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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 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 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 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

MATHEMATICS T(MATEMATIK T)

PAPER 1(KERTAS 1)

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.

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. CONFIDENTIAL* *Kertas soalan ini SULIT sehingga peperiksaan kertas ini tamat SULIT*

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.

PERCUBAAN STPM 2009

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1. If yixz += and iiz 343 =−+ , show that .0168622 =+−++ yxyx

[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

).12(log24log21)23(log 33

23 −+<+− xxx [6 marks]

4. Given that baxxx +−+ 23 4 is exactly divisible by )2( +x but leaves a

remainder 3a when divided by )( ax − , calculate the values of a and b . [5 marks]

5. The sum of first n terms of a series is ab

aba nnn

−−− )(2

, ab ≠ .

Find the thn term of the series. [4 marks]

Hence, show that the series is a geometric series. [2 marks]

6. Given that baxxxxf ++−= 23 10)( , where ba and are constants. If one of

the zeroes is 23 + , find the values of . and ba [4 marks]

Hence, (a) show that the equation xxf 312)( −= does not have any real roots other than 4. [3 marks] (b) find the remainder when )(xf is divided by .42 −x [2 marks]

7. (a) Expand x31− in ascending powers of x , up to and including the term

in 3x . State the range of values of x for which the expansion is valid. [5 marks]

(b) Find the term that is independent of x in the expansion24

231

−

xx

.

[4 marks]

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1. Jika yixz += dan iiz 343 =−+ , tunjukkan bahawa

.0168622 =+−++ yxyx [4 markah]

2. Dengan menggunakan kaedah prinsip pertama, tunjukkan bahawa nilai terbitan

pertama bagi f(x) = x3- 2 ,x∈IR pada x = a ialah 3a2. [5 markah]

3. Cari set nilai x yang memuaskan ketaksamaan berikut

).12(log24log

21)23(log 33

23 −+<+− xxx [6 markah]

4. Diberi bahawa baxxx +−+ 23 4 boleh dibahagi tepat oleh )2( +x tetapi

meninggalkan baki 3a apabila dibahagi oleh )( ax − , carikan nilai a dan b . [5 markah]

5. Hasil tambah bagin sebutan yang pertama suatu siri ialah ab

aba nnn

−−− )(2

, ab ≠ .

Cari sebutan ke- n bagi siri tersebut. [4 markah]

Seterusnya tunjukkan bahawa ia adalah suatu siri geometri. [2 markah] 6. Diberi bahawa baxxxxf ++−= 23 10)( , di mana ba dan adalah pemalar .

Jika salah satu pensifar ialah 23 + , cari nilai b dan a . [4 markah]

Seterusnya, (a) tunjukkan bahawa persamaan xxf 312)( −= tidak mempunyai sebarang punca nyata selain 4. [3 markah] (b) cari baki apabila )(xf dibahagi oleh .42 −x [2 markah]

7. (a) Kembangkan x31− dalam kuasa x menaik, sehingga dan termasuk

sebutan dalam 3x . Nyatakan julat nilai x supaya kembangan ini sah. [5 markah]

(b) Cari sebutan yang bebas daripada x dalam kembangan 24

231

−

xx .

[4 markah]

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8. A curve has the equation ( )( )123+−

−=

xxxy .

(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]

9. Show that the equation 082 =−+ myxm is a tangent to the curve 16=xy for all values of .0 , ≠mm

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 ( ) .64222 xyyx =+ [10 marks] 10. Matrices A, B and C are given as

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

11. (a) Express ( )( )421245

2

2

++++

xxxx

in partial fractions.

Hence, evaluate ( )( ) .42

12453

12

2

∫ ++++ dx

xxxx

[7 marks]

(b) Using trapezium rule, with 5 ordinates, evaluate ( )∫ +2

1

21ln dxx ,giving

your answer correct to three decimal places. [4 marks]

A =

−

−

21213211k

, B =

651432321

, C =

−−−

03451338660170

.

Show that 1=k if the determinant of A is -17. [2 marks] Find BC and ABC. Deduce A-1. [6 marks]

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8. Suatu lengkung mempunyai persamaan ( )( )123+−

−=

xxxy .

(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]

9. Tunjukkan bahawa persamaan 082 =−+ myxm ialah tangen kepada lengkung

16=xy untuk semua nilai .0 , ≠mm 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 ( ) xyyx 64222 =+ apabila m berubah-ubah [10 markah] 10. Matrik A, B dan C diberkan oleh

A =

−

−

21213211k

, B =

651432321

, C =

−−−

03451338660170

.

