I-E+

53 BP-55

CEDAR GIRLS' SECONDARY SCHOOLPreliminary Examination 201 8Secondary Four

ADDITIONAL MATHEMATICSPaper 1

AdditionalMaterials: AnswerPaper

4047t0117 August 2018

2 hours

READ THESE INSTRUCTIONS FIRST

Write your Centre number, index number and name on all the work you hand in.Write in dark blue or black pen on both sides of the paper.You may use a HB pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer all the questions.Write your answers on the separate Answer Paper provided.Give non€xact numerical answers conect to 3 significant figures, or 1 decimal place in the case ofangles in degrees, unless a different level of accuracy is specifed in the question.The use of an approved scaentific calculator is expected, where appropriate.You are reminded of the need for clear presentation in your answeG.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets I I at the end of each question or part question.The total number of marks for this paper is 80.

This document consists of q printed pages and 1 cover page

flurn over

Quadratic EEaation

For the equation cr' + bx + c - 0.

54

2Mothematical Formulae

I. ALGEBRA

2a

BP-56

Binomial expansion

(a+b)n =0'1 +

Identities

Formulue for A.ABC

n\ n(n-l)...(n-r+l)rlr l(n - r)l

2, TRIGONOMETRY

sin'I+cos'zl=1

sec' A =1+ tan' A

cosec2A-1+cot2 A

sin(,4 t 8) = sin ,4 cos B t cos ,4 sin B

cos(l t B) = cos I cosB + sin ,4 sin I

(:)

- tan A+:lan Btan(AtRt=_I + tan ltan B

2tan Alan2A=-| - ianz A

n

I

')"-,'.(\), ,u, . .(;)u ,b, + +b,,,

where n is a positive integer and

sin2A = 2sin AcosA

cos2l = cos: A-slnl A=2cos'A-1-1-2sin'1 A

a b c

sin I sin -B sir C

a' =b'+c' -2bccosA

A - labsin C2

-tx"E -q*

BP-5755

3

Ansu€r all the questions.

I A cone has curved surface ur"u ,(ll - "l-t) "m2 and slant heigfrt (z-:rE)cm.Without using a calculator, find the diameter of the base of the cone. in cm, in theform of a + bJl , where a and 6 are integers. t4l

2 The roots ofthe quadratic equation 5x'?-3x+t=0 u." 1 undd

Find a quadratic equation with roots a' and S .

3 (i) Show that 2x2 +1 is a factor of 2x'-4x'+x-2.

lh'-5x:-ll(ii) Express --;:-;:j--jj;. in partial fractions.' 2x'-4x'+t-2

Ip

tsl

4 (i) Sketch the graPh of Y =4

G for x>0. t2)

(iD Find the coordinates of the point(s) of intersection of y = 4r and y'? = 31;r. l4lJ The diagram shows a cylindet ofheight lr cm and base radius,'cm inscribed

in a sphere of radius 35 cm.

1225-12 . 12)

I6l

L2l

t4l

(i) Show that the height of the cylinder, i cm, is given by ft = 2

(iD Given tlat r can vary. find the maximum volume of the cylinder

[Turn over

56 BP-58

1

2 -sec2 x cos x -sin x6 (D Show-that2 tan x + sec' .r cosr+sln.T t3l

t3l

t3l

t4l

(ii) Hence find, for 0 ( .;r ( 2z , the values of .r for which - 6-3sec'?I

=1.2tan x+ sec2 x 2'

d'y= 2dx' ezx-1

7 A curve is such that and the pointP(1.5. 2) lies on the curve.The gradient ofthe normal to the curve at P is 10. Find the equation ofthe curve. t61

3

E The diagram shows the graph of y = xl -4;r which passes through the origin O andcuts the r-axis at the pointl(16. 0). Tangents to the curve at O and I meet at thepoint B.

v

rAo

B

(i) Shou that B is the point (ti. ,,j)

(iD Find the area ofthe shaded region bounded by the curve and the lines OBand AB.

