MARKING SCHEME - Mathematics & Additional Mathematics · 1. x = a b b ac 2 2 - ± - 4 8. a b b c c a log log ... 3472/2 Additional Mathematics Paper 2 ... [ Round off your answer

SULIT

3472/2 [Lihat sebelahSULIT

2

The following formulae may be helpful in answering the questions. The symbols given are the onescommonly used.

ALGEBRA

1. x =a

acbb2

42 -±- 8.abb

c

ca log

loglog =

2. aaa nmnm +=´ 9. dnaT n )1( -+=

3. aaa nmnm -=¸ 10. ])1(2[2

dnanS n -+=

4. aa mnnm =)( 11. 1-= nn arT

5. nmmn aaa logloglog +=12.

rra

rraS

nnn -

-=

--

=1

)1(1

)1( , r ≠ 1

6. log log loga a am m nn= - 13.

raS-

=¥ 1 , r < 1

7. mnm an

a loglog =

CALCULUS

1. y = uv,dxduv

dxdvu

dxdy

+=4 Area under a curve = ò

ba dxy or

= òba dyx

2. y =vu ,

2vdxdvu

dxduv

dxdy -

=

5. Volume of revolution = ò

ba dxy 2p or

= òba dyx2p

3.dxdu

dudy

dxdy

´=

GEOMETRY

1. Distance = 212

212 )()( yyxx -+- 4. Area of triangle

= 1 2 2 3 3 1 2 1 3 2 1 3

1( ) ( )

2x y x y x y x y x y x y+ + - + +

2. Mid point

( x , y ) = ÷øö

çèæ ++

2,

22121 yyxx

5. 22 yxr +=

3. Division of line segment by a point

( x , y ) = ÷÷ø

öççè

æ++

++

nmmyny

nmmxnx 2121 ,

6.2 2

ˆxi yj

rx y

+=

+% %

j2kk

http://mathsmozac.blogspot.com

http://tutormansor.wordpress.com/

SULIT

3472/2 [Lihat sebelahSULIT

3

STATISTICS

1.N

xx å= 7åå=

i

ii

WIWI

2.åå=

ffxx 8

)!(!rn

nPrn

-=

3.N

xxå -=

2)(s = 2

2x

Nx

-å 9!)!(

!rrn

nCrn

-=

4.å

å -=

fxxf 2)(

s = 22

xf

fx-

åå 10 P(AÈB) = P(A) + P(B) – P(AÇB)

11 P ( X = r ) = rnrr

n qpC - , p + q = 1

5. m = L + Cf

FN

m ÷÷÷

ø

ö

ççç

è

æ -21 12 Mean , m = np

13 npq=s

6. 1000

1 ´=QQI 14 Z =

sm-X

TRIGONOMETRY

1. Arc length, s = rq 8. sin ( A ± B ) = sin A cos B ± cos A sin B

2. Area of sector, A = q221 r 9. cos ( A ± B ) = cos A cos Bm sin A sin B

3. sin ² A + cos² A = 110 tan ( A ± B ) =

BABA

tantan1tantan

m

±

4. sec ² A = 1 + tan ² A11 tan 2A =

AA2tan1

tan2-

5. cosec ² A = 1 + cot ² A12

Cc

Bb

Aa

sinsinsin==

6. sin 2A = 2sin A cos A 13 a² = b² + c² – 2bc cos A7. cos 2A = cos ² A – sin ² A = 2 cos ² A – 1 = 1 – 2 sin ² A

14 Area of triangle = 1 sin2

ab C

j2kk



SULIT September, 2009

3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

4

Section ABahagian A[ 40 marks ]

[ 40 markah ]

Answer all questions.Jawab semua soalan.

1. Solve the simultaneous equations 3 2 0x y- - = and ( )2 1 3x y - = .Give your answers correct to three decimal places.

[5 marks]Selesaikan persamaan serentak 3 2 0x y- - = dan ( )2 1 3x y - = .Beri jawapan anda betul kepada tiga tempat perpuluhan.

[5 markah]

2.

In the diagram, the gradient and y-intercept of the straight line PQare 2 and 3 respectively. R is a point on the x-axis.

