ms-bank-0606

2002 – 2011

Compiled & Edited By

Dr. Eltayeb Abdul Rhman

www.drtayeb.tk

First Edition

2011

MS BANK

ADDITONAL MATHEMATICS

I.

2.

3.

y+2x 7 T=xY·l ........ 3y'-7y+2=0 Solution

Complete elimination ofxory. or 6x'-35x+SO=O

- (JL~l and (2Y,,2)

Uses (0,3) to fmd C

(i) A= (2+3V2)(j-2..J2) ,. -2+t!V2

(i!) D1 =(Hl2V2+IS)+(25-20V2+8)

= 55 - s..J2

(A"' IO+S,il8 - 8-'---J2 -12) still needs I" tv,.· a steps

Ml

Al DMI Al

4

Either x or y must be completely removed

Any multiple of this- needn't be =0. Correct method of solution- see end. All correct. (not 0.34)

Bl Bl Co.

Ml Needs to bring in Lhe constant in an AI integrated expression.

4 All correct

M I M I Anywhere. 4: .J2 could be 8 :...J2 if multiplying first.

Al

Ml Al

5

Com:ct only.

Reasonable squaring v.iili Pythagoras. Correct onlv.

Decimal work gets no credit anyv.•here. Possible to get 4 marks on (ii) alone.

B l Correct only. nb i,j throughout is ok.

M t A I Complete method for M. A mark co.

[oc Ml

A2J Ml

Al' j

A I for 50Q-40P

co.

Completely correct method. Follow through directly on his OR.

,,,r __ _ (i) Graph ofy=Y, cos2x B2. I BO unless 2 full cvcles_ Starts and fmishes at max_ ±Y, shown somewhere_

I~~---

-'1 _, ~

\''~h J l(

_y_ . --- ;,L.

Graph ofy=Y. + sinx 82,1

(ii) Equate the y's Ml V1 coo2x='!. + sinx.

-+ 2cos2x=i+4sinx ....... k=4 Al

6

80 unless I full cycle_ Starts and ends

above the origin. I y, and-'!. shown.

Independent of graphs_ For stating only. Collld find ~H-alue from accurate graph and substitute. (not from sketch)

Correct only_

Mark Scheme Syllabus Pilper IGCSE Examinations June 2002 0606 1

6. (a) Number 5 " 5! or i of6! 600 MIA I Correct method.

(b) Total=9C4=(126) ,\11 No need for 126. Needs one step past 9C4 Total with no women= 5C4 = 5 81 Needs 5_ -. Number of ways= 126-S"' 121 MIA! For ·total- 0 women ·. Ind of lstM.

O<

(or Ways with IW 3M= 4Cl><SC3 =4" 10 = 40 Ml For \W,3M or 2W 2M or3W,1M Ways with 2W 2M=4C2><5C2 = 6" 10 = 60 Needs one product of 2 nCr's. Ways with JW 1M =4C3xSCl =4" 5 = 20 81 For 1.. Ways with 4W =I. =>Toaal = 121. ) MIA! Adding 4(or 3) events. CoiTeet only.

6

7.(i) (2~x")S - 25 + Sx2"(~x2) + I 0><2'><(~x2)' etc Powers of2 and (±x2) more or less com:ct. Ml Corro:t use of powers- even if no (-)s. Binomial coefficients used com::ctly. Ml Com::ct use of binomial coeffs. - 32- 8Qxl + 8Qx4- 4Qx6 +JQx8 - XIO AI All correct.

(ii) (I+x')'"' 1 +2x2 +x4 Bl Independent of anything else.

Attempt to multiply and pick out 3 renns Ml Reasonable attempt with 3 terms..

AI Corro:t oaly. --+(-40+ I60-80)x6 => 40 6

8. OlS~ 6SRY Y,*(l-qx) 81 Correct only unsimplif1ed A=~><qx><(l-x) 81 CoiTect only - unsimplified

!., ').. A=l-'llx-ASRY-t.XYQ Ml For ''square- 3 triangles" attempted

P •~ =Y1(I-x+qx') No A mark since answer given.

(ii) dAidx = Y.(-1 + 2qx) Ml Attempt at differentiation_ = 0 when qx= Y, ie QY =YR DMI Putting his differential to 0_

AI Beware fortuitous answers- ans given_

Minimum A= V. ( 1- Il(2q) + q!(4q"} 81 1 1 CoiTect only- any unsimplificd form ok.

=--- 7 2 8q

9. (i) d/dx(-J(2x+5) y, " (2x+sr"' " 2 Ml Must have ''x2" iefnofafu_

dy/dx = -J(2x+S) + (x-5) >< "above ans" Ml AI Must use product rule coiTectly- M mark - k=3 AI is independent of first M mark.

(ii) Oy, [dy/dx],.10 "Ox = ± 6p M!At..J Needs numencal dy/dx - 1\x=±p. not 10-p for the M mark.

(iii) dy/dt = dy/dx >< dxldt Ml Use of chain rule- must be for 3-+-dy/d,_x_

=> 3=6dx!dt dx/dt = 0.5 unit/s AI CoiTect only_ Ignore units_ 8

10. (i)

[~0 0 400 500 ~OJ (so 75 IOO)x 300 0 0 300 600 MIA! Could also be (5x3) x (3x I). Must be

400 600 600 0 400 compatible but needn't be in COITCCt order.

13

[ 400 0 400 500

600 l 7 [16500]

MIA! Could also be (I x5) x (5 x3) ie

300 0 0 300 6QQ X 10 "' 10200 (16500 10200 18600)_ Order needed forM.

600 0 18600 81 Allow 81 for arithmetic or wrong order ..

400 600 400 5

' [ '"] {"'00]

Could be (!6500 10200 I86oo)x 3_00 (iii) ((2.10 3.00 3.75 10200 = SIJS 000 MIA!-./

3.75 !8600 I"' 8 B mark anyhow_

II. 2x'-lh±5 2 (' 2)' . 3 3• 61 Bla2Blfor 2 in bracket. 81 for-)_

(i) Domain O::Ox...'.':S Range off1s -3 to J5_ BIBIV 81 for \5_ Bl'l' for bottom limit of"c".

(ii) fisnotltol. Bl' Correct explanatwn for h1s ..... alues.

(iii) k=x-valuecorr torning="-b"-+k=2 Bl' k =''his -b··

(iv) Put )=2(x-2)'- 3 and make x the subject. Ml Knows what to do

'ff Ml Reasonable order of operations Replace xbyy-g- = -

2-+2. AI Co

(or reverse order ie +3 . .,.z, 'i, +2) 10 12. EITHER (a) 27=ax2.25" or lg2T=lga+nlg2.l5

64=ax4" or lg 64 =lga +nlg4 First M I is for completely eliminating a _.Elimination of a or lga or n MI or lga or Ina or n. --->Solve for n (or a) MI The sc.;ond M 1 is for solving rhe resulting

n=\.50 eqn- needs to be powers or logs. a=8.00 AI Both needed.

->Substitute back for p = 8(6.25)' = 125 MIA! Substitution into eqn or log eqn. Co.

(b) Plots In m against t • *See p4 for graph

t 10 20 30 40 50 m 40_2 27_0 18_0 12_2 B.I MIA\ Must be In graph- not lg graph.

l"m 3_69 3_30 2_89 2.50 2.09

Gradient "' -k ~ k"" 0_04 (0 038-0_042) Bl Bl Intercept -== In m.:. ~ m.,,.60 ( 57-63) BIB\

II 120R. .~ (i) (BC) y-11 Y,(x-4) MIA\ M needs gradient + eqn. (2}=x+l8)

"' /, Gradient of CD is -2 Ml Foruseofm1xm2=-l IndoflstM_

' ' \ y-I0=-2(x-17) AI' -.J on his value of perpendicular (y+ 2x==44) (CD)

l=d Solution of sim eqns DMI Both Ms needed_ . --+C(l4,16) AI Correct only

' . (ii) Line rotio BC=o/o AD f'l {i,l => BE"'Y. AE = J7',AB Ml Completely corre<:t method that leads toE {i,ll ~ E (9,26) AI Correct only.

D ,, or{AB) y=3x-l (CD) y--2x+44 => E (9,26) (MlAlJ (completely correct method)

(iii) Ratio of small tl to large similar 6 = 5 : 8 Ml '( steps - or lengths- or y-steps

Ratio of areas of small 6 to large 8 = 25 : 64 Ml Squarmg process

Ratio of 8EBC: trapezium= 25: (64-25) Correct only ::::) 25 : 39 AI

(or area oflarge 8"' Y,x-.J320x-.J320 = 160 (MI Pythagoras for '·length+ Y,bh"'

area ofsmall8"' Y,x-.JJ25x-.JJ25 "'62.5 Ml Or matrix method correctly used.

Ratio:62.5 :97.5= 125: 195=25: 39 AI) Subtractzon. Correct only. Any form ok,

II including 62.5: 97.5.

OM\ for quadratic equation. (a) Formula must be used correctly on "eqn-=0". (b) Factors must be on oon"'O. Coeffs ofx' ,X0 ok

' 1,

,.,

o-.;

Mark Scheme IGCSE Examinations June 2002

( \,) .

Syllabus Paper 0606 1

m:-k :: o.o,O

Q ll'lo -: l..·o1

,.,,; S"~·b.

~------------------~-----~~ \\) ~ '1:.0 ~0

1 ("l ····r· - ,, l 81 81 5 J •

-4 2S ,. A-lA-~ "' (: ']• ( 5 -

5') "'1-01 => k= to M1 A1

5 -·

2 ("l . 'f Ill """"" y "'txl+i Bt

~Vi y = 12x-3.1 82.1~

(lljZ 81 ~_ ... .,

3 [5J ,,, HnP "' (ii) Pcll "' PnM=P "' Puii=M at

(iii) SludenLs sludying MatRemalicsGAI'f 81

(tv) &OOellls sludyioQ' HiSkiPf Of Ma&taematics Of bo&h tM AOt PhysiCs B2,t,O-

4 [6] ...... f (-2}= 0 => •=--.2 MtAt

Divide by ,., => :i-61'+1- M1Af

SOlve x: 6±.JJ6 -4 2

:3±2.5 DM1A1

5 (6] Combine (121lm + 24(1)1- 4(251Jr + 160D • 2IJUJ -40UJ .M1 A1

" y.: (i20Di + 240D - czsor + 16UD • 5lli -10UJ

Square, add and square-roof components Speed"' 112 f.f1 A1

Tan -t(ratio of components} Bearing"' 3.33(.4)~ M1A1

6[7! (iL~.( cos x } _, {1- sin x X· sin x ~-cos x(- cos x} dx\1-sinx (1-Binx)'

82,1,0

Use Pythagoras on numerator k=1 M1A1

fiij -../2 J2 CO& X M1 J dx ~ 1--...x t-sin.x

[ ~·· ... both"""'"''"' => 2 DM1 A1

7[7! lx 10 (i) LAOB=tan-11016=1.03 81

~ (iii Arc AB = a :.. 1.03 ~·,/to'-6' -~ Mt

Peri11'1Ser = 6.18 + 5.66 + 10"' 21.8 n1 AI

(iii) SeQar AOB=1/2x 82 x 1.113 Area XAB=1/2x 10x 6- SeQor AOB = 11-.S '" M1A1

1[7] A (I) A8=6skHI AS= Sltan o 81 61

D~ (II) 6sin6 = Sllan6"' ScosOtslno 811'

5 B ScosO= 69in:z-9 = 6{1-<:052 0) M1

Solve or factorise oos8"'% (or-1.5) 6= 48.2 M1 A1 A1

0[7] (a) Eliminate x or y "" ?=4{y-k)+6 "' (x+k)2=<4JC+8 M1

"" l-.4y+ (41c-8)= 0 Of x2+{2k-4)x+ ~-8)=0

use discriminant "" 16=4(4k-8} " (2k-4f = 4(.1(! -8) M1 A1 ......... "" lc= 3 A1

(b) (X- 2){11: + 4) 81

<~-2)(.x+ 4)> 0 :4~+2x>8 "" a=2,.b:8 UtA1

10[7] (1) lg 2x -lg (x - 3) .. lg {2J:I(x - l}} 1 = lg 10 111 81

Solw 2Jrl(x - 3) "" 1 0 "" x=3.75 A1

(ii) ~:: 1Aogy3 (= Ut) "' log,.3"'- 1!1ogl)' (• "" '"

Substilul&

"" u/-4u.+4=0 "' 4-u:!z -4uz + 1 =- 0 ttl

""""' 11!=2 or":!=% "" y=& DM1 AI

11[9] (I) t·, X+7 12 & make y the subject 81 M1 >H-- X•--. 3 y • 2

9' · I 14 .:!_! + 2 , not defined fOrK = 0 A1 X

(il) 38 !g(x)•--7•X x-2

., fg(x) =X ~ g{x) = f "1(x) 12 X+ 7 o>--·--x-2 3

M1

Simplify & solve Ji-+ Sx-50=0 :::::. x = -10 or 5 DM1 A1

(iii) y f.,

Sketch off 81 (0,""

(-7,0)/

Sketch off-1 81

(0, 2%) X All intersedions

/ with ID(es shown 081

I (0, -7)

Mark Scheme Syllabus Paper IGCSE Examinations June 2002 0606 2

12(11) '\p d • = 2x. 6 = 0 :=o x = 3 :=o P i9 (3,1) M1 A1 -(x·- ex+ 10) ,, ......

Equl!llon or PO is y -1 = -2(x -3) M1

.i'-4x+3=0 or i-By+5=0 ' Eliminate y => >=> M1

..... => Qis{1,5) M1 Af •

.A..-4of rectangle wtth OQ as diagonal = 1•5

J<x'-Bx+10)clx = ~x 3 -3x 2 +f0r Mf Af

E. ... t ..... t~~, [ Y" Dnl ' ' Area required = red.angle + [ ] ~ = 5 + ((12)- (7%)} ,. "' Mf A1

o. (i) A!B f; 5 v= 1531225= 15 81

AtC vo= 0 t= T •20 B1

(il) 8=~ .. _!_(20 ·t) 2 x(-1) M1 Af d.t 225

[a]t.t4 = • 3 X 36

225 - -0.48 A1

(iii) v st111ight Une BU

15 V\ ;f..---.;T

Curve Bf.t

0 --..- -= t

(W) AB = Y.t (5" 15) = 37Y.t 81~

J~20 -t) 3 dt = -1-(20 ·tl4 {·1) M1 Af

225 . 900 .

AC .. 37%+[]~ = 37%+{0-(-225)}"' 93% A1 4

Mark Scheme Syllabus Paper

IGCSE EXAMINATIONS – JUNE 2003 0606 1

© University of Cambridge Local Examinations Syndicate 2003

1. x or y eliminated completelyUses the discriminant b2-4ac on aquadratic set to 0

Arrives at k = 0 from 32k = 0Correct answer k�0.

M1M1

A1A1

[4]

Allow as soon as x or y eliminated.Condone poor algebra – quadraticmust be set to 0 – b2-4ac = 0, <0, >0all ok.For k and 0.For k�0.

2. Length = (1 + √ 6)� (√ 2 +√ 3)Multiplying top and bottom by�(√ 3 -√ 2)�√ 3 + √ 18 - √ 2 - √ 12Reduces √ 18 to 3√ 2 or √ 12 to 2√ 3

�2√ 2 - √ 3

�√ 8 - √ 3

M1

M1

DM1

A1[4]

Multiply both top and bottom by�(√ 3 -√ 2).

Allow wherever this comes – notDM.Dependent on first M – collects √ 2and √ 3.Co.

3. (i) 32 – 80x + 80x2

(ii) (k + x) � (i)Coeff. of x is –80k + 32Equated with –8�k = ½ or 0.5

B1 x 3

M1A1√

[5]

Allow 25 for 32 (if whole series isgiven, mark the 3 terms).

Must be 2 terms considered.For solution of k = (-8 - a)� (b)

4. Liner travels 54km or relative speedof lifeboat is 60km/h.

Correct vel./distance triangle

Use of cosine rule in triangleV2 = 602 + 362 – 2.60.36cos45 ord2 = 902 + 542 – 2.90.54cos45.

V = 42.9 or d = 64.4�V = 42.9

B1

B1

M1

A1

A1[5]

Anywhere.

Triangle must be correct with 54,45o, 90 or 36, 45o, 60 or even 36,45o, 90.Allow for other angles.

Unsimplified and allow for 135o aswell as 45o.Co.

www.xtremepapers.net





5. Elimination of x or y.�4x2 + 6x – 4 = 0 ory2 – 12y + 11 = 0Solution of quadratic = 0.

� (0.5, 11) and (-2, 1)

Length = √(2.52 + 102) = 10.3

M1A1

DM1

A1

M1A1[6]

x or y eliminated completely.Correct equation – not necessarily =0Usual method for solving quadratic =0

All correct. Condone incorrectpairing if answers originally correct.Must be correct formula correctlyapplied.

6. A2 = ��

��

� �

10

32

��

��

� �

10

32

��

��

� �

10

94

A-1 = ½ � ��

��

�

20

31

B = A2 - 4A-1 = ��

��

�

�

�

30

152

M1A1

B1B1

M1A1[6]

Do not allow M mark if all elementsare squared. If correct, allow bothmarks. If incorrect, some working isneeded to give M mark.

B1 for ½, B1 for matrix.

M mark is independent of first M.Allow M mark for 4A-1 - A2.

7. f(x) = 4 – cos2x

(i) amplitude = �1. Period = 180o or�

(ii)

Max (90o, 5) and (270o, 5)

B1B1

B2,1

B1B1[6]

Independent of graph. Do not allow“4 to 5”.

Must be two complete cycles. 0/2 ifnot. Needs 3 to 5 marked or implied.Needs to start and finish atminimum. Needs curve not lines.

Independent of graph (90, 270 getsB1). Allow radians or degrees.






8.

(i) O, P, S correct

(ii) 34, 35, 36, 37 correct

O�S = odd squares� 4O�S = odd and even squares

� 49 + 5 = 54

B2,1

B2,1

B1

M1A1[7]

Give B1 if only one is correct.

These 2 B marks can only beawarded only if B2 has been givenfor part (i).

Co.

Any correct method. Co.

9. (i) log42 = ½ log864 = 2�2x + 5 = 91.5 �x = 11

(ii) Quadratic in 3y

Solution of quadratic = 0

�3y = 5 or –10

Solution of 3y = k

y = 1.46 or 1.47

B1B1M1A1

M1

DM1

M1

A1[8]

Anywhere.Forming equation and correctlyeliminating “log”. Co.

Recognising that the equation isquadratic.Correct method of solving theequation = 0.

Not dependent on first M1. Correctmethod.Co. (not for log5� log3). Ignore ansfrom 3y = -10.






10.

(i) Plots xy against x2 or x2 againstxy to get a line

c = 12 to 12.5 or -7.25 to -7.75m = 1.55 to 1.65 or 0.62 to 0.63xy = 1.6x2 + 12

or x2 = 0.625xy - 7.5� y = 1.6x + 12/x

(ii) Reads off at xy = 45 � x = 4.5 to 4.6

M1A2,1

B1B1

M1A1

M1A1[9]

Knows what to do.Points accurate – single line withruler

Allow if y = mx + c used.

Allow if y = mx + c used.Must be xy = mx2 + c orx2 = mxy + c.

Algebra is also ok as long as xy = 45is solved with an equation given M1above.

11. y = xe2x

(i) d/dx(e2x) = 2e2x

dy/dx = e2x + x.2 e2x

sets to 0�x = -0.5

(ii) d2y/dx2 = 2 e2x + [2 e2x + 4x e2x] = 4 e2x(1 + x) �k = 4

(iii) when x = -0.5, d2y/dx2 is +ve (0.74) � Minimum

B1

M1M1A1

M1A1A1

M1A1[9]

Anywhere – even if dy/dx = 2x e2x

or 2 e2x.Use of correct product rule.Not DM mark. Allow for stating hisdy/dx = 0.

Use of product rule needed.Allow if he reaches 4e2x(1 + x).

No need for figures but needscorrect x and correct d2y/dx2.

12. EITHER

At A, y = 4dy/dx = 2cosx - 4sinxdy/dx = 0 when tanx = ½

At B, x = 0.464 or 26.6o

B1M1A1M1A1

A1

Anywhere.Any attempt at differentiation.Sets to 0 and recognises need fortangent.Co. Accept radians or degrees here.

x 2 3 4 5 6 y 9.2 8.8 9.4 10.4 11.6 xy 18.4 26.4 37.6 52.0 69.6 x2 4 9 16 25 36






� (2sinx + 4cosx)dx = -2cosx + 4sinx

Area under curve = [ ]0.464 – [ ]0� -(-2) = 2.

Reqd area = 2 - (4 � 0.464) = 0.144(5 or 6).

M1A1

DM1

M1A1[11]

Any attempt with trig. functions.

x-limits used correctly. If “0” ignoredor automatically set to 0, give DM0.

Plan mark – must be radians for bothM and A.

12. OR

dy/dx = ½(1 + 4x)-½ � 4At P, m = 2/3

Eqn of tangent y - 3 = 2/3(x - 2)At B, x = 12/3

� √ (1 + 4x)dx = (1 + 4x)1.5 � 2/3 � 4

Area under curve = [ ]2 – [ ]0 = 41/3

Shaded area =Area of trapezium - 41/3 = 1/3

Or Area undery = 2/3x + 12/3 - 41/3 = 1/3

[or � xdy = � (¼y2 - ¼)dy= y3/12 - y/4

area to left of curve = [ ]3 – [ ]1 = 12/3shaded area =

12/3 – triangle (½.2.11/3)= 1/3]

M1A1

M1A1

M1A1A1

DM1A1

M1

A1

[M1A1A1

DM1A1

M1A1][11]

Any attempt with dy/dx – not for√ (1 + 4x) = 1 + 2√x. A mark needseverything.Not for normal. Not for “y + y1” or form on wrong side. Allow A forunsimplified.

Any attempt at integration with(1 + 4x) to a power. Other fn of xincluded, M1 only.Use of limits 0 to 2 only. Mustattempt a value at 0.

Plan mark independent of M marks.

A1 co.

Attempt at differentiation. A1 foreach term.

Must be limits 1 to 3 used correctly.

Plan mark independent of other Ms.

DM1 for quadratic equation. Equation must be set to 0.Formula - must be correctly used. Allow arithmetical errors such as errors over squaringa negative number.Factors – must be an attempt at two brackets. Each bracket must then be equated to 0and solved.Completing the square – must result in (x�k)2 = p. Allow if only one root considered.






1

[7]

Put x = -b/2 (or synthetic or long division to remainder)�3b3 + 7b2 - 4 = 0 AG

Search �b = -1 [or b = -2] (1st root or factor)

Attempt to divide�3b2 + 4b - 4 (or 3b2 + b - 2) or further search�b = -2 [or b = -1]

Factorise (or formula) [3 term quadratic] or method for 3rd value�b = -2, -1 or 2/3

M1 A1

M1 A1

M1

DM1 A1

2 (i)

(ii)

[6]

AB = OB - OA = �(9i + 12j)

Unit vector = AB 22129 �� = �(0.6i + 0.8j) [Accept any equivalent

unsimplified version of column vectors, � ��

��

�

12

9, � ��

�

��

�

8.0

6.0]

AC = 2/3 AB = 6i + 8j (or CB = 1/3 AB = 3i + 4j)

OC = OA + AC (or OB - CB ) = 12i + 5j (or equivalent)

M1

M1 A1

M1

M1 A1

3

[6]

)23( 5.05.0 �� xx dx = 3x1.5/1.5 + 2x0.5/0.5(one power correct sufficient for M mark)

��8

1

(2 x 8√ 8 + 4√ 8) – (2 + 4) Must be an attempt at integration

Putting √ 8 = 2√ 2 (i.e. one term converted √ to k 2 )� -6 + 40√ 2

M1 A1 A1

M1

B1√ A1

4

[4]

16x+1 = 24x+4 or 16 x 24x or 16 x 42x or 16 x 16x

20 (42x) = 20(24x) or 5(24x+2) or 20 x 16x

2x-3 8x+2 = 2x-3 23x+6 = 24x+3 or 8 x 24x or 8 x 42x or 8 x 16x

Cancel 24x+2 or 24x and simplify �4.5 or equivalent

B1 B1

B1

B1

5 (i)

(ii)

(iii)

[7]

f(0) = ½ f2(0) = f(½) = (√ e + 1)/4 � 0.662 (accept 0.66 or better)

x = (ey + 1)/4 � ey = 4x - 1 � f-1 : x � In(4x - 1)

Domain of f-1 is x�½ Range of f-1 is f-1�0

B1 M1 A1

M1 A1

B1 B1






6 (i)

(ii)

(iii)

[7]

x2 – 8x + 12 = 0 Factorise or formula �Critical values x = 2, 6x2 – 8x + 12 > 0 � � �2: �xx � � �6: �xx

x2 – 8x = 0 � Must be an attempt to find 2 solutionsx2 – 8x < 0 � � �80: �� xx

Solution set of │x2 – 8x + 6│< 6 is combination of (i) and (ii)� �20: �� xx � �86: �� xx

M1 A1A1

M1A1

B1 B1(one foreachrange)

7 (i)

(ii)

(iii)

(iv)

(v)

[8]

6! = 720

M … � 5! = 120

4! 48

6!/4! 2! = 15 Accept 6C4 or 6C2 = 15

5!/3! 2! = 10 (or, answer to (iv) less ways M can be omitted)(Listing – ignoring repeats � 8 [M1] � 10 [A1])

B1

M1 A1

M1 A1

B1

M1 A1

8 (i)

(ii)

[8]

Collect sin x and cos x � sin x = 5 cos xDivide by cos x � tan x = 5 (accept 1/5 – for M only)x = 78.7o or (258.7o) i.e. 1st solution + 180o

Replace cos2 y by 1 – sin2 y3sin2 y + 4sin y - 4 = 0 Factorise (or formula) (3 term quadratic) � sin y = 2/3 (or -2)

y = 0.730 (accept 0.73 or better) or (2.41) i.e. � (or 7

22) less 1st solution

M1M1A1 A1√

B1

M1

A1 A1√

9 (i)

(ii)

[8]

� � )12( 2tt dt = 6t2 – 1/3t3

From t = 0 to t = 6 distance = �6

0

= 144

Max. speed = 36 � from t = 6 to t = 12 distance = 36 x 6 (= 216)

During deceleration distance = (02 – 362) � 2(- 4) = 162Area of � is fine for M mark but value of t must be from constant

acceleration not 12 – 2t = �4

Total distance = 144 + 216 + 162 = 522

v

t

M1 A1

A1

B1

M1

A1

B2, 1, 0






10 (i)

(ii)

(iii)

[9]

dx

dy =

2)2(

1)42(2)2(

�

��

x

xx

= 2)2(

8

�

�

x

� k = -8

Must be correct formula for M mark (accept 2)2(

8

�

�

x

as answer)

When y = 0, x = -2 (B mark is for one solution only) NB. x = 0, y = -2

mtangent = -8/16 = -1/2 � mnormal = +2(M is for use of m1 m2 = -1, whether numeric or algebraic)

Equation of normal is y - 0 = 2(x + 2)(candidate’s mnormal and [x]y=0 for M mark)

When y = 6, x = 4

��

dt

dx

dx

dy

dt

dy�

�

�

2)2(

8

x

0.05 = �

�

4

80.05 = -0.1 (accept �)

i.e. 4�

��

��

�

xdx

dy x 0.05 for M mark.

√ is for error in k only. (Condone S � dx

dyx S)

M1 A1

B1

M1

M1 A1

B1

M1 A1√

11 EITHER y D (13½, 11)

B

A C (7, 4) (3, 2) O x

E

(i) mAC = (4 - 2)/(7 - 3) = ½

mBD = ½

mBC = -2

Equation of BD is y - 11 = ½(x - 13.5) i.e. 4y = 2x + 17

Equation of BC is y - 4 = -2(x - 7) i.e. y = -2x + 18

Solving y = 7, x = 5.5

B1

B1√

B1√

M1

M1

M1 A1






[10]

(ii) EAC

EBD

�

� = (ratio of corresponding sides or x- or y- steps)2 = 4/1

Quadrilateral ABDC/�EBD = 3/4

[Or, find E(1/2, -3) and then use array method to find one of:

area quadrilateral ABDC = 22.5 area �EBD = 30Find other area and hence ratio = 3/4 or equivalent]

M1 A1

A1

M1 A1A1

11

[10]

OR

B

6 7

P 5 Q

(i) (r + 6)2 + 52 = (r + 7)2

Solve � r = 6

tan AOB = 5/12 AOB = 0.395 or 22.6o

Length of arc AB = 6 x 0.395 = 2.37 or better

(ii) Sector AOB = ½ x 62 x 0.395 = 7.11

Shaded area = ½ x 5 x 12 - 7.11

All figures in sector and triangle correct √

22.9 or better

M1

M1 A1

M1

M1 A1

M1

M1

A1√

A1

O

r r

A






1. x or y eliminated completelyUses the discriminant b2-4ac on aquadratic set to 0

Arrives at k = 0 from 32k = 0Correct answer k�0.

M1M1

A1A1

[4]

Allow as soon as x or y eliminated.Condone poor algebra – quadraticmust be set to 0 – b2-4ac = 0, <0, >0all ok.For k and 0.For k�0.

2. Length = (1 + √ 6)� (√ 2 +√ 3)Multiplying top and bottom by�(√ 3 -√ 2)�√ 3 + √ 18 - √ 2 - √ 12Reduces √ 18 to 3√ 2 or √ 12 to 2√ 3

�2√ 2 - √ 3

�√ 8 - √ 3

M1

M1

DM1

A1[4]

Multiply both top and bottom by�(√ 3 -√ 2).

Allow wherever this comes – notDM.Dependent on first M – collects √ 2and √ 3.Co.

3. (i) 32 – 80x + 80x2

(ii) (k + x) � (i)Coeff. of x is –80k + 32Equated with –8�k = ½ or 0.5

B1 x 3

M1A1√

[5]

Allow 25 for 32 (if whole series isgiven, mark the 3 terms).

Must be 2 terms considered.For solution of k = (-8 - a)� (b)

4. Liner travels 54km or relative speedof lifeboat is 60km/h.

Correct vel./distance triangle

Use of cosine rule in triangleV2 = 602 + 362 – 2.60.36cos45 ord2 = 902 + 542 – 2.90.54cos45.

V = 42.9 or d = 64.4�V = 42.9

B1

B1

M1

A1

A1[5]

Anywhere.

Triangle must be correct with 54,45o, 90 or 36, 45o, 60 or even 36,45o, 90.Allow for other angles.

Unsimplified and allow for 135o aswell as 45o.Co.




5. Elimination of x or y.�4x2 + 6x – 4 = 0 ory2 – 12y + 11 = 0Solution of quadratic = 0.

� (0.5, 11) and (-2, 1)

Length = √(2.52 + 102) = 10.3

M1A1

DM1

A1

M1A1[6]

x or y eliminated completely.Correct equation – not necessarily =0Usual method for solving quadratic =0

All correct. Condone incorrectpairing if answers originally correct.Must be correct formula correctlyapplied.

6. A2 = ��

��

� �

10

32

��

��

� �

10

32

��

��

� �

10

94

A-1 = ½ � ��

��

�

20

31

B = A2 - 4A-1 = ��

��

�

�

�

30

152

M1A1

B1B1

M1A1[6]

Do not allow M mark if all elementsare squared. If correct, allow bothmarks. If incorrect, some working isneeded to give M mark.

B1 for ½, B1 for matrix.

M mark is independent of first M.Allow M mark for 4A-1 - A2.

7. f(x) = 4 – cos2x

(i) amplitude = �1. Period = 180o or�

(ii)

Max (90o, 5) and (270o, 5)

B1B1

B2,1

B1B1[6]

Independent of graph. Do not allow“4 to 5”.

Must be two complete cycles. 0/2 ifnot. Needs 3 to 5 marked or implied.Needs to start and finish atminimum. Needs curve not lines.

Independent of graph (90, 270 getsB1). Allow radians or degrees.




8.