BP-5957

5

9 A tram, moving along a straight road, passes station O with a velocity of 975 mlmin.Its acceleration, a mlminz,l mins after passing through station O, is given bya=2t-80.The tram comes to instantaneous rest, first at station I and later at station B.Find

(i)

(iD

(iiD

t3l

L2l

t21

the acceleration ofthe tram at station I and at station B.

the greatest speed ofthe tram as it travels from station ,4 to station B,

the distance between station I to station B.

10 (D By considering the general term in the binomial expansion of r'--L')1_.,: lwhere ft is a positive constant, explain why there are only even powers of;in this expansion.

(ii) Given that the term independent ofx in this birromial expansion i, ], nna27the value offt.

(iiD Using the value oft found in part (ii), hence obtain the coefficient of xr8/ I \6

in (z-:,')1,'-+ I\ rax .i

l2l

t2|

I4l

t6l

12)

l1 Mand Nare two points on the circumference ola circle, where M is the point (6, 8)andNisthepoint(10. 16).Thecentreof the circle lies on the line y=)aa1 .

O Findthe equation ofthe circle inthe form x2 + y2 +ax+by+c=0, wherea, b atd c are constants.

(ii) Explain whether the point (9, l0) lie inside the circle. Justifr your answerwith mathematical calculations.

lTurn over

58 BP-60

6

12

A F

D

In the diagram. t$,o circles intersect at B and,F. BC is the diameter ofthe largercircle and is the tangent to the smaller circle at B.Point I lies on the smaller circle such that IFEC is a straight line.PointD lies on the larger circle such that BHED is a straight line.Prove that

(i)

(iD

(iiD

(iv)

CD is parallel to AH.

lB is a diameter of the smaller circle,

triangles IBC and .BFC are similar,

AC2*AB2=CFxAC.

End of Paper

t3l

12)

12)

L2)

9I

-

BP-6159

CEDAR GIRLS' SECONDARY SCHOOLPreliminary ExaminationSecondary Four

ADDITIONAL MATHEMATIGS

AdditionalMaterials: AnswerPaperGraph paper (1 sheet)

4047t0220 August 20'18

2 hours 30 minutes

READ THESE INSTRUCTIONS FIRST

Write your Centre number, index number and name on all the work you hand inWrite in dark blue or black pen.You may use a pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or corection fluid.

Answer all questions.Write your answers on the separate Answer Paper provided.Give non-exact numerical answers corect to 3 significant figures, o|I decimal place in the case ofangles in degrees, unless a different level of accuracy is specined in the question.The use of a scientifc calculator is expected, where appropriate.You are reminded of the need for clear presentation in your answers.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part questionThe total number of marks for this paper is 100.

This document consists of I printed pages and 'l cover page

ffurn over

f{@F

60 BP-62

Mathematicol Fornulae

I. ALGEBRAQuqdratic Equation

For the equation qr'+bx+c=0.

-brJb'4*2a

Binomial expansion

(q +b)n = a't +

Identities

Formulae .f

ot3

BP'63

Answer all the questions.

r (u) Given that lt""(*}fi) = z+zlgx-lgy ,where.;r and y are positive numbers,express, in its simplest form, y in terms of x.

(b) Given that p =logrq, express, in terms ofp,

(i) r"s-l,ll.\s)

(i0 log,4q.

2 (i) Show that 1d.r

(sin .rcos r) = 2 cos'2 :r - I .

(iD Hence, without using a calculator. find the value ofeach ofthe constants aand b for which

li cos', d, = a + bz.Jo

3 The variables -t andy are such that when uulu"s of 1+1u.e plotted against l,yxxa straight line with gradient m is obtained. It is given that , = 1 *6.n t = I

and thar v =1 *h"n , =f.'22

(i) Find the value ofrz.

