Version hij. General Certificate of Education. Mathematics MPC1 Pure Core 1. Mark Scheme examination - January series

Transcription

1 Version.0 00 hij General Certificate of Education Mathematics 660 MPC Pure Core Mark Scheme 00 examination - January series

2 Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner analyses a number of candidates scripts: alternative answers not alrea covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this Mark Scheme are available to download from the AQA Website: Copyright 00 AQA and its licensors. All rights reserved. COPYRIGHT AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance. The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 6447) and a registered charity (registered charity number 074). Registered address: AQA, Devas Street, Manchester 5 6EX Dr Michael Cresswell Director General

3 MPC - AQA GCE Mark Scheme 00 January series Key to mark scheme and abbreviations used in marking M m or dm A B E mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation or ft or F follow through from previous incorrect result MC mis-copy CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A, or (or 0) accuracy marks NOS not on scheme x EE deduct x marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

4 MPC - AQA GCE Mark Scheme 00 January series MPC p ( ) ( ) p NOT long division (a) ( ) = must attempt ( ) = = 0 x + is factor A shown = 0 plus statement Deleted: XMCA Q... [] (b) ( x )( x bx c) ( x x 4) ( x )( x 4)( x ) Full long division, comparing coefficients or by inspection either b= orc= 4 or A for either ( x 4) or ( x+ ) obtained A clearly found using factor theorem + + A CSO; must be seen as a product of factors NMS full marks for correct product Total 5 7 (a)(i) grad AB = = (must simplify 4/) A SC B for ( x+ )( x 4)( ) Δy Δx or ( x+ )( x+ )( ) or ( x+ )( x+ 4)( x ) NMS correct expression, possibly implied 7 9 (ii) grad BC = = + 4 Condone one slip NOT Pythagoras or cosine rule etc grad AB grad BC = ABC = 90 or AB & BC perpendicular A convincingly proved plus statement SC B for /(their grad AB) or statement that mm = for perpendicular lines if M0 scored (b)(i) M ( 0, 6 ) B B + B each coordinate correct (ii) ( AB = ) ( ) + ( 7 ) ( BC = ) ( + ) + ( 7 9) (iii) AB = + 4 or BC = 4 + or 0 found as a length AB = BC AB = BC or AB = 0 and BC = grad BM = 0 or /(grad AC) attempted A A either expression correct, simplified or unsimplified Must see either AB =.., or BC =..., ft their M coordinates BM has equation = A correct gradient of line of symmetry y= x+ 6 A CSO, any correct form Total 4

5 MPC - AQA GCE Mark Scheme 00 January series MPC (cont) (a)(i) one term correct 4t 4t 4 A another term correct dt 8 A all correct (no + c etc) unsimplified (ii) (b) (c)(i) d y t = 4 ft one term correct dt 8 A correct unsimplified (penalise inclusion of +c once only in question) t = ; = Substitute t = into their dt d t = 0 stationary value dt A CSO; shown = 0 plus statement d y d y t = ; = 6 4= Sub t = into their dt d t y has MINIMUM value A 4 CSO t = ; = Substitute t = into their dt d t = A OE; CSO NMS full marks if d y correct dt (ii) 0 dt > (depth is) INCREASING E allow decreasing if states that their d y 0 dt < Reason must be given not just the word increasing or decreasing Total (a) 50 = 5 ; 8 = ; 8 = or or or At least two of these correct (b) ( 7 )( 7 5) ( 7 + 5)( 7 5) 5 + A Answer = 4 A CSO any correct expression all in terms of or with denominator of 8, 4 or simplifying numerator eg numerator = m expanding numerator ( condone one error or omission) denominator = B (seen as denominator) Answer = 4 7 A 4 Total 7 OE 8 5