(i) O, P, S correct

(ii) 34, 35, 36, 37 correct

O�S = odd squares� 4O�S = odd and even squares

� 49 + 5 = 54

B2,1

B2,1

B1

M1A1[7]

Give B1 if only one is correct.

These 2 B marks can only beawarded only if B2 has been givenfor part (i).

Co.

Any correct method. Co.

9. (i) log42 = ½ log864 = 2�2x + 5 = 91.5 �x = 11

(ii) Quadratic in 3y

Solution of quadratic = 0

�3y = 5 or –10

Solution of 3y = k

y = 1.46 or 1.47

B1B1M1A1

M1

DM1

M1

A1[8]

Anywhere.Forming equation and correctlyeliminating “log”. Co.

Recognising that the equation isquadratic.Correct method of solving theequation = 0.

Not dependent on first M1. Correctmethod.Co. (not for log5� log3). Ignore ansfrom 3y = -10.




10.

(i) Plots xy against x2 or x2 againstxy to get a line

c = 12 to 12.5 or -7.25 to -7.75m = 1.55 to 1.65 or 0.62 to 0.63xy = 1.6x2 + 12

or x2 = 0.625xy - 7.5� y = 1.6x + 12/x

(ii) Reads off at xy = 45 � x = 4.5 to 4.6

M1A2,1

B1B1

M1A1

M1A1[9]

Knows what to do.Points accurate – single line withruler

Allow if y = mx + c used.

Allow if y = mx + c used.Must be xy = mx2 + c orx2 = mxy + c.

Algebra is also ok as long as xy = 45is solved with an equation given M1above.

11. y = xe2x

(i) d/dx(e2x) = 2e2x

dy/dx = e2x + x.2 e2x

sets to 0�x = -0.5

(ii) d2y/dx2 = 2 e2x + [2 e2x + 4x e2x] = 4 e2x(1 + x) �k = 4

(iii) when x = -0.5, d2y/dx2 is +ve (0.74) � Minimum

B1

M1M1A1

M1A1A1

M1A1[9]

Anywhere – even if dy/dx = 2x e2x

or 2 e2x.Use of correct product rule.Not DM mark. Allow for stating hisdy/dx = 0.

Use of product rule needed.Allow if he reaches 4e2x(1 + x).

No need for figures but needscorrect x and correct d2y/dx2.

12. EITHER

At A, y = 4dy/dx = 2cosx - 4sinxdy/dx = 0 when tanx = ½

At B, x = 0.464 or 26.6o

B1M1A1M1A1

A1

Anywhere.Any attempt at differentiation.Sets to 0 and recognises need fortangent.Co. Accept radians or degrees here.

x 2 3 4 5 6 y 9.2 8.8 9.4 10.4 11.6 xy 18.4 26.4 37.6 52.0 69.6 x2 4 9 16 25 36




� (2sinx + 4cosx)dx = -2cosx + 4sinx

Area under curve = [ ]0.464 – [ ]0� -(-2) = 2.

Reqd area = 2 - (4 � 0.464) = 0.144(5 or 6).

M1A1

DM1

M1A1[11]

Any attempt with trig. functions.

x-limits used correctly. If “0” ignoredor automatically set to 0, give DM0.

Plan mark – must be radians for bothM and A.

12. OR

dy/dx = ½(1 + 4x)-½ � 4At P, m = 2/3

Eqn of tangent y - 3 = 2/3(x - 2)At B, x = 12/3

� √ (1 + 4x)dx = (1 + 4x)1.5 � 2/3 � 4

Area under curve = [ ]2 – [ ]0 = 41/3

Shaded area =Area of trapezium - 41/3 = 1/3

Or Area undery = 2/3x + 12/3 - 41/3 = 1/3

[or � xdy = � (¼y2 - ¼)dy= y3/12 - y/4

area to left of curve = [ ]3 – [ ]1 = 12/3shaded area =

12/3 – triangle (½.2.11/3)= 1/3]

M1A1

M1A1

M1A1A1

DM1A1

M1

A1

[M1A1A1

DM1A1

M1A1][11]

Any attempt with dy/dx – not for√ (1 + 4x) = 1 + 2√x. A mark needseverything.Not for normal. Not for “y + y1” or form on wrong side. Allow A forunsimplified.

Any attempt at integration with(1 + 4x) to a power. Other fn of xincluded, M1 only.Use of limits 0 to 2 only. Mustattempt a value at 0.

Plan mark independent of M marks.

A1 co.

Attempt at differentiation. A1 foreach term.

Must be limits 1 to 3 used correctly.

Plan mark independent of other Ms.

DM1 for quadratic equation. Equation must be set to 0.Formula - must be correctly used. Allow arithmetical errors such as errors over squaringa negative number.Factors – must be an attempt at two brackets. Each bracket must then be equated to 0and solved.Completing the square – must result in (x�k)2 = p. Allow if only one root considered.




1

[7]

Put x = -b/2 (or synthetic or long division to remainder)�3b3 + 7b2 - 4 = 0 AG

Search �b = -1 [or b = -2] (1st root or factor)

Attempt to divide�3b2 + 4b - 4 (or 3b2 + b - 2) or further search�b = -2 [or b = -1]

Factorise (or formula) [3 term quadratic] or method for 3rd value�b = -2, -1 or 2/3

M1 A1

M1 A1

M1

DM1 A1

2 (i)

(ii)

[6]

AB = OB - OA = �(9i + 12j)

Unit vector = AB 22129 �� = �(0.6i + 0.8j) [Accept any equivalent

unsimplified version of column vectors, � ��

��

�

12

9, � ��

�

��

�

8.0

6.0]

AC = 2/3 AB = 6i + 8j (or CB = 1/3 AB = 3i + 4j)

OC = OA + AC (or OB - CB ) = 12i + 5j (or equivalent)

M1

M1 A1

M1

M1 A1

3

[6]

)23( 5.05.0 �� xx dx = 3x1.5/1.5 + 2x0.5/0.5(one power correct sufficient for M mark)

��8

1

(2 x 8√ 8 + 4√ 8) – (2 + 4) Must be an attempt at integration

Putting √ 8 = 2√ 2 (i.e. one term converted √ to k 2 )� -6 + 40√ 2

M1 A1 A1

M1

B1√ A1

4

[4]

16x+1 = 24x+4 or 16 x 24x or 16 x 42x or 16 x 16x

20 (42x) = 20(24x) or 5(24x+2) or 20 x 16x

2x-3 8x+2 = 2x-3 23x+6 = 24x+3 or 8 x 24x or 8 x 42x or 8 x 16x

Cancel 24x+2 or 24x and simplify �4.5 or equivalent

B1 B1

B1

B1

5 (i)

(ii)

(iii)

[7]

f(0) = ½ f2(0) = f(½) = (√ e + 1)/4 � 0.662 (accept 0.66 or better)

x = (ey + 1)/4 � ey = 4x - 1 � f-1 : x � In(4x - 1)

Domain of f-1 is x�½ Range of f-1 is f-1�0

B1 M1 A1

M1 A1

B1 B1




6 (i)

(ii)

(iii)

[7]

x2 – 8x + 12 = 0 Factorise or formula �Critical values x = 2, 6x2 – 8x + 12 > 0 � � �2: �xx � � �6: �xx

x2 – 8x = 0 � Must be an attempt to find 2 solutionsx2 – 8x < 0 � � �80: �� xx

Solution set of │x2 – 8x + 6│< 6 is combination of (i) and (ii)� �20: �� xx � �86: �� xx

M1 A1A1

M1A1

B1 B1(one foreachrange)

7 (i)

(ii)

(iii)

(iv)

(v)

[8]

6! = 720

M … � 5! = 120

4! 48

6!/4! 2! = 15 Accept 6C4 or 6C2 = 15

5!/3! 2! = 10 (or, answer to (iv) less ways M can be omitted)(Listing – ignoring repeats � 8 [M1] � 10 [A1])

B1

M1 A1

M1 A1

B1

M1 A1

8 (i)

(ii)

[8]

Collect sin x and cos x � sin x = 5 cos xDivide by cos x � tan x = 5 (accept 1/5 – for M only)x = 78.7o or (258.7o) i.e. 1st solution + 180o

Replace cos2 y by 1 – sin2 y3sin2 y + 4sin y - 4 = 0 Factorise (or formula) (3 term quadratic) � sin y = 2/3 (or -2)

y = 0.730 (accept 0.73 or better) or (2.41) i.e. � (or 7

22) less 1st solution

M1M1A1 A1√

B1

M1

A1 A1√

9 (i)

(ii)

[8]

� � )12( 2tt dt = 6t2 – 1/3t3

From t = 0 to t = 6 distance = �6

0

= 144

Max. speed = 36 � from t = 6 to t = 12 distance = 36 x 6 (= 216)

During deceleration distance = (02 – 362) � 2(- 4) = 162Area of � is fine for M mark but value of t must be from constant

acceleration not 12 – 2t = �4

Total distance = 144 + 216 + 162 = 522

v

t

M1 A1

A1

B1

M1

A1

B2, 1, 0




10 (i)

(ii)

(iii)

[9]

dx

dy =

2)2(

1)42(2)2(

�

��

x

xx

= 2)2(

8

�

�

x

� k = -8

Must be correct formula for M mark (accept 2)2(

8

�

�

x

as answer)

When y = 0, x = -2 (B mark is for one solution only) NB. x = 0, y = -2

mtangent = -8/16 = -1/2 � mnormal = +2(M is for use of m1 m2 = -1, whether numeric or algebraic)

Equation of normal is y - 0 = 2(x + 2)(candidate’s mnormal and [x]y=0 for M mark)

When y = 6, x = 4

��

dt

dx

dx

dy

dt

dy�

�

�

2)2(

8

x

0.05 = �

�

4

80.05 = -0.1 (accept �)

i.e. 4�

��

��

�

xdx

dy x 0.05 for M mark.

√ is for error in k only. (Condone S � dx

dyx S)

M1 A1

B1

M1

M1 A1

B1

M1 A1√

11 EITHER y D (13½, 11)

B

A C (7, 4) (3, 2) O x

E

(i) mAC = (4 - 2)/(7 - 3) = ½

mBD = ½

mBC = -2

Equation of BD is y - 11 = ½(x - 13.5) i.e. 4y = 2x + 17

Equation of BC is y - 4 = -2(x - 7) i.e. y = -2x + 18

Solving y = 7, x = 5.5

B1

B1√

B1√

M1

M1

M1 A1




[10]

(ii) EAC

EBD

�

� = (ratio of corresponding sides or x- or y- steps)2 = 4/1

Quadrilateral ABDC/�EBD = 3/4

[Or, find E(1/2, -3) and then use array method to find one of:

area quadrilateral ABDC = 22.5 area �EBD = 30Find other area and hence ratio = 3/4 or equivalent]

M1 A1

A1

M1 A1A1

11

[10]

OR

B

6 7

P 5 Q

(i) (r + 6)2 + 52 = (r + 7)2

Solve � r = 6

tan AOB = 5/12 AOB = 0.395 or 22.6o

Length of arc AB = 6 x 0.395 = 2.37 or better

(ii) Sector AOB = ½ x 62 x 0.395 = 7.11

Shaded area = ½ x 5 x 12 - 7.11

All figures in sector and triangle correct √

22.9 or better

M1

M1 A1

M1

M1 A1

M1

M1

A1√

A1

O

r r

A


ADDITIONAL MATHEMATICS– JUNE 2004 0606 1

© University of Cambridge International Examinations 2004

1. (i) y=(3x-2)Q(x2+5)

dy/dx = (x2+5)3 - (3x-2)2x (x2+5)2

(ii) Num = 15 + 4x - 3x2 = 0 when

→ x = -5/3 or x = 3

M1 A1 M1 A1

[4]

Formula must be correct - allow unsimplified. Setting to 0 + attempt to solve. Both correct.

2. x3 = 5x-2

x3 - 5x + 2 = 0 Tries to find a value x = 2 fits

Q=(x-2) → x2 + 2x - 1 = 0 Solution → x = –1 ±√2

M1 A1 M1 DM1 A1

[5]

Equating + attempt at a value by TI Co - allow for (x-2) or for f(2) Must be Q by (x-his value) As by quadratic scheme Co

3. (i) y = |2x+3|

-ve then +ve slope Vertex at (-h,0) y = 1 – x Line, -ve m, (k,0) (ii) x + 2x + 3 = 1 → x = -⅔ (-0.65 to -0.70) x - (2x+3) = 1 → x = -4 (-3.9 to -4.1)

B1 DB1 B1

[3] B1 M1 AI

[3]

Must be 2 parts – ignore -2 to -1 V shape-Vertex on -ve x-axis + lines -ve slope, crosses axes at x,y +ve – allow if only in 1st or 2nd quadrants From graph, or calculation or guess B2 if correct. M mark for any method. Squares both sides M1 quadratic A1 Answers A1

4. x = asin(bx)+c

(i) a = 2 and b = 3 (ii) c = 1

(iii) 3 cycles (0 to 360)-1 to 3

Period 120° +

all correct.

B1 B1 B1 B1 B1 DB1 [6]

Wrong way round - no marks. No labels - allow B1 if both correct. Co Even if starting incorrectly. Needs to be marked - allow for any trig graph. Everything in relatively correct position - needs both B's






5. xy + 24 = 0 and 5y + 2x = 1 Makes x or y the subject and subs → 5y2 = y + 48 or 2x2 – x = 120 Solution of quadratic = 0 → (8,-3) and (-7.5,3.2) d = √(15.52+6.22) = 16.7

M1 A1 DM1 A1 M1 A1 √

[6]

x or y removed completely – condone poor algebra. A1 co.

By scheme for quadratic = 0 Co M mark ind of anything before. A1√ on his 2 points.

6.

( )

( )

( ) ( )

2.5

5240300

8

6

4

54186300

240

300

1.1.

4.3.

5.6.

864

8

6

4

1.

1.

4.

3.

5.

6.240300

or

or

Final answer → $2748

B2,1.0 M1 A1

M1 B1

[6]

For 3 correct matrices – independent of whether they are conformable – allow with or without the factor of 100. 1st product. Co. Matrices must be written in correct order – for M mark, the 2x3 or 3x2 must be used.

2nd product. By any method, inc numerical. Omission of 100 loses last B1 only.

7.

sinα = sin135 7 12 → α = 24.4° = 20.6°. Bearing is 020.6°

B1 M2 A1 A1

[5]

Correct triangle of velocities - must be 7,12 and 135° opposite 12. Sine rule used in his triangle. If 45° or 135° between 7 and 12, allow M1 for cos rule, M1 for sine rule Co. Co. Allow 21°.






8. y = (ax+3)lnx

On x-axis, y = 0 ax + 3 = 0 → x is -ve →no soln

But lnx = 0 → x =1 dy/dx = alnx + (ax+3).(1/x)

Use of m1m2 = -1 Gradient of tangent = -1 Q (-1/5)

→ a = 2

M1 A1 M1 B1 M1 A1 A1

[7]

Needs an attempt at solution. Ignore other solutions at this stage. Correct use of "uv" formula. For d/dx(lnx), even if M0 given above. Could equate m with -1 Q (dy/dx) Co. Co.

9. (a) 18

52

1

−

x

x

18C15 (x)15 (1/2x5)3 → 18.17.16(–⅛) ÷ 6 → –102 (b) (1 + kx)n

Coeff of x2 = nC2k2

Coeff of x3 = nC3k3

Equating and changing to factorials → k = 3/(n–2) or equivalent without factorials

B1

B1

B1 [3]

B1 B1 M1 A1

[4]

For 18C3 or 18C15

For (±½)3 – even if in (1/2x)3

Co Co. Co. Needs attempt at nCr Co

10. (i) Area = ∆ – sector

BCA = π – 1.4 or height = 20sin0.7 ∆ = ½.202sin(π –1.4) or ½bh = 197.1

Sector = ½2020.7 = 140 → Area = 57.1

(ii) DC = 20 x 0.7 (=14) AB = 2 x 20cos0.7 or cos rule BD = AB – 20 = 10.6 → Perimeter = 44.6 Could be [5] + [3] if AB used in part (i)

M1 M1 M1 A1

[4]

M1 M1 M1 A1

[4]

Award for either of these. Correct method for area of ∆ Use of ½r2θ Co Use of s = rθ Correct trig – could gain this in (i) Co






11. (i) m = –a/x3 → y = ½ax-2 (+c)

Puts in (2, 3.5) → 28 = a + 8c Puts in (5, 1.4) → 70 = a + 50c Solution → a = 20, c = –1 (ii) ∫(10x-2 + 1)dx = –10x-1 + x A = [ ]P – [ ]2 = –10/p + p + 3 B = [ ]5 – [ ]p = 10/p – p + 3 P = √10 or 3.16

M1 A1 DM1 M1 A1

[5] M1 A1√ M1 M1 A1

[5]

Any attempt to integrate. Co. Substitutes one of his points – even if +c missing Correct method of soln. Both co. (beware fortuitous ans. a = 20 given) N.B: assumes a = 20 without checking that both points work (M1A0DM1M0A1) Integrates his "curve" Use of limits correctly in either A or B or in A+B (2 to 5). Award M1 for each. (Can get these if only one integration) co

12 EITHER

12 questions – 3 trig, 4 alg, 5 calc Answer 8 from 12.

(a) (i) 12C8 = 495

(ii) T and A → 0 T and C → 1 A and C → 9 Total = 10

8 dresses, A → H

(b) (i) 8P5 = 6720 (ii) ⅛ of (i) = 840 or 7P4 (iii) ⅝ of (i) = 4200

or 5 x (ii) or 8P5 – 7P5

M1 A1 M1 A1

[4] M1 A1 M1 A1√ M1 A1√

[6]

12C8 gets M1. Answer only gets both marks. Needs to have considered 2 of the possibilities. Must be 8P5 for M1 – co for A1. Any method ok. √ on (i) if appropriate Any method ok. √ on (i) or (ii)






12 OR

x 2 4 6 8 10 y 9.8 19.4 37.4 74.0 144.4

lgy 0.99 1.29 1.57 1.87 2.16

(i) Finds values of lgy

Draws graph accurately.

(ii) lgy = lgA + xlgb

m = lgb → b = 1.4 (± 0.05) c = IgA → A = 5.0 (± 0.2)

(iii) lgy = xlg2 i.e Straight line Y = 0.301x x = 4.5 (± 0.2) Use of simultaneous eqns in part (ii) gets B1 only, unless both points used are on his line, in which case allow marks if to correct accuracy.

M1 A1

[2] B1 M1 A1 M1 A1

[5]

B1 M1 A1

[3]

Knows what to do. Don't penalise incorrect scale. Points correct to ½ small square. Anywhere – even if no graph Gradient measured + equated to lgb. Intercept measured + equated to lgA. Even if no line – give if line correct. Must be a line. To this accuracy.

DM 1 for quadratic equation. Equation must be set to 0 if using formula or factors. Factors Must attempt to put quadratic into 2 factors. Each factor then equated to 0.

Formula Must be correct – ignore arithmetic and algebraic slips.






1 [4] ( i – 7j ) + λ( 0.6i + 0.8j ) = 4i + k j M1 A1 1 + 0.6λ = 4 ⇒ λ = 5 – 7 + 0.8λ ⇒ – 7 + 0.8 × 5 = – 3 = k M1 A1 2 [4] Attempt at cos-1 0.3 ⇒ [72.5° A0] = 1.266 [ 5.017, 7.549 ] M1 A1 accept 1.3 x + 1 = 2.532, 10 034, 15.098 ⇒ x = 14.1 or better M1 A1 3 [4] (i) Some vegetarians in the college are over 180 cm tall [or equivalent] B1 (ii) No cyclists in the college are over 180 cm tall [or equivalent] B1 (iii) B ∩ C ⊂ A/ [or equivalent] B1 B1

4 [4]

−

+

θ

θ

θθ sin

cos

sin

1

cos

11

⇒ θθ

θ

sincos

cos12

− M1 M1

1 – cos2θ ≡ sin2

θ θθθ

θtan

sincos

sin2

→ B1 A1

Must be useful use of Pythagoras

5 [5] x = ( )35

2

242020±=

×−± or 2

1220 ± M1 A1

35

1

35

1

−

+

+

[or 1220

2

1220

2

−

+

+

] M1

rationalising each fraction or bringing to common denominator

Denominator = 2 [or 8 ] ⇒ 511

=+

dc

A1 A1

6 [6] (a) 2x

2 – 3x – 14 = 0 ⇒ ( 2x – 7 )( x + 2 ) = 0 ⇒ x = –2, 3.5 M1 A1 {x : x < – 2 }∪ {x : x > 3.5} A1

(b) Eliminate y ⇒ x2 + 4( 8 – kx ) = 20 [ or x 2048

2

=+

−⇒ y

k

y ] M1

x2 – 4kx + 12 = 0 [ or y2 + ( 4k

2 –16)y + ( 64 – 20k2) = 0 ]

Apply “ b2 = 4ac “ 16k

2 = 48 [ or 16k4 = 48k

2 ] ⇒ k = ±√3 M1 A1






7 [6] (i) e2x-3 (= 7) ⇒ x = ½ ( 3 + ln 7 ) ≈ 2.47 ~ 2.48 (not 2.5) M1 A1 (ii) h = 2ex – 3 (x, y or) h > – 3 accept ≥ B1 B1 (iii) h−1 (or y) = ln {½ ( x + 3 )} or ln(x + 3) – ln2 or lg{½(x + 3)}/lge M1 A1 but ln{½(y + 3)} M1 A0 lg (or log) {½(x + 3)} M1 A0 (M1 for logs taken in valid way

8 [8] (i) log 3 ( 2x + 1) – log 3 ( 3x – 11) = log 3 113

12

−

+

x

x [Or, later, give M1 for M1

log + log=log(product) log 3 ( ) = 2 ⇒ ( ) = 32 B1 2x + 1 = 9( 3x – 11) ⇒ x = 4 DM1 A1

(ii) log 4 y = yy

2

2

2 log2

1

4log

log=

[ or log 2 y = yy

4

4

4 log22log

log=

] M1 A1

½ log 2 y + log 2 y = 9 [or log 4 y + 2log 4 y = 9] ⇒ y = 26 or 43 = 64 DM1 A1 9 [8] 6 + 4x – x2 ≡ 10 – ( x – 2)2 M1 A1 (i) x = 2 y = 10 Maximum B1√B1√B1 (ii) f(0) = 6, f(2) = 10, f(5) = 1 ⇒ 1 ≤ f ≤10 M1 A1 [alternatively 1 ≤ B1, ≤ 10 B1] (iii) f has no inverse; it is not 1:1 B1 10 [10] (i) mBC = 3/5 Equation of AD is y – 4 = 3/5( x + 2) B1 M1 A1 mAC = – ¼ Equation of CD is y – 2 = 4( x – 6) B1 M1 A1 (ii) Solve x = 8, y = 10 M1 A1

(iii) Length of AC = Length of CD = 68 M1 A1 11 [10] (i) d/dx ( 2x – 3)3/2 = ( 2x – 3)1/2 × 3/2 × 2 M1 A1 dy/dx = 1 × ( 2x – 3)3/2 + ( x + 1) × { candidate’s d/dx ( 2x – 3)3/2 } M1 = )}1(3)32{(32 ++−− xxx = 325 −xx ⇒ k = 5 A1

(ii) δy ≈ dy/dx × δx = (dy/dx)x=6 × p = 90p M1 A1 ( y )x=6+p = ( y )x=6 + δy = 189 + 90p A1√

(iii) dxxx∫ − 32 = 1/5 ( x + 1)( 2x – 3)3/2 M1

[ ] 62

= 1/5 ( 189 – 3) =37.2 DM1 A1






12 [11] (i) a = dv/dt = 5e-1/2 t M1 A1

EITHER v = 8 = 10(1 – e-1/2 t ) ⇒ e-1/2 t = 0.2 ⇒ a = 1 M1 A1

(ii) s = ∫ tv d = ( ) tt

de10102/∫ −

− = 10t + 20e-t/2 M1 A1

[ ]6

0 = ( 60 + 20e-3 ) – ( 20 ) ≈ 41 DM1 A1

v (iii) 10 (iv) 10 B1 B2,1,0 t

12 [11] (i) d/dθ {( cosθ )-1} = – ( cosθ )-2 ( – sinθ ) = sinθ/cos2

θ M1 A1 OR

(ii) AX = 2secθ PX = 2tanθ B1 B1

T = 5

tan210

3

sec2 θθ −+

M1 A1

(iii) θ

θ

θ

θ

2

2sec

5

2

cos

sin

3

2

d

dT−=

B1 B1√

= 0 when 5sinθ = 3 ⇒ sinθ = 3/5 M1 A1 PX = 2tanθ = 2 × ¾ = 1.5 A1




IGCSE – JUNE 2005 0606 1


1

−=

−=

03

33

11

122

2A

(A²)–1 =

−

33

30

9

1

or A–1 first B1 B1 followed by squaring B1√B1√

B2,1 B1√,B1√ [4]

One off for each error. B1√ for 1÷9. B1√ for rest. √ from his attempt at A².

If

11

14 used, could get last 2

marks.

2 9 CDs →4 Beatles, 3 Abba, 2 Rolling

(i) 8C3 = (8×7×6)÷(3×2×1) = 56 (ii) 2B 2A 4C2×3C2 = 18 2B 2R 4C2×1 = 6 2A 2R 3C2×1 = 3 → Total of 27

M1 A1 [2]

M1 M1 A1 [3]

2 if correct without working 9C3 M0. 4×8C3 gets M1 A0 One correct product with nCrs 3 products added – even if nPr CAO

3 θθ2

sin1cos −= = 3

2

cs

s

−

=

3

1

3

2

3

1

−

= 12

1

−

× top and bottom by )12( +

→ 1 + 2

M1 A1 M1

M1

A1 [5]

Use of s² + c² = 1 to obtain cos as a surd – or correctly from 90o triangle. Correct algebra – getting rid of √3 Correct technique used to rationalise the denominator. This form ok. No need for a =, b = . (decimals get no credit)

4 OA =

−

−

1

3, OB =

2

1, AB =

3

4

AC = 5

3AB =

59

512

OC =OA + AC =

−

−

1

3+

59

512

=

−

54

53

OC = ( )25

1625

9+ = 1

M1 A1 M1 A1 M1 A1 [6]

Use of b – a or a – b – not for a + b CAO – not for negative of this.

Could be implied by correct OC .

Any correct method ok. CAO

Correct method on his OC. Answer was given.




IGCSE – JUNE 2005 0606 1


5 f(x) = A + 5cosBx

(i) A = −2 (ii) Amplitude = 5 (iii) B = 3 (iv) Range 3 to –7

B1 B1 B1 B1 B2,1 [6]

CAO CAO CAO −3 to 7 implied somewhere – table ok – even if no graph Needs 1½ oscillations – over-rides rest. √ on 3 and –7 Start at max – finishes at second min. Curves – but be tolerant

6 (i) −7 ≤ f(x) ≤ 8

(i) 0 ≤ g(x) ≤ 8 (ii) −7 ≤ h(x) ≤ 2

f yes g no h no

B1 B1 B1 B1 B1 B2,1 [7]

CAO Allow < for ≤ CAO As above CAO As above Loses one for each wrong decision. (answer f on its own – allow B2)

7 (a) tll )1(0 α+=

Subs and divides 1.031 = 1.0025t

t = lg 1.031÷ lg 1.0025 = 12.3

(b) 1 = log 10 LHS = lg 10(8 − x) 80 − 10x = 3x +2 → x = 6

M1 M1 A1 [3] B1 M1 M1 A1 [4]

Sub + division before taking logs. (or lgl = lgl0 + tlg(1+α) + use) Taking logs. CAO to 3 sf or more. Anywhere in the question. Putting any 2 logs together Complete elimination of 3 logs CAO

8 lgx 1 2 3 4 lgy 3.28 2.40 1.49 0.60

(i) Knows what to do. Pts within ½

square. (ii) Gradient = ±n n = −0.88 to − 0.92

log k = y-intercept k = 14 000 to 16 000

M1 A2,1 [3] B1 A1 B1 A1 [4]

For part (ii) – use of sim eqns is ok if points used are on line, not from table. Knows what to do. Accuracy within ½ square. B1 even if just stated without graph. B1 even if just stated without graph.




IGCSE – JUNE 2005 0606 1


9 (i) x² + 2x + k = 3kx − 1 → x² + (2 – 3k)x + (k + 1) = 0 Uses b² − 4ac =, > or < 0 → 9k² − 16k

End-points of 0 and 16/9 Use of b² − 4ac < 0 Solution set 0 < k < 16/9

(ii) Same case with k = 1 No intersection since k inside the range

Special case. Solves simultaneous.

eqns → √−7. B1

M1 A1 DM1 M1 A1 [5]B1 B1√ [2]

Any use of b² − 4ac This quadratic only. Solution of this quadratic →2 values Definite recognition of − ve. CAO NB No intersection on its own without k = 1 gets no credit.

10 (i) x = –a → 12132223

+++− aaa x = a → 121322

23+−+ aaa

12132223

+++− aaa = 3( 121322

23+−+ aaa )

2a³ + a² – 13a + 6 = 0

(ii) Tries a = 2 : fits ok. ( or −3, ½) ÷ (x – 2) → 2a² + 5a – 3 Solution → a = –3 and ½

If factors left as final answer, loses the last 2 marks.

M1

M1 A1 [3] M1A1 M1 M1 A1 [5]

For either of these – ignore simple algebraic and numeric slips

Allow M1 if 3 wrong side. Answer given. Tries a search for first value Must be (x – ) for M. CAO for A mark. CAO for both. T & I : M1 A1 for first value, A1 for second value, A2 for third.

11 a = −2 − 2t (i) v = −2t − t² (+ c)

v = 0 when t = 4 → c = 24 if t = 0, v = 24 ms–1

(ii) s = −t² −t³/3…+…(24t)…

Put t = 4 → 3

258 m

(iii)

M1 A1 DM1 A1 [4]M1A1√

A1 [3] B1 [1]

T Attempt at ∫. Ignore omission of c Attempt at c. CAO Attempt at ∫. “24t” not needed.

CAO Curve necessary.




IGCSE – JUNE 2005 0606 1


12 EITHER y = 8 − e–2x

Tangent crosses y-axis at (3½, 0) y = 0, x = ½ln8 or 1.04 Area of triangle = ½×3.5×7 = 12.25 ∫ curve = [8x − ½e2x] From 0 to his “x” [4ln8 − 4] − [0 − 0.5] 12.25 − (4ln8 − 3.5) = 7.43

M1 A1 M1 A1 [4] B1 M1 M1 A1

DM1 A1 [6]

For differential. CAO for gradient of −2. Any method ok providing calculus used. Numeric gradient for M1. Anywhere in the question. Even if no integration later. Attempt at ∫. CAO

DM0 if value at 0 assumed to be 0. CAO

12 OR

(i) Perimeter of square + circumference = 2 m → 4x + 2π r = 2

→ r = π

x21−

→ A = x² + 2

21

−

π

π

x

→ π

π 14)4( 2+−+

=xx

A

(ii) )482(1

dx

dA−+= xxπ

π

= 0 when x = 82

4

+π

= 0.28 m

A = 0.14

(iii) )82(1

dx

d2

2

+= π

π

A +ve → MIN

M1

A1 M1

A1 [4]

M1 A1

DM1

A1 [4]

M1 A1 [2]

Allow for π d or π r and for 2x or 4x

CAO – in any form Needs π r² and l² (both)

CAO – answer given

Attempt at diff. A0 if π missing, but can then gain rest of marks.

Sets his differential to 0.

CAO – 2 sig figures sufficient.

Any valid method ok. Needs correct

algebraic2

2

dx

d Afor A mark.

