Stpm Trial 2009 Matht Q&a (johor)

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JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN 954/1 PERCUBAAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN STPM JOHOR 2009 JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN MATHEMATICS T(MATEMATIK T) JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PAPER 1(KERTAS 1)JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

JABATAN PELAJARAN NEGERI JOHOR SIJIL TINGGI PERSEKOLAHAN MALAYSIA Instructions to candidates: DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO. Answer all questions. Answers may be written in either English or Bahasa Melayu. All necessary working should be shown clearly. Non-exact numerical answers may be given correct to three significant figures, or one decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Mathematical tables, a list of mathematical formulae and graph paper are provided. Arahan kepada calon: JANGAN BUKA KERTAS SOALAN INI SEHINGGA ANDA DIBENARKAN BERBUAT DEMIKIAN. Jawab semua soalan. Jawapan boleh ditulis dalam bahasa Inggeris atau Bahasa Malaysia. Semua kerja yang perlu hendaklah ditunjukkan dengan jelas. Jawapan berangka tak tepat boleh diberikan betul hingga tiga angka bererti, atau satu tempat perpuluhan dalam kes sudut dalam darjah, kecuali aras kejituan yang lain ditentukan dalam soalan. Sifir matematik, senarai rumus matematik, dan kertas graf dibekalkan.

This question paper consists of 7 printed pages and 1 blank page. (Kertas soalan ini terdiri daripada 7 halaman bercetak dan 1 halaman kosong) STPM 954/1 *This question paper is CONFIDENTIAL until the examination is over. *Kertas soalan ini SULIT sehingga peperiksaan kertas ini tamat

CONFIDENTIAL* SULIT*

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1.

If z = x + yi and z + 3 − 4i = 3i , show that x 2 + y 2 + 6 x − 8 y + 16 = 0. [4 marks]

2. By using first principles, show that the value of the first derivative of f(x) = x3 – 2, x ∈ IR at x = a is 3a2. [5 marks]

3.

Find the set of values of x that satisfies the inequality

log 3 ( x 2 − 3 x + 2) <

4.

1 log 3 4 + 2 log 3 (2 x − 1). 2

Given that x 3 + 4 x 2 − ax + b is exactly divisible by ( x + 2) but leaves a

remainder a 3 when divided by ( x − a ) , calculate the values of a and b .

5.

6.

[6 marks]

The sum of first n terms of a series is

[5 marks]

a 2−n ( b n − a n ) , b ≠ a. b−a

Find the n th term of the series.

[4 marks]

Hence, show that the series is a geometric series.

[2 marks]

Given that f ( x ) = x 3 − 10 x 2 + ax + b , where a and b are constants. If one of

the zeroes is 3 + 2 , find the values of a and b.

[4 marks]

Hence, (a) show that the equation f ( x ) = 12 − 3 x does not have any real roots other than 4. find the remainder when f (x ) is divided by x 2 − 4.

(b)

7.

[3 marks]

(a)

Expand 3

[2 marks]

1− 3 x in ascending powers of x , up to and including the term

in x . State the range of values of x for which the expansion is valid.

[5 marks] 24

(b)

1   Find the term that is independent of x in the expansion  x −  3x 2  .  [4 marks]

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1.

Jika z = x + yi dan z + 3 − 4i = 3i , tunjukkan bahawa

x 2 + y 2 + 6 x − 8 y + 16 = 0.

2.

[4 markah]

Dengan menggunakan kaedah prinsip pertama, tunjukkan bahawa nilai terbitan

pertama bagi f(x) = x3- 2 ,x ∈ IR pada x = a ialah 3a2.

3.

Cari set nilai x yang memuaskan ketaksamaan berikut

log 3 ( x 2 − 3 x + 2) <

4.

[5 markah]

1 log 3 4 + 2 log 3 (2 x − 1). 2

[6 markah]

Diberi bahawa x 3 + 4 x 2 − ax + b boleh dibahagi tepat oleh ( x + 2) tetapi

meninggalkan baki a 3 apabila dibahagi oleh ( x − a ) , carikan nilai a dan b .

5.

6.

Hasil tambah bagi n sebutan yang pertama suatu siri ialah

[5 markah]

a 2−n ( b n − a n ) b ≠ a. b−a ,

Cari sebutan ke- n bagi siri tersebut.

[4 markah]

Seterusnya tunjukkan bahawa ia adalah suatu siri geometri.

[2 markah]

Diberi bahawa f ( x ) = x 3 − 10 x 2 + ax + b , di mana a dan b adalah pemalar . Jika salah satu pensifar ialah 3 + 2 , cari nilai a dan b .

[4 markah]

Seterusnya, (a) tunjukkan bahawa persamaan f ( x ) = 12 − 3 x tidak mempunyai sebarang punca nyata selain 4.

7.

[3 markah]

(b)

cari baki apabila f (x ) dibahagi oleh x 2 − 4.

(a)

Kembangkan 3

[2 markah]

1− 3 x dalam kuasa x menaik, sehingga dan termasuk

sebutan dalam x . Nyatakan julat nilai x supaya kembangan ini sah.

[5 markah] 24

(b)

1   Cari sebutan yang bebas daripada x dalam kembangan  x −  . 3x 2   [4 markah]

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8.

9.

A curve has the equation y =

x −3 . (2 − x )(x + 1)

(a)

Write the equations of the asymptotes of this curve.

[2 marks]

(b)

Find the turning points and determine its nature.

[5 marks]

(c)

Sketch the curve.

