2009 PSPM Kedah AddMath 12 w Ans
Transcript of 2009 PSPM Kedah AddMath 12 w Ans
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PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH
NEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2009
Kertas soalan ini mengandungi 17 halaman bercetak
For Examiners use only
Question Total MarksMarks
Obtained
1 2
2 4
3 3
4 3
5 3
6 4
7 4
8 3
9 3
10 3
11 3
12 4
13 3
14 3
15 2
16 3
17 4
18 419 3
20 2
21 4
22 3
23 3
24 3
25 4
TOTAL 80
M A T E M A T I K T A M B A H A N
Kertas 1
Dua jam
J A N G A N B U K A K E R T A S S O A L A N IN I
S E H I N G G A D I B E R I T A H U
1 This question paper consists of 25 questions.
2. Answerall questions.
3. Give on ly on e answer for each quest ion.
4. Write your answers clearly in the spaces provided in
the question paper.
5. Show yo ur working . I t may help you to get marks .
6. If you wish to change you r answer, cross out the work
tha t you have done. Then write down the new
answer.
7. The diagrams in the questions provided are not
drawn to scale unless stated.
8. The marks a l loca ted for each quest ion and sub- art
of a question are shown in brackets.
9. A l ist of formulae is provided on pag es 2 to 3.
10 . A boo klet of four-figure ma thematical tables is
provided .
.
11 You m ay use a non-programm able scien t if iccalculator.
12 This question paper must be handed in at the end of
the examinat ion .
Name : ..
Form : ..
3472/1
Additional Mathematics
Paper 1
Sept 2009
2 Hours
ADDITIONAL MATHEMATICSPaper 1
Two hours
3 to 4.
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BLANK PAGE
HALAMAN KOSONG
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The following formulae may be helpful in answering the questions. The symbols given are the ones
commonly used.
ALGEBRA
12 4
2
b b acx
a
- -=
2 am an = a m + n
3 am an = a m - n
4 (am)
n= a
nm
5 log a mn = log a m + log a n
6 log an
m= log a m log a n
7 log a mn
= n log a m
8 logab =
a
b
c
c
log
log
9 Tn = a + (n1)d
10 Sn = ])1(2[2
dnan
-+
11 Tn = arn-1
12 Sn =r
ra
r
rann
-
-=
-
-
1
)1(
1
)1(, (r 1)
13
r
aS
-
=1
, r
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STATISTIC
1 Arc length,s = rq
2 Area of sector ,A = 21
2rq
3 sin 2A + cos2A = 1
4 sec2A = 1 + tan
2A
5 cosec2A = 1 + cot
2A
6 sin 2A = 2 sinA cosA
7 cos 2A = cos2A sin
2A
= 2 cos2A 1
= 1 2 sin2A
8 tan 2A =A
A2
tan1
tan2
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TRIGONOMETRY
9 sin (A B) = sinA cosB cosA sinB
10 cos (A B) = cosA cosB m sinA sinB
11 tan (A B) =BA
BA
tantan1
tantan
m
12C
c
B
b
A
a
sinsinsin==
13 a2 = b2 + c
2 2bc cosA
14 Area of triangle = Cabsin2
1
1 =N
x
2 =
f
fx
3 s =N
xx - 2)(= 2
2
xN
x-
4 s =
-f
xxf 2)(= 2
2
xf
xf-
5 m = Cf
FN
Lm
-
+ 2
1
6 1
0
100Q
IQ
=
71
11
w
IwI
=
8)!(
!
rn
nPrn
-=
9!)!(
!
rrn
nCr
n
-=
10 P(A B) = P(A)+P(B) P(A B)
11 P (X= r) = rnrrn qpC - , p + q = 1
12 Mean = np
13 npq=s
14 z =s
m-x
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Answer all questions.
Jawabsemua soalan.
1. Diagram 1 shows the relation between set A and set B.
Rajah 1 menunjukkan hubungan antara set A dan set B.
a) State the image of 9.Nyatakan imej bagi 9.
b) Find the value ofx.
Cari nilai x. [ 2 marks]
[2 markah]
Answer/Jawapan : (a) ..
(b) ...
2. Given5
3:
1 xxf-
- , find the value of
Diberi5
3:
1 xxf
-- , cari nilai bagi
(a) )3(-f ,
(b)p if 7)( -=pf .[ 4 marks ]
[4 markah]
Answer/ Jawapan : (a) ..
(b) ...4
2
2
1
Forexaminers
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Set A Set B
49
9
9
7
3
Diagram 1
Rajah 1
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3. Given that function axx +2: and 9:2 -bxxg .
Diberi fungsi axg +2: dan 9:2 -bxxg .
Find the value of a and ofb
Cari nilai bagi a dan b .
[3 marks]
[3 markah]
Answer/Jawapan : a =.........................b =.........................
4. Given that the straight line 14 +=y is a tangent to the curve kxy += 2 .
Find the value of k.
Diberi garis lurus 14 += xy ialah tangen kepada lengkung kxy += 2 .
Cari nilai k.
[ 3 marks]
[3 markah]
Answer/Jawapan : k=......
For
examiners
use only
3
4
3
3
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5.
4)2(3 2 --= xy
Diagram above shows the graph of the function 4)2(3 2 --= xy . Q is the minimum pointof the curve and the curve intersects the y-axis at point P. Find the equation of the straightline PQ.
[ 3 marks ]
Rajah di atas menunjukkan graf bagi fungsi 4)2(3 2 --= xy . Q ialah titik minimum bagi
lengkung itu dan lengkung tersebut bersilang dengan paksi-y di titik P. Cari persamaan garislurus PQ.
[3 markah]
Answer /Jawapan: ........................
___________________________________________________________________________
6. Given that aand b are the roots of the quadratic equation 0792 =+- x .
Find the value of
Diberi a danb adalah punca bagi persamaan kuadratik 0792 =+- x . Cari nilai
bagi
(a) ba+(b) ab
(c)22 ba + [ 4 marks ]
[4 markah]
Answer/Jawapan : (a).............................
(b)............................
(c).............................
3
5
4
6
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x
y
P
Q
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7. Solve the equation:
Selesaikan persamaan:
xxx 51 4)8(2 =- .
