Kelantan mtambahan + skema

Name : ……………………………………………………
0

Form : …………………………………………………….

SMKA NAIM LILBANAT 15150 KOTA BHARU KELANTAN.
“SEKOLAH BERPRESTASI TINGGI”

PEPERIKSAAN PERCUBAAN SPM 2013
ADDITIONAL MATHEMATICS
Kertas 1

3472/1
2 Jam

2 Jam

Untuk Kegunaan Pemeriksa
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

Markah
Penuh
2
3
3
3
2
3
3
3
4
2
3
3
4
3
3
4
4
4
3
3
4
3
4
3
4

JUMLAH

80

Soalan
Arahan:
1. Kertas soalan ini mengandungi 25 Soalan.
2. Jawab semua soalan.
3. Tulis jawapan anda dalam ruang yang
disediakan dalam kertas soalan.
4. Tunjukkan langkah-langkah penting dalam
kerja mengira anda. Ini boleh membantu
anda untuk mendapatkan markah.
5. Anda dibenarkan menggunakan kakkulator
saintifik.

3472/1

3472/1

Kertas soalan ini mengandungi 11 halaman bercetak.

http://edu.joshuatly.com/
http://fb.me/edu.joshuatly

Markah
Diperoleh

1
(Answer all questions)
Jawab semua soalan

1.

Diagram 1 shows the graph of function f ( x)  2 x  4 for domain 0  x  5
Rajah 1 menunjukkan graf bagi fungsi f ( x)  2 x  4 , untuk domain 0  x  5 .

y
● (5, 6)
4●

O

y = f(x)

●
2
Diagram 1 / Rajah 1

x

State

Nyatakan

(a)
(b)

f (0),
range of f(x) corresponding to the given domain.
julat f(x) berdasarkan domain yang diberi.

Answer / Jawapan :
(a)

[2 markah]

1
(b)
2
2.

Given the function g ( x)  3x  2 and gh( x)  12 x  4 ,
Diberi fungsi g ( x)  3x  2 dan gh( x)  12 x  4 ,
find / cari
(a) h(x),
(b) value of x such that h(x) maps onto itself.

nilai x dengan keadaan h(x) memeta kepada dirinya sendiri.

Answer / Jawapan:
[3 markah]

(a)

2
3

3472/1

(b)


2
3.

Diagram 3 shows the function f : x  kx  5 , where k is a constant.
Rajah 3 menunjukkan fungsi f : x  kx  5 , dengan keadaan k ialah pemalar.
f

x

kx+5

3

14
Diagram 3 / Rajah 3
Find the value of k
Cari nilai k.

[3 markah]

Answer / Jawapan :
3
3
4.

Straight line y  2x  1 is tangent to the curve y  x 2  p.
Garis lurus y  2x  1 ialah tangen kepada lengkung y  x 2  p.
Find the value of p
Cari nilai p.

[3 markah]

Answer / Jawapan :

4
3

3472/1


3
5.

Sketch the graph of the function f ( x)  ( x  3)2  4 on the given axes for 0 ≤ x ≤ 5
Lakar pada paksi-paksi yang diberi, graf fungsi kuadratik f ( x)  ( x  3)2  4 ,
untuk 0 ≤ x ≤ 5.
f (x)
Answer / Jawapan :

5

O

x
[2 markah]

2
6.

Given that f(x) = x( x  3)  10 . Find the range of values of x when f (x) ≥ 0.
Diberi f(x) = x( x  3)  10 . Cari julat nilai x apabila f (x) ≥ 0.
[3 markah]
Answer / Jawapan :

6
3
7.

Solve the equation

Selesaikan persamaan 4 x + 2 – 4 x + 1 = 6.

[3 markah]

Answer / Jawapan :

7
3
8.

Solve the equation log3 6 x  2  log3 (3x  1)
Selesaikan persamaan log3 6 x  2  log3 (3x  1)
.

8
:

Answer / Jawapan

3

3472/1


[3 markah]

4
9.

