give full credit for integer multiple of equation 0

24 7 Mark Scheme June 008 7(i) r+ ( r ) ( )( ) r r+ r r+ Attempt at correct method ( r )( r+ ) [] Correct, without fudging 7(ii) n n = r r+ r r+ ( )( ) r= r= Attempt to use identity = n n + = n + A [] Terms in full (at least two) Attempt at cancelling if factor of missing, max if answer not in terms of n Section A Total: 6

25 7 Mark Scheme June 008 Section B 8(i) x =, x =, y = [] 8(ii) 8(iii) + Large positive x, y (e.g. consider x = 00 ) Large negative x, y (e.g. consider x = 00 ) [] Evidence of method required Curve Central and RH branches correct Asymptotes correct and labelled LH branch correct, with clear minimum [] y = 8(iv) < x < x 0 B [] B max if any inclusive inequalities appear B for < x < 0 and 0< x <,

26 7 Mark Scheme June 008 9(i) 9(ii) + j and j B [] mark for each B [] mark for each correct pair 9(iii) ( x j)( x + j)( x+ + j)( x+ j) ( x x 8)( x x ) = B Attempt to use factor theorem Correct factors, minus each error if only errors are sign errors One correct quadratic with real coefficients (may be implied) = x + x + x x 8x 8x+ 8x + 6x+ 6 = x x + x + 8x+ 6 Expanding A=, B=, C = 8, D= 6 OR A [7] Minus each error, if only errors are sign errors α = αβγδ = 6 * * * * * * αβ = αα + αβ + αβ + ββ + βα + β α * * * * * * αβγ = αα β + αα β + αββ + α ββ αβ =, αβγ = 8 A=, B=, C = 8, D= 6 OR A [7] Both correct Minus each error, if only errors are sign errors Attempt to substitute in one root Attempt to substitute in a second root Equating real and imaginary parts to 0 Attempt to solve simultaneous equations A=, B=, C = 8, D= 6 A [7] Both correct Minus each error, if only errors are sign errors

27 7 Mark Scheme June 008 Qu Answer Section B (continued) 0(i) n n n r r+ = r + r ( ) r= r= r= ( ) ( )( ) ( ) ( ) ( ) = n n+ + n n+ n+ 6 = n n+ n n+ + n+ ( )( 7 ) = n n+ n + n+ ( )( )( ) = n n+ n+ n+ Mark E [] Comment Separation of sums (may be implied) One mark for both parts Attempt to factorise (at least two linear algebraic factors) Correct Complete, convincing argument 0(ii) n r = ( + ) = ( + )( + )( + ) r r n n n n n =, LHS = RHS = Assume true for n = k k r r k k r = k k k + r = r ( + ) = ( + )( + )( + ) ( r+ ) k( k )( k )( k ) ( k ) ( k ) ( k )( k ) [ k( k ) ( k ) ] = = ( k )( k )( k k ) = ( k )( k )( k )( k ) ( k ) (( k ) ) (( k ) ) ( ( k ) ) = = But this is the given result with k + replacing k. Therefore if it is true for k it is true for k +. Since it is true for k =, it is true for k =,, and so true for all positive integers. E E E [8] must be seen Assuming true for k (k + )th term Attempt to factorise Correct Complete convincing argument Dependent on previous and previous E Dependent on first and previous E Section B Total: 6 Total: 7

28 76 Mark Scheme June (FP) Further Methods for Advanced Mathematics (a)(i) x = r cosθ, y = r sinθ (M0 for x = cos θ, y = sinθ ) ( r cos θ + r sin θ ) r = ( r cosθ )( r sinθ ) = r cosθ sin θ r = cosθ sin θ ag (ii) Loop in st quadrant Loop in th quadrant (b) 0 OR x Put 0 dx = arcsin = arcsin π = x x = sinθ x dx = π 0 dθ 0 π = (c)(i) ln( + x ) = x x + x x + x... (ii) ln( x ) = x x x x x + x ln = ln( + x) ln( x) x = x + x + x Fully correct curve Curve may be drawn using continuous or broken lines in any combination For arcsin x For and Exact numerical value Dependent on first (A0 for 60 / ) Any sine substitution For dθ dependent on first Accept unsimplified forms Obtained from two correct series Terms need not be added If M0, then for x + x + x

29 76 Mark Scheme June 008 (iii) r= 0 (r + ) r = + + = + + = ln (i) z = 8, arg z = π (ii) z * = 8, arg z* = z w = 8 8 = 6 π arg( z w) = π + π = π 6 z w arg( z w 8 = = 8 z w ) = = 7 7 ( π π = π cos( = π ) + jsin( π ) = j, a = b (iii) r = 8 = θ = π π kπ θ = + 7 = π, π (iv) θ 7 π j 7 π j ) = ln w * = 8 e, so e = k = z * = 8 e π j = 8 e π j k = So e = z * j w = 8 e ( π + 7 π ) j π j π j = 8 e π j w* π j ( = 8 e, so e = j w k = ) +... ag ft ft ft ft ft ft ft ft ft 7 Terms need not be added For x = seen or implied Satisfactory completion Must be given separately Remainder may be given in exponential or r cjsθ form (B0 for 7 π ) (B0 if left as 8/8) If M0, then for and Accept 8 Implied by one further correct (ft) value Ignore values outside the required range Matching w* to a cube root with 7 argument π and k = or ft r ft is 8 Matching z* to a cube root with argument π May be implied r ft is z * Matching jw to a cube root with argument π May be implied OR for arg( j w) = π + arg w (implied by π or π ) r ft is 8 6

30 76 Mark Scheme June 008 (i) k + Q = k k k 6 When k =, Q = 8 6 Evaluation of determinant (must involve k) For ( k ) Finding at least four cofactors (including one involving k) Six signed cofactors correct (including one involving k) Transposing and dividing by det Dependent on previous Q correct (in terms of k) and result for k = stated After 0, SC for Q when k = obtained correctly with some working (ii) 0 0 P = 0, D = M = PDP = = = 9 (iii) Characteristic equation is ( λ )( λ+ )( λ ) = 0 λ λ λ+ = 0 M = M + M I M = M + M M = (M + M I) + M M = 0M 9I a = 0, b = 0, c = 9 B A For B, order must be consistent Give for M = P DP 6 or 0 8 Good attempt at multiplying two matrices (no more than errors), leaving third matrix in correct position 7 Give for five elements correct Correct M implies BA -8 elements correct implies B In any correct form (Condone omission of =0 ) M satisfies the characteristic equation Correct expanded form (Condone omission of I ) 7