DM1 for quadratic equation. Equation must be set to 0 if using formula or factors. Formula Factors Must be correct Must attempt to put quadratic into 2 factors – ignore arithmetic and algebraic slips. Each factor then equated to 0.

x = 0, y = 7

dy/dx = −2e2x

At = x = 0, m = −2




IGCSE – June 2005 0606 2


1

(i)

dy/dx = (2x –1) –2 × ( – 8) × 2

B3, 2, 1

(ii) dy/dt = [dy/dx]x = −0.5 × dx/dt ⇒ 0.2 = – 4 × dx/dt ⇒ dx/dt = – 0.05

M1 A1

[5]

2

(i) ( )

5

1240300 or ( )

40

300512 = ( 3800) B1 B1

(ii)

4

10

150400

40180 or ( )

15040

400180410 =

4600

1960 or ( )46001960 M1 A1

(iii) 3800 + 1960 + 4600 = 10 360 B1 [5]

3

x + y = 12 AP2 – BP2 = AB2 ⇒ x2 – y2 = 60 Solve for y [via (12 – y)2 – y2 = 60 or using x – y = 5] BP = 3.5

B1

M1 A1

M1 A1 [5]

4

g2 (2.75) = g (2.5) or via [2(2x – 3) – 3]x=2.75 = 2

g-1(x) = 2

3+x g-1f(x) =

2

3sin +x x = π /2

M1 A1

M1 M1 A1

[5]

5

(i)

xd

d (x ln x – x) = ln x + (x × 1/x) – 1 (= ln x) M1 A1

(ii) ∫ −= xxxxx lndln

y = 0, x = 1 [ ]31

= (3ln3 – 3 ) – (0 – 1) ≈ 1.30

M1

B1 M1 A1

[6]




IGCSE – June 2005 0606 2


6

(i)

d/dx (e2x) = 2e2x

x

xx

xx

xxx

2

222

sin

cose)e2(sin

sin

e

d

d −=

or (sin x)−1(2e2x) + e2x (sin x)-2 (-cos x)

= 0 when 2sin x – cos x = 0

B1

B2, 1, 0√

B1 (ii) tan x = 0.5 ⇒ x ≈ 0.464 (26.6°)

M1 A1

[6]

7

Express in powers of 5 ⇒ 53x = 52+y [or xlg125 = lg25 + ylg 5] Express in powers of 7 ⇒ 7x – 2y = 70 or 7x = 72y [or x lg7 – y lg 49 = 0] Solve 3x = 2 + y [or 2.10 x = 1.40 + 0.70 y ⇒ x = 0.8

x = 2y 0.845 x – 1.69 y = 0] y = 0.4

B2, 1, 0

B2, 1, 0

M1 A1 [6]

8

(i)

Insert k in C∩D 6k in C∩D′ 3k in C′∩D n (C∪D) = 5/6 n(ε ) = 10k ⇒ Insert 2k in (C′∩D′)

B1B1B1

M1 A1√

(ii) 11k = 165 000 ⇒ n(ε ) = 12k = 180 000

M1 A1

[7]

9

X Correct ∆ of velocities α 300 V 60° β Y 120 sin α = (120sin 120°)/300 ⇒ α ≈ 20.3° [β ≈ 39.7°] V = 300 sin 39.7°/sin 120° [or 120 sin 39.7°/sin 20.3° or cos rule] ≈ 221 T ≈ 720/221 ≈ 3.24~6 [or via components 300 sin β = V cos 30°, 300 cos β = V cos 60° + 120 Square, add and solve for V, T = 12/(√22 –1)]

B1

M1 A1

M1 A1DM1 A1

[7]




IGCSE – June 2005 0606 2


10

(a)

Replace tan2 x by sec2 x –1 4sec2 x + 15sec x – 4 = 0 ⇒ (4sec x –1)(sec x + 4) = 0 (cos x = 4), cos x = – 0.25⇒ x = 104.5° or 255.5°

B1

M1

A1 A1√

(b) tan-15 ≈ 1.37 (or 78.7°), or any correct value of tan−1(−5) Any 2 correct values, or the specific value, of tan−1 (−5) = 1.77 (101.3°), 4.91 (281.3°), 8.05 (461.3°) Add 2 [or 114.6°], divide by 3 [consistent] y ≈ 3.35 [one answer only]

B1

B1

M1 A1

[8]

11

(a)

(i) 32 + 80x + 80x2 + 40x3 + 10x4 + x5

All coefficients to be resolved B3, 2, 1

(ii) x = √3 ⇒ x3 = 3√3, x5 = 9√3

32 + 80√3 + 240 + 120√3 + 90 + 9√3 = 362 + 209√3

B1 B1

B1 (b) …+ x4(– 4/x )3… × 7C4 (or 7C3 ) = 35 = – 2240 M1 A1 A1

[9]




IGCSE – June 2005 0606 2


12E

DC = BD [or D (5, 6) midpoint of C (x, y), B(8, 8)] ⇒ C is (2, 4) mDE = mAC = 7/4 mCE = –1/mAC = – 4/7 Equation of DE is y – 6 = 7/4 (x – 5) Equation of CE is y – 4 = – 4/7 (x – 2) Solve for E ⇒ x = 3.4, y = 3.2 Complete method for entire area → 15.6

M1 A1

B1√ B1√

A1√

M1(either)

A1√

M1 A1

M1 A1 [11] 12O

(i)

∠ BOD = 2sin−1 0.8 ≈ 1.855 (106.3°) or ∠ BOE = 0.927 (53.1°)

∠BAD = ½ ∠BOD ≈ 0.927 (53.1°) or ∠ BAE = 0.464 (26.6°)

[or O to BD = √( 102 – 82 ) = 6, ∠BAD = 2tan−1

8/16] AB = 8/sin(½ ∠BAD) ≈ 17.9 [or via √(82 + 162)]

M1 A1

M1 A1

M1 A1

(ii) Perimeter = (10 × 1.855) + (17.89 × 0.927) [or degrees × π /180] ≈ 35.1 M1 A1

(iii) Use of ½ r2θ or ½ r2 ( θ – sinθ )

(radians or degrees × 180

π

)

Complete plan Segment BCDB = ½ 102 × 1.855 – ½ 16 × 6 ≈ 44.75 Segment BEDB = ½ 17.92 × 0.927 – ½ 16 × 16 ≈ 20.3~5 Area ≈ 24.2~5

M1

M1

A1 [11]




IGCSE – May/June 2006 0606 01






© UCLES 2007

1. (i)

(ii) A ∩ B’ ∩ C (iii) (X∪Y)’ X’∪Y’

B1 [1] B1 [1] B1 B1 [2]

co co co co.

2. 2

42

−

+=

x

xy dy/dx =

2)2(

)42(2)2(

−

+−−

x

xx

If x = 4, dy/dx = −2 Perpendicular has m = ½ If x = 4, y = 6 → Eqn y − 6 = ½(x − 4) [2y=x+8]

M1 A1 M1 B1 A1 [5]

Formula must be completely correct co. (may be implied) Independent of first M mark. Anywhere in the question.

3. 1823 += yx 05022322

=++− yxx

→ 016102

=+− xx or 01832

=−+ yy M1A1 Complete elimination of x/y for M. → → (2, −6) and (8, 3) DM1 A1 Correct method of solution of quad.

Vector moves or other → P (4, −3) M1A1√ [6]

Any valid method.

4. (i) 5)2( u+ = 32 + 80u + 80u2

(ii) Replaces u by 2x − 5x2 –400 from ‘u’ term or +320 from ‘u2’ term Also … +80 22 )52( xx − → − 400 + 320 = −80

B2,1,0 [2] M1 B1

M1

A1√ [4]

One lost for each error Recognises and uses the link. Co (may be implied by answer)

Needs to look at 2 terms for x²

From his original expansion.





© UCLES 2007

5. x

xy9

+=

(i) x

y

d

d=

2

3

2

9

2

1

xx

−

2

2

dx

d y=

2

3

4

1

x

−

+2

5

4

27

x

(ii) If x = 9 , dx

dy = 0

.

B1 B1 B1 B1 [4] B1 [1]

Accept all these B marks if given as negative powers of x Answer given.

(iii) If x = 9, 2

2

dx

d y > 0. Minimum

M1 A1

[2]

Looks at sign of 2

2

dx

d y. Needs all

correct for the A mark.

6. (i) In 1.8s , alien goes 27 cm up. In 1.3 s missile goes 39 up. But alien starts at 12 up. → 39 – 27 = 12 (ii) In 1.8s. alien goes 72 across In 1.3 s, missile goes 1.3k 72 = 1.3k + 46 → k = 20.

B1 B1 M1 A1 [4] B1 M1 A1 [3]

Equates 2 vertical displacements. Equates 2 horizontal displacements.

7. (a) )5(485 1 xx −+

+= → 1485

−

+= uu → 5u2 − 8u − 4 = 0 → u = 2 or − 0.4 Soln of 5x = 2 → x = lg2 ÷ lg5 → x = 0.431 (b) qpqp loglog)log( −=−

B1 B1 M1 M1 A1 [5]

B1 for 5u and B1 for 4u–1 Solution of a quadratic. Allow for any soln of 5x = k. co.

= log (p/q) B1 co. p − q = p/q M1 Eliminating lg + good algebra.

→ 1

2

−

=

q

qp A1 co.

[3]

8. (a) 03cos51 =+ x cos3x = −0.2 3x = cos–1(−0.2) → x = 0.59 or 1.50 (b) .cos3tan5sec yyy =+ secy = 1/cosy and × cosy

uses cos2 = 1 − sin2 → 3sin2y + 5siny − 2 = 0 + solution → sin y = ⅓ y = 19.5º and 160.5º.

M1 A1 A1 [3] M1 M1 DM1 A1 A1√ [5]

Looks up cos before ÷ 3 co.co. Needs both of these. Needs correct link. Solution of quadratic co. √ for 180 − (first ans)





© UCLES 2007

9. (i)

1/x 10 8 6.25 5 2.5 1/y 20 15.6 11.8 9.0 3.5

(ii) Gradient 2.2 (±0.05) Intercept = −2(±0.1)

21.2.2

1−=

xy

→ x

xy

22.2 −

=

(iii) y = 0.15 1/y = 6.7 → 1/x = 4 → x = 0.254 (±0.010)

M1 A2,1,0 [3] B1 B1 M1 A1√ [4] M1 A1 [2]

Knows what to do. Accuracy. Within given range – graph needed Uses Y = mX + c

Correct form with his m and c. Uses 1/y and 1/x correctly – or solves equation from part (ii). co within range.

10

(i) AC = cos–14/5 = 0.6435 rads BCE = 2×BAC = 1.287

(ii) arc BD = 8×0.6435 =(5.148) arc BE = 5×1.287 = (6.435)

DE = 10−8 → Perimeter = sum of these = 13.6 m.

M1 A1 [2] M1 B1 DM1 A1 [4]

Complete method inc radian use. co – answer given. Any use of s=rθ Anywhere Sum of three parts. co.

(iii) Area of∆ ABC = 3×4 or ½absinC=12 M1 Correct method for triangle. Area of sect CBE = ½×25×1.287= (16.09) M1 Any use of A=½r2θ Area of sect ABD = ½×64×0.6435=(20.59) → shaded area = 12+16.09−20.59 M1 Must be linked correctly. Not DM. → 7.50 m2 A1 Correct to 3 sf. [4]





© UCLES 2007

Q 9





© UCLES 2007

11 EITHER

(i) dy/dx = 3cosx − 4sinx

= 0 when 3cosx − 4sinx=0 tanx = ¾

→ x = 0.644 → y = 5.00

(ii) A = ∫ +

2

0

.cos4sin3

π

dxxx

M1 A1 DM1 DM1 A1 [5] M1

Attempt at differentiation. co. Sets differential to 0. Arrives at tanθ = k. Both x and y needed. Any attempt to integrate

= [–3cosx + 4sinx] A1 A1 Each term. = [0+4}–[–3+0] DM1 Correct use of limits – DM0 if “0” left

→ 7 A1 co [5]

11 OR

2)23(

12

+

=

x

y

(i) dy/dx = −24 × (3x+2)–3 × 3 When x = 0 , dy/dx = −9 At A, x = 0 and y = 3 → B: x = ⅓

(ii) A = ∫+

3/1

0

2

.)23(

12dx

x

= [ −12(3x+2)−1 ÷ 3 ] = −4/3 − −2 = ⅔ Area of triangle = ½×3×⅓ = ½ → A = 1/6

B1 B1

B1 B1 [4] M1 A1 A1 DM1 M1 A1 [6]

For −24 × (3x+2)–3, for × 3

co. co Attempt needed to integrate For −12(3x+2)−1). For ÷ 3. Not given if bottom limit ignored. Anywhere. co

DM1 for quadratic equation. Equation must be set to 0 if using formula or factors. Formula. Factors Must be correct Must attempt to put quadratic into 2 factors. – ignore arithmetic and algebraic slips. Each factor then equated to 0.





© UCLES 2007


15

2•

2~+

In>(!,>:

~ = +

3 ll 1 ;.: +

" A' « 'If=

4 ··4 2

LHS = 2'

X "'

,1-n flt

1,

A'

ll1

M1

B1

A1

A1 M1

A1

8181

AI

A1

M1 A1

OM! A1

M1All>"




© UCLES 2007


~-~ +<I lOr " 4]

• 4+ = + 211: ~ 18 + +3:6 OJ

Met!:£ 4{ 4 f " +36 Ml

2 AI

Whiiin k 2 6• = Or +40

St:tl~?<t tt20 AI

5 5 At

Bl Ml

m " 10MIAI

Bl

g Mi

Ml

il<)




© UCLES 2007


11 14

-9t+ 14m Ml AI' "•

Ml AI

12 a, AB -) a

20"11 l 6 6

"""""""""""""""""~"

121 o' A81s ' 1)

"' Bf'ill! 4 ~ 3 2

= 5,

4 FC iZ =4

vin 24 4

e>IA£1~ y-41 " -:;L 15 M1A1

"'" p 6 p

1:;:0 - + + c R • I) 41

1 + +II+ if! Ml

-9-12 = c

Sea ron 1 + 2 Jr=-1 81

•• •" + 1111 AI

• "2]J<piJ 73




© UCLES 2007


1 (3] Y:. x ( 13 - 2x) ( >) 3 81

2>1- 13x + 6 => ( 2x - 1 )( x - 6 ) Y:.<x<6 M1 A1 l\CLl.{'t ~

2 [3] (i) c = 3 (i i) b = 1 (iii) a= 2 81 81 81

3 [5] J28 ±-J28 -8 M1 A1 X=

2

=>.fi ± JS ..fi +../5 Fl+.JS A1 M1 X

../7 -JS .J7 +..J5 => 6 + J35 · A1

[Or ( J28 +J20 x 2 JxJ28+J20 => 48 + 2J560 l [ M1 A1]

2 J28- J20 J28 + J20 8

4 (5] (i) ,oc4 = 210 M1 A1

(i i) 4w => sC4 = 15 3w + 1o => sC3 X 4c, = 80 Total = 95 81 M1 A1

- 5 1 ~ 6

_1_ = 22 - -x 5 (5] (i) J32

(ii) (64)-; : 2X (iii) LHS = 2X 81 8181 "

6 5 2?- 5x - 12 = 0 ( 2x + 3 )(x- 4 ) = 0 x=4or-1.5 M1 A1 -- X= - - => => =>

X 2

6 [6] (i) d/dx (?In x) = 2x In x +>1-tx M1 A1

(ii) J(2xln x + x}jx = x 2 1nx => J x2 2xlnxdx=x 2 1nx - 2 M1 A1

f4xln XdX = [2x 2 1n X - x2 r = ( 2e2 - e2) - ( 0 - 1) = e2 + 1 DM1 A1

7 [7) (i) A2 = ( 2 :1)( -

22

3 ) (- 2 -35) X= A2 + 2( B 10) = c 4 23) M1A1B1 V - 2 - 1 = - 2 -4 2 -10 - 1

( - 1 -23) 1 Y= (8 1;)xA1

= G -41) (ii) A.1 = 2

x - 8181M1A1 4 - 4




© UCLES 2007


8 (8]

9 [9]

10 [9]

(i) Eliminate y =;> 2( 3x- k ) = x' + 4

x' - 6x + ( 4 + 2k ) = 0

Meet if 36 - 4( 4 + 2k ) 2: 0

=;> k $ 2.5 =:> k=2

( Or x =;> 2y = ( y; k r + 4]

f Or I + ( 2k - 18 )y + ( It + 36 ) = 0 ]

[ Or ( 2k - 18 )2 - 4( ~ + 36 ) 2: 0 ]

(i i) When k = - 2 i - 6x = 0 [ Or 1- 22y + 40 = 0 ]

M1

M1

A1 A1

Solve (0,2).(6,20) M1 A1

Midpoint ( 3: 11 ) Lies on y = 2x + 5 since 11 = 6 + 5

(i) PQ = ( : } PR =(:}oR = ( _4

2) IPOI = JBO.jPRI = 10.!0RI =Eo

(.J80) + (Eo)2 = 102 =;> by Pythagoras L PQR= go•

(ii) PR + IPRI =:> ( 8i + 6j ) /10

(iii) ( 151) = m(;) + n( : ) =;> 5 = m + 9n, 11 =3m+ 9n Solve m = 3, n = 2/9

(i ) f '1 : X H Y, ( X+ 2 )

7y -a x= - - =:> xy+x=7y-a

y + 1 · 1 a + x

=;> 9 : XH--7-X

.. -• 28- a (11 )~9 (4) =-- ·•l.

5 .. , 7x- a "' 2

o ·r ·~ <j LX) ~ - -~,_l _ _ :)

(x;t7)

81 M1

M1 A1

M1A1 --J

tv11DM1A1

81

M1 A1

M 1 ,

£'V:.t."-t<t G. 7.... a. .... ~~ 'l'«.c\...'-L ~"' \"i a ~t• o·.,., ?~C\•Q.rl L'j 28 -a

[ Or g(4) = f(2) =;> -5

- = 4]

Dt\ \

=:> a= a btAUA1

(.11 .. 1) 7x - 9 = 9 + x . 9 1 2 [Or. w1th a= , g (x) = x, or g · (x) = x, or g (x) = x ]

x + 1 7 - x

=:> x2 - 6x + 9 = 0 =;> ( x- 3 )2 = 0 =:> x = 3 only

LC ·r s h.o ... .~s ~'l - ltt.\.c...: o 'l

M1

A1 A1




© UCLES 2007


11 [10]

12E [10]

120 [10]

(i) v= Jadt = f-9t (+c)

Solve f - 9t + 14 = 0

(ii) When a = 0. t = 4.5

3 9 2 (iii) s = Jvdt = !._ - - 1- + 14t (+c)

3 2

2 SA = 12 -

3

1 sa= -8-

6

~

~

~

(i) Equation of AB is 1

y-6= -(x + 1) 3

Whent=O, v=14 ~ c=14 M1 A1

A : t = 2, a = - 5 : B : t = 7, a= 5 M1 A1 ~ .,,

v = (-) 6.25 M1 A1

M1A2,1,0

Distance AB = ( - ) 20 ~ ( 20.8) A1 6

81

4-6 1 mAE"' --= - -

3 + 1 2 => maE= 2 => Equation of BE is y - 4 = 2( x - 3 ) B 1M 1 A 1-J

Solve x = 5,y=8

(ii) Height of 6 EBC = 4 => 'h x 4 x EC = 24 => EC = 12 ~ Xc = 15. Yc = 4

[Or via Iii! : : !ll = 24)

(iii) Equation of AE is y- 4 = - 'h ( x- 3 ) => Yo = - 2, Xo = 15

- ( 4)- (12) [ Or via AE - _ 2

, ED = p => p = - 6 ]

(a) (i) - 8 + 4a - 2b + c = ( R ) = 8 + 4a + 2b + c

(ii) 1 + a + b + c = ( R ) = 4a + c

(i ii) 27-9- 12 +c = 4

(b) Search - 1+3 - 2=0

Divide or factorise

x = - 2 ± J4;8 ~ -2. 73or0. 73 2

=> b= -4

=> a= - 1

~ c= - 2

=> x=-1

( x + 1 )( ;( + 2x - 2 )

A1

B1 -iM1A1

M1A1.f

M1 A1

M1 A1

81

M1 A1

J)M1 A1




© UCLES 2008

1

( )( )

4 3 28 3 2

4 3 2 4 3 2

32 12 2 24 2 18

16 18

50 36 2

2

−−

+ −

− − +

−

−

−

a = –25, b = 18

M1 DM1 A1 [3]

M1 for attempt to rationalise DM1 for attempt to expand out and simplify Allow A1 at this stage

2 (i) 10

5252C =

(ii) 4 women, 1 man: 6 3 women, 2 men: 4 6

3 2C C×

= 60 Total = 66

B1 [1]

M1 B1 B1

A1 [4]

M1 for a plan B1 for 6 B1 for 60

A1 for total Allow marks for other valid methods

3 (i)

( )( )

2

2 2

4 16 0

4 5 144 0

x kx

y ky k

+ + =

− + + =

M1 M1 for attempt to get a quadratic in

terms of one variable

16 256, ,422

±=== kkacb DM1, A1 [3]

DM1 for use of b2 – 4ac A1 for both

(ii) using 2

bx

a= − , or equivalent

When k = –16, (2, –10) When k = 16, (–2, 10)

B1 B1

[2]

B1 for each pair Allow B1 for x values only

4 (i) gradient = 2, equation of line of form Y = mX + c, where c = 0.6 e 0.6

y∴ =

(ii)

2

2

e 2 0.6

ln(2 0.6)

yx

y x

= +

∴ = +

M1 A1

[2]

A1

M1 A1

[3]

M1 for attempt to get equation of straight line

A1 for correct form (allow if seen in (i))

M1 for attempt to take ln





© UCLES 2008

5 2

2

d tan sec

d tan

y x x x

x x

−

=

When π d π, 1

4 d 2

yx

x= = −

Using d d d d

, 2 πd d d d

y y x y

t x t t= × = −

(-1.14)

M1 A1 M1 M1 A1

[5]

M1 for correct attempt to differentiate a quotient A1 all correct

M1 for attempt to sub π

4x = in to their

d

d

y

x

M1 for attempt to use rates of change

6 3 22 3 17 12 0x x x+ − + =

f(1) = 0, (x – 1) is a factor

( )( ) 0125212

=−+− xxx

( )( )( ) 034321 =+−− xxx

4 2

3 1, −= ,x

M1 M1 M1

DM1 B1,A1

[6]

M1 for simplification M1 for attempt to find a root M1 for attempt to get quadratic factor

DM1 for factorising on all previous M marks B1 for solution from first root A1 for the other pair

7 (i) 214 10

2θ = , leading to

θ = 1.25 rads (ii)

5

4 tan1.25, 12.038

44, 8.685

cos1.25

AB

AC AC

BC BC

=

= =

= − =

Perimeter = 25.7, allow 25.8

M1

A1 [2]

B1

M1 M1

A1 [4]

M1 for use of 21

2r θ

M1 for attempt to get AC M1 for attempt to get BC

8 (i) 1

2a =

(ii) 1

3b = (allow 0.33 or better)

(iii) 8loglog 333

=+ yx

2loglog

33=+ yx

27 3,log3

== xx

3

1 1,log

3=−= yy

Allow solutions using index notation

B1 [1]

B1 [1]

M1 DM1 A1 A1

[4]

M1 for reducing equations to terms of base 3 logs DM1 for dealing with simultaneous equations and logs to get final answers A1 for each





© UCLES 2008

9 (i) π

sin 22

y x c

= − +

c = 2

(ii) at 3π d

, 24 d

yx

x= = −

Grad of normal = 1

2

When 3π

4x = , y = 2

normal 1 3π

22 4

y x

− = −

M1 A1 M1, A1

[4]

M1

M1

M1, A1

[4]

M1 for π

sin 22

x

−

A1 correct M1 for attempt to get c Allow A1 for c = 2

M1 for attempt to get d

d

y

x

and for ⊥ gradient

M1 for attempt to obtain y using 3π

4x = in answer to (i)

M1 for attempt to obtain normal, must be using ⊥ gradient – allow unsimplified

10 (i) v = ( )

2

215ji +

v = 15i + 15j

(ii) (2i + 3j) + (15i + 15j)1.5 24.5i + 25.5j (iii) (2 + 15t)i + (3 + 15t)j

Allow (2i +3j) + (15i + 15j)t

(iv) relative velocity (15i + 15j) – 25j = 15i – 10j

(v) relative displacement (47i – 27j) – (2i + 3j) = 45i – 30j

Time taken = 3 hours Position vector at interception 47i + 48j

or

( )

( )

2 3 15 15

47 27 25

t

t

+ + + =

− +

i j i j

i jor equivalent

Allow solutions to (v) by drawing

M1 A1

[2] B1

[1] M1, √A1

[2] M1, A1

[2] M1 A1

[2]

M1 for attempt to get a direction vector Answer given M1 for use of their velocity vector with 2i + 3j.

Follow through on their velocity vector

M1 for a difference of velocities M1 for attempt to get relative displacement or other valid method. M1 for equating like vectors and attempt to get t





© UCLES 2008

11 (i) 5

tan3

x = −

o o

121.0 ,301.0x =

(ii) 23sec sec 4 0y y− − =

( )( )

o o o

3sec 4 sec 1 0

3cos , 1

4

41.4 ,318.6 ,180

y y

y

y

− + =

= −

=

(iii) 2 0.6 0.9273,2.2143z − =

z = 0.764, 1.407 (allow 1.41)

M1 A1, √A1

[3] M1 M1 M1 B1, A1

[5]

M1 M1 A1, A1

[4]

M1 for use of tan and attempt at one solution A1 for each, √ on first solution for x M1 for use of correct identity and formation of a 3 term quadratic in one variable. M1 for factorising a 3 term quadratic M1 for all terms in terms of cos

B1 for 0180 , A1 for the other pair

M1 for correct order of operations M1 for a valid attempt at a second solution A1 for each

12 EITHER

(i) ( 3,0)± allow

(ii) ( )2d3 e e 2

d

x xy

x xx

− −

= − − +

( )2e 2 3

x

x x−

= − +

d

d

y

x= 0 , 2

2 3 0x x− − =

leading to x = 3, –1 and y = 6e–3( 0.299), –2e (5.44)

(iii) ( ) ( )2

2

2

de 2 2 e 2 3

d

x xy

x x xx

− −

= − − − +

When x = 3, 2

2

d

d

y

x is –ve, max

When x = –1, 2

2

d

d

y

x is +ve, min

B1, B1

[2] M1, A1

M1 A1 A1

[5] M1 B1 B1

[3]

M1 for a correct attempt to differentiate a product or a quotient A1 allow unsimplified

M1 for attempting to solve d

d

y

x= 0

A1 for each pair

M1 for attempt at second differential or use of gradient method

B1 for each





© UCLES 2008

12 OR

(i) 0

1, 1

1v v

t

= =

+

(ii)

( )1 1

2 2 1v

t t

= −

− +

4 4

1 1 1;

4 5 20v v= − = ( 0.05)

(iii)

( ) ( )

42 2

1 1 17;

2002 2 1a a

t t

= − + = −

− +

(–0.085)

(iv) ( )1 1

2 2 1t t

−

− +

= 0, t = 5

(v)

3ln 4s = (1.386)

4

16 2ln

5s = (1.509)

In 4th sec, 4 2

ln5

s = (0.123)

(allow 0.124)

M1, A1 [2]

M1 A1

[2] M1, A1

[2] DM1, A1

[2] M1 A1

[2]

M1 for attempt to differentiate M1 for attempt to differentiate M1 for attempt to differentiate DM1 for equating v to zero M1 for attempt to find s3 and s4





© UCLES 2008

1 (i) a = 8 B1

b = –13 B1 (ii) (–4, –13) B1√ [3] 2 (a) (i)

E

A B

B1 (ii)

E

E

C

D

oe B1 (b) ( ) ( )YXYX ∩∪∩ '' B2, 1, 0

or ( ) ( )YXYX ∪∩∩ '

or ( ) ( )YXYX ∪∩∪ ''

or ( ) ( )( )''YXYX ∪∪∩

or ( )( ) ( )( )YYXXYX ∩∩∪∩∩ '' [4] 3 Eliminates x or y M1 021147

2=−− xx or 0105147

2=−+ yy oe A1

Solve 3 term quadratic M1 ( )( )31 −+ xx A1 (3,3) and (–1,–5) A1

[or 2

162 ±=x M1

–1 and 3 A1 (3,3) and (–1,–5) A1] [5]





© UCLES 2008

4

(i) straight line, +ve gradient, – ve intercept B1 idea of modulus (V shape on axis) B1 meets axes in correct places DB1 (ii) 6 B1 4 B1 [5] 5 (i) evidence of 27 or 56 in correct place B1 1512 B1 (ii) 28 x 9 B1 complete plan M1 504 A1 [5]

6 (a) uses cos2 x = 1 – sin2 x or sec2 x = x

2cos

1or sec2 x = 1 +

x

x

2

2

cos

sin B1

2

1

1

p−

or ( )( )pp −+ 11

1 or

2

2

11

p

p

−

+ B1

(b) express LHS in terms of sine and cosine B1 uses common denominator B1 uses sin2 A + cos2 A = 1 in useful way B1 correct conclusion B1 [6]





© UCLES 2008

7 (i) x2 +

+=

4

2

2

2

3224

x

x

x

B1

(ii) 2x –5

128

x

B1 + B1

(iii) equates to 0 and attempts to solve M1 x = 2 A1

OP = 6 or 2.45 A1 [6]

8 (i) (2x + 1)log2 = log20 or 2xlog2 = log10 M1 attempt at valid solution M1 1.66 A1 (ii) express in powers of 5 (or 25 or 125) M1

y

y

y

y

24

93

2

14

5

5

5

5

−

+−

= A1

4y – 1 – 2y = 3y + 9 – (4 – 2y) M1 –2 A1 [7] 9 (i) Matrix multiplication M1

−

−

46

1812 A1

(ii) Matrix multiplication M1

2

7 A1

(iii) 10

11=

−

A

−

−

42

13 B1 + B1

premultiply

−

20

53 M1

−

−

1.80.6

1.70.9 A1 [8]

10 (a) (i) k(2x – 1)–3 +(c) M1 k = –2 A1 (ii) multiplies out and integrates M1

4 3 22( )

4 3 2

x x x

c− + +

A2, 1, 0





© UCLES 2008

(b) (i) uses product rule M1

4

)5(42

+

−++

x

x

x oe A1

correct completion A1 (ii) ( ) 45 +− xxk M1

3

2=k oe A1 [10]

11 (i) ≥ 2 B1 (ii) 51 B1 (iii) method for inverse M1 ( ) 12 −−x A2, 1, 0

(iv) solve 21

20=

+x

or g –1 : x 120

−

x

a M1

9 A1 (v) finds expression for fg(x) DM1

211

202

+

+

+x A1

equate to 38 and solve quadratic M1 3 A1 [or (v) g(x) = f –1 (38) M1 g(x) = 5 A1

solve 51

20=

+x

M1

3 A1] [or (v) x = g–1 f –1 (38) M1 x = g–1 (5) A1 evaluate M1 3 A1] [11] 12 EITHER

(i) A (4, 0) B1

x

y

d

d= 4 – 2x M1

gradient = 2 A1 y – 3 = 2(x – 1) or y = 2x + 1 DM1

correctly reaches x = –2

1 A1





© UCLES 2008

(ii) 2x2 –3

3x

B1

uses limits of 4 and 1 M1 9 A1 area triangle = 9/4 B1 11.25 A1 [10] OR M(1, 3) B1 grad AB = ⅓ B1 uses m1m2 = –1 M1 ( )133 −−=− xy or 63 =+ yx A1 grad BC = –2 B1 ( )222 +−=− xy or 22 −=+ yx M1 solve equation of MD with equation of AD M1 x = 8, y = –18 A1 method for area M1 77 A1 [10]



Mark Scheme: Teachers’ version Syllabus Paper


© UCLES 2009

1 (i) 12 15θ= , 0.8θ = rads M1, A1 [2] M1 for use of s rθ=

(ii) Area = ( )2115 0.8

2

leading to 90 (cm2)

M1

A1 [2]

M1 for use of 21

2A r θ=

2 x3 = 8, leading to x = 2

23

d

dx

x

y= leading to grad of

12

1−

for normal

( )212

10 −−=− xy

+−=6

1

12

1xy

B1 M1 DM1 A1 [4]

B1 for finding where curve crosses the x axis M1 for attempt to differentiate and use of m1m2 = –1 DM1 for attempt at equation of normal Allow unsimplified

