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3472/2 Matematik Tambahan Kertas 2 September 2 Y2 jam

MAJLIS PENGETUA SEKOLAH MENENGAH MALAYSIA CAWANGAN NEGERI SEMBILAN DARUL KHUSUS

PROGRAM PENINGKATAN AKADEMIK TINGKATAN 5

. SEKOLAH-SEKOLAH MENENGAH NEGERI SEMBILAN 2014

MATEMATIK TAMBAHAN

Kelias 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1.

Kertas so alan ini adalah dalam dwibahasa.

2.

Soalan dalam Bahasa lnggeris mendahului soalan yang sepadan dalam Bahasa Melayu.

3.

Calon dikehendaki membaca arahan di halaman belakang.

4.

Calon dikehendaki menceraikan halaman 20 dan ikat sebagai muka hadapan bersama-sama dengan kertasjawapan.

Kertas soalan ini mengandungi 20 halaman bercetak dan 1 halaman kosong. . [Lihat halaman sebelah SULIT

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The following fonnulae may be helpful in answering the questions. The symbols given are the ones commonly used. Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-sirnbol yang diberi adalah yang biasa digunakan. ALGEBRA

-b ±.Jb2 -4ac

1. x= - - - - - ­

2a

2. a m

an

X

= a m+n

= a + (n -

9. Tn

l)d

n 10. Sn =-[2a+(n-l)d] 2 n I 11. T" = ar -

_ a(rn -1) _ a(1 - r") 12 . S r n r-1 1-r '

5. loga mn = loga m + loga n m

6. logo -

= logo m -logo n

13. S""

n

*1

= ~, Irl < 1 1-r

CALCULUSIKALKULUS dy dv du 1. y=uv, -=u-+vdx dx dx

4. Area under a curve Luas di bawah lengkung b

b

a

a

fydx or (atau) = fxdy

=

5. Volume of revolution du dv v--u ­ lsi padu kisaran dx 2. y == ~ , dy = dx b 2 V dx v = 7r y2 dx or (atau)

f a

a

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347212

STATISTICS / STATISTIK

-

"WI

7. I = L.

I

I

IW;

8. nPr 9.

==~­ (n-r)!

nc _ r

10. peA

5.

-N-F . [1 1 m~Lj-l2f. C

n! (n-r)!r!

---­

UB) =

peA) + PCB) - peA n B)

12. Mean 1Min, jJ..

a=~npq

13.

6. I=ii x100 Qo

= np

X-jJ..

14. Z ==

(J

GEOMETRY 1GEOMETRI

5.ld =~X2 + y2

I.Distance I Jarak

=~(X2-Xl)2 -J .

2. Midpoint I Titi,7 :..ngah

6.

" ~ Y2J ( X,Y ) _(Xl+X7 - - 2 -' - 2 3. A point dividir Titik yang men

(X,y)=(nx1 +

h. "Z.:

m -l l'

4. Area oftriang

1 2

r= -

X!+ yj

~X2 + y2

ent of a line suatu tembereng garis

:~l +mY2J

m +n

L ULls segitiga

=-1(x1Y2 + X2Y3 + X3Yl) - (X 2Yl

+ X3Y2 + X1Y3 ~

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TRlGONOMETRY 1 TRIGONOMETRI 1. Arc length, s = rB Panjang lengkok, s

=je

8. sine A ± B) = sin A cos B ± cos A sin B sin(A ± B) = sin A kos B ± kos Asin B

1 2 2. Area of sector, A =- r B 2

9. cos(A ± B) = cos A cos B + sin Asin B kos (A ± B) = kosA kosB + sin A sin B

Luas sector, L

=-1 j .2e 2 A =1

3. sin 2 A + cos 2 sin 2 A + kos 2 A

tan A ± tanB 1O. tan(A ± B) = - - - - ­ l+tan AtanB

=1

4. sec 2 A = 1 + tan 2 A sek 2 A ,;,;, 1 + tan 2 A 5.

co'sec- A = 1 + cot 2

11. tan2A

A kasek 2 A = 1 + kat 2 A

2

I-tan A

12. - = - sinA sinB

6. sin2A = 2sinAeosA sin 2A = 2 sin A kosA 2

2tanA

abc

2

7. cos 2A = cos 2 A - sin

=

13. a 2

=-­

= b2 + C2

2

2

sine -

2bc cos A

2

a =b +c -2bc kosA A

= 2eos 2 A-I = 1- 2 sin 2 A

14. Area of triangle 1 Luas segitiga 1 b sm . C' =-a

2

kos2A = kos 2 A -sin 2 A = 2 kas 2 A-1 = 1- 2sin 2 A

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Section A BahagianA

[40 marks] [40 markah] Answer all questions. Jawab semua saalan.

Solve the following simultaneous equations: Selesaikan persamaan serentak berikut. 2x = 5 + y 3x 2 -y 2 =14- 2 x - y

[5 marks] [5 markah]

2

Given that the quadratic function [(x) = a(x - p)2 + 3 with a minimum point(2, q), where a, p and q are constants. Point A(O, 7) lies on the curve y = j(x). Diberifungsi kuadratik [ex) = a(x - p)2 + 3 mempunyai titik minimum (2,q), dengan keadaan a, p dan q adalah pemalar. Titik A (0, 7) berada pada lengkung y = j(x).

