Tutormansor.files.wordpress.com 2012 09 Add Math f4 Final 2011 Melaka p2 Ans

8/12/2019 Tutormansor.files.wordpress.com 2012 09 Add Math f4 Final 2011 Melaka p2 Ans

1/30

SULIT 3472/211

3472/2Form 4Additional MathematicsPaper 22011

221 hours

PEPERIKSAAN SELARAS AKHIR TAHUNSEKOLAH-SEKOLAH MENENGAH NEGERI MELAKA

KelolaanPEJABAT PELAJARAN DAERAH

JASIN * ALOR GAJAH * MELAKA TENGAHDengan kerjasama :

JABATAN PELAJARAN NEGERI MELAKATINGKATAN 4 [ 2011 ]

ADDITIONAL MATHEMATICSPaper 2

22

1hours

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. This questions paper consists of three sections : Section A , Section B andSection C.

2. Answer all questions in Section A , four questions from Section B and twoquestions from Section C.

3. Give only one answer / solution to each question..4. Show your working. It may help you to get marks.5. The diagram in the questions provided are not drawn to scale unless stated.6. The marks allocated for each question and sub-part of a question are shown in

brackets..7. A list of formulae is provided on pages 2 to 3.8. A booklet of four-figure mathematical tables is provided.9. You may use a non-programmable scientific calculator.

Kertas soalan ini mengandungi 17 halaman bercetak


2/30

SULIT 3472/22

The following formulae may be helpful in answering the questions.The symbols givenare the ones commonly used.

Rumus-rumus berikut boleh digunakan untuk membantu anda menjawab soalan. .Simbol-simbol yang diberi adalah yang biasa digunakan .

ALGEBRA

1. x =a

acbb2

42

2 a m a n = a m + n

3 a m a n = a m - n

4 (am

)n

= anm

5 log a mn = log am + log a n

6 log a nm

= log am - log a n

7 log a m n = n log a m

8 log ab =ab

c

c

loglog

9 T n = a + (n-1)d

10 S n = ])1(2[2

d nan

11 T n = ar n-1

12 S n =r r a

r r a nn

1)1(

1)1(

, (r 1)

13r

aS

1 , r


3/30

SULIT 3472/23

STATISTICS

TRIGONOMETRY

1 x = N

x

2 x = f fx

3 = N

x x 2)( =

2 _ 2

x N

x

4 =

f

x x f 2)( =

22

x f

fx

5 M = C f

F N L

m

21

6 1000

1

P P

I

x N

x

x f

fx

N

x x 2)(

2 _ 2

x N

x

f

x x f 2)(

22

x f

fx

C f

F N L

m

21

1000

1

P P

I

71

11

w I w

I

8)!(

!r n

n P r n

9. !)!(

!r r n

nC r

n

10 P(A B)=P(A)+P(B)-P(A B)

11 p (X=r) = r nr r n q pC , p + q = 1

12 Min(mean) = np

13 npq

14 z =

x

1 Arc length , s = r

2 Area of a sector, L = 221

r

3 sin 2A + cos 2A = 1

4 sec 2A = 1 + tan 2A

5 cosek 2 A = 1 + cot 2 A

6 sin2A = 2 sinAcosA

7 cos 2A = cos 2A sin 2 A= 2 cos 2A-1= 1- 2 sin 2A

8 tan2A = A

A2tan1

tan2

9 sin (A B) = sinAcosB cosAsinB

10 cos (A B) = cos AcosB sinAsinB

11 tan (A B) = B A B A

tantan1tantan

12C

c

B

b

A

a

sinsinsin

13 a 2 = b 2 +c 2 - 2bc cosA

14 Area of triangle = C ab sin21


4/30

SULIT 3472/24

Section A

[ 40 Marks ]

Answer all questions from this section

Jawab semua soalan dari bahagian ini

1. The straight line 3 + x = 3y intersects the curve 1 + 2x 2 = y 2 + 3xy

at two points. Find the coordinates of the two points.

Garis lurus 3 + x = 3y menyilang lengkung 1 + 2x 2 = y 2 + 3xy pada

dua titik. Cari koordinat bagi dua titik tersebut.

[ 5 marks ]

2. The functions f and g are defined by f ( x ) = x1

6 , x 1 and

g ( x ) = kx + 10 , where k is a constant.