 4m 
Given that log 2 m  a and log 2 n  b , express log8 
 in terms of a and b.
 n 
Diberi log 2 m  a dan log 2 n  b , ungkapkan log8  4m  dalam sebutan a dan b.


 n 
Answer / Jawapan :

[4 markah]

9
4
10.

The first three terms of an arithmetic progression are x - 27, 12, y .
Tiga sebutan pertama bagi suatu janjang aritmetik ialah, x - 27, 12, y .

Find the value of x + y.
Cari nilai x +y

[2 markah]

Answer / Jawapan :

10
2
11.

The sum of n first terms of arithmetic progression is given by 3(2n + 1).

Hasil tambah n sebutan pertama suatu janjang aritmetik diberi oleh 3(2n + 1).
Find / Cari

(a)

first term

(b)

common different

sebutan pertama,

beza sepunya.

[3 markah]

Answer / Jawapan :
(a)

11
(b)
3
3472/1


5
12. The first term of geometric progression is 6 and the common ratio is 2.
Find the sum from the forth term to the eight term of the progression.
Sebutan pertama bagi janjang geometri ialah 6 dan beza sepunya ialah 2.
Cari hasil tambah dari sebutan keempat hingga sebutan kelapan janjang itu.

Answer / Jawapan :

[3 markah]

12
3
13. The variables x and y are related by the equation y  8h x , where h is a constant.
Diagram 13 shows the straight line obtained by plotting log 2 y against x
Pembolehubah x dan y dihubungkan oleh persamaan y  8h x , dengan keadaan h
ialah pemalar.
Rajah 13 menunjukkan graf garis lurus yang diperolehi dengan memplotkan log 2 y
melawan x.
log 2 y

 (2, 7)

(0, k)
O

x
Diagram / Rajah 13

Find the value of
Cari nilai

(a) k
(b) h .
13

Answer / Jawapan :
4

(a)
(b)

3472/1


[4 markah]

6
14.

Diagram 14 shows the graphs of a straight line.
Rajah 14 menunjukkan graf bagi satu garis lurus.
y

●

A(-4, 6)
C(5, 3)
●

●

O

B(11, 1)
x

Diagram / Rajah 12

x

Given that point C divides the line segment AB such that

AC m
 , find m : n.
CB n

Diberi titik C membahagi garis lurus AB dengan keadaan AC  m , cari m : n.

Answer / Jawapan

CB

n

[3 markah]

14
3
15.

Given that u = 2i + 3j and v = 2i + kj , where k is a constant, find the values of k
if 2u  v = 10.
Diberi u = 2i + 3j dan v = 2i + kj , dengan keadaan k ialah pemalar, cari nilai-nilai k
apabila 2u  v = 10.

[3 markah]

Answer / Jawapan :

15
3

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7
16.

Diagram 16 shows a triangle PQR.

Rajah 16 menunjukkan segitiga PQR.

y

.

Q(-1, 2)
R(k, m)
O

x

P(-3, -2)
Diagram / Rajah 16
Given PQ  QR  5i  3 j , find the value of k and of m.
Diberi PQ  QR  5i  3 j , cari nilai k dan nilai m.

Answer / Jawapan :

[4 markah]

16
4
17.

Solve the equation 4 sin x cos x = 1, for 0≤ x ≤ 360.
Selesaikan persamaan 4 sin x kos x = 1, untuk 0≤ x ≤ 360.
Answer / Jawapan

17
4

3472/1


[4 markah]

8
18.

Diagram 18 shows a sector OAB of a circle with centre O and radius is 10 cm
Rajah 18 menunjukkan suatu sektor OAB dengan pusat O dan berjejari 10 cm.

B

10 cm


O

P

A

Diagram / Rajah 18
Given P, A and B are the points such that OP = PA and  OPB = 90o, find,
Diberi P, A dan B adalah titik-titik dengan keadaan OP = PA dan  OPB = 90o , cari
[Use / Guna  = 3.142 ]
(a)  AOB , in radian
dalam radian,
(b) area, in cm2, of the shaded region.
luas, in cm2 , kawasan berlorek.

Answer / Jawapan :
(a)

[4 markah]

18
(b)
4
19.