(ii)11

Find the value of x when - +- -3.v.x

(iiD Express y in terms of .r.

12)

tzl

t2l

t3l

t4)

t4l

t2)

tzl

Cedar Girls' Secondary School 1047 /\z/SllPrclim 2018 [Turn over

6i

The equation ofa cun'e is y = 163 l Otz , wherep is a positive constant.

(D Show that the origin is a stationary point on the curve and find the.x-coordinate ofthe other stationary point in terms ofp.

(ii) Find the nature ofeach of the stationary points.

Another curve has equation y = x' + px' + px .

t3l

t3l

(iii) Find the set ofvalues ofp for which this curve has no stationary points. t3l

A quadratic function f(.:r) is given by f(.r) =761r-2;: -(x-3)(;r+2), where I is aconstant and I +1.

(i) Find the value offr such that the graph of l, = f(r) touches the r-axis at onepoint. t3l

Find the range ofvalues olftforwhich the function possesses a maximumpoint. trl

(iiD Find the range ofvalues of& for which the value ofthe function neverexceeds 18. t3l

6 (a) A substance is decaying in such a way that its mass, rn kg, at a time / yearsfrom now is given by the formula

m = 240e-o *'

(i) Find the time taken for the substance to halve its mass 12)

(iD Find the value oft for which the mass is decreasing at a rate of2.1 kgper year. t3l

(b) The noise rating. N and its intensity, l are connected by the formula

BP-64

1

:l

(,rf), *r,*" o,s a constant.

(ii)

N=10

A hot water pump has a noise rating of 50 decibels.A dishwasher, however, has a noise rating of 62 decibels.

lntensiw of the noise from the dishwasherl- ind the value ol -

Intensity ofthe noise from the hot water pumpt3l

Cedar Girls' Secondary School 404'l l02l54 ?rclim 12018

BP-65

7

bJ5

v=(6x+2)'(1,2)

The diagram shows the curve I =(ax+\! and the pointl (1,2) which lies on thecurve. The tangent to the curve at I cuts the y-axis at B and the normal to the curveat I cuts the x-axis at C.

C -I

(i)

(ii)

(iiD

Find the equation ofthe tangent lB and the equation ofthe normal ,{C.

Find the length ofBC.

Find the coordinates ofthe point ofintersection, E. of OA and BC.

t4l

12)

t41

12)

Ljl

tll

L2l

8 It is given that lt =tanx and y2-lcos2x+l .

(i) State the period, in radians, of y, and the amplitude of yr.

For the interval 03x32r,

(iD sketch, on the same diagram, the graphs of y, and y,

(iii) state the number of roots of the equation ltan .rl - 2 cos 2r = I ,

(iv) find the range(s) ofvalues of.r for which y, and y, are both increasing asxlncreases.

Cedar Girls' Secondary School 1017 l02l SllPrelim 20 I I lTurn over

646

M

v

BP-66

9 (a)

(b)

x

o

The diagram shows paft ofthe curve,Y = lan x cos2x ,

and its maximum point M.

Show that 9 = 4cost x -sec'x- 2.dx

Hence find the x-coordinate of M.

(i)

(it)

A particle moves along the line .v = tn.,E in such a way that the' \x-2

t5l

t3l

[3]

x-coordinate is increasing at a constant rate of 0.4 units per second.Find the rate at which they-coordinate ofthe particle is increasing at lheinstant when r = 2.5.

Cedar Girls' Secondary School 1047/02/54 Prelim i20l 8

BP-67657

I0 (a) The function fis defined forall real valuesof;rby f1x1=s" -3.2'

(i) Show that f '(x)>0 forall values of.x

(iD Show that f "(;r)=ftf1r), *1lere ft isan integer.

(iii) Find the value of:r for which f ' (.x) = 0 in the form x = plnq.where p and q are rational numbers.