Fungsi f dan g ditakrifkan sebagai f ( x ) = x1

6 , x 1 dan

g ( x ) = kx + 10 , dengan k ialah pemalar.

a ) Find the value of k if f -1g ( -2 ) = -

Cari nilai k jika f -1g ( -2 ) = -

b ) Find g 2 ( x )

Cari g 2 ( x )

[ 6 marks ]

3. A set of scores p 1, p 2, p 3, p 4 and p 5 has mean 6 and a standard deviationof 2.3

Suatu set skor p 1, p 2, p 3, p 4 and p 5 mempunyai min 6 dan sisihan piawai 2.3

(a) Find

Cari

(i) the sum of the score, p

hasil tambah skor, p


5/30

SULIT 3472/25

(ii) the sum of the squares of the scores, 2 p

hasil tambah kuasa dua skor , 2 p

[ 3 marks ]

(b) Each score is multiplied by 3 and the 2 is added to it. Find, for the new

set of scores,

Setiap skor didarab 3 dan ditambah 2 , cari nilai baru bagi set skor

(i) the mean

min

(ii) the variance varians .

[ 4 marks ]

4. (a) Given y = x x

1, find

2

2

dx

yd .

Diberi y = x x1 , cari 2

2

dx yd .

(b) Given y =1619

81 x is the equation of normal to the curve

y = ( 4x 5 ) 2 at P . Find the coordinates of P.

Diberi y =1619

81 x adalah persamaan normal kepada lengkung

y = ( 4x 5 ) 2 di P. Cari koordinat- koordinat di P.

[ 6 marks ]


6/30

SULIT 3472/26

5. The straight line k x y 4 is the normal to the curve 212 2 x y at point A.

Garis lurus k x y 4 adalah normal kepada lengkok 212 2 x y pada titik AFind

Cari

a) the gradient of the tangent at point A

kecerunan bagi tangen pada titik A

[ 2 marks ]

b) the coordinates of point A and the value of k

koordinat tititk A dan nilai bagi k

[ 4 marks ]

c) the equation of tangent at point A

persamaan bagi tangen pada tititk A

[ 2 marks ]

6. a) Given that a x 10log and b y 10log , express 210100

log y

xin terms of a and b

Diberi a x 10log dan b y 10log , bentukkan 210100

log y

x dalam bentuk a dan b

[ 3 marks ]

b) Solve the equation 622 105.2 x x x

Selesaikan persamaan 622 105.2 x x x

[ 2 marks ]

c) Solve the equation 4loglog 327 x x

Selesaikan persamaan 4loglog 327 x x

[ 3 marks ]


7/30

SULIT 3472/27

Section B

Bahagian B

[ 40 marks ]

Answer four questions from this sectionJawab empat soalan dari bahagian ini

7. Given A(5,-2) and B(2,1) are two fixed points. Point Q moves such that the

ratio of AQ and QB is 2 : 1

Diberi A(5,-2) dan B(2,1) adalah dua titik tetap. Titik Q bergerak dengan nisbah bagi AQ

kepada QB adalah 2 : 1

a) Show that the equation of the locus of point Q is x 2+y2-2x-4y-3=0.

Tunjukkan persamaan lokus bagi titik Q ialah x 2 +y 2 -2x-4y-3=0.

[ 2 marks ]

b) Show that point C(-1,0) lies on the locus of point Q.

Tunjukkan bahawa titik C(-1,0) melalui lokus titik Q.

[ 2 marks ]

c) Find the equation of the straight line AC

Cari persamaan bagi garis lurus AC

[ 3 marks ]

d) Given the straight line AC intersects the locus of point Q again at point D,

find the coordinates of point D.

Diberi garis lurus AC bersilang dengan titik lokus Q pada titik D, cari titik bagi D

[ 3 marks ]


8/30

SULIT 3472/28

8. In the Diagram 8 ,0

90 ABC and the equation of the straight line BCis 2y+x+8=0.

Rajah 8 menunjukkan 090 ABC dan persamaan garis lurus BC ialah

2y+x+8=0 .

a) Find

Cari

(i) The equation of the straight line AB,

Persamaan garis lurus AB

(ii) The coordinates of B

Koordinat B

[ 5 marks ]

b) The straight line AB is extended to a point D such that AB : BD = 2 : 3.