3

2 2

2 2

1 cos sin

sec 1 tan

θ θ

θ θ

−

=

−

M1 M1

M1 for use of 2 21 cos sinθ θ− =

M1 for use of 2 2sec 1 tanθ θ− =

2

cos θ= M1 M1 for attempt to simplify

=2

1 sin θ− A1 [4]

Alt Scheme

2 2

22

2

1 cos sin

1 cossec 1

cos

θ θ

θθ

θ

−

=

−−

2 2

2

sin cos

sin

θ θ

θ=

2

cos θ=

=

21 sin θ−

M1

M1

M1

A1

M1 for use of 2 21 cos sinθ θ− =

M1 for attempting to get all in terms of cos

M1 for attempt to simplify

4 (i) 53352

+−=− xkxx M1 M1 for equating line and curve equations

0882

=+− xkx

using 042

=− acb , k = 2

(Alt scheme: 5 2 3kx= − , 4

xk

=

20 16 12

3 5k k k− = − +

leading to 2k = )

DM1, A1 [3]

DM1 for use of 24b ac− on resulting

quadratic (Alt scheme: M1 for attempt to differentiate quadratic and equate to 5 DM1 for simplification and solution using resulting quadratic

(ii) leading to x = 2, y = 7 M1, A1 [2]

M1 for obtaining x and y coords





© UCLES 2009

5 (a) ( )2 2 1 3

3 3x x−

=

4 2 3x x− =

2x =

(b) 2a b− or

2

b

a (allow here)

1 2, =−= qp

B1

B1

B1 [3] B1 B1 [2]

B1 for ( )2 2 1

3x−

B1 for 33

x

B1 for 2x = B1 for each

6 f (3), f ( 5) or f (0.5) 0− = spotted

Either ( )( )152122

−+− xxx

Or ( )( )37252

+−+ xxx

Or ( )( )59232

−+− xxx

B1

M1

A1

M1

B1 for spotting one root

M1 for attempt to obtain quadratic factor

A1 all correct

M1 for solution of quadratic

x = 3, –5, 0.5 A2,1,0 [6]

A2 for all 3 solutions (–1 each error) Correct factors only – lose 1 A mark

7 (i) 3 3 33 e e e

x x x

x + −

= 33 e

x

x

M1, A1, B1

[3]

M1 for attempt to differentiate a product. A1 for correct product. B1 for 3

ex

−

(ii) 3

3 31 ee d e

3 3

x

x x

x x x

= −

∫

DM1 DM1 A1 [3]

DM1 for recognition of the ‘reverse’ to (i) DM1 for dealing with ‘3’ A1 all correct (condone omission of c)

8 (i) ( ) ( )

( )22

2

9

2229

d

d

+

−+=

x

xxx

x

y

( )22

2

9

218

+

−=

x

x

, turning points,

x = ± 3

B2,1,0 M1 A1 [4]

Attempt to differentiate a quotient –1 each error M1 for correct attempt to find the turning points. A1 for both

(ii) d

2d

x

t

=

d 16

2d 100

y

t

= ×

= 0.32 or 8

25

B1

M1

A1 [3]

B1 for use of d

2d

x

t

=

M1 for use of rates of change





© UCLES 2009

9 (i) 1 1

10 2 10 10

2 2

+ = +

i j i j

M1 A1 [2]

M1 for attempt at a correct direction vector A1 all correct

(ii)

( ) ( )4 8 20 20 16 28− + + + = +i j i j i j M1 A1 [2]

M1 for valid attempt A1 all correct

(iii) ( ) ( )10 10 8 6 2 4+ − + = +i j i j i j M1 A1 [2]

M1 for attempt at vector difference A1 condone negative

(iv) displacement of

( ) ( )19 34 16 28 3 6+ − + = +i j i j i j

time =1330 hours (accept 1.5 hours)

at 31 43+i j Alternative scheme:

( ) ( )( ) ( )19 34 8 6

16 28 10 10

t

t

+ + + =

+ + +

i j i j

i j i j

or equivalent

M1 A1 A1 [3]

M1 for displacement and attempt to obtain time A1 for correct time A1 for correct position vector M1 for attempt to equate like vectors A marks as above

10 (i) mAB = 0.75 line AB )4(75.00 +=− xy

mPQ = 3

4−

line PQ ( )410 1

3y x− = − −

intersection at C (4, 6) Q (8.5 0) (ii) AC = 10, CQ = 7.5 Area = 37.5

M1 A1 M1

A1 M1 A1 √ B1 [7] M1 A1 [2]

M1 for attempt at mAB and line AB M1 for use of ‘m1m2 = –1’ and attempt at line PQ

M1 for attempt at solving simultaneous equations Ft on their line PQ M1 for attempt at lengths and area





© UCLES 2009

11 (i) ktns lnlnln += ln t 1.6 2.7 3.4 4.2 4.6 ln s 7.2 5.9 5 4 3.6

Plot sln against tln (ii) grad n = –1.2 (–1.4 to –1.0) Intercept = ln k, leading to k = 7900 – 10 000 (iii) when t = 50, ln t = 4.4 leading to s = 80 (72 – 92) Alternative method

(i) ktns lglglg +=

lg t 0.7 1.2 1.5 1.8 2 lg s 3.1 2.5 2.2 1.7 1.6

M1, A1 M1 A1 [4]

M1, A1 M1, A1 [4] M1 A1 [2]

M1for attempt to take logs A1 for correct form M1 for attempt to plot correct graph A1 for a reasonable straight line

M1 for use of grad = n M1 for use of intercept = ln k M1 for attempt to obtain s

Same scheme applies

1 2 3 4 5

1

2

3

4

5

6

7

8

9

10

ln t

ln s





© UCLES 2009

12 EITHER

(i) amplitude = 1

(ii) period = 6π, 18.8

(iii) 2

1

3sin =

x,

2

5,

2

ππ

=x

(iv) Area under curve

x

x

d3

sin1

2

π5

2

π

∫

+ =

2

5π

2

π3cos3

−

x

x

B1 [1] B1 [1] M1 A1, A1 [3] M1

B1, B1

M1 for attempt to solve correctly A1 for each (allow degrees here) M1 for attempt to integrate

B1 for x, B1 for 3cos3

x

−

leading to 2π + 3 3

Area of rectangle = 2

3

2

π

2

π5×

−

= 3π

Shaded area = 3 3 – π (2.05)

DM1 M1

A1 [6]

DM1 for correct use of limits M1 for attempt at rectangle plus subtraction – must be working in radians

Alternative solution: Shaded area

x

x

d5.03

sin

2

π5

2

π

∫

− =

5π

2

π

2

0.5 3cos3

x

x− −

M1 M1

B1, B1

DM1, A1

M1 for subtraction (must be using radians) M1 for attempt to integrate

B1 for 0.5x− , B1 for 3cos3

x

−

DM1 for correct use of limits





© UCLES 2009

OR

(i) π

8t =

(ii) 4 sin 4a k t= −

(iii) 3π

12 4 sin2

k= − leading to

3k =

(iv)

1 2 3

–4

–3

–2

–1

1

2

3

4

t

v

(v) ∫=

24

π

0

d.4cos3 tts

π

24

0

3sin 4

4t

=

leading to 3

8

B1 [1]

M1, A1 [2]

M1

A1 [2] B1 √ B1 [2]

M1, √ A1

DM1, A1 [4]

M1 for attempt to differentiate

M1 for attempt to substitute into their acceleration equation B1 for correct shape B1 ft on their value for k

M1 for attempt to integrate Ft on their value for k

DM1 for application of limits or equivalent





© UCLES 2009

1 (a) ''or)( YXYX ∩∪ ' B1 (b) (i)

B1 (ii) 9 B1 (iii) 11 B1 4

2

−

−=−

73

64

10

11A B2,1,0

premultiply

3

17 M1

x = 5 , y = –3 A1 4

3 Understands modulus M1 Curve from x I 2 to x K 6 A1 Cusp A1 Position correct A1 4

4 (i) 15 or 24 B1 240 B1

(ii) 160 B1

( ) ( )1604

1240 ×

−+ M1

200 A1 5

10 4 7

P E

9





© UCLES 2009

5 (i) 4

9612

x

x − oe B1+B1

(ii) uses xx

yy ∂×=∂

d

d with x = 2 M1

Substitute ∂x = 0.04 DM1 0.72 A1 5

6 (i) f(x) A 2 B1 (ii) Method for inverse M1 (x – 2)2 + 3 or x2 – 4x + 7 A1

(iii) g(5) or gf(x) ( )

2

32

12+

−+

=

x

B1

4.4 only B1 5

7 (i) 4.5 B1 (ii) –9 B1 (iii) log X + log Y M1 15 A1

(iv) Y

X

log

log M1

1.5 A1 6

8 (i) 27 – t2 B1 2t × (27 – t2) B1

(ii) 2

654d

dt

t

A−= B1

Solve 0d

d=

t

A M1

t = 3 A1 (iii) Substitute for t in expression for A M1 A = 108 only A1 completely correct method and maximum B1 8





© UCLES 2009

9 (i) )1( 234

6789

×××

×××

M1

126 A1

(ii) )1( 2

34

×

×

B1

23)1( 2

34××

×

× M1

36 A1 (iii) adds number of arrangements of 2,1,1 and 1,2,1 and 1,1,2 only M1 multiplies for each selection M1 (36) + 4 × 3 × 2 + 4 × 3 (× 1) A1√ 72 A1 9

10 Eliminate y (or x) M1 x

2 + 12x + 32 = 0 (or y2 – 48y + 560 = 0) A1 solve 3-term quadratic M1 x = –4 and x = –8 (or y = 20 and y = 28) A1 (–4, 20) and (–8, 28) A1√ M (–6, 24) M1 Grad AB = –2 B1

Uses grad perpendicular = AB

m

1− and coordinates of a point M1

272

1+= xy or )6(

2

124 +=− xy or 0542 =+− yx A1

9

11 (a) (i) tan x = 1.5 B1 56.3 B1 236.3 B1√ (ii) uses sin2 y = 1 – cos2 y M1 2cos2 y + 3cos y – 2 = 0 A1 solve 3-term quadratic M1 60 A1 300 A1√ (b) 1.12 B1 0.06 or 0.0599 B1 0.51 or 0.511 B1 11





© UCLES 2009

12 EITHER (i) integrates to find y M1

)(e2 2

1

c

x

+ A1

substitute (0, 5) into ckyx

+=2

1

e M1 c = 3 A1 (y-coord of Q is ) 2e + 3 A1√ (ii) uses y = mx + c M1

y = x + 5 A1 y = ex + 3 A1 solve linear equations M1

1e

2

−

=x A1

10 OR

(i) A (0, 6) B1

x

x

y2

1

e2

1

d

d= B1

uses m1m2 = –1 M1 B (3, 0) B1√ (ii) Integrates for area below curve M1

x

x

52e 2

1

+ A1 uses limits of 0 and xB M1 13 + 2e1.5 or 21.96 or 22 A1 Area rectangle = 3(e1.5 + 5) or 28.4(4..) M1 Area = e1.5 + 2 or 6.45 to 6.5 A1 10




IGCSE– May/June 2010 0606 11

© UCLES 2010

1 (i) 2

1

32 )1(32

1 −

+ xx

(ii) 2x cos2x – 2x2 sin2x

B1,B1

[2] M1 A2,1,0

[3]

B1 for 2

1

3 )1(2

1 −

+ x

B1 for × 3x2

M1 for attempt to differentiate a product –1 each error

2 (i) 1 + 18x + 135x2… (ii) (1 × –5) + (18 × –3) + (135 × 1) = 76

B1,B1 [2]

M1,A1ft A1

[3]

B2, 1, 0 –1 for each error M1 for a correct method using their (i) A1ft on their 3 terms unsimplified

3 (k – 2)2 – 4(2k – 4) k2 – 12k + 20 = 0 critical values 2 and 10 k Y 2 and k [ 10

M1 A1 M1 A1 A1

[5]

M1 for use of discriminant for 3 term quadratic in k M1 for attempt to solve quadratic A1 for critical values A1 for range

4 (i), (ii) and (iii)

(b) (i) {9,10,11,12,13,14} (ii) {5,6,7,8,9,10,11,12,13,14,15,16,17,18,19, 20} (iii) x or { }

B1 B1 B1

[3]

B1 B1 B1

[3]

B1 for each correct Venn diagram Or equivalent Or equivalent

5 3x3 + 17x2 + 18x – 8 = 0 f(–2) = 0 (or other roots) (x + 2)(3x2 + 11x – 4)(= 0) (x + 2)(3x – 1)(x + 4)(= 0) x = –2, –4,

31

M1 M1 M1 DM1 B1, A1

[6]

M1 for simplification = 0 M1 for attempt to find a root M1 for attempt to obtain quadratic factor DM1 for obtaining linear factors or use of quadratic formula B1 for first solution A1 for the other pair

www.XtremePapers.net

http://www.xtremepapers.net



© UCLES 2010

6 (i) 2

1x + 2y

(ii) y – 1

(iii) 2log

log

2log

64log

8

8

8

8p

+

= 6 + 3x

B1 B1

[2]

M1 A1

[2]

M1

B1 A1

[3]

B1 for each term

M1 for difference of 2 logarithms M1 for attempt at a valid method

B1 for 6 A1 for + 3x

7 (i) f [ –3

(ii) f–1 = 2

13 −+x

(iii) 13311

32

2

=−

+

+ x

161

72

=

+

+

x

x

x = 1

B1 [1]

M1 M1 A1

[3] M1

A1 M1 B1

[4]

M1 for correct order of operations M1 for ‘interchange’ of x and y M1 for correct order

A1 for correct simplification M1 for solution B1 for one solution only

8 (a) 23 – 4x 22x + 8 = 2 3 – 4x + 2x + 8 = 1 x = 5 (b) (i) 32

(ii) ( )( )2525

2553

+−

++

leading to 1

1155 +

M1 DM1 A1

[3] M1 A1

[2]

M1

A1 A1

[3]

M1 for to obtain powers of 2, 4 or 8 DM1 for attempt to equate powers of 2, 4 or 8, using addition M1 for attempt to obtain each term in terms of 3

M1 for attempt to rationalise

A1 for numerator A1 for denominator (can be implied)





© UCLES 2010

9 (a) (i) x

y

d

d = 5 + 4e–x

(ii) when x = 0, x

y

d

d = 9

use of dy ≈ x

y

d

ddx leading to

dy ≈ 9 p

M1 A1

[2]

M1

A1 [2]

M1 for attempt to differentiate M1 for attempt to use small changes

(b) td

dA = 0.5

A = x2, x

A

d

d = 2x, x = 3

t

x

d

d = 0.5 ×

x2

1

= 12

1

B1

M1

DM1

A1 [4]

M1 for attempt to get x

A

d

d

DM1 for correct use of rates of change

10 (i) tan x = 0.25 x = 14.0°, 194.0° (ii) 3 + sin y = 3(1 – sin2 y) 3sin2 y + sin y = 0 sin y(3sin y + 1) = 0

sin y = 0, sin y = –3

1

y = 180°, y = 199.5°, 340.5°

(iii) cos4

1

3=

z

3

z

= 1.3181 leading to

z = 3.95 Allow 3.96, 1.25π, 1.26π

M1 A1,√A1

[3] M1

DM1

B1 A1 √A1

[5] B1

M1 A1

[3]

M1 for use of tan M1 for use of correct identity and attempt to simplify

DM1 for attempt to solve quadratic

B1 for 180° A1 for 189.5°

Ft on their 189.5°

B1 for cos4

1

3=

z

or equivalent in terms

of cos M1 for a correct order of operations (allow π)





© UCLES 2010

11 EITHER

(i) x

y

d

d =

4

12 ln2)(

x

xxxx−

3

ln21

x

x−

=

when x

y

d

d = 0, ln x =

21 , x = e 2

1

, y = e

21

,

y = e2

1

(ii) ( ) ( )

6

223

2

23ln21

d

d

x

xxx

x

yx

−−−

=

= 4

ln65

x

x+−

(iii) when x = e 21

, 2

2

d

d

x

yis –ve

(= 2

e

2−

), max

B3,2,1,0

M1 A1

A1 [6]

M1

A1, A1 [3]

M1

A1 [2]

–1 each error

M1 for attempt to solve x

y

d

d = 0

M1 for attempt at 2nd derivative

A1 for a, A1for b

M1 for a correct method

A1 must be from correct working only

11 OR

(i) y = 3sin ( ) ( )cx ++

2π2

5 = 3sin π + c, c = 5 y = 3sin ( )

2π2 +x + 5

(ii) cos ( )2

π2 +x = 0

x = 0, 2

π , π

(iii) when x = 4

3π , y = 5

x

y

d

d = 6

normal y – 5 = –6

1( )

43π

−x

+−= 39.5

6

1xy

M1, A1 M1,

A1 [4]

M1

A2,1,0 [3]

M1

M1

DM1

A1 [4]

M1 for sin ( )

2π2 +x

M1 for attempt to find c

M1 for attempt to solve x

y

d

d = 0

M1 for attempt to obtain y

M1 for attempt to obtain x

y

d

d and perp

gradient DM1 for attempt at straight line

(Must have (i) correct)





© UCLES 2010

1 24x2 – 6x = 0 (or y2 + 3y + 2 = 0)

leading to (0, 1) and

−2,

4

1

M1 M1

DM1 A1,A1

[5]

M1 for attempt to get an equation in one variable. M1 for attempt to get 2 or 3 term quadratic = 0

DM1 for attempt to solve A1 for each pair of values

2 6(–2)3 + a(–2)2 – (a + 1)(–2) + b = 15 6a + b = 61 when x = –1, 2a + b = 29 leading to a = 8 and b = 13

M1 A1 A1 M1 A1

[5]

M1 for substitution of x = –2 or –1, or verification A1 for each correct (allow unsimplified) M1 for attempt to solve A1 for a = 8, b = 13

3 (i)

−−

=

25

17

5

4BA

r

−=

20

21

unit vector =

−29

20

2921

or equivalent

(ii)

−=

−−

20

213

25

17CO

r

−=

35

46CO

r

B1

M1, A1 [3]

M1

A1 [2]

B1 for BA

r

M1 for magnitude of BA

r

M1 for BA

r

325

17+

−

4 (i) gradient = –2 y2 = –2sec x + c leading to y2 = –2sec x + 6.4

(ii) when y = 2, cos x = 6

5

B1 M1 A1

[3] DM1 A1

[2]

B1 for gradient M1 for correct attempt to link y2 and sec x DM1 for attempt to solve their equation using y = 2

5 2

3

d

d

xx

y= ,

gradient at A = 3

1,

normal grad = –3 coords of A (3, 5) normal y – 5 = –3(x – 3)

when y = 0, x = 3

14

M1

DM1 B1 DM1

A1 [5]


DM1 for use of perp grads DM1 for attempt at normal





© UCLES 2010

6 (a) (i)

2 4 6

−2

−1

1

2

3

x

y

(ii) 4 (b) (i) 5

(ii) 3

π2

B1

B1

B1

[3] B1

[1] B1

[1] B1

[1]

B1 for y = cos x

B1 for either a translation of

1

0 or 2

cycles B1 for correct curve

7 (i)

lgv 1 1.70 2.04 2.36

lgp 3.15 2.18 1.72 1.28

(ii) gradient = n = –1.37 (allow 1.32 to 1.42) (iii) p = 30 (allow 28 to 32)

M1 A2,1,0

[3]

M1 A1

[2]

M1 A1

[2]

M1 for attempt to take logs and plot graph –1 for each error either in table or on graph. M1 for use of gradient M1 for use of graph or their equation

8 (i)

− 21

916

(ii)

−

−

− 41

32

38

1

(iii) X = AB

=

−

80

125

B1 B1

[2] B1 B1

[2] M1 A2,1,0

[3]

B1 at least 2 correct B1 all correct B1 for determinant B1 for matrix M1 for attempt at valid method –1 each error





© UCLES 2010

9 (i) 5 + 5 + 3θ + 8θ = 15.5 θ = 0.5 (ii) ½(3)2 θ : ½(8)2 θ – ½(3)2 θ = 9 : 55

M1, DM1 A1

[3] M1 DM1 DM1, A1

[4]

M1 for use of arc length DM1 for attempt to find perimeter M1 for a sector area M1 for attempt to find area of XABY M1 for attempt to obtain ratio

10 (i) 10C7 = 120 (ii) 6C5 × 4C2 = 36 (iii) Need (6C + 1M) + (5C + 2M) + (4C + 3M) 4 + (ii) + (6C4 × 4C3) = 100

B1 [1]

B1, B1

[2] M1 B1, B1 A1

[4]

B1 for 6C5 × 4C2, B1 for 36 M1 for a correct method B1 for 4, B1 for 60

11 (i) 48 = 12 ln (2t + 3) 2t + 3 = e4 t = 25.8 (ii) x = 12 ln (2t + 3)

32

24

+

=

t

v

when t = 1, v = 4.8

(iii) 2)32(

48

+

−=

t

a

when t = 1, a = –1.92

M1 DM1 A1

[3] B1

B1

B1 [3]

B1

√B1 B1

[3]

M1 for attempt to deal with logs DM1 for attempt to solve

B1 32

1

+t

B1 24

B1 for 4.8

B1 for 2)32(

1

+t

√B1 on ‘24’ B1 for –1.92





© UCLES 2010

12 EITHER

(i) y = 4 sin 2x + c

passes through

7,

4

π, c = 3

(ii) 5 = 4 sin 2x + 3 0.5 = sin 2x

x = 12

π5,

12

π

(iii) ∫12

12

5

π

π

4 sin 2x + 3dx

[ ]12

12

5

32cos2π

π

xx +−

= π + 2 3

Shaded area = π + 2 3 – 3

5π

(= 1.37)

M1 M1 A1

[3] M1 M1

A1 √A1

[4] M1 A1 DM1 M1

A1 [5]

M1 for attempt to integrate M1 for attempt to get c provided a function of sin 2x is used M1 for attempt to equate to 5 and solve M1 for a correct method to find x

√A1 on first solution M1 for attempt to integrate DM1 for correct use of limits M1 for area of rectangle

12 OR

(i) y = 2e3x – 12x + c Passes through (0, 1), so c = –1 (ii) 6e3x – 12 = 0

leading to x = 3

1ln 2 and y = 3 – 4 ln 2

(allow (0.231, 0.227)

(iii) 2

2

d

d

x

y = 18e3x, always +ve so min

(iv) at (0, 1), gradient = –6 tangent : y – 1 = –6(x – 0)

when y = 0, x = 6

1

M1, A1 M1, A1

[4] M1

A1, A1

[3] M1, A1

[2] M1 DM1 A1

[3]

M1 for attempt to integrate, condone omission of c

M1 for attempt to obtain c M1 for attempt to solve

M1 for a complete, correct method M1 for attempt to get equation of tangent at (0, 1) DM1 for substitution of y = 0





© UCLES 2010

1 Gradient –2 B1 y intercept 15 B1

1523

2+−= x

x

y M1

25152 xxy +−= A1√ [4]

2 (i) 40320 B1

(ii) ( )12345

45678

××××

××××

or !3!5

!8

×

M1

56 A1

(iii) uses 5, 4 and 3 only M1 60 A1 [5] 3 (i) f(1) = 0 M1 b = 5 – a only A1 (ii) Finds f(2) and f(3) M1 substitute b = 5 – a or a = 5 – b M1 2a – 22 or a – 11 (–2b – 12 or –b – 6) A1 both correct A1 [6] 4 (a) p2 + (2p)2 = 1 or cot x = 2 B1

5

1=p or 5 cosec cot1 cosec

22=⇒+= xxx B1

5 cosec =x or 5

1=p B1√

(b) cot2 x – tan2 x oe B1 cosec2 x – 1 – (sec2 x – 1) or other relevant use of Pythagoras M1

Correctly reaches conclusion xx

22cos

1

sin

1− A1 [6]

5 21

322−

+ xx B1+B1

23

162−

− x B1+B1 Equate to 0 and solve M1 4=x A1 144=y A1 [7]





© UCLES 2010

6 Eliminate x or y M1 4x2 + 4x – 15 = 0 or 4y2 – 28y + 33 = 0 A1 Factorise 3 term quadratic or use formula M1

2

3=x and

25− A1

2

11=y and

23 A1√

2244 + M1

32 or 24 or 5.66 A1 [7] 7 Midpoint (1, 8) B1

Gradient BC = 3

2 B1

Uses m1m2 = –1 and equation of perpendicular bisector M1

( )12

38 −−=− xy or 3x + 2y = 19 A1

Solve with y = 5 M1 D (3, 5) A1 Complete method for area M1 15 A1 [8]

8 (a) (i)

− 8102

1464 B1


4464

812 A1

(iii) Matrix multiplication M1

− 80622

1815 A1

(b) (i)

−

−=−

27

16

5

1C

1 or B1+B1

(ii) X = DC–1 M1

− 05

23

5

1 or

− 01

4.06.0 A1 [9]





© UCLES 2010

9 (i) 2

π=t B1

∫ += ctktv 2cosd only M1

t2cos2− A1 Uses both limits M1 4 m A1

(ii) tkt

va 2cos

d

d== only M1

t2cos8 A1 7.68 ms–2 A1 [8]

10 (i)

=

24

20OP B1

=

24

7PL or

−

−=

24

7LP B1

2524722=+ M1A1

−

−=

t

tPL

636

823 oe B1

(ii) ( ) ( ) 222

25636823 =−+− tt M1

Solve 3 term quadratic ( )[ ]01281002

=+− tt M1 6 – 2 = 4 hours A1 [8]





© UCLES 2010

11E (i) Sector angle = 1.2π B1 12=OD B1

π8.0cos6122612222

××−+=AD M1 2.17=AD A1 Uses ( ) ( )ππ 2.72.16 =×=s (or 22.6) M1

Complete plan ( )6++ θrAD or ( )62.72.17 ++ π M1 45.8 A1

(ii) π8.0sin1262

1 ××=∆ AOD M1

21.2 A1

Uses ( )π2.162

1 2××=A M1

21.6π or 67.8 or 67.9 A1 89.0 or 89 A1 [12]

11O (i) 8123d

d 2+−= xx

x

y M1A1

gradient tangent = –1 A1 tangent y – 8 = –1(x – 1) or x + y = 9 A1 Valid method (e.g. substitute x = 4 in both) M1 Valid conclusion (e.g. y = 5 in both) A1

(ii) xxxx

xy 5424

d 23

4

++−=∫ M1A1

Uses limits F(4) – F(1) M1 12.75 A1 Subtracts from area of trapezium (19.5) B1 6.75 A1 [12]





© UCLES 2010

1 ( )2

5 12

2 2

x

x c

x

+ + +

−

oe B1 + B1 + B1 [3]

2 (a)

B1

(b) YX ∪' , )''( YX ∩ , )(' YXX ∩∪ , )'( YXY ∪∪ , or )''( YXY ∩∪ oe B1 (c) 18 + 16 + 2 = 30 + x M1 6 A1 [4]

3 2

π4d

dr

r

V= B1

r = 6 B1

Uses π

1π

3

4

d

d

d

d 3×

= r

rt

V M1

144 A1 [4]

4 (i) Evidence of rationalising 2

16 or

37

1 or

67

16 M1

21

68oe A1

(ii) ( )22

37

2

16+

M1

275 A1 115 A1 [5]

A B

C





© UCLES 2010

5 Search for first root or factor M1

x = –2 or 2

1 or 3 or (x + 2) or (x – 3) or (2x – 1) A1

Attempt to factorise cubic M1 (x + 2)(2x

2 – 7x + 3) or (x – 3)(2x

2 + 3x – 2) A1 or (2x – 1)(x2 – x – 6) Solve 3 term quadratic M1

x = –2 and 2

1 and 3 A1 [6]

6 (i) A = xy M1 A = 12x – 2x

2 oe A1

(ii) xx

A412

d

d−= B1√

equate to 0 and solve M1 x = 3 A1√

(iii) A = 18 A1 Completely correct method and maximum B1√ [7]

7 (i) Idea of modulus correct M1 Shape and position completely correct A1 (0, 9) (–3, 0) indicated on graph A1 (ii) Straight line with +ve gradient and +ve y intercept, correct position B1 (iii) 3x + 9 = x + 6 ⇒ x = –1.5 B1 Solve –(3x + 9) = (x + 6) or (3x + 9)2 = (x + 6) 2 M1 x = –3.75 A1 [7]

8 (a) (i) 1 – 21x + 189x2 – 945x

3 B3, 2, 1, 0 (ii) 2 × (189) and 5 × (–945) M1 –4347 A1√

(b) Identifies relevant (x2)3

6

2

x B1

Multiplies by 84 M1 5376 A1 [8]





© UCLES 2010

9 (i) ( ) ( ) 41242

1124

d

d2

12

1

×+=+

−

xx

x

or ( ) ( ) 41242

1124

d

d2

32

1

×+−

=+

−−

xx

x

B1

Uses quotient rule or product rule M1

( ) ( )( )

1 12 24 12 2 2 4 12

4 12

x x x

x

−

+ − + +

+

or ( ) ( )( )31

2 24 12 2 2 4 12x x x

− −

+ − + + A1

Express with common denominator of ( )4 12n

x + M1

( )

( )32

2 4

4 12

x

x

+

+

or 2k = A1

(ii) ( )2 1

4 12

x

kx

+

×

+

M1

uses both limits correctly on ( )2

4 12

C x

x

× +

+

M1

9

16 oe A1 [8]

10 (a) (i) 2log log

p pX Y− or 8log

pp M1

8 A1

(ii) log

log

p

p

X

Y A1

1.5 A1 (b) (i) 2

log 32 3= or 5

2 3z−

= or 2 2

log 32 log 3+ B1

52 3× or

2log 96 M1

96 A1

(ii) ( )92512 2= or ( )

92512 4= or

log 512

log 4x = B1

944 B1 [9]

11 (a) 0.85 B1 Subtract from π M1 2.29 A1





© UCLES 2010

(b) 4 cos

6sinsin sin

yy

y y= + B1

Uses 2 2

sin 1 cosy y= − to reach quadratic in cos y M1 Solve 3 term quadratic DM1

2 1cos and cos3 2

y y= = − A1

48.2 and 120 A1 311.8 and 240 A1√ [9] 12E (i) Rearranges to ( )2

0ax bx c+ + = and uses b2 * 4ac M1

k

2 – 16k + 48 * 0 A1 Solve 3 term quadratic M1 4 and 12k = A1 4 < < 12k A1

(ii) 2

5 152

2 4x

+ +

or 2a = , 5

2b = , 15 or 3.75

4c = B1 + B1 + B1

(iii) minimum 15 or 3.754

B1√

54

x−

= or 1.25− B1√ [10]

12O (i) (f) 1≥ B1

(ii) gf(3)=g(10) or ( )2gf ( ) 2 1 5x x= + − M1

15 A1 (iii) g

–1(15) = 10 B1 Finds fg(x) M1

( )2 22 5 1 or 4 20 26x x x− + − + A1

Solves ( )2

2 5 1 10x − + = or 24 20 16 0x x− + = M1

1 and 4 A1 (iv) complete, labelled exponential curve B1 complete logarithmic curve not cutting exponential curve B1 [10]






1 ( ) ( )

( )( )

1 cos 1 cos

1 cos 1 cos

θ θ

θ θ

+ + −

+ −

= 2

2

1 cos θ−

= 2

2

sin θ

= 2 cosec2 θ

M1

M1 A1

[3]

M1 for attempt to deal with fractions

M1 for attempt at simplification and use of 2 2

1 cos sinθ θ− = in denominator

2 3lg lg1000ab −

= 3

lg1000

ab

B1B1 B1

[3]

B1 for 3lg ab , B1 for lg1000

3 (a) (i)

(ii)

(iii)

(b) n(P) = 3

B1 B1 B1

B1

[4]

B1 for each region shaded correctly

4 (a) Powers of 2: ( ) ( )4 3 2 3 2x x− =

or equivalent for powers of 4, 8 or 16

4

3x = , allow 1.33

(b) p = 1, q = –5

4

M1

A1

A1

B1B1 [5]

M1 for powers of 2, 4, 8 or 16

A1 for all powers correct

A

C

B


http://www.xtremepapers.com




5 (i)

45 90 135 180

−2

−1

1

2

3

4

x

y

(ii)

45 90 135 180

−2

−1

1

2

3

4

x

y

(iii) 5

B1 B1 B1

√B1

√B1

[5]

B1 for shape B1 for 1 cycle between 4 and –2 B1 all correct √B1 for modulus of (i) √ on their graph

6 (i) 2 23 2 20 20x x x= − + − and verification.