The radius of a circle decreasing by 0.5 cm s 1 . Find the rate of area of circle
when the radius is 4 cm.
Jejari suatu bulatan menyusut dengan kadar 0.5 cm s 1 . Cari kadar perubahan bagi
luas bulatan apabila jejari bulatan itu ialah 4 cm.

[3 markah]

Answer / Jawapan :

19
3

3472/1


9
20.

Given f ( x)  2 x3  px 2  5x , where p is a constant.
Diberi f ( x)  2 x3  px 2  5x , dengan keadaan p ialah pemalar.
1
Find the value of p when f "    4 p
2
Cari nilai p apabila f "  1   4 p .
 
2

[3 markah]

Answer / Jawapan :

20
3
21.

Given

d
4
, find the value of
(3  2 x) 2 =
dx
(3  2 x)3

Diberi d (3  2 x)2 =
dx

4
, cari nilai
(3  2 x)3



2

1
2



1
dx
(3  2 x)3

1
dx
(3  2 x)3

Answer / Jawapan :

21
4

3472/1

2

1
2


[4 markah ]

10
A set of data 1, 3, 4, 5, 7, 8,10 and 11.

22.

Suatu set data terdiri daripada 1, 3, 4, 5, 7, 8,10 dan 11.

Determine,
Tentukan

(a) median,/ median
(b) interquartile range

julat antara kuartil bagi data itu

Answer / Jawapan
(a)
.

(b)

[3 markah]

22
3
A group of students which consists of 3 boys and 5 girls to be arrange in a row.
Calculate the number of possible ways if,

23.

Sekumpulan murid yang terdiri daripada 3 orang murid lelaki dan 5 orang murid
perempuan hendak disusun dalam satu baris. Hitungkan bilangan cara susunan berlainan
yang mungkin jika.

(a) no condition is imposed
tiada syarat dikenakan

(b) all the boys sit next to each other.

semua murid lelaki duduk bersebelahan antara satu sama lain

[4 markah]

Answer / Jawapan :
(a)
(b)

.
.

23
4

3472/1


11
24.

A bag contains four yellow marbles and six green marbles. Two marbles are drawn
at random from the bag one after another without replacement. Find the probability
that the two marbles drawn are of difference colour.
Sebuah bag menagndungi 4 biji guli kuning dan 6 biji guli hijau. Dua biji dipilih secara rawak
daripada bag itu, satu demi satu tanpa pengembalian. Cari kebarangkalian dua biji yang dipilh
adalah berlainan warna.

Answer/ Jawapan:

[3 markah]

24
3
25. In a Mathematics test, 30% of the students who sat the test failed to obtain 50 marks.
If 8 students are selected from those who sat for the test, find the probability that
Dalam suatu ujian Matematik, 30% daripada pelajar gagal mencapai 50 markah.
Jika 8 orang pelajar dipilih, hitung kebarangkalian bahawa

(a) half of them failed to obtain 50 marks.

separuh daripada pelajar itu gagal mencapai 50 markah.

(b) at least 7 of them failed to obtain 50 marks.

sekurang-kurannya 7 orang gagal mencapai 50 markah.

Answer / Jawapan :
[4 markah]

25
4

3472/1


12
SEK. MEN. KEB. AGAMA NAIM LILBANAT
SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 1
No
Solution and Mark Scheme
(a) 4
1

Sub Marks
1

0  f ( x)  6

1

(a)
2

(b)

4x  2

2

2

1

2
3

Total Marks

(b)



3

B1 : 4 x  2  x
3
3

3

B2 : 3k  5  14

3

B1 : f (3)  14
2
4

3
2

B2 : (2)  4(1)( p  1)  0
2

3

2

B1 : x  p  2 x  1 or x  2 x  p  1  0
B1 : bentuk minimum.
B1 : melalui mana-mana dua titik
f(x)

5

2

(0, 13) 



(5, 8)

(3, 4)
O

x

x  2, x  5
B2 :
6

3472/1

( x  2)( x  5)  0

B1 :

x 2  3x 10  0

3


3

13


7

1
2

B2 :
B1 :