[4 marks][4 markah]
Answer/Jawapan :x =................................
8. The set of positive integers 2, 5, 7, 9, 11,x,y has a mean 8 and median 9. Find the
values ofx and ofy ify >x.
[3 marks]
Satu set integer positif 2, 5, 7, 9, 11, x, y mempunyai min 8 dan median 9. Cari
nilai-nilai bagi x dan y jika y > x.
[3 markah]
Answer/Jawapan : ...................................
4
7
3
8
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9. Given that 2loglog 93 =+ yx . Expressy in terms ofx.
[ 3 marks ]
Diberi 2loglog 93 =+ yx . Ungkapkan y dalam sebutan x.
[3 markah]
Answer/Jawapan : ......................................
10. The sixth and eleventh terms of an arithmetic progression are 12 and 37 respectively.Find the value of the sixteenth term of this arithmetic progression.
[3 marks]
Sebutan keenam dan kesebelas bagi suatu janjang aritmetik ialah 12 dan 37 masing-masing. Cari nilai bagi sebutan keenambelas bagi janjang aritmetik ini.
[3 markah]
Answer/Jawapan : ......
3
10
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9
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11. The first three terms of a geometric progression are 36, 36 p and q. If the
common ratio is3
1- , find the value of
Tiga sebutan pertama suatu janjang geometri ialah 36, 36 p dan q. Jika nisbah
sepunya ialah31- , cari nilai bagi
(a) p ,
(b) q.
[ 3 marks ][3 markah]
Answer/Jawapan: a) p = ...........
b) q =...............................
12. The first term of a geometric progression is a and the common ratio is r. Given that
096 =+ ra and the sum to infinity is 32, find the value of a and of r.
[ 4 marks ]
Sebutan pertama bagi suatu janjang geometri ialah a dan nisbah sepunya r.Diberi 096 =+ ra dan hasil tambah hingga sebutan ketakterhinggaan ialah 32,cari nilai a dan r.
[4 markah]
Answer/Jawapan:a=...........
r=...............................4
12
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11
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13. Given that A, B )4,4(- , C )7,2( are collinear and 3AB=BC, find the coordinates of A.[ 3 marks ]
Diberi A, B )4,4(- ,C )7,2( adalah segaris dan 3AB=BC, carikan koordinat titik A.[3 markah]
Answer/Jawapan : ...
14. Diagram below shows the graph of y10log against x .
Rajah di bawah menunjukkan graf y10log lawan x.
The variablesx andy are related by the equation2310 -= xy
Find the value ofh and ofk.
Pembolehubah x dan y dihubungkait dengan persamaan2310 -= xy .
Cari nilai h dan k.[3 marks]
[3 markah]
Answer/Jawapan : h=
k=....
3
13
3
14
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),0( k
)2,(hy10log
x
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15.
In the diagram above, jniOA += 8 and jiAB 23 -= . Given
unitsOA 10= , find
Dalam rajah di atas, jniOA += 8 dan jiAB 23 -= . Diberi
unitOA 10= , cari
(a) the value ofn.
nilai n.
(b) coordinates of B.
koordinat B.
[2 marks][2 markah]
Answer/Jawapan : (a) n =
(b)..
16 Given that jipa 2+= and jib --= 2 , find the value ofp if ba - is parallel
to j .
[ 3 marks ]
Diberi jipa 2+= dan jib --= 2 , cari nilaip jika ba - selari dengan j .
[3 markah]
Answer/Jawapan :..
2
15
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3
16
y
O
A
B
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17. Solve the equation 0cotcos2 =+ x for 00 3600 x .
[ 4 marks ]
Selesaikan persamaan 02 =+ kotxkosx bagi 00 3600 x .
[4 markah]
Answer/Jawapan: ............
18. Diagram below shows a circle with centre O.
Rajah di bawah menunjukkan satu bulatan dengan pusat O.
Given that the minor angle POQ is p3
2radian and the area of the shaded
region is212 cmp . Find the length of the minor arc PQ.
Diberi sudut minor POQ ialah p3
2radian dan luas sektor berlorek
ialah212 cmp . Cari panjang lengkok minor PQ.
[4 marks][4 markah]
Answer/Jawapan:
_
p3
2_1
O
P
4
17
4
18
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Q
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19. Find the gradient of the curve2
5
2
34
2 -+=x
xy at the point )3,1( .
[ 3 marks ]
Cari kecerunan kepada lengkung
2
5
2
34
2 -+=x
xy pada titik )3,1( .
[3 markah]
Answer/Jawapan:
20. Differentiate
12
14 2
+
-xwith respect tox .
[2 marks]
Bezakan12
14 2
+
-
x
xterhadapx.
[2 markah]
Answer/Jawapan: ............
3
19
2
20
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examiners
use only
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21. Given that the gradient function of a curve passing through the point (1, 2) is
2)12(
3
-x+ 2x , determine the equation of the curve.
Fungsi kecerunan bagi suatu lengkung yang melalui titik ( 1, 2) ialah
2)12(3-x
+ 2x, tentukan persamaan bagi lengkung ini.
[4 marks]
[4 markah]
Answer/Jawapan: ..
22. Given that10
)32( 5-=x
y andx is increasing at the rate of 2 units per second, find the
rate of change ofy when2
1=x . [ 3 marks ]
Diberi10
)32( 5-=
xy dan x bertambah dengan kadar 2 unit sesaat, cari kadar
perubahan bagi y apabila2
1=x . [3 markah]
Answer/Jawapan: .
4
21
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3
22
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23. 90 percent of the students of Form 5 Euler passed the April mathematics test. Amongthose who passed, 20 percent score with distinction.
90 peratus pelajar Tingkatan 5 Euler lulus ujian matematik bulan April. Antaramereka yang lulus, 20 peratus skor dengan cemerlang.
(a) If a student of Form 5 Euler was selected at random, find the probability that hepassed the April mathematics test with distinction.
Jika seorang pelajar dari tingkatan 5 Euler dipilih secara rawak, cari
kebarangkalian dia lulus ujian matematik bulan April dengan cemerlang.