31 76 Mark Scheme June 008 (i) x x x x cosh x = [ (e + e )] = (e + + e ) x x x x x = = + sinh [ (e e )] (e e ) x x cosh sinh = ( + ) = OR x x x x x cosh x+ sinh x = (e + e ) + (e e ) = e x x x x x x x x x cosh sinh = (e + e ) (e e ) = e ag For completion cosh x sinh x = e e = Completion (iii) (ii) ( + sinh x) + 9sinh x = sinh x+ 9sinh x 9 = 0 OR sinh x =, x = ln, ln( + 0 ) x x x x e + 9e e 9e + = 0 x x x x (e e )(e + 6e ) = 0 e =, + 0 x = ln, ln( + 0 ) ft dy = 8cosh xsinh x+ 9cosh x dx = cosh x(8sinh x+ 9) = 0 only when sinh x = 9 x = + = 8 6 cosh ( ) 9 7 y = + 9 ( ) = ft 6 (M0 for sinh x ) Obtaining a value for sinh x Exact logarithmic form Dep on Max if any extra values given Quadratic and / or linear factors Obtaining a value for e x Ignore extra values Dependent on Max if any extra values given Just x = ln earns M0A0A0A0 Any correct form 9 7 or y = (sinh x+ ) +... ( ) 6 Correctly showing there is only one solution Exact evaluation of y or cosh x or cosh x Give B (replacing ) for.06 or better (iv) ln 0 ( + cosh x + 9sinh x)dx [ x sinhx 9coshx ] ln = = ln = ln+ 8 0 A ag Expressing in integrable form Give for two terms correct sinh(ln ) = ( ) Must see both terms for Must also see cosh(ln ) = ( + ) for 8

32 76 Mark Scheme June 008 OR ln x x 9 x x (e + + e + (e e ))dx 0 ln x x 9 x 9 x e x e e e 0 = A = + ln = ln+ ag 8 (i) λ = 0. λ = λ = Expanded exponential form (M0 if the is omitted) Give for three terms correct e ln ln and e = = both seen Must also see e = and e = for ln ln (ii) Ellipse (iii) y = cos( θ π ) Maximum y = when θ = π ag or sin( θ + π ) dy OR = sinθ + cosθ = 0 when θ = π dθ y = + = (iv) + = cos cos sin + sin x y λ θ θ θ θ λ + cos θ + cosθ sinθ + sin θ = ( λ + )( sin θ) + ( + )sin θ λ = + λ + ( λ )sin θ When λ sin θ = 0, x + y = + λ When sin θ =, x + y = + λ Since 0 sin θ, distance from O, x + y, is between + and + λ λ ag ag 6 Using cos θ = sin θ (v) When λ =, x + y = Curve is a circle (centre O) with radius 9

33 76 Mark Scheme June 008 (vi) B A, E at maximum distance from O C, G at minimum distance from O B, F are stationary points D, H are on the x-axis Give ½ mark for each point, then round down Special properties must be clear from diagram, or stated Max if curve is not the correct shape 0

34 77 Mark Scheme June (FP) Further Applications of Advanced Mathematics (i) 6 0 uuur uuur AB AC = 8 = 0 (ii) ABC is x+ y 0z = x+ y 0z+ 9= 0 Distance is = ( ) ( = ) (iii) 6 0 uuur uuur AB CD = 8 = 0 [ = 0 ] 0 0. uuur Distance is AC.nˆ = + + = (iv) Volume is 6 (AB uuur uuur AC). uuur AD 8 = = ( ) (v) E is ( + 0 λ, λ, + λ) ( + 0 λ) ( + λ) + = 0 λ = 7 F is ( + 8 μ, μ, + 6 μ) ( + 8 μ) ( + 6 μ) + = 0 μ = Since 0 λ Since 0 μ < <, E is between A and C < <, F is between A and D B ft Ignore subsequent working Give for one element correct SC for minus the correct vector For x + y 0z Accept x+ y 0z = 99 etc Using distance formula (or other complete method) Condone negative answer Accept a.r.t..8 uuur uuur Evaluating AB CD or method for finding end-points of common perp PQ or P (,, ) & Q ( 7 8 8, 9, 6) uuur PQ = (,, ) or Scalar triple product Accept a.r.t. 7.

35 77 Mark Scheme June 008 uuur uuur uuur (vi) VABEF = (AB AE). AF 6 uuur uuur uuur = λμ (AB AC). AD 6 = λμvabcd = V ABCD Ratio of volumes is : 7 = :7 (i) g = 6z ( x + y + z) = x y x g = ( x + y + z) y g = 6x 6( x + y + z) = y 8z z (ii) g g g At P, = 6, =, = 6 x y z 7 Normal line is r = 7. + λ 9 (iii) δ g 6δx δ y+ 6δz uuur If PQ = λ, 9 δ g 6( λ) ( λ) + 6(9 λ) ( = 9 λ ) (iv) (v) h = δ g, so h 9λ uuur h PQ, so 9 n = g g Require = = 0 x y x y = 0 and x+ y+ z = 0 x+ y = 0 and z = 0 g( x, y, z ) = 0 0 = 0 Hence there is no such point on S g Require = 0 z g g and = y x x 8y z = ( x y) ag ft ft 6 ( ) ft if numerical 6 Finding ratio of volumes of two parts SC for : 7 deduced from without working Partial differentiation Any correct form, ISW Evaluating partial derivatives at P All correct Condone omission of r = Alternative: M for substituting x = 7+ λ, into g = + h and neglecting λ ft for linear equation in λ and h for n correct Useful manipulation using both eqns Showing there is no such point on S Fully correct proof g g Implied by = λ, = λ x y This can be awarded for x y = and x 8y z =

