8/13/2019 Ujian Mac Math t4 2014 (k2)

1/14

1

[100 marks]Answer all questions from this section.

Jawab semua soalan daripada bahagian ini.

1 Solve the equation x2 x 16 = 2(1 x)

Selesaikan persamaan x2

x 16 = 2(1 x)

[4 marks]Answer:

2 Solve the equation 4( x2 + 3) 16 x = 0Selesaikan persamaan 4( x2 + 3) 16 x = 0 .

[4 marks]Answer:

3 Solve the equation 2 t =

4t + 23 .

Selesaikan persamaan 2t =4t 2 + 2

3 .

[4 marks]Answer:

8/13/2019 Ujian Mac Math t4 2014 (k2)

2/14

2

4 Calculate the value of x and of y that satisfy the following simultaneous linear equations: Hitungkan nilai x dan nilai y yang memuaskan persamaan linear serentak berikut:

3 x 2 y 12 = 07 x + 6 y + 4 = 0

[5 marks]Answer:

5 Diagram 1 shows two sectors OFG and OHI with the same centre O. OFGJ is a quadrant of acircle with centre O. OGH and OJI are straight lines.

Rajah 1 menunjukkan dua sektor bulatan OFG dan OHI yang sama-sama berpusat O. OFGJ ialah sukuanbulatan berpusat O. OGH dan OJI ialah garis lurus.

Diagram 1 OJ = JI = 14 cm and HOI = 45.OJ = JI = 14 cm dan HOI = 45.

Using =227 , calculate

Dengan menggunakan =227 , hitungkan

(a) the perimeter, in cm, of the whole diagram,erimeter, dalam cm, seluruh rajah itu,

(b) the area, in cm 2, of the shaded region.luas, dalam cm 2 , kawasan yang berlorek.

[6 marks]Answer:

8/13/2019 Ujian Mac Math t4 2014 (k2)

3/14

3

6 Diagram 2 shows two sectors OAB and OCDE with the same centre O. OFE is a semicircle withdiameter OE and OE = 2 AO. AOE and OBC are straight lines.

Rajah 2 menunjukkan dua sektor bulatan OAB dan OCDE yang sama-sama berpusat O. OFE ialah semibulatan dengan OE sebagai diameter dan OE = 2AO. AOE dan OBC ialah garis lurus.

Diagram 2

AO = 14 cm and AOB = 45. AO = 14 cm dan AOB = 45.

Using =227 , calculate

Dengan menggunakan = 227 , hitungkan

(a) the perimeter, in cm, of the whole diagram,erimeter, dalam cm, seluruh rajah itu,

(b) the area, in cm , of the shaded region.luas, dalam cm 2 , kawasan yang berlorek.

[6 marks]Answer:

7 The Venn diagram in the answer space shows sets P , Q and R such that the universal set = P

Q R.Gambar rajah Venn di bawah menunjukkan set P, Q dan R di mana set universal = P Q R.

On the diagrams in the answer space, shade Pada gambar rajah di ruang jawapan di bawah, lorekkan

(a) the set P Q,

(b) the set ( Q R ) P.

[3 marks]

8/13/2019 Ujian Mac Math t4 2014 (k2)

4/14

4

Answer:(a)

(b)

8 Each of the following Venn diagrams shows sets A, B and C . On a separate diagram, shade theregion which representsSetiap gambar rajah Venn berikut menunjukkan set-set A, B, dan C. Pada rajah yang berasingan, lorekkan

rantau yang mewakili (a) A ' B ' (b) A U B U C

(b) A B C '

[6 marks]

8/13/2019 Ujian Mac Math t4 2014 (k2)

5/14

5

9 Given that the universal set Diberi set semesta = { x : 25 x 45, x is an integer}, = { x : 25 x 45 , x ialah suatu integer } , set X = { x : x is a multiple of 6},

set X = { x : x ialah suatu nombor gandaan 6} , set Y = { x : x is a factor of 100}, and

set Y = { x : x ialah suatu faktor bagi 100} , dan set Z = { x : x is a number such that the sum of its digits is greater than 6}.

set Z = { x : x ialah suatu nombor di mana hasil tambah digit-digitnya adalah lebih besar daripada 6}. (a) List the elements of

Senaraikan unsur-unsur bagi (i) X ,(ii) Y ,

(b) FindCari (i) n(Y Z ),

(ii) n[( X Z )'],[6 marks]

Answer:

10 Given that the universal set = X Y Z such that X = { g , h, j, q, t , v, x, y}, Y = {b, d , e, g , h, q,r , s, x} and Z = { i, k , l , q, y}.