(a) Find the value of h and k .

(b) Given that PQ is perpendicular to QR, find the x-intercept of QR.

(c) Calculate the area of ∆ PQR.

[3 marks]

[3 marks]

[2 marks]

Dalam rajah, kecerunan dan pintasan-y bagi garis lurus PQ masing-masing ialah 2 dan 3 . R ialah titik pada paksi-x.

(a) Cari nilai h dan nilai k .

(b) Diberi bahawa PQ berserenjang dengan QR, cari pintasan-x bagi QR.

(c) Hitungkan luas ∆ PQR.

[3 markah]

[3 markah][2 markah]

x

y

0

Q(4, k)

P(h, – 3)R

j2kk





5

3. (a) Sketch the graph of 2 sin 2y x= - for 0 2x p£ £ .

(b) Hence, using the same axes, sketch a suitable straight line to find

the number of solutions for the equation sin 22x xp= for

0 2x p£ £ . State the number of solutions.

(a) Lakar graf bagi 2 sin 2y x= - untuk 0 2x p£ £ .

(b) Seterusnya, dengan menggunakan paksi yang sama, lakar satu garis lurus yang sesuai untuk mencari bilangan penyelesaian

bagi persamaan sin 22x xp= untuk 0 2x p£ £ .

Nyatakan bilangan penyelesaian itu.

[4 marks]

[3 marks]

[4 markah]

[3 markah]

4. Given that ( )2 1x x - is the gradient function of a curve which passesthrough the point )1,1(-P . Find

(a) the gradient of the tangent to the curve at P,

(b) the equation of the curve,

(c) the coordinates of the turning point at x = 1 . Hence determine whether the turning point is a maximum or a minimum point.

[1 mark]

[3 marks]

[3 marks]

Diberi ( )2 1x x - ialah fungsi kecerunan bagi suatu lengkung yangmelalui titik )1,1(-P . Cari

(a) kecerunan tangen kepada lengkung itu di P,

(b) persamaan lengkung itu,

(c) koordinat bagi titik pusingan pada x = 1 . Seterusnya tentukan sama ada titik pusingan itu adalah titik maksimum atau titik minimum.

[1 markah]

[3 markah]

[3 markah]

j2kk





6

5. A set of fifty numbers, 1 2 3 50, , ,......x x x x , has a mean of 11 and astandard deviation of 8.

(a) Find

(i) xå ,

(ii) 2xå

(b) If each of the numbers is multiplied by 1.8 and then increased by 5, find the new value for the

(i) mean, (ii) variance.

[3 marks]

[3 marks]

Suatu set yang terdiri daripada lima puluh nombor,1 2 3 50, , ,......x x x x , mempunyai min 11 dan sisihan piawai 8.

(a) Cari

(i) xå ,

(ii) 2xå

(b) Jika setiap nombor didarab dengan 1.8 dan ditambah dengan 5, cari nilai yang baru untuk

(i) min , (ii) varians .

[3 markah]

[3 markah]

j2kk





7

6.

A strip of metal is cut and bent to form some semicircles. Thediagram shows the first four semicircles formed. The radius of thesmallest semicircle is 5 cm. The radius of each subsequentsemicircle is increased by 3 cm.

(a) If the radius of the largest semicircle is 104 cm, find the number of semicircles formed.

(b) Calculate the total cost needed to form all the semicircles in (a) if the cost of the metal strip is RM4 per meter. [ Round off your answer to the nearest RM ]

Satu jalur logam dipotong dan dibengkok untuk membentukbeberapa semi bulatan. Rajah menunjukkan empat semi bulatanpertama yang telah dibentukkan. Jejari semi bulatan yang terkecilialah 5 cm. Jejari semi bulatan yang berikutnya bertambahsebanyak 3cm setiap satu.

(a) Jika jejari bagi semi bulatan yang terbesar ialah 104 cm, caribilangan semi bulatan yang telah dibentuk.

(b) Hitungkan jumlah kos yang diperlukan untuk membentuk semuasemi bulatan dalam (a) jika kos jalur logam ialah RM4 semeter.