36 77 Mark Scheme June 008 y = z and x = z 6( zz ) ( z) = z =± Points are (, 7., ) and (, 7., ) (i) x& + y& = ( t ) + (8t 8 t ) = 76t + t 88t + 6t 6 = t + 88t + 6t = (8t+ 8 t ) Arc length is 6 (8t+ 8 t ) dt 0 = 9t + t = 68 (ii) Curved surface area is π yds (iii) = π (9t t )(8t+ 8 t )dt = π (t + 7t t )dt = π 8t + t t = 00 π ( 67 ) 0 xy &&& &&& xy ( t )(8 t ) (8 t)(8t 8 t ) κ = = ( x& + y& ) (8t+ 8 t ) 8 t (9 t t ) 8 t (9 + t ) = = 8 t (9+ t ) 8 t (9+ t ) 6 = t(t + 9) ft ag ag or z = y and x = y or y = x and z = x 0 or x =, y =, z = 8 λ λ λ or x: y: z = 0: : Substituting into g( x, y, z ) = Obtaining one value of x, y, z or λ Dependent on previous ft is minus the other point, provided all M marks have been earned Note (8 + 8 t ) dt = 8t+ t = earns A0A0 Using d s = (8t+ 8 t )dt Correct integral expression including limits (may be implied by later work) Using formula for κ (or ρ ) For numerator and denominator Simplifying the numerator

37 77 Mark Scheme June 008 (iv) 6 When t =, x = 8, y = 7, κ = 69 ρ = ( ) 6 dy y& 8t 8t 0 = = = x x& t d n ˆ = 8 69 c = Centre of curvature is (8, 9) 6 (i) Commutative: x y = y x (for all x, y) Associative: ( x y) z = x ( y z) (for all x, y, z) (ii) ( x+ )( y+ ) = xy+ x+ y+ = xy+ x+ y = x y (iii)(a) If x, y S then x > and y > + y+ >, so x y S x+ > 0 and y+ > 0, so ( x+ )( y+ ) > 0 ( x )( ) (B) 0 is the identity since 0 x = 0+ x+ 0= x If x S and x y = 0 then xy + x + y = 0 x y = x + y + = > 0 (x + ) so y S (since x > ) S is closed and associative; there is an identity; and every element of S has an inverse in S (iv) If x x = 0, x + x+ x = 0 x = 0 or 0 is the identity (and has order ) is not in S B ag 7 6 Finding gradient (or tangent vector) Finding direction of the normal Correct unit normal (either direction) Accept e.g. Order does not matter Give for a partial explanation, e.g. Position of brackets does not matter Intermediate step required or x y ( + )( + ) = 0 or y + = ( x + ) Dependent on

38 77 Mark Scheme June 008 (v) 6= = 8 = 6 + = (vi) (vii) So o 6= Element 0 6 Order 6 6 {0}, G {0, 6} {0,, } ag B Give B for correct for correct Condone omission of G If more than non-trivial subgroups are given, deduct mark (from final ) for each non-trivial subgroup in excess of

39 77 Mark Scheme June 008 Pre-multiplication by transition matrix (i) P = (ii) P = P(B used on 7th day) = 0.8 (iii) (iv) = 0.96 P = = 0.67 (v) Q = , 0.86, 0.89 are the long-run probabilities for the routes A, B, C (vi) a b 0. = c (vii) a = 0., so a = b = 0., so b = c = 0., so c = 0. After C, routes A, B, C will be used with probabilities 0., 0, = 0.6 B A Give for two columns correct 6 7 Using P (or P ) For matrix of initial probabilities For evaluating matrix product Accept 0.8 to 0.8 Using diagonal elements from P Correct method Accept a.r.t For evaluating P Using diagonal elements from P Correct method Accept a.r.t. 0.6 Deduct if not given as a ( ) matrix Deduct if not dp Accept equilibrium probabilities Obtaining a value for a, b or c Give for one correct Using long-run probs 0., 0., 0. Using diag elements from new matrix 6

40 77 Mark Scheme June 008 Post-multiplication by transition matrix (i) P = (ii) (iii) (iv) ( 6 ) P = ( ) P(B used on 7th day) = = 0.96 P = = 0.67 (v) Q = (vi) (vii) 0.89, 0.86, 0.89 are the long-run probabilities for the routes A, B, C = a b c ( ) ( ) a = 0., so a = b = 0., so b = c = 0., so c = 0. After C, routes A, B, C will be used with probabilities 0., 0, = 0.6 B A Give for two rows correct 6 7 Using P (or P ) For matrix of initial probabilities For evaluating matrix product Accept 0.8 to 0.8 Using diagonal elements from P Correct method Accept a.r.t For evaluating P Using diagonal elements from P Correct method Accept a.r.t. 0.6 Deduct if not given as a ( ) matrix Deduct if not dp Accept equilibrium probabilities Obtaining a value for a, b or c Give for one correct Using long-run probs 0., 0., 0. Using diag elements from new matrix 7

41 78 Mark Scheme June Differential Equations (i) ( ) && x = g 8 x+ 0.g kv NL equation with all forces using given expressions for tension and resistance Weight positive as down, tension negative as up. Resistance negative as opposes motion. && x+ kx& + x = 0 E Must follow correct NL equation (ii) x = Acos t+ Bsin t t = 0, x = 0. A= 0. Find the coefficient of cos x& = Asint+ Bcost so t = 0, x& = 0 B = 0 Find the coefficient of sin x = 0.cos t cao (iii) α + α + = 0 Auxiliary equation α = ± j CF for complex roots t x = e ( Ccos t+ Dsin t) F CF for their roots t = 0, x = 0. C = 0. Condition on x t x& = e ( Ccos t+ Dsin t) t + e Csin t+ Dcos t Differentiate (product rule) ( ) 0= C+ D Condition on x& 0. D = t ( ) x = 0.e cos t+ sin t cao Curve through (0,0.) with zero gradient Oscillating Asymptote x = 0 (iv) k > 0 Use of discriminant Correct inequality (As k is positive) k > Accept k < in addition (but not k > ) Curve through (0,0.) Decays without oscillating (at most one intercept with positive t axis) 8