Or ( )2

2 0x − = , x = 2

(ii) for OA, d

2d

yx

x=

when x = 2, grad = 4

for other curve , d 4 20

d 3 3

yx

x= − +

when x = 2, grad = 4

Or 4 20

23 3

x x= − +

leading to x = 2 (iii) tangent ( )4 4 2y x− = −

B1

B1

B1

M1

A1

M1 M1 A1

B1 [5]

substitution of x = 2

B1 for solution of equation

B1 for grad at A from OA

M1 for attempt to differentiate the other curve and substitute x = 2 M1 for differentiation of both M1 for equating and attempt to solve






7 Grad of AB = –2, perp grad = 1

2

Eqn of perp ( )1

15 22

y x− = +

C(0, 16)

Area = 1

125 52

= 12.5

(or ( )2 3 0 21 1

38 1315 5 16 152 2

− −

= − )

B1M1

M1 A1 M1

A1 [6]

B1 for grad AB M1 use of

1 21mm = −

M1 for correct attempt to find the equation of AC and hence to find C

M1 for a valid method to find area

8 (a) AB, AC

(b) Either: Y = X

+−

+−

637

6312

yx

yx

+−

+−=

482471

361852

4

32

yx

yx

yx

yx

leading to y = 12 and x = 4 Or

−

−

− 23

45

1210

1Y =

+−

+−

637

6312

yx

yx

+−

+−=

−−

−−

637

6312

4

6

2

1

yx

yx

yx

yx

leading to y = 12 and x = 4

B2,1,0

M1

M1A1 M1 A1A1

B1 B1 M1 M1

A1A1 [8]

–1 each one incorrect or extra M1 for pre-multiplying by X M1 for multiplication of matrices

A1 for correct product M1 for equating like elements

B1 for determinant for inverse B1 for ‘matrix part’ of inverse M1 for multiplication of matrices M1 for equating like elements

9 (i) 5 (ii) 20sin 4a t= − sin 4 0.5t = −

7π

24t = ( allow 0.916)

(iii) 5sin 4 ( )

4s t c= +

When t = 5, s = 1.14

B1

M1A1 DM1

A1

M1A1 DM1 A1

[9]

M1 for attempt to differentiate DM1 for attempt to solve for 4t

M1 for attempt to integrate DM1 for substitution of t in radians






10 (a) (i) 2 3a= − , a = 5 (ii)

1 35

x

y e x c−

= − − +

10c =

1 35 10

x

y e x−

= − − +

(b) (i) ( )4

31 3

7 87 4

x +

(ii) ( )8

4

3

0

37 8

28x

+

=180

7 or 25.7

B1

√B1

B1B1

M1 A1

B1B1

M1A1

[10]

√B1for first term using their a B1 for 3

x− , B1 for c+

M1 for attempt to find c

B1 for 1

7, B1 for ( )

4

33

7 84

x +

M1 for use of limits

11 (i) ( )2

2 2 3x − −

(ii) 2x ≥ or equivalent (b) (i) ( )g 4x ≥ , ( )-1

h 0x ≥

(ii) Correct sketch

(iii) ( )g 4 25 85x − =

( )2

4 25 4 85x − + =

17

2x = , x = 4

Discarding x = 4

B1B1

√B1

B1B1

B1 B1 B1

M1

DM1

A1

B1 [12]

B1 for –2, B1 for –3

√ on their ‘–2’

B1 for each

B1 for g(x) B1 for g-1(x) B1 for idea of symmetry

M1 for correct order

DM1 for attempt to solve

A1 for both

B1 for discarding x = 4






12 EITHER

(i) 2d

3 14 8d

yx x

x= − +

When d

0d

y

x= ,

2,4

3x =

2

2

d6 14

d

yx

x= − ,

2

3x = max, 4x = min

(ii) Use of d d d

d d d

y y z

t z t= × , leading to

d

d

y

t=

5

6− allow –0.833

(iii) Use of d d d

d d d

y y x

t x t= × leading to

d 5

d 48

x

t

=

M1A1

M1A1

M1

A1

M1

A1

M1

√A1 [10]


M1 for attempt to equate to zero and solve

M1 for attempt to differentiate (or other valid method)

A1 correct from correct working for both



ft on d

d

y

t

12 OR

(i) 22 72x y = , 2

4 6A x xy= +

leading to given answer

(ii) 2

d 2168

d

Ax

x x= −

When d

0d

A

x= , 3

27x = = 3

Dimensions are 3 by 6 by 4

(iii) Use of d

d

AA x

x∂ ≈ ×∂ leading to

38A p∂ = − , decrease

B1M1

A1

M1A1

M1

A1

M1

A1√A1 [10]

B1 for 2 2

150x y = ,

M1 for 24 6A x xy= +


M1 for attempt to equate to zero and solve

A1 for dimensions

M1 for attempt to use small changes

A1 for – 38 p, √A1 on their A∂






1 ( ) ( )2 22 10 5 0x k x k+ + + + =

( ) ( )2 22 10 4 5k k+ = +

2k = −

(or ( ) ( )d2 2 10 , 5

d

yx k x k

x= + + = − +

( ) ( )( )2 20 5 2 10 5 5k k k k= + − + + + +

leading to 2k = − )

(or ( ) ( )2 2 22 10 5x A x k x k+ = + + + +

( ) 2 25 , 5A k A k= + = +

( )2 25 5,k k+ = + leading to 2k = − )

(or by completing the square

( )( ) ( ) ( )2 2 25 5 5y x k k k= + + − + + +

( )2 25 5k k+ = +

leading to 2k = − )

M1

M1

A1 [3]

M1 M1

A1

M1

M1

A1

M1 M1

A1

M1 for equating to zero and use of 2

4b ac= M1 for solution

M1 for differentiation and attempt to equate to zero. M1 for attempt to substitute in for x in terms of k, for y = 0 and for attempt at solution.

M1 for approach

M1 for equating and attempt at solution

M1 for approach M1 for equating last 2 terms to zero and attempt to solve

2 2

5 2 3 4

3 22 (10)

9

aC a C=

1

6a =

B1B1

M1

A1

[4]

B1 for 5 2 3

32C a , B1 for

2

4

2

9

aC

M1 for a relationship between the 2 coefficients and attempt to solve

3 (a) k = 2, m = 3, p = 1 (b) (i) 5

(ii) 2π

3

B3 B1 B1 [5]

B1 for each

There must be evidence of working without a

calculator in all parts

4 (i) ( )( )

( )( )

4 2 1 2

2 2

1 2 1 2

+ −

=

+ −

(ii) Area = ( ) ( )14 2 2 1 2

2× + × +

= 4 3 2+ (iii) Area = 2

AC

( ) ( )2 2

4 2 2 1 2= + + +

= 27 18 2+

M1A1

M1 A1

M1

A1 [6]

M1 for attempt to rationalise and attempt to expand

M1 for attempt at area using surd form and attempt to expand

M1 for attempt at 2AC or AC in surd

form, with attempt to expand






5 (i) 1 1 1

2 5 10 48 4 2

− + −

= 0

(ii) ( )( )22 1 2 4x x x− − +

For ( )22 4x x− + , ‘ 2

4b ac< ’

so only one real root of x = 0.5

M1 A1

M1A1

M1

A1 [6]

M1 for substitution of x = 0.5 or attempt at long division

M1 attempt to obtain quadratic factor A1 for correct quadratic factor M1 for correct use of discriminant or solution of quadratic equation = 0

A1, all correct with statement of root.

6 (i) ( )1lg 3 5

5y x− = −

(ii) Either 1

5b =

1

25

10

x

y

+

= ,

1

2510 10x

= 100a = Or lg lg lg10bx

y a= +

lg lgy a bx= + , lg 2a =

100a =

1

5b =

Or ( )5310 10

b

a=

( )15510 10

b

a=

1

5b = , 100a =

B1M1 A1

B1

M1

A1 [6]

M1

A1 B1 M1

B1, A1

B1 for gradient, M1 for use of straight line equation

B1 for 1

5b =

M1 for use of powers of 10 correctly to obtain a A1 for a M1 for use of logarithms correctly to obtain a

A1 for a

B1 for 1

5b =

M1 for simultaneous equations involving powers of 10

B1 for 1

5b = , A1 for 100a =

7 (i) 14

63003C =

(ii) 8 6

4 2C C×

1050=

(iii) 8 8

6 56 364C C+ =

B1

B1B1

B1

B1B1

B1 [7]

B1 for 84

C or 6

2C

B1 for × by 62

C or 84

C B1 for 1050 B1 for 8

6C or equivalent

B1 for 8

56 C or equivalent

B1 for 364






8 (i)

−2 −1 1 2 3 4

−10

−5

x

y

(ii) ( )1, 9−

(iii)

−2 −1 1 2 3 4

5

10

x

y

B1 B1 B1 B1

B1

√B1 B1

[7]

B1 for x = – 0.5 B1 for x = 2.5 B1 for y = –5 B1 for shape

√B1 on shape from (i) B1 for a completely correct sketch

9 (i) : 2 π3

OBAθ

θ

∆ + =

(ii) 3π

9π5

r= ×

r = 15

(iii) Area = 2 21 3π 1 3π15 15 sin

2 5 2 5

× × − × ×

=105

M1 A1

M1

A1

M1M1

A1 [7]

M1 for using angles in an isosceles triangle

M1 for use of s rθ=

M1 for use of 21

2r θ or

1

2rs

M1 for use of 21sin

2r θ or other correct

method

10 (i) 29 5 24

13 6 7

− =

− − −

Magnitude = 25, unit vector 241

725

−

(ii) 2 3AC AB=

uuur uuur

or 2 2 3AB BC AB+ =

uuur uuur uuur

leading to

36

10.5AC

=

−

uuur

OC OA AC= +

uuur uuur uuur

or 2 2OB OA OC OB− = −

uuur uuur uuur uuur

leading to 41

16.5OC

=

−

uuur

(equivalent methods acceptable)

M1

M1 A1

M1

M1

A1

A1 [7]

M1 for subtraction

M1 for attempt to find magnitude of their vector

M1 for attempt to find ACuuur

– may be part of a larger method

M1 for attempt to find OCuuur

A1 for each






11 (i) 22cosec 5cosec 3 0x x− − =

( )( )2cosec 1 cosec 3 0θ θ+ − =

leading to sin x = 3

1, x = 19.5°, 160.5°

(ii) 5

tan 24

y =

2y = 51.34°, 231.34°

y = 25.7°, 115.7°

(iii) π 2π 4π

,6 3 3

z

+ =

2π π

3 6z = −

4π π

3 6

−

π 7π,

2 6z = allow 1.57, 3.67

M1A1

DM1 A1√A1

M1 M1

A1,√A1

M1

A1, A1 [12]

M1 for use of correct identity or attempt to get in terms of sin x

DM1 for attempt to solve

√ o

180 – their x M1 for attempt to get in terms of tan M1 for dealing correctly with double angle

√ 90° their y M1 for dealing with order correctly and attempt to solve

12 EITHER

(i) 2d9 4 5

d

yx x

x= + −

when x = –1, x

y

d

d = 0

tangent y = 5, A (0, 5) (ii) B (0, 1)

At B, d

5d

y

x= −

normal 1

15

y x− = C (–5, 0)

At D 1

1 55x + = , D (20, 5)

Area = 1

20 52× × ,

= 50

M1

DM1 A1

B1

M1A1

M1A1

M1

A1 [10]

M1 for differentiation and substitution of x = –1

DM1 for attempt at equation of tangent and coordinates of A

B1 for B

M1 for attempt at normal and C, must be from differentiation and using correct point

M1 for attempt to obtain D, equating normal and tangent equations

M1 for valid attempt at area






12 OR

2d

3 12 9d

yx x

x= − +

When x

y

d

d = 0, x = 1, 3

( )1,4P

Area = 8 – ∫3

1

3x – 6x

2 + 9x dx

=

34 2

3

1

98 2

4 2

x x

x

− − +

27 11

84 4

= − +

= 4

M1

M1 A1 A1 √B1M1

A2,1,0

DM1

A1 [10]

M1 for differentiation and equating to 0, can be using a product

M1 for attempt to solve A1 for both x values A1 for y coordinate √B1 on y coordinate for area of rectangle M1 for attempt to integrate

–1 each error

DM1 for application of limits






1 ( ) 320373252

+=+ B1

( )32

32

32

32037

−

−×

+

+ M1

3314 + A1 + A1√ [4]

2 (i) 220 or 8

1± B1

(ii)

–27.5 oe 16.5(x2) Correct method for collecting terms (66 + (i)) 38.5 oe

B1 B1 M1 A1√

[5]

3 AB = 6i + 24j (or AC = 4i + 16j) B1

ABOAOC

3

2+= ( )

++− jiji 246

3

24 M1

OC = 5i + 12j A1

22125 +=OC M1

13 A1 [5]

4

Eliminates y x2 + kx – 2x + 16 (= 0) Uses b2 – 4ac k2 – 4k – 60*0 or (k – 2)*± 8 k = –6 or 10 k < –6 or k > 10

M1 A1 M1 A1 A1 A1

[6]

5 (i)

(ii)

f(1) = 1 + 8 + p – 25 (= p – 16) f(–2) = –8 + 32 – 2p – 25 (= –2p – 1) p – 16 = 2p + 1 oe p = –17 Evaluates f(–3) or divides by (x + 3) to remainder 71 (= 20 – 3p)

B1 B1 M1 A1 M1 A1√

[6]

6 (a)

(i) Evidence of 8, 7, 6, 5, 4, 3, 2, 1 or 8! 40320 (ii) Evidence of 5! (120) or 4! (or 24) 2880

M1 A1 B1 B1

(b) )35()1(23

567=

××

××

and )10()1(2

45=

×

×

B1

Multiply 350

M1 A1

[7]

7 (i)

(ii)

m = 2.5 c = 2 lg y = 2.5lg x + 2 2 = lg100 or lg102 2.5lg x = lg x2.5 y = 100 x2.5 Solve 2.5lg x = lg3 or x2.5 = 3 correctly 1.55

B1 B1 M1 B1√ B1√ A1 M1 A1

[8]






8 (i)

(ii)

(iii)

70 39.7 55e–0.1t = 25 – 15 oe

B1 B1 B1

0.1t = ln

10

55 oe M1

17(.0)

A1

(iv)

t

T

d

d= ke–0.1t M1

k = –5.5 oe –1.11

A1 A1

[8]

9 (i) Either

B1

10 or 45 found Uses cosine rule D2 = 102 + 302 – 2 × 10 × 30 × cos60 or V2 = 152 + 452 – 2 × 15 × 45 × cos60 39.7 or 39.8 or 715

B1 M1 A1

A1

(ii) VD /

60sin

15/10

sin=

α

(or D

60sin

30

sin=

β and use β ) M1

α = 19.1 or β = 101 251

A1 A1√

[8]

9 (i) Or

B1

10 Dsinα = 10sin60 and Dcosα = 25 or Vsinα = 15sin60 and Vcosα = 37.5 Solve equations V = 39.7 or 39.8

B1 B1 M1 A1

(ii) 25

60sin10tan =α M1

α = 19.1 251

A1 A1√

[8]

30/45

D/V 10/15

60

β

α

30/45

10/15

60

D/V β

α






10 (i)

tanx = –1.33 126.9 306.9

B1 B1 B1√

(ii) 13cos

6cos6 =+

yy or 13sec6

sec

6=+ y

y B1

(iii)

Forms quadratic in cosy or secy (6cos2y – 13cosy + 6 = 0) Solve 3 term quadratic 48.2 311.8 2z – 3 = 0.775 (or 2.37) radians Solves for z using radians 1.89 and 2.68

M1 M1 A1 A1√ B1 M1 A1

[11]

11

(i)

EITHER

9.0cos

12=OA M1

(ii)

AC = 19.3 – 12 = 7.3 Complete method for major arc (2π – 1.8) × 12 53.8 AB = 2 × 12tan0.9 or cosine rule 30.2 Complete plan (53.8 + 30.2 + 2 × 7.3) 98.6

A1 M1 A1 M1 A1 M1 A1

(iii) Complete method for major sector ( )8.12122

1 2−×× π M1

8.1sin3.19

2

1 2×× or 122.30

2

1×× M1

323 or 181 504

A1 A1

[12]

11

(i)

OR

Uses product rule

+= xxx

x

ycossin

d

d M1

At x =

2

π

gradient = 1 A1

Uses m1m2 = –1 M1

Correctly reaches conclusion. e.g.

−−=−2

12

ππ

xy with y = 0 A1

(ii) xxxxxxx cosdsindcos =−∫∫

xxxxxxx cosdcosdsin −= ∫∫

M1

M1

sinx – xcosx

A1

(iii) Uses limits of π and 2

π

M1

2.14 or π – 1 A1

Area triangle = 222

1 ππ

×× or 8

2π

or 1.23 B1

Subtracts area of triangle

0.908 (allow 0.906 or 0.907) or 0.91 or π − 1 – 8

2π

M1

A1

[12]






1 (i) d

3cos3d

yx

x= B1

(ii) Uses xx

yy ∂×

=∂

d

dat attempt with

9x

π

= and x p∂ = . M1

1.5p A1√ on k [3]

2 (a) (i) n(E) = 72 or ( )n W B R 72∪ ∪ = B1

(ii) R ⊂ W or R W R∩ = or R W W∪ = or /R W∩ = ∅ B1

(b) (i)

B1

(ii) ( )′∩′ YX or ( ) YYX ′∪∩ or ( ) XYX ∪′

∪ or ( ) XYX ∪′∩′ B1 [4]

3 (i)

−=+

62

13IA B1

−=+

32

16

20

1)( 1-IA B1 + B1√

(ii)

=+=

4

14)( 1-IAX evaluated to matrix with 2 entries M1

4

2

A1 [5]






4 (a) Expresses with common denominator M1

2

2sin

cos

x

x

A1

sin 1

2cos cos

x

x x

=2 tan secx x or 1

2 tancos

x

x

= 2 tan secx x A1 ag

(b) 2

cos 1x p= − B1

1

cosec2sin 2

x

x

= B1

2

1

2 1p p−

B1 [6]

5 (i) Uses product rule M1

2 15

2 15

x

x

x

+ +

+

A1

( )

( )3 5

3

2 15

xk

x

+

⇒ =

+

A1

(ii) 1

2 15x xk

+ M1

Uses limits on 2 15Cx x + M1

34

3 A1 [6]

6 Eliminates y (or x) M1 2

3 10 0x x+ − = (or 227 72 0y y+ + = ) oe A1

Factorises 3 term quadratic or solves using formula M1 5 and 2x = − (or 24 and -3y = − ) A1

24 and -3y = − (or 5 and 2x = − ) A1√ Uses Pythagoras M1

22.1 or 490 or 7 10 A1 [7]






7 (i) 3.75 oe B1

(ii) 3)43(d

d

+==

t

k

t

va M1

k = –360 oe A1 –0.36 oe A1√

(iii) ( )3 4

ks c

t= +

+

M1

20k = − oe A1 Substitutes t = 0, s = 0 into k(3t + 4)n M1

20

5

3 4t

−+

or 15

3 4

t

t +

A1 [8]

8 (a) (i) log3 log 200x = or 3

log 200 B1

4.82 B1

(ii) 2 log 25= or 22 log5= B1

( ) ( )5 40

log 5 40 log 2 log2

yy y

y

++ − + =

+ oe M1

Deals with logs correctly and solves DM1 0.5y = − A1

(b) a = 4 B1 b = –2 B1 c = 5 B1 [9] 9 2

ABm = B1

1 2

1mm = − M1

AC: ( )1

2 42

y x− = − − or 1

42

y x= − + or 2 8x y+ = A1

C(14, –3) A1 Midpoint M(6, 1) B1 + B1 D(14, 12) B1 Complete method for area M1 150 √ on

Dy A1√ [9]






10 (a) (i) a = 50 B1

b = –2 OR ( )2

50 2 4x− − B1

c = –4 B1

(ii) (4, 50) B1√

(iii) Correct shape B1 Maximum and y intercept in correct quadrant, B1

(b) (i) Method for inverse M1

7 3x + − A1 (ii) g(0) = 2 B1

Solves g–1(x) = 2 or solves x = g(g(0)) = g(2) M1 18 A1 [11]

-20

-10

0

10

20

30

40

50

60

-2 3 8






11 EITHER

(a) cos 0.5x = B1

Uses Pythagoras (to find 3 ) M1

2

3sin =x A1

(b) (i) PS x y= + B1

60 3

2

x

y

−= B1

(ii) height = 3

2

x

B1

Substitutes height and y into ( )height

Area2

PS y= × + M1

or height into ( )height

Area 60 22

x= × −

Correctly reaches ( )23

302

x x− A1 ag

(iii) ( )d 3

30 2d 2

Ax

x= − B1

Equates to 0 and solves M1 15x = A1 Completely correct for method with 15x = leading to maximum B1 [12]

OR

(i) 3 22

28803

r r hπ π π= + B1

2

2880 2

3h r

r

= − or 3

2

8640 2

3

rh

r

−

= B1

(ii) 2

3 2A r rhπ π= + B1 substitute for h M1

correctly reaches 25 5760

3A r

r

π

π= + A1 ag

(iii) 2

d 10 5760

d 3

Ar

r r

π

π= − B1 + B1

equate to 0 and solve M1 12r = A1

(iv) substitute in formula for A DM1

720π A1

(v) Completely correct method with 12r = leading to minimum. B1 [12]




Page 1 Mark Scheme Syllabus IGCSE Examinations - November 2002 0606

1.

2.

3.

4sine+ 3cose=o => tane = -0.75

e = 143.1° or 323.1°

Complete elimination ofy (or x) - x2=4(mx-9) Use ofb2-4ac on his quadratic=O 16m2=144

m=±3

10t-t~5

Solo of t2-1 Ot+5=0 - t=S±--./20

Difference t2-tt = 2--./20 = 4--./5

4. ; j

5.

~~?P' I

---:-I jl

Areaundercurve=[ ] 1 - [] -t (4.701) Area of rectangle = 2( e+e- 1

) ( 6.172)

Req'd area= 1.47 or 1.48 (or 4e- 1)

Large matrix either 3x4 or 4x3

(

3 2 2

2 0 3

0 I 6 ~}oc[~ ~ ~] Displayed compatible for x, with row or col mat

Eg (20 30 15) X 1st oc lstx m]

Product eg (120 55 220 135) or [~~~] 580

Multiplied by the 3rd matrix => $27 60

Ml

AI AH

M1

Ml

AI AH

Ml

DMl Al

Ml Al

M1 AI

DMl Ml DM1 AI

BI

Bl--./

3

4

5

6

Ml AI

Ml Bl 6

Use oft=s+c (allow ±4/3 or+%)

Co For 180°+ his value and no other values.

y or x must go completely or m=Y2x.

quadratic must = 0

m=3 gets AI only for the- value from m2=k.

Setting quadratic to 0

Correct form of solution for Quad=O. Correct only - decimals ok here.

Difference between the 2 values Decimal check is not acceptable.

Knowing to integrate + attempt with "e" co

Value at I - Value at -1 in his integral Anywhere- numeric or in terms of e Subtraction of two areas - on first M. co.

Anywhere

Or for (30 40 50 80) x 2cnd. Order ok.

ili fur 2cnd X m) Order ok.

Must be compatible forM. co for A.

Must be compatible forM. co for B 1 - even if no matrices.



6.

7.

8.

9.

Mark Scheme Syllabus Page2 IGCSE Examinations - November 2002 0606

y = f 6(2x-3)-2dx = -3(2x-3)- 1 (+C) M1 A1

Passes through (3,5) :::::::> C = 6 M 1 A 1

On x-axis, y=O :::::::> x=l.75

(i) y=x3+x-1 :::::;> dy/dx=3x2+1 Puts dy/dx=O- realises there is no solution And therefore no max or min.

Realises the function is 1 to l and f-'exists.

(ii) f- 1(9) is value ofx such that x3+x-1=9 Search- finds x==2.

(i) ex=k ~ k(2k-1)==10 2k2-k-l0=0 :::::;> k=2.5 (or-2) Soln of ex = 2.5 :::::;> x= In2.5 = 0.92 ok

(ii) RHS = 2 = log525 LHS =logs [(8y-6)+(y-5)]

Soln of [(8y-6)+(y-5)] = 25 :::::::> y = 7

Eliminate x or y :::::;> 3x2-6x-24=0 or 2y2+6y-36=0 . ,

Solution ~~ ~ -+ (4,3)and(-2.-6)

Gradient of line joining = ~ (-t;-/,

Gradient of perpendicular= -%

Midpoint = (I, - i) Eqn of perp bisector y+i = - %(x- 1)

Ml Al 6

Ml DMl A1

AI--./

Ml Ml AI

7

Ml DMI

A1

Bl Bl M1 Al

Ml AI DMl AI

Ml

7

Ml M1Al

8

Must bt~ f2x-3)k + k -no other f(x). + 2 and k==-1. Uses both coordinates in an integral.

Puts y=O into his equation obtained by integration.

Knows to use calculus for M. Puts dy/dx==O + attempt to solve. Correct conclusion from 3x2+l=O only.

Realises link between no soln and 1: 1 -only from kx2+ 1.

Realises need to solve f(x)=9 Tries values forM. Correct answer.

Realisation that eqn is quadratic in ex Solution of quadratic

Co ln2.5 is enough. Ignore soln from -2.

co. co. Putting the two logs together. co.

Needs complete elimination. Correct equation . Method of solution=O. (needs 3 term) Co- all 4 values.

Uses (~(x1 +x2), ~(y1 +y2)) In any form - unsimplified or 6y+4x+5=0. (no mid-point or no perp- max 5/8)

10. (i) rp = t(20i+10j) + 50j

r0 = t(-1 Oi+30j) + 80i + 20j

(ii) t=2 ::::::> PQ = (60i+80j)-(40i+70j) = 20i+l0j

IPQI = 1005 = 22.4 km.

(iii) across 20t =-lOt +80 t = 2% up 10t+50==30t+20 t = ~ . Therefore no interception. p 1--J"' j~o.J

.l.;,

•ct:. Q

Ml DMl Al

M 1 for txvelocity vector in either Adding on constant vector in either For both expressions.

M 1 Knowing to subtract, with or without t=2.

Ml AI Use of Pythagoras. co.

Ml AI

8

Any valid method- could find t form one and substitute into the other.



Page 3 Mark Scheme Syllabus IGCSE Examinations- November 2002 0606

11. (i)

(ii)

dy (x + 2)2 -(2x -6)1 -= dx (x+ 2)2

10 =

(x+ 2)2 k= 10

Qis{0,-3) P) is (3,0)

gradient at x=3 is 2/5 gradient of normal = -5/2

Eqn of normal is y=-5/2(x-3)

R is (0, 7"h.) IRQI = IO"h.

12. EITHER _- .iL (i) OPQ bisects angle MON

(~~. '\ J anglePMO=anglePN0=90° ~ . ~~ r

·:'Eft"- R = OP + PQ = -.-+ r :,j smB ~

'C (ii) 9=30° => R = 3r

Total area= ~R2x(1t/3) = ~1tr2

Fraction with roses = (xr)+dxr) = ¥3

(iii) Arc MN = Sx21t/3 OM=ON=5+tan30 or v75

Perimeter= l0(~1t+03) = 27.8m

12. OR

I X

I 50 I 100

I 150

I 200

I 250

y/x 74 110 144 180 214

(i) Plotting y/x against x Accuracy and line

(ii) y=x(Ax+B) => y/x =Ax+ B => A = gradient, B=intercept

Intercept 38--+40 Gradient 0.68 to 0.72

(iii) Line ofy/x against x drawn on the graph. Value ofx at point of intersection makes the rectangle into a square.

(iv) As x-+oo, ratio of2 sides -+A or 1/A ie 0.7 or 1.43

I

M1 AI Must use correct formula for quotient or product. A mark for unsimplified.

Al co.

BIB!

Ml

Ml

AI Btv

9

Both these anywhere in the question.

Getting the perpendicular gradient- but must be using - 1 +dy/dx Correct unsimplified. -needs use of x=O

AI for 7K B I v for "his 7"h.- "his -3". Allow ±10.5

B 1 Used somewhere- on diagram ok B 1 Used somewhere -on diagram ok

M I A l OP needs to be trig function with sine

B I Co- in (ii) only or at end of (i)

M1 Use of "h.r9 with his Rand his 9

MIA 1 Needs xr and a ratio.

Ml Use ofs=r9 with his rand his 9 Ml For r+ tan(his9) or Pythagoras etc

Al l1

Ml AlAI

B1Bl BlBl

Ml A1 Bl

Blv 11

co

Knowing what to do AI accuracy. AI line.

Stating m=A and c=B. For numerical values of m and c only.

M 1 line of gradient 1 A 1 accurately drawn Square or sides of rectangle equal.

For his gradient or reciprocal of gradient.



Page 1

1 [4}

2 [4]

3 [5]

4 [6]

Mark Scheme Syllabus IGCSE Examinations- November 2002 0606

( 6 -3) Inverse = -7 4

1 x-

3

(X) 1(6 -3)(-7) (2) y = 3 -7 4 -16 = -5

+ 6 X 25 X X

= 64 + 192x + 240x2

Replace x by x - x2 coefficient of x2 = - 192 + 240 = 48

(i) Either or

(ii)

Simplify

1 1 2-.J3 2-../3 -= X =---p 2+.J3 2-$ 4-3

p _ _!_= 2 + .J3- (2-.J3)= 2.J3 p

Or P _ _!_ = 2 + .J3 _ 1 == 6 + 4 .J3 p 2+../3 2+.J3

Multiply by 2- .J3 2- .J3

Solving inequalities: A X< 3.5 B x2 -x-2=0 => (x-2)(x+1)=0 => x=-1,2

x2 -x-2>0 => x<-1, x>2

Required values -5 <X< -1

2 <X< 3.5

81 81

M1 A1

82, 1,0 r~l <.>..ll .. ·.~c•<Y.; r\ .01iolr-~"iJ"'Sl~i.\J,·~t2 N.~

M1 A1 C.S.Q.

M1

A1

M1A1.f

A1

M1 A1

81 M1 A1

A1 M1

A1



5 [6]

6 [6]

7 [6]

8 [6]

9 [7]

Page2 Mark Scheme IGCSE Examinations- November 2002

(a) Either sC3 = 5 x 4 x 3 1x2x3

Product = 10 x 6 = 60

Syllabus 0606

81

M1 A1

(b) Either, ending in 1 {or 3) ~ 2 x 5 x 4 or, ending in 5 (or 7) ~ 3 x 5 x 4 81

Adding all 4 cases ~ 40 + 40 + 60 + 60 = 200

(i) f (x) = - (x- 1 )(x - 2) {x - k)

f {3) = - 2 X 1 X (3 - k) : 8

(ii) f(-3) = -(-4)(-5)(-10) = 200

d . ) . - (X Sin X = Sin X + X COS X dx

(i)

(ii) J X COS X d X = X sin X - J sin X d X

J sin x dx = ::cos x

k = 7

X Sin X +COS X 7t 2- 1 = 0.571

(i)

Make In x the subject ~ In x = - % (x- 2) ~ line is y= 1 -x/2

(a) Correct combination of indices

Or

Sum = a + b2

22x+2 = 4 X 22x ~·l.."t. ~ ''""...:.._

(b)

: 10x = 4/5

M1 A1

M1 A1

M1 A1

M1 A1

M1 A1

M1

]>M.1

82,1,0

f\l

A\

M1 A1

M1

A1

A1

8 J.,l.O J" .... t .:t-·u·~_r. (~ot«.Ct~..-~c:.r ~.oar r-,..~-.... ~i·•'(.\ t·~.,;~ ...... )

M1 A1



10 [9]

11 [11]

12[10] Either

Or

Page3 Mark Scheme Syllabus IGCSE Examinations November 2002 0606

(i) AP = bl3 - a OM= al2 + bl2

(ii) OQ = A. (al2 + bl2)

(iii) OQ = OA + Jl AP = a + Jl (bl3 - a)

(iv) Comparing coefficients A. 12 = 1 - Jl and A./2 = Jl 13

Solving A.= Yz J.l=%

[V}t= o = 20 :::::> C = 20 [V}t=4 = 12-24+20 = 8

(ii)

AB = [ t = 16 - 48 + 80 = 48

(iii) v8 = 8 , Vc = 20 :::::> lac = (20 - 8} I 2 = 6

(iv) curve

straight line ~ t

A= 1tf2+1trl => 1 = (120-1tl)!1tr

60r- Y:z 1[ r dVIdr = 60-31CI12

1" = 0 when I = 40hr.,/,:= 3.57

Stationary value of V = 143 (1r,.2..75)

d2 VI dl = - 3 1t r < 0 for r > 0 :::::> maximum [ or any valid method ]

(i) dy I dx = x2 x 11x + 2x In x

At Q, y = 0 => In x = 0 :::::> x = 1 [dyldx]x=1 = 1 c.s.c

(ii) At P, dy I dx = 0 => x(1 + 2 In x} = 0 => In x = - Yz

=> x = e·% = 1 t..Je ( ..,._ o. &Ot.5'l (AG)

(iii) d2y I dx2 = d (x + 2x In x) I dx = 1 + 2 In x + (2x x1/x}

= 3 + 21n x

81 M1 A1

B1..J

M1 A1..J

M1

M1 A1

M1 A1

A1 A1

M1 A1..J

A1

M1A1/"

81

81J'

13 1t\ 1

M1 A1

81 M1 A1

A1

M1 A1

M1 A1

81 A1

M1 A1

A1

M1 A!