(a) Find the value of a, of p and of q.

[4 marks] [4 markah]

Tentukan nilai a, p dan q.

(b) Sketch the graph of the function y =j(x).

[3 marks] [3 markah]

Lakarkan grafbagi fungsi y = j(x).

(c) State the function of the curve when the graph in 2(b) is reflected on the x-axis. Nyatakan fungsi bagi lengkung apabila grafdi 2(b) dipantulkan pada paksi-x.

[1 mark] [1 markah]

3

Given that the function y Diberi fungsi y

= x(x -

= x(x -

3), find the values of x that satisfy the equation

3), cari nilai-nilai x yang memenuhi persamaan

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Solution by scale drawing is not accepted. Penyelesaian secara lukisan berskala tidak diterima.

Diagram 4 shows a trapezium OEFG. The line OE is perpendicular to the line EF which intersects with y-axis at the point H. EO is parallel to FG. Given that the equation of OE is y = 2x and the equation of EF is x + 2y + 15 =

o.

Rajah 4 menunjukkan trapezium OEFG. Garis lurus OE adalah berserenjang dengan garis lurus EF yang bersilang dengan paksi-y pada titik H. EO adalah selari dengan FG. Diberi persamaan OE ialah y = 2x dan persamaan EF ialah x + 2y + 15 = O.

G

x

E

Diagram 4 Rajah 4 (a)

Find the coordinates of E. Cari /wordinat E.

(b)

GivenEH:HF=3:1,find Diberi EH: HF = 3 : 1, carikan (i) the coordinates of F, /wordinat F, (ii) the equation of the straight line FG.

persamaan garis lurus FG.

[2 marks] [2 markah]

[4 marks] [4 markah] (c)

A point J moves such that 2JE = JF. Find the equation of the locus of 1. Titik J bergerak dengan keadaan 2JE = JF. Carikan persamaan lokus 1.

[2 marks] [2 markah]

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Table 5 shows the frequency distribution of the ages of workers in a factory. Jadual 5 menunjukkan taburan frekuensi bagi umur pekerja di sebuah kilang. Age (years) Umur (tahlln) 25 - 29 30 - 34 35 - 39 40 ­ 44 45 -49 50 ­ 54

I

Number of workers Bilangan pekerja

5 8 x 15.

8 4 Table 5 Jadual5

(a) Given the mean age of the workers is 39.5 years, find the value of x . Diberi min umur bagi pekerja-pekerja itu ialah 39.5 tahun, cari nilai x.

[3 marks] [3 markah]

(b) Without drawing an ogive, find the median of age . Tanpa melukis ogif, cari median bagi umur.

[3 marks] [3 markah]

6 (a) Prove that Buktikan bahawa 2 tanx 2

1 + tan x

. 2

= Sltl

[2 marks]

X

[2markah] (b) Sketch the graph of y =

Lakarkan graf bagi y

~ sin 2x 2

=i

2

for 0 ~ x

~ ~ Jl' .

sin 2x bagi 0 ~ x

2

~i

2

Jl' .

Hence, using the same axes, draw a suitable straight line to find the number of solutions . 4 3 2tanx 3 for the equatIon ~ 3 x --2 = 2 for 0 ~ x ~ ­ Jr . Jl' 1+ tan x 2 State the number of solutions .

Seterusnya, dengan menggunakan paksi yang sama, lukiskan salu garis lurus yang 4 3 2tanx sesuai untuk mencari bilangan penyelesaian bagi persamaan - x - - = - - ­ 3Jl' 2 1+ tan 2 x

~x~i

Jl' . 2 Nyatakan bilangan penyelesaian ilu.

bagi 0

[6 marks] [6 markah] [Lihat halaman sebelah

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Section B Ballagian B [40 marks] l40 markah] Answer any four questions from this section.

Jawab mana-mana empat .malan daripada bahagian ini.

7

Diagram 7 shows a triangle PQR. The point S lies on PR and the point T lies on PQ.

The straight line QS intersects the straight line RT at the point U.

Rajah 7 menunjukkan segi tiga PQR. Titik S terletak pada P R dan titik T terletak pada PQ.

Garis lurus QS bersilang dengan garis funts RT pada tilik U.

P T '-..,

S

Q

R Diagram 7 Rajah 7

It is given that LPRQ = 90°, PR = 9~, RQ = 12r, PR : sR = 3 : 1 and PQ : pt = 5 : l. Diberi bahawa LPRQ = 90 °, PR = 9~, RQ = 12r. PR : sR = 3 : 1 dan PQ : (a) Express in terms of~ and

pt = 5:

1.

r'

Ungkapkan dalam sebutan ~ dan Y,..J (i) QS (ii) FIT

[3 marks] [3 markah]

(b) Using RU = hll and QU and of k.

=

kQS, where hand k are constants, find the value of h

Dengan rnenggunakan R-v- = hR! dan QU = kQS, dengan keadaan h dan k ialah pernalar, cari nilai h dan k. [5 marks] [5 rnarkah] (c) Given that I~I

=

5 units and

Diberi I~I = 5 unit dan

Irl

l!: 1 = 2 units, find IPQI.

[2 marks]

IPQI.

[2 rnarkah]

= 2 unit, cari

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SULIT Use graph paper to answer this questi on.