(b) If 5 students of Form 5 Euler were selected at random, find the probability that
only one of the five students selected passed the April mathematics test with
distinction.
Jika 5 orang pelajar dari tingkatan 5 Euler dipilih secara rawak, cari
kebarangkalian hanya seorang daripada lima pelajar terpilih lulus ujian
matematik bulan April dengan cemerlang.
[3 marks]
[3 markah]
Answer/Jawapan: (a) ..
(b) ..
24. Five cards are numbered 1, 2, 3, 4 and 5 respectively. How many different odd numbers
can be formed by using four of these five cards?
[ 3 marks ]
Lima kad masing-masing ditulis dengan nombor 1, 2, 3, 4 dan 5. Berapa nombor ganjil
boleh dibentuk dengan menggunakan empat daripada lima kad ini ?[3 markah]
Answer/Jawapan: 3
24
3
23
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25. A random variable X is normally distributed with mean 370 and standard deviation10. Find the value of
Satu pembolehubah rawak X bertaburan normal dengan min 370 dan sisihan piawai10. Cari nilai bagi
(a) the z-score if X = 355.skor z jika X = 355
(b) )367(
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Additional Mathematics
Kertas 2September, 2009
2 jam 30 minit
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAHNEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2009
ADDITIONAL MATHEMATICS
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. This question paper consists of three sections : Section A, Section B andSection C.
2. Answerall questions in Section A, four questions from Section B andtwo questionsfrom Section C.
3. Give only one answer/solution to each question.
4. Show your working. It may help you to get your marks.
5. The diagrams provided are not drawn according to scale unless stated.
6. The marks allocated for each question and sub - part of a question are shown inbrackets.
7. You may use a non-programmable scientific calculator.
8. A list of formulae is provided in page 2 and 3.
This question paper consists of19 printed pages and 1 blank page.
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2
The following formulae may be helpful in answering the questions. The symbols given are the onescommonly used.
ALGEBRA
1. x =a
acbb
2
42 -- 8.
a
bb
c
ca
log
loglog =
2. aaa nmnm += 9. dnaT n )1( -+=
3. aaanmnm -= 10. ])1(2[
2dna
nSn -+=
4. aamnnm =)( 11.
1-= nn arT
5. nmmn aaa logloglog +=12.
r
ra
r
raS
nn
n-
-=
-
-=
1
)1(
1
)1(, r 1
6. log log loga a am
m nn
= - 13.r
aS
-=
1, r < 1
7. mnm ana loglog =
CALCULUS
1. y = uv,dx
duv
dx
dvu
dx
dy+=
4 Area under a curve
= b
adxy or
= b
adyx
2. y = v
u
, 2v
dx
dvu
dx
duv
dx
dy-
=
5. Volume of revolution
= b
adxy 2p or
= b
adyx2p
3.dx
du
du
dy
dx
dy=
GEOMETRY
1. Distance = 2122
12 )()( yyxx -+-4. Area of triangle
=1 2 2 3 3 1 2 1 3 2 1 3
1( ) ( )
2x y x y x y x y x y x y+ + - + +
2. Mid point
(x , y ) =
++
2,
2
2121 yyxx5. 22 yxr +=
3. Division of line segment by a point
(x , y ) =
+
+
+
+
nm
myny
nm
mxnx 2121 ,
6.2 2
xi yj
ry
+=
+% %
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STATISTICS
1.N
xx
= 7
=i
ii
W
IWI
2.=
f
fxx 8
)!(
!
rn
nPr
n
-=
3.N
xx -=2)(
s = 22
xN
x- 9 !)!(
!
rrn
nCr
n
-=
4.
-=
f
xxf 2)(s = 2
2
xf
fx-
10 P(AB) = P(A) + P(B) P(AB)
11 P (X = r) = rnrrn qpC - ,p + q = 1
5. m = L + Cf
FN
m
-21 12 Mean , m= np
13 npq=s
6. 10001 = Q
QI 14 Z = s
m-X
TRIGONOMETRY
1. Arc length,s = rq 8. sin (A B ) = sinA cosB cosA sinB
2. Area of sector, A = q2
2
1r
9. cos (A B ) = cosA cosBm sinA sinB
3. sin A + cosA = 110 tan (A B ) =
BA
BA
tantan1
tantan
m
4. sec A = 1 + tan A11 tan 2A =
AA2
tan1tan2
-
5. cosec A = 1 + cot A12
C
c
B
b
A
a
sinsinsin==
6. sin 2A = 2sinA cosA 13 a = b + c 2bc cosA7. cos 2A = cos A sin A
= 2 cos A 1= 1 2 sin A
14 Area of triangle =1
sin2
ab C
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3472/2 Additional Mathematics Paper 2 [Lihat sebelah
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Section A
Bahagian A
[ 40 marks ][ 40 markah ]
Answerall questions.
Jawab semua soalan.
1. Solve the simultaneous equations 3 2 0x y- - = and ( )2 1 3x y - = .
Give your answers correct to three decimal places.[5 marks]
Selesaikan persamaan serentak 3 2 0x y- - = dan ( )2 1 3x y - = .
Beri jawapan anda betul kepada tiga tempat perpuluhan.[5 markah]
2.
In the diagram, the gradient andy-intercept of the straight line PQ
are 2 and 3 respectively.R is a point on thex-axis.
(a) Find the value ofh and k.
(b) Given thatPQ is perpendicular to QR, find thex-intercept ofQR.
(c) Calculate the area ofPQR.
[3 marks]
[3 marks]
[2 marks]
Dalam rajah, kecerunan dan pintasan-y bagi garis lurus PQ masing-masing ialah 2 dan 3 . R ialah titik pada paksi-x.
(a) Cari nilai h dan nilai k .
(b) Diberi bahawa PQ berserenjang dengan QR, cari pintasan-x
bagi QR.
(c) Hitungkan luas PQR.
[3 markah]
[3 markah]
[2 markah]
x
y
0
Q(4, k)
P(h, 3)
R
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3.(a) Sketch the graph of 2 sin 2x= - for0 2x p .
(b) Hence, using the same axes, sketch a suitable straight line to find
the number of solutions for the equation sin22
xx
p= for
0 2x p . State the number of solutions.