42 78 Mark Scheme June 008 (i) Ae t x = Any valid method t = 0, x = 8 A= 8 Condition on x x = 8e t (ii) y& + y = 6e t Substitute for x α + = 0 α = Auxiliary equation CF y = Be t ae t t t t PI y = ae + ae = 6e Differentiate and substitute a = 6 cao t t GS y = 6e + Be Their PI + CF (with one arbitrary F constant) t = 0, y = 0 B = 6 Condition on y t t ( ) y = 6 e e F Follow a non-trivial GS Alternative mark scheme for first 7 marks: Substitute for x I = e t Attempt integrating factor IF correct d(y e t )/dt = 6e t Integrate y e t = 6e t + B cao y = 6e t + Be t F Divide by their I (must divide constant) (iii) t t y 6e ( e ) = Or equivalent (NB e t > e t needs justifying) 6e t 0 t t > < hence y > 0 E Complete argument Starts at origin General shape consistent with their solution and y > 0 Tends to zero 9 (iv) (v) d ( ) ( ) ( ) ( ) 0 d x + y + z = t x + x y + y = Consider sum of DE s x + y+ z = c E Hence initial conditions x+ y+ z = 8 E z = 8 x y Substitute for x and y and find z t t t ( ) ( ) z = 8 e + e = 8 e E Convincingly shown (x, y must be correct) ( ) = 8 e t Correct equation (any form) t = or.98 99% is Z after.0 hours Accept value in [.9,.] 9

43 78 Mark Scheme June 008 k (i) y& + y = Divide by t (condone LHS only) t k k I = exp( dt) = exp( kln t t ) = t Attempt integrating factor Integrating factor k k k t y& + kt y = t F Multiply DE by their I d k k ( yt ) = t dt LHS k k yt = t dt Integrate t k + + k + k + = A cao (including constant) k y = t+ At F Divide by their I (must divide constant) k+ k+ t =, y = 0 0 = + A A= Use condition k + k ( ) y = t t F Follow a non-trivial GS (ii) y = ( t t ) Shape consistent with their solution for t > Passes through (, 0) Behaviour for large t 0 (iii) yt = t dt Follow their (i) = ln t+ B cao y = t( ln t+ B) F Divide by their I (must divide constant) t =, y = 0 B = 0 y = tln t cao (iv) dy = + t sin y dt Rearrange DE (may be implied) t y dy/dt 0 Use algorithm y(.) y(.) (v) 0.8 as smaller step size Must give reason Decreasing step length has increased estimate. Assuming this estimate is more accurate, decreasing step length further will Identify effect of decreasing step length increase estimate further, so true value likely to be greater. Hence underestimates. Convincing argument Alternative mark scheme for last marks: dy/dt seems to be increasing, hence Euler s method will underestimate true value + sketch (or explanation). Identify derivative increasing Convincing argument 0

44 78 Mark Scheme June 008 (i) && x = x& 6y& 9cost Differentiate first equation = x& 6(x y 7sin t) 9cost Substitute for &y y 6 ( ) y in terms of x, x & && x = x& 8x+ (x x& 9sin t) + sint 9cost Substitute for y && x + x& x = sint 9cost E LHS E RHS (ii) α + α = 0 Auxiliary equation α = or t t CF x = Ae + Be F CF for their roots PI x = acost+ bsin t PI of this form ( ac bs) + ( as + bc) ( ac + bs) = s 9c Differentiate twice and substitute a+ b a = 9 Compare coefficients ( equations) b a b = Solve ( equations) a =, b = 0 cos e t x = t+ A + Be Their PI + CF (with two arbitrary F constants) (iii) y = ( x x t ) & 9sin 6 y in terms of x, x & t cos e e t t sin e e t = t+ A + B + t A + B 9sin t Differentiate x and substitute 6 ( ) cos sin e t t y = t t+ A + Be (iv) x bounded A = 0 Constants must correspond with those in x Identify coefficient of exponentially growing term must be zero y bounded E Complete argument (v) t = 0, y = 0 0 = B+ B = Condition on y x = cost e t, y = cost sint e t Follow their (non-trivial) general F solutions x = cost cao y = cost sint cao 6 9

45 76 Mark Scheme June Mechanics Q mark comment sub (i) NL = 00a NL. Accept F = mga and no weight Weight correct (including sign). Allow if seen. a = 0. so 0. m s - upwards Accept ± 0.. Ignore units and direction (ii) TBA 980 = NL. F = ma. Weight present, no extras. Accept sign errors. so tension is 060 N (iii) T BA cos0 = 060 Attempt to resolve their (ii). Do not award for their 060 resolved unless all forces present and all resolutions needed are attempted. If start again allow no weight. Allow sin cos. No extra forces. Condone sign errors FT their 060 only T BA = so 0 N ( s. f.) cao 8 Q m a r k comment sub (i) Sketch. O, i, j and r (only require correct quadrant.) Vectors must have arrows. Need not label r. (ii) + ( ) Accept = or 6.0 so 6.0 ( s. f.) Need 80 arctan ( ) Or equivalent. Award for arctan ( ± ) or arctan ( ± ) or equivalent seen without 80 or so cao (iii) i j or Do not award for magnitude given as the answer. Penalise spurious notation by mark at most once in paper 6

46 76 Mark Scheme June 008 Q mark comment sub Penalise spurious notation by mark at most once in paper (i) F = so = 0 0 N Use of NL in vector form Ignore units. [Award for answer seen] [SC for or equiv seen] (ii) s = + + Use of s= tu+ 0.t aor integration of a. Allow s 0 omitted. If integrated need to consider v when t = 0 Correctly evaluated; accept s 0 omitted. 6 6 s = so 9 9 m Correctly adding s 0 to a vector (FT). Ignore units. 8 [NB 6 seen scores ] Q mark comment sub (i) (ii) The distance travelled by P is t The distance travelled by Q is 0t Accept 0t + if used correctly below. Meet when 0.t = + 0t Allow their wrong expressions for P and Q distances Allow ± or omitted F Award for their expressions as long as one is quadratic and one linear. Must have with correct sign. so t 0t 00 = 0 Solving Accept any method that yields (smaller) + ve root of their term quadratic t = 0 (or -0) cao Allow ve root not mentioned Distance is 0. 0 = 6 m cao [SC 00 m seen] 7