6 MPC - AQA GCE Mark Scheme 00 January series MPC (cont) 5(a) x 8x+ 5+ B Terms in x must be collected, PI their x ( 4 ) ( + k) ft ( x p) ( x 4) = + A for their quadratic ISW for stating p = 4 if correct expression seen (b)(i) y shape in any quadrant (generous) 7 O 4 x A correct with min at (4, ) stated or 4 and marked on axes condone within first quadrant only B crosses y-axis at (0, 7) stated or 7 marked on y-axis (ii) y = k y = constant y = A Condone y= 0x+ (c) Translation (not shift, move etc) E and no other transformation 4 One component correct with vector or ft either their p or q CSO; condone 4 across, up; or two A separate vectors etc Total 6

7 MPC - AQA GCE Mark Scheme 00 January series MPC (cont) 6(a)(i) terms correct 4x 9 6x A all correct (no + c etc) when x =, = m gradient = 5 A 4 CSO (ii) grad of normal y+ 6= their x 5 = 5 B ft their answer from (a)(i) ( ) or y = their x+ cand c evaluated 5 using x = and y = 6 ft grad of their normal using correct coordinates BUT must not be tangent condone omission of brackets x+ 5y+ 8 = 0 A CSO; condone all on one side in different order (b)(i) one term correct 9 4 x x x A another term correct 4 A all correct (ignore +c or limits) 8 8 F attempted = m ( ) = 4 A 5 CSO; withhold A if changed to +4 here (ii) Area Δ= 6 = 6 B condone 6 Shaded region area = 4 6 difference of ± ±Δ = 8 A CSO Total 5 7

8 MPC - AQA GCE Mark Scheme 00 January series MPC (cont) 7(a)(i) x = ± or y = ± 6 or ( x ) + ( y+ 6) (ii) ( ) C (, 6) A correct r = (RHS = ) their ( ) + their (6) 5 r = 5 A Not ± 5 or 5 (b) explaining why yc > r ;6 > 5 E Comparison of y C and r, eg = or indicated on diagram full convincing argument, but must have correct y C and r E Eg highest point is at y = scores E E: showing no real solutions when y = 0 +E stating centre or any point below x- axis (c)(i) ( PC ) ( 5 ) ( k 6) = + + ft their C coords = 9+ k + k+ 6 and attempt to multiply out PC = k + k + 45 A AG CSO (must see PC = at least once) (ii) PC > r PC > 5 k + k+ 0 > 0 k + k+ 45 > 5 B AG Condone k + k+ 0 > 0 (iii) ( k )( k 0) + + Correct factors or correct use of formula k=, k= 0 are critical values A Use of sketch or sign diagram: May score, A for correct critical values seen as part of incorrect final answer with or without working If previous A earned, sign diagram or sketch must be correct for, otherwise may be earned for an attempt at the sketch or sign diagram using their critical values. k>, k< 0 A 4 k, k 0 loses final A mark Condone k> OR k< 0 for full marks but not AND instead of OR Take final line as their answer Total TOTAL 75 Answer only of k>, k> 0 etc scores, A, M0 since the critical values are evident. Answer only of k>, k< 0 etc scores M0, M0 since the critical values are not both correct. Deleted: 8

9 Page 4: [] Deleted cway /05/009 09:0:00 XMCA (a) x = B Seeing OE (b)(i) 4 p(.5)= (.5) + (.5) 8(.5) 4(.5) Attempting to evaluate p(.5 or p(.5) p(.5) = = 0 ] so (x + ) is a factor of p(x) ] A CSO Need both the arithme to show = 0 and the conclusion. x 4x =0 x(x Dividing throughout by x OE 4) =0 x 4 = x x = + 4 x = + 4 (since x>0) A CSO x x (ii) x =. B AWRT. x =.46 B AWRT.46 x 4 =.49 B CAO Total 8 (a) 5 + x A B Either multiplication by = + denominator or cover up rule ( x)( + x) x + x attempted. 5 + x = A( + x) + B ( x) Substitute x = ; Substitute x = m Either use (any) two values o to find A and B or equate to solve A B= and A+B=5 A =, B = A (b)(i) ( x ) = + ( )( x) + px p 0 = + x + x... A (ii) x x ( )( ) x x + = + ( ) [ + ( ) + kx ]! A x Correct expn of x ( x)( + x) = ( x) + (+x) Using (a) with powers. P = ( x + x...) x x m Dep on prev Ms = x +.5x + AF 5 Ft only on wrong integer valu for A and B, ie simplified [Award equivalent marks for other valid methods.] Total 0 Page Break