Find the coordinates of D.

Garis lurus AB memanjang ke titik D dengan AB : BD = 2 : 3. Cari titik D

[ 2 marks ]

c) A point Q moves such that it s distance from point A is always 6 units.

Find the equations of the locus of Q.

Titik Q bergerak dengan jarak 6 unit dari titik A. Cari persamaan lokus bagi Q

[ 3 marks ]

B

x

yA(5,11)

0

C

2y+x+8=0

Diagram 8 / Rajah 8


9/30

SULIT 3472/29

9. In Diagram 9, is a right-angled triangle. PR is an arc of a circle with centre

O and radius 8 cm. ORST is a quadrant of a circle with centre O. The

ratio of OP : PQ = 2 : 1

Rajah 9, sebuah segitiga bersudut tegak. PR adalah lengkok bulatan dengan

berpusat O dan berjejari 8 cm. ORST ialah sukuan bulatan berpusat O. Nisbah

bagi OP : PQ = 2 : 1

. CalculateKirakan

a) The value of in radians.

Nilai dalam radian,

[ 2 marks ]

b) The perimeter, in cm, of the shaded region,

Perimeter, dalam cm . bagi kawasan berlorek,

[ 5 marks ]

c) The area, in cm 2 , of the shaded region.

Luas, dalam cm 2 , bagi kawasan berlorek.

[ 3 marks ]

Diagram 9/ Rajah 9


10/30

SULIT 3472/210

10. The Diagram 10 shows a right-angled triangle. LMN described in asemicircle with centre O. LM = 4 cm and MN = x cm.

Rajah 10 menunjukkan sebuah segitiga bersudut tegak. LMN adalah semibulatan

berpusat O. LM = 4 cm dan MN = x cm.

a) Show that the area of the shaded region, A cm 2 is given by

A = x x 28

2 2

Tunjukkan bahawa kawasan berlorek, A cm 2 diberi sebagai A= x x 28

2 2

.

[ 5 marks ]

b) Find the value of x such that A is a minimum. Hence, find the minimum of A

in terms of .

Cari nilai bagi x dimana A adalah minimum. Seterusnya, cari nilai bagi A dalamsebutan

[ 5 marks ]

L

M

NO

Diagram 10/ Rajah 10


11/30

SULIT 3472/211

11. Table 11 shows the frequency distribution of the height of 40 children in a

kindergarden.

Jadual 11 menunjukkan taburan kekerapan bagi tinggi 40 orang kanak-kanak disebuah tadika.

Height / Tinggi ( cm ) Number of children

Bilangan kanak-kanak

100 104 8

105 109 5

110 - 114 10

115 119 6

120 124 7

125 129 4

( a ) Based on Table 11 , construct a cumulative frequency

distribution table.

Berdasarkan Jadual 11 bina jadual kekerapan longgokan bagi

taburan itu.

[ 2 marks ]

( b ) Without using ogive, calculate the first quartile of the distribution.

Tanpa menggunakan ogif, hitung kuartil pertama bagi taburan itu.

[ 3 marks ]( c ) Without using ogive calculate the median height of the children.

Tanpa menggunakan ogif Hitung tinggi median kanak-kanak itu.

[ 3 marks ]

Table 11 / Jadual 11


12/30

SULIT 3472/212

Section C

Bahagian C

[ 20 marks ]

Answer two questions from this section.

Jawab d u a soalan daripada bahagian ini.

12. Table12 shows the prices, price indices and percentages of usage of four

items, A, B, C and D,which are the main ingredients in the manufacturing of a

type of soap.

Jadual 12 menunjukkan harga, indeks harga dan peratus penggunaan empat

barangan A,B,C dan D yang merupakan bahan utama bagi penghasilan sejenis

sabun.