(b) The function g is defined for all real values of;r by g\x) = e" +3e-"The curve / = g(r) and rhe lin. * = 1 ln 3 intersect at poinl 0.

4

Show that the y-coordinate of O is ,tr6 , where I is an integer.

I I Solutions to this question by accurate drawing will not be accepted.

l2l

12)

l2l

12)

vA(2,7)

F

E

(6, s)

xo B(r.0)

The diagram, which is not drawn to scale, shows a triangle IBC with verticesA(2,7), B(1, 0)andC(6, 5) respectively. EandFare pointsonBCandlCrespectively for which lE is perpendicular to BC and BF bisects AC.G is the point of intersection of lines AE and BF.Find

(D

(iD

(iiD

t4)

I2l

t2l

the coordinates of G,

the coordinates of the point D such that ABCD is a parallelogram,

the area of ABCD.

Cedar Girls' Secondary School 1047 /02/S4iPrcl1m 2018 ITurn over

668

bm

om

o

The diagram above shows a quadrilateral in lvhich PX= n m and QX - b m.Angle OQX = Angle OPX -- 0" and OQ is perpendicular to OP.

(D Showthat OP = acos9 +bsin9.

(iD It is given that the maximum length of OP is rEm and the correspondingvalue of d is 63.43 ".By using OP = Rcos(.0 -a), where R > 0 and d is acute, find the valueofa and ofb.

(iiD Given that OP = 2.15 m, find the value of d.

End of Paper

BP-68

t2

X

t3l

t5l

12)

Cedar Girls' Secondary School 4047/02i 54 Prelim /2018

BP-69

67Cedar Girls' Secondary School

2018 Prelfuninary Examination 2Additional MathGmatics 4(X7 Paper 1Solutions

Mati ernatic s Deparftnent

Qn Working

I *r = n(rt -.,11)

r(rz-f)w(rz-S)74^E

Xt*d3t*t6

1 1 0 + 44.,fr

22

r = s+2J1Diameter: =10+4J5cm

2

11-3-+-= --ap s

J

5

l1ap5d9=51,1 d+P-+-=-a0 aBd+p 3

55d+ P=3

a'+ fi -toapt(t'z-o8+p'z)=q@1 p\z.3dpl-3(3)1 3 (5) I: _18

a3p3 =(ap)3

-125

Equation: x'? +78x +725 = O

Qn Working

BP-70

Cedar Girls' Secmda-y School

Za3 -4a.2 "ya-2=(2xz +l)(x-Z)It is divisible by 2r'?+1 withno ranainder

-5.r1 +11"r-11 A Bt+C2x3-4x2+x-2 --,!-x-2 2i +1

-5:r2 + 1 1r- 1 1 = A(zrc) +l) + (Bx + C)Q - 2)When::2,A:-lCornparing x2: -5= 2A+ B-5--2+BB=-jConparing corstar : -11-- A-ZC-lt= -l-2CC=5

68

.I

Mathernatics DepartrneDt

3ii

-5.r'+11*-11 1 5-3*2x3 -4x2 + t-2 t-2 2x2 +l

1

v

l= 4x z

0

BP-71

69Cedar Girls' Seconda'y School Mathanatics Department

4ii

l!=sr,I

81r'? = 16

4r=1-9

4I=-

9

Y=6

Point ofintersection:

5i lt+ rz =352 (Py'thagoras' Theorem)

2

L=1225-r'4

hz =4(.1225-11)

h= 225 - rz(shown)

.-u v -rr'12^@1v =2rr'?(1225-r\2

4 = z o r, r! cz r)lt 225 -,, 1-i 1 * 1t zzs -,, tl G r r)dr 2'll

=-2nrtQ225-r'z) 2 +4nr(1225 -r?)lll

-2Tr3 (1225 - r2)'7 + 4nr(1225 - r'1)i - 0r3 ='2r(1225-r2l3r3 =2450 r

r =28.577 ( reject r = 0 and -ve r )

Usirg First Derivative Test,

4

G - 81r

r4 _)[;''J

r 28.s77 (-j '28.577 28.577 (+)dV

sicn of -"&

+ve 0

siope

Z is maximum at r : 28.5ii

BP.7?