Gunakan kulas graf untuk menjawab soalun into

Table 8 shows the values of two variable.'. x and y, obtained from an experiment. The 6x

variables x and yare related b I the equation y = L q

, where p and q are constants.

Jadual 8 menunjukkan nilai-nilai bag i dua pemboleh ubah, x dan y, yang diperoleh 6x

daripada satu eksperimen. Pemboleh ubah x dan y dihubungkan oleh persamaan y =

L

q

,

dengan keadaan p dan q ialah pernalar.

x y

4.3

1.5 5.1

2 6.2

2.5 7.2

3 8.9

3.5 10.7

Table 8 }adual8

(a) Based on Table 8, construct a ta ble for the values of logio y.

[1 mark]

[1 markah]

Berdasarkan Jadual 8, bina satu jadual bagi nilai-nilai 10giO y .

(b) Plot 10g iO y against x, by using a sc,li e of 2 em to 0.5 units on the x-axis and 2 cm to 0.1 units on the log lo y-ax is. Hence, draw the line of best fit. [3 marks] Plot logJO y melawan x, dengan menggunakan skala 2 em kepada 0.5 unit pada paksi-x dan 2 em kepada 0.1 unit pada paksi-IogJo y . Seterusnya, lukis garis lurus penyuaian terbaik.

[3 markah]

(c) Use the graph in 8(b) to find the value of Gunakan grafdi 8(b) untuk rn encari nilai

y when x = 2.7, y apabila x = 2.7, (ii) p, (iii) q.

(i)

[6 marks] [6 markah]

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9 a)

Given y =

:2

j~

find the value of

when x

3472/2

=

3. Hence, find the approximate value of

7

(2.98)2 .

Diberi y =

:2

cari nilai bagi

:

apabUa x

=

3. Seterusnya cari nilai hampir bagi

7

[3 marks]

3 markah]

Diagram 9 Rajah 9

b) Diagram 9 shows the curve y = x 2 + 1, and the straight line AB which intersect at point P (1, 2). Calculate Rajah 9 menunjukkan lengkung y = x 2 + 1, dan garis lurus AB yang bersilang pada titik P (1, 2). Hitung

(i) the area of the shaded region, luas kawasan berlorek,

[3 marks] [3 markah]

(ii) the volume of revolution, in terms of 1[, when the region bounded by the curve, the y-axis and the straight line y = 3 is revolved 360 0 about the y-axis. [4 marks] isi padu janaan, dalam sebutan 7[, apabila rantau yang dibatasi oleh lengkung itu, paksi-y dan garis lurus y = 3, dikisarkan melalui 360 0 pada paksi-yo [4 markah]

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Diagram 10 shows A BC is a semicircle with centre 0 and radius 6 cm. APD is a sector of a circle with centre P and radius 8 cm. Rajah 10 menunjukkan AOEe ialah sebuah semibulatan berpusat 0 dengan berjej,;;ri 6 em. APD ialah sebuah bulalan berpusat P dengan jejari 8 em.

D

o

A

p

B

Diagram 10 Rajah 10 It is given that OC lS perpendicular to AOB . Diberi bahawa OC bersererenjang dengan AOB.

[UselGuna 7r = 3.142]

Calculate Hitung

e,

(a) the value of in radians, nilai dalam radian,

[2 marks] [2 marlwh]

(b). the perimeter, in cm, of the shaded region. perimeter, dalam em, rantau berlorek.

[4 marks] [4 marlwh]

(c) the area, in cm 2 , of the shaded region. luas, dalam em 2 , rantau berlorek.

[4 marks] [4 marlwh]

e,

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11 (a) It is fo und that 40% of the swan in a zoo have white feathers. If5 swans from the zoo are eho 'en at rand om, calcu late the probability that Didapati bahawa 40 % angsa di sebuah zoo mempunyai bulu putih. Jika 5 ekor angsa dari zoo itu dipilih secara rawak, hitungkan kebarangkalian bahawa (i)

exactly 3 of the swans have white feathers, tepat 3 ekor angsa mempunyai bulu putih,

[2 marks] [2 markah]

(ii)

less than 4 of the swans have white feathers. k1!ran, dorip ada 4 ekor angsa mempunyai bulu putih.

[3 marks] [3 markah]

(b) The masses of students in a college follow a nonnal distribution with a mean of 48 kg and a standard deviation of 16 kg, find Jisim pelajar di sebuah kolej mengikut taburan normal dengan min 48 kg dan sisihan piawai 16 kg , cari . .

(i) the probability that a stud ent chosen randomly from the college has a mass of not more than 57.4 kg, kebarangkalian seorang pelajar yang dipilih secara rawak dari kolej mempunyai jisim yang tidak melebihi 57.4 kg, vaJu~ of m if 70% of the students from the college have a mass of more than mkg. nilai mjika 70% daripada pelajar-pelajar sekolah itu mempunyaijisim melebihi m kg. [5 marks] [5 markah]

(ii) the

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Section C

Bahagian C

[20 marks] [20 markah] Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini.

12

A particle moves along a straight line and passes through a fixed point O. Its velocity, vms-I, at time t seconds after leaving 0 is given by v = 7t - 6 -

r.