Item

Barangan

Price per unit

( RM )

Harga seuni t

( RM )

Price index in 2008

based on 2004

Indeks harga pada

2008 berasask an

2004

Percentage of usage

( % )

Peratus penggu naan

( % )

2004 2008

A p 54 120 3k

B 36 q 125 2k

C 40 32 80 20

D 40 42 r 5k

Table 12 / Jadual 12

a) Find the values of p, q and r

Cari nilai-nilai p, q dan r

[ 3 marks ]


13/30

SULIT 3472/213

b) State the value of k. Hence, calculate the composite index for the cost

of manufacturing of the soap in the year 2008 based on the year 2004

Nyatakan nilai k. Seterusnya, hitung indeks gubahan bagi kos penghasilan

sabun itu pada tahun 2008 berasaskan tahun 2004 .[ 3 marks ]

c) Calculate the price of a box of soap in the year 2004 if the

corresponding price in the year 2008 is RM250.

Hitung harga sekotak sabun pada tahun 2004 jika harga yang sepadannya

pada tahun 2008 ialah RM250

[ 2 marks ]

d) The cost of manufacturing of the soap is expected to increase by 30%

from the year 2008 to the year 2012. Find the expected composite

index in the year 2012 based on the year 2004.

Kos penghasilan sabun itu dijangka meningkat sebanyak 30% dari tahun

2008 kepada tahun 2012. Cari indeks gubahan yang dijangka pada tahun

2012 berasaskan tahun 2004.

[ 2 marks ]


14/30

SULIT 3472/214

13 The Diagram 13 shows a bar chart indicating the weekly cost of the

iterms J, K, L, M and N for the year 2001. The table shows the prices

and the price indices for the items.

Diagram 13 menunjukkan carta bar merujuk kepada perbelanjaan mingguan

bagi iterm J, K, L, M and N pada tahun 2001. Jadual pula menunjukkan harga

dan indeks harga barang-barang tersebut.

Item

(Barangan)

Price in 2001

(Harga pada

2001)

Price in 2004

(Harga pada

2004)

Price index in 2004 based on2001

(Harga pada 2004 berasaskan

pada 2001)

J x RM2.40 120K RM3.20 RM4.00 125

L RM4.00 RM5.20 y

M RM6.00 RM9.00 150

N RM2.40 z 110

Weekly cost (RM)

50

36

30

24

10

Items

Diagram 13 / Rajah 13

J LK M N


15/30

SULIT 3472/215

a) Find the value of

Cari nilai bagi

i) x

ii) yiii) z

[ 3 marks ]

b) Calculate the composite index for the items in the year 2004 based on the

year 2001.

Kirakan indeks gubahan bagi barang-barang itu bagi tahun 2004 berdasarkan

tahun 2001.

[ 2 marks ]

c) The total weekly cost of the items in the year 2001 was RM800. Calculate the

corresponding total weekly cost for tha year 2004.

Jumlah perbelanjaan mingguan barang-barang itu pada tahun 2001 ialah RM800.

Kirakan jumlah perbelanjaan mingguan yang sepadan bagi barang-barang itu pada

tahun 2004.

[ 2 marks ]

d) The cost of the items increases by 25% from the year 2004 to the year 2007.

Find the composite index for the year 2007 based on the year 2001.

Perbelanjaan barang-barang itu telah meningkat 25% dari tahun 2004 kepada tahun

2007. Cari indeks gubahan pada tahun 2007 berasaskan tahun 2001.

[ 3 marks ]


16/30

SULIT 3472/216

14. Calculate

Hitung

(i) the length of AB

panjang AB

[ 3 marks ]

(ii) the angle of BAC

Sudut BAC

[ 3 marks ]

(iii) the area of a new triangle of ABC if AC is extended while the length

of AB , the length of BC and the angle of BAC are maintained

luas bagi segitiga baru ABC jika AC dipanjangkan manakala panjang AB,

BC dan sudut BAC tidak berubah.

[ 4 marks ]

Diagram 14 / Rajah 14

Figure 14 shows a triangle ABCRajah 14 menunjukkan segitiga ABC


17/30

SULIT 3472/217

15. Diagram 15 shows a quadrilateral PQRS with a triangle ABC. D, E and F are

points the sight AB, BC and AC respectively such that AB:DB= 1:2 and BE = 8

cm. It also given that AB = 12 cm and AC = 7 cm.

Rajah 15 menunjukkan sebuah segi tiga ABC. D, E dan F masing-masing ialah

titik ada sisi AB, BC dan AC dengan keadaan AD:DB=1:2 dan BE = 8 cm. Diberi

juga bahawa AB = 12 cm dan AC = 7 cm.