(a) Lakar graf bagi 2 sin 2y x= - untuk 0 2x p .
(b) Seterusnya, dengan menggunakan paksi yang sama, lakar satugaris lurus yang sesuai untuk mencari bilangan penyelesaian
bagi persamaan sin22
xx
p= untuk 0 2x p .
Nyatakan bilangan penyelesaian itu.
[4 marks]
[3 marks]
[4 markah]
[3 markah]
4. Given that ( )2 1x x - is the gradient function of a curve which passes
through the point )1,1( -P . Find
(a) the gradient of the tangent to the curve atP,
(b) the equation of the curve,
(c) the coordinates of the turning point at x = 1 . Hence determine
whether the turning point is a maximum or a minimum point.
[1 mark]
[3 marks]
[3 marks]
Diberi ( )2 1x x - ialah fungsi kecerunan bagi suatu lengkung yang
melalui titik )1,1( -P . Cari
(a) kecerunan tangen kepada lengkung itu di P,
(b) persamaan lengkung itu,
(c) koordinat bagi titik pusingan pada x = 1 . Seterusnya tentukansama ada titik pusingan itu adalah titik maksimum atau titikminimum.
[1 markah]
[3 markah]
[3 markah]
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5. A set of fifty numbers, 1 2 3 50, , ,......x x x x , has a mean of 11 and a
standard deviation of 8.
(a) Find
(i) ,(ii)
2
(b) If each of the numbers is multiplied by 1.8 and thenincreased by 5, find the new value for the
(i) mean,
(ii) variance.
[3 marks]
[3 marks]
Suatu set yang terdiri daripada lima puluh nombor,
1 2 3 50, , ,......x x x x , mempunyai min 11 dan sisihan piawai 8.
(a) Cari
(i) ,
(ii)2
(b) Jika setiap nombor didarab dengan 1.8 dan ditambahdengan 5, cari nilai yang baru untuk
(i) min ,
(ii) varians .
[3 markah]
[3 markah]
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7
6.
A strip of metal is cut and bent to form some semicircles. Thediagram shows the first four semicircles formed. The radius of the
smallest semicircle is 5 cm. The radius of each subsequentsemicircle is increased by 3 cm.
(a) If the radius of the largest semicircle is 104 cm, find the number
of semicircles formed.
(b) Calculate the total cost needed to form all the semicircles in (a) if
the cost of the metal strip is RM4 per meter.[ Round off your answer to the nearest RM ]
Satu jalur logam dipotong dan dibengkok untuk membentukbeberapa semi bulatan. Rajah menunjukkan empat semi bulatan
pertama yang telah dibentukkan. Jejari semi bulatan yang terkecilialah 5 cm. Jejari semi bulatan yang berikutnya bertambah
sebanyak3cm setiap satu.
(a) Jika jejari bagi semi bulatan yang terbesar ialah 104 cm , caribilangan semi bulatan yang telah dibentuk.
(b) Hitungkan jumlah kos yang diperlukan untuk membentuk semua
semi bulatan dalam (a) jika kos jalur logam ialah RM4 semeter.[ Bundarkan jawapan anda kepada RM yang terdekat]
[3 marks]
[4 marks]
[3 markah]
[4 markah]
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8
Section B
Bahagian B
[ 40 marks ][ 40 markah ]
Answerfour questions from this section.
Jawab empatsoalan daripada bahagian ini.
7.
The table shows the values of two variables, x andy, obtained from
an experiment. Variables x and y are related by the equation
1=+yxlm , where mand lare constants.
(a) Ploty
1against
1, using a scale of 2 cm to 0.1 unit on both
axes. Hence draw the line of best fit.
(b) Use your graph in 7(a) to find the value of
(i) m,
(ii) l.
[5 marks]
[5 marks]
Jadual menunjukkan nilai-nilai bagi dua pembolehubah, x dan y,
yang diperoleh daripada satu eksperimen. Pembolehubah x dan y
dihubungkan oleh persamaan 1=+yx
lm, dengan keadaan mdan l
adalah pemalar
(a) Ploty
1melawan
x
1, dengan menggunakan skala 2 cm
kepada 0.1 unit pada kedua-dua paksi. Seterusnya, lukis
garis lurus penyuaian terbaik.(b) Gunakan graf di 7(a) untuk mencari nilai
(i) m,
(ii) l.
[5 markah]
[5 markah]
x 1.5 2.0 2.5 4.0 5.0 10.0
y 0.96 1.2 1.4 2.0 2.2 3.0
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9
8.
The diagram shows straight linex +y = 4 that intersects with thecurve y = (x 2 )2
at points Pand Q.
Find
(a) the coordinates of Q,
(b) the area of the shaded regionA,
(c) the volume generated, in terms of , when the shaded region
B is revolved through 360o
about thex-axis.
[2 marks]
[4 marks]
[4 marks]
Rajah menunjukkan garis lurus x + y = 4 bersilang denganlengkung y = (x 2 )
2pada titik P dan Q.
Cari
(a) koordinat Q,
(b) luas rantau berlorek A,
(c) isipadu janaan, dalam sebutan , apabila rantau berlorek B
dikisarkan melalui 360opada paksi-x.
[2 markah]
[4 markah]
[4 markah]
O x
y = (x 2 )2
P
QAB x + y = 4
y
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10
9.
In the diagram, PQRS is a quadrilateral. The diagonalsPR and QS
intersect at point T. It is given PQ = 2~
x , PS = 3~
y and
SR =~
x ~
y .
(a) Express in terms of~
x and~
y :
(i) QS
(ii) PR .
(b) Given that QT = m QS , PT = n PR , where m and n are
constants, express
(i) QT in terms of m,~
x and~
y ,
(ii) PT in terms of n,~
x and~
y .
(c) Using PQ = PT + TQ , find the value of m and of n.
[10 marks]Dalam rajah, PQRS ialah sebuah sisiempat. Pepenjuru-pepenjuru
PR dan QS bersilang di titik T . Diberi PQ = 2~
x , PS = 3~
y
dan SR =~
x ~
y .
(a) Ungkapkan dalam sebutan~
x dan~
y :
(i) QS
(ii) PR .