[3 marks]

Show that the equation m 2 x + y − 8m = 0 is a tangent to the curve xy = 16 for

all values of m, m ≠ 0. A perpendicular line from the origin meets this tangent at R. Find the coordinates of R in terms of m .

(

Show that the equation of the locus of R as m varies is x 2 + y 2

)

2

= 64 xy . [10 marks]

10.

Matrices A, B and C are given as

 k 1 − 1   A = 2 − 3 1  , 2 1 2  

11.

1 2 3   B = 2 3 4 ,  1 5 6  

− 17 0   0   C =  − 66 − 38 3  .  51 34 0  

Show that k = 1 if the determinant of A is -17.

[2 marks]

Find BC and ABC. Deduce A-1.

[6 marks]

Hence, solve the simultaneous equations x + y – z = -1 -2x + 3y – z = 3 2x + y + 2z = 25

[3 marks]

(a)

Express

5 x 2 + 4 x + 12 in partial fractions. (x + 2 ) x 2 + 4

(

)

5 x 2 + 4 x + 12 ∫1 (x + 2) x 2 + 4 dx. 3

Hence, evaluate

(

)

[7 marks]

∫ ln(1 + x )dx ,giving 2

(b)

Using trapezium rule, with 5 ordinates, evaluate

2

1

your answer correct to three decimal places.

[4 marks]

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Suatu lengkung mempunyai persamaan y =

8.

x −3 . (2 − x )(x + 1)

(a) Tuliskan persamaan-persamaan asimptot bagi lengkung ini.

[2 markah]

(b) Cari titik-titik pemusingan dan tentukan sifat semula jadinya.

[5 markah]

(c) Lakar lengkung tersebut.

[3 markah]

Tunjukkan bahawa persamaan m 2 x + y − 8m = 0 ialah tangen kepada lengkung

9.

xy = 16 untuk semua nilai m, m ≠ 0. Satu garis dari asalan yang berserenjang bertemu dengan tangen ini di R . Cari koordinat-koordinat R dalam sebutan m .

(

Tunjukkan bahawa persamaan lokus bagi R ialah x 2 + y 2

)

2

= 64 xy apabila m

berubah-ubah

10.

[10 markah]

Matrik A, B dan C diberkan oleh

 k 1 − 1   A = 2 − 3 1  , 2 1 2  

1 2 3   B = 2 3 4 ,  1 5 6  

− 17 0   0   C =  − 66 − 38 3  .  51 34 0  

Tunjukkan bahawa k = 1 jika penentu bagi A ialah − 17 .

[2 markah]

Cari BC dan ABC. Deduksikan A-1.

[6 markah]

Oleh yang demikian, selesaikan persamaan-persamaan serentak x + y – z = -1 -2x + 3y – z = 3 2x + y + 2z = 25

11. (a)

Ungkapkan

[3 markah]

5 x 2 + 4 x + 12 dalam pecahan separa. (x + 2 ) x 2 + 4

(

)

5 x 2 + 4 x + 12 ∫1 (x + 2) x 2 + 4 dx . [7 markah] 3

Dengan yang demikian, tentukan nilai

(b)

(

)

Gunakan petua trapezium dengan 5 ordinat, untuk menentukan nilai

∫ ln(1 + x )dx , berikan jawapan anda betul sehingga tiga 2

2

tempat perpuluhan.

1

[4 markah]

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12.

The function f is defined as

 5 − x +1 , − 6 < x < 1 f (x) =  2 x − 6 x + 8 , 1 ≤ x < 4 . (a)

Find lim− f ( x ) and lim+ f ( x ) . x →1

x →1

Hence, determine whether f is continuous at x = 1 .

[3 marks]

(b)

Sketch the graph of f .

[4 marks]

(c)

Determine the range of f .

[2 marks]

(d)

Determine the set of values of x so that f(x) > 2 – x.

[5 marks]

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12. Fungsi f ditakrifkan oleh

 5 − x +1 , − 6 < x < 1 f (x) =  2 x − 6 x + 8 , 1 ≤ x < 4 . (a)

Cari lim− f ( x ) dan lim+ f ( x ) . x →1

x →1

Dengan yang demikian, tentukan sama ada f adalah selanjar di x = 1. [3 markah] (b)

Lakarkan graf f .

[4 markah]

(c)

Tentukan julat f .

[2 markah]

(d)

Tentukan set nilai x supaya f(x) > 2 – x.

[5 markah]

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JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN 954/2 PERCUBAAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN STPM JOHOR 2009 JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN MATHEMATICS T(MATEMATIK T) JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PAPER 2(KERTAS 2)JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

JABATAN PELAJARAN NEGERI JOHOR SIJIL TINGGI PERSEKOLAHAN MALAYSIA Instructions to candidates: DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO. Answer all questions. Answers may be written in either English or Bahasa Melayu. All necessary working should be shown clearly. Non-exact numerical answers may be given correct to three significant figures, or one decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Mathematical tables, a list of mathematical formulae and graph paper are provided. Arahan kepada calon: JANGAN BUKA KERTAS SOALAN INI SEHINGGA ANDA DIBENARKAN BERBUAT DEMIKIAN. Jawab semua soalan. Jawapan boleh ditulis dalam bahasa Inggeris atau Bahasa Malaysia. Semua kerja yang perlu hendaklah ditunjukkan dengan jelas. Jawapan berangka tak tepat boleh diberikan betul hingga tiga angka bererti, atau satu tempat perpuluhan dalam kes sudut dalam darjah, kecuali aras kejituan yang lain ditentukan dalam soalan. Sifir matematik, senarai rumus matematik, dan kertas graf dibekalkan.