Diagram 15/ Rajah 15

Calculate

Kirakan

a) (i) the length, in cm, of EC,

Panjang dalam cm , bagi EC.

[ 2 marks ]

(ii) ACB

[ 2 marks ]

b) It is given that AF = AC.

Diberi bahawa AF = AC.

(i) Calculate the area, cm 2, of DEF.

kirakan luas, dalam cm 2 , bagi DEF.

[ 3 marks ]

(ii) Hence, find the perpendicular distance, in cm , from D to EF.

Oleh yang demikian , cari jarak serenjang, dalam cm, dari D ke EF.

[ 3 marks ]

END OF QUESTION PAPER / KERTAS SOALA N TAMA T

110 o

E

A

C

B

F

D


18/30

SULIT 3472/218


19/30

1

3472/2Form 4Additional MathematicsPaper 22011

PEPERIKSAAN SELARAS AKHIR TAHUNSEKOLAH-SEKOLAH MENENGAH NEGERI MELAKA

Kelolaan

PEJABAT PELAJARAN DAERAHJASIN * ALOR GAJAH * MELAKA TENGAH

Dengan kerjasama :

JABATAN PELAJARAN NEGERI MELAKATINGKATAN 4

2011

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 12 printed pages.

http://tutormansor.wordpress.com/


20/30

2

Number Solution and marking scheme Submarks Fullmarks

1 x = 3y - 3

8 y - 27 y + 19 = 0

y =)8(2

)19)(8(4)27()27( 2

y = 2.375 , y = 1

x = 4.125 , x = 0

( 4.125, 2.375 ) , ( 0, 1 )

atau setara

1

1

1

1, 1 5

2 a) g (-2) = -2k + 10

f 1 (-2k + 10) =)102(

6)102(k

k

2

1 =

)102(

)42(

k

k

k = 3

b) g 2 ( x ) = g ( 3x + 10 )

= 3(3x + 10) + 10

= 9x + 40

1

1

1

1

1

1 63. (a) (i) Mean = 6

65 p

p = 30

(ii) standard deviation, = 2.3

11



21/30

3

22

65

p= 2.3

29.5365

2 p

p2 = 206.45

b) (i) New mean = 6(3) + 2 = 20

(ii) new variance = 2.3 x 3 2 = 20.7

1

1

1

1

17

4(a) 12 x

dxdy

2

2

dx yd

2x-3

= 32

x

(b) The normal, m2 =81

m1 = 8Gradient of tangent = 8Given y = ( 4x 5 ) 2

dx

dy = 8 ( 4x 5 )

x =23

, y = 1

P (23

, 1 )

1

1

1

1

1

16

5.a)

41

N m

4T m

b) (-1,3))2)(12(2 x

dxdy

4)12(4 x 3 y atau 2)1)1(2( 2 y

1

1

111

1



22/30

4

c) ))1((43 x y 14 x y

11 8

6.

a) y

x21

10

10log

y x 101010 loglog21

10log

ba2

1

b) x210 62 x x

6 x

c)27log

log

3

3 x atau 3log12 3

3log 3 x 27 x

1

1

1

1

1

1

1

1 8

7. a) AQ = 2QB

[( x - 5) 2 + ( y + 2) 2]2 = [2( x - 2) 2 + ( y 1) 2]2

3 x2 + 3 y

2 6 x - 12 y - 9 = 0

x 2 + y 2 2 x - 4 y - 3 = 0 - terbukti

b) subtitude (-1,0) into locus Q

(-1) 2 + (0) 2 2(-1) 4(0) - 3 = 0 terbukti

c) m AC =1502

=31

y 0 =31

( x (-1))

1

1

1,1

1

1



23/30

5

y =31

x31

d) y =31

x31

3 y = - x 1

x = -1 3 y .........(1)

x 2 + y 2 2 x - 4 y 3=0.....(2)

substituted (1) into (2)

(-1-3 y )2 + y 2 2(-1-3 y ) - 4 y 3=0

10 y 2 + 8 y =0

y (10 y + 8)=0

y = 0, y = -54

Substituted y = -54

into (1)

x = - 1 3(-54

)

=57

D ( 57

,- 54

)