(b) Diberi QT = m QS , PT = n PR , dengan keadaan m dan nialah pemalar, ungkapkan
(i) QT dalam sebutan m,~
x dan~
y ,
(ii) PT dalam sebutan n,~
x dan~
y .
(c) Dengan mengguna PQ = PT + TQ , cari nilai m dan
nilai n. [10 markah]
P
T
S
R
Q
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10.
The diagram shows the cross-section of a cylindrical roller with
centre O and radius 20 cm resting on a horizontal groundPQ.OAB is a straight line that represents the handle of the roller and
OA :AB = 1 : 3.
Calculate
(a) POA in radian,
(b) the perimeter, in cm, of the shaded region,
(c) the area, in cm2
, of the shaded region.
[3 marks]
[3 marks]
[4 marks]
Rajah menunjukkan keratan rentas sebuah penggelek berbentuk
silinder dengan pusat O dan jejari 20 cm yang terletak di atas
lantai mengufuk PQ. OAB ialah garis lurus yang mewakilipemegang penggelek itu dan OA : AB = 1 : 3.
Hitungkan
(a) POA dalam radian,
(b) perimeter, dalam cm, kawasan berlorek,
(c) luas, dalam cm2
, kawasan berlorek.
[3 markah]
[3 markah]
[4 markah]
QP
O
A
B
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11(a) In a survey carried out in a school, it is found that 40% of its
students are participating actively in co-curricular activities.
(i) If 6 students from that school are chosen at random,calculate the probability that at least 4 students are
participating actively in co-curricular activities.
(ii) If the variance of the students who are active in co-curricularactivities is 288, calculate the student population of the
school.[5 marks]
(b) The masses of chicken eggs from a farm has a normal distribution
with a mean of 62 g and a standard deviation of 8 g. Any egg thathas a mass exceeding 68 g is categorised as grade double-A.
(i) Find the probability that an egg chosen randomly from the
farm has a mass between 60 g and 68 g.
(ii) If the farm produces 3000 eggs daily, calculate the number ofeggs with grade double-A.
[5 marks]
(a) Dalam satu tinjauan yang dijalankan ke atas murid-murid di sebuah sekolah, didapati 40% daripada murid-murid sekolah itu
mengambil bahagian secara aktif dalam aktiviti kokurikulum.
(i) Jika 6 orang murid daripada sekolah itu dipilih secara rawak,
hitungkan kebarangkalian bahawa sekurang-kurangnya 4orang murid adalah aktif dalam aktiviti kokurikulum.
(ii) Jika varians murid-murid yang mengambil bahagian secaraaktif dalam aktiviti kokurikulum ialah 288, hitungkan bilangan
murid dalam sekolah itu.[5 markah]
(b) Jisim telur ayam dari sebuah ladang adalah mengikut satu taburannormal dengan min 62 g dan sisihan piawai 8 g. Sebarang telur
dengan jisim melebihi 68 g dikategorikan sebagai telur greddouble-A
(i) Cari kebarangkalian bahawa sebiji telur yang dipilih secara
rawak dari ladang itu mempunyai jisim di antara 60 g dan68 g.
(ii) Jika ladang itu menghasilkan 3000 biji telur setiap hari,
hitungkan bilangan telur yang mempunyai gred double-A.
[5 markah]
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13
Section C
Bahagian C
[ 20 marks ]
[ 20 markah ]
Answertwo questions from this section.Jawab dua soalan daripada bahagian ini.
12. A particle P starts from a fixed point O and moves in a straight
line so that its velocity, v ms-1
, is given by v = 8 + 2t t2, where
t is the time, in seconds, after leaving O.[Assume motion to the right is positive.]
Find
(a) the initial velocity, in ms-1 , of the particle,
(b) the value of t at the instant when the acceleration is 1 ms-2
,
(c) the distance of P from O when P comes to instantaneousrest,
(d) the total distance, in m, travelled by the particle P in the first
5 seconds.
Suatu zarah P mula dari suatu titik tetap O dan bergerak di sepanjang garis lurus. Halajunya vms
-1 , diberi oleh
v = 8 + 2t t2 , dengan keadaan t ialah masa, dalam saat, selepas
melalui O.[Anggapkan gerakan ke arah kanan sebagai positif]
Cari
(a) halaju awal, dalam ms-1
, bagi zarah itu,
(b) nilai bagi t apabila pecutannya ialah 1 ms-2,
(c) jarak P dari O apabila P berada dalam keadaan rehatseketika,
(d) jumlah jarak yang di lalui, dalam m , oleh zarah P dalam
5 saat yang pertama.
[1 mark]
[2 marks]
[3 marks]
[4 marks]
[1 markah]
[2 markah]
[3 markah]
[4 markah]
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14
13. The pie chart shows five items,A, B, C, D and E used in makingcakes. The table shows the prices and the price indices of these
items.
E
D
C
B
A
30
75
120
Items
Bahan
Price (RM)
per kg for theyear 2003
Harga (RM)per kg padatahun 2003
Price (RM)
per kg for theyear 2006
Harga (RM)per kg padatahun 2006
Price index for the
year 2006 based onthe year 2003
Indeks harga pada
tahun 2006berasaskan tahun
2003
A 0.40 x 150
B 1.50 1.65 110C 4.00 4.80 y
D 3.00 4.50 150
E z 2.40 120
(a) Find the value of(i) x,
(ii) y,(iii) z.
(b) Calculate the composite index for the cost of making these
cakes in the year 2006 based on the year 2003.
(c) The total expenditure on the items in the year 2006 isRM 5000. Calculate the corresponding total expenditure
in the year 2003.
(d) The price of each item increases by 20 % from the year 2006to the year 2008. Find the composite index for total
expenditure on the items in the year 2008 based on the year2003.
[3 marks]
[3 marks]
[2 marks]
[2 marks]
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15
Carta pai menunjukkan lima bahan, A, B, C, D dan E yang
digunakan untuk membuat sejenis kek. Jadual menunjukkan hargabahan dan nombor indeks bagi kelima-lima bahan tersebut.
(a) Carikan nilai
(i) x,(ii) y,
(iii) z.