This question paper consists of 7 printed pages and 1 blank page. (Kertas soalan ini terdiri daripada 7 halaman bercetak dan 1 halaman kosong) STPM 954/2 *This question paper is CONFIDENTIAL until the examination is over. *Kertas soalan ini SULIT sehingga peperiksaan kertas ini tamat

CONFIDENTIAL* SULIT*

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1.

Find the angle between vectors a and b if a = 5, b = 3 and a − b = 7. [4 marks]

2.

Express 3 sin x + 4 cos x in the form r sin(x + α ) where r is positive and α is an

acute angle, giving the value of α to the nearest 0.10.

[4 marks]

Hence, find the minimum and maximum values of 6 sin x + 8 cos x + 5. [2 marks]

3. A ship P moves due east towards a target with a speed of 30 km h-1. A ship Q moves with a speed of 60 km h-1 along the course 2100. a) Find the magnitude and direction of the velocity of Q relative to P.

[5 marks]

b) If initially ship P is at 20 km to the west of ship Q, find the shortest distance between ship P and ship Q. [2 marks] 4.

Write an expression for 2sin2 x in terms of cos 2x.

[1 marks]

Show that sin 3x = 3 sin x - 4 sin3 x.

[2 marks]

Based on the two expressions above, express 8 sin5 x in the form of a sin x + b sin 3x + c sin 5x, where a, b and c are constants which must be evaluated. [4 marks] 5.

C

B

D

A In the diagram above, BC and BA are tangents to the circle. a) Prove that i) BA = BC. ii) ∠ BAC = ∠ ADC.

[4 marks] [4 marks]

b) If AC = AD, prove that triangle ACD and triangle BCA are similar triangles. [4 marks]

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1.

Cari sudut di antara vektor a dan b jika a = 5, b = 3 dan a − b = 7. [4 markah]

2.

Ungkapkan 3 sin x + 4 cos x dalam bentuk r sin(x + α ) dengan r adalah positif

dan α

adalah su dut t irus, den gan m emberikan nilai α sehingga 0.10 yang paling

hampir

[4 markah] Dengan

demikian,

carikan

nilai

minimum

dan

nilai

maksimum

6 sin x + 8 cos x + 5.

bagi

[2 markah]

3. Sebuah kapal P bergerak tepat ke timur menuju ke satu sasaran dengan laju 30 km j-1. Sebuah kapal Q bergerak dengan laju 60 km j-1 mengikut haluan 2100. a) Carikan magnitud dan arah halaju kapal Q relatif kepada kapal P. [5 markah] b) Jika pada mulanya kapal P berada 20 km tepat ke barat kapal Q, carikan jarak terdekat di antara kapal P dan kapal Q. [2 markah] 4.

Tuliskan satu ungkapan bagi 2sin2 x dalam sebutan cos 2x.

[1markah]

Tunjukkan bahawa sin 3x = 3 sin x - 4 sin3 x.

[2 markah]

Berdasarkan dua un

gkapan di at as, un gkapkan 8 si n5 x dalam bentuk

a sin x + b sin 3x + c sin 5x, di mana a, b dan c adalah pemalar yang perlu ditentukan nilainya.

[4 markah]

5.

C

B

D

A Dalam rajah di atas, BC dan BA adalah tangen kepada bulatan. a) Buktikan, bahawa i) BA = BC. ii) ∠ BAC = ∠ ADC.

[4 markah] [4 markah]

b) Jika AC = AD, buktikan bahawa segitiga ACD dan segitiga BCA adalah segitiga serupa. [4 markah]

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6.

a) Find the solution of the differential equation

t = 0.

dp 3 + p = 12, for which p = 0 when dt 2 [5 marks]

Sketch a graph to show the relationship between p and t if t ≥ 0.

[2 marks]

b) The acceleration of a moving particle is given by g – gk2v2, where g and k are positive constants and v is the speed of the particle. Write dow n a di fferential eq uation r elating v and t ime t. H ence, express v explicitly in terms of t if the particle moves from rest. [7 marks] 7. The r andom variable X has a P oisson di stribution w ith m ean λ. Given t hat 2P( X = 1) = P( X = 2 ).

8.

a) Find the value of λ.

[2 marks]

b) Find P( X > 3 ).

[3 marks]

Given t hat X and Y are t wo e vents with t he following pr obabilities P(X) =

P(X|Y’) =

2 , 7

3 1 and P(X ∩ Y) = . 7 21

a) Find P(X U Y).

[4 marks]

b) Determine whether X and Y are independent events.

[2 marks]

5

The probability distribution function of random variable X is P(X = x) = m   for  x x = 0, 1, 2, 3, 4, 5 where m is a constant. 9.

a) Determine the value of m.

[2 marks]

b) Tabulate P(X = x).

[2 marks]

c) Find E(5X – 3)

[3 marks]

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6.

a) Cari peny elesaian ba gi per samaan pem bezaan

p = 0 bila t = 0.

dp 3 + p = 12 , diberi bahawa dt 2 [5 markah]

Lakarkan graf untuk menunjukkan hubungan antara p dan t jika t ≥ 0. [2 markah] b) Pecutan sebutir zarah yang bergerak diberikan oleh g – gk2v2, di mana g dan k adalah pemalar positif dan v adalah laju zarah itu. Tuliskan sa tu per samaan pem bezaan y ang menghubungkan v dengan m asa,t. Dengan dem ikian, Ungkapan v secara explisit dalam se butan t jika z arah i tu ber mula dari keadaan rehat. . [7 markah] 7. Pembolehubah r awak X mempunyai t aburan Poisson dengan m in λ. Diberi bahawa 2P( X = 1) = P( X = 2 ).