Tunjukkan bahawa 1=k 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 [3 markah]

11. (a) Ungkapkan ( )( )421245

2

2

++++

xxxx

dalam pecahan separa.

Dengan yang demikian, tentukan nilai ( )( )∫ ++++3

12

2

421245 dx

xxxx

. [7 markah]

(b) Gunakan petua trapezium dengan 5 ordinat, untuk menentukan nilai

( )∫ +2

1

21ln dxx , berikan jawapan anda betul sehingga tiga tempat perpuluhan.

[4 markah]

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12. The function f is defined as

<≤<<−

+−+−

=.41

16,,

8615

)( 2 xx

xxx

xf

(a) Find )(lim1

xfx −→

and )(lim1

xfx +→

.

Hence, determine whether f is continuous at 1=x . [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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(c) Tentukan julat f . [2 markah] (d) Tentukan set nilai x supaya f(x) > 2 – x. [5 markah]

12. Fungsi f ditakrifkan oleh

<≤<<−

+−+−

=.41

16,,

8615

)( 2 xx

xxx

xf

(a) Cari )(lim

1xf

x −→ dan )(lim

1xf

x +→ .

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

(b) Lakarkan graf f . [4 markah]

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954/2


PAPER 2(KERTAS 2)

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.

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. CONFIDENTIAL* *Kertas soalan ini SULIT sehingga peperiksaan kertas ini tamat SULIT*

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.

PERCUBAAN STPM 2009

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1. Find the angle between vectors a and b if a = 5, b = 3 and ba − = 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. In the diagram above, BC and BA are tangents to the circle.

a) Prove that

i) BA = BC. [4 marks] ii) ∠BAC = ∠ADC. [4 marks]

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

D

A

B

C

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1. Cari sudut di antara vektor a dan b jika a = 5, b = 3 dan ba − = 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 bagi

6 sin x + 8 cos x + 5. [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. Dalam rajah di atas, BC dan BA adalah tangen kepada bulatan.

a) Buktikan, bahawa

i) BA = BC. [4 markah] ii) ∠BAC = ∠ADC. [4 markah]

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

D

A

B

C

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6. a) Find the solution of the differential equation p23

+dtdp = 12, for which p = 0 when

t = 0. [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 ).

a) Find the value of λ. [2 marks]

b) Find P( X > 3 ). [3 marks]

8. Given t hat X and Y are t wo e vents with t he following pr obabilities P(X) = 72

,

P(X|Y’) = 73

and P(X∩Y) = 211

.

a) Find P(X U Y). [4 marks] b) Determine whether X and Y are independent events. [2 marks]

9. The probability distribution function of random variable X is P(X = x) = m

x5

for

x = 0, 1, 2, 3, 4, 5 where m is a constant. 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 penyelesaian ba gi per samaan pem bezaan p23

+dtdp = 12 , diberi bahawa

p = 0 bila t = 0. [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 ).

a) Carikan nilai λ. [2 markah]

b) Cari P( X > 3 ). [3 markah]

8. Diberi X dan Y adalah dua peristiwa dengan keadaan P(X) = 72

, P(X/Y’) = 73

dan

P(X∩Y) = 211

.

a) Cari P(X U Y). [4 markah] b) Tentukan sama ada peristiwa X dan Y adalah merdeka. [2 markah] 9. Fungsi t aburan ke barangkalian pembolehubah r awak X diberikan sebagai

P(X = x) = m

x5

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.

a) Find f(x). [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 Mean Standard deviation Standard bottle 0.760 0.008 Large bottle 1.010 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 2 31 – 40 5 41 – 50 11 51 – 60 23 61 – 70 47 71 – 80 20 81 – 90 9

91 – 100 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]

⅔

2 3 1

f(x)

x

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

a) Carikan f(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 Min Sisihan Piawai Sedang 0.760 0.008 Besar 1.010 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 pelajar-

pelajar 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]

⅔

2 3 1

f(x)

x

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954/1


PAPER 1(KERTAS 1)

PERCUBAAN STPM 2009

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No. Marks 1 [4]

x+y i + 3 − 4 i = 3 i

22 )4y()3x( −++ = 3 x2 + 6x + 9 + y2

−8y +16 = 9

x2 + y2

+ 6x −8y +16 = 0

1 1 1 1

Using his in the modulus Equation correct. For squaring CA

2 [5]