A1



IGCSE EXAMINATIONS – NOVEMBER 2003 0606 1


1. x + 3y = k and y ²=2x + 3 Elimination of x or y → y² + 6y −(2k+3)=0 or →x² − (2k + 18)x + (k² − 27) = 0 Uses b² − 4ac → k < −6

M1 A1 M1 A1 [4]

x or y must go completely, but allow for simple arithmetic or numeric slips co Any use of b²−4ac, even if =0 or >0 co

2. 8-x = 2-3x 4½x = 2x Attempts to link powers of 2 → x −3− (−3x) = 5 − (x) → x = 1.6 or 8/5 etc [ log 8-x = −3xlog2, log 4½x = xlog2 equate coefficients of log 2]

B1 B1 M1 A1

[4] [B1B1 M1A1]

Wherever used Needs to use xa

÷xb=xa-b

co

3. x³ +ax² +bx - 3 Puts x=3 → 27+9a+3b−3=0 Puts x=−2 → −8+4a−2b−3=15 (9a+3b=−24 and 4a−2b=26) Sim equations → a = 1 and b = −11

M1A1 M1A1 A1

[5]

Needs x=3 and =0 for M mark Needs x=−2 and =15 for M mark (A marks for unsimplified) co

4. (√3-√2)² = 5 − 2√6 or 5−2√2√3 Divides volume by length²

625

625

625

3324

+

+×

−

−

Denominator = 1 Numerator = 20√2−15√3+8√12-6√18 But √12 = 2√3 and √18 = 3√2 → 2√2 + √3

B1 M1 M1 M1 A1

[5]

Co anywhere V÷l² used × by denominator with sign changed Correct simplification somewhere with either of these co

5 y=0 when 3x + ¼π = π → x = ¼π ∫6sin(3x+π/4)dx = −6 cos (3x+π/4) ÷ 3 Between 0 and π/4 → 2 + √2 or 3.41

B1 M1 A2,1 DM1 A1

[6]

Co. Allow 45° Knows to integrate. Needs “cos”. All correct, including ÷3, ×6 and −ve Uses limits correctly – must use x=0 In any form – at least 3sf

6 Wind 50i− 70j V(still air) = 280i −40j (i) Resultant velocity = vair + w → 330i − 110j tan-1(110/330) = 18.4° → Bearing of Q from P = 108° (ii) Resultant speed = √(330²+110²) Time = 273 ÷ resultant speed = 47 minutes Scale drawings are ok.

M1 A1 DM1 A1 M1 A1√

[6]

Connecting two vectors (allow −) Co (Could get these 2 marks in (ii) ) For use of tangent (330/110 ok) co Use of Pythagoras with his components For 273 ÷ √(a²+b²)






7

×

×

60

50

50

40

2233

2345

5668

)5.02.06.0(

= ( )

×

60

50

50

40

4.42.59.53.7

or ( )

×

490

670

1220

5.02.06.0

→ $1111

B2,1,0 M1 A1 M1 B1 [6]

Wherever 3 matrices come – as row or column matrices – as 3 by 4 or 4 by 3 – independent of whether they are compatible for multiplication or not. Correct method for multiplying any 2 of the 3 - co for A mark. Correct method for remaining two. Co – even if from arithmetic.

8 (i) d/dx(lnx) = 1/x

2)32(

2)(ln1

)32(

+

×−×+

=

x

xx

x

dx

dy

(ii) δy = (dy/dx) × δx = 0.2p (iii) dy/dt = dy/dx × dx/dt

→ dx/dt = 0.6

B1 M1A1√ M1A1 M1 A1√

[7]

Anywhere, even if not used in “u/v” Uses correct formula. All ok. Could use product formula. A mark unsimplified. Allow if δy mixed with dy/dt. M mark given for algebraic dy/dx × p. Allow if dy/dt mixed with δy √ for 0.12 ÷ his dy/dx. Condone use of δx etc

9 (a) Uses sec²x = 1+tan²x → quad in sec or ×c² then uses s²+c²=1 → quad in cos → 4sec²x+8secx−5=0 → −5cos²x+8cosx+4=0 →secx = −2.5 (or0.5) or cosx=−0.4 (or2) → x = 113.6° or 246.4° (b) tan(2y+1) = 16/5 = 3.2 Basic angle associated with 3.2 = 1.27 Next angle = π + 1.27 and 2π + 1.27 (Value − 1) ÷ 2 → 3.28 (others are 0.134 and 1.705)

B1 M1 A1A1√ B1 M1 M1A1

[8]

Co. Sets to 0 and uses correct method for solution of a 3 term quadratic in sec or cos. A1 co. A1√ for 360°−”first ans” only. Anywhere (allow 72.6°) Realising the need to add on π and/or 2π Correct order used ie −1, then ÷2 for any correct value. Allow if all 3 values are given, providing none are over 4. (degrees – max 2/4 B1, M0, M1, A0)






10 f(x) = 5−3e½x

(i) Range is <5 (ii) 5−3e½x = 0 → e½x = 5/3 Logs or calculator → x = 1.02 (iii) (1.02, 0) and (0, 2) (iv) e½x = (5 − y)÷3 x/2 = ln[(5-y)/3] f-1(x) = 2ln[(5−x)/3]

B1 M1A1 B1 B1√ M1 M1 A1

[8]

Allow ≤ or < Normally 2,0 but if working shown, can get M1 if appropriate Shape in 1st quadrant. Both shown or implied by statement. Reasonable attempt e½x

as the subject. Using logs. All ok, including x, y interchanged.

11 (i) y=½x and y=3x-15 → C(6,3) OB=OC+CB → B(8,9) m of OC = ½, m of AD = −2 eqn of AD is y−6=−2(x−2) or y=−2x+10 Soln of y=½x and eqn of AD → D(4,2) (ii) Length OC = √45, OA = √40 Perimeter of OABC = 2(√45+√40) → 26.1

M1 A1 M1 A1√ M1 A1 M1A1 M1 M1 A1 [11]

Soln of simultaneous eqns Co (or step method if B done first) Vectors, step or soln of y=½x+5 and y=3x-15 From his C

use of m�m�=−1 (M0 if perp to y=3x) Co – unsimplified. Sol of simultaneous eqns. co. Once. Adding OA,AB,BC,CO Co.

12 EITHER (i) 125 = πr + 2x + 2(5r/4) → x = ½(125 − πr − 5r/2) h = 3r/4 Area of triangle = ½ × 2r × 3r/4 = 3r²/4 A = ½πr² + 2rx + ….. = 125r − ½πr² -7r²/4

(ii) dA/dr = 125 − πr –7r/2 Solved = 0 to give → r = 250 / (2π + 7) or 18.8

M1 A1 M1 M1 B1 A1 M1A1 DM1 A1 [10]

Attempt at 4/5 lengths. Co. Anywhere in the question –independent of any other working Use of ½bh with h as function of r Correct ½πr² + 2rx. Answer given – beware fortuitous ans. Any attempt to differentiate. Co. Setting his differential to 0. Any correct form.






12 OR (i) h / (12−r) = 30 / 12 → h = 5(12-r) / 2 Uses V=πr²h to give → V = π(30r² -5r³/2) (ii) dV/dr = π(60r − 15r²/2) = 0 when r = 8 → h = 10 → V = 640π or 2010 Volume of cone = ⅓π×12²×30 → 1440π or 4520 Ratio of 4 : 9 or 1 : 2.25 (3 sf)

M1 A1 M1 A1 M1A1 DM1 A1 M1 A1 [10]

Use of similar triangles – needs ¾ lengths correct. Correct in any form – needs h as subject Needs correct formula Beware fortuitous answers (AG) Any attempt to differentiate. co Setting his dV/dr to 0 + attempt. Correct to 3 or more sig figures Anywhere Exactly 4:9 or 2.25 to 3 sig figures

DM1 for quadratic equation (1) Formula. Sets the equation to 0 Formula must be correct and correctly used. Condone simple slips in sign.

(2) Factors Sets the equation to 0 Attempts to obtain brackets Solves each bracket to 0.






1 [4] Eliminate x or y ⇒ y2 – 8y + 15 = 0 x2 – 10x + 9 = 0 Factorise or formula ⇒ (1, 3) and (9, 5) Midpoint is (5, 4)

M1 DM1 A1 B1 √

2 [4] cos

( )

−=

−=

−

−−+

θ

θθ

θ

θθ

θ

θθθ

222sin1

cossin2

sin1

sin2cos

sin1

sin1sin1

M1 A1

Use of Pythagoras 2tan2cos

cossin2

2=⇒=⇒ kθ

θ

θθ

B1 A1

3 [4] log2x = 2log4x or log4 (x – 4) =

2

1 log2(x – 4)

B1

2log4x – log4 (x – 4) = 2 or log2x –2

1 log2(x – 4) = 2

Eliminate logs 164

2

=

−x

x or 4

4

=

−x

x

M1 A1

Solve for x ⇒ x = 8 A1

4 [4] (i)

B2 B1 B1

(ii)

(iii)

'' CBA ∩∩

)( CAB ∩∪

5 [5] (i)

(ii)

243x5 –405x4 +270x3 Coefficient of x4 = (–405 x 1) + (270 x 2) = 135

B1 B1 B1 M1 A1

6 [6] At B, v = 40 (e–t – 0.1) = 0 ⇒ e–t = 0.1 ⇒ t = ln 10 (=2.30) M1 A1 ( ) ( )∫ −−=−

−−

tttt

1.0e40d1.0e40 M1 A1

( ) ( )∫ ≈−=

−−

−−==

10log

0

8.2610ln94110

10ln

10

140AB

DM1 A1






7 [7] Dealing with elements

−

−

−

22

13

43

21and

M1

A–1= –

−

−

43

21

2

1 B–1 =

−

22

13

8

1

A1 A1

(i) C = B – 2A–1 =

−

−=

−

−+

− 75

13

43

21

32

12

M1 A1

(ii) D = B–1A =

=

−

614

59

8

1

13

24

22

13

8

1

M1 A1

8 [7] (i) 210

4321

7891010=

×××

×××

=

6!4!

!

M1 A1

(ii)

(iii)

No pink selected i.e. any 6 from (5 + 2) = 7 All selections contain at least 1 red

B1

No yellow selected i.e. any 6 from (3 + 5) = 28

8=

6!2!

!

M1 A1

At least 1 of each colour – 120 – (7 + 28) = 175

M1 A1

9 [8] (i) ( ) ( ) 42

13434

d

d2

1

××−=−−

xx

x

M1 A1

( ){ } ( ) 342

34

2323432

d

d−+

−+=−+ x

x

xxx

x

M1 A1 √

12

34

12=⇒

−

= k

x

x

A1

(ii) ( )∫ ×−+=

− 12

13432d

34xxx

x

x

M1 A1

( )∫ =−=

7

1 3

26585

2

1

A1

10 [10] (i) ∠AOB = 19.2 + 16 = 1.2 M1 A1

(ii) DE = 8 sin 1.2 ≈ 7.46 M1 A1

(iii) ∠DOE = sin–1 (7.46 ÷ 16) ≈ 0.485 (AG) M1 A1

(iv) Sector DOB =

2

1 x 162 x 0.485 = 62.08 M1

Length OE = √ (162 – 7.462) ≈ 14.2 M1

∆DOE =

2

1 x 7.46 x 14.2 ≈ 52.97 M1

Shaded area ≈ 9.1 – 9.3 (9.275) A1






11 [10] v 5 10 15 20 25 (i) Plotting lg R against lg v M1

R 32 96 180 290 420 Accuracy of points: Straight line A2, 1, 0

lg v 0.70 1.00 1.18 1.30 1.40 (ii) R =

βkv ⇒ lg R = lgk + β lg v B1

lg R 1.51 1.98 2.26 2.46 2.61 β = gradient ≈ 1.55 - 1.60 M1 A1

lg k = lg R intercept ≈ 0.4 ⇒ k ≈ 2.4 - 2.6 M1 A1 (iii) lg R = lg 75 ≈ 1.88 ⇒ from graph lg v ≈ 0.92 - 0.96 ⇒ v ≈ 8.3 - 9.1

[Or by solving e.g., 75 = 2.5v1.58 or 1.88 = 0.4 + 1.58 lg v]

M1 A1

12 EITHER

[11]

(i) gf(x) =

( )232

4

−− x

B1

Solve 2

34

4=

− x

[or solve fg(x) = 3

− x2

4 – 2 = 2]

M1

⇒ x = 2/3 A1

(ii) f(x) = g(x) ⇒ 3x – 2 =

x−2

4 ⇒ 3x2 – 8x + 8 = 0

Discriminant = 64 – 96 < 0 ⇒ No real roots M1 A1

(iii) f–1 : x a (x + 2) ÷ 3

B1

y = 4 / (2 – x) ⇒ x = 2 – 4/y ⇒ g–1 : x a 2 – 4/x

M1 A1

(iv)

Lines intersect at (1, 1)

B1 B1 B1






12 OR[11]

(i) 1 – x2 + 6x ≡ a – (x + b)2 = a –x2 – 2bx – b2 ⇒ a – b2 = 1 and – 2b = 6 M1 A1

[or 1 – x2 + 6x ≡ 1 – (x2 – 6x) ≡ 1 – {(x –3)2 –9} ] ⇒ b = –3, a = 10 A1 (ii) 1 – x2 + 6x ≡ 10 – (x – 3)2 ⇒ Maximum at (3, 10)

∴ Single-valued for x [ 3 and hence for x [ 4

M1 A1

(iii) y = 10 – (x – 3)2 ⇒ (x – 3)2 = 10 – y ⇒ x – 3 = √ (10 – x) M1 ⇒f–1 : x a 3 + √ (10 – x) A1 (iv) When x = 2, g(x) = 9 and when x = 7, g(x) = –6 B1 Range of g is –6 Y g Y 10 B1

(v)

B 2, 1, 0






1 OA = i + 9j OB = 5i – 3j OC = k(i + 3j)

AB or BA = 4i – 12j AC or CA = (k – 1)i + (3k – 9)j CB or BC = (5 – k)i – (3k + 3)j

M1 A1

For | one relevant vector |. Must be “–” Any 2 of these correct unsimplified.

Ratio of i to j is the same → k = 2 M1 A1 Using ratio idea (not DM). Co. If use of AB = a–b and correct answer obtained, allow full marks.

[or m = –3 = (3k + 3) ÷(k – 5) = (3k – 9) ÷ (k – 1) → M1 A1 → k = 2 M1 A1]

[4]

[or (1, 9) → (5, –3) y = –3x + 12 M1 A1 subs (k, 3k) or solve with y = 3x M1A1]

Drawing M2 A2

2 (i)

TDP ′∩′∩

or ( )′∪TDPI or TD ′∩′ or ( )′∪TD

B1 B1 Co.co

(ii)

TDP ∩′∩

B1 B1

[4]

Co.co

3 Change to powers of 2. 23x ÷ 2y = 26 → 3x – y = 6

M1 A1

Needs to try all terms as powers of 2 and have ±. Co

Change to powers of 3. 34x × 3-2y+2 = 34 → 4x – 2y + 2 = 4

M1 A1

Needs to try all terms as powers of 3 and have ±. Co with bracket sorted.

→ x = 5 and y = 9 A1

[5]

Co

4

Let OR = r ½r2 (4/3) – ½72 (4/3) = 48

M1 A1

Any use of ½r2θ for M mark. A1 unsimplified, but complete.

→ r = 11 A1 Co.

If x + 7 for r allow A1 for x = 4.

Perimeter = 11 × (4/3) + 7 × (4/3) + 2 (11 – 7)

M1 A1√

M1 for any use of s = rθ. A1√ unsimplified and √ on his r or x only.

→ 32 A1 [6] Co – allow anything rounding to 32.0






5 (a + x) (1 – 2nx + …) = 3 – 41x + bx2 term in x3 = nC2 (± 2x)2

B1

Wherever it comes.

→ a = 3 B1 Co – anywhere

1 – 2an = –41 → n = 7 M1 A1 Must use 2 terms.

Coeff of x2 is 3 × 84 – 1 × 14 M1 Must use sum of 2 products.

→ 238 A1

[6]

Co.

6 f(x) = 5 + 3cos4x

(i) a = 3, period = ½π B1 B1 Co. allow 90° for period.

(ii) max/min x = π/4 or 2π/4 or 3π/4 → max of 8 → min of 2

B1 B1

When “8” is used as stationary value. When “2” is used as stationary value.

(π/4, 2) (2π/4, 8) (3π/4, 2) B2, 1√

[6]

√ for 5 ± his “a”. [B0 if degrees here]

Ignore inclusion of max/min at 0 or π.

7 (a) 8 × 8! or

9

8 × 9! or 9! – 8!

→ 322 560

M1

A1 [2]

Must be nCr – knows what to do. Ans only is ok for 2 marks.

(b) 2G, 1B 5C2 × 3C1 = 10 × 3 = 30 3G, 0B 5C3 = 10

M1 A1 B1

Needs to be a product of nCr’s. Co. Anywhere.

total = sum of these = 40 A1

[4]

Co.

8 (i) y = (3x + 11)/(x – 3)

Makes x the subject.

f -1(x) = (3x + 11)/(x – 3)

f and f -1 are the same functions.

M1

A1

Good algebra in making x the subject.

→ Graph has y = x as line of symmetry.

B1

[3]

Co accept any mention of y = x.

(ii) g(x) = ½(x – 3) g -1(x) = 2x + 3 → 2x + 3 = (3x + 11)/(x – 3) → 2x2 – 6x – 20 = 0 → x = –2 or 5

B1 M1 A1

[3]

Anywhere. Algebra must lead to quadratic. Co.

(iii) gf(x) = –2 → f(x) = g -1 (–2) → x = –2

B1

[1]

However obtained.






9 (a) s2 = 3c2 + 4s Use of s2 + c2 = 1 → 4s2 – 4s – 3 = 0 → s = –½ or 3/2 → x = 210° and 330°

M1 DM1 A1 A1√

[4]

Used to eliminate cos completely. Scheme for quadratic. Co. √ for 2nd value from incorrect sine. (A1√ not given for extra values in the range, but could be given if soln of quadratic led to 2 values of sine<1.)

(b) cot = 1/tan used tan 2y = 4 2y = 1.326 → y = 0.66 or 2y = π + 1.326 or 2π + 1.326

M1 A1

Use of cot = 1/tan even if “2” removed incorrectly. Not for tan and 2y split. Co (must be radians) – not for 0.67.

→ y = 2.23 → y = 3.80 or 3.81

A1√ A1√

[4]

For (i) + ½π For (i) + π or (ii) + ½π

[S–1 for extra values in the range] [sc All answers in degrees B1.]

10 y = x3lnx

(i) dy/dx = 3x2lnx + x3 (1/x)

= 3x2lnx + x2

M1

A1 [2]

M1 correct “uv”. A1 ok unsimplified.

(ii) dy/dx = 0 lnx = –⅓ M1 A1

[2]

Not DM – setting his dy/dx to 0 + attempt to solve.

(ii) δy = dy/dx × δx = (e2 + 3e2)p = 4e2p or 29.6p

M1 A1

[2]

Use of small increases. Allow for use of dy/dt. ∆x = p essential for M mark. Alg expression with “p” ok for M1.

(iii) d/dx (x3lnx) = x2 + 3x2 lnx M1 A1 ∫ is reverse of diff used. A1 needs ⅓x3

Integrating → x3lnx = ⅓x3 + ∫3x2lnxdx

∫x2lnxdx = ⅓(x3lnx – ⅓x3)

A1

[3]

co

Integration by parts ok. M1 A1 A1.

11 4y = 3x + 1 and xy = 28x – 27y Sim equations. → x2 – 10x + 9 = 0 or y2 – 8y + 7 = 0 → (9, 7) [(1, 1) was given]

M1 DM1 A1

Complete elimination of x or y Soln of quadratic (by scheme) Co for (9, 7)

P(1, 1), Q(9, 7) → gradient of PQ = ¾ Gradient of perp bisector is –4/3 M (mid-point of PQ) = (5, 4)

M1 M1

Use of m1m2 = –1 with his PQ Use of (½(x1 + x2), ½(y1 + y2))

Eqn of perp bis. y−4 = – 4/3(x – 5)3y + 4x = 32 meets y = 4x at R(2, 8) Area of ∆PQ = ½ × PQ × MR = ½ × 10 × 5 (or matrix method) → 25

A1 M1 M1

A1 [9]

Co – unsimplified ok. Simulataneous eqns. Must be with a perp Any correct method. Co.






12

(a) (i)

EITHER

N = 20 000e -0.05n n = 10, N = 12 130 or more places.

B1 [1]

Co

(ii) 2000 = 20 000e -0.05n e -0.05n = 0.1

Take logs n = 45.1 → 2006

M1

M1 A1

[3]

Isolating exponential – or taking logs to get 3 terms.

Taking logs. Co. needs 2006, not 2005.

(b) Put y = 3x 3x + 1 = 3y or 3x – 1 = ⅓y 3y – 2 = 8y/3 → y = 6

M1 A1 A1 A1

Used. For each expression. Co.

3x = 6, xlg3 = lg6 → x = 1.63 [or ÷ by 3x – 1 M1 A1 3x – 1 = 2 M1 A1] [or ÷ by 3x + 1 M1 A1 3x + 1 = 18 M1 A1]

M1 A1

[6]

Taking logs for his 3x. co.

12 OR

y = e 2

x

+ 3e 2

x−

(i) dy/dx = 22 e

2

3e

2

1xx −

−

= 0 when ex = 3 y = √3 + 3 ÷ √3 = 2√3

B1 B1

M1 A1

[4]

Anywhere –

Setting his dy/dx to 0 and reasonable attempt at making ex the subject. Co. Decimal check – A0.

(ii) d2y/dx2 = 0e

4

3e

4

122

>+

−xx

, MIN M1 A1

[2]

M1 Reasonable attempt by any method. A1 Correct deduction but needs second differential correct.

(iii)

∫

+

−

22 e3e

xx

dx = 2 22 e6e

xx −

−

[ ] at 1 – [ ] at 0 = 4 + 2√e - 6√e = 3.66

M1 A1

M1 A1

[4]

Knowing to integrate for area + any attempt with exponentials. A1 co. DM0 if [ ] at 0 is ignored. Co.

DM1 for quadratic equation.

Formula Equation must be set to 0. Formula must be correct and correctly used, but allow for numerical and algebraic errors.

Brackets Equation must be set to 0. Must be an attempt to get two linear brackets. Each bracket must then be equated to 0 and solved.






1 [4]

−=−

25

341

A × 23

1

−=

−

−

−=

2

1

13

4

25

34

23

1

y

x

B1 B1

M1 A1

2 [4] 34

34

34

34

13−=

−

−×

+

or 3819

169

34

132

+=

+

( ) 3819342

−=− or 3819

3819

3819

3819

169−=

−

−×

+

OR ( )( )

=+

=+⇒=++

1692419

081916938193

ba

abba solve M1

−=

=⇒

8

19

b

aA1

M1 A1

M1 A1

3 [5] Integrate –3/2 cos 2x + 4 sin x

[ ] ( ) 75.15.52/

0=−−=

π

Must use both limits properly, not assume cos0 = 0, not use

2π degrees.

M1 A1 A1

M1 A1

4 [5] Eliminate y → (x + 2)2 + (x + k)2 or x → x2 + (y – 2 + k)2

2x2 + (4 + 2k)x + (2 + k2) = 0 or 2y2 + (2k – 4)y + (k2 – 4k + 2) = 0

Apply “b2 – 4ac” ⇒ 16k – 4k2

⇒ 40

40

≤≤

=

k

k or

OR

≤

≥

14

10

B

B

k

k

Solving quadratic in k to 2 solutions – condone <

M1

M1 A1

M1

A1

5 [6] log4 (3x) + log4 (0.5) = log4 (1.5x)

log16 (3x – 1) = ( )16log

13log

4

4 −x

For change of base – also to base

10,16, 2

½ log4 (3x – 1) = log4 13 −x or 2log4 (1.5x) = log4 (2.25x2) Changing k log z to log zk

3x – 1 = 2.25x2

9x2 – 12x + 4 = 0 ⇒ (3x – 2)2 = 0 ⇒ x = 3

2

Solving 3 term quadratic Accept 0.66 or 0.67 or better

B1

M1

M1

A1

M1 A1






6 [6] (i) 3sinθ – 2cosθ = 3cosθ + 2sinθ ⇒ sinθ = 5cosθ ⇒ tanθ = 5

OR, squaring + Pythagoras ⇒ sinθ = 26

5 or cosθ = 26

1 for M1

θ = 78.7° or 1.37 rad or better (acute angle only accepted)

(ii) x2 + y2 = (9 sin2θ - 12sinθ cosθ + 4 cos2

θ) + (9cos2

θ + 12sinθ cosθ + 4 sin2θ)

= 13 sin2θ + 13cos2

θ = 13 Pythagoras

M1 A1

A1

B1

M1 A1 c.s.o

7 [7] Put x = a ⇒ 6a3 + 5a2 – 12a = –4 or divide by x – a to remainder

Search 6(–2)3 + 5(–2)2 = 12(–2) + 4 = 0 ⇒ a = –2

(at least 2, if unsuccessful, for M1) similarly, if a = 2

1or

3

2 is found

6a3 + 5a2 – 12a + 4 ≡ (a + 2) (6a2 – 7a + 2) OR, finding 2nd root

6a2 – 7a + 2 ≡ (3a – 2) (2a – 1) = 0 ⇒ a = 2

1, 3

2

OR, finding 3rd root

M1

M1 A1

M1 A1

M1 A1

8 [7]

BAX = tan-1 200/150 = tan-1 4/3 ≈ 53.13°, or 36.87°, or 250

ABX = sin-1 {(3sinBAX) ÷ 6} = sin-1 0.4 ≈ 23.58°

Incorrect obtuse-angled ∆s – allow M1 for use of sine or cosine rule)

AXB = 180° – (53.13° + 23.58° ) = 103.29°

V = (6sin103.29°) ÷ sin 53.13° = 7.3 [or via cosine rule]

[or VACROSS = 6sin 76.71° ≈ 5.84 or VDOWN = 3 + 6cos76.71° ≈ 4.38]

Time = 250 ÷ 7.3 [or 200 ÷ 5.8 or 150 ÷ 4.4] ≈ 34 s (accept 34 ∼ 34.5)

2 stages can be combined by applying cosine rule to velocities:

36 = V 2 + 9 – 6V cos 53.13° M1 ⇒ 10 V 2 – 36V – 270 = 0 A1

Solve M1 V = 2.3 A1

3 stages can be combined by applying cosine rule to displacements:

(6t)2 = (250)2 + (3t)2 – 6t cos 53.13° M2 ⇒27t 2 + 900t – 62500 = 0 A2

Solve DM1 t = 34.3 A1

First 5 marks by vector method: V = (150i + 200j )/t B1

VBOAT = (150i + 200j )/t – 3i M1A1

B1

M1 A1

M1 A1

DM1 A1






| VBOAT | =| (150 – 3t)i/t + 200j/t | =6 M1 ⇒ 27t 2 + 900t – 62500 = 0 A1

By scale drawing: Construct 53.13 with 200, 150

⇒ Velocity ∆ M1A1 ⇒ V = 7.3 ± 0.1 M1A1 ⇒ T = 34 DM1A1

9 [7] (i) Y = log y, X = x m = log b, c = log a

(ii) Y = log y, X = log x m = k, c = log A

(iii) Y = 1/y, X = 1/x

−=

=

pm

pc

/9

/1

[Other valid alternatives acceptable

Y y x

y x y

x x

1

X x

y y y

x x y

1

m q q

1 p p

1 q

p−

c p q

p− q p

q− q

1 ]

B1 DB1

B1 DB1

M1 A1 A1






10 [9] (i) Let y = x2 – 8x + 7 dy/dx = 2x – 8 = 0 at x = 4

d2y/dx2 = 2 ∴min at x = 4

OR via completing the square: y = (x – 4)2 – 9 ⇒ min –9 at x = 4

∴ f(x) has maximum at x = 4, corroborated by argument re

reflection of –9 or by graph

(ii)

Judge by shape, unless values clearly incorrect.

Ignore curve outside domain.

Cusp needed at x-axis.

Accept straight line for right-hand arm, but curvature, if shown,

must be correct.

(iii) 0 ≤ f (x) ≤ 9 [condone <]

(iv) k = 4

M1

A1

B2, 1, 0

B2, 1, 0

B1 B1

B1






11 [10]

Let A be (x, y) i.e. (x, 3x)

Length of OA = 250922=+ xx ⇒ x = 5, A is (5, 15)

( 25022=+ yx enough for M1)

Gradient of AB is –3

1

Equation of AB is y – 15 = −3

1 (x – 5) ⇒ B is (0, 16

32 )

AND substitute x = 0 for M1 Decimals 16.6 or 16.7, – 1 p.a.

Gradient of BC is 3

Equation of BC is y = 3x + 163

2

Meets y + 2x = 0 when –2x = 3x + 163

2 ⇒ x = –3

3

1,

C is (–33

1 , 63

2 ) but accept (–3.32, 6.64), (–3.34, 6.68)

In essence, scheme is 3 marks for each of A, B, C. Possible to find B before A e.g.

13

2161565.71sin/2501565.713tanˆ 1

AMB ⇒===− oo

OBXOA

Gradient of AB is –3

1 B1 Solve y −16

3

2 = –

3

1x with y = 3x M1 ⇒

(5,15) A1

B1

M1 A1

B1

M1 A1

B1

M1

M1

A1

12 [10]

EITHER

(i) v = ∫ +−= 5.03.04.12ttta d

At rest v = 0 ⇒ 3t 2 – 14t – 5 = 0 ⇒ (3t + 1) (t – 5) = 0 ⇒ t = 5

OR, by verifying [1.4t – 0.3t 2 + 0.5]t = 5 = 0

(ii) s = ∫ +−= ttttv 5.01.07.032

d

[s] t = 5 = 7.5

[s] t = 10 = –25 OR s 10 – s 5 = –32.5

Total distance = (2× 7.5) + 25 = 40

OR 7.5 + 32.5

M1 A2,1,0

A1

M1 A1√

A1√

A1M1A1c.s.o.