70Cedar Girls' Seconda'v School

Maxirmm voiurne.

v = n(28.577)1(

= 103688

= lO4 000: 104 000 cm3 (3 s.f.)

Mathanatics D€pa.rtrcnt

1225-(28.5'77)?\)

Qn Worklng

6i

6ii

LHS.2 - sec? x 2-(tan':+l)

2 tan;r + sec2 r 2 tan:+ (tan2 x+1)I - tan2 .r

2tan x+tart2 x+L(1 - tan *)(1 + tan *)

(tan.t + l)'l-tanr1+ tanr- srn.rl__

co6.r- sln.rl+-

cos,cosr-sln, oosl

=

-x-

cos .t cos .r + s1n rcos.r - sin-rcos -n + sln Jr

(shown)

^ cQsr - sin.r 3--tx- = -oosx + sin;r Zcos.x - sinr 1cos,r + sin:r 2zcosjc -zsin x= cos x + sin x

cos.ir=3sinx

Itiar.r = -t:0.322 or : = 3.46 (3 s.f.)

BP-73

71Cedar Girls' Seccndry School Mathematics Departnert

Qn

7dy ^ 3-2xdaz

dy =21-l ".r,1* "da2

dv--:-:-e"'+c&t

Gradiort at tangent atP : -l

-e3-1t I

10

when.r:1.59

10

dv 9

10dr1

'2]-1r

e9+-r+c

10

2:Lnt-21's12

3

20

9(1.5) + c

l0

Eqt: y1 3- 2r

e93+-')r+-

2 l0 z0

'W'orking

8P.74

7t2

Cedar Girls' Secqrda-y School Mathernatics Dcpaftnent

Qn Working8i

It

av J?-_:_ = _tt _4dazAto.t:0- 9=-4' '&

Equation O3: y = 4t ..,(l)

AtA.x:16.4=2'dxY=bt+c0 = 2(16)+ c

c=-32EquationlB: y= 2x-32

2x-32=4t_lx=5-

3Subinto (Q,

n=-2r!J

n=(sL[ 3'-zr])an

*">

Area of cuwe : [',; - * *l = [.,,*

-,, ],'

-102.4 units2

Area of trian.sle OAB:1x16^211"23

=1702 rrLits23

Area of shaded r eglon =fiOZ-rc2.a:68.3 urits2 (3 s.f)

BP-75

73Cedar Girls' Seconday School Matiernatics Departnert

Qn 'Working

9i

9ii

9iii

a -2t -80v={ -8at + ct:A,v:975975 = (o)'1 -80(0) + cc :975v = t2 -8Ot +9'75

Whan v:0,t2 -8Ot+975=A(r- 15)(r- 6, = 0t:15,t-65

Accelerationat a= 2(15) - 80- - 50 m/min2

Acceleration at a: 2(65) - 80: 50 m/min2

When a :0,15+65

2

r:40v=(40)'] -80(40)+975v:-625rnlmnGreatest speed: 625 ln/frir:-

DstancerB: ll,l,)r -ro,*nrt

o4

=[i-"'.'"'1,1

= 20833!m3

: 20 800 m ( 3.s.0: 203 km

BP-76

Cedar Gids' Seconda'y School

General Term :

Mathfiutics Departnent

jrr'II, ,ro-0,

( |( irI l-;\r

(6

[,I

--xk2

(:)

Since 6r is an eve[ nurnbe\ 24- 6r will be even.

