mat tamb k2 sep

SULIT 3472/2

3472/2MatematikTambahanKertas 2September 2004

2 jam

JABATAN PENDIDIKAN SABAH

UJIAN SIMILAR 2004

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS QUESTION INI SEHINGGA GIVEN THATTAHU

1. Kertas question ini mengandungi tiga Part : Part A, Part B and Part C.

2. Jawab semua question dalam Part A, empat question daripada Part B and dua question daripada Part C.

3. For setiap question berikan SATU answer/solutions sahaja.

4. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu anda for mendapatkan marks.

5. Diagram yang mengiringi question tidak dilukiskan mengikut skala kecuali distate.

6. Marks yang diperforkan for setiap question and ceraian question ditunjukkan dalam kurungan.

7. Satu senarai rumus disediakan di halaman 2 hingga 4.

8. Sebuah buku sifir matematik empat angka disediakan.

9. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram.

Kertas question ini mengandungi 12 halaman bercetak.

3472/2 © 2004 Hak Cipta Jabatan Pendidikan Sabah

3472/2 SULIT

SULIT 3472/2

Rumus-rumus berikut boleh membantu anda menjawab question. Simbol-simbol yang Given that is yang biasa digunakan.

ALGEBRA

1 x =

2

3

4

5

6

7

8

9 Tn = a + (n -1) d

10 Sn =

11 Tn = ar n - 1

12 Sn =

13

KALKULUS

1 y = uv,

2 y = ,

3

4 Luas di bawah lengkung

or

5 Isipadu janaan

or

STATISTIK

3472/2 SULIT

2

SULIT 3472/2

3472/2 SULIT

3

1 =

2 =

3

4

5

6

7

8 nPr =

9 nCr =

10

11

12 Min = np

13

14

GEOMETRI

1 Jarak =

2 Point tengah

3 Point yang membahagi suatu tembereng garis

4 Luas segitiga

5

6

TRIGONOMETRI

1 Panjang lengkok,

2 Luas sektor,

3 sin2 A + kos2 A = 1

4 sek2 A = 1 + tan2 A

5 kosek2 A = 1 + kot2 A

6 sin 2A = 2 sin A kos A

7 kos 2A = kos2 A – sin2 A

= 2 kos2 A – 1

= 1 – 2 sin2A

8 sin (A B) = sin A kos B kos A sin B

9 kos (A B) = kos A kos B sin A sin B

10 tan (A B) =

11 tan 2A =

12

13 kos A

14 Luas segitiga

Part A

[40 marks]

Answer all the question in this part.

1 Solve the simultaneous equation x – 2y = 1 and + = 6. [5 marks]

2 Given that the function f(x) = - 5 + 7x – 2x 2 .

(a) Without sketching the graph or method of differentiation,

(i) find the coordinate of maximum point.

(ii) state the axis of symmetry . [4 marks]

(b) Sketch the graph of function g(x) = for domain -1 x 4 and state the corresponding range of g(x). [4 marks]

3 Given that the gradient of the curve at the point (1, 2) is 4 and this curve passes through the point (0,1).

(a) Find the value of a, b and c. [5 marks]

(b) Find the normal of equation at the point (1, 2). [3 marks]

A O B

P

Q

Diagram 1

4 Diagram 1 shows two sectors of the circle, OAP and OBQ, with centre at O. Given that rad, OB = 3AO and the ratio of arc AP to the arc BQ is 2 : 3.

(a) Calculate in degree. [4 marks]

(b) Find the perimeter of whole diagram if AB is 12 cm. [3 marks]

5 (a) It is given that X is the number of set 7, 14, 15, 18, 36, 47 and 52. If another number is added to that X, the mean is unchanged. Find the standard deviation of the new set of number. [3 marks]

Height

(cm)120 - 129 130 - 139 140 - 149 150 - 159 160 - 169 170 - 179

Numberof

students8 20 41 60 23 8

Table 1

(b) Table 1 shows the distribution of height which is obtained from 160 students. Find the median of the distribution. [3 marks]

6 Given that = and = .

(a) Calculate . [1 marks]

(b) Find the unit vector in the direction parallel with . [1 marks]

(c) Given that = h - k , and R is point (-2, 9) . Find the value of h

and k. [4 marks]

Part B

[40 marks]

Answer four question from this part.

7 Use the graph paper provided to answer this question.

Table 2 shows the relation between the rates of chemical reaction K, with its temperature T.