8.

a) Carikan nilai λ.

[2 markah]

b) Cari P( X > 3 ).

[3 markah]

Diberi X dan Y adalah dua peristiwa dengan keadaan P(X) =

P(X ∩ Y) =

9.

3 2 , P(X/Y’) = dan 7 7

1 . 21

a) Cari P(X U Y).

[4 markah]

b) Tentukan sama ada peristiwa X dan Y adalah merdeka.

[2 markah]

Fungsi t aburan ke barangkalian pembolehubah r awak X diberikan sebagai

5  x

P(X = x) = m   dengan x = 0, 1, 2, 3, 4, 5 di mana m adalah pemalar. a) Tentukan nilai m.

[2 markah]

b) Jadualkan nilai P(X = x).

[2 markah]

c) Carikan E(5X – 3)

[3 markah]

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10. The continuous random variable X has probability density function, f(x) as shown in the diagram below. f(x) ⅔

1

a) Find f(x).

2

x

3

[3 marks]

b) Find the cumulative distribution function of X and sketch its graph.

[4 marks]

c) Find P(0.9 < X ≤ 2.1).

[3 marks]

11. Mineral water is sold in bottles of two sizes, standard and large. For each size, the content, in litres, of a randomly chosen bottle is normally distributed with mean and standard deviation as given in the table below: Size of the bottle Standard bottle Large bottle

Mean 0.760 1.010

Standard deviation 0.008 0.009

a) Find the probability that a randomly chosen standard bottle contains less than 0.750 litres of mineral water. [2 marks] b) Find the probability that a box of 10 randomly chosen standard bottles contains at least three bottles whose contents are each less than 0.750 litres. [3 marks] c) Find the probability that there is more mineral water in four randomly chosen standard bottles than in three randomly chosen large bottles. [5 marks] 12. The following table shows the range of marks for chemistry test obtained by 120 students. Marks Number of students 21 – 30 31 – 40 41 – 50 51 – 60 61 – 70 71 – 80 81 – 90 91 – 100

2 5 11 23 47 20 9 3

a) Calculate the mean and standard deviation for the marks obtained by the students, giving your answer correct to one decimal place. [5 marks] b) Plot the cumulative frequency graph, hence, estimate the i) median.

[4 marks]

ii) percentage of students whose marks are within the range of one standard deviation from mean. [3 marks]

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10. Pembolehubah rawak selanjar X mempunyai fungsi ketumpatan kebarangkalian, f(x) seperti ditunjukkan dalam gambarajah di bawah. f(x) ⅔

a) Carikan f(x).

1

2

3

x [3 markah]

b) Carikan fungsi taburan longgokan bagi X dan lakarkan grafnya.

[4 markah]

c) Carikan P(0.9 < X ≤ 2.1).

[3 markah]

11. Air mineral dijual dalam dua botol yang bersaiz, sedang dan besar. Kandungan dalam liter setiap botol yang dipilih secara rawak itu mempunyai taburan normal dengan min dan sisihan piawai seperti ditunjukkan dalam jadual di bawah. Saiz Botol Sedang Besar

Min 0.760 1.010

Sisihan Piawai 0.008 0.009

a) Carikan kebarangkalian bahawa sebuah botol sedang yang dipilih secara rawak mempunyai kandungan kurang daripada 0.750 liter air mineral. [2 markah] b) Carikan kebarangkalian bahawa dalam sebuah kotak yang mengandungi 10 botol sedang air mineral yang dipilih secara rawak, mempunyai sekurang-kurangnya tiga botol yang mempunyai kandungan kurang daripada 0.750 liter. [3 markah] c) Carikan kebarangkalian bahawa jumlah kanduangan air mineral dalam empat botol sedang yang dipilih secara rawak adalah melebihi jumlah kandungan air mineral dalam tiga botol besar yang dipilih secara rawak. [5 markah] 12. Jadual di bawah menunjukkan julat markah satu ujian kimia yang diperolehi oleh 120 orang pelajar. Markah Bilangan pelajar 21 – 30 2 31 – 40 5 41 – 50 11 51 – 60 23 61 – 70 47 71 – 80 20 81 – 90 9 91 – 100 3 a) Hitungkan min dan sisihan piawai bagi markah yang diperolehi oleh pelajarpelajar itu, dengan memberikan jawapan anda betul sehingga satu tempat perpuluhan. [5 markah] b) Plotkan graf kekerapan longgokan. Dengan demikian anggarkan i) median. [4 markah] ii) peratusan pelajar yang memperolehi markah dalam lingkungan satu sisihan piawai dari min. [3 markah]

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JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN 954/1 PERCUBAAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN STPM JOHOR 2009 JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN MATHEMATICS T(MATEMATIK T) JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PAPER 1(KERTAS 1)JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

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No. 1 [4]

2 [5]

Marks x+y i + 3 − 4 i = 3 i 1 1

Using his in the modulus Equation correct.