10 XMCA (cont) (a)(i) Modulus graph A Correct shape including cusp at (π, 0). Ignore any part of graph beyond 0 x π. (ii) k = B (b) Two branch curve, genera shape correct. A Min at (α, ) Max at (β, with α roughly halfway between 0 and π, and β roughly halfway between π and π and curve asymptotic to x = 0, x = π and x = π. 4(a) Total 5 x x d y ( x + )e e () B = ( x + ) A (e x ) = e x Quotient rule OE (b) 0 0 6e e 5 Attempt to find / at x= When x = 0, = = AF 4 A 0, B Equation of tangent at A: 5 y = ( x 0) 4 A 4 ACF Total 7 Page Break XMCA (cont) 5 V = π cos( x ) cos( x ) 0 A Correct limits. (Condone kπ or missing π until the final mark)

11 Applying Simpson s rule to cos( x ) 0 x B PI Y=y (47) () (4) 0.540(0) [πy vals..45(9).54(5).049().6575(5).6974(0)] B PI 0.5 { Y (0) + Y () + 4[ Y (0.5) + Y (0.75)] + Y (0.5)} Use of Simpson s rule V = π So V =.846 (to 4 d.p.) A 6 CAO Total 6 Page Break (B for general shape in s quadrants; ln not required XMCA (cont) 6(a)(i) B,,0 B correct sketch-no part o curve in nd, rd or 4 th quadrants and ln quadrant, ignore other (ii) Range of f: f(x) ln A ln or >ln or f ln Allow y for f(x). (b)(i) y = f (x) f(y) = x ln(y + ) = x x y at any stage y + = e x m Use of ln m = N m = e N x e f (x) = A ACF-Accept y in place of f (x) (ii) Domain of f is: x ln BF ft on (a)(ii) for RHS (c) d (x + ) [(ln(x + ) ] = A /(x+) (d)(i) P, the pt of intersection of y = f(x) and y = f (x), must lie on the line y = x ; ; so P has coordinates (α, α). f(α) = α OE eg f (α) = α ln(α +) = α α + = e α A A.G. CSO

12 (ii) x [ f ( x) ] = e d BF x e Product of gradients = x + At P(α, α), the product of the gradients α e α + is = = α + α + B AG CSO Total 5 e = α e α = α + α Page Break XMCA (cont) 7(a) d y Product rule OE. = x e x + e x. A At stationary point(s) e x (x + ) = 0 m e x > 0 E OE eg accept e x 0 Only one value of x for st. pt. Curve has exactly one st pt A CSO with conclusion. Stationary point is (, e ) A 6 (b) Stationary point is (, k e ) BF Or E for y = x e x to y = x e x + k is a vertical translation of k units. St. pt is on x-axis, so k = e. B Total 8 8 y y d cos x = 6 + sin x Separating variables with intention to then integrate. ln y = ln (6 + sin x) (+c) A A A for each side. Condone missing +c For correct alternatives to the stretch ln = ln 6 + c m Substituting x = 0, y = to find c ln y = ln (6 + sin x) + ln ln 6 so y = ( 6 + sin x) A 5 Correct simplified form not involving logs Total 5 9(a) y = e x e x 6e x. Reflection; in the y-axis ;A Stretch, Stretch with either (I) or (II). (I) parallel to y-axis, (II) scale factor A 4 6. after writing y = e x+ln6 award B for translation in x-dirn. and B for the correct vector (OE) noting order of transformations. (b)(i) Area of rectangle/shaded region below B x-axis = k