Temperature, T (83.3 33.3 21.7 15.1 12.5

The rate of chemical reaction, K

(mol s-1)

39.8 20.0 12.0 5.25 3.17

Table 2

It is known that the rates of chemical reaction, K and its temperature, T, is related

by an equation where A and b are constant.

(a) Covert the equation to the linear form of equation.[1 marks]

(b) Draw a graph against . [5 marks]

(c) From your graph, find the value of a and the value of b. [4 marks]

8 (a) Prove that tan ( 1 + sec 2 ) = tan 2 . [4 marks]

(b) (i) Sketch the graph of y = 2 cos 2x – 1 for 0 x .

(ii) Find the equation of a suitable straight line to solve the equation

cos 2x = .

Hence, on the same axes, sketch the straight line and state the number of

solutions of the equation cos 2x = . [6 marks]

9 (a) A piece of wire of length 60 cm is cut to the form of a circle. When the wire is heat, its length increases with the rate of 0.1 cm s-1.

(use

(i) Calculate the rate of change of the radius of the circle.

(ii) Hence, calculate the radius of the circle after 4 seconds. [5 marks]

(b) (i) Given that where g(x) is a linear function. Find the value

of . [2 marks]

y

y = g(x)

(2, 9)

0 x

Diagram 2

(ii) Diagram 2 shows the graph of a straight line which intersects with the curve at point (2, 9).

Find the area where is bounded by the straight line of , the curve , x-axis and y-axis. [3 marks]

10 (a) A dice is thrown, it is considered as success if once thrown show number 1 or 4. The dice is thrown for five times. If x is a random variable of representing the number of succeed. Calculate

(i) the probability of obtaining at least 4 times of succeed

(ii) the mean,

(iii) the variance. [5 marks]

(b) The height, in cm, of a student in school is obtained as normally distributed with the mean of 160 cm and the variance of 225 cm2. Find

(i) the z- score of height 180 cm,

(ii) the height of a student who corresponding with z-score -0.8,

(iii) If a student is randomly chosen, calculate the height of that student is within the range 160 cm and 180 cm. [5 marks]

11 Solution to this question by scale drawing will not be accepted..

y T (4,6) U(2,4)

0 V (k,0) xDiagram 3

Diagram 3 shows a triangle TUV with T(4, 6), U(2, 4) and V(k, 0).

(a) If T is 90 o , find the value of k. [3 marks]

(b) Point W(x, y) moves along the arc of a circle which centered at U and passing through T.

(i) Find the equation of locus of point W. [3 marks]

(ii) If locus of W cuts with y-axis at two points, state the coordinates of that two points. [2 marks]

(iii) Calculate the area of quadrilateral which formed from the points T , U and 2 points of intersection in question (b) (ii). [2 marks]

Part C[20 marks]

Answer two questions from this part.

12 A particle moves along a straight line in such an acceleration, a ms -2 , is given by a = 2t – 6 with t is the time in seconds after passing through the fixed point O . It is known that the initial velocity of the particle is 5 ms -1 .Find

(a) the velocity of the particle when its acceleration is zero, [4 marks]

(b) the time when the particle is rest momentously, [2 marks]

(c) the total distance of the particle in first three seconds. [4 marks]

13 (a)

Television Year 2000 Year 2002 Price Index

A RM 2150 RM 2260 P

B RM 1960 Q 115

C S RM 1890 120

Table 3

Table 3 shows the price of three types of television A, B and C in the year 2000 and 2002 and the price index for the year 2002 taking year 2000 as the base year. Calculate the value of P, Q and S. [5 marks]

(b)

Food Price Index Weightage

Mi goreng 130 5

Bread 126 3

Milk 128 2

Nasi lemak 115 4

Fruits 120 1

Table 4

Table 4 shows the price index and the weightage of some types of food in a school canteen for the year 2000 taking year 1997 as the base year.

(i) Calculate the composite index for the year 2000 taking the year 1997 as the base year.

(ii) If the composite index continually increases at the same rate from the year 2000 until the year 2003 as the year 1997 until the year 2000, calculate the composite index for the year 2003 taking the year 1997 as the base year.

[5 marks]

14 Gunakan kertas graf for menjawab question ini.

Pak Su sells two types of mi goreng. One pack of common mi goreng needs 120 prawn and 300 g meat, In the other hand, one pack of special mi goreng needs 240 g prawn and 200 g meat. Pak Su only has 8.4 kg prawn and 12 kg meat for making x pack of common and y pack of special. The number of common pack is not more than three times of number of special pack.

(a) Write three inequalities apart from x 0 and y 0 which satisfy the above constraints. [2 marks]

(b) Using a scale of 2 cm to 10 unit at the x-axis and 2 cm to 5 unit at the y-axis , draw the graphs of the three inequalities. Hence, construct and shade the feasible region of R that satisfies the given constraints.