x2 + 6x + 9 + y2 −8y +16 = 9

1

For squaring

x2 + y2 + 6x −8y +16 = 0

1

CA

( x + 3) 2 + ( y − 4) 2 = 3

f (x) = x 3 − 2

f ( x + δx ) = ( x + δx ) 3 − 2

f ( x + δx ) = ( x + δx ) 3 − 2 = x + 3 x δx + 3 x (δx ) + (δx ) − 2 3

2

2

( x + δx ) 3 − x 3 δx →0 δx

f ' ( x ) = lim

3 x 2 δx + 3 x (δx ) + (δx ) δx →0 δx 2 = 3x 2

1

correct formula substitute with limit

1

Expand

1

CA CA

3

= lim

f ' (a ) = 3a 2 3 [6]

1

3

1

For log 3 ( x 2 − 3 x + 2) and 2 log 3 (2 x − 1) to be defined, x 2 − 3 x + 2 > 0 and 2 x − 1 > 0 ⇒

x < 1 or x > 2 , and

⇒

1 < x < 1 or 2

x>

1

both conditions (with “and”)

1 2

x>2

log 3 ( x 2 − 3 x + 2) < log 3 2(2 x − 1) 2

1

1

Correct way of using laws

x 2 − 3 x + 2 < 2(2 x − 1) 2

1 x<0

∴

4 [5]

x>

or

5 7

1

5   the solution set is  x : < x < 1 or x > 2 7  

. − 8 + 16 + 2a + b = 0 a3 + 4 a2 − a2 + b = a3 2a − 3a2 + 8 = 0

1 1 1

CA CA

1

Solving

(3a + 4 ) ( a − 2 ) = 0 a=− a=2,

4 3

, b=−

b = − 12

16 3

1 1

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5 [6]

S n-1 =

a 3− n (b n −1 − a n −1 ) b−a

1

a 2− n (b n − a n ) a 3− n (b n −1 − a n −1 ) − b−a b−a 2− n n 2 3− n n −1 2 a b −a −a b +a = b−a 2 − n n −1 a b (b − a ) = b−a = a2− n bn−1.

T n = S n − S n-1 =

Tn a 2− n b n −1 = 3− n n − 2 Tn −1 a b b = (independent of n) a ⇒ the series is a geometric series. 6 [9]

1

Use his S n-1 In the formula

1

factorisation

1

1

use his to find r (do division)

1

with correct reason

f ( x) = x 3 − 10 x 2 + ax + b  3 + 2 is a zero, ∴ 3 − 2 ia also a zero.

[x − (3 + 2) ][x − (3 − 2) ] = x (

2

)

Equating coefficient of x :

b ) 7

b − 6 = −10 ⇒ b = −28 7 6b 7− = a ⇒ a = 31 7

a = 31, b = −28

(a)

Conjugate must be correct

1

Use factorisation or long division

1

Method to find a or b

1

both a and b correct

1

factorisation

1

* followed by **

1

Conclusion

− 6x + 7

x 3 − 10 x 2 + ax + b = x 2 − 6 x + 7 ( x + Equating coefficient of x 2 :

1

f ( x) = 12 − 3 x x 3 − 10 x 2 + 31x − 28 = 12 − 3 x ( x − 4)( x − 6 x + 10 ) = 0 2

x = 4 or x 2 − 6 x + 10 = 0 For x 2 − 6 x + 10 = 0 * Discrimina nt = ( −6) − 4(1)(10 ) = −4 < 0 * * 2

Hence, the only real root is 4.

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(b)

(

)

x 3 − 10 x 2 + 31x − 28 = x 2 − 4 Q( x ) + Ax + B When x = 2, 2 = 2 A + B........(1) When x = −2, − 138 = −2 A + B.....( 2) (1) + (2), 2B = −136 ⇒ B = −68

1

Method

1

CA

(1) − (2), 4 A = 140 ⇒ A = 35 ∴ the remainder is 35 x − 68. 7 [9]

1

1

7(a)

1 − 3 x = (1 − 3 x ) 2

1 1 1 1 3 ( )( − ) ( )( − )( − ) 1 2 ( −3 x ) 2 + 2 2 2 ( −3 x ) 2 + ... = 1 + ( −3 x ) + 2 2 2! 3! 3 9 2 9 3 = 1− x + x − x + ... 2 8 16 1 1  Range :  x : − < x <  3 3  (b)

1   x − 2  3x  

1, 1

2 correct (1) 3 correct(1+1)

1

1

24

 24  1   U r +1 =   x 24 −r  −  2  3x   r 

r

1

r

 24  r  1 =  (− 1)   x 24 −3 r r 3  

24 − 3r = 0 r =8

1

The term in x must be correct

1

His = 0

8

8 [10]

 24  8 1 The term independen t of x =  (− 1)   x 0 3 8 81719 = 729 x−3 x−3 = y= 2 (2 − x)( x + 1) − x + x + 2 (a)

Asymptotes : x = 2, x = −1, y = 0.

(

)

dy − x 2 + x + 2 − (x − 3)(−2 x + 1) = 2 dx − x2 + x + 2 =

(

x − 6x + 5 2

(− x

2

+x+2

)

2

)

1

2

2 correct = 1 3 correc = 2

1

formula

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dy =0 dx x 2 − 6x + 5

(− x

2

+x+2

)

2

=0 1

His = 0

1

Correct answer in coordinate form

1

method

1

nature of both the turning points correct

1

all parts correct

1

points

1

asymptotes and perfect

1

quadratic eq.

1 1

b2-4ac=0 or other method Conclusion

1

Equation of OR

1

substitution

x − 6x + 5 = 0 ( x − 1)( x − 5) = 0 2

x = 1, 5.