Suatu zarah bergerak di sepanjang suatu garis lurus dan melalui satu titik tetap 0. Halaju zarah itu, vms-I, pada masa t saat selepas melalui 0, diberi oleh v = 7t - 6 - 1. [Assume motion to the right is positive.J [Anggapkan gerakan ke Ctroh kanan sebagai positif.] Find Cari (a) (i)

[1 mark] [1 markah]

the initial velocity of the particle, halojll awol zarah ilu,

(ii) the time interval during which the particle moves towards the right. selang maso opabila zarah itu bergerak ke arah kanan. (b)

[2 marks] [2 markah] [3 marks] [3 markah]

the maximum velocity of the particle, in m S-I, halaju maksimum zarah itu, dalam m S-I,

(c) (i) Sketch the velocity-time graph of the motion of the particle for the first 6 seconds. [2 marks] Lakarkan graf halaju-masa bagi pergerakan zarah itu bagi tempoh 6 saat pertama. [2 markah] (ii) the distance travelled by the particle in the sixth second after leaving O. [2 marks]

jarak yan.g dilalui oleh zarah itu dalam saat keenam selepas melalui 0. [2 markah]

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A particular kind of biscuits is made by using four ingredients A, B, C and D. abl e 13 shows the prices of the ingredients. Sejems biskut diperbuat daripada bahan-bahan A, B, C dan D. Jadual 13 menunjukkan harga bagi bahan-bahan tersebut. Price (RM) per kilogram in the year Ingredients

Bahan-bahan

Harga (RM) per kilogram pada tahun 2008

2009

A

5.00

w

B

4.00

5.00

C

x

D

6.00

y I

7.80

Table 13 Jadual13 (a) The price index of ingredient A in the year 2009 based on the year 2008 is 120. Calculate the value of w. [2 marks] lndeks harga bahan A pada tahun 2009 berasaskan tahun 2008 ialah 120. Hilung nilai w.

[2 markah] (b) The price index of ingredient C in the year 2009 based on the year 2008 is 150. The price per ki logram of ingredient C in the 'year 2009 is RM3.00 more than its corresponding price in the year 2008. Calculate the value of x and of y. [3 marks] lndeks harga bahan C pada tahun 2009 berasaskan tahun 2008 ialah 150. Harga bahan C sekilogram pada tahun 2009 ialah RM3.00 lebih daripada harganya yang sepadan pada tahun 2008. Hilung nilai x dan y. [3 markah]

(c) The composite index for the cost of making the biscuit in the year 2009 based on the year 2008 is 128. Find lndeks gubahan bagi kos penghasilan biskut pada tahun 2009 berasaskan tahun 2008 ialah 128. Cari (i) the price of a packet of biscuit in the year 2008 if its corresponding price in the year 2009 is RM35.00. [2 marks] harga sebungkus biskut pada tahun 2008 jika harganya yang sepadan pada tahun [2 markah] 2009 ialah RM35.00. (ii) the value o f m if the quantities of ingredients A, B, C and D used are in ratio of 3 : m : 1 : 4. [3 marks] nilai mjika kuantiti bahan A, B, C dan D digunakan dalam nisbah 3 : m : 1 : 4. [3 markah]

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Use the graph paper to answer this question. Gunakan kertas grajuntuk menjawab soalan ini. Mathematics Society organizes a visit to National Science Ce'Q.tre by renting x buses and y vans. The rental for a bus is RM750 and the rental for a van is RM250. The rental for the vehicles is based on the following constraints: Persatuan l'v1atematik menganjurkan lawatan ke Pusat Sains Negara dengan menyewa x buah bas and y buah van. Sewaan sebuah bas ialah RM750 dan sewaan sebuah van ialah RM250. Sewaan kenderaan adalah berdasarkan kekangan berikut : The total number of vehicles to be rented is not more than 8. Jumlah kenderaan yang disewa adalah tidak melebihi 8 buah.

II

The num ber of buses.is at most twice the number of vans. Bilangan bas selebih-lebihnya dua kali ganda daripada bilangan van.

III

The maximum allocation for the rental of the vehicles is RM2 250.

Perzmtuknn maksimum untuk sewaan kenderaan ialah RM2 250.

(a) Write three inequalities, other than x ~ 0 and y ~ 0 which satisfy all the above [3 marks] constraints. Tulis tiga ketaksamaan, selain x ~ 0 dan y ~ 0 yang memenuhi semua kekangan di [3 markah] atas. (0') Using a scale of 2 cm to I vehicle on both axes, construct and shade the region R which satisfies all the above constraints. [3 marks] Dengan menggunakan skala 2 em kepada 1 kenderaan pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan di atas. [3 markah] (c) Using the graph constructed in 14(b), find

Gunakan grafyang dibina di 14(b) untuk meneari

(i) the minimum number of vans rented if2 buses are rented

bilangan minimum van yang disewajika 2 buah bas disewa.

(ii) the maximum number of members that can be accommodated into the rented vehicles if a bus can accommodate 40 passengers and a van can accommodate 10 passengers. Bilangan maksipmum ahli yang boleh dimuatkan ke dalam kenderaan yang disewa jika sebuah bas boleh dimuatkan dengan 40 penumpang dan sebuah van boleh dimuatkan dengan 10 orang penumpang. [4 marks] [4 markah]

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Diagram 15 shows a quadrilateral PQRS . LQRS is obtuse.

Rajah 15 menunjukkan sisiempat PQRS. LQRS ialah sudut eakah.