3

1
2
4 x (16)  4 x (4)  6
3

3
7
B2 :

8

3

4x 

 6x 
2

3
3x  1 


3

 6x 
B1 : log 3 
2
 3x  1 

(a)

4

2ab
3

B3 :
B2 :

9

2  log 2 m  log 2 n
3

log 2 4m  log 2 n
log 2 8

4

 4m 
log 2 

 n 
B1 :
log 2 8
51
10

2
2

B1 : 12  ( x  27)  y 12
(a)

11

9

1

(b)

3

2

B1 :

1488

d  T2  T1 or T2  S 2  S1
3

B2 :

S8  S3  1530  42

B1 :

6(23  1)
S3 
2 1

12

3472/1

3

or

6(28  1)
S8 
2 1


3

14
(a)

3
B1 :

13

(b)

1
4
B1 :

2

log 2 y  log 2 8  x log 2 h
2

(2,7) : 7  3  2(log 2 h) or

gradient =

73
=2
20

3:2
B2 :

3

2m  3n

14

3
B1 :

11m  (4)n
1m  6n
3
 5 or
mn
mn

2 and -14

3

6 2  (6  k ) 2  10

B2 :
15

4

2 or  14

B1 :

6i  (6  k ) j  10

or

k  2 and m  1

3

4

B3 :

k  2 or m  1

B2 :

k  3  5 or m  2  3


16

B1 :

PQ  2i  4 j

or



QR  (k  1)i  (m  2) j

3472/1


4

15

150 , 750 , 195o , 255o

4

B3 : betul mana-mana 3 nilai sudut
B2 : betul mana-mana 2 nilai sudut or

sin(2 x)  0.5

17

(a)

1.047 rad
B1 :

18

(b)

cos  

2

5
10

30.699 or 30.70
B1 :

B2 :

3472/1

B1 :

2

4

1
(10) 2 (1.047)
2
1
(5)(10) sin 60
or
2

 4

19

4

B1 : betul salah satu nilai sudut or
30o, 150o or 390o, 510o

3

dA dA dr

x
dt dr dt
dA
 (8 ) x(0.5)
dt
dA
 2r or
dr
dr
 0.5
dt


3

16
3
B2 :

20

B1 :

3
16
B3 :
21

B2 :

B1 :

22

(a)
(b)

B1 :
(a)

23

6
5.5

4320
B1 :

8
15
B2 :
24

B1 :

(a)
25

(b)

1
12( )  2 p  4 p
2
f ' ( x)  6 x 2  2 px  5 or
f "( x)  12 x  2 p

1
1
1
[(
)(
)]
4 (3  2(2)) 2
(3  2(0.5)
1
1
1
dx  [
 (3  2 x)3 4 (3  2 x) 2 ]

4

4
1
dx 
 (3  2 x)3 (3  2 x) 2
1
2

Q1  3.5 or Q3  9

3

2

8! or 8P8

3!x5!x6 or 3! x6!

2

4

3

12
30
12
30
)  ( ) or ( )  ( )
90
90
90
90
4 3
6 5
x or
x
10 9
10 9
4 6
6 4
or
x or
x
10 9
10 9

1 (

0.1361
4
4
B1 : 8C 4 (0.3) (0.7)
0.001290

7
1
8
0
B1 : 8C7 (0.3) (0.7) or 8C8 (0.3) (0.7)

3472/1

3

4

40 320
B1 :

(b)

3


3

2
2

4

17

3472/1


0
Name : ……………………………………………………
Form : …………………………………………………….

SMKA NAIM LILBANAT 15150 KOTA BHARU KELANTAN.
“SEKOLAH BERPRESTASI TINGGI”


3472/2

ADDITIONAL MATHEMATICS
Kertas 2
2 ½ Jam

2 ½ Jam

Untuk Kegunaan Pemeriksa
Soalan
1

6
10

8

10
10
10
10

12

10

13

10

14

10

15

10

JUMLAH

3472/2
3472/2

8

11

C

8

7
B

4

10

3. Write your answers on the paper sheets
provided.

6

6

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

3

5

A

7

9

1. This question paper consists of three sections:
Section A, Section B and Section C.

5

2

Arahan:

Markah
Penuh

100

Kertas soalan ini mengandungi 7 halaman bercetak.