1 Turning points are (1, − 1) and (5, − ). 9

(b)

d 2 y (2 x − 6)( − x 2 + x + 2)2 − ( x 2 − 6 x + 5)(2)( − x 2 + x − 2)( −2 x + 1) = dx 2 ( − x 2 + x + 2)4 =

(2 x − 6)( − x 2 + x + 2) − ( x 2 − 6 x + 5)(2)( −2 x + 1) ( − x 2 + x + 2)3

d 2y < 0 ⇒ local maximum point dx 2 1 d 2y  Point  5, − , > 0 ⇒ local minimum point 9 dx 2 

Point (1, -1),

(c) y

-1 0

2

3

(1, - 1) -3/2

9 [10]

5

x

1 (5, − ) 9

x(8m − m2 x ) = 16 m2 x2 – 8mx + 16 = 0 b2 − 4ac = 64m2 − 4 (m2)(16) =0 2 ∴ m x + y − 8m = 0 is a tangent to xy = 16. Eqn. of OR is y =

1 m2

m x + y − 8m = 0 2

(1) → (2) m2 x +

………….(2) 1 m2 3

x=

x …………(1)

8m 1+ m4

x − 8m = 0

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8m 1+ m4 8m 8m 3 R( , ) 4 1+ m4 1+ m m2 = x

1

x or y correct

1 1

Coordinates R m or m2 in terms of x and y.

1

elimination of m

y=

y

(m2 x + y )2 = 64m2 ( x × x + y )2 = 64 ( x ) y

y

2

2 2

(x + y ) = 64xy 10 [11]

A  = k

1

−3 1 2 1 2 −3 −1 −1 1 2 2 2 2 1

= k (−6−1)−(4−2)−(2+6) = −7k −2 −8 = −7k −10 −7k −10 = −17 7k = 7 k=1 1 2 3

− 17 0

  BC =  2 3 4   − 66 − 38 3   1 5 6   51 34 0  0



 21 9 6  =  6 − 12 9     − 24 − 3 15 

1

1

*1

(

=

1

2   − 24 − 3 15 

 51 0 0   0 51 0     0 0 51

ABC = 51 I  21  6   − 24  7 / 17 =  2 / 17  − 8 / 17 

A−1 =

1 51

9 6 − 12 9  − 3 15  3 / 17 2 / 17  − 4 / 17 3 / 17  − 1 / 17 5 / 17 

 1 1 − 1  x   − 1  2 − 3 1    =      y   − 3   2 1 2   z   25   − 1 x A  y  =  − 3      25  z

)(

) is seen

1

CA (if 1 is not obtained here, multiplication process must be seen to obtain *1)

1

A( *)



 1 1 − 1  21 9 6 ABC =  2 − 3 1   6 − 12 9      2

CA

1

1

His BC

1

CA

1

CA

* his BC

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x 9 6   − 1  21  y = 1    − 3 51 6 − 12 9       z  − 24 − 3 15   25  1 51

=

1

His inverse

1

CA

1

CA

1

Method to find A, B &C

1

CA

 102   2   255  = 5     408    8 

x = 2. y = 5, z = 8 11 [11]

5x + 4 x + 12 Bx + C A ≡ + 2 2 x+2 ( x + 2)( x + 4) x +4 2 2 5x + 4x + 12 = A(x +4) + (Bx+C)(x+2) x= −2, 20 − 8+12 = 8A A=3 2 coef. of x , 5 = A + B B=2 x = 0, 12 = 4A + 2C C=0 2 5x + 4 x + 12 2x 3 ∴ ≡ + 2 2 ( x + 2)( x + 4) x + 2 x +4 2

a)

∫

5 x 2 + 4 x +12 ( x + 2 )( x 2 + 4 )

3 1

∫

dx

3

2x 3 + 2 dx x+2 x +4 = [3 ln(x+2) + ln (x2 + 4 ) ] 3 =

1

1

= 3ln (5) + ln(13) − 3ln(3) − ln(5) 325 = ln 27 2 −1 b) h = = 0.25 4 x 1 1.25 1.5 1.75 2

∫

2 1

y 0.6931 0.9410 1.1787 1.4018 1.6094

ln( 1 + x 2 ) dx = 0.25 [ 0.6931+1.6094 + 2

2(0.9410+1.1787+1.4018)] = 1.168

1, 1

First 1, ln is seen 2nd 1, all correct

1

his substitution

1

CA

1

CA

1

All correct

1

Correct formula with his value

1

CA

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12 [14]

(a)

lim f ( x) = lim− (5 − x − 1 ) = 3

1

CA

lim+ f ( x) = lim+ ( x 2 − 6 x + 8) = 3

1

CA

1

With reason **

1, 1

Parts of y = 5 − x +1

1

parabola

1

points,  is marked at end-points

x→1−

x→1

x →1

x →1

f (1) = 1 − 6 + 8 = 3 lim− f ( x ) = lim+ f ( x ) = f (1). * * x →1

x →1

⇒ f is continuous at x = 1.

(b)

y 5 4 (1, 3) −6

(c)

(d)

-1 0 1 2 3 4

x

y = 9 − 18 + 8 = −1

1

Range of {y : −1 ≤ y ≤ 5}

1

5 − ( − x − 1) = 2 − x x − 2 = x 2 − 6x + 8

1 1

x = −2, 2, 3,

{x : −2 < x < 2} ∪ {x : 3 < x < 4} If other method is used, mark accordingly.

1

1,1

method to find minimum point

CA

CA CA

All correct Correct line is see( if no line is seen, no mark) Ignore shading If 1 set is correct, I mark only.