Diberi set semesta = X Y Z di mana X = { g , h, j, q, t , v, x, y} , Y = { b, d , e, g , h, q, r , s, x} , dan Z = { i,k , l , q, y}. (a) List the elements of set Y Z .

Senaraikan unsur-unsur bagi set Y Z. (b) Find

Cari (i) n( X Y Z ),(ii) n( Z ').

[6 marks]

Answer:

8/13/2019 Ujian Mac Math t4 2014 (k2)

6/14

6

11 Solve the quadratic equation

2( x + 4)5 = 2 x.

Selesaikan persamaan kuadratik2( x2 + 4)

5 = 2 x.

[4 marks]

Answer:

12 Diagram 6 shows a cylindrical solid. A hemisphere shown by the shaded region, is removed fromthe solid.

Rajah 6 menunjukkan sebuah pepejal berbentuk silinder. Kawasan berlorek yang berbentuk hemisfera telahdikeluarkan dari pepejal itu.

Diagram 6

Given that the diameter of the hemisphere is 4 cm, calculate the volume, in cm 3, of the remaining

solid. (Use =227 )

Diberi diameter hemisfera itu ialah 4 cm, Hitung isi padu pepejal yang tinggal, dalam cm 3.

(Guna =22

7 )

[4 marks]Answer:

8/13/2019 Ujian Mac Math t4 2014 (k2)

7/14

7

13 Diagram 7 shows a composite solid comprises of a cylinder and a right cone. Rajah 7 menunjukkan sebuah pepejal gubahan yang terdiri daripada sebuah silinder dan sebuah kon tegak.

Diagram 7

The height of the cylinder is 14 cm while the height of the cone is 7 cm. Find the volume of the solid.

(Use =227 )

Tinggi silinder itu ialah 14 cm manakala tinggi kon itu ialah 7 cm. Cari isi padu bagi pepejal itu.

(Guna =227 )

[4 marks]

Answer:

14. The Venn diagram in the answer space shows sets P, Q and R such that the universal set = P Q R.

On the diagram in the answer space, shade

(i) the set P Q,

(ii) the set (P Q ) R .

[3 marks]

Answer:(i)

Q P R

8/13/2019 Ujian Mac Math t4 2014 (k2)

8/14

8

(ii)

Q P R

20 Table 2 shows the values of two variables, x and y, of a fuction. Rajah 2 menunjukkan nilai-nilai dua pemboleh ubah x dan y bagi suatu fungsi.

x -3 -2 -1 0 1 2 3y 75 -15 -35 -15 15 25 -15

Table 2 (a) By using scales of 2 cm to 1 unit on the x-axis and 2cm to 10 units on the y-axis, label both axes.

Dengan menggunakan skala 2 cm kepada 1 unit pada paksi-x dan 2 cm kepada 10 unit pada paksi-y, labelkankedua-dua paksi.

(b) Based on the table, plot the points on a graph paper. Berdasarkan jadual itu, plotkan titik-titik itu pada kertas graf.

(c) Draw the graph of the function. Lukiskan graf fungsi itu.