P~----------------~----------~--~Q

15 em

R Diagram 15 Rajah 15 Given that the area of triangle' QRS is 28 em 2. Find

Diberi bahawa luas segitiga QRS ialah 28 em 2 . Cari

(a) the value of nilai e,

B,

[2 marks] [2 markah]

(b) the length, in em, of panjang, dalam em, (i)

(ii)

QS, PQ.

[5 marks] [5 markah]

(c) the area, in em 2 ,of quadrilateral PQRS. luas , dalam em 2 , sisiempat PQRS.

[3 marks] [3 markah]

END OF QUESTION PAPER KERTAS SOALAN TAMAT

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THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(O, I) KEBARANGKALIAN HUJUNG A TAS Q(z) BAGI TABURAN NORMAL N(O, 1) 1

2

3

0.4641

4

8

12

16

20

0.4286

0.4247

4

8

12

16

20

0.3936

0.3897

0.3859

4

8

12

15

19

0.3594

0.3557

0.3520

0.3483

4

7

11

15

0.3228

03192

0.3156

0.31 21

4

7

11

15

0.2912

0.2877

0.2843

0.2810

0.2776

3

7

10

14

0.2578

0.2546

0.2514

0.2483

0.2451

3

7

10

13

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

3

6

9

12

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

3

5

8

0.1762

0.1736

0.1711

0. 1685

0.1660

0.1635

0.1611

3

5

8

z

0

I

2

3

4

5

6

7

8

,

9

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681 '

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.2

0.4207

0.4168

OAl29

OA090

0.4052

0.4013

0.3974

0.3

0.38 21

0.3783

0.3745

0.3707

0.3669

0.3632

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.6

0..2743

0.2709

0.2676

0.2643

0.2611

0.7

0.2420

0.2389

0.2358

0.2327

0.8

0.2119

0.2090

0.2061

0.9

0.1841

0.181 4

0.1788

4

5

6

7

8

9

24

28

32

36

24

28

32

36

23

27

31

35

19

22

26

30

34

18

22

25

29

32

17

20

24

27

31

16

19

23

26

29

15

18

21

24

27

11

14

16

19

22

25

10

13

15

18

20

23

Minus / Tolak

1.0

0.1587

0.1562

0.1539

0. 1515

0.1492

01469

0.1446

0.1423

0.1401

0.1379

2

5

7

9

12

14

16

19

21

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1170

2

4

6

8

10

12

14

16

18

1.2

0.1151

0.1131

0.1 "'2

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

0.0985

2

4

6

7

9

11

13

15

17

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

0.0838

0.0823

2

3

5

6

8

10

11

13

14

lA

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0,0708

0.0694

0.0681

1

3

4

8

10

11

13 11

1.5

0.0656

0. 0655

0.0643

0.0630

0.0618

00606

0.0594

0.0582

0.0571

0.0559

1

2

4

6 .. - 5

7 6

7

8

10

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

0.. 0475

0.0465

0.0455

1

2

3

4

5

6

7

8

9

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

0.0392

0.0384

0.0375

0.0367

1

2

3

4

4

5

6

7

8

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0322

0.0314

0.0307

0.0301

0.0294

1

1

2

3

4

4

5

6

6

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

0.0256

0.0250

0.0244

0.0239

0.0233

1

1

2

2

3

4

4

5

5

1.0

0.0128

0.0122

0.0117

0.0112

00107

0.0101

0.0197

0.0192

0.0188

0.0183

0

1

1

1

1

3

3

4

4 4

.-

-

,

1.1

0.0179

0.0174

0.0170

0.0166

0.0161

0.0158

0.0154

0.0150

0.0146

0.0143

0

1

1

2

1

1

3

3

11

0.CD9

0.0135

0.0132

0.0119

0.0115

0.0111

0.0119

0.0116

0.0113

0.0110

0

1

1

1

1

1

1

3

3

2. 3

O .O~O7

0.0104

0.0102

0

1

I

1

1

1

1

1

1

0.00990

0.0 0964

0.00939

0.00914 0. 00889

1.4

0.00820

0.0 07 g8

0.00776

0.00755

0.00866

0.00842

0.007 34

3

5

8

10

13

15

18

10

23

2

5

7

9

11

14

16

16

11

2

4

6

8

11

13

IS

17

19

7

9

11

13

15

17

0.00714

0.00695

0.00676

0.00657

0.00639

2

4

6

1.5

0.00621

0.00604

0.00581

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2

3

5

6

8

9

11

11

14

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

1

2

3

5

6

7

9

9

10

L7

0.00347

0.00336

0.00316

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00164

1

1

3

4

5

6

7

8

9

2.8

0.00256

0.00248

0.00240

0.00 233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

1

1

2

3

4

4

5

6

6

2.9

0.00187

0.00181

0.00175

0.00169

0.00164

0.001 59

0.00154

0.00149

0.00144

0.00139

0

1

1

2

2

3

3

4

4

0.00135

0. 00131

0.0 0116

0.00121

0.0 011 8

0.00114

0.00111

0.00107

0.00104

0.00100

0

1

I

2

2

2

3

3

4

3.0

I

(z)

Example / Contoh:

1 (12 J

f(z)=--exp --z 2

If X ­ N(O, I), then

.J2Tr

Jika X ­ N(O, 1), maka 00

Q(z)

J

= fez) dz

P(X > k)