[ Bundarkan jawapan anda kepada RM yang terdekat ]

[3 marks]

[4 marks]

[3 markah]

[4 markah]

j2kk





8

Section BBahagian B

[ 40 marks ][ 40 markah ]

Answer four questions from this section.Jawab empat soalan daripada bahagian ini.

7.

The table shows the values of two variables, x and y, obtained froman experiment. Variables x and y are related by the equation

1=+yxlm , where m and l are constants.

(a) Ploty1 against

x1 , using a scale of 2 cm to 0.1 unit on both

axes. Hence draw the line of best fit.

(b) Use your graph in 7(a) to find the value of

(i) m,

(ii) l.

[5 marks]

[5 marks]

Jadual menunjukkan nilai-nilai bagi dua pembolehubah, x dan y,yang diperoleh daripada satu eksperimen. Pembolehubah x dan y

dihubungkan oleh persamaan 1=+yxlm , dengan keadaan m dan l

adalah pemalar

(a) Ploty1 melawan

x1 , dengan menggunakan skala 2 cm

kepada 0.1 unit pada kedua-dua paksi. Seterusnya, lukis garis lurus penyuaian terbaik.(b) Gunakan graf di 7(a) untuk mencari nilai

(i) m,

(ii) l.

[5 markah]

[5 markah]

x 1.5 2.0 2.5 4.0 5.0 10.0y 0.96 1.2 1.4 2.0 2.2 3.0

j2kk





9

8.

The diagram shows straight line x + y = 4 that intersects with thecurve y = ( x – 2 ) 2 at points P and Q.

Find(a) the coordinates of Q,

(b) the area of the shaded region A,(c) the volume generated, in terms of π , when the shaded region

B is revolved through 360o about the x-axis.

[2 marks][4 marks]

[4 marks]

Rajah menunjukkan garis lurus x + y = 4 bersilang denganlengkung y = ( x – 2 ) 2 pada titik P dan Q.Cari

(a) koordinat Q,(b) luas rantau berlorek A,

(c) isipadu janaan, dalam sebutan π , apabila rantau berlorek Bdikisarkan melalui 360o pada paksi-x.

[2 markah][4 markah]

[4 markah]

O x

y = ( x – 2 ) 2P

QAB x + y = 4

y

j2kk





10

9.

In the diagram, PQRS is a quadrilateral. The diagonals PR and QSintersect at point T . It is given PQ = 2

~x , PS = 3

~y and

SR =~x –

~y .

(a) Express in terms of~x and

~y :

(i) QS (ii) PR .

(b) Given that QT = m QS , PT = n PR , where m and n areconstants, express

(i) QT in terms of m,~x and

~y ,

(ii) PT in terms of n,~x and

~y .

(c) Using PQ = PT + TQ , find the value of m and of n.

[10 marks]Dalam rajah, PQRS ialah sebuah sisiempat. Pepenjuru-pepenjuruPR dan QS bersilang di titik T . Diberi PQ = 2

~x , PS = 3

~y

dan SR =~x –

~y .

(a) Ungkapkan dalam sebutan~x dan

~y :

(i) QS

(ii) PR .

(b) Diberi QT = m QS , PT = n PR , dengan keadaan m dan n ialah pemalar, ungkapkan (i) QT dalam sebutan m,

~x dan

~y ,

(ii) PT dalam sebutan n,~x dan

~y .

(c) Dengan mengguna PQ = PT + TQ , cari nilai m dan nilai n. [10 markah]

P

T

S

R

Q

j2kk





11

10.

The diagram shows the cross-section of a cylindrical roller withcentre O and radius 20 cm resting on a horizontal ground PQ.OAB is a straight line that represents the handle of the roller andOA : AB = 1 : 3.Calculate(a) Ð POA in radian,(b) the perimeter, in cm, of the shaded region,(c) the area, in cm2 , of the shaded region.

[3 marks]

[3 marks]

[4 marks]

Rajah menunjukkan keratan rentas sebuah penggelek berbentuksilinder dengan pusat O dan jejari 20 cm yang terletak di ataslantai mengufuk PQ. OAB ialah garis lurus yang mewakilipemegang penggelek itu dan OA : AB = 1 : 3.Hitungkan

(a) Ð POA dalam radian,(b) perimeter , dalam cm, kawasan berlorek,(c) luas, dalam cm2 , kawasan berlorek.