Markah
Diperoleh

1
Section A
[40 marks]
(Answer all questions)
1.

Solve the simultaneous equations

and

Give your answers correct to three decimal places.
[5 marks]
(a)

Skatch the graph of y  2  3 sin x for 0  x  2

(b)

2.

[4 marks]
By using the same axes, sketch a suitable straight line to find the number of the solutions
1
of the equation 3 sin x  x  0 .



[3 marks]

3.

A curve has a gradient function x 2  kx  p with turning points (2, 0) and (1,
and p are constants.
Find
(a)
the value of k and of p,
(b)

4.

9
) , where k
2

[3 marks]

equation of curve.

[3 marks]

Diagram 4 shows of n siries of circle. The total of perimeter of all circle is 144  cm. Given the
radius of smallest circle is 2cm and radius of the biggest circle is 16 cm. The arrangement of
circles form a arithmetic progression.

16 cm

2 cm

Diagram 4
Find
(a)

the value of n,
[3 marks]

(b)

perimeter of the sixth circle in term of .
[3 marks]

(c)

total perimeter for the first six circles in term of π
[2 marks]

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2
5.

Diagram 5 shows two triangles ABC and ABD. Point E lies on the straight line AD.
C

10 y

D

E

~

6y
~

A

B

8x
~

Diagram 5






It is given that AC  10 y , BD  6 y and AB  8 x
(a)

~

~

~

Express in terms of x and y
~

~



(i) AD


(ii) BC
[2 marks]
(b)











It is given that AE  m AD and EC  k BC , Express AE in terms of
(i) m, x and y
~

~

(ii) k , x and y
~

~

[3 marks]
(c)

Hence, find the value of m and of k.
[3 marks]

6.

The mean of set numbers k, (k+1), (2k-1), and (2k+4) is 7.
(a)

Find the value of k
[2 marks]

(b)

Each number in the set is divided by 7 and then 1 added to it.
Find
(i) new mean
(ii) the new standard deviation.
[4 marks]

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3
Section B

[40 marks]
(Answer any four questions from this section)
7.

Use graph paper to answer this question.
Table 7 shows the values of two variables, x and y, obtained from an experiment. It is known
k p
that x and y are related by the equation y   2 , where p and k are constants.
x x
x
1.0
1.5
2.0
2.5
3.0
3.5
y
3.80
5.20
4.95
4.48
4.00
3.57
Table 7
1
1
Plot xy against by using a scale to 2 cm to 0.2 units on -axis and 2 cm to 2 unit on
(a)
x
x
xy - axis.
Hence, draw the line of best fit.
[4 marks]
(b)

Use the graph in 7(a) to find the value of
(i) k and p.
(ii) y when x = 1.72.
[6 marks]

8.

Diagram 8 shows a part of curve y 

4
which passes through A (-1, 4)
(2x  1) 2

y

A (-1, 4)
y

4
(2x  1) 2

Diagram 8
(a)

-3

-2

O

x

Find the equation of tangent at point A.
[4 marks]

(b)

3472/2

The shaded region is baounded by x-axis, straight line x  3 and x  2 .
(i) Find the area of the shaded region..
(ii) Find the volume revolution, in term of π , when the shaded region is rotated through
360o about the x - axis
[6 marks]

4
9.

Solution by scale drawing is not accepted.
Diagram 9 shows a trapeziuem ABCD. Given the equation of straight line DC is 3y  2x  2
y
C(7, 9)

D

B(k, 0)

O

x

A(4, -1)
Diagram 9

Find,
(a)

the value of k
[2 marks]

(b)

the equation of straight line AD.
[2 marks]

(c)

coordinates of D
[3 marks]

(d)

equation of the perpendicular bisector of AC.
[3 marks]

3472/2


5
10.