=

Q(k)

k

P(X > 2.1) = Q(2.l) = 0.0179

o 347212 © 2014 MPSM Cawangan Negeri Sembilan

k

z [Lihat halaman sebelah SULIT

SULIT

18

3472/2

BLANK PAGE

HALAMAN KOSONG

3472/2 © 2014 MPSM Cawangan Negeri Sembila

. [Lihat halaman sebelah SULIT

347212

19

SULIT

Nama: ................................................................................................. . Kelas: ............................................................................................................... .. Arahan Kepada Cal on 1 Tulis nama dan kelas anda pada ruang yang disediakan. 2 Tandakan (,j ) untuk soalan yang dijawab. 3 Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan buku jawapan. Bahagian

Soalan

Soalan Dijawab

Markah Diperolehi (Untuk Kegunaan Pemeriksa)

Markah Penuh

1

5

2

8

"

5

3 A

B

C

.

I

4

8

5

6

6

8

7

10

8

10

9

10

10

10

11

10

12

10

13

10

14

10

15

10

-- ..

Jumlah

[Lihat halaman sebelah 3472/2 © 2014 MPSM Cawangan Negeri Sembilan

SULIT

SULIT

20

347212

INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON

. \

1. This question paper consists of three sections: Section A, Section B and Section C. Kertas sa alan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C

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

Jawab semua sa alan dalam Bahagian A, mana-mana empat saalan daripada Bahagian B dan mana-mana dua soalan daripada Bahagian C

3. Show your working. It may help you to get marks.

Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini baleh membantu anda untuk mendapatkan markah.

4. The diagrams in the questions provided are not drawn to scale unless stated.

Rajah yang mengiringi sa alan tidak dilukis mengikut skala kecuali dinyatakan.

5. The marks allocated for each question and sub-part of a question are shown in brackets.

Markah yang diperuntukkan bagi setiap saalan dan ceraian so alan diturijukkan dalam kurungan.

6. The Upper Tail Probability Q(z) for The Normal Distribution N(O, 1) Table is provided on page 17. Jadual Kebarangkalian Hujung Atas Q(z) bagi Taburan Narmal N(O, 1) disediakan di

halaman 17. 7. A list of formulae is provided on pages 2 to 4.

Satu senarai rumus disediakan di halaman 2 hingga 4 . 8. Graph paper is provided.

Kertas grafdisediakan. 9. You may use a scientific calculator.

Anda dibenarkan menggunakan kalkulator saintifik. 10. Tear out page 19 and tie it together with the answer papers and graph papers. Ceraikan halaman 19 dan ikat bersama dengan kertasjawapan dan kertas graph.

3472/2 © 2014 MPSM Cawangan Negeri Sembila

SULIT

MAlLIS PENGETUA SEKOLAH MENENGAH MALAYSIA

CAW ANGAN NEGERI SEMBILAN DARUL KHUSUS

PROGRAM PENINGKATAN AKADEMIK TINGKATAN 5

SEKOLAH-SEKOLAH MENENGAH NEGERI SEMBILAN 2014

!

ADDITIONAL

MATHEMATICS

2

Solution and marking scheme

I Number

I

y=2x-5

Kl

32 -(2x-5f =14-2x-(2x-5) (x-2Xx-2~=O x=2,x=22

(y-39XY+l)=O y=39,y=-1 x=22,x=2

y=3~y=-1

KI

NI

Nl

5

PI

PI

p=2 q=3 7 = a(O - 2)2 + 3

KI

a=l

NI

Y ... ~ (b)

Full Marks

PI

5+y 2

X=­

{5~y)2 _y2=14_ {5~Y)_y

2 (a)

Sub Marks

\

(0, 7)

PI titik (0,7)

PI (2,3) I

Titik (2,3)

....

y

x

0

(c)

3

f(x)

dy

=

-(x - 2)2 - 3

= 2x-3

dx

PI Bentuk graf

Nl

8

Kl

2

d x -=2 dy2

I

(x 2 -3xp+x(2x-3)+5=O

Kl Kl Kl

(4x- 5Xx -1)= 0

x

5

1 4'

=-

Nl

5

2

3

Solution and marking scheme

Number 4(a)

2x E

(b)(i)

(ii)

1 15 = --x -2 2

Nl

= (-3, -6)

C(

(0 -7 ~) = -3)+3x 1( -6)+3 Y )

'24' 4

KI

F= (1, -8)

Nl

y = 2x + c or -8

= 2(1) + c

Kl NI

2J(x + 3)2 + (y + 6)2

= JCx -

1)2 + (y + 8)2

3x2 + 3/ + 26x + 32y + 115 = 0

Sea)

27(5) + 32(8) + 37 x J.. 42(15) + 47(8) + 52( 4) 5+8+x+15+8+4

.