[3 markah]

[3 markah]

[4 markah]

QP

OA

B

j2kk





12

11(a) In a survey carried out in a school, it is found that 40% of itsstudents are participating actively in co-curricular activities.

(i) If 6 students from that school are chosen at random,calculate the probability that at least 4 students areparticipating actively in co-curricular activities.

(ii) If the variance of the students who are active in co-curricularactivities is 288, calculate the student population of theschool.

[5 marks](b) The masses of chicken eggs from a farm has a normal distribution

with a mean of 62 g and a standard deviation of 8 g. Any egg thathas a mass exceeding 68 g is categorised as grade ‘double-A’.

(i) Find the probability that an egg chosen randomly from the farm has a mass between 60 g and 68 g.

(ii) If the farm produces 3000 eggs daily, calculate the number of eggs with grade ‘double-A’.

[5 marks](a) Dalam satu tinjauan yang dijalankan ke atas murid-murid di

sebuah sekolah, didapati 40% daripada murid-murid sekolah itumengambil bahagian secara aktif dalam aktiviti kokurikulum.

(i) Jika 6 orang murid daripada sekolah itu dipilih secara rawak, hitungkan kebarangkalian bahawa sekurang-kurangnya 4 orang murid adalah aktif dalam aktiviti kokurikulum.

(ii) Jika varians murid-murid yang mengambil bahagian secara aktif dalam aktiviti kokurikulum ialah 288, hitungkan bilangan murid dalam sekolah itu.

[5 markah](b) Jisim telur ayam dari sebuah ladang adalah mengikut satu taburan

normal dengan min 62 g dan sisihan piawai 8 g. Sebarang telurdengan jisim melebihi 68 g dikategorikan sebagai telur gred‘double-A’

(i) Cari kebarangkalian bahawa sebiji telur yang dipilih secara rawak dari ladang itu mempunyai jisim di antara 60 g dan 68 g.

(ii) Jika ladang itu menghasilkan 3000 biji telur setiap hari, hitungkan bilangan telur yang mempunyai gred ‘double-A’.

[5 markah]

j2kk





13

Section CBahagian C

[ 20 marks ][ 20 markah ]

Answer two questions from this section.Jawab dua soalan daripada bahagian ini.

12. A particle P starts from a fixed point O and moves in a straightline so that its velocity, v ms-1 , is given by v = 8 + 2t – t2, wheret is the time, in seconds, after leaving O.[Assume motion to the right is positive.]

Find

(a) the initial velocity, in ms-1 , of the particle,

(b) the value of t at the instant when the acceleration is 1 ms-2,

(c) the distance of P from O when P comes to instantaneousrest,

(d) the total distance, in m, travelled by the particle P in the first5 seconds.

Suatu zarah P mula dari suatu titik tetap O dan bergerak disepanjang garis lurus. Halajunya v ms-1, diberi olehv = 8 + 2t – t2, dengan keadaan t ialah masa, dalam saat, selepasmelalui O.[Anggapkan gerakan ke arah kanan sebagai positif]

Cari

(a) halaju awal, dalam ms-1 , bagi zarah itu,

(b) nilai bagi t apabila pecutannya ialah 1 ms-2,

(c) jarak P dari O apabila P berada dalam keadaan rehat seketika,

(d) jumlah jarak yang di lalui, dalam m, oleh zarah P dalam5 saat yang pertama.

[1 mark]

[2 marks]

[3 marks]

[4 marks]

[1 markah]

[2 markah]

[3 markah]

[4 markah]

j2kk





14

13. The pie chart shows five items, A, B, C, D and E used in makingcakes. The table shows the prices and the price indices of theseitems.

E

D

C

B

A

30°

75°

120°

Items

Bahan

Price (RM)per kg for the

year 2003

Harga (RM)per kg padatahun 2003

Price (RM)per kg for the

year 2006

Harga (RM)per kg padatahun 2006

Price index for theyear 2006 based on

the year 2003

Indeks harga padatahun 2006

berasaskan tahun2003

A 0.40 x 150B 1.50 1.65 110C 4.00 4.80 yD 3.00 4.50 150E z 2.40 120

(a) Find the value of(i) x,(ii) y,(iii) z.