47 76 Mark Scheme June 008 Q mark comment sub either Overall, NL 9 = ( +)a Use of NL. Allow F = mga but no extra forces. Allow 9 omitted. a = so m s - For A, NL NL on A or B with correct mass. F = ma. All T 9 = relevant forces and no extras. so 6 N cao or T = a * equation in T and a. Allow sign errors. Allow F = mga T 9 = a Both equations correct and consistent Solving Dependent on M* solving for T. T = 6 so 6 N cao. Q 6 mark comment sub (i) 0 0.6t t Use of s = ut + 0.at with a = ± 9.8, ± 0. Accept 0 or for u. = t t Any form (ii) either Need zero vertical distance so t t = 0 Equate their y to zero. With fresh start must have correct y. so t = 0 or t =.8 Accept no reference to t = 0 and the other root in any form. FT their y if gives t > 0 or Time to highest point, T Allow use of u = 0 and Award even if half range found. 0 = T so T =. and time of flight is.8 May be awarded for doubling half range later. range is =.6 Horiz cpt. Accept 0.6 instead of 0.8 only if consistent with expression in (i). FT their t. so m ( s. f.) cao [NB Use of half range or half time to get 76.8 (g = 0) or 78.6 (g = 9.8) scores ] [If range formula used: sensible attempt at substitution; allow sinα wrong sinα correct all correct cao] 6

48 76 Mark Scheme June 008 Q 7 mark comment sub (i) Continuous string: smooth ring: light string E E One reason Another reason [(ii) and (iii) may be argued using Lami or triangle of forces] (ii) Resolve : 60cosα 60cos β = 0 Resolution and an equation or equivalent. Accept s c. Accept a correct equation seen without method stated. Accept the use of T instead of 60. (so cosα = cos β ) and so α = β E Shown. Must have stated method (allow seen). (iii) Resolve Resolution and an equation. Accept s c. Do not award for resolution that cannot give solution (e.g. horizontal) 60 sinα 8g = 0 Both strings used (accept use of half weight), seen in an equation (iv) so α = 0.79 so 0.8 ( s. f.) Resolve 0 + T cos T cos = 0 QC PC Resolve T sin + T sin 8g = 0 PC QC Solving sinα or equivalent seen in an equation All correct Recognise strings have different tensions. Resolution and an equation. Accept s c. No extra forces. All forces present. Allow sign errors. Correct. Any form. Resolution and an equation. Accept s c. No extra forces. All forces present. Allow sign errors. Correct. Any form. * A method that leads to at least one solution of a pair of simultaneous equations. T CQ = so. N ( s. f.) cao either tension T CP = so 80. N ( s. f.) F other tension. Allow FT only if * awarded [Scale drawing: st then, for answers correct to s.f.] 7 8

49 76 Mark Scheme June 008 Q 8 mark comment sub (i) 0 (ii) v = 6 + 6t 6t Attempt at differentiation (iii) a = 6 t Attempt at differentiation F (iv) Take a = 0 Allow table if maximum indicated or implied so t = 0. FT their a and v = 7. so 7. m s - cao Accept no justification given that this is maximum (v) either Solving 6 6t 6t 0 + = A method for two roots using their v Factorization or formula or of their expression so t = - or t = E Shown or Sub the values in the expression for Allow just substitution shown v Both shown to be zero E Both shown A quadratic so the only roots Must be a clear argument then x(-) = - cao x() = 9 cao (vi) x() x(0) + x() x() Considering two parts = Either correct = 98 so 98 m cao [SC for s() s(0) = 6] (vii) At the SP of v Or any other valid argument e.g find all the zeros, sketch, consider sign changes. Must have some x(-) = - i.e. < 0 and x() = 9 i.e. > 0 working. If only a sketch, must have correct shape. Also x(-) = > 0 and x(6) = -98 < 0 Doing appropriate calculations e.g. find all zeros; sketch cubic reasonably (showing roots); sign changes in range so three times times seen 9 6

50 76 Mark Scheme June Mechanics Q mark comment sub (a) (i) In i direction: 6u = 8 Use of I-M so u = i.e. i m s - Accept 6u = 8as total working. Accept E instead of i. either In i direction: 0.v + = 0. Use of I-M Use of + i or equivalent v = so i m s - Accept direction indicated by any means or v = PCLM v = Allow only sign errors so i m s - Accept direction indicated by any means (ii) Using NEL: = e Use of NEL. Condone sign errors but not reciprocal expression F FT only their (even if +ve) e = (0.) & 9 F FT only their and only if ve (allow s.f. accuracy) (iii) In i direction: 7= 0.v 0. Use of I = Ft Use of I = m(v u ) v = 7 so 7 i m s - For ± 7 cao. Direction (indicated by any means) or i = 0. a Use of F = ma so a = i m s - For ± v = i i 7 Use of uvast v = 7 so 7 i m s - cao. Direction (indicated by any means) (b) u i + ev j For u For ev v ev Use of tan. Accept reciprocal argument. Accept tanα =, tan β = u u use of their components v tan β = e = etanα u E 7 Both correct. Ignore signs. Shown. Accept signs not clearly dealt with. 7

51 76 Mark Scheme June 008 Q m a r k comment sub (i) x 6 0 ( + 6) y = Method for c.m. Total mass correct x y = = For any of the st RHS terms For the th RHS term x =.6 E y =. cao [If separate cpts, award the nd for x- terms correct and rd for 7in y term] 6.6. D.6. G 6 O vertical (ii) Diagram showing G vertically below D.6 and their. correctly placed (may be implied).6.6 arctan = arctan Use of arctan on their lengths. Allow reciprocal + (6.).6 of argument. Some attempt to calculate correct lengths needed + (6 their.) seen so 8.07 so 8.0 ( s. f.) cao (iii) moments about D moments about D. No extra forces.6= 6 T BP so tension in BP is N F FT their values if calc nd Resolve vert: + TDQ = Resolve vertically or moments about B. so tension in DQ is N F FT their values if calc nd (iv) We require x-cpt of c.m. to be zero A method to achieve this with all cpts either ( 0 + L) x = 0.6 L or 6 ( 0. 6) L = 0 For the 0. L All correct L = 9 8