2)( 3 −= xxf

( ) ( ) 233

2)()(3223

3

−+++=

−+=+

xxxxxx

xxxxf

δδδ

δδ

( ) ( )

2

322

0

33

0

3

33lim

)(lim )('

xx

xxxxxx

xxxxf

x

x

=

++=

−+=

→

→

δδδδ

δδ

δ

δ

23)(' aaf =

1

1 1 1

1

2)()( 3 −+=+ xxxxf δδ

correct formula substitute with limit Expand CA CA

3 [6]

2 or 12 1

21 and , 2 or 1

012 and 023

defined, be to )12(log2 and )23(log For2

32

3

><<⇒

>><⇒

>−>+−

−+−

xx

xxx

xxx

xxx

75 or 0

)12(223

)12(2log)23(log

22

23

23

><

−<+−

−<+−

xx

xxx

xxx

><<∴ 2 or 1

75 : is set solution the xxx

1 1 1 1 1 1

both conditions (with “and”) Correct way of using laws

4 [5]

. − 8 + 16 + 2a + b = 0

a

3 + 4 a2 − a2 + b = a

2a − 3a

3 2

(3a + 4 ) ( a − 2 ) = 0

+ 8 = 0

a = − 34 , b = − 3

16

a = 2 , b = − 12

1 1

1

1 1

CA CA Solving

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5 [6] Sn-1 ab

)ab(a 1n1nn3

−− −−−

=

Tn = Sn− Sn-1 ab)ab(a nnn2

−−−

= − ab

)ab(a 1n1nn3

−− −−−

= ab

abaaba 21nn32nn2

−+−− −−−

= ab

)ab(ba 1nn2

−−−−

= a2− n bn−1

.

1n

n

TT

−

= 2nn3

1nn2

baba

−−

−−

= ab (independent of n)

⇒ the series is a geometric series.

1 1 1 1 1 1

Use his SIn the formula

n-1

factorisation use his to find r (do division) with correct reason

6 [9]

baxxxxf ++−= 23 10)(

23 + is a zero, 23 −∴ ia also a zero. [ ][ ] 76)23()23( 2 +−=−−+− xxxx

( ) )7

(7610 223 bxxxbaxxx ++−=++−

Equating coefficient of 28 1067b :2 −=⇒−=− bx

Equating coefficient of 31 76b 7 : =⇒=− aax

28 ,31 −== ba

(a) xxf 312)( −=

**04)10)(1(4)6( ntDiscrimina

*0106 For

0106 or 4

0)106)(4(

312283110

2

2

2

2

23

<−=−−=

=+−

=+−=

=+−−

−=−+−

xx

xxx

xxx

xxxx

Hence, the only real root is 4.

1 1 1 1 1 1 1

Conjugate must be correct Use factorisation or long division Method to find a or b both a and b correct factorisation * followed by ** Conclusion

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(b)

( )

68.35 is remainder the 351404 (2),(1)68B 1362 ),2()1(

)2.....(2138 ,2 When)1........(22 ,2 When)(4283110 223

−∴=⇒=−−=⇒−=+

+−=−−=+==

++−=−+−

x A A

BBAx

BA xBAxxQxxxx

1 1

Method CA

7 [9] 7(a) ( )2

13131 xx −=−

...169

89

231

...)3(!3

)23)(

21)(

21(

)3(!2

)21)(

21(

)3(211

32

22

+−+−=

+−−−

+−−

+−+=

xxx

xxx

Range :

<<−

31

31: xx

(b) 24

231

−

xx

( ) rr

r

rr

r

xr

xx

rU

324

224

1

311

24

3124

−

−+

−

=

−

=

80324

==−

rr

( )

72981719

311

824

of tindependen term The 08

8

=

−

= xx

1

1, 1 1 1 1 1 1 1

2 correct (1) 3 correct(1+1) The term in x must be correct His = 0

8 [10] 2

3)1)(2(

32 ++−−

=+−

−=

xxx

xxxy

(a) Asymptotes : 0.y ,1 ,2 =−== xx

( ) ( )( )

( )22

2

22

2

2

56

2

)12(32

++−

+−=

++−

+−−−++−=

xx

xxxx

xxxxdxdy

2 1

2 correct = 1 3 correc = 2 formula

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( )