(b) Hitungkan nombor indeks gubahan bagi kos penghasilan kekitu pada tahun 2006 berasaskan tahun 2003.
(c) Jumlah kos bahan-bahan tersebut pada tahun 2006 ialah
RM 5000. Hitungkan jumlah kos yang sepadan pada tahun2003.
(d) Harga bagi setiap bahan bertambah sebanyak 20% daritahun 2006 ke tahun 2008. Cari nombor indeks gubahanbagi jumlah kos ke atas bahan-bahan tersebut pada tahun
2008 berasaskan tahun 2003.
[3 markah]
[3 markah]
[2 markah]
[2 markah]
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16
14. Use graph paper to answer this question.
A factory produces two components, S and T for a digitalcamera, by using machines P and Q. The table shows the time
taken to produce components S and T respectively.
In any given week, the factory produces xunits of component S
and y units of component T. The production of the componentsper week is based on the following constraints:
I : MachineP operates not more than 2000 minutes.
II : Machine Q operates at least 1200 minutes.
III : The number of component Tproduced is not more than threetimes the number of component Sproduced.
(a) Write three inequalities, other than x 0 and y 0, which
satisfy all the above constraints.
(b) Using a scale of 2 cm to 10 units on both axes, construct andshade the region R which satisfies all of the above
constraints.
(c) Use your graph in 14(b) to find
(i) the maximum number of component S that could be
produced, if the factory plans to produce only 30 unitsof component T,
(ii) the maximum profit per week if the profit from a unit
of component S is RM20 and from a unit of component T is RM30.
Time taken (minutes)Masa diambil(minit)
Component
Komponen Machine PMesin P
Machine QMesin Q
S 40 15
T 20 30
[3 marks]
[3 marks]
[4 marks]
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17
Gunakan kertas graf untuk menjawab soalan ini.
Sebuah kilang menghasilkan dua komponen, S dan T bagi sesuatu kamera digital dengan menggunakan mesin P dan Q.
Jadual menunjukkan masa yang diambil untuk menghasilkankomponen-komponen S dan T.
Dalam mana-mana satu minggu, kilang tersebut menghasilkan x
unit bagi komponen S dan y unit bagi komponen T. Penghasilan komponen-komponen tersebut adalah berdasarkan
kekangan berikut:
I : Mesin P beroperasi tidak melebihi2000 minit.
II : Mesin Q beroperasi sekurang-kurangnya1200 minit.
III : Bilangan komponen T yang dihasilkan tidak melebihi tigakali ganda bilangan komponen S yang dihasilkan.
(a) Tuliskan tiga ketaksamaan, selain x 0 dan y 0 , yang
memenuhi semua kekangan di atas.
(b) Menggunakan skala 2 cm kepada 10 unit pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua
kekangan di atas.
(c) Gunakan graf anda di 14(b) untuk mencari
(i) bilangan maksimum bagi komponen S yang bolehdihasilkan jika kilang tersebut bercadang untuk
menghasilkan 30 unit komponen T sahaja,
(ii) keuntungan maksimum seminggu jika keuntungan yangdiperoleh dari satu unit komponen S ialah RM20 dan
dari satu unit komponen T ialah RM30.
[3 markah]
[3 markah]
[4 markah]
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18
15. The diagram shows a cyclic quadrilateral PQRS with PQ = 9 cm,
PR = 11 cm and QR = 7 cm.
(a) Find PQR.
(b) Given that PS= 6 cm, find the length ofRS.
(c) Calculate the area ofPQRS.
S R
Q
P
11 cm
7 cm
9 cm
[3 marks]
[4 marks]
[3 marks]
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19
Rajah menunjukkan satu sisiempat kitaran PQRS denganPQ = 9 cm , PR =11 cm , dan QR = 7 cm.
(a) Cari PQR.
(b) Diberi PS = 6 cm, cari panjang RS.
(c) Hitungkan luas bagi PQRS.
[3 markah]
[4 markah]
[3 markah]
END OF QUESTION PAPER
KERTAS SOALAN TAMAT
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SULIT JPNKd/2006/3472/1
Additional Mathematics paper 1
Nama Pelajar : Tingkatan 5 : .
3472/1AdditionalMathematicsPaper1
September 2009
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAHNEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2009
ADDITIONAL MATHEMATICS
MARKING SCHEME
Paper 1
.
SULIT 3472/1
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SULIT 3472/12
SPM Trial Examination 2009 Kedah Darul Aman
Marking Scheme
Additional Mathematics Paper 1
Question Solution/ Marking Scheme Answer Marks
1 (a) 3
(b) 81
1
1
2
(a) B1: 35
3
xor xxf 53)(
(b) B1: 753 p or p
5
)7(3
(a) 18
(b) 2
2
2
3
B2: 934 aorb
B1: aax )2(2
a = 3
b = 4
3
4
B2: 410)1)(1(4)4( 2 kork
B1 : or0142 kxx 142 xkx
5 3
5
B2: 620
)4(8
orm
B1: )4,2()8,0( QorP
86 xy 3
6
(c) B1: )7(292
(a) 9(b) 7(c) 67
1
1
2
7
B3: xxx 1033
B2 :xx or 1033 22
B1: xxx 5213 )2()2(2
x =2
1 4
3472/1 Additional Mathematics Paper 1
SULIT
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Question Solution/ Marking Scheme Answer Marks
8
B2: 22 yx
B1: 87
119752
yx
12,10 yx
13,9 yx 3
9 B2: 2log4log 292
3 yxoryx
B1: 2log3log
log2
9log
loglog 9
9
9
3
33 y
xor
yx
2
81
xy 3
10
B2: 513 danda
B1: 3710125 daorda 62 3
11
(a) B1:3
1
36
36
p
(a) 48
(b) 4
2
1
12B3: 48
2
1 aorr
B2: 32
961
096)1(32
a
aorrr
B1: 321
r
a
21,48 ra 4
13
B2 : 44
374
4
23
yand
x
B1 : 44
374
4
23
yor
x
( 6, 3) 3
14
B2 : 23230
22
hor
h
kork
B1 : 23log10 xy
h =3
4
k= 2
3
15(a) 6
(b) (11, 4)
1
1
3472/1 Additional Mathematics Paper 1
SULIT
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Question Solution/ Marking Scheme Answer Marks
16B2: 02 p
B1: jip 3)2( p = 2 3
17 B3:
)42(330,210,270,900000
ofoutany
B2 : )(2
1sin,0cos bothxx
B1 : 0sin
coscos2
x
xx
0
0
00
330
,270
,210,90
4
18B3 : )
32(6 S
B2 : 6r
B1: 12)3
2(
2
1 2 r
12.57 cm
( 4 ) cm4
19B2:
2)1(2
3)1(8
B1:2
2
38
xx
dx
dy
2
13
3
20 B1: 12 xy
or
2
2
)12(
)14(2)8)(12(
x
xxx
dx
dy 2 2
21
B3 : c
12
32
B2:2
2
2
)12(321
xand
x
B1:2
2
2
)12(321
xor
x
2
5
)12(2
3 2
x
xy
4
3472/1 Additional Mathematics Paper 1
SULIT
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3472/1 Additional Mathematics Paper 1
SULIT
5
Question Solution/ Marking Scheme Answer Mark
22
B2 : 210
2)32(54
x
dt
dy
B1 : 210
2)32(5 4
dt
dxor
x
dx
dy
32 units per
second
3
23(b) B1:
411
5)82.0()18.0(C (a)0.18
(b)0.40691
2
24 B2: 34
13
PP
B1: 34
13
PorP
72 3
25(a) B1 :
10
370355
(b) B1 :10
370367
(a) 1.5
(b) 0.3821
2
2
END OF MARKING SCHEME
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SULIT JPNKd/2006/3472/2
Nama Pelajar : Tingkatan 5 : .