(b) Calculate the composite index for the cost of making thesecakes in the year 2006 based on the year 2003.

(c) The total expenditure on the items in the year 2006 is RM 5000. Calculate the corresponding total expenditure in the year 2003.

(d) The price of each item increases by 20 % from the year 2006to the year 2008. Find the composite index for totalexpenditure on the items in the year 2008 based on the year2003.

[3 marks]

[3 marks]

[2 marks]

[2 marks]

j2kk





15

Carta pai menunjukkan lima bahan, A, B, C, D dan E yangdigunakan untuk membuat sejenis kek. Jadual menunjukkan hargabahan dan nombor indeks bagi kelima-lima bahan tersebut.

(a) Carikan nilai(i) x,(ii) y,(iii) z.

(b) Hitungkan nombor indeks gubahan bagi kos penghasilan kek itu pada tahun 2006 berasaskan tahun 2003.

(c) Jumlah kos bahan-bahan tersebut pada tahun 2006 ialahRM 5000. Hitungkan jumlah kos yang sepadan pada tahun2003.

(d) Harga bagi setiap bahan bertambah sebanyak 20% daritahun 2006 ke tahun 2008. Cari nombor indeks gubahanbagi jumlah kos ke atas bahan-bahan tersebut pada tahun2008 berasaskan tahun 2003.

[3 markah]

[3 markah]

[2 markah]

[2 markah]

j2kk





16

14. Use graph paper to answer this question.

A factory produces two components, S and T for a digitalcamera, by using machines P and Q. The table shows the timetaken to produce components S and T respectively.

In any given week, the factory produces x units of component Sand y units of component T. The production of the componentsper week is based on the following constraints:

I : Machine P operates not more than 2000 minutes.

II : Machine Q operates at least 1200 minutes.

III : The number of component T produced is not more than three times the number of component S produced.

(a) Write three inequalities, other than x ³ 0 and y ³ 0, whichsatisfy all the above constraints.

(b) Using a scale of 2 cm to 10 units on both axes, construct andshade the region R which satisfies all of the aboveconstraints.

(c) Use your graph in 14(b) to find

(i) the maximum number of component S that could beproduced, if the factory plans to produce only 30 unitsof component T,

(ii) the maximum profit per week if the profit from a unitof component S is RM20 and from a unit ofcomponent T is RM30.

Time taken (minutes)Masa diambil (minit)

Component

Komponen Machine PMesin P

Machine QMesin Q

S 40 15

T 20 30

[3 marks]

[3 marks]

[4 marks]

j2kk





17

Gunakan kertas graf untuk menjawab soalan ini.

Sebuah kilang menghasilkan dua komponen, S dan T bagisesuatu kamera digital dengan menggunakan mesin P dan Q.Jadual menunjukkan masa yang diambil untuk menghasilkankomponen-komponen S dan T.

Dalam mana-mana satu minggu, kilang tersebut menghasilkan xunit bagi komponen S dan y unit bagi komponen T.Penghasilan komponen-komponen tersebut adalah berdasarkankekangan berikut:

I : Mesin P beroperasi tidak melebihi 2000 minit.

II : Mesin Q beroperasi sekurang-kurangnya 1200 minit.

III : Bilangan komponen T yang dihasilkan tidak melebihi tiga kali ganda bilangan komponen S yang dihasilkan.

(a) Tuliskan tiga ketaksamaan, selain x ³ 0 dan y ³ 0, yangmemenuhi semua kekangan di atas.

(b) Menggunakan skala 2 cm kepada 10 unit pada kedua-duapaksi, bina dan lorek rantau R yang memenuhi semuakekangan di atas.

(c) Gunakan graf anda di 14(b) untuk mencari

(i) bilangan maksimum bagi komponen S yang bolehdihasilkan jika kilang tersebut bercadang untukmenghasilkan 30 unit komponen T sahaja,

(ii) keuntungan maksimum seminggu jika keuntungan yangdiperoleh dari satu unit komponen S ialah RM20 dandari satu unit komponen T ialah RM30.

