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Peperiksaan pertengahan tahun t4 2012 (2)

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Peperiksaan pertengahan tahun t4 2012 (2)

  1. 1. [100 marks] Answer all questions from this section. Jawab semua soalan daripada bahagian ini.1 Calculate the value of s and of t that satisfy the following simultaneous linear equations: Hitungkan nilai s dan nilai t yang memuaskan persamaan linear serentak berikut: s t=6 s+t=4 [4 marks] Answer:2 Solve the equation 4(x2 + 3) 16x = 0 Selesaikan persamaan 4(x2 + 3) 16x = 0 . [4 marks] Answer:3 4t2 + 2 Solve the equation 2t = . 3 4t2 + 2 Selesaikan persamaan 2t = 3 . [4 marks] Answer: 1
  2. 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: 3x 2y 12 = 0 7x + 6y + 4 = 0 [5 marks] Answer:5 Diagram 1 shows two sectors OFG and OHI with the same centre O. OFGJ is a quadrant of a circle 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 sukuan bulatan berpusat O. OGH dan OJI ialah garis lurus. Diagram 1 OJ = JI = 14 cm and ∠HOI = 45°. OJ = JI = 14 cm dan ∠HOI = 45°. 22 Using π = , calculate 7 22 Dengan menggunakan π = , hitungkan 7 (a) the perimeter, in cm, of the whole diagram, perimeter, dalam cm, seluruh rajah itu, (b) the area, in cm2, of the shaded region. luas, dalam cm2, kawasan yang berlorek. [6 marks] Answer: 2
  3. 3. 6 Diagram 2 shows two sectors OAB and OCDE with the same centre O. OFE is a semicircle with diameter OE and OE = 2AO. 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°. 22 Using π = 7 , calculate 22 Dengan menggunakan π = , hitungkan 7 (a) the perimeter, in cm, of the whole diagram, perimeter, dalam cm, seluruh rajah itu, (b) the area, in cm2, of the shaded region. luas, dalam cm2, kawasan yang berlorek. [6 marks] Answer:7 (a) Complete the following statement with the quantifier "all" or "some" to form a true statement. Lengkapkan pernyataan berikut dengan pengkuantiti "semua" atau "sebilangan" untuk membentuk suatu pernyataan benar. __________ right prisms have cross- sections in the form of trapezium. __________ prisma tegak mempunyai keratan rentas dalam bentuk trapezium. (b) 1 The area of a triangle = × base × height. 2 1 Luas sebuah segi tiga = × tapak × ketinggian. 2 STU is a triangle with base 45 cm and height 84 cm. STU ialah sebuah segi tiga dengan tapak 45 cm dan ketinggian 84 cm. 3
  4. 4. Special conclusion: Kesimpulan khas: (c) Complete the Premise 1 in the argument below: Lengkapkan Premis 1 dalam hujah berikut: Premise 1: Premis 1: Premise 2: 4xx + 2x + 10 is a quadratic equation in x. Premis 2: 4xx + 2x + 10 ialah suatu persamaan kuadratik dalam x. Conclusion: The value of x is 2. Kesimpulan: Nilai x adalah 2. [5 marks] Answer:8 Each of the following Venn diagrams shows sets A, B and C. On a separate diagram, shade the region which represents Setiap 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]9 Given that the universal set Diberi set semesta ξ = {x : 25 ≤ x ≤ 45, x is an integer}, ξ = {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}, 4
  5. 5. 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) Find Cari (i) n(Y ∩ Z), (ii) n[(X ∪ Z)], [6 marks] Answer:10 Given that the universal set ξ = 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 ξ = 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:11 (a) Complete the following statement using the quantifier "all" or "some" to make it a true statement. Lengkapkan pernyataan berikut dengan menggunakan pengkuantiti "semua" atau "sebilangan" untuk membentuk suatu pernyataan benar. __________ quadratic equations have negative roots. __________ persamaan kuadratik mempunyai punca yang negatif. (b) Write down Premise 2 to complete the following argument: 5