12 [10]

OR (i) ∫ ∫ −=

+=

x

xx

xxxy

2

2

3d

23d

2

2 accept

x

bax−

2

2

One term correct sufficient for M1

[ ] ( ) 5.18162

124

4

2=−−

−=

(ii) (2, 3) on curve ⇒ 3 = 2a + 4

b

3

2

d

d

x

ba

x

y−= 0

40

d

d

2

=−⇒=

=

ba

x

y

x

Solving a = 1, b = 4

y = x + 42

2

32

24

d

d81

d

d4

xx

y

xx

y

x=⇒−=⇒ > 0 when

x = 2 ∴min [or any equivalent method]

M1 A1

DM1 A1

B1

M1 M1

A1

M1 A1c.s.o




IGCSE – November 2005 0606 1





IGCSE – November 2005 0606 2





IGCSE - OCT/NOV 2006 0606 01

© UCLES 2006


(5 8 El1

()

M1 A1

81 8 rU!N milllm! If II !:fli<IMI'l

milllm! lllid \.!ial 11111'!>1. if:!! - llfJIIm 'f ilrithmm[IC hal Dun U!IIKL

Ci:lel'ici!lm I!! ~98 2

M1A1 wilh if:!!,

Cl:le!Pioom of w" ill ' .m::z = 150 ~

' k

6{)11; 96. fl4 21etm•.

- k"3 il'lt:Orred

7 II

Minimum ill :c;'J\ M1A1 Corflld fl(!llr!M lor r -m of

(ll 10 B1 B1 81 ~slue. ~

-11) Minimum lin For

11) MIDirM.Im frl:lm his of

8 -xl

I ie1

I ·00 or dlli':i mal ;.; H

,j,

M1

M1 at IJOIM!Ifll 2 - c "' A1 QIJ,



IGCSE - OCT/NOV 2006 0606 01

© UCLES 2006


10

llmll 1

v~t;;ret4 f!l li >rl111 '~~>ill<>

M1A1

M1 AI

Ml

A1

Ute of correct 1\:lmmill!t •alue fot l'l'l,i!'Nrlli!OL Prr•l'lh&r:l

AlltJw if CCI'lCIIM ~!1illlot Nil ban cbtainl!d for

Ml full formula o1t A1 ~rrect dedudkln - nel!d!l

AI

AI

M1A1 M1 AI



IGCSE - OCT/NOV 2006 0606 01

© UCLES 2006


12 EITHER

;r" 10 ~ =1 ~~· oo aist x = to

81

81

!.Ill 1.111 A1

U1 AI

00,

ea.

Ua d Yllilh For IX(

Mull use ~~ltilla d vli!UM of lr oo othiir Mark

to



IGCSE - OCT/NOV 2006 0606 02

© UCLES 2006

M1 A1 cos

2

1−x

7 = 8 + 2a + b B1

Solving for a and b

=

−=⇒

17

9

b

a




IGCSE - OCT/NOV 2006 0606 02

© UCLES 2006

and

M1 A1

M1 B1 A1

of relevant two terms only = 910

[≈0.736]

or MN = rsecθ




IGCSE – October/November 2008 0606 01

© UCLES 2008

1 (i) correct diagram

(ii) correct diagram

(iii) correct diagram

B1 B1 B1 [3]

2 (2x + 1)2 > 8x + 9 4x

2 – 4x – 8 > 0 x

2 – x – 2 > 0 (x + 1)(x – 2) > 0

Leads to critical values x = –1,2 x < –1 and x > 2

M1 DM1 A1 √A1 [4]

M1 for simplification to 3 term quadratic DM1 for factorisation A1 for critical values Follow through on their critical values.

3

LHS = AA

AAA

sin)cos1(

cos2cos1sin22

+

+++

= AA

A

sin)cos1(

cos22

+

+

= Asin

2 leading to 2cos ecA

M1 A1

M1 A1 [4]

M1 for attempt to deal with fractions and attempt to obtain numerator A1 correct

M1 for use of sin2 A + cos2 A = 1

4 Substitution of x = 1 leading to a + b + 4 = 0

Substitution of 2

1−=x leading to

–a +2b – 28 = 0

Leading to a = –12, b = 8

M1

M1 A1 M1 A1 [5]

M1 for substitution of x = 1 and equated to 3

M1 for substitution of x = 2

1− and equated to 6

A1 for both correct M1 for solution A1 for both

5 (i) a = 13

1

(5i – 12j)

(ii) q(5i –12j) + pi + j = 19i – 23j 5q + p = 19 –12q + 1 = –23 Leading to q = 2, p = 9

M1, A1 [2]

M1 M1 A1 [3]

M1 for a valid attempt to obtain magnitude.

M1 for equating like vectors M1 for solution of (simultaneous) equations A1 for both

6 (i) 2t

2 – 9t – 5 = 0 (2t + 1)(t – 5) = 0

2

1

=t , t = 5

(ii) 5 0.5,2

1

−=x

x = 0.25, 25

M1 DM1

A1 [3]

M1 A1,A1 [3]

M1 for attempting to form a quadratic in t DM1 for attempt to solve a 3 term quadratic

A1 for both

M1 for realising that x0.5

is equivalent to t (or valid attempt at solution)





© UCLES 2008

7 (i) y = 4x2 – 12x + 3

y = (2x – 3)2 – 6

(ii)

−6,

2

3

(iii) f ≥ –6

B1 B1 B1 [3] √B1, √B1 [2]

√B1 [1]

B1 for 2 (part of linear factor) B1 for –3 (part of linear factor) B1 for –6 Follow through on their a, b and c Allow calculus method.

Follow through on their c

8 )(e2

d

d 2c

x

yx

+−=−

When 5,0,3

d

d=∴==

1cx

x

y

5e2

d

d 2+−=

− x

x

y

y = e–2x + 5x(+c2

) When x = 2, y = e–4 ∴c2 = –10 y = e–2x +5x – 10

B1

M1 A1 B1

M1 √A1 [6]

B1 for –2e–2x

M1 for attempt to find c1

B1 for –2e–2x

M1 for attempt to find c2

√ –2 times their c1

9 (i) 25 + 5C1

24(–3x) + 5C2

23(–3x)2 32 – 240x + 720x

2 (ii) 32a = 64, a = 2 32b – 240a = –192, b = 9 –240b + 720a = c c = –720

B1 B1 B1 [3] B1 M1 A1 M1 A1 [5]

B1 for 32 or 25 B1 for –240 B1 for 720. B1 for a = 2 M1 for equation in a and b equated to ±192 A1 for b = 9 M1 for equation in a and b equated to c A1 for c = –720

10 (a) (i) fg(x) = f

+ 2x

x

= 3 – 2+x

x

(ii) 3 –2+x

x= 10

leading to x = –1.75 (b) (i) h(x) > 4 (ii) h–1(x) = ex – 4

h–1(9) = e5 (≈ 148) or 4 + lnx = 9, leading to 5

ex =

(iii) correct graphs

M1 A1 [2]

DM1

A1 [2] B1 [1] M1 A1 [2]

B1 B1 B1 [3]

M1 for order

DM1 for dealing with fractions sensibly

M1 for attempting to obtain inverse function

B1 for each curve B1 for idea of symmetry





© UCLES 2008

11 (i) tan22x = 3

tan2x = (±) 3 2x = 60°, 120°, 240°, 300° x = 30°, 60°, 120°, 150° (ii) 2cosec2 y + cosec y –3 = 0 (2cosec y + 3)(cosec y – 1) = 0

cosec y = –2

3, 1

sin y = –3

2, 1

y = 221.8°, 318.2°, y = 90°

(iii) cos

+

2

π

z = –2

1

3

4

,

3

2

2

πππ

=+z

66

5 ,ππ

=z , allow 0.52, 2.62 rads

M1

DM1 A1, A1 [4] M1, A1 M1 A1, A1 [5] M1 A1,A1 [3]

M1 for an equation in tan22x

M1 for attempt to solve using 2x correctly A1 for any pair M1 for correct use of identity or other valid method A1 for a correct quadratic M1 for solution of quadratic and attempt to solve correctly A1 for 221.8°, 318.2°, A1 for 90° M1 for dealing with sec and order of operations A1 for each

12 EITHER

(i) 2

2

)1(

2)1(

d

d

+

−+

=

x

xxx

x

y

2)1(

)2(

+

+

=

x

xx

0

d

d

=x

y, x = 0, –2

y = 0, –4

(ii) gradient of normal = 3

4−

normal y = 3

4− x +

6

11, leads to

M (1.375,0) N (0, –4) Area = 2.75

M1 A1 DM1

A1,A1 [5] M1 A1 √ B1 B1 M1 √ A1 [6]

M1 for attempt to differentiate a quotient A1 correct allow unsimplified DM1 for equating to zero and an attempt to solve

A1 for each pair (could be x = 0 and x = –2) M1 for attempt to obtain gradient of the normal A1 for a correct (unsimplified) normal equation Follow through on their normal B1 for N

M1 for attempt to get area of triangle Ft on their M and N (must be on axes)





© UCLES 2008

12 OR

(i) 22

e

d

d

−

−

=

x

x

y

22

e 0,

d

d

=

−

=

x

x

y

x = 2 + ln2 (2.69) y = 4 – 2ln2 (2.61)

2

2

2

e−

=x

dx

yd, always +ve ∴min

(ii)

[ ]∫ +=+ −−

−−

3

0

3

0

2226ed)62e( xxxx

xx

= (e – 9 + 18) – (e–2) = e – e–2 + 9 k = 9

B1 B1 M1

A1 A1

B1 [6] M1, A1 M1 A1 B1 [5]

B1 for 2e

x− B1 for –2 only M1 for equating to zero and attempt to solve

A1 for x A1 for y

B1 for conclusion from a valid method M1 for attempt to integrate M1 for correctly applying limits A1 for e – e–2 B1 for k





© UCLES 2008

1

−

−=−

137

64

10

11A B1+B1

evaluate

−

24

411

A M1

x = 2 , y = 2.5 A1 [4]

2 ( )22 9k x − M1

( )26 2 9x − A1

substitute d

7 and 4d

xx

t

= = intot

x

x

y

t

y

d

d

d

d

d

d×= M1

600 A1 [4]

3 eliminate y M1 use 2

4b ac− or 24b ac∗ DM1

210 39 0m m+ − ∗ or ( )25 64m+ ∗ A1

factorise 3 term quadratic in m or take square root M1 13 3m− < < A1 [5]

4 (i) ( )d 1ln

dx

x x

= B1

1 ln x+ B1 (ii) ( )1 ln d ln ( )x x x x c+ = +∫ M1

ln d ln 1 d ( )x x x x x c= − +∫ ∫ M1

( )lnx x x c− + A1

[5]

5 (i) express as powers of 2 (or 4 or 8) M1 applies rules of indices [ ]2 (5 ) 4 3( 3)x x x x− − = − − DM1

7 A1 (ii) ( ) ( ){ }lg 2 10 lg lg 2 10y y y y+ + = + or 2 lg100= B1

22 10 100y y+ = oe B1

5 only B1 [6]





© UCLES 2008

6 (a) 10, 3 and 15 B1 multiply 3 values M1 450 A1 (b) ( )4 5 4 3× × × B1+B1

240 B1 [6]

7 (i) speed of travel = 4.8 or distance downstream = 14 B1

draw right angle triangle with 1.4 and (4.8) at o

90 B1

2 21.4 (4.8)+ M1

5 A1

(ii) 1 (4.8)tan1.4

− oe M1

73.7 or 1.29 radians A1 [6]

8 (i) 5 B1 (ii) 180 or π B1 (iii) 8 and 2− B1+B1 correct start and endpoints B1 2 cycles in 0 to 2π B1 correct max and min points B1 [7]

(4.8)

1.4

OR

(4.8)

1.4





© UCLES 2008

9 eliminate y (or x) M1 2

7 42 35 0x x− + = (or 27 42 49 0y y+ − = )oe A1

solve 3 term quadratic M1 x = 1 and 5 (or y = –7 and 1) A1 find second coordinates M1 find mid-point M1 use

ABm ,

1 21mm = − and coordinates of a point M1

1

3 ( 3)2

y x+ = − − or 2 3 0x y+ + = or1 3

2 2y x= − − A1

[8]

10 (i) 16163d

d2

+−= xxx

y B1

equate to 0 and solve 3 term quadratic M1 4, 0x y= = A1 AG

27

256or

27

139

3

4== yx or 9.48 or 9.5 A1

(ii) integrate M1

4 3

288

4 3

x x

x− + A1

use limits of 4 (and 0) DM1

3

121 or 21.3 A1

[8]

11 (i) plot xy against 1/x with linear scales M1 xy 4.5 3.24 2.82 2.64 1/x 0.5 0.25 0.17 0.125 A2, 1, 0 (ii) attempt at gradient using plotted points DM1 5 0.2± A1 intercept 2 0.1± B1 (or A1 if calculated from y mx c= + ) use Y mX c= + in correct way M1

2

5 2y

x x= + or

2

5 2xy

x

+= or

1 52y

x x

= +

A1√

(iii) read from graph or substitute in formula to find x M1 2.5 0.2x = ± A1 1.6 0.1y = ± A1 [11]





© UCLES 2008

12 EITHER

(i) cos0.62

OC= or 2cos0.6OC = or

2

sin 0.97sin

2

OC

π

= M1

1.65 A1

2sin 0.6CD = or 2 2CD OD OC= − M1

1.13 A1 (ii) 6 0.6× B1

complete plan CD + 4 + rθ +(6 – 1.65) M1 13.1 A1

(iii) 216 0.6

2× × B1

complete plan 21 1

2 2r OC CDθ − × × M1

9.87 A1 [10]

OR

(i) 22 12 16t t− + B1+B1+B1

equate to 0 and solve quadratic for 2 values M1 2 and 4 A1 (ii) ds v t= ∫ M1

3 226 16

3t t t− + A 2, 1, 0√

use limits and subtract DM1

3

22 or 2.67 A1

[10]




IGCSE– October/November 2009 0606 01

© UCLES 2009

1 (i) 2a3 – 7a2 + 7a2 + 16 = 0 leading to a3 = –8, a = –2

(ii) 162

114

2

17

2

12

23

+

−−

−−

−

= 21

M1 A1

[2]

M1

A1 [2]

M1 for use of x = a and equated to zero, maybe implied

M1 for substitution of x = –2

1into their

expression or f(x)

2 (i) (ii)

=

22

35

32

43

1

2

3

5

7221

0552

3423

2136

B1, B1

[2]

B2, 1, 0 [2]

B1 for each matrix, must be in correct order –1 for each error

3 4(2k + 1)2 = 4(k + 2) 4k2 + 3k – 1 = 0

leading to k = 4

1, –1

M1 A1

M1 A1

[4]

M1 for use of ‘b2 – 4ac’ Correct quadratic expression

M1 for correct attempt at solution A1 for both values

4 (13 – 3y)2 + 3y2 = 43

(or x2 + 3

)13( 2x−

= 43)

6(2y2 – 13y + 21) = 0 (or 2(2x2 – 13x + 20) = 0) (2y – 7)(y – 3) = 0 (or (2x – 5)(x – 4) = 0

y = 3 or

= 4or

2

5

2

7x

(or x = 4 or

= 3or

2

7

2

5y )

M1

A1

DM1

A1,A1

[5]

M1 for eliminating one variable

A1 for correct quadratic

DM1 for correct attempt at solving quadratic

A1 for each correct pair

5 (i) ( ) ( ) 22232322

=−++ AC = 22

(ii) tan A = 23

23

+

−

( )( )( )( )2323

2323

−+

−− =

7

2611−

M1

A1 [2]

M1

M1, A1

[3]

M1 for use of Pythagoras Use of decimals M1, A0

M1 for correct ratio

M1 for rationalising 2 term denominator





© UCLES 2009

6 (i) 3x2 – 10x – 8 = 0 (3x + 2)(x – 4) = 0

critical values –3

2, 4

A = {x : –3

2 Y x Y 4}

(ii) B = {x : x [ 3} A ∩ B = {x : 3 Y x Y 4}

M1

A1

√A1 [3]

B1 B1

[2]

M1 for attempt to solve quadratic

A1 for critical values

Follow through on their critical values. B1 for values of x that define B. B1 (beware of fortuitous answers)

7 (i) 13C8 = 1287

(ii) 7 teachers, 1 student : 6 6 teachers, 2 students 7C6 × 6C2 : 105 5 teachers, 3 students 7C5 × 6C3 : 420 531

M1, A1 [2]

B1 B1 B1 B1

[4]

M1 for correct C notation

8 (i) When t = 0, N = 1000

(ii) t

N

d

d = –1000ke–kt

when t = 0, t

N

d

d = –20 leading to

k = 50

1

(iii) 500 = 1000e–kt

t = –50ln2

1 leading to 34.7 mins

B1 [1]

M1

DM1

A1 [3]

M1

M1 A1

[3]

M1 for differentiation

DM1 for use of t

N

d

d = ±20

M1 for attempt to formulate equation using half life M1 for a correct attempt at solution (beware of fortuitous answers)

9 (i) 20 × –2(1 – 2x)19

(ii) x2

2

1 + 2x ln x

ISW

(iii) 2

2 )12tan())12(sec2(

x

xxx +−+

ISW

B1,B1

[2] M1 B1 A1

[3] M1 B1 A1

[3]

B1 for 20 and (1 – 2x)19

B1 for –2 provided (1 – 2x)19 is present M1 for attempt to differentiate a product.

B1 for x

1

A1 all other terms correct

M1 for attempt to differentiate a quotient. B1 for differentiation of tan (2x + 1) A1 all other terms correct





© UCLES 2009

10 (i) x

y

d

d = 9x2 – 4x + 2

at P grad = 7 tangent y – 3 = 7(x – 1)

(ii) at Q, 7x – 4 = 3x3 – 2x2 + 2x

leading to 3x3 – 2x2 – 5x + 4 = 0 (x – 1)(3x2 + x – 4) – 0 (x – 1)(3x + 4)(x – 1) = 0

leading to x = –3

4, y = –

3

40

M1 A1 DM1 A1

[4] M1 B1 DM1 DM1

A1 [5]

M1 for differentiation A1 for gradient = 7 and y = 3 DM1 for attempt to find tangent equation. M1 for equating tangent and curve equations B1 for realising (x – 1) is a factor DM1 attempt to factorise cubic DM1 for attempt to solve quadratic

A1 for both

11 (a) tan θ + cot θ = θ

θ

θ

θ

sin

cos

cos

sin+

= θθ

θθ

sincos

cossin22

+

= θθ sincos

1

= cos ec θ sec θ (b) (i) tan x = 3sin x

x

x

x

sin3cos

sin=

sin x – 3sin x cos x = 0

leading to cos x = 3

1, sin x = 0

x = 70.5°, 289.5° and x = 180° (ii) 2 cot2 y + 3 cosecy = 0 2(cosec2y – 1) + 3 cosecy = 0 2 cosec2y + 3 cosecy – 2 = 0 (2 cosecy – 1)(cosecy + 2) = 0

leading to sin y = –2

1, y =

6

π11 ,

6

π7

allow y = 3.67, 5.76

B1

B1

B1

[3]

M1

A1√A1 B1

[4] M1 M1 M1 A1,A1

[5]

B1 for attempt to obtain one fraction

B1 for use of an appropriate identity

B1 for simplification Scheme follows for alternative proofs

M1 for use of tan x = x

x

cos

sin and correct

attempt to solve

√A1 on their x = 70.5° B1 for x = 180° M1 for use of correct identity M1 for attempt to solve quadratic M1 for dealing with cosec/cot Scheme follows for alternative solutions





© UCLES 2009

12 EITHER

(i) πr2h = 1000, leading to

h = 2

π

1000

r

(ii) A = 2πrh + 2πr2 leading to given answer

A = 2πr2 + r

2000

(iii) r

A

d

d = 4πr –

2

2000

r

when r

A

d

d = 0, 4πr =

2

2000

r

leading to r = 5.42

(iv) 2

2

d

d

r

A = 4π +

3

4000

r

+ ve when r = 5.42 so min value Amin = 554

M1

A1 [2]

B1

A1 [2]

M1 A1

DM1

A1 [4]

M1 A1 B1

[3]

M1 for attempt to use volume

B1 for A = 2πrh + 2πr2 GIVEN ANSWER

M1 for attempt to differentiate given A A1 all correct

DM1 for solution = 0

M1 for second derivative method or gradient method’ A1 for minimum, must be from correct work

12 OR

(i) y = x + cos 2x

x

y

d

d = 1 – 2 sin 2x

when x

y

d

d = 0, sin 2x =

2

1

leading to x = 12

5 ,

12

ππ

(ii) Area = ∫ +

12

5

12

d.2cos

π

π

xxx

= 12

5

12

2

2sin2

1

2

π

π

+ x

x

= 12

2π

M1

A1

DM1 DM1 A1,A1

[6]

M1 A1,A1 DM1

A1 [5]


DM1 for setting to 0 and attempt to solve DM1 for correct order of operations

M1 for attempt to integrate A1for each term correct DM1 for correct use of limits – must be in radians

(Trig terms cancel out)





© UCLES 2009

1 (i) 2a3 – 7a2 + 7a2 + 16 = 0 leading to a3 = –8, a = –2

(ii) 162

114

2

17

2

12

23

+

−−

−−

−

= 21

M1 A1

[2]

M1

A1 [2]

M1 for use of x = a and equated to zero, maybe implied

M1 for substitution of x = –2

1into their

expression or f(x)

2 (i) (ii)

=

22

35

32

43

1

2

3

5

7221

0552

3423

2136

B1, B1

[2]

B2, 1, 0 [2]

B1 for each matrix, must be in correct order –1 for each error

3 4(2k + 1)2 = 4(k + 2) 4k2 + 3k – 1 = 0

leading to k = 4

1, –1

M1 A1

M1 A1

[4]

M1 for use of ‘b2 – 4ac’ Correct quadratic expression

M1 for correct attempt at solution A1 for both values

4 (13 – 3y)2 + 3y2 = 43

(or x2 + 3

)13( 2x−

= 43)

6(2y2 – 13y + 21) = 0 (or 2(2x2 – 13x + 20) = 0) (2y – 7)(y – 3) = 0 (or (2x – 5)(x – 4) = 0

y = 3 or

= 4or

2

5

2

7x

(or x = 4 or

= 3or

2

7

2

5y )

M1

A1

DM1

A1,A1

[5]

M1 for eliminating one variable

A1 for correct quadratic

DM1 for correct attempt at solving quadratic

A1 for each correct pair

5 (i) ( ) ( ) 22232322

=−++ AC = 22

(ii) tan A = 23

23

+

−

( )( )( )( )2323

2323

−+

−− =

7

2611−

M1

A1 [2]

M1

M1, A1

[3]

M1 for use of Pythagoras Use of decimals M1, A0

M1 for correct ratio

M1 for rationalising 2 term denominator





© UCLES 2009

6 (i) 3x2 – 10x – 8 = 0 (3x + 2)(x – 4) = 0

critical values –3

2, 4

A = {x : –3

2 Y x Y 4}

(ii) B = {x : x [ 3} A ∩ B = {x : 3 Y x Y 4}

M1

A1

√A1 [3]

B1 B1

[2]

M1 for attempt to solve quadratic

A1 for critical values

Follow through on their critical values. B1 for values of x that define B. B1 (beware of fortuitous answers)

7 (i) 13C8 = 1287

(ii) 7 teachers, 1 student : 6 6 teachers, 2 students 7C6 × 6C2 : 105 5 teachers, 3 students 7C5 × 6C3 : 420 531

M1, A1 [2]

B1 B1 B1 B1

[4]

M1 for correct C notation

8 (i) When t = 0, N = 1000

(ii) t

N

d

d = –1000ke–kt

when t = 0, t

N

d

d = –20 leading to

k = 50

1

(iii) 500 = 1000e–kt

t = –50ln2

1 leading to 34.7 mins

B1 [1]

M1

DM1

A1 [3]

M1

M1 A1

[3]

M1 for differentiation

DM1 for use of t

N

d

d = ±20

M1 for attempt to formulate equation using half life M1 for a correct attempt at solution (beware of fortuitous answers)

9 (i) 20 × –2(1 – 2x)19

(ii) x2

2

1 + 2x ln x

ISW

(iii) 2

2 )12tan())12(sec2(

x

xxx +−+

ISW

B1,B1

[2] M1 B1 A1

[3] M1 B1 A1

[3]

B1 for 20 and (1 – 2x)19

B1 for –2 provided (1 – 2x)19 is present M1 for attempt to differentiate a product.

B1 for x

1

A1 all other terms correct

M1 for attempt to differentiate a quotient. B1 for differentiation of tan (2x + 1) A1 all other terms correct





© UCLES 2009

10 (i) x

y

d

d = 9x2 – 4x + 2

at P grad = 7 tangent y – 3 = 7(x – 1)

(ii) at Q, 7x – 4 = 3x3 – 2x2 + 2x

leading to 3x3 – 2x2 – 5x + 4 = 0 (x – 1)(3x2 + x – 4) – 0 (x – 1)(3x + 4)(x – 1) = 0

leading to x = –3

4, y = –

3

40

M1 A1 DM1 A1

[4] M1 B1 DM1 DM1

A1 [5]

M1 for differentiation A1 for gradient = 7 and y = 3 DM1 for attempt to find tangent equation. M1 for equating tangent and curve equations B1 for realising (x – 1) is a factor DM1 attempt to factorise cubic DM1 for attempt to solve quadratic

A1 for both

11 (a) tan θ + cot θ = θ

θ

θ

θ

sin

cos

cos

sin+

= θθ

θθ

sincos

cossin22

+

= θθ sincos

1

= cos ec θ sec θ (b) (i) tan x = 3sin x

x

x

x

sin3cos

sin=

sin x – 3sin x cos x = 0

leading to cos x = 3

1, sin x = 0

x = 70.5°, 289.5° and x = 180° (ii) 2 cot2 y + 3 cosecy = 0 2(cosec2y – 1) + 3 cosecy = 0 2 cosec2y + 3 cosecy – 2 = 0 (2 cosecy – 1)(cosecy + 2) = 0

leading to sin y = –2

1, y =

6

π11 ,

6

π7

allow y = 3.67, 5.76

B1

B1

B1

[3]

M1

A1√A1 B1

[4] M1 M1 M1 A1,A1

[5]

B1 for attempt to obtain one fraction

B1 for use of an appropriate identity

B1 for simplification Scheme follows for alternative proofs

M1 for use of tan x = x

x

cos

sin and correct

attempt to solve

√A1 on their x = 70.5° B1 for x = 180° M1 for use of correct identity M1 for attempt to solve quadratic M1 for dealing with cosec/cot Scheme follows for alternative solutions





© UCLES 2009

12 EITHER

(i) πr2h = 1000, leading to

h = 2

π

1000

r

(ii) A = 2πrh + 2πr2 leading to given answer

A = 2πr2 + r

2000

(iii) r

A

d

d = 4πr –

2

2000

r

when r

A

d

d = 0, 4πr =

2

2000

r

leading to r = 5.42

(iv) 2

2

d

d

r

A = 4π +

3

4000

r

+ ve when r = 5.42 so min value Amin = 554

M1

A1 [2]

B1

A1 [2]

M1 A1

DM1

A1 [4]

M1 A1 B1

[3]

M1 for attempt to use volume

B1 for A = 2πrh + 2πr2 GIVEN ANSWER

M1 for attempt to differentiate given A A1 all correct

DM1 for solution = 0

M1 for second derivative method or gradient method’ A1 for minimum, must be from correct work

12 OR

(i) y = x + cos 2x

x

y

d

d = 1 – 2 sin 2x

when x

y

d

d = 0, sin 2x =

2

1

leading to x = 12

5 ,

12

ππ

(ii) Area = ∫ +

12

5

12

d.2cos

π

π

xxx

= 12

5

12

2

2sin2

1

2

π

π

+ x

x

= 12

2π

M1

A1

DM1 DM1 A1,A1

[6]

M1 A1,A1 DM1

A1 [5]


DM1 for setting to 0 and attempt to solve DM1 for correct order of operations

M1 for attempt to integrate A1for each term correct DM1 for correct use of limits – must be in radians

(Trig terms cancel out)





© UCLES 2009

1 (i) > e–1 or > 0.37 B1 (ii) Uses ln function properly M1 1 + ln x A1 (iii) > e–1 B1√ [4]

2 (i) 64 – 96x + 60x2 – 20x3 B1+B1+B1+B1 (ii) 1 × (–20) + 2 × (60) + 1 × (–96) M1 –20 + 120 – 96 = 4 A1 [6] 3 (i) Plots x2y against x with linear scale. M1

x 2 4 6 8 10

x2y 24.96 45.12 64.44 85.12 105

A2,1,0 (ii) x2y = bx + a B1

Calculates gradient M1 b = 10 ± 0.4 A1 a = 5 ± 2 from intercept or substitution B1 [7] (ii) Alternative last 3 marks Equates intercept to a(5 ± 2) B1 Uses a to find b M1 b = 10 ± 0.4 A1

4 4563d

d 2 −+

= xx

x

y B2, 1, 0

Equates x

y

d

d to 0 and solves 3 term quadratic M1

x = 3 and x = –5 A1 (3, –21) and (–5, 235) A1 Complete method for max/min M1 minimum when x = 3 and maximum when x = –5 A1 [7]

5 (i) 22247 + M1

25=OA A1

(ii)

−=

4

3AB B1

5=AB B1





© UCLES 2009

(iii)

−==

20

155ABAC M1

ACOAOC += used DM1

4

22 A1 [7]

6 (i) Uses product rule M1

2

1

)124(42

1124

−

+×++ xxx A1

Expresses with common denominator M1 k = 6 A1

(ii) 1243

+xxk

M1

Uses limits of 6 and –2 in 124 +xCx M1 20 A1√ [7] 7

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

(i) Attempt at sine curve M1 Correct position at multiples of 45° A2, 1,0 (ii) 2cos x – 1 B1

Attempt at cosine curve M1 (0, 1), (90, –1), (180, –3), (270, –1), (360, 1), A1 (iii) 2 B1√ [7]

8 (i) Matrix multiplication M1

−

−

1210

60 A1


10

11 A1





© UCLES 2009

(iii) A–1 =

− 24

13

10

1 B1+B1

X = A–1B stated M1

−

120

95

10

1 A1 [8]

9 (i) 1.25 B1

(ii) 3)42(d

d

+

==

t

k

t

va M1

Substitutes 3 into t

v

d

d M1

–0.08 A1

(iii) s = 42

d+

=∫t

ktv M1

42

10

+

−

t

A1

Correct use of limits of 0 and 8 only on attempt at ∫ tvd

or finds c from s = 0, t = 0 and substitutes t = 8 M1 2 A1 [8] 10 (a) 2 lg 5 = lg25 or lg52 B1 2 = lg 100 or lg102 B1 Uses rules of logs correctly (lg(175x – 75) = lg(100x + 300)) M1 5 A1 (b) Substitutes and express as equation in u M1 3u2 – 28u + 9 = 0 A1 Solves 3 term quadratic M1

u = 3

1and 9 A1

x = –1 and 2 A1 [9]





© UCLES 2009

11 EITHER

(i) AB = 3 or 3

6sin

3

sinπ

=∠APQ

B1

Correct use of trigonometry to APB = 3

2π B1

(ii) Uses s = rθ M1

3.14 (π) or 3.63

3

32 π

A1

6.77

+

3

32 π

π A1

(iii) Uses θ2

2

1r or rs

2

1 M1

Uses 2

2

1r sin θ or area kite M1

Either 4.71 (1.5π) and 3.14 (π),

or 3.90

4

39 and 1.30

4

33 or 5.20 ( )33 A1

Complete plan DM1 2.65 to 2.66 ( )335.2 −π A1 [10] OR

(i) Method for D M1 (–4, 9) A1 (ii) Method for E M1 (–1,7) A1 (iii) Finds area parallelogram (= 80) M1 Area trapezium = 120 A1 Height trapezium = 6 B1 Uses Area = ½ × (6) × (AB + EF) M1 EF = 30 A1 F (29, 7) A1 [10] (iii) alternative last 4 marks Array method complete (with only one variable) M1 F (k, 7) A1 3k + 33 = 120 oe A1 F (29, 7) A1





© UCLES 2010

1 (i) a = –12, b = –4 B1, B1 [2]

B1 for each

(ii) –4 √ B1 [1]

Follow through on their y value

2 (i) Graphs B1 B1

[2]