5. ,10)5)(1(

056

02

56

0

2

22

2

==−−

=+−

=++−

+−

=

xxx

xxxx

xxdxdy

(b) Turning points are ).91 ,5( and )1 ,1( −−

32

22

42

2222

2

2

)2()12)(2)(56()2)(62(

)2()12)(2)(2)(56()2)(62(

++−+−+−−++−−

=

++−+−−+−+−−++−−

=

xxxxxxxx

xxxxxxxxxx

dxyd

Point (1, -1), ⇒< 02

2

dxyd local maximum point

Point ,91 ,5

− ⇒> 02

2

dxyd local minimum point

(c)

1 1 1 1 1 1 1

His = 0 Correct answer in coordinate form method nature of both the turning points correct all parts correct points asymptotes and perfect

9 [10]

x(8m − m2

m x ) = 16

2 x2

b – 8mx + 16 = 0

2 − 4ac = 64m2 − 4 (m2

= 0 )(16)

∴ m2

Eqn. of OR is y =

x + y − 8m = 0 is a tangent to xy = 16.

2m1

x …………(1)

m2

(1) → (2) m

x + y − 8m = 0 ………….(2) 2

2m1

x + x − 8m = 0

x = 4

3

m1m8+

1 1 1 1 1

quadratic eq. b2

Conclusion

-4ac=0 or other method

Equation of OR substitution

y

x -1 0 2 3 5

-3/2 1)- ,1( )

91 ,5( −

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y = 4m1m8+

R ( 4

3

m1m8+

, 4m1m8+

)

m2

yx =

(m2 x + y )2 = 64m (

2

yx × x + y )2

yx = 64 ( )

(x2 + y2 )2

= 64xy

1 1 1 1 1

x or y correct Coordinates R m or m2

in terms of x and y.

elimination of m

10 [11] A = k 21

13− − 1 2212 −1 12

32 −

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

BC =

651432321

−−−

03451338660170

=

−−−

1532491266921

ABC =

−

−

212132111

−−−

1532491266921

=

510005100051

ABC = 51 I

A−1511 =

−−−

1532491266921

=

−−−

17/517/117/817/317/417/217/217/317/7

−

−

212132111

zyx

=

−−

2531

A

zyx

=

−−

2531

1 1 *1 1 1 1 1 1 1

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

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dx

zyx

= 511

−−−

1532491266921

−−

2531

= 511

408255102

=

852

x = 2. y = 5, z = 8

1 1

His inverse CA

11 [11] a)

)4x)(2x(12x4x5

2

2

++++ ≡

2xA+

+ 4xCBx

2 ++

5x2 + 4x + 12 = A(x2

x= −2, 20 − 8+12 = 8A +4) + (Bx+C)(x+2)

A = 3 coef. of x2

B = 2 , 5 = A + B

x = 0, 12 = 4A + 2C C = 0

∴ )4x)(2x(

12x4x52

2

++++

≡ 2x

3+

+ 4x

x22 +

∫3

1 )4x)(2x(12x4x5

2

2

++++

= ∫3

1 2x3+

+ 4x

x22 +

dx

= [3 ln(x+2) + ln (x2

13 + 4 ) ]

= 3ln (5) + ln(13) − 3ln(3) − ln(5)

= ln 27325

b) h = 4

12 − = 0.25

∫ +2

1

2 dx)x1ln( = 225.0 [ 0.6931+1.6094 +

2(0.9410+1.1787+1.4018)] = 1.168

x y 1 0.6931 1.25 0.9410 1.5 1.1787 1.75 1.4018 2 1.6094

1 1 1 1, 1 1 1 1 1 1 1

CA Method to find A, B & C CA First 1, ln is seen 2nd

1, all correct

his substitution CA CA All correct Correct formula with his value CA

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12 [14]

(a) 3)15(lim)(lim11

=−−=−− →→

xxfxx

3)86(lim)(lim 2

11=+−=

++ →→xxxf

xx

.1 at continuous is

**).1()(lim)(lim3861)1(

11

=⇒

===+−=

+− →→

xf

fxfxff

xx

(b)

(c) 18189 −=+−=y Range of { }51: ≤≤− yy (d)

,3 ,2 ,2

862

2)1(52

−=

+−=−

−=−−−

x

xxx

xx

{ } { } 43:22: <<∪<<− xxxx If other method is used, mark accordingly.