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JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN 954/2 PERCUBAAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN STPM JOHOR 2009 JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN MATHEMATICS T(MATEMATIK T) JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PAPER 2(KERTAS 2)JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

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Q 1

Steps

Marks

Notes

1

Diagram with arrow in correct direction

1

using cosine rule

1

CA

1

conclusion

a− b ~

b

~

θ

~

a ~

52 + 32 − 72 2(5 )(3) 0 θ = 120

cosθ =

Thus, angle between a and b is 1200 ~

1

~

4

Alternative method:  x m a =   , b =   ~ ~  y n 2 2 x + y = 25 and m2 + n2 = 9  x − m  a − b =  ~ ~  y −n (x – m)2 + (y – m)2 = 49 x2 + y2 +m2 +n2 – 2mx – 2ny = 49 15 mx + ny = − 2 a . b =| a || b | cos θ ~ ~

~

1

Either one correct

1

Dot product

1 1

CA

~

 x   m  .  = 5 x 3cosθ  y  n  xm + yn = 5x3cosθ 15 = 5x3cosθ − 2 1 cosθ = − 2 0 θ = 120 Hence, the angle between a and b is 1200 ~

~

conclusion

4

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2

3sinx + 4cos x = r sinx cosα + rcos x sin α rcos α = 3 , rsin α = 4 r = 5 or 32 + 4 2 seen 4 tanα = 3

both rcos α = 3 rsin α = 4 seen, if not, -1

1 1

α = 53.10 Hence, 3sinx + 4cos x ≡ 5 sin(x + 53.10)

1

6 sin x + 8 cos x + 5 = 10 sin(x + 53.10 ) + 5

1

for 10 sin(x + 53.10 ) + 5

-5 ≤ 10 sin(x + 53.10 ) + 5 ≤ 15 Max. value = 15 Min. value = -5

1

For his logic inequality

1

Both values correct

6

Ν

3(a)

θ 300

v P Q

Diagram with correct arrows

1 Ν

vQ

1200 vP QvP=

302 + 602 – 2(30)(60)cos 1200

1

Cosine rule based on his diagram

1

CA

1

His sine rule

Direction of Q v P is S 49.10 W

1

CA

Shortest distance = 20 sin 40.90 = 13.09 km

1 1

Using his angle CA

QvP

= 30 7 or 79.37 km

sinθ sin 120 0 = 30 79.373 θ = 19.110 / 19.10

(b)

7

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4.

2 sin2 x = 1 – cos 2x

1

CA

sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x = 2 sin x cos2x + ( 1 – 2sin2x)sin x = 2 sin x(1 – sin2 x) + sin x – 2 sin3 x

1

Identity(either one)

1

CA

1

Using above ans.

1

Either one of the factor formula used

1

CA

1

conclusion

= 3 sin x – 4 sin3 x 8 sin 5 x = (2sin2 x)(4sin3 x) = (1 – cos 2x)(3 sin x – sin 3x) = 3 sin x – sin 3x – 3cos 2x sin x + sin 3xcos 2x = 3 sin x –sin 3x – 3[ ½ sin 3x – ½ sin x] + ½ sin 5x + ½ sin x 5 = 5 sin x - sin 3x + ½ sin 5x 2 Hence, a = 5, b = -

5 1 ,c= 2 2

7

5(a) C

Β

O D Α

∠BCO = ∠BAO = 900(radius perpendicular to tangent) BO = BO (common line) OC = OA (radii of circle) ∴∆BAO ≡ ∆BCO (RHS)

1

∴BC = AB

1

1 1

Must provide at least 1 correct reason, if not -1

Must have above 3 statements even without RHS ∆BAO ≡ ∆BCO seen

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(b)

C

E x0

900-x0

0

D

Mark accordingly to his labelling

B

O x

x0 A

Let AOE be the diameter of the circle, ∠CAE = 900 – x0(tangent ⊥ to radius) ∠ECA = 900(angle in semicircle) ∠CEA = 1800 – (900- x0) - 900 = x0 ∠CDA = ∠CEA = x0(angle in same segment) ∴∠BAC = ∠ADC

Must provide at least 1 correct reason, if not -1

1 1

1 1

All the above 3 statements correct

C

(c)

Β D Α

∠BAC = ∠ADC = x0(from above) ∠ACD = x0 (base angle of isosceles ∆) ∠BCA= x0(base angle of isosceles ∆/alternate segment) ∴∠ACD = ∠BCA ∠DAC = 900 – 2x0 = ∠ABC Hence, ∆ ACD ≅ ∆ADC

either one

Must provide at least 1 correct reason, if not -1

1 1

1 1 12

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6a dp 3 1 = 12 − p = (24 − 3 p ) dt 2 2 dp 1 ∫ 24 − 3 p = ∫ 2 dt 1 1 − ln(24 − 3 p ) = t + c 3 2 1 1 − ln[24 − 3(0)] = (0) + c 3 2 1 c = − ln 24 3 1 1 1 − ln(24 − 3 p ) = t − ln 24 3 2 3  24  3  = t ln  24 − 3 p  2 − t 24 − 3 p =e 2 24 3

p = 8 – 8e

3 − t 2

= 8(1- e

3 − t 2

)

1

Any correct separation

1

Integrate

1

Correct subst. for finding c

1

Get rid of ln

1

CA

1

Shape of curve & 8 seen.