3472/2Additional
Mathematics
September 2009
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAHNEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2009
ADDITIONAL MATHEMATICS
Paper 2
.
MARKING SCHEME
SULIT 3472/2
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2
MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2
SPM TRIAL EXAMINATION 2009
N0. SOLUTION MARKS
1
2
2
2
3 23 2 (3 2 1)
0 6 6 3
2 2 1 0
2 ( 2) 4(2)( 1)
2(2)
1.366 0.366
3(1.366) 2 3( 0.366) 2
2.098 3.098
y xx x
x x
x x
x
x or x
y y
= -= - -
=> = - -
- - =
- - -=
= = -
= - = - -
= = -
P1K1 Eliminate y
K1 Solve quadratic
equation
N1
N1
5
2
(a)
(b)
(c)
( )
2 3
( , 3) 3 2 3
3
(4, ) 2 4 3
11
y x
P h h
h
Q k k
k
= +
- - = +
= -
= +
=
OR
3 32 3
0
32 11
4 0
hh
kk
+= = -
--
= =-
1 2
12
2
11 1
4 2
1: 13
2
0 26
m m
y
x
QR y x
y x
= = -
-= -
-
= - +
= =
OR
11 0 1
4 2
111 2
2
26
x
x
-= -
-
= - +
=
x-intercept = 26
Area
3 4 26 31
3 11 0 32
133 78 12 286
2
1385
2
192.5
- -=
- -
= - - + -
= -
=
K1 Use equation or
gradient
N1
N1
P1 For 21
2m = -
K1 Use equation or
gradient
N1
K1 Use formula and
find the area
triangle
N1
8
2kk
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3
N0. SOLUTION MARKS
3
(a)
(b) 2 sin 2 22
xx
p- = -
or
22
xy
p= -
Draw the straight line 22
yp
= -
Number of solutions = 4 .
P1 Negative sineshape correct.
P1 Amplitude = 1
[ Maximum = 3
and Minimum =
1 ]
P1 Two full cycle in
0 x 2p
P1 Shift up the graph
N1 For equation
K1 Sketch the
straight line
N1
7
3
12
2
xy
p= -
y
2
p 2p xO p
2 sin 2x= -
2
2
3p
2kk
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4
N0. SOLUTION MARKS
4
(a)
(b)
(c)
( ) ( )2
1 1 1 2dy
dx= - - - = -
3 2
4 3
4 3
( )
4 3
1 11
4 3
5
12
5
4 3 12
y x x dx
x xy c
c
c
x xy
= -
= - +
= + +
=
= - +
( )2
22
2
2
2
1 0
0@1
3 2
1 1 0
1 1 5 1
4 3 12 3
dy
x xdx
x
d yx x
dx
d yx
dx
y
= - ==
= -
= = >
= - + =
11,
3
Minimumpoint.
N1
K1 Integrate gradient
function
K1 Substitute (1,1)into equation y
N1
K1 Find2
2
d y
dx
N1 N1
7
2kk
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5
N0. SOLUTION MARKS
5
(a)
(i)
(ii)
(b)
(i)
(ii)
11 55050
xx= => =
2
2
2
11 850
50(64 121)
9250
x
x
- =
=> = +
=
New mean = 11 1.8 5 24.8 + =
New variance =2
(8 1.8) 207.36 =
K1 Use formula of
mean and/or
standard deviation
N1
N1
K1 Find new mean
and/or new
variance
N1
N1
6
6
(a)
(b)
( )
5
3
5 1 3 104
1 33
34
l r
a
d
n
n
n
p
p
p
p p p
=
=
=
+ - =
- =
=
( ) ( )3434
2 5 33 32
S p p = +
1853
5821
58.21
cm
m
p=
=
=
Total cost
4 58.21
232.8
233
RM
RM
RM
=
=
=
P1 Value of a and/or d
K1 Use Tn = 104p
N1
K1 Find S34
N1
K1 RM4 S34
N1
7
2kk
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6
N0. SOLUTION MARKS
7
(a)
(b)(i)
(ii)
ll
m 111+-=
xy
y
11.04 0.83 0.71 0.50 0.45 0.33
10.67 0.5 0.4 0.25 0.2 0.1
l
1=y-intercept
l = 5
l
m- = gradient
m = 6.25
P1
N1
N1
K1 for correct axes
and scale
N1 for all points
plotted correctly
N1 for line of
best-fit
K1 for y-intercept
N1
K1 for gradient
N1
10
y
1
x
10.2
0
2kk
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7
N0. SOLUTION MARKS
8
(a)
(b)
(c)
(x 2 )2 = 4 x
Q(3, 1 )
A = [ ]dxxx ---3
0
2)2()4(
= dxxx -3
0
2)3(
=3
0
32
32
3
-
xx
=2
9
Note : If use area of trapezium and dxy , give marks accordingly.