[3 markah]

[3 markah]

[4 markah]

j2kk





18

15. The diagram shows a cyclic quadrilateral PQRS with PQ = 9 cm,PR = 11 cm and QR = 7 cm.

(a) Find Ð PQR.

(b) Given that PS = 6 cm, find the length of RS.

(c) Calculate the area of PQRS.

S R

Q

P

11 cm

7 cm

9 cm

[3 marks]

[4 marks]

[3 marks]

j2kk





19

Rajah menunjukkan satu sisiempat kitaran PQRS denganPQ = 9 cm, PR = 11 cm, dan QR = 7 cm.

(a) Cari Ð PQR.

(b) Diberi PS = 6 cm, cari panjang RS.

(c) Hitungkan luas bagi PQRS.

[3 markah]

[4 markah]

[3 markah]

END OF QUESTION PAPERKERTAS SOALAN TAMAT

j2kk



SULIT JPNKd/2006/3472/2

Nama Pelajar : ………………………………… Tingkatan 5 : ……………………. 3472/2 Additional Mathematics

September 2009

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH

NEGERI KEDAH DARUL AMAN

PEPERIKSAAN PERCUBAAN SPM 2009 ADDITIONAL MATHEMATICS

Paper 2

.

MARKING SCHEME SULIT 3472/2

j2kk



2

MARKING SCHEMEADDITIONAL MATHEMATICS PAPER 2

SPM TRIAL EXAMINATION 2009

N0. SOLUTION MARKS1

2

2

2

3 23 2 (3 2 1)

0 6 6 32 2 1 0

2 ( 2) 4(2)( 1)2(2)

1.366 0.3663(1.366) 2 3( 0.366) 2

2.098 3.098

y xx x

x xx x

x

x or xy y

= -= - -

=> = - -

- - =

± - - -=

= = -= - = - -= = -

P1K1 Eliminate y

K1 Solve quadraticequation

N1

N1

52

(a)

(b)

(c)

( )

2 3( , 3) 3 2 3

3(4, ) 2 4 3

11

y xP h hhQ k kk

= +- Þ - = +

= -

Þ = +

=

OR

3 3 2 30

3 2 114 0

hh

k k

+= Þ = -

--

= Þ =-

1 2122

11 14 2

1: 132

0 26

m m

yx

QR y x

y x

= Þ = -

-= -

-

= - +

= Þ =

OR

11 0 14 2

111 22

26

x

x

x

-= -

-

= - +

=

x-intercept = 26

Area3 4 26 313 11 0 32

1 33 78 12 28621 3852192.5

- -=

- -

= - - + -

= -

=

K1 Use equation orgradient

N1

N1

P1 For 212

m = -

K1 Use equation orgradient

N1

K1 Use formula andfind the areatriangle

N1

8

j2kk



3

N0. SOLUTION MARKS3

(a)

(b) 2 sin 2 22xxp

- = -

or

22xyp

= -

Draw the straight line 22xyp

= -

Number of solutions = 4 .

P1 Negative sineshape correct.

P1 Amplitude = 1[ Maximum = 3and Minimum =1 ]

P1 Two full cycle in0 £ x £ 2p

P1 Shift up the graph

N1 For equation

K1 Sketch thestraight line

N1

7

3

1 22xyp

= -

y

2p 2p xO p

2 sin 2y x= -

2

23p

j2kk



4

N0. SOLUTION MARKS4

(a)

(b)

(c)

( ) ( )21 1 1 2dydx

= - - - = -

3 2

4 3

4 3

( )

4 31 114 35

125

4 3 12

y x x dx

x xy c

c

c

x xy

= -

= - +

= + +

=

= - +

ò

( )2

22

2

2

2

1 0

0@1

3 2

1 1 0

1 1 5 14 3 12 3

dy x xdxx

d y x xdx

d yxdx

y

= - =

=

= -

= Þ = >

= - + =

11,3

æ öç ÷è ø

Minimum point.