[3 marks]

(c) Based on your graph, answer the following questions:

(i) If Pak Su cooks special mi goreng 10 packs more than common mi goreng, state the maximum number of common and special pack of mi goreng can be cooked by Pak Su?

(ii) If the selling price of one pack of common mi goreng and special mi goreng are RM 5 and RM 7 respectively, calculate the maximum profit that can be obtained? [5 marks]

15

Q R

9 cm 40 0

P 8 cm S

Diagram 4

Diagram 4 shows a quadrilateral PQRS , where QR is parallel with PS and the area of PRS is 31.176 cm 2 . Find

(a) the length of SP [2 marks]

(b) the length of PR [2 marks]

(c) PRQ [2 marks]

(d) the area of PRQ [4 marks]

KERTAS QUESTION TAMAT

UJIAN SIMILAR SEPTEMBER 2004

PERATURAN PEMARKSAN KERTAS 2

Question Solutions and marking scheme Marks

1 x - 2y =1 x = 2y + 1 ………….(i)

+ = 6

y 2 + 5x 2 = 6xy ……………(ii)

Substitute (i) dalam (ii)

y 2 + 5(2y + 1) 2 = 6y(2y + 1)

y 2 + 5(4y 2 + 4y + 1) = 12y 2 + 6y

9y 2 + 14y + 5 = 0

( y + 1)(9y + 5) = 0

y = -1 ,-

x = - 1, -

5

2 (a) (i)

(ii)

f(x) = -5 + 7x – 2x 2

f(x) = -2( x 2 - x) -5

= -2[ ( x - ) 2 – ( ) 2 ] -5 or similar

= -2( x - ) 2 +

max of f(x) at ( 4

7, )

Axis of symmetry : x = 4

P1

K1

K1

N1

N1

K1

K1

N1

N1

(b)shape of curve f(x) and turning point is correct (

4

7, )

Shape of curve g(x) is correct

passing through (-1,14) and(4,9)

range 0 14 4

3 (a)

Substitute point (0, 1) in the curve

Substitute point (1, 2) in the curve

………………(i)

Given that m = 4 at (1, 2)

……………(ii)

(ii) – (i) a = 3

Substitute a = 3 in (i)

b = 2

Hence, a = 3, b = 2 and c = 1. 5

P1

P1

P1

N1

P1

K1

K1

N1

N1

(b)Gradient of normal at the point (1, 2) =

Equation of normal at the point (1, 2) is

3

4 (a)

(b)

Given that OB = 3AO

let AO = j, hence OB = 3j

or

Given that

use the ratio theoram

j = 3 cm

3 0.5=1.5 or 9 0.25 = 2.25

Perimeter = 12 + 3 + 9 + 1.5 + 2.25

= 27.75 cm

4

3

5 (a) = 27

the added number is 27 or similar

baru = 15.41

3

K1

K1

N1

P1

K1

K1

N1

K1

K1

N1

N1

P1P1

(b) median class = 150 – 159

Median = Substitute in the formulae:

= 149.5 +

= 149.5 + 1.833

= 151.33 cm

3

6 (a)

(b)

(c)

= 5

i - j or similar

= h

3 = -2h – k or - 4 = 9h – 5k

-19 = 19 h (eliminate h or k)

h = -1

k = -1

6

7 (a)

L = 149.5

F = 69

K1

N1

N1

P1

N1

K1

K1

N1

N1

(b)

(c)

or similar

0.012 0.030 0.046 0.066 0.080

1.60 1.30 1.08 0.72 0.50

Use the axes and scale correctly.

Plot all points correctly.

The best fit line.

* Refer to attachment A for a straight line graph.

Use the y-intercept to find A

or c = 1.8

A = 63.10

Use the gradient to find the value of b.

1

5

4

8 (a)tan ( 1 + sec 2 ) = tan (1 + )

= tan

= tan

K1

K1

P1

N1

N1

K1

N1

N1

K1

N1

K1

N1

(b) (i)

(b) (ii)

= tan or =

=

= =

= tan 2 or = tan 2

or similar.

Graf y = cos 2x

Graf y = 2 cos 2x

Graf y = 2 cos 2x – 1

Recognise the equation of y = - 1

Draw graph

number of solutions = 2 (from the graph)

*Refer to the Attachment B for graph.