1

All correct

p 8

0

t

7

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6(b

dv = g − gk 2 v 2 dt dv ∫ 1 − k 2 v 2 = ∫ gdt 1 A B = + By writing (1 − kv )(1 + kv ) 1 − kv 1 + kv 1 = A(1 + kv) + B(1 – kv) 1 1 A= ,B= 2 2 ⇒∫

1 1 + dv = ∫ gdt 2(1 − kv ) 2(1 + kv ) 1  1 + kv  ln  = gt + c 2k  1 − kv  1 ln(1) = g (0) + c 2k  1 + kv  ln  = 2kgt  1 − kv   1 + kv  2 kgt  =e  1 − kv  1 + kv = (1 − kv )e 2 kgt

e 2 kgt − 1 v= k + ke 2 kgt

1 Any correct separation

1

1

Correct A & B

1

His correct integration base on his A&B Correct subst.

1

Correct subs. for finding c

1

Get rid of ln

1

CA

7

7(a)

(b)

2e − λ λ1 e − λ λ2 = 1! 2! −λ −λ 2 4e λ - e λ = 0 e − λ (4 - λ) = 0 e − λ ≠ 0, λ = 4

1

Forming correct equation

1

CA

P(X > 3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]  42 43   = 1 - e − 4  1 + 4 + + 2! 3!  

1

Complement

1

At least 3 correct terms with his λ.

= 0.5666

1

CA

5

papercollection

8

1 2 , P(X ∩Y) = 21 7 3 P(X|Y’) = 7 P( X ∩ Y ' ) 3 = P(Y ' ) 7 3 P(X ) –P(X ∩ Y) = [1 – P(Y)] 7 2 1 3 − = [1 − P(Y )] 7 21 7 4 P(Y) = 9 P(X) =

(a)

2 4 1 + 7 9 21 43 = 63

P(X ∪ Y) =

(b)

P(X).P(Y) =

8 2 4 x = 7 9 63

1

Expansion of P(X|Y’) correctly.

1

Seen/applied in the equation

1

Correct formula with his value

1

CA

1

Shown 2 7

x

4 or his 9

value

P(X).P(Y) ≠ P(X ∩ Y) ∴X and Y are not independent events. 9(a)

(b)

(c)

1 6

5 5 5 5 5 5 m   + m   + m   + m   + m   + m   = 1 0  1 2 3 4 5 m + 5m + 10m + 5m + m = 1 1 m= 32 x P(X = x)

0 1 32

1 5 32

2 10 32

3 10 32

4 5 32

19 or 2

9.5

63

1

At least one of the equation seen

1

CA

5

1

1 32

At least 4 prob. correct based on his m.

1

All correct with table

1 1

Finding E(X) based on his m For correct usage of formula 5E(X) - 3

1

CA

E(5X – 3) = 1 5 10 5 10 1 )+1( )+2( )+3( )+4( )+5( )] - 3 5[0( 32 32 32 32 32 32 =

Both statements correct and 8 seen

7

papercollection

10 (a)

(b)

 31 x  f ( x ) = − 23 x + 2  0 

,0 ≤ x ≤ 2

1

For 3

,2 ≤ x ≤ 3 , otherwise

1 1

For

1

At least one quadratic function correct

1

All correct

1

Curves correct shape

1

All correct

1

App. of formula/integration

1

Correct subst.

1

CA

,x <0  0  x2 ,0 ≤ x < 2  6 F ( x) =  - 1 ( x 2 − 6 x + 6), 2≤x <3  3  1 ,x ≥3

1

x

seen

− 23 x + 2

seen

All correct

F(x) 1 2

3

0

(c)

x 1

2

3

P(0.9 < X ≤ 2.1) = F(2.1) – F(0.9) ( 0 .9 ) 2 1 = − [(2.1)2 – 6(2.1) + 6] 6 3 = 0.73 – 0.135 = 0.595

10

11 (a)

(b)

0.750 − 0.760   P(S < 0.750) = P  z <  0.008   = P(Z < -1.25) = 0.1057

1

Standardization

1

CA

X~B(10, 0.1057)

1

Binomial distrb/implied

P(X ≥ 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1- [90.8943]10 + (10(0.1057)(0.8943)9 + 45(0.1057)2(0.8943)8]

1

Correct subst of his binomial

1

CA

= 0.0803

papercollection

(c)

F = S1 + S2 + S3 + S4 Let T = F – G

G = L1 + L2 + L3 Correct linear combination(4S-3L is accepted if his var(T) is correct)

1

E(T) = 4(0.760) – 3(1.010) = 0.01

1

Var(T) = 4(0.0082) +3(0.0092) = 0.000499

1

P(T > 0)  0 − 0.01   = P  Z > 0.000499  

All correct

All correct

1

= 0.6728

Standardization based on his values

1

CA

10

12 (a)

−

∑ x ÷120

1

His

1

CA

1 1

510020

1

CA

1

Boundaries

1

Axes & >6 pts correct

1

All correct with smooth curve

(i) From the graph, median 64.5 ± 0.5

1

Shown in graph by dotted line

(ii) Range for one deviation from mean for marks = [50.11, 77.39] Range for one deviation from mean for number of students = [14, 107]

1

Answer in this range

7650 120 = 63.75

x=

510020  7650  σ = −  120  120  = 13.64

2

Correct formula

(b) 120 110

105

Number of students

100 90 80 70 60

median = 64.5 ± 0.5

50 40 30 20

16.3

10 0 20.5

30.5

40.5

50.5

60.5

70.5

80.5

90.5

100.5

Marks

papercollection

105 − 16 × 100 120 = 74.2%

% students in this range =

1

Correct formula based on his value

1

Accept 74% - 78%

12