[5 marks]Answer:

END OF THE QUESTIONS

8/13/2019 Ujian Mac Math t4 2014 (k2)

9/14

9

SMK SULTAN ABDUL SAMAD, PETALING JAYASKEMA PEMARKAHAN MATEMATIK TINGKATAN 4

PEPERIKSAAN PERTENGAHAN TAHUN 2012

1 s t = 6 s = 6 + t -------------- (1) s + t = 4 ------------- (2)1(6 + t ) + t = 4 .........................(1)

6 + t + t = 42t = 2t = 1.......................... (1)

s = 6 + ( 1).................(1) s = 5 ..........................(1) s = 5, t = 1

2 4( x2 + 3) 16 x = 04 x2 16 x + 12 = 0 ...........................(1) 4( x2 4 x + 3) = 04( x 3)( x 1) = 0 ..........................(1)( x 3)( x 1) = 0

x = 3 or x = 1 ..............................(2)

3 2t =

4t + 23

6t = 4 t 2 + 24t 2 6t + 2 = 02t 2 3t + 1 = 0 .........................(1) (t 1)(2 t 1) = 0 .....................(1) (t 1) = 0 or (2 t 1) = 0

t = 1 or t =1

2 .............................(2)

4 3 x 2 y 12 = 0 ------------ (1)7 x + 6 y + 4 = 0 ------------ (2)(1) 3, 9 x 6 y 36 = 0

9 x 6 y = 36 ---- (3) @ (1)

(2) 1, 7 x + 6 y + 4 = 07 x + 6 y = 4 ------------- (4)(3) + (4)9 x + 7 x = 36 + ( 4).............................(1) 16 x = 32

x = 2 ......................................(1) 3(2) 2 y 12 = 0 ...............(1) 2 y = 6 122 y = 6

= 3.......................(1) x = 2, y = 3

5 (a) Length of arc FG

=45

360 2 227 14

= 11 cmLength of arc HI (1)

=45

360 2 227 28

= 22 cmPerimeter= 11 + 22 + 14 4 ..................(1)

= 89 cm .......................(1)(b) Area of sector OFG

8/13/2019 Ujian Mac Math t4 2014 (k2)

10/14

10

=45

360 227 14

2

= 77 cm 2 Area of sector OHI @ (1)

=45

360

22

7 28 2

@

= 308 cm 2 Area of sector OGJ

=45

360 227 14

2

= 77 cm 2 Area of the shaded region= 77 + 308 77..................(1)

= 308 cm 2................................(1)

6 (a) Length of arc AB

=

45

360 2

22

7 14= 11 cmLength of arc CDE @ (1)

=135360 2

227 28

= 66 cmPerimeter= 11 + 66 + 14 + 28 .....................(1) = 119 cm .....................(1)

(b) Area of sector AOB

=45

360

22

7 14 2

= 77 cm 2 Area of sector OCDE @

=135360

227 28

2 (1)

= 924 cm 2@

Area of semicircle OFE

=12

227 14

2

= 308 cm 2 Area of the shaded region= 77 + 924 308 ....................(1)

= 693 cm 2................................(1) 7 (a) Some right prisms have cross-sections

in the form of trapezium.. ..............(1)(b)

The area of triangle STU is12 45 84

that is 1 890 cm 2.......................(2)(c) Premise 1:

If 4 x x + 2 x + 10 is a quadratic equationin x, then the value of x is 2. ..............(2)

8 (a)

(2)

(2)

(c)

(2)

9 = {25, 26, 27, 28, 29, 30, 31, 32, 33, 34,35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45}

X = {30, 36, 42}Y = {25}

Z = {25, 26, 27, 28, 29, 34, 35, 36, 37, 38,39, 43, 44, 45}(a) (i) X = {30, 36, 42} .......................(1)

(ii) Y = {25} ........................(1)(b) (i) Y Z = {25} ...............(1)

n(Y Z ) = 1 .................(1)(ii) X Z = {25, 26, 27, 28, 29, 30,

34, 35, 36, 37, 38, 39, 42, 43, 44,45}( X Z )' = {31, 32, 33, 40,41} .........................(1) n[( X Z )'] = 5 ..............(1)

10 (a) Y Z = { q} .........................(1)(b) (i) X Y Z = { b, d , e, g , h, i, j, k , l ,

q, r , s, t , v, x, y}.......................(1) n( X Y Z ) = 16 ...................(1)