1

1 1 1, 1 1 1 1 1 1 1 1 1,1

CA CA With reason ** Parts of

15 +−= xy parabola points, is marked at end-points 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.

y

x 4 3 2 1 0 1- 6−

4 (1, 3)

5

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954/2


PAPER 2(KERTAS 2)

PERCUBAAN STPM 2009

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Q Steps Marks Notes 1

~a

~b

~~ba−

θ

cosθ = ))(( 352735 222 −+

0120=θ Thus, angle between

~a and

~b is 120

0

1 1 1 1

4

Diagram with arrow in correct direction using cosine rule CA conclusion

1 Alternative method:

=

yx

a~

,

=

nm

b~

x2 + y2 = 25 and m2 + n2

= 9

−−

=−nymx

ba~~

(x – m)2 + (y – m)2

x = 49

2 + y2 +m2 +n2

mx + ny =

– 2mx – 2ny = 49

215

−

θcos||||.~~~~baba =

nm

yx

. = 5 x 3cosθ

xm + yn = 5x3cosθ

215

− = 5x3cosθ

cosθ = 21

−

θ = 120Hence, the angle between

0

~a and

~b is 120

0

1 1 1 1

4

Either one correct Dot product CA

conclusion

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2 3sinx + 4cos x = r sinx cosα + rcos x sin α rcos α = 3 , rsin α = 4 r = 5 or 22 43 + seen

tanα = 34

α = 53.1Hence,

0

3sinx + 4cos x ≡ 5 sin(x + 53.10

)

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

) + 5

-5 ≤ 10 sin(x + 53.10

Max. value = 15 ) + 5 ≤ 15

Min. value = -5

1 1 1 1 1 1

6

both rcos α = 3 rsin α = 4 seen, if not, -1 for 10 sin(x + 53.10

) + 5

For his logic inequality Both values correct

3(a) (b)

vQ

vP

PvQ

Ν

Ν

300

1200

θ

QvP= 302 + 602 – 2(30)(60)cos 120

0

QvP 7 = 30 or 79.37 km

37379120

30

0

.sinsin

=θ

θ = 19.110 / 19.1

0

Direction of QvP is S 49.10

W

Shortest distance = 20 sin 40.9 = 13.09 km

0

1 1 1 1 1 1 1

7

Diagram with correct arrows Cosine rule based on his diagram CA His sine rule CA Using his angle CA

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4. 2 sin2

x = 1 – cos 2x

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

= 2 sin x(1 – sinx)sin x

2 x) + sin x – 2 sin3

x

= 3 sin x – 4 sin3

x

8 sin 5 x = (2sin2 x)(4sin3

= (1 – cos 2x)(3 sin x – sin 3x) x)

= 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 sin x - 25 sin 3x + ½ sin 5x

Hence, a = 5, b = - 25 , c =

21

1 1 1 1 1 1 1

7

CA

Identity(either one) CA Using above ans. Either one of the factor formula used CA conclusion

5(a)

Α

Β

C

D

O

∠BCO = ∠BAO = 900

BO = BO (common line) (radius perpendicular to tangent)

OC = OA (radii of circle) ∴∆BAO ≡ ∆BCO (RHS) ∴BC = AB

1 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)

A

C

D

E

O B

x0x0

x0

900-x0

Let AOE be the diameter of the circle, ∠CAE = 900 – x0

∠ECA = 90(tangent ⊥ to radius)

0

∠CEA = 180(angle in semicircle) 0 – (900- x0) - 90

= x0

∠CDA = ∠CEA = x

0

0

∴∠BAC = ∠ADC (angle in same segment)

Α

Β

C

D

∠BAC = ∠ADC = x0

∠ACD = x(from above)

0

∠BCA= x (base angle of isosceles ∆) either one

0

∴∠ACD = ∠BCA (base angle of isosceles ∆/alternate segment)

∠DAC = 900 – 2x0

Hence, ∆ ACD ≅ ∆ADC = ∠ABC

1 1 1 1 1 1 1 1

12

Mark accordingly to his labelling Must provide at least 1 correct reason, if not -1

All the above 3 statements correct Must provide at least 1 correct reason, if not -1

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6a

pdtdp

2312 −= = )( p324

21

−

∫ ∫=−dt

pdp

21

324

ctp +=−−21324

31 )ln(

c+=−− )()](ln[ 0210324

31

2431 ln−=c

2431

21324

31 ln)ln( −=−− tp

tp 2

3324

24=

−

ln

tep 2

3

24324 −

=−

p = 8 – 8t

e 23

−= 8(1-

te 2

3−

)

p

t0

8

1 1 1 1 1 1 1

7

Any correct separation Integrate Correct subst. for finding c Get rid of ln CA Shape of curve & 8 seen. All correct

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6(b

22vgkgdtdv

−=

∫ ∫=−gdt

vkdv

221

By writing kv

Bkv

Akvkv +

+−

=+− 1111

1))((

1 = A(1 + kv) + B(1 – kv)