For independent tenn, 24-6r =0= r= 4Ie\r r \' 5[4J1. ft] 27155

k=+

27

F-=r(asr>o)(z -lrt )( ... + Term in r'E + Term in *'2 *...)For term rn )B , 24 6r =18 = r =7," /6\' l\,,Therel'ore. term in ./' I - ,,xIrJ\ 3'For term in J'?, 24-6r = 12 => r = 2

Therefore, term in l=

Z:.-tE

[6][-!]',,, =1,,,[2]( 3/ 3

Coefficient of l= 2 (-2) + (-3) /s\t_t[:.,1

9

Qn Working

10(i)

(iii)

(iD

Qn Working

111

11ii

LetMNte a chord of circle.

Midooinr or,{a^r' - f lE ltg)""""'-( z 2 )= (8, 12)

cradient of,\ZM= 16-810- 6

Gradient ofperpendictlar bisector = -l2

Equation of perpendicular bisector ofMN:

,-tz = -1(r-s)' 2'I __

v = --r+ l6"2

-l;+16 = 2*+12

x=6Y =13Cente of cirole : (6, 13)

Radius:13-8: 5 urfts

Equation ofcircle:(r- 6)'? + (y -13)'? = 5?x' + y' -l2x-26l+ 180 = 0

Lergth of point to centre ofcircle

9-6)? +(10-13)z

=.f8= 4.24 units< 5 ( radius)

Yes, the point lies inside the circle as its length fromthe centre of the circle is less than the radius.

BP-77

Cedar Girls' Secmdary School Mathanatics Departnent

Qn Working

tzii

l2iii

12iv

t2i

BP-78

76Cedar Girls' Seccndary School Nfath€rnatics Departrnent

Z-BDC = 90" (l in semichcle)ZBFC - 90o (l in same segment) or (Z in semicircle)LBFA = l8A" - 90q (ailj Zs oir shaigtrt Lne)

ono

Z-B!A4 = /BFA = 90o (Z in same segment)L4HD = 180" - 90o (adj Zs on straight the)= 900

/AHD -- IBDC - IHDC (altemate angles),.,CD /I AH

/tsIA = IBFA=90" (Z ir mme segment)l3 is a diameter of the smaller ctrcle(Z in semicircle) .

Since LB and BC are tangents to the smaller and bigger circlerespectively, ABC =90" (tan L n0lAtsC = ZBFCZ-BCA= IFCB (conrnon 1)Triangle l.BC is similar to tiangle BFC as all coresponding angles are

equal.

BC AC=: = :j:- (raUo of simrlar hrangles)FC CBBC'z=CFxAC

BC'? = AC? - LB? (g,t}agoras' Theorem)

:. ACz - AB' = CF xlC (shoqrD

BP-79

77Cedar Girls' Seccnday School

2018 Preliminary Examlnation 2Additional Mathemalics 4A€l 12Solutions

Matiernatics Depa-rtsnent

Working Marks Total Remarks

la

b(t)

b(t0

*(,ati)=z+zts*-rcy3lgl+lg y = 2a 219 t-lgYlgt+ 27gy =2tc("y')=z

ry'z--70'1=l}ohoo

'=l;10 l0.EJ, a

1og, l= 1eg, 1-1*, ,q

-A*p=-p

lqr4q =1ogr 4+lqrq

^ log,qloEEz

:2+3p

Total

I3I

121

t4

m

2(i)

(ii)

d

dr(sinxcos:)

= sin*( -sinr)+ cos*(cos:)

= cos2 .r -sinz .x:cos' .r-(1 - cos' ,)=2cos2x-I

Jl( z "o.' ,)a, = [srn.'cos.r] J

Jl( z "or',) a, - f r ar = [sin"rcos r] J

J'Jl1= -x-= -nna

2J,' ( zcos' x)d, = i+ [r] o'

iii*,,,1a,= ] - t -, =i, u:*Total

t21

I4l

t6I

Qn

an rfforking NIarks 'f otal Remarks

3(r)

(ii)

(iii)