V = -2
0
4
)2( dxxp
=2
0
5
5
)2(
-xp
= p
5
32
K1 Solve for x
N1
K1 use
dxyy - )( 12
K1 integrate
correctly
K1 correct limit
N1
K1 integrate
dxy2
p
K1 integrate
correctly
K1 correct limit
N1
10
2kk
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8
N0. SOLUTION MARKS
9
(i)
(ii)
(b)
(i)
(ii)
(c)
QS= QP + PS
= 2~
x + 3~
y
PR = PS + SR
=~
x + 2~
y
QT = m QS
= m ( 2~
x + 3~
y )
= 2m~
x + 3m~
y
PT = n PR
= n (~
x + 2~
y )
= n~
x + 2n~
y
PQ = PT + TQ
2~
x = n~
x + 2n~
y + 2m~
x 3m~
y
= (n + 2m)~
x + (2n 3m)~
y
n + 2m = 2
2n 3m = 0
m =7
4
n =7
6
K1 for using vector
triangle
N1
N1
N1
N1
K1 for substituting
& grouping into
components
K1 for equating
coefficients
correctly
K1 for eliminating
m or n
N1
N1
10
2kk
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9
N0. SOLUTION MARKS
10
(a)
(b)
(c)
1
1cos
4
1cos
4
POA
POA-
=
=
= 75.52 @75 31"o o
= 1.318 rad.
Arc PA = 20 ( 1.318 ) = 26.36
PQ2
= 802 20
2
PQ = 77.46
Perimeter
= 60 + 26.36 + 77.46
= 163.82 cm
Area OPQ = ( ) ( )1
20 77.46 774.62
=
Area sector POA = ( ) ( )21
20 1.318 263.62
=
Area of shaded region
= 774.6 263.6
= 511 cm2
K1 Use correctly
trigonometric
ratio
N1
K1 Use s rq=
K1 Use Pythagoras
Theorem
K1
N1
K1 Use formula
1
2A bh=
K1 Use formula
21
2A r q=
K1
N1
10
2kk
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10
N0. SOLUTION MARKS
11
(a)
(i)
(ii)
(b)
(i)
(ii)
p = 0.4 q = 0.6 n = 6
P ( )4X
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )4 2 5 1 6 06 6 6
4 5 6
4 5 6
0.4 0.6 0.4 0.6 0.4 0.6
0.13824 0.036864 0.004096
0.1792
P X P X P X
C C C
= = + = + =
= + +
= + +
=
2
288npqs = =
( )( )288
12000.4 0.6
n = =
62m= 8s =
( )
( )
( ) ( )
60 68
60 62 68 62
8 8
0.25 0.75
1 0.25 0.75
1 0.4013 0.2266
0.3721
P X
P Z
P Z
P Z P Z
<
= - -
=
( ) ( )68 0.75 0.22663P X P Z > = > =
0.22663 3000 = 679.89
= 679 @ 680
P1 Value of p and/or q
AND p + q =1
K1 Use P(X = r)
=n
Cr prq
nr
N1
K1
N1
K1 Use Z =s
m-X
K1
N1
K1
N1
10
2kk
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11
N0. SOLUTION MARKS
12
(a)
(b)
(c)
8 ms-1
dva
dt
= =02 2t = 1
t = 12
8 + 2t t2 = 0
(t 4 ) (t + 2) = 0
t = 4 t = 2 (not accepted)
s v dt vdt + 4 50 4
=t t
t t t t + - + + -
4 53 3
2 2
0 4
8 83 3
= ( ) + - - + + - - + -
64 125 64 32 16 0 40 25 32 16
3 3 3
= + -80 103 3
= 30 m
N1
K1
N1
K1
K1
N1
(for t = 4 only)
K1
(for and 4 5
0 4
)
K1
(for integration)
K1
(for summation)
N1
10
2kk
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12
N0. SOLUTION MARKS
13
(a)
(b)
(c)
(d)
.
x= 150 1000 4
(or formula finding y /z)
x = RM 0.60
y = 120
z = RM 2.00
45o
( ) ( ) ( ) ( ) ( )x x x x xI
+ + + += 150 30 110 90 120 75 150 120 120 45360
=46800
360
= 130
P03 = ( )100
5000130
= RM 3846.2
/ .I x08 03 130 1 2 (or 130 + 130x0.2)
= 156
N1
N1
N1
P1
K1
N1
K1
N1
K1
N1
10
2kk
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13
N0. SOLUTION MARKS
14
(a)
(b)
(c)
(d)
40x + 20y 2000 or
2x + y 100
15x +30y 1200 or
x + 2y 80
y 3x
(20, 60)
(35, 30)
2x + y = 100y = 3x
x + 2y = 80
y = 30
100
90
80
70
60
50
40
30
20
10
10080604020 9070503010
x
y
At least one straight line is drawn correctly from inequalities
involving x and y.
All the three straight lines are drawn correctly
Region is correctly shaded
35
Maximum point (20, 60)
Maximum profit = 20(20) + 30(60)
= RM 2200
N1
N1
N1
K1
N1
N1
N1
N1
K1
N1
10
2kk
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N0. SOLUTION MARKS
15
(a)
(b)
(c)
112
= 92
+ 72
2(9)(7)cosPQR
cosPQR =9
126
PQR = 85o 54
PSR = 180o 85o 54
= 94o 6
'
sin sinPRS = 094 66 11
PRS = 32o 57
\ RPS = 180o 94o 6 32o 57= 52o 57
' 'sinsin
oo
RS = 1152 57
94 6
RS = 8.802 cm
Area = ' '( )( )sin ( )( . )sino
o +1 19 7 85 54 6 8 802 94 6 2 2
= 31.42 + 26.34
= 57.76
K1
K1
N1
P1
K1
K1
N1
K1, K1
(for using
area= absinc
and summation)
N1
2kk