4

6

9 (a)

(i) ( L = circumference)

K1

K1

P1

N1

P1

K1

N1

K1

P1

(ii)

(b) (i)

(ii)

= 0.01592 cm s-1

circumference after 4 seconds= 60 cm + 0.1 (4)

= 60.4 cm

= 60.4

j =

j = 9.613 cm

OR

j = 9.5493 cm

radius after 4 seconds = 9.5493 4(0.1592)

= 9.613 cm

use

area =

5

K1

N1

N1

K1

K1

N1

K1

N1

K1

=

= 23 unit2

5

10 (a)

(i)

(ii)

(iii)

(b) (i)

p = the probability to get number 1 or 4

p = q =

the probability to get at least 4 succeed

=

=

=

= 0.04527

mean = np

=

=

variance = npq

=

=

min = 160 cm

variance = 225 cm2

When x = 180 cm,

=

= 1.333

5

K1

N1

K1

K1

N1

P1

P1

P1

(ii) Given that

x – 160 =

x = 148 cm

(iii) P(height of student is in the range 160 cm and 180 cm)

=

=

=

= 0.4087

5

11 (a)

(b) (i)

(ii)

= -1

k = 10

WU = TU

x 2 – 4x + y 2 – 8y + 12 = 0

Ganti x = 0 and get (y -6)(y -2) = 0

3

3

K1

N1

K1

N1

K1

K1

K1

P1

N1

P1

N1

(iii)

(0,2) and (0,6)

area =

= 8 unit 2

2

2

12 (a)

(b)

(c)

a = 0 , t = 3 s

v =

= t 2 – 6t + c

c = 5 , find c.

v = - 4 ms -1

v = 0

t 2 -6t + 5 = 0

t = 5 s , 1 s

s = (integral)

= (limit correct) or similar

=

= 7

4

2

4

13 (a)

K1

N1

N1

P1

K1

K1

N1

K1

N1

K1

K1

N1

N1

(b) (i)

= 105.1

RM 2254

= RM 1575

Price Index (I) Weightage (w) Iw

130 5 650

126 3 378

128 2 256

115 4 460

120 1 120

5

N1

N1

K1

K1

N1

K1K

1

(ii)

= 124.3

(year 2003 taking year 1997 as the base year)

=

= 154.5

5

14 (a)

(b)

(c) (i)

x 3y all correct

x + 2y 70 only two correct

3x + 2y 120 only one correct

draw at least 1 straight line is correct

draw all 3 straight lines are correct

region and feasible lines are correct

x mak = 16 , ymak = 26 (integer only)

profit mak at (24, 23)

2

3

N1

N1

K1

P2

P1

P0

K1

N1

N1

N1

N1

N1K1

k mak = 5(24) + 7(23)

= 281

* Refer to attachment C for graph.

5

15 (a)

(b)

(c)

(d)

(8)(9) sin RSP = 31.176

RSP =

PR 2 = 8 2 + 9 2 – 2(8)(9)cos RSP

PR 2 = 8 2 + 9 2 – 2(8)(9)cos

PR = 8.544cm

sin RPS =

sin RPS =

PRQ = 65.82 0 or 65 0 49’

RQP = 180 – 40 - 65.82 = 74.18 0

QR =

QR =

QR = 5.708 cm

2

2

2

N1

K1

N1

N1

K1

K1

N1

K1

N1

area PRQ = (PR)(QR) sin PRQ

= (8.544)(5.708) sin 65.82 0

= 22.25 cm 2 (accept 22.24)

OR

PQ =

PQ =

= 8.101 cm

area of PRQ = (PR)(PQ) sin 40 0

= (8.544)(8.101) sin 40 0

= 22.25 cm 2

4

ATTACHMENT A

Question 7(b)

K1

N1

N1

N1

K1

K1

ATTACHMENT B

Question 8 (b)

ATTACHMENT C

Question 14 (b)

UJIAN SIMILARMATEMATIK TAMBAHAN SPM

R

Penyelaras: Judy Lian Yee LingSektor Pengurusan Akademik, Jabatan Pendidikan Sabah

Ahli Panel:1. Cheong Nyok Tai, SMK (P) Likas, Kota Kinabalu2. Chung Su Kiong, SMJK Tiong Hua, Sandakan3. Phoon Chiew Fun, SMK Tenom, Tenom4. Shirney Chua, SMK Holy Trinity, Tawau5. Surianih Sewan, SMK Tamparuli, Tamparuli6. Theresa Lee Choon Moi, SMK St. Michael Penampang, Penampang7. Ting Kai Chu, Maktab Sabah, Kota Kinabalu8. Yeoh Sek Goh, SMK Agaseh, Lahad Datu