.11lhe lmear equaton rs -+ --t1yx

ISutst v=- and r= I-56+7 = m+c =>rn+c=7

l1$ubst v=:-ard a=:'222+2=2m+c=2m+c=4m=-3 and c=10

f l)L_l[ *,]

+c

sin "1+3 =3=+l*.!=r,yxya2l

1=Ja1g31=lr3

11yx

r1)t-[ ,,]

+ 10

t+y _1+10xr.y a

,T

v l0-r 4

Total

l4l

t21

t8l

BP.8O

78Cedar Girls' Secondarv School Matlernatics Departrnent

t2I

BP-81

79Cedar Girls' Secmda'y School Matiematics Deparfrnent

Qn Worklng Marks TotaI Remarks

4(D

(iD

0ii)

y= a3 + px?

*=rf +2px=tQx+2p)

For stationary point, fr = 0

:.t=os1 *=-22When r=0, /=0.Therefore, (0, 0) is a stationary point.

The other r-coordinate olstationary point is2p

x1

1'\ =u,rzo(1r-

wtren r=0. {l*=2orors o>o' dx'Therefore, (0, 0) is a minimurn point

'r\rhen x = -2dlv-4=6OI

5

2p +2p=-2O.9*OrO3

Therefore, there is a maximurn point at x = -22

y=x3+pxz+ptc

9=3r'*2or*odr

si*. fl+ o, b2 - 4dc

BP-82

80Cedar Girls' Secondary School Mathetniitics Departnent

Qn Working Marks Total Remarks

s(i)

(ir)

Gii)

f(x) = *q* - 212 -(.r - 3)(x+ 2)=k(x'1-4x+4)-(r'-r-O=b2 - 4,g+ 4k- x2 +x+6= (t-l)l + (1- 4k)x+4k+6

Since it touches the r-axis at one point, b2 - 4ac = A(t - 4q'1 - 4

BP-83

81

Cedar Gids' Seccndary School Mathernatics Depa-rtsnent

Qn Working Marks Total Remarks

6a(i)

a(tt)

b

When ,: 0, m = 240Wher. 24Oe-o u' =12Ae-o

u' = 0.5

In0.5

-{.04t =17 .3No. of years : 17.3

@ = roO,+ OAl e-oo4 = -9.6e-o''dt-9.6e-ooo' -- -2.1

ln7.t9 6

38 0-0 04

ror"(!)"( r,l =50=IP

k= 10t

where 1, - irtensity ofpurnp

u!n=9-6.2=i'5]= ro,,"k 10 \.k/where -I, = intensity of dishwasher

1r=lou.'e=ts.aI P 10'k

Total

t2l

t3l

l3l

t8l

BP-8462

Cedar Gids' Seconda'y School Matiernatics D€partrnent

Qn.Working

Marks Total Remarks

7(i)

7ii

Tiii

Iy=(sx+2fdv L- -.-1 - 2---:- - _l oJ.i 2l i.o=_dr 3- ^.i(or+rr:rWhen 1

dy

dx

2 12

.x2

(0(r) + z[

Eqt of AB: y - z = |(x -1) +, = | x + |EEr of AC: y - 2 = -2(x -I)= y : -Za. + 4

When.r:0, y=1.5Coordinates of B : (0,1.5)When y= 0, -2x+4=0=> x=2Coordinates of C: (2, 0)BC= 1.52+2'=2.5rmits

tt.tGradteftof OA - ' " =21-0Therefore, egr of OA '. y = la

15 1Gradient of BC:-= _:

-24 "r3Therefore- eon ofBC: y = -1a a:-nAtE,

??2r---1-r+142

11x364 - 2-'- tt

(5

)t?1

- 11 -' 11!:? 11

Coordinates ofE : 6 I1l' 11

Total

t4I

t4

14I

tl0I

Use of chain rule

Correct substitr-lticl

i

BP-85

83Cedar Girls' Seconda-y School Mathernatics Departnent

Qn Working Marks Total Remarks

8i

II

TY

Period of / = r radiansAmplitude of y, = 2

tan r

n-

BP-86

84Cedar Girls' Secryrdary School Mathfiratics Depar8nent

Qn Working l\larks Remarks

9a(i)