52 76 Mark Scheme June 008 Q mark comment sub (a) (i) D T AD T BD A B C 0 60 T AB T BC Internal forces all present and labelled All forces correct with labels and arrows (Allow the internal forces set as tensions, thrusts or a mixture) L N L N (ii) Equilibrium equation at a pin-joint attempted st ans. Accept + or. A Second equation attempted T AD sin 0 L = 0 so T AD = L so L N (T) F nd ans. FT any previous answer(s) used. A T AB + T AD cos 0 = 0 so T AB = L so L N (C) Third equation attempted B T BD sin 60 L = 0 rd ans. FT any previous answer(s) used. so T BD = L so L N (T) B Fourth equation attempted T BC + T BD cos 60 T AB = 0 F th ans. FT any previous answer(s) used. so T BC = L so L N (C) E All T/C consistent [SC all T/C correct WWW] 9 (b) Leg QR with frictional force F moments c.w. about R U sin60 l Wlcos60= 0 Horiz equilibrium for QR F = U Accept only leg considered (and without comment) Suitable moments equation. Allow force omitted a.c. moments c.w. moments A second correct equation for horizontal or vertical equilibrium to eliminate a force (U or reaction at foot) [Award if correct moments equation containing only W and F] E * This second equation explicitly derived Hence W = F Correct use of nd equation with the moments equation and so F = W E Shown. CWO but do not penalise * again

53 76 Mark Scheme June 008 Q mark comment sub (a) (i) Tension is perp to the motion of the sphere (so WD, Fd cosθ = 0) E (ii) Distance dropped is cos0 = Attempt at distance with resolution used. Accept sin cos E Accept seeing cos0 GPE is mgh so = J Any reasonable accuracy (iii) v = Using KE + GPE constant so v =.087 so.0 m s - ( s. f.) F FT their GPE (iv) 0. ( v. ) Use of W-E equation (allow KE term or GPE term omitted) (b) KE terms correct 0 = π 60 WD against friction WD against friction correct (allow sign error) v =.0678 so.06 m s - ( s. f.) cao NL down slope: g sin0 F = g 8 Must have attempt at weight component Allow sign errors. 9g so F = (=.0) 8 R = g (=.6 ) μ = F R Use of F = μ R E Must be worked precisely 8 6 0

54 76 Mark Scheme June Mechanics (a)(i) [ Velocity ] (ii) (iii) (iv) = L T [ Acceleration ] [ Force ] = M L = M L T = L T [ Force ] M L T [ λ ] = = [ v ] (L T ) U g (L T = L T ) λu mg = L (M L )(L T = M (L T ) M L T = = L M L T [ H ] = L ; all terms have the same dimensions (M L ) (L T β = + α + γ = α γ = 0 α = 6 γ = α ) M β ) (L T γ ) = L cao cao cao E cao cao cao (Deduct mark if kg, m, s are consistently used instead of M, L, T) (Condone constants left in) Dependent on At least one equation in α, γ One equation correct

55 76 Mark Scheme June 008 (b) EE is ( = ) (PE gained) = (EE lost) + (KE lost) h = h = h = 0. OA = 0 h = 9. m F Equation involving PE, EE and KE Can be awarded from start to point where string becomes slack or any complete method (e.g. SHM) for finding v at natural length If B0, give for v = 88. correctly obtained or 0 = s ( s =.) Notes used as EE can earn B0FA used as EE gets B0M0

56 76 Mark Scheme June 008 (i) T cosα = mg.9 cosα = cosα = 0.7 Angle is. (0.7 rad) Resolving vertically (Condone sin / cos mix for M marks throughout this question) (ii) T sin α = v m r v.9 sin α = 0.. sin α Speed is.9 m s Force and acceleration towards centre (condone v /. or.ω ) For radius is.sin α ( =.778) Not awarded for equation in ω unless v = (. sin α) ω also appears (iii) v T mg cosθ = m a 8. T cos 60 = 0.. Tension is 6. N Forces and acceleration towards O (iv) mv mg. cosθ = v m cosθ = v = cosθ mg. cos 60 E For ( ) mg. cosθ in PE Equation involving mv and PE (v) v ( T ) mg cosθ = m a cosθ ( T ) m 9.8 cosθ = m. String becomes slack when T = 0 9.8cosθ = cosθ 7 cosθ = 9. θ = 0 (.8 rad) Force and acceleration towards O Substituting for v Dependent on first No marks for v = 0 θ =

57 76 Mark Scheme June 008 (i) TPB = ( x.) [ = x ] (ii) TBQ = (6. x.8) = (.7 x) [ =. x ] d x TBQ + mg TPB = m dt d x (.7 x) ( x.) =. dt A d x 60 0x =. dt d x = 6 6x dt E Finding extension of BQ Equation of motion (condone one missing force) Give for three terms correct (iii) d x At the centre, = 0 dt x = (iv) ω = 6 Period is π = π 6 =.7 s (v) Amplitude A =. = 0. m Maximum speed is A ω = 0. =.6 m s (vi) x = + 0. cos t v = ( ).6sint 0.9 When v = 0.9, sin t =.6 t = π Time is 0.9 s OR 0.9 = 6(0. y ) y = 0.07 y = 0. cos t 0.07 cos t = 0. t = π Time is 0.9 s cao ft cao cao Seen or implied (Allow for ω = 6 ) Accept π ft is. (iii) For v = C sin ωt or C cosωt This can be earned in (v) Fully correct method for finding the required time e.g. 0.9 arcsin period.6 + Using v = ω ( A y ) and y = A cosωt or Asin ωt For y = ( ± ) 0. and y = 0. cos t