N1

K1 Integrate gradientfunction

K1 Substitute (－1,1)into equation y

N1

K1 Find2

2d ydx

N1 N1

7

j2kk



5

N0. SOLUTION MARKS5

(a)(i)

(ii)

(b)(i)

(ii)

11 55050

xx= => =å å

22

2

11 850

50(64 121)9250

x

x

- =

=> = +

=

å

å

New mean = 11 1.8 5 24.8´ + =

New variance = 2(8 1.8) 207.36´ =

K1 Use formula ofmean and/orstandard deviation

N1

N1

K1 Find new meanand/or newvariance

N1

N1

66

(a)

(b)

( )

53

5 1 3 1041 3334

l rad

nnn

ppp

p p p

===

+ - =

- ==

( ) ( )3434 2 5 33 32

S p pé ù= +ë û

1853582158.21

cmm

p===

Total cost4 58.21232.8233

RMRMRM

= ´==

P1 Value of a and/or d

K1 Use Tn = 104p

N1

K1 Find S34

N1

K1 RM4 ´ S34

N1

7

j2kk



6

N0. SOLUTION MARKS7

(a)

(b)(i)

(ii)

llm 111

+´-=xy

y1

1.04 0.83 0.71 0.50 0.45 0.33

x1

0.67 0.5 0.4 0.25 0.2 0.1

l1 = y-intercept

l = 5

lm

- = gradient

m = – 6.25

P1

N1

N1

K1 for correct axes and scale

N1 for all points plotted correctly

N1 for line of best-fit

K1 for y-intercept

N1

K1 for gradient

N1

10

y1

x10.2

0

j2kk



7

N0. SOLUTION MARKS

8(a)

(b)

(c)

( x – 2 )2 = 4 – xQ(3, 1 )

A = [ ]dxxxò ---3

0

2)2()4(

= dxxxò -3

0

2 )3(

=3

0

32

323

úû

ùêë

é-

xx

=29

Note : If use area of trapezium and ò dxy , give marks accordingly.

V = ò -2

0

4)2( dxxp

=2

0

5

5)2(úû

ùêë

é -xp

= p5

32

K1 Solve for xN1

K1 use

dxyyò - )( 12

K1 integrate correctly

K1 correct limit

N1

K1 integrate

ò dxy 2p

K1 integrate correctlyK1 correct limit

N1

10

j2kk



8

N0. SOLUTION MARKS9(i)

(ii)

(b)(i)

(ii)

(c)

QS = QP + PS

= – 2~x + 3

~y

PR = PS + SR

=~x + 2

~y

QT = m QS

= m (– 2~x + 3

~y )

= – 2m~x + 3m

~y

PT = n PR

= n (~x + 2

~y )

= n~x + 2n

~y

PQ = PT + TQ

2~x = n

~x + 2n

~y + 2m

~x – 3m

~y

= (n + 2m)~x + (2n – 3m)

~y

n + 2m = 2

2n – 3m = 0

m =74

n =76

K1 for using vector triangleN1

N1

N1

N1

K1 for substituting & grouping into components

K1 for equating coefficients correctly

K1 for eliminatingm or n

N1

N1

10

j2kk



9

N0. SOLUTION MARKS10(a)

(b)

(c)

1

1cos4

1cos4

POA

POA -

Ð =

Ð =

= 75.52 @75 31"o o

= 1.318 rad.

Arc PA = 20 ( 1.318 ) = 26.36

PQ2 = 802 － 202

PQ = 77.46

Perimeter

= 60 + 26.36 + 77.46

= 163.82 cm

Area △OPQ = ( )( )1 20 77.46 774.62

=

Area sector POA = ( ) ( )21 20 1.318 263.62

=

Area of shaded region

= 774.6 － 263.6

= 511 cm2

K1 Use correctlytrigonometricratio

N1

K1 Use s rq=

K1 Use PythagorasTheorem

K1

N1

K1 Use formula12

A bh=

K1 Use formula21

2A r q=

K1

N1

10

j2kk



10

N0. SOLUTION MARKS11(a)(i)

(ii)

(b)(i)