In diagram 10, OKLM is a sector of circle with centre O and radius 17 cm.
L

S

M
17 cm

K

8 cm
O
Rajah 10

Calculate
(a)

MOK, in radian
[3 marks]

(b)

the perimeter, in cm, of the shaded region.

(c)

the area , in cm , of the shaded region.

[3 marks]
2

[4 marks]

11.

(a)

The result of a survey in an urban area shows the the probability of a student having a
mobile phone is k. The mean and variance of n students chosen at random having a
mobile phone are 360 and 72 recpectively.
Find the value of n and of k.

(b)

[5 marks]
A group of worker are given medical check up. The blood pressure of the workers
have a normal distribution with a mean of 140 mmHG.and a standard deviation of 10
mmHg. Blood pressure that is more than 150 mmHg is classified as
”high blood presure”
(i) A worker is chosen at random from the group. Calculate the probability that the
worker has a blood pressure between 135 mmHg and 145 mmHg.
(ii) It is found that 80 workers have ”high blood pressure” . Find the total number of
workers in the group.
[5 marks]

3472/2


6
SECTION C
[20 marks]
(Answer any two questions from this section)
12.

Diagram 12 shows a quadrilateral PQRS.
S
R

110
7 cm

P

o

6.3cm

52 o
Q

18.2 cm
Diagram 12

(a)

Calculate
(i) the length, in cm , of QS
(ii) PQS
[4 marks]

(b)

Point Q ' lies on the QS such that Q ' R  QR
(i) Sketch a triangle Q ' RS
(ii) Calculate the area, in cm 2 , of triangle Q ' RS
[6 marks]

13.

Table 13 shows the price of four items P, Q, R and S in the year 2002 and year 2005. Price
index and weightage for the year 2005 based on the year 2002.
Items
P
Q
R
S
(a)

Price on 2002
(RM)
6.00
5.50
5.60
8.00

Price on 2005
(RM)
7.20
7.70
y
8.80
Table 13

Price index for the
year 2005 based
2002
120
x
125
110

weightage
4
2
3
1

Find the value of
(i) x
(ii) y
[2 marks]

(b)

3472/2

Calculate the composite index of the price of those items for the year 2005 based on
the year 2002.
[3 marks]

7
(c)

The total cost of all items in the year 2002 is RM 9600. Calculate the corresponding
cost of items in the year 2005.
[2 marks]

(d)

The price of items P and Q are increase by 10 % and the price of items R and S are
increase by 5% fom the year 2005 to the year 2007. Find the composite index for the
year 2007 based on the year 2002.
[3 marks]

14.

Use graph paper to answer this question.
The member of a Naim’s Teacher Club plan to organise a picnic. They agree to rent x bus
and y van. The rental of a bus is RM 900 and the rental of a van is RM400. The rental of the
vehicles for the trip is based on the following constrains.
I : The total number of vehicles to be rented is not more than 9.
II : The number of bus is at most twice the number of vans.
III : The minimum allocation for the rental of the vehicles are RM3 600
(a)

Write three inequalities, other than x  0 and y  0 , which satisfy all the above
constrain.
[3 marks]

(b)

Using a scale of 2 cm to 1 vehicle on both axes, construck and shade the region R
which satisfies all the above constrains.
[3 marks]
Using the graph constructed in 14(b), find
(i) the maximum number of van rented if 4 buses are rented.
(ii) the maximum number of members that can be accommodated into the rented
vehicles if a bus can accommodate 45 passengers and a van can accommodate 20
passengers.
[4 marks]

(c)

15.