(ii)

b

y=tantcos2J

$= tan, (-z sin 2"r ) + cos 2r (sec' ,;r )dr\=ffi {-, * r rlr,.x cosr) + (2cos': *- 1)= *4sin2 x+ 2-sec2:= -4(1- cos2 :)+ 2- sec2 x= 4cos2 r*secz x* 2

I

cos 7 .ir

When 9=0.dr

4cos2-r-sec2r-2=04cosa*-2cos2;r-i=0

cos' l =2! 4-4(4X-1)

8:0.80902cosx = 0.89945* = 0.452 or 25.9oThe *-coordinate of M is 0.452

Iv :l[ln 5:r - In(r - 2)l- 1t ' .rdv1 5 \ lfl__LI rt

1

dr2 5.x x-Z1l2x 2(t-2)

dy dy dr-=-x-dt dx dtwhen.x= 2s 9=/l- I

.],.04

' dt [5 2(05)JThe Rate is - 0.32 wits per secand..

8-0.32

25

Total

I5I

I3I

I3l

[1 1]

Total

BP-87

85Cedar Girls' Secmday School

f (t) = g2' -3ea'f '(x:) = 2ex' + 6e-2'since e" >o and t2' > o, f '(r)>o

f ,(x)=41'-12C-2t =4(e2* -3e-2')Therefore l"(*)=all*;

2t ^ -2xe -3e =u3

Mathematics Departnent

e1x

e

e =54xlte =ln3

,=!k,:4

g(x) = el' +3e-7' ,

wh"., r=!h3,r,-ltn r, -:,!u:,g(x)-2'e ala2 t

=5.fr=z.E

l. -e?

!rn:+ke 7

Therefore the y-coordinate is 2r[

Qn Working Marks Total Remarks

10(a)(i)

(ti)

(iir)

(b)

Total

t21

t2t

121

I2I

I8I

BP-88

86Cedar Girls' Secsrday Schoot Mathanatics Depaftnent

Qn Worklng Marks Total Remarks

11i

(ii)

(iti)

Md-poirtof.4c, F: 2+6 7+5 = (4,6)2 2

Gradient ofBF = 6-0

=24-1.

Eqn ofBF: y-0= 2(r-i)= y=2x-2

Gradient ofBC= 5-0

=l6-lGradient ofLS: -rEqn of AE: y - 7 = -1(x - 2) = y = -r+ 9

-x+9 =2t-2*=34.'..v=-3++9:5+G(3+,s\)

Let (.r, y) be coordinates ofD.(l+, o+ v)I z' ;)=t4'6t+x=7,y=12Coordinates ofD : (7,12)

xeaot,tnco: !12 I 6217 0 572127 = 30 sq udts

Total

t4l

121

t2t

t8l

I

Qn Working Marks Total Remarks

t2

(D

(ii)

(iiD

bm

fim

s

*s6r-8=,sp = acosea

g5r=9[=65 =bsmob

OP = SP+OS

OP = ecos? +bs;n9.

fi=W -a2+b2 =5

Max value of OP occus at 0 :63.43" .cos(6-a) = | 3 0-a=0,+a= 0=61.43

1*,o =!--b =tan63.43 = I .999 6 =+ b = r 9996aaaSr:bst 6 = 1.9996 a it a2 +b2 =5a2 + (7-9996a)1 = 5= d = 1.00.'. D = 2.00

cosA+2sin A= 2.15

Scos(€-63.a)= 2.15

(e - e:.+a) = ",r- 1ff)0 ' 79.4 or 47.5

Total

t3l

tsl

12t

I10l

BP-89

87Cedar cirls" Secondary School Mathematics Departnent