58 76 Mark Scheme June 008 (a)(i) V = π x = π dy = π ( 8 [ y y ] = 6π = π V y = π y x 8 dy = π y ( 0 y = y) dy y) dy 8 8 [ y y ] = π = 8 π 6π 8 6 (.67 ) 0 7 π may be omitted throughout Limits not required for M marks throughout this question Dependent on (ii) CM is vertically above lower corner tan θ = = ( = ) y 8 θ = 6.9 ( = 0.6 rad) Trig in a triangle including θ Dependent on previous Correct expression for tan θ or tan( 90 θ ) Notes tanθ = implies cand's y cand's y tanθ = implies tanθ = without further cand's y evidence is M0M0

59 76 Mark Scheme June 008 (b) A= (8 x ) dx 6 = 8x x = May use 0 x throughout or 8 0 () y d y Ay= (8 x ) dx 6 = x x + x = 0 y = = =. or 8 0 Dependent on first two s 7 () y y d y (M0 if ½ is omitted) 6 For x x + x Allow one error or 8 y ( y) ( y ) or 6 6 ( y) + ( y ) 6

60 76 Mark Scheme June Mechanics (i) If δ m is change in mass over time δ t PCLM mv = ( m + δ m)( v + δv) + δm ( v u) [N.B. δ m < 0 ] Change in momentum over time δt δv δm dv dm Rearrange to produce DE ( m + δ m) + u = 0 m = u δt δt dt dt Accept sign error dm = k m= m0 kt dt Find m in terms of t dv ( m0 kt) = uk d t E Convincingly shown (ii) uk v = dt m0 kt Separate and integrate = uln( m0 kt) + c cao (allow no constant) t = 0, v = 0 c = uln m0 Use initial condition m 0 v = uln m0 kt All correct (iii) m = m Find expression for mass or time 0 m0 kt = m 0 Or t = m 0 / k v = uln (i) P = Fv Used, not just quoted dv Use NL and expression for = mv v dx acceleration dv mv = m( k v ) dx Correct DE v dv = k v dx Rearrange k dv = k v dx E Convincingly shown k dv = dx k v Separate and integrate ln k + k v v x c = + k v LHS x = 0, v = 0 c = 0 Use condition ln k + x k v = v k v cao (ii) Terminal velocity when acceleration zero v = k.9 v 0.9k x kln = = 0.9 k = ( ln9 0.9 ) k 0. F 0.7k Follow their solution to (i) 9 7

61 76 Mark Scheme June 008 (i) a Use circular elements (for M or I) M = k( a+ r) π rdr 0 Integral for mass a Integrate (for M or I) = kπ ar + r For [ ] 0 = kπ a E a ( ) π d Integral for I 0 I = k a+ r r r r a 0 = kπ ar + r 9 For [ ] = 0 kπ a cao 7 0 Ma = E Complete argument (including mass) 9 (ii) I =. Seen or used (here or later) = Iω Use angular momentum Use moment of impulse ω. cao (iii) && 0. θ =.8 0 Find angular acceleration Couple = Iθ Use equation of motion 8.7 F Follow their ω and I (iv) Iθ = θ Allow sign error and follow their I (but not M) d & θ I = & θ dt Set up DE for & θ (first order) d & θ dt & θ = I Separate and integrate ln t θ = +c. ln ( multiple of & θ ) seen & /. θ = Ae t Rearrange, dealing properly with constant t = 0, & θ = 0 A= 0 Use condition on & θ & /. θ = 0e t (v) Model predicts θ & never zero in finite time. 7 8

62 76 Mark Scheme June 008 (i) E mg V = ( aθ ) mgacosθ + (relative to centre 0a EPE term of pulley) Extension = aθ GPE relative to any zero level (± constant) dv mg = a θ mgasinθ dθ 0a Differentiate dv mga ( θ 0 dθ sinθ ) 6 (ii) (iii) dv θ = 0 = mga ( 0 ( 0) sin0) = 0 Consider value of d V dθ dθ hence equilibrium E d V Differentiate again mga ( cosθ 0 ) dθ V (0) = 0.9mga < 0 Consider sign of V hence unstable E V must be correct If the pulley is smooth, then the tension in the string is constant. Hence the EPE term is valid. (iv) Equilibrium positions at θ =.8, One correct θ = 7. All three correct, no extras and θ = 8. Accept answers in [.7,.0), [7,7.], [8.,8.] From graph, V (.8) = mgaf (.8) > 0 Consider sign of V or f hence stable at θ =.8 Accept no reference to V for one V (7.) = mgaf ( 7.) < 0 unstable at θ = 7. conclusion but other two must relate V (8.) = mgaf 8. > 0 stable at θ = 8. to sign of V, not just f. (v) ( ) 6 6 P in approximately correct place P B B in approximately correct place (vi) If θ < 0 then expression for EPE not valid hence not necessarily an equilibrium position. 9

63 766 Mark Scheme June Statistics Q (i) Mean = 7. (or better) Standard deviation: (awfw) Bcao Bcao fx =. fx = 96. Allow s =.6 to.68 () for variance s.o.i.o Allow rmsd =.6.66 (awfw) () for rmsd After B0, B0 scored then if at least correct mid-points seen or used.{.,, 6, 8., } fx Attempt of their mean =, with 0 fx 6 and fx () mid-points () 6.8 mean 7.86 (ii) Q (i) strictly from mid-points not class widths or top/lower boundaries. Upper limit = =.7 or their sensible mean + their sensible s.d. So there could be one or more outliers P(W) P(C) = = 0.0 P(W C) = 0.06 (given in the question) Not equal so not independent (Allow or p (W C) so not independent). ( with s.d. < mean) Edep on B, B earned and comment TOTAL 6 for multiplying or 0.0 seen (numerical justification needed) (ii) W C G for two overlapping circles labelled G for 0.06 and either 0. or 0. in the correct places 0.69 The last two G marks are independent of the labels G for all correct probs in the correct places (including the 0.69) NB No credit for Karnaugh maps here (iii) P(W C) = P( W C ) = 0.06 = 6 = 0. P(C) (awrt 0.) for 0.06 / 0.7 cao 60