A particle moves in a straight line and passes through a fixed point O. Its acceleration, a ms-2,
given by a  2t  8 , where t is the time, in s, after passing through O. The initial velocity is
12 ms-1.
Find
(a)

the minimum velocity, in ms-1, of the particle.
[4 marks]

(b)

the time, in s, at which the particle in instantaneously at rest.
[2 marks]

(c)

the total distance, in m, travelled by the particle in the first 4 seconds.
[4 marks]

3472/2


8
SEK. MEN. KEB. AGAMA NAIM LILBANAT
SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2
NO
1.

SOLUTIONS

MARKS TOTAL
P1

y 3
y4
y = 2x + 3
or
x
x
2
2
2
 y 3
 y 3
2
x2-2(2x +3)2 –x(2x+3)+27= 0 or 
 y  27  0
  2y  
 2 
 2 
x2 +3x – 1 = 0
or y2 = 13

x

K1

 3  32  4(1)(1)
2(1)

or y 

K1

(2)  (2)2  4(1)(28)
2(1)

x = 0.303 , – 3.303 (both)
y =3.6060, – 3.606 (both)
2(a)

5

N1
N1

y
5

2

x

0
-1

shape of sine curve
1 cycle for 0  x  2

P1

maximum = 5 and minimum = -1

(b)

P1

P1

straight line y  2 

1



N1
K1 N1

x

No. of solutions = 2

3(a)
3472/2

7

N1

K1

dy
 x 2  kx  p
dx


6

9
(2,0) : 4  2k  p  0
9
(1, ) : 1  k  p  0
2

K1

k  1 and p  2
(b)

y   ( x2  x  2) dx



4(a)

N1

x3 x 2
  2x  c
3 2

K1

9
gantian (2, 0) or (1, ) :
2
3
2
x
x
10
y

 2x 
3
2
3
a  2 (2)  4

l  2 (16)  32

K1
N1
P1

or

n
4  32   144
2
n8

Sn 

(b)

K1
N1
K1
N1

T8  4  7d  32
d  4
T6  4  5(4 )

N1

 24

(c)

5(a)

6
2(4 )  5(4 
2
 84

S6 





K1
N1



(i) AD  AB  BD
 8 x 6 y
~



N1

~




(ii) BC  BA AC

 8 x 10 y
~

3472/2

N1

~


8

10
5.(b)





(i) AE  mAD
 8m x 6m y
~



P1

~



(ii) EC  kBC






EA AC  k BC


EA  8k x 10k y  10 y
~

~

~



N1

AE  8k x  (10  10k ) y
~

(c)

K1

~

8

8m  8k
mk

K1

or 6m  10  10k

N1

5
8
5
n
8
m

6(a)

N1

k  k  1  2k  1  2k  4
7
4
k4

(i) New mean 

7
1  2
7

K1
N1
N1

(ii) 4, 5, 7, 12

4 2  5 2  7 2  12 2
 (7 ) 2
4
 3.08

K1
N1

3.08
 0.44
7

N1

Standard deviation 

New standard deviation 

3472/2


6

11
7 (a)

xy

3.80

7.80

9.90

11.20

12.00

12.50

1
x

1.00

0.67

0.50

0.40

0.33

0.29

N1
N1

Both axes correct (at least plotting 1 point)
Plotting all 6 points – correct
line of the best fit - correct.
(b)

K1
K1
N1

p
(i) xy  k  x
k  16

P1
N1

p  gradient  

(ii) x  1.72 ,

16
 12.31
1.3

1
 0.58
x

N1
K1

xy  9
9
y
 5.23
1.72
8 (a)

10

N1

4
(2 x  1) 2
dy
16

dx
(2 x  1) 3
dy
A(1, 4);
 16
dx
y

K1
N1

Equation of tangent :

y  4  16( x  1)
y  16 x  20
(b)

K1
N1

(i) Area of the shaded region:
1

Integrate

 4(2 x  1)

2

0



10

dx

4(2 x  1) 1
2(1)

K1

2

4(2x  1) 1
4


2(1) 3
15


3472/2

4
unit 2 @ 0.267
15


K1
N1

12
(ii) Volume:
2

Integrate = π 16(2 x  1)  4 dx

K1

3

(c)

2



8
= 
3
 3(2 x  1)  3

N1

= 0.077 or 784  unit 3 or 0.243

N1

10125

9 (a)

3y  2x  2
2
2
y  x
3
3
m DC 

2
3

0 1 2

k 4 3
11
k
2

 m AB 

(b)

m AD  

K1
N1

3
2

3
 equation of AD : y  1   ( x  4)
2
2y  3x  10
(c)