(ii)

p = 0.4 q = 0.6 n = 6

P ( )4X ³

( ) ( ) ( )( ) ( ) ( )( ) ( )( )4 2 5 1 6 06 6 6

4 5 6

4 5 6

0.4 0.6 0.4 0.6 0.4 0.60.13824 0.036864 0.0040960.1792

P X P X P X

C C C

= = + = + =

= + +

= + +=

2 288npqs = =

( )( )288 1200

0.4 0.6n = =

62m = 8s =

( )

( )( ) ( )

60 68

60 62 68 628 8

0.25 0.75

1 0.25 0.751 0.4013 0.22660.3721

P X

P Z

P Z

P Z P Z

< <

- -æ ö= < <ç ÷è ø

= - < <

= - > - >

= - -=

( ) ( )68 0.75 0.22663P X P Z> = > =

0.22663 ⅹ 3000 = 679.89

= 679 @ 680

P1 Value of p and/or qAND p + q =1

K1 Use P(X = r)= n Cr prqn–r

N1

K1

N1

K1 Use Z =s

m-X

K1N1

K1

N1

10

j2kk



11


(b)

(c)

8 ms-1

dvadt

= =0

2 – 2t = 1

t =12

8 + 2t – t2 = 0

(t – 4 ) (t + 2) = 0

t = 4 t = –2 (not accepted)

s vdt vdt= +ò ò4 5

0 4

= t tt t t té ù é ù

+ - + + -ê ú ê úê ú ê úë û ë û

4 53 32 2

0 4

8 83 3

= ( )é ù é ùæ ö æ ö æ ö+ - - + + - - + -ç ÷ ç ÷ ç ÷ê ú ê úè ø è ø è øë û ë û

64 125 6432 16 0 40 25 32 163 3 3

= + -80 103 3

= 30 m

N1

K1

N1

K1

K1

N1(for t = 4 only)

K1

(for andò ò4 5

0 4

)

K1(for integration)

K1(for summation)

N1

10

j2kk



12


(b)

(c)

(d)

.x

= ´150 1000 4

(or formula finding y /z)

x = RM 0.60

y = 120

z = RM 2.00

45o

( ) ( ) ( ) ( ) ( )x x x x xI + + + +=

150 30 110 90 120 75 150 120 120 45360

= 46800360

= 130

P03 = ( )100 5000130

= RM 3846.2

/ .I x=08 03 130 1 2 (or 130 + 130x0.2)

= 156

N1

N1

N1

P1

K1

N1

K1

N1

K1

N1

10

j2kk



13


(b)

(c)

(d)

40x + 20y £ 2000 or2x + y £ 100

15x +30y ³ 1200 orx + 2y ³ 80

y £ 3x

(20, 60)

(35, 30)

2x + y = 100y = 3x

x + 2y = 80

y = 30

100

90

80

70

60

50

40

30

20

10

10080604020 9070503010x

y

· At least one straight line is drawn correctly from inequalitiesinvolving x and y.

· All the three straight lines are drawn correctly

· Region is correctly shaded

35

Maximum point (20, 60)

Maximum profit = 20(20) + 30(60)

= RM 2200

N1

N1

N1

K1

N1

N1

N1

N1

K1

N1

10

j2kk



14


(b)

(c)

112 = 92 + 72 – 2(9)(7)cosÐPQR

cosÐPQR = 9126

ÐPQR = 85o 54’

ÐPSR = 180o – 85o 54’

= 94o 6’

'sin sinPRSÐ

=094 6

6 11

ÐPRS = 32o 57’

\ ÐRPS = 180o – 94o 6’ – 32o 57’ = 52o 57’

' 'sin sinoo

RS=

1152 57 94 6

RS = 8.802 cm

Area = ' '( )( )sin ( )( . )sinoo +

1 19 7 85 54 6 8 802 94 62 2

= 31.42 + 26.34

= 57.76

K1

K1

N1

P1

K1

K1

N1

K1, K1(for usingarea= ½absincand summation)

N1

10

END OF MARKING SCHEME

j2kk



Documents

MARKING SCHEME - Mathematics & Additional Mathematics · 1. x = a b b ac 2 2 - ± - 4 8. a b b c c a log log ... 3472/2 Additional Mathematics Paper 2 ... [ Round off your answer