KI NI

= 39.5

KI

10

NI

Seen 39.5

PI

--23

r

15

1 5

8

KI

1605 + 37 x = 39 .5(40 + x)

39.5+ 50 2

Full Marks

Kl

cuba selesaikan

y=2x-lO

(c)

Sub Marks

KI Nl

6

40.17

3

4

Solution and marking scheme

Number

6(a)

Seen sec x or 2sin x cos x

Sub Marks

Full Marks

Kl Nl

sin 2x

1.5 Shape

of Sine

PI

0.75

6 (b)

Amplitud PI

- 1.5 -2.25

Period

PI

4x 3 . 2 --= Sln x 3rr 2

KI

Y = -2x - -9 or \.me on t he grap h

PI

rr

4

Number of solutions = 3

NI

8

4

5

~D

'QS = QR + liS

or Iff =

RQ + QT

or Iff =

RQ + ~ QP 5

-3~

ii)

Sub Marks

Solution and marking scheme

Number

-

12~

KI NI

Nl

36 12 --x+-y 5 -

Full Marks

5 _

KI

KI -3k= ­ 36 h or

-12k=gh -12 5

5

KI

5

h=­ 13

k=

Nl

g

13

NI

KI

(c) 51 unit

8 (a)

Nl

I ~----~----~--~~--~----~----~-----, ' loglO y I 0.6335 I 0.7076 Ii 0.7924 I 0.8573 0.9494 1.0294 1

!

Sekurang-kurangnya dua tempat perpuluhan

(b)

PI loglO y

KI

= 0.9

Nl

y = 7.943 (iii)

Nl

Rujuk graf

(c)(i)

(ii)

!

10

6 loglOP loglOP

=

m

-log lO q = Y-intercept

or

= 0.02583

or

P = 0.9612 ­ 1.1612 q = 0.2388 - 0.4388

or

Kl

loglO q = -0.47

Nl Nl

10

5

6

Solution and marking scheme

Number

Sub Marks

Full Marks

Kl

9(a)

Kl Nl 0.7881

K1

(b)

Kl Nl

K1

(c)

Kl Kl

lO (a)

(b)

(c)

2IT

Nl

tan B = 6/2

Kl

Nl

e=

1.2492 rad

1.571 x 6 == 9.2426 AD == 1.2492 x 8 = 9.9936 9.2426 + 9.9936 + 1.6754 20.9 116

Kl Kl Kl Nl

or

1. (8)(8)(1.2492)

Kl

2 ~ 2-(6)(8)(1.571) 2 ,

K1

1. (8)(8)(1.2492) - ~ (6)( 6)(1.571) - ~ (6)(2) 2

10

2

2

Kl Nl

10

5.6964

6

7

Solution and marking scheme

Number 11 a)(i)

(ii)

I

Sub Marks

P(X = 3) = :>C 3(OA)"\0.6)L 0.2304

Kl Nl

P(X= 0) + P(X= 1) + P(X= 2) + P(X= 3) or 1 - P(X = 4) - P(X = 5) 5Co (0.4)o(0.6)5 + 5C[(0.4)1(0.6)4 + 5C 2(0.4i(0.6)3 + 5C 3(0.4)3(0.6)2 or 1 - 5C4 (0.4)4(0.6)[ + 5C s(OA)\0.6)o

Kl

Full Marks

Kl Nl

0.9130 (b )(i)

Kl Nl

P(Z:S 57.4 - 48 ) 16

0.7216

(ii)

Kl

P(Z> 0.524) = 0.3

= -0.524

m - 48 16

Kl Nl

m = 39.62 12 (a)(i) (ii)

(b)

-6

PI

(t-6)(t-l)<0 l
Kl Nl

a = -2t + 7 vmax = 7 (7.-) 2

_6_

10

Kl Kl

(?)2 2

6.25

Nl Shape of graph Point (0, - 6) and (1, 0) or (6, 0)

(c )(i)

(ii)

PI PI

s=- (5)3 + 7(5)2 -6(5)= 95 m

S

3

2

(6)3

7(6)2

=- -

3

+

6

~--

2

6(6) = 18m

95

13

6

6

Distance = 18-- =-m

or

Kl 10

Nl

7

8

Solution and marking scheme

Number

13 (a)

(b)

N1

K1

Nl N1

128 = 35.00 x 100

(ii)

K1

Q2008 Q2008

= RM27.34

N1

125 or 130

120(3) + 125m + 150(1) + 130(4)

= 128 8+m

PI Kl

.

10

m=2

x 750x (b) (c) i) ii)

Nl

N1 Nl

x+y$8

14 a)

~

1

2y or y>-x -2

+ 250y

~

Full Marks

K1

w

120 = -xl00 5 w= RM6.00 x+3 150 = --x100 x x =6, y=9

(c )(i)

Sub Marks

2250

or 3x

+y

~

N1 9

Refer to graph 1 titik (2, 3) 2(40) + 3 (10) 110

PI Kl N1 Nl

10

8

9

Solution and marking scheme

Number I 5(a)

(b)(i)

(ii)

2 LB=153.61°

N1

QS2 =142 +9 2 -2(l4)(9)cosI53.61° QS= 22.42

Kl NI

sinLP 22.42

Kl

sin35

=-15­

or

Full Marks

KI

~(l4)(9) sin B = 28

LP = 59.020

PQ sin85.98°

(c)

Sub Marks

-

15 sin35°

KI

PQ= 26.09 em

Nl

. 1 -(22.42)(26.09)sin 35 2

Kl

Area quadrilateral = l(22.42)(26.09)sin 35+28 2 2 = 195.75crn

Kl N1

10

9

10

Question No. 14

y

P2

all 3 straight lines correct any 2 straight lines correct correct shaded region

PI Nl

9

...