64 766 Mark Scheme June 008 (iv) Children are more likely than adults to be able to speak Welsh or proportionally more children speak Welsh than adults Do not accept: more Welsh children speak Welsh than adults Q (i) ( A) p+ q = so p+ q = 0. ( B) p+ q = 0.67 so p+ q = 0. ( C) from above p+ q = 0.0 so q = 0.0, p= 0. EFT Once the correct idea is seen, apply ISW p + q in a correct equation before they reach p + q =0. TOTAL 8 p + q in a correct equation before they reach p + q = 0. (ii) Q (i) (ii) E(X ) = =.0 Var(X) = their = 0.60 (awrt 0.6) (, can be earned with their p + and q + but not A mark) X ~ B(8, 0.0) 8 ( A) P( X = 0) = 0.9 = or better Or using tables P( X = 0) = ( B) P( X = ) = = 0.79 P( X > ) = ( ) = 0.07 Or using tables P(X > ) = 0.98 = 0.07 Expected number of days = =. awrt () for any correct answer B for both correct answers Σx p (at least non zero terms correct) dep for ( 0.67²), provided Var( X ) > 0 cao (No n or n- divisors) TOTAL CAO Or B (tables) for P(X = )(allow 0.8 or better) for P(X ) must have both probabilities cao ( ) for P(X ) 0.98 for P(X ) cao (must end in ) for 0 x prob(b) FT but no rounding at end TOTAL 7 6

65 766 Mark Scheme June 008 Q (i) Let p = probability of remembering or naming all items (for population) (whilst listening to music.) H 0 : p = 0. H : p > 0. for definition of p for H 0 for H H has this form since the student believes that the probability will be increased/ improved/ got better /gone up. (ii) Let X ~ B(, 0.) Either: P(X 8) = = 0. > % Or < 9% So not enough evidence to reject H 0 (Accept H o ) Edep on p>0. in H 0 In words not just because p > 0. Either: for probability (0.) dep for comparison dep Conclude that there is not enough evidence to indicate that the probability of remembering all of the items is improved / improved/ got better /gone up. (when listening to music.) Or: Critical region for the test is {9,0,,,,,} 8 does not lie in the critical region. So not enough evidence to reject H 0 Conclude that there is not enough evidence to indicate that the probability of remembering all of the items is improved / improved/ got better /gone up. (when listening to music.) Edep on all previous marks for conclusion in context Or: for correct CR(no omissions or additions) dep for 8 does not lie in CR dep Edep on all previous marks for conclusion in context Or: The smallest critical region that 8 could fall into is {8, 9, 0,,,,, and }. The size of this region is > % So not enough evidence to reject H 0 Conclude that there is not enough evidence to indicate that the probability of remembering all of the items is improved (when listening to music) Or: for CR{8,9, }and size = 0. dep for comparison dep Edep on all previous marks for conclusion in context TOTAL 8 6

66 766 Mark Scheme June 008 Section B Q6 (i) (A) P(both rest of UK) = = 0.0 (B) Either: All case P(at least one England) = (0.79 x 0.0) + (0.79 x 0.0) + (0.0 x 0.79) + (0.0 x 0.79) + (0.79 x 0.79) = = 0.99 Or P(at least one England) = P(neither England) = (0. 0.) = 0.0 = 0.99 or listing all = { ( 0. x 0.) + (0. x 0.0) + (0.0 x 0.0) + (0.0x 0.0)} = (**) = { ) = 0.0 = 0.99 for multiplying cao for any correct term (case or case) for correct sum of all (or of all ) with no extras cao (condone 0.96 www) Or for or for (**) fully enumerated or 0.0 seen dep for ( st part) cao (ii) Or: All case P(at least one England) = = = = 0.99 (C)Either 0.79 x x x x 0. = Or = Or { 0.79 x x x x } = = P(both the rest of the UK neither overseas) P(the rest of the UK and neither overseas) = P(neither overseas) 0.0 = = answer( A) {Watch for: as evidence of method (p <)} answer( C) See above for case for sight of all correct terms summed cao (condone 0.98 www) or for 0.99 x 0.99 cao Or for everything {..} cao for numerator of 0.0 or their answer to (i)(a) for denominator of or their answer to (i) (C) FT (0 < p < ) 0.0 at least 6

67 766 Mark Scheme June 008 (iii) (A) Probability = 0.79 = = 0.69 (accept awrt 0.69) see additional notes for alternative solution for 0.79 or for 0.79 dep CAO (B) 0.79 n > 0.9 EITHER: 0.79 n > 0.9 or 0.79 n < 0. (condone = and throughout) but not reverse inequality log 0. n >, so n > log Minimum n = 0 Accept n 0 for equation/inequality in n (accept either statement opposite) (indep) for process log a of using logs i.e. log b CAO OR (using trial and improvement): Trial with or = (< 0.9) or = 0.98 (> 0.) = 0.90 ( > 0.9 ) or = (< 0.) (indep) for sight of or 0.98 ( indep) for sight of 0.90 or dep on both M s cao Minimum n = 0 Accept n NOTE: n = 0 unsupported scores SC only TOTAL 6 6

68 766 Mark Scheme June 008 Q7 (i) Positive (ii) Number of people = 0 ( 000) + 8 (000 ) = 660 ( 000 ) + 90 (000) = first term (indep) second term cao NB answer of 90 scores MA0 (iii) (A) a = = 0 (B) Median = age of 8 (000 th ) person or 8. (000) Age 0, cf = 0 (000); age 0, cf = 80 (000) Estimate median = (0) Median =. years (...) If no working shown then. or better is needed to gain the. If. seen with no previous working allow SC for sum cao 0 or 0 thousand but not 000 for 8 (000) or 8. for attempt to k interpolate 0 70k (. or better suggests this) cao min dp (iv) Frequency densities: 6, 6, 77, 9,, 7 (accept. and 7. for and 7) for any one correct for all correct (soi by listing or from histogram) Note: all G marks below dep on attempt at frequency density, NOT frequency G Linear scales on both axes (no inequalities) G Heights FT their listed fds or all must be correct. Also widths. All blocks joined G Appropriate label for vertical scale eg Frequency density (thousands), frequency (thousands) per 0 years, thousands of people per 0 years. (allow key). OR f.d. 6