B1 for one correct curve B1 for a second correct curve consistent with the first curve

(ii) 3 √ B1

[1]

Follow through on number of clear points of intersection

3 ( ) ( )

x

xxxx

2sin1

sin1cossin1cos

−

−++

x

x

2cos

cos2

xsec2

M1 DM1 M1 A1

[4]

M1 for attempt to get in terms of a single fraction DM1 simplifying numerator M1 simplifying denominator

4 x = –1 or 7 or 2

1− seen

Either ( )( )713212

−−+ xxx

or ( )( )13272

++− xxx

or ( )( )76122

−−+ xxx

leading to ( )( )( )1271 +−+ xxx

M1 DM1 A1 DM1, A1

[5]

M1 for attempt to find a root DM1 for attempt to obtain quadratic factor A1 correct quadratic factor DM1 attempt to factorise quadratic factor

5 (i) 3

π

π +=a , 3

4π=a

B1

[1]

Must be in terms of π

(ii) xxxx

ysin2cos2

d

d+=

at P, 2d

d=

x

y, ⇒ grad of normal =

2

1−

normal:

−−=−22

1

3

4 ππ

xy

−= xy

6

192

π

M1, A1 M1 M1, A1

[5]

M1 for attempt to differentiate a product M1 for m1m2 = –1, must have used differentiation M1 for attempt at a normal equation, must have used differentiation, allow unsimplified

www.XtremePapers.com




© UCLES 2010

6 (i) 264 960 6000x x− +

B1, B1, B1

[3]

B1 for each correct term, allow 62

(ii) 1 × (their x term) + 2

10 × (their 64)

–960 + 320 = –640

M1 B1 A1

[3]

M1 for 2 terms

B1 for 2

10 or 5

7 (a) (i) x = 30°, 150° B1, B1 [2]

B1 for each

(ii) x – 30° = 120°, 240° x = 150°, 270° A∪B = {30°, 150°, 270°}

B1 √ B1

[2]

B1 for x = 150°, 270° only Follow through on their A and B

(b) 13cos ±=x or 03tan =x 3x = 0°, 180°, 360°, 540° x = 0°, 60°, 120°, 180° n(C) = 4

M1 A1 √ B1

[3]

M1 for dealing with sec and 3x

A1 for all solutions correct Follow through on their number of solutions

8 (i) and (ii) Gradient = – 0.5 Use of ratios or ln y = –0.5 ln x + c ln y = 6.8 ln y = b ln x + ln A ( )6.8their

e=A A = 898, b = –0.5

M1 M1 A1 B1 M1 A1, A1

[7]

M1 for attempt at gradient M1 for attempt at y intercept A1 for ln y = 6.8 B1 for ln y = b ln x + ln A M1 for use of e A1 for A and A1 for b

9 (i) 2xA = , x

x

A2

d

d=⇒

B1

[1]

(ii) When x = 5, 10d

d=

x

A

d 0.003

d 10

x

t

=

= 0.0003

√ B1 M1

A1 [3]

Follow through on their x

A

d

d

M1 for 0.003 ÷ their 10

(iii) 34xV = , 2

12d

dx

x

V=

0003.012d

d 2×= x

t

V

= 0.09

B1, B1 M1

A1 [4]

B1 for each





© UCLES 2010

10 (i) PA

4

6tan =

π

, 34=PA

4

6sin

4+=

π

PB , PB = 12

allow equivalent methods

B1 B1

[2]

B1 for PA (answer given) B1 for PB (answer given)

(ii) Sector area = 3

122

1 2 π

×

Area of kite = 4342

12 ×××

Shaded area = 47.7

√ B1 M1, A1 A1

[4]

√ B1 sector area, ft on their PB M1 for attempt to find area of kite or appropriate triangle

(iii) ( ) ( )42341223

12 +−+

×=π

P

= 30.7

B1, B1, B1 B1

[4]

B1 for each of the 3 terms B1 for final answer

11 (i) ( ) ( )cx ++ 2

1

12

M1, A1 [2]

M1 for ( )21

1 x+ , A1 for 2

(ii)

( )

x

xxx

x

y

+

+−+

=

−

1

12

1212

d

d2

1

( ) ( )311

2

x

x

x +

−

+

=

M1 A2, 1, 0 A1

[4]

M1 attempt at differentiation –1 each error A1 all correct

(iii) ( ) ( ) x

x

x

x

x

x

x

+

−

+

=

+

∫∫1

2d

1

2d

13

( )cx

x

x ++

−+=1

214

( )

x

x

xd

1

3

0 3∫

+

= (8 – 3) – (4), = 1

M1

A1

M1, A1

[4]

M1 for idea of using (ii) ‘in reverse’

A1 all correct

M1 for attempt evaluation





© UCLES 2010

12 EITHER

(i) ( )cxx

y +−= 9

3

43

when x = 3, y = 1, so c = –8

M1, A1 M1, A1

[4]

M1 for attempt to integrate M1 for attempt to find c

(ii) 4x2 – 9 = 0, leads to x = ±1.5

Points (1.5, –17), (–1.5, 1)

M1

A1, A1 [3]

M1 for attempt to solve 0d

d=

x

y

A1 for each pair

(iii) Midpoint AB: (0, –8)

Gradient of AB = –6, perp grad = 6

1

Equation: x – 6y = 48

M1 M1 M1, A1

[4]

M1 for attempt to find midpoint M1 for attempt to find grad of perp M1 must be working with perp

12 OR

(i) 50 = A + B

xx

BAx

y−

−= ee2d

d 2

–20 = 2A – B leads to A = 10 and B = 40

B1 M1 A1 DM1 A1

[5]

M1 for attempt to differentiate A1 all correct DM1 for attempt to solve equations.

(ii) xx

x

y−

−= e40e20d

d 2 , xx −

= e40e202

e3x = 2

2ln3

1=x or 0.231

y = 47.6

M1 M1 M1 A1

[4]

M1 for equating to zero and attempt at solution M1 for dealing with exponentials M1 for attempt to obtain x A1 for both

(iii) xx

x

y −+= e40e40

d

d 2

2

2

Always +ve, so min

M1 A1

[2]

M1 for attempt at second derivative or other valid method A1 for a correct conclusion





© UCLES 2010

1 (i) a = –12, b = –4 B1, B1 [2]

B1 for each

(ii) –4 √ B1 [1]

Follow through on their y value

2 (i) Graphs B1 B1

[2]

B1 for one correct curve B1 for a second correct curve consistent with the first curve

(ii) 3 √ B1

[1]

Follow through on number of clear points of intersection

3 ( ) ( )

x

xxxx

2sin1

sin1cossin1cos

−

−++

x

x

2

cos

cos2

xsec2

M1 DM1 M1 A1

[4]

M1 for attempt to get in terms of a single fraction DM1 simplifying numerator M1 simplifying denominator

4 x = –1 or 7 or 2

1− seen

Either ( )( )713212

−−+ xxx

or ( )( )13272

++− xxx

or ( )( )76122

−−+ xxx

leading to ( )( )( )1271 +−+ xxx

M1 DM1 A1 DM1, A1

[5]

M1 for attempt to find a root DM1 for attempt to obtain quadratic factor A1 correct quadratic factor DM1 attempt to factorise quadratic factor

5 (i) 3

π

π +=a , 3

4π=a

B1

[1]

Must be in terms of π

(ii) xxxx

ysin2cos2

d

d+=

at P, 2d

d=

x

y, ⇒ grad of normal =

2

1−

normal:

−−=−22

1

3

4 ππ

xy

−= xy

6

192

π

M1, A1 M1 M1, A1

[5]

M1 for attempt to differentiate a product M1 for m1m2 = –1, must have used differentiation M1 for attempt at a normal equation, must have used differentiation, allow unsimplified





© UCLES 2010

6 (i) 264 960 6000x x− +

B1, B1, B1

[3]

B1 for each correct term, allow 62

(ii) 1 × (their x term) + 2

10 × (their 64)

–960 + 320 = –640

M1 B1 A1

[3]

M1 for 2 terms

B1 for 2

10 or 5

7 (a) (i) x = 30°, 150° B1, B1 [2]

B1 for each

(ii) x – 30° = 120°, 240° x = 150°, 270° A∪B = {30°, 150°, 270°}

B1 √ B1

[2]

B1 for x = 150°, 270° only Follow through on their A and B

(b) 13cos ±=x or 03tan =x 3x = 0°, 180°, 360°, 540° x = 0°, 60°, 120°, 180° n(C) = 4

M1 A1 √ B1

[3]

M1 for dealing with sec and 3x

A1 for all solutions correct Follow through on their number of solutions

8 (i) and (ii) Gradient = – 0.5 Use of ratios or ln y = –0.5 ln x + c ln y = 6.8 ln y = b ln x + ln A ( )6.8their

e=A A = 898, b = –0.5

M1 M1 A1 B1 M1 A1, A1

[7]

M1 for attempt at gradient M1 for attempt at y intercept A1 for ln y = 6.8 B1 for ln y = b ln x + ln A M1 for use of e A1 for A and A1 for b

9 (i) 2xA = , x

x

A2

d

d=⇒

B1

[1]

(ii) When x = 5, 10d

d=

x

A

d 0.003

d 10

x

t

=

= 0.0003

√ B1 M1

A1 [3]

Follow through on their x

A

d

d

M1 for 0.003 ÷ their 10

(iii) 34xV = , 2

12d

dx

x

V=

0003.012

d

d 2×= x

t

V

= 0.09

B1, B1 M1

A1 [4]

B1 for each





© UCLES 2010

10 (i) PA

4

6tan =

π

, 34=PA

4

6sin

4+=

π

PB , PB = 12

allow equivalent methods

B1 B1

[2]

B1 for PA (answer given) B1 for PB (answer given)

(ii) Sector area = 3

122

1 2 π

×

Area of kite = 4342

12 ×××

Shaded area = 47.7

√ B1 M1, A1 A1

[4]

√ B1 sector area, ft on their PB M1 for attempt to find area of kite or appropriate triangle

(iii) ( ) ( )42341223

12 +−+

×=π

P

= 30.7

B1, B1, B1 B1

[4]

B1 for each of the 3 terms B1 for final answer

11 (i) ( ) ( )cx ++2

1

12

M1, A1 [2]

M1 for ( )21

1 x+ , A1 for 2

(ii)

( )

x

xxx

x

y

+

+−+

=

−

1

12

1212

d

d2

1

( ) ( )311

2

x

x

x +

−

+

=

M1 A2, 1, 0 A1

[4]

M1 attempt at differentiation –1 each error A1 all correct

(iii) ( ) ( ) x

x

x

x

x

x

x

+

−

+

=

+

∫∫1

2d

1

2d

13

( )cx

xx +

+

−+=

1

214

( )

x

x

x

d

1

3

0 3∫

+

= (8 – 3) – (4), = 1

M1

A1

M1, A1

[4]

M1 for idea of using (ii) ‘in reverse’

A1 all correct

M1 for attempt evaluation





© UCLES 2010

12 EITHER

(i) ( )cxx

y +−= 93

43

when x = 3, y = 1, so c = –8

M1, A1 M1, A1

[4]

M1 for attempt to integrate M1 for attempt to find c

(ii) 4x2 – 9 = 0, leads to x = ±1.5

Points (1.5, –17), (–1.5, 1)

M1

A1, A1 [3]

M1 for attempt to solve 0d

d=

x

y

A1 for each pair

(iii) Midpoint AB: (0, –8)

Gradient of AB = –6, perp grad = 6

1

Equation: x – 6y = 48

M1 M1 M1, A1

[4]

M1 for attempt to find midpoint M1 for attempt to find grad of perp M1 must be working with perp

12 OR

(i) 50 = A + B

xx

BAx

y−

−= ee2d

d 2

–20 = 2A – B leads to A = 10 and B = 40

B1 M1 A1 DM1 A1

[5]

M1 for attempt to differentiate A1 all correct DM1 for attempt to solve equations.

(ii) xx

x

y−

−= e40e20d

d 2 , xx −

= e40e202

e3x = 2

2ln3

1=x or 0.231

y = 47.6

M1 M1 M1 A1

[4]

M1 for equating to zero and attempt at solution M1 for dealing with exponentials M1 for attempt to obtain x A1 for both

(iii) xx

x

y−+= e40e40

d

d 2

2

2

Always +ve, so min

M1 A1

[2]

M1 for attempt at second derivative or other valid method A1 for a correct conclusion





© UCLES 2010

1 x

x

xx coscos

1cossec −=−

x

x

cos

cos12−

=x

xxcos

sinsin=

xx tansin=

M1 M1 A1

[3]

M1 for dealing with sec and fractions M1 for use of trig identity

(Alt: xxx

x

x

x

x

xcostan

cos

sin

sec

tan

sec

1sec22

==−

)

M1 M1 A1

M1 for dealing with sec and fractions M1 for use of trig identity

2 (i) 84047

=P

B1, B1 [2]

B1 for 4

7P only

(ii) 4 × 3

6P or 840

7

4×

480

M1 A1

[2]

M1 for a valid method

3 181222

++=+ xxmx

( ) 016122

=+−+ xmx

( ) 164122

×=−m leading to m = 4, 20 Alt scheme: 122 += xm

( ) 181221222

++=++ xxxx x = ±4 so m = 4, 20

M1 M1 M1, A1

[4] M1 M1 M1 A1

[4]

M1 for equation in x only, allow unsimplified

M1 for use of ‘ acb 42− ’

M1 for solution of quadratic M1 for equating gradients M1 for elimination of m M1 for x and subsequent calculation for m

4 f(2) = 8 + 4k – 10 – 3 f(–1) = –1 + k + 5 – 3 (4k – 5) = 5(k + 1) leading to k = –10

M1 M1 M1 A1

[4]

M1 for use of x = 2 M1 for use of x = –1 M1 for attempt to link the two remainders

5 a = b2, 2a – b = 3 2b2 – b – 3 = 0 or 4a2 – 13a + 9 = 0

leading to 4

9=a ,

2

3=b

B1, B1 M1 A1, A1

[5]

M1 for solution of equations leading to a quadratic. Final A1 – correct pair only.





© UCLES 2010

6 x = 2 or –4 or 3

1−

Either (x – 2)(3x2 + 13x + 4) or (x + 4)(3x2 – 5x – 2) or (3x + 1)(x2 + 2x – 8) (x – 2)(x + 4)(3x + 1)

x = 2, –4, 3

1−

B1

M1 A1 M1, A1

A1

[6]

B1 for spotting a solution

M1 for attempt to get quadratic factor A1 for correct quadratic factor M1 for dealing with quadratic factor

A1 for correct factors A1 for all solutions

7 (i) Graph of modulus function B1 B1 B1

[3]

B1 for shape B1 for 5 marked on y axis

B1 for 3

5marked on x axis

(ii) Straight line graph B1 [1]

B1 for straight line with greater gradient

(iii) 8x = ±(3x – 5)

leading to 11

5=x or 0.455 only

M1

M1, A1 [3]

M1 for attempt to deal with modulus

M1 for solution

8 (a) (i) fmin = –10, occurs when x = –2

B1 B1

[2]

(ii) e.g. x [ – 2 B1

[1]

Allow any suitable domain that makes f a 1:1 function

(b) (i)

−= 1

2

yx , leading to

g–1(x) = 2(x + 1)

M1 A1

[2]

M1 for a valid method of finding the inverse function

(ii) ( )1212

2

+=−−

x

xx

leading to x2 – 5x – 6 = 0 solution x = 6 and –1

M1

DM1 A1

[3]

M1 for correct order

DM1 for solution of quadratic





© UCLES 2010

9 (a) ( )cxxxxxx ++−=+−∫ 9

2

9

5

3d96

3

4

3

5

3

1

3

2

M1 A2,1,0

[3]

M1 for expansion and attempt to integrate –1 each error

(b) (i)

+++=

62

2

6

d

d

2

2

x

x

xx

x

y

M1 A2,1,0

[3]

M1 for attempt to differentiate a product. –1 each error

(ii) 62

1d

6

3 2

2

2

+=

+

+

∫ xxx

x

x

M1 A1

[2]

M1 for use of their answer to (i)

10 (i) 1e5

−=t or 52e1=+t

t = 12.1

B1 B1

[2]

(ii) distance = ln 10 – ln 5 = ln 2 or 0.693

M1 A1

[2]

M1 for s3 – s2

(iii) 1

2

2+

=

t

tv , v = 0.8

M1, A1

[2]


(iv) ( ) ( )

( )22

2

1

2221

+

−+=

t

ttta

When t = 2,25

6−=a , or –0.24

M1, A1

A1 [3]

M1 for attempt to differentiate a product or quotient

A1 all correct, allow unsimplified

11 (i) 3

4tan =x , x = 53.1°, 233.1°

M1 A1, √A1

[3]

M1 for an equation in tan Follow through on their first answer +180°

(ii) 11 sin y + 1 = 4(1 – sin2 y)

(4 sin y – 1)(sin y + 3) = 0

4

1sin =y , y = 14.5°, 165.5°

M1 M1 A1,√A1

[4]

M1 for use of correct identity M1 for dealing with quadratic Follow through on their 14.5

(iii) 2

1

32cos −=

+π

z

3

2

3

2ππ

=+z , 3

4π so

6

π

=z , 2

π

B1 M1 A1, A1

[4]

M1 for correct order of operations





© UCLES 2010

12 EITHER

(i) 4

cos6

sin3ππ

BA += , BA

2

1

2

1

3 +=

xBxAx

y3sin32cos2

d

d−=

π

π

sin3

3

2

cos24 BA −=−

A = 4, B = 2

M1 A1

M1

A1 A1, A1

[6]

M1 for attempt at substitution A1 for correct equation


A1 for all correct A1 for each

(ii) ∫ +=3

0

d 3cos2sin4

π

xxBxA

3

0

3sin3

2cos2

π

+−= xB

x

( )2sin33

2cos2 −−

+−= π

π B

, = 3

M1

A2,1,0

DM1,A1 [5]

M1 for attempt to integrate

–1 each error

DM1 for use of limits

12 OR

(i) 268

d

d

xxx

y−=

Grad at A = 2, perp grad = 2

1−

At A, y = 2

Equation of normal: ( )12

12 −−=− xy

C (0, 2.5)

M1 M1

B1

DM1

A1 [5]

M1 for differentiation M1 for use of m1m2 = –1

B1 for y coordinate

DM1 for finding equation of normal

A1 answer given

(ii) B (2,0)

( ) ∫ −++=

2

1

32d24125.2

2

1xxxA

2

1

43

23

4

25.2

−+=xx

= 12

49

or 4.08

B1

M1

M1 A1 DM1

A1 [6]

B1 for coords of B

M1 for area of trapezium

M1 for attempt to integrate A1 all integration correct DM1 for correct use of limits





© UCLES 2010

1 –1.5 Solve 2x + 10 = –7 or (2x + 10)2 = 49 –8.5

B1 M1 A1

[3]

2 Find f(2) or f(–3) or long division to remainder 8 + 4a – 30 + b = 0 or 4a + b = 22 –27 + 9a + 45 + b = 75 or 9a + b = 57 Solve simultaneous equations a = 7, b = –6

M1 A1 A1 M1 A1

[5]

3 (i) Solve 0.5 = e–34k using ln or log correctly

k = 0.0204 or 2ln

34

1

M1 A1

(ii) ekt = 5 or e–kt = 0.2 with k numerical

5ln

1

kt = with k numerical

79

B1

M1

A1 [5]

4 ( )

−

5.0

3.0

2.0

012

531

567

215 or ( )

−

2

1

5

055

136

217

5.03.02.0

Matrix multiplication using 3 × 3 matrix

( )303132 or

7.0

6.3

7.5

(or transposed)

Matrix multiplication of 1 × 3 with 3 × 1 30.7

B1

M1

A1

DM1 A1

[5]

5 Eliminate y x2 + (8 – m)x + 9 = 0 Use b2 * 4ac Reach (8 – m) * ±6 or solves m2 – 16m + 28 * 0 (* is either > or =) m = 2 and 14 m < 2, m > 14

M1 A1 DM1 DDM1 A1 A1

[6]





© UCLES 2010

6 (i) 7 × 6 ×5 × 4 840

B1 B1

(ii) 2 × 6 × 5 × 4 or ( )8407

2×

240

M1

A1

(iii) 2 × 5 × 4 × 2 or ( )2406

2× or clear indication of method

80

M1

A1 [6]

7 (i) ( )( )xxxV −−= 60245

Correctly reach 3221652700 xxxV +−=

M1

A1 ag

(ii) 2

63302700d

dxx

x

V+−

=

Solve 3 term quadratic expression for 0

d

d=

x

V.

10 only

B2,1,0

M1

A1 [6]

8 (i) 9lg3lg2 = or 23lg3lg2 =

10lg1 = Use lg rules correctly to eliminate logs (e.g. 9(5x + 10) = 10(4x + 12)) x = 6

B1 B1 M1 A1

(ii) Express in powers of 3

=

−

+

− )2(3

34

7

4

3

3

3

3

y

y

y

y

Correctly use rules of indices y = 4

M1

M1 A1

[7]





© UCLES 2010

9

250

60sin

80

sin=

α

α = 16.1 β = 104

βcos25080225080222

×××−+=v or

==

αβ sin

80

60sin

250

sin

v

v = 280(.2…)

v

t500

=

1 hour 47 minutes or 107 mins

M1 A1 A1√ DM1 A1

DM1

A1 [7]

10 (i) 5

1=

ABm

Use m1m2 = –1 in equation for BC [y – 5 = –5(x – 6) or 5x + y = 35] C (7,0) Use mCD = mAB and point C in equation of line

CD: y(–0) = 5

1(x – 7) or x – 5y = 7

B1

M1 A1 M1

A1

(ii) At D x = 1 At D y = –1.2 Method for area not involving measuring 28.6

M1 A1 M1 A1

[9]

11 (i) tan x = 0.6 31(.0) or 30.96(…) 211 (= 31 + 180)

B1 B1 B1√

(ii) Use cos2 y = 1 – sin2 y 2sin2 y + sin y – 1 = 0 Solve 3 term quadratic for sin y 30 and 150 270

M1 A1 M1 A1 B1

(iii) cos z = 0.3 1.27 5.02 or 5.01 (= 2π – 1.27)

B1 B1 B1√

[11]

OR





© UCLES 2010

12E

(i) fg(9) = f(4) evaluated or fg(x) 411

532

−

+

−

+=

x

x

21

M1 A1

(ii) Method for f–1(x) f–1(x) 14 −+= x

Put 1

53

−

+=

x

xy and rearrange

g–1(x)3

5

−

+=

x

x

M1

A1

M1

A1

(iii) Rearrange two of xx

x

x

x=

−

+=

−

+

3

5

1

53 to quadratic equation

2(x2 – 4x – 5) = 0 Solve 3 term quadratic 5 only

M1 A1 M1 A1

[10]

12O

(i) 4 B1

(ii) Differentiate v to find an expression for a 6 – 8 sin 2t Substitute t = 5 10.3 to 10.4

M1 A1 DM1 A1

(iii) 14 B1

(iv) Integrate v to find an expression for s s = 3t2 + 2 sin 2t Use limits 4 and 5 23.9

M1 A1 DM1 A1

[10]





© UCLES 2010

1 –1.5 Solve 2x + 10 = –7 or (2x + 10)2 = 49 –8.5

B1 M1 A1

[3]

2 Find f(2) or f(–3) or long division to remainder 8 + 4a – 30 + b = 0 or 4a + b = 22 –27 + 9a + 45 + b = 75 or 9a + b = 57 Solve simultaneous equations a = 7, b = –6

M1 A1 A1 M1 A1

[5]

3 (i) Solve 0.5 = e–34k using ln or log correctly

k = 0.0204 or 2ln34

1

M1 A1

(ii) ekt = 5 or e–kt = 0.2 with k numerical

5ln1

kt = with k numerical

79

B1

M1

A1 [5]

4 ( )

−

5.0

3.0

2.0

012

531

567

215 or ( )

−

2

1

5

055

136

217

5.03.02.0

Matrix multiplication using 3 × 3 matrix

( )303132 or

7.0

6.3

7.5

(or transposed)

Matrix multiplication of 1 × 3 with 3 × 1 30.7

B1

M1

A1

DM1 A1

[5]

5 Eliminate y x2 + (8 – m)x + 9 = 0 Use b2 * 4ac Reach (8 – m) * ±6 or solves m2 – 16m + 28 * 0 (* is either > or =) m = 2 and 14 m < 2, m > 14

M1 A1 DM1 DDM1 A1 A1

[6]





© UCLES 2010

6 (i) 7 × 6 ×5 × 4 840

B1 B1

(ii) 2 × 6 × 5 × 4 or ( )8407

2×

240

M1

A1

(iii) 2 × 5 × 4 × 2 or ( )2406

2× or clear indication of method

80

M1

A1 [6]

7 (i) ( )( )xxxV −−= 60245

Correctly reach 32

21652700 xxxV +−=

M1

A1 ag

(ii) 2

63302700d

dxx

x

V+−

=

Solve 3 term quadratic expression for 0d

d=

x

V.

10 only

B2,1,0

M1

A1 [6]

8 (i) 9lg3lg2 = or 23lg3lg2 = 10lg1 = Use lg rules correctly to eliminate logs (e.g. 9(5x + 10) = 10(4x + 12)) x = 6

B1 B1 M1 A1

(ii) Express in powers of 3

=

−

+

− )2(3

34

7

4

3

3

3

3

y

y

y

y

Correctly use rules of indices y = 4

M1

M1 A1

[7]





© UCLES 2010

9

250

60sin

80

sin=

α

α = 16.1 β = 104

βcos25080225080222

×××−+=v or

==

αβ sin

80

60sin

250

sin

v

v = 280(.2…)

v

t500

=

1 hour 47 minutes or 107 mins

M1 A1 A1√ DM1 A1

DM1

A1 [7]

10 (i) 5

1=

ABm

Use m1m2 = –1 in equation for BC [y – 5 = –5(x – 6) or 5x + y = 35] C (7,0) Use mCD = mAB and point C in equation of line

CD: y(–0) = 5

1(x – 7) or x – 5y = 7

B1

M1 A1 M1

A1

(ii) At D x = 1 At D y = –1.2 Method for area not involving measuring 28.6

M1 A1 M1 A1

[9]

11 (i) tan x = 0.6 31(.0) or 30.96(…) 211 (= 31 + 180)

B1 B1 B1√

(ii) Use cos2 y = 1 – sin2 y 2sin2 y + sin y – 1 = 0 Solve 3 term quadratic for sin y 30 and 150 270

M1 A1 M1 A1 B1

(iii) cos z = 0.3 1.27 5.02 or 5.01 (= 2π – 1.27)

B1 B1 B1√

[11]

OR





© UCLES 2010

12E

(i) fg(9) = f(4) evaluated or fg(x) 411

532

−

+

−

+=

x

x

21

M1 A1

(ii) Method for f–1(x) f–1(x) 14 −+= x

Put 1

53

−

+=

x

xy and rearrange

g–1(x)3

5

−

+=

x

x

M1

A1

M1

A1

(iii) Rearrange two of x

x

x

x

x

=

−

+=

−

+

3

5

1

53 to quadratic equation

2(x2 – 4x – 5) = 0 Solve 3 term quadratic 5 only

M1 A1 M1 A1

[10]

12O

(i) 4 B1

(ii) Differentiate v to find an expression for a 6 – 8 sin 2t Substitute t = 5 10.3 to 10.4

M1 A1 DM1 A1

(iii) 14 B1

(iv) Integrate v to find an expression for s s = 3t2 + 2 sin 2t Use limits 4 and 5 23.9

M1 A1 DM1 A1

[10]





© UCLES 2010

1 ( ) 44

d

d−

+= xkx

y

k = –30

Use xx

yy ∂×=∂

d

d with x = 6 and px =∂

–0.003p or 1000

3−p

M1

A1

M1

A1√ [4]

2 Integrate 3x2 – 6x (y =)x3 – 3x2 (+ c) Substitute x = 4, y = 22 y = x3 – 3x2 + 6

M1 A1 M1 A1

[4]

3 (a) (i) 4 B1

(ii) 3 B1

(iii) 5 B1

(b) (i) π or 180 B1

(ii) 6 B1 [5]

4 (i) 1 + 6x + 15x2 + 20x3 B2,1,0

(ii) Substitute x = p – p2 Multiply out brackets to obtain terms in p3. (–30 p3 + 20p3) –10

M1 M1 A1

[5]

5 (a) ( )154 − B1+B1

(b) (i)

−−=

−

32

54

2

11

C or

−−−=

−

32

54

2

11

C B1+B1

(ii) X = C–1D evaluated

−− 89

1217

M1 A1

[6]





© UCLES 2010

6 (a)

B1

(b)

B1+B1

(c) 24 – x + x + 18 – x + 3x = 50 or 24 + 18 – x + 3x = 50 Solve for x (4) 12

B1 M1 A1

[6]

7 Eliminate y (or x) 4x2 + 12x – 160 = 0 (or y2 + 18y – 88 = 0) Factorise 3 term quadratic x = 5 and –8 (or y = –22 and 4) y = –22 and 4 (or x = 5 and –8) Use Pythagoras 29.1 or 845 or 513

M1 A1 M1 A1 A1√ M1 A1

[7]

R

P

B A

Q





© UCLES 2010

8 (i) ( ) 18230252352

++=+ or 1821521525 +++ B1 AG

(ii) ( ) ( )2352230432 +=+

256 +

M1

A1

(iii) ( )235 − B1

(iv) 235

1

+

235

235

235

1

−

−×

+

7

235 −

alternative for last 3 marks

23043

23043

23043

1

−

−×

+

49

23043−

7

235 −

B1

M1

A1

(M1)

(A1)

(A1) [7]

9 (i) 22

158 + AO = 17

M1 A1

(ii) ( )15

8arctan2−= πAOB or

××

−+

17172

301717arccos

222

AOB = 2.16

M1 A1

(iii) Complete, correct plan with s = rθ 82.7

M1 A1

(iv) Complete, correct plan with θ2

2

1rA =

432

M1 A1

[8]





© UCLES 2010

10 Use product rule 2xex + x2ex Evaluate gradient of tangent at P

e + 2e = 3e

Equation tangent y – e = 3e(x – 1) or y = 3ex – 2e At A, e20 −=⇒= yx or –5.44

Use m1m2 = –1, equation normal is ( )1e3

1e −−=− xy

At B, 1e302+=⇒= xy or 23.2

Area OAB = ee33+ or 63(.0) or 63.1

M1 A1 M1 A1 M1 A1 M1 A1 A1

[9]

11 (i) abAQ −= 3

AQOAOX µ+=

( )aba −+ 3µ

M1

M1

A1

(ii) baBP −= 2

BPOBOX λ+= ( )bab −+ 2λ

M1

M1

A1

(iii) Equate vectors and solve

−=

=−

λµ

λµ

13

21

2.0=µ 4.0=λ

M1 A1 A1

[9]

12E (i) Plot with (attempt at) linear scales v 2 4 6 8 v

2p 24.9 45.4 65.9 86.4

M1 A1

(ii) v2p = a + bv

Valid attempt at gradient b = 10.2 to 10.3 a = 4 to 4.5

B1 M1 A1 B1

(iii) Attempt to rearrange to ...+=

v

aY

pv on y-axis

M1

A1

(iv) Gradient is a y intercept is b

A1 A1

[10]





© UCLES 2010

12O (i) Plot with (attempt at) linear scales ln t 0.69 2.08 3.18 3.99 ln r 3.09 4.90 6.33 7.38

M1 A1

(ii) Gradient is 1.3 Intercept is 2.2 ln r = 1.3ln t + 2.2 ln r = 1.3ln t + ln 9 ln r = ln t1.3 + ln 9 r = 9t

1.3

B1 B1 M1 M1 M1 A1√

(iii) Gradient is 1.3 Intercept 0.95 or lg9

B1√ B1√

[10]



2002 – 2011

Compiled & Edited By

Dr. Eltayeb Abdul Rhman

www.drtayeb.tk

First Edition

2011

MS BANK

ADDITONAL MATHEMATICS

ms-bank-0606

Documents

Transcript of ms-bank-0606