A = 21 , B =

21

⇒∫ ∫=++

−gdtdv

kvkv )()( 121

121

cgtkvkv

k+=

−+

11

21 ln

( ) cgk

+= )(ln 0121

kgtkvkv 2

11

=

−+ln

kgtekvkv 2

11

=

−+

kgtekvkv 211 )( −=+

kgt

kgt

kekev 2

2 1+

−=

1 1 1 1 1 1 1

7

Any correct separation Correct A & B His correct integration base on his A&B Correct subst. Correct subs. for finding c Get rid of ln CA

7(a) (b)

!! 212 21 λλ λλ −−

=ee

4 λλ−e - 2λλ−e = 0 λ−e (4 - λ) = 0 λ−e ≠ 0, λ = 4

P(X > 3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= 1 -

+++−

!! 34

2441

324e

= 0.5666

1 1 1 1 1

5

Forming correct equation CA

Complement

At least 3 correct terms with his λ.

CA

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8 (a) (b)

P(X) =72 , P(X ∩Y) =

211

P(X|Y’) = 73

73

=∩

)'()'(

YPYXP

P(X ) –P(X ∩ Y) = 73 [1 – P(Y)]

[ ])(YP−=− 173

211

72

P(Y) = 94

P(X ∪ Y) = 72 +

94 -

211

= 6343

P(X).P(Y) = 72 x

94 =

638

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

1 1 1 1 1 1

6

Expansion of P(X|Y’) correctly. Seen/applied in the equation Correct formula with his value CA Shown

72 x

94 or his

value Both statements correct and

638 seen

9(a) (b) (c)

m

05

+ m

15

+ m

25

+ m

35

+ m

45

+ m

55

= 1

m + 5m + 10m + 5m + m = 1

m = 321

x 0 1 2 3 4 5 P(X = x)

321

325

3210

3210

325

321

E(5X – 3) =

5[0(321 )+1(

325 )+2(

3210 )+3(

3210 )+4(

325 )+5(

321 )] - 3

= 2

19 or 9.5

1 1 1 1 1 1 1

7

At least one of the equation seen

CA At least 4 prob. correct based on his m. All correct with table Finding E(X) based on his m For correct usage of formula 5E(X) - 3 CA

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10 (a) (b) (c)

≤≤+−

≤≤=

otherwise , 0 3x,2

2x,0 23

231

xx

xf )(

≥

<≤+−

<≤

<

=

3x , 1

3x2 31-

2x,0

0x ,

),()(

666

0

2

2

xx

x

xF

F(x)

x0

1

32

1 2 3 P(0.9 < X ≤ 2.1) = F(2.1) – F(0.9)

= 31

− [(2.1)2

690 2).( – 6(2.1) + 6] -

= 0.73 – 0.135 = 0.595

1 1 1 1 1 1 1 1 1 1

10

For x31 seen

For 232 +− x seen

All correct At least one quadratic function correct All correct Curves correct shape All correct App. of formula/integration

Correct subst. CA

11 (a) (b)

P(S < 0.750) = P

−

<0080

76007500.

..z

= P(Z < -1.25) = 0.1057 X~B(10, 0.1057) 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

]

= 0.0803

1 1 1 1 1

Standardization CA

Binomial distrb/implied Correct subst of his binomial CA

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(c)

F = S1 + S2 + S3 + S4 G = L1 + L2 + LLet T = F – G

3

E(T) = 4(0.760) – 3(1.010) = 0.01 Var(T) = 4(0.0082) +3(0.0092

= 0.000499 )

P(T > 0)

= P

−>

00049900100

..Z

= 0.6728

1 1 1 1 1

10

Correct linear combination(4S-3L is accepted if his var(T) is correct) All correct All correct Standardization based on his values CA

12 (a) (b)

1207650

=−

x

= 63.75

2

1207650

120510020

−=σ

= 13.64

20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.50

10

20

30

40

50

60

70

80

90

100

110

120

Num

ber o

f stu

dent

s

Marks

median = 50564 .. ±

105

16.3

(i) From the graph, median 50564 .. ± (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 1 1 1 1 1 1 1 1 1

His ∑ x ÷120

CA 510020 Correct formula CA Boundaries Axes & >6 pts correct All correct with smooth curve Shown in graph by dotted line Answer in this range

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% students in this range = 100120

16105×

−

= 74.2%

1 1

12

Correct formula based on his value Accept 74% - 78%

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