8

7

6

5

3x + Y

9

x+ y

\

8

y

Ylx

4

3

R

2

1

x

40x + lOy

k

11

loglO y

I Question No.8

1.0

0.9

0.8

0.6

0.4

0.3

all 6 points plotted correctly uniform scale for axis Iine of best fit

PI PI PI

0.2

0.1

o

x

1 ANALISIS JADUAL SPESIFIKASI UJIAN MATEMATIK TAMBAHAN 2014

PROGRAM PENINGKATAN AKADEMIK TINGKATAN 5

SEKOLAH-SEKOLAH MENENGAH NEGERI SEMBILAN 2014

S A1

Paper 2

Paper 1

CONTEXTS

M

Section B S M

Section A S M

Function IF4) 1 2 3

-

.~.

1

2 (R) I 3 (S)

I

3(R)

' ­ ;oJ ' -

4

4 (S)

5 6

3 (R) 3 (S)

2

8 (R)

1

5 (R)

7

3 (S) 4 (R) 4 (S)

8

9 10 11

1

I

3 (R) 3 (R)

13

3(R)

8

10 (S)

12

4 (R)

10

10 (S)

7

10 (T)

.'

14 15

!

4 (R) 2 (R)

.1 1

16

3 (T)

6

8 (T)

17

3(R)

4

6(R)

18

3 (R) 3 (T)

3

6 (S)

19

~,}:;. ;" - "

20 21

-

- ..

3 (S) 2(S) _~ -'

22

3 (S)

23

4 (R)

24

4 (R)

5

,

7 (S)

.~

"

~

I'



I

Section C

S

M

2

Paper 1

CONTEXTS

S4

Probability Distributions (F5) Binomial Distributions Normal Distributions

Pa.Q.er 2 Section B 111a)

25

I

Section C

5 (S)

41R)

Solutions Of Triangles lF4)

AST1

15

10 R)

12

10 (S)

13

10 (R)

14 4

10 (R)

Motion On the Straight Line (F5)

AST2 ASS1

Section A

II

Index Number (F4)

ASS2 Linear Programmin_9 (F5)

J

JUMLAH

25

80

I

6

, 40

5

50

Nota : A - Komponen Algebra G - Komponen Geometri S - Komponen Statistik T - Komponen Trigonometri K - Komponen Kalkulus ASS - Aplikasi Sains Sosial AST -Aplikasi Sains Teknologi

RSTSM-

Aras Rendah Aras Sederhana Aras Tinggi Soalan Markah

BILANGAN SOALAN MENGIKUT ARAS KESUKARAN:

KERTAS

1

2

ARAS KESUKARAN

BILANGAN SOALAN

RENDAH SEDERHANA TINGGI

15

JUMLAH SOALAN

25

RENDAH SEDERHANA TINGGI

6 5 4

JUMLAH SOALAN

15

NISBAH

8

2 6:3:1

4:3:3

40

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NS K2 Skema.pdf
(iv) Jika rumusan kurang daripada 120 patah perkataan, markah tidak. dipolong. [Markab bahasa maksimum : to markahl. lI03fl & 1103/1 0 20U Hol Cipla MPSM ...

NS K1.pdf
I Pengaruh media massa pada alaf ini banyak mengubah cara berfikir rakyat Malaysia. ... anda ialah 'Peranan Ibu Bapa untuk Menangani Gejala Sosial dalam ...

NS B-L4.pdf
the serpent in the desert, so must the Son of. Man be lifted up, so that ... Jesus also tells us about God's immense love for. us. God loved us so much and ... to provide us with the wisdom to. make the choices which align with. Your light and steer

NS B-E1.pdf
buried just three days ago. As she approaches. the tomb, she discovers the stone blocking the. entrance of the tomb removed. Fearing the. worse, she runs to find Simon Peter and John,. the disciple whom Jesus loved. Mary tells the. disciples that the

Uppercase-Ns-on-Four-Lines.pdf
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NS B-O2.pdf
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2010-4-Fabiano-NS-_Corte_di_giustizia._Leccezione__prioritaria ...
2010-4-Fabiano-NS-_Corte_di_giustizia._Leccezione__p ... tituzionalit_francese_deve_cedere_il_passo_alla_.pdf. 2010-4-Fabiano-NS-_Corte_di_giustizia.

NS B-Most Holy Trinity.pdf
Page 1 of 2. WEEK B - MOST HOLY TRINITY Matthew 28: 16-20. What does it mean to you when you make the sign of the cross, "in the. name of the Father, and of the Son, and of the Holy Spirit"? What would you. say to people who have asked you what does

QRC Add Users.pdf
If you need a student added to your course, please contact the IT Help Desk. at 702-895-0777. Changing the role of a User in Your WebCampus Course.

CN Add on.pdf
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NS Skema K2.pdf
o mudah akses. - tL03l2(PP) 6 zoi+ rneSM CawanganNegeri sembilan. Page 3 of 21. NS Skema K2.pdf. NS Skema K2.pdf. Open. Extract. Open with. Sign In.

G'ville NS 2:17.pdf
muscle energy techniques, hydrotherapies, and cross fiber friction to. help identify and eliminate pain and dysfunction. You will have the. opportunity to dialogue ...

ns-2 Tutorial Exercise
Work in group of two. At least one people in each group must hav e an account on the CS department UNIX servers. • Start Exceed on the machine at your desk. Login on one of the CS compute servers (mamba.cs, viper.cs, etc). • Set up your environme