(ii Z ' = { b, d , e, g , h, j, r , s, t , v, x}(1) n( Z ') = 11 ..........................(1)

11 (a) Some quadratic equations havenegative roots. . ......................(1)

(b) Premise 2:79 is not divisible by 4. . ...................(2)

(c) 3(2) n + n ........................(2)(d) Implication 1:

Ifn2 >

n9, then n > 0. ......................(1)

Implication 2:

8/13/2019 Ujian Mac Math t4 2014 (k2)

11/14

11

If n > 0, thenn2 >

n9......................(1)

12 (a) 110 >

15 or 5

2 =1

25.........................(2)

(b) Conclusion:The right prism does not have a cross-sectional area of 70 cm 2 and a height of15 cm. ............................(2)

(c) All the numbers of 4, 7, 12, 19, ... can be written in the form n2 + 3, n = 1, 2,3, 4, ... ..........................(2)

13 2( x2 + 4)5 = 2 x

2( x2 + 4) = 10 x

2 x2

+ 8 = 10 x 2 x2 10 x + 8 = 0 x2 5 x + 4 = 0 .............................(1) ( x 4)( x 1) = 0 .........................(1) ( x 4) = 0 or ( x 1) = 0

x = 4 or x = 1 .................................(2)

14 (a) YZ = 8 7 ................................(1) = 56 ..........................(1)

q = 56(b) Gradient of XY

= (7)(9) ................(2)

= 79 .......................(1)

15 (a) x-intercept of P Q = -4 ..............(1)

(b) Gradient of QR

=3 (8)0 (4) ...(1)

=54 ...........(1)

=54 .............(1)

16 (a) Length of AC = 4 1..............(1) = 3

AC = CB p = 3 ..........(1)

(b)

intercept- yintercept- x

= 54 ..............(1)

intercept- y = (54

4)= 5 ...(1)

17 Volume of cylinder= 924 cm 3

........................(1)

Volume of hemisphere

= 161621 cm

3......................(1)

Volume of remaining solid

= 924 161621 ......................(1)

= 9075

21 cm3........................ (1)

18. Volume(1) (1)

=227 13

2 14 +13

227 6

2 7

=227 169 14 +

13

227 36 7

= 7 436 + 264 ........................(1)

= 7 700 cm 3...........................(1)

19 .Skala .............................(2)Plot ..............................(2)

Sambung titik...................(2)

8/13/2019 Ujian Mac Math t4 2014 (k2)

12/14

12

20. Skala .............................(2)Plot ..............................(2)

Sambung titik...................(2)

SULIT NAMA :TINGKATAN :

PEPERIKSAAN AKHIR TAHUN 2012 1449/2TINGKATAN 4MATHEMATICSKertas 2Mei2 jam Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHUUntuk Kegunaan Pemeriksa

Kod Pemeriksa :Bahagian Soalan Markah

PenuhMarkah

Diperolehi

SMK SULTAN ABDUL SAMADPETALING JAYA

8/13/2019 Ujian Mac Math t4 2014 (k2)

13/14

13

1. Kertas soalan ini mengandungi hanya SATU bahagian: Bahagian A

2. Jawab semua soalan.

3. Tulis jawapan pada ruang yang disediakandan tunjukkan kerja mengira anda untukmembantu mendapatkan markah.

4. Satu senarai formula disediakan.

5. Anda dibenarkan menggunakan kalkulatorsainstifik

Kertas soalan ini mengandungi 11 halaman bercetak

A

1 42 43 44 55 66 67 58 69 6

10 611 712 613 414 515 416 417 4

18 4

19 520 5

Jumlah 100

Disemak Oleh:

..........................................(Cik Rafeidah Bt YaacobGuru Matematik Ting. 5

Disediakan oleh:

.....................................(Pn Nor Mala Bt Mahadi)Ketua Panitia Matematik

Disahkan Oleh:

.....................................(Pn. Naik Soo Fong)GKMP Sains Dan Math

8/13/2019 Ujian Mac Math t4 2014 (k2)

14/14

14