1

1

1

1 10

8. a) (i) m 1m 2 = -1



24/30

6

-21

m 2 = -1

m 2 = 2

y 11 = 2( x 5 )y = 2x + 1

(ii)y = 2x + 1.....(1)

2y+x+8=0......(2)

Substituted (1) into (2)

2(2x+1)+x+8=0

x= -2

Substituted x= - 10 into (1)

y = 2(-2)+1

= -3

B(-2,-3)

b)

52)11(3

,5

152 y x= (-2,-3)

5152 x

= -2 ,5

2)11(3 y= -3

x =225

, y = - 24

c)

01102210

36121222510

6)11()5(

22

22

22

y x y x

y y x x

y x

1

1

1

1

1

1

1

1

1

1 10



25/30

7

9. a)

= 0.8411rad

b) Arc PR = 8 x 0.8411= 6.769 cm

QR 2 = 12 2 8 2 or RT 2 = 8 2 + 8 2 = 8.944 cm = 11.31 cm

Arc RST= 8 x

= 12.57 cm

Perimeter = 4+ 8.944+6.729+11.31+12.57= 43.56 cm

c) Area of PQR =(

( )

= 8.858 cm 2 or

Area of RST =(

= 18.27 cm 2

Area of shaded region = 8.858 + 18.272= 27.13 cm 2

1

1

1

1

1

1

1

1

1

1 10

10. a) LN = 4 + xLN =

ON =2

16 2 x

Area of sector OLMN= 2

=

Area of

= 2 x Area of shaded region = 2 x

=

1

1

1

11



26/30

8

b) ,

when A has a stationary value,,

hence A is a minimum when ,

Amin =

= cm 2.

1

1

1

1

1

10

11a)

Height(cm)

Number ofchildren

Cumulativefrequency

100-104 8 8

105-109 5 13

110-114 10 23

115-119 6 29

120-124 7 36

125-129 4 40

[ 2 m ]

Note: at least 1 mistake, give 1mark

2

b) Q1 = 104.5 + (5

8)40(41

) ( 5 )

= 104.5 + 2

= 106.5

1

1

1



27/30

9

b)Q2 = 109.5 + (10

13)40(21

) ( 5 )

= 109.5 + 3.5

= 113

1

1

1 812. a) p = 45

q = 45

r = 105

b) 3k + 2k +20 + 5k = 100

k = 8

100)105(5)80(20)125(2)120(3 k k k

100)105(40)80(20)125(16)120(24

= 106.8

c) RM 250 x

8.106

100

= RM 234.08

d)106.8 x100130

= 138.84

1

1

1

1

1

1

11

1

1

10

13. a)

64.2130

00.2

RM z y

RM x

11

1



28/30

10

b)

150)30110()50150()36130()10125()24120(

01/04 I

= 130.73c)

84.1045

73.130100800

04

04

RM P

P

d)

41.163

100125

73.130

01/07

01/07

I

I

12504/07 I

1

1

1

1

1

1

1

10

14 (i) AB = 6 + 8.1 2(6)(8.1) cos 130 AB = 12.81

(ii) 0130sinsin6 AB BAC

Sin BAC =

AB

Sin 01306

=81.121306 0Sin

BAC = 21 02

(iii) The area =21

(6)(8.1)sin 130 +21

(6)(6)sin80

= 36.34 cm 2

1,11

1

1

1

1,1,11

10



29/30

11

15 AD : DB = 1 : 2 AD = = 4 cm

BD = 12 4 = 8 cm.

a) (i) Using the Cosine Rule in ABC, BC 2= 12 2 + 7 2 2(12)(7)Cos 110 o BC = 15.8259EC = 15.8259 8

= 7.8259= 7.526 cm

(ii) Using Sin Rule in ABC,

Sin


30/30

12

(ii) Using the Cosine Rule in CEF, EF 2 = 3.5 2 + 7.8259 2 2(3.5)(7.8259)Cos45.44 o EF = 5.9209 cm

Let the perpendicular distance from D to EF be h cm.

Area of DEF = 9.8305 (5.9209)(h) = 9.8305h = 3.3206

= 3.321 cm

1

1

1

10

Documents

Tutormansor.files.wordpress.com 2012 09 Add Math f4 Final 2011 Melaka p2 Ans