3 y  2 x  2 ..... (1)
2 y  3x  10 ..... (2)

N1
K1 K1
N1

D(2, 2)
(d)

K1

 4  7 1  9 

,

2 
 2
Mid-point AC
 11 
  , 4
2


P1

Equation bisector AC:

3
11
(x  )
10
2
20y  6 x  113

K1

y4  

N1

3472/2


10

13
10 (a)

(b)

(c)

11(a)

1
 8
cos MOK  
2
 17
1
MOK  61.9275 o
2
123.855 x
MOK 
 2.162 rad
180
2

2

Perimter  2( 17  8 )  17(2.162)
 66.754 cm

1
1
(17 2 )(2.162)  (17 2 ) sin(123.855)
2
2
 312.409  120
 192.409

Area 

  nk  360
  nkq  72
2

nk  360
360
n
0.8
 450

N1
K1 K1
K1
N1

N1

  140   10

145  140 
 135  140
P
z

10
 10

P 0.5  z  0.5

10
K1
N1

 0.3829
(ii)

150  140 

(ii ) P z 

10


P z  1.0
 0.1587
80
 0.1587
N

10

K1

 2(0.1915)

3472/2

K1 K1

N1

 k  0.8

(c)

N1

K1

q  0.2

(i)

N1

P1

and

360q  72

(b)

K1

K1

 N  504 / 505

K1
N1


14
12(a)

(i ) QS  7 2  18.2 2  2(7)(18.2) cos(52 o )
 223.3695
 14.946 cm

K1
N1

sinPQS sin 52 o

7
14.946
sin PQS  0.3691
(ii )

PQS  21.66 o / 21o 40 '

(b)

K1
N1

(i)

S

N1

R

Q’

6.3 cm
Q2

(ii)

sin QSR sin 110

6.3
14.946
QSR  23 o 20 '

o

2
cm

K1
N1

 SQR  180 o  110 o  23 o 20 '

and

 46 o 40 '

 QRQ '  180 o  2(46 o 40 ' )

N1

 86 o 40 '

1
1
Area of Q ' RS  ( 14.946 )( 6.3 )( sin 46 o 40' ) - ( 6.3)(6.3) sin 86 o 40 ' )
2
2
 34.2447  19.8114
 14.433 cm 2

3472/2


K1

N1

10

15
13 (a)

K1

Q1
100
Qo
x  140,
y7

Use I 

N1

120(4)  140(2)  125(3)  110(1)
4  2  3 1
 124.5

(b)

I 05 / 02 

K1
N1
N1

(c)

P05
x100  124.5
9600
P05  RM 11 952

K1

(d)

120x1.1= 132
140x1.1=154
125x1.05=131.25
110x1.05=115.5

4
2
3
1

132 x 4  154 x 2  131.25 x3  115.5 x1
10
 134.525

I 07/02 

14 (a)

x y 9`
x  2 y or y 

P1

K1
N1

N1

1
x
2

N1

900 x  400 y  3600 or 9 x  4 y  36
(b)

N1

Draw correctly all three straight line which involves x and y.

K1

Region shaded correctly
(c)

N1 N1

(i) x  4, y  5

N1

(ii) Use 45x  20 y for point in the shaded region
(6, 3) : 45(6)  20(3)
 330

3472/2

10

w

I 07 / 02
P
Q
R
S

N1

K1
N1
N1


10

16
15(a)

v   2t  8 dt

K1

 t 2  8t  12

V minimum, a =0
2t  8  0
t4

N1

Vmin  (4 2 )  8(4)  12

K1
N1

 4ms 1
(b)

rest, v =0
t 2  8t  12  0

K1

(t  2)(t  6)  0
t  2, t  6

(c)

N1

s   t 2  8t  12 dt
t3
  4t 2  12t
3

K1

2

total distance   vdt 
0



4

2

vdt
2

t 3

t3
   4t 2  12t    4t 2  12t
3
0 3
32
16


3
3
 16 m

3472/2

4

K1 K1

2


N1

10

17

3472/2


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