13 Area of shaded region above x-axis k x = 6 e 0 k e x = e k ( ) = [ ] 0 B A F(k) F(0) following an integration. ACF Total area of shaded region = k e k + = 4 k e k = 0 (k )e k = 0 A 6 AG CSO. (ii) Let f(k) = (k )e k f(0.6) = 0.8e. = 0.(4..)<0 f(0.7) =.e.4 =.(46..)>0 Both f(0.6) and f(0.7) [or better] attempted Since change of sign (and f continuous), 0.6 < k < 0.7 A AG Note: Must see the explicit reference to 0.6 and 0.7 otherwise A0 Total Page Break XMCA (cont) 0(a) 5 (b) (c)(i) AB = 0 = Line AB: r = 0 + λ 0 4 = = 9 4 A for ± ( OB OA ) OE for BA BF OE Ft on AB ± AB direction vector of l evaluated = 6 ; BF Either; Ft on either of c s vectors + + = cosθ = 9 Use of a b cos θ = a b cos θ = 9 9 = = cos θ = = 9 A 4 AG CSO

14 (ii) Award equivalent marks for alternative + p 5 p BP = = p p A p 4 p 4 Condone one slip ± BP direction vector of l = 0. BP = 0; 6p = 9 p =.5 A Condone one slip P (.5,,.5) is mid point of BC A 5 x C + 5 y + =.5 C z + 4 = C =.5 C (, 5, ) A Condone written as a column vector. valid methods. Total 4 Page Break XMCA (cont) (a) sin(x + x) = sinx cosx + cosx sinx = [sinxcosx]cosx + [ sin x]sinx B;B B for each [ ]. Accept alternative correct forms for cosx = sinx( sin x) + ( sin x)sinx m All in terms of sin x = sin x sin x sin x sin x sinx = sinx 4sin x. A 5 CSO (b) sinx = cosx (sinx 4sin x) = cosx Using (a) (sinx 4sin x) = ( sin x) A Equation in sin x sinx ( sinx 4sin x) = 0 [sinx = 0] ( 4sinx)( + sin x) = 0 m Factorising/solving quadratic in sin (a)(i) (ii) sin x = 0 ; x = 80º B sin x = 0.75 ; x = 48.6º,.4º A Ignore solns outside 0º<x<60º throughout sin x = ; x = 70º A 7 Total d v Attempt to use parts formula in the u = x and = sec x correct direction d u = and v = tan x A PI.. = x tan x tan x A = x tan x ln (sec x) (+ c) A 4 OE CSO (Condone absence of +c) x tan x = x(sec x ) Use of identity + tan x = sec x

15 (b) = [x tan x ln (sec x)] x (+ c) AF [ ] ft on (a)(i) x = sinθ, = cosθ dθ = f(θ ) dθ OE m 4 x = 4( sin θ ) cosθ dθ A Eliminating all x s = 4cos m θ dθ = (cos θ + ) dθ Use of cosθ to integrate cos θ. = sin θ + θ (+ c) AF Ft a slip = sinθ sin θ + θ (+ c) = x x x + sin (+ c) 4 A 6 ACF (accept unsimplified) Total Page Break XMCA (cont) x = t + t y = 8 t = + t = 6t Both attempted and at least one dt dt correct. 6t = + t A At P( 4, 5), t = B At P( 4, 5), 6 = = + Gradient of normal at P is Eqn of normal at P: y 5 = ( x + 4) A ACF y + x = Normal cuts curve C when 8 t + t + t = t t + t A ( t + )( t 4t + 7) = 0 (*) m ( 4t + 7) = 0 t has no real solutions since ( 4) < 4()(7). t = is only real solution of (*) so normal only cuts C at P, where t = ie the normal does not cut C again. E Total Chain rule.