  6. 6. Tulis Premis 2 untuk melengkapkan hujah berikut: Premise 1: If Y is divisible by 8, then Y is divisible by 4. Premis 1: Jika Y boleh dibahagi dengan 8, maka Y boleh dibahagi dengan 4. Premise 2: Premis 2: Conclusion: 79 is not divisible by 8. Kesimpulan: 79 tidak boleh dibahagi dengan 8.(c) Make a general conclusion by induction for the sequence of numbers 7, 14, 27, ... which follows the following pattern. Buat satu kesimpulan umum secara aruhan bagi urutan nombor 7, 14, 27, ... yang mengikuti pola berikut. 7 = 3(2)1 + 1 14 = 3(2)2 + 2 27 = 3(2)3 + 3 ... = ..........(d) Write down two implications based on the following statement: Tulis dua implikasi berdasarkan pernyataan berikut: n n > if and only if n > 0. 2 9 n n > jika dan hanya jika n > 0. 2 9 [7 marks]Answer: 6
  7. 7. 12 (a) Combine the following statements to form a true statement. Gabungkan pernyataan-pernyataan berikut untuk membentuk suatu pernyataan benar. Statement 1: Pernyataan 1: 1 1 > 10 5 Statement 2: Pernyataan 2: 1 5−2 = 25 (b) Write a conclusion based on the following premises. Tulis satu kesimpulan berdasarkan premis-premis berikut. Premise 1: If a right prism has a cross-sectional area of 70 cm2 and a height of 15 cm, then the volume of the prism is 1 050 cm3. Premis 1: Jika sebuah prisma tegak mempunyai luas keratan rentas 70 cm2 dan ketinggian 15 cm, maka isipadu prisma itu adalah 1 050 cm3. Premise 2: The volume of a right prism is not 1 050 cm3. Premis 2: Isipadu sebuah prisma tegak bukan 1 050 cm3. Conclusion: Kesimpulan: (c) Make a conclusion by induction for the number sequence 4, 7, 12, 19, ..., according to the pattern below. Bina satu kesimpulan secara aruhan bagi jujukan nombor 4, 7, 12, 19, ..., mengikut pola berikut. 4 = 12 + 3 7 = 22 + 3 12 = 32 + 3 19 = 42 + 3 ... = .......... [6 marks] Answer:13 2(x2 + 4) Solve the quadratic equation 5 = 2x. 2(x2 + 4) Selesaikan persamaan kuadratik = 2x. 5 [4 marks] Answer: 7
  8. 8. 14 Diagram 3 shows two straight lines XY and YZ drawn on Cartesian plane. Rajah 3 menunjukkan dua garis lurus XY dan YZ yang dilukis pada satah cartesan . Diagram 3 Given that YZ = 8XO. Find Diberi YZ = 8XO. Cari (a) the value of q, nilai q, (b) the gradient of the straight line XY. kecerunan garis lurus XY. [5 marks] Answer:15 Diagram 4 shows two straight lines PQ and QR drawn on Cartesian plane. Rajah 4 menunjukkan dua garis lurus PQ dan QR yang dilukis pada satah cartesan. Diagram 4 8
  9. 9. Given that the length of OR is 3 units. Find Diberi panjang OR ialah 3 unit. Cari (a) the x-intercept of the straight line PQ, pintasan-x untuk garis lurus PQ, (b) the gradient of the straight line QR. kecerunan garis lurus QR. [4 marks] Answer:16 Diagram 5 shows a isosceles triangle ABC and a straight line AD. Rajah 5 menunjukkan segi tiga sama kaki ABC dan garis lurus AD . Diagram 5 5 Given that the x-intercept and the gradient of the straight line AD are 4 and − . Find 4 5 Diberi pintasan-x dan kecerunan garis lurus AD ialah 4 dan − . Cari 4 (a) the value of p, nilai p, (b) the y-intercept of the straight line AD. pintasan-y bagi garis lurus AD. [4 marks] Answer: 9
  10. 10. 17 Diagram 6 shows a cylindrical solid. A hemisphere shown by the shaded region, is removed from the solid. Rajah 6 menunjukkan sebuah pepejal berbentuk silinder. Kawasan berlorek yang berbentuk hemisfera telah dikeluarkan dari pepejal itu. Diagram 6 Given that the diameter of the hemisphere is 4 cm, calculate the volume, in cm3, of the remaining 22 solid. (Use π = 7 ) Diberi diameter hemisfera itu ialah 4 cm, Hitung isi padu pepejal yang tinggal, dalam cm 3. 22 (Guna π = ) 7 [4 marks] Answer:18 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. 10
  11. 11. Diagram 7The height of the cylinder is 14 cm while the height of the cone is 7 cm. Find the volume of the solid. 22(Use π = 7 )Tinggi silinder itu ialah 14 cm manakala tinggi kon itu ialah 7 cm. Cari isi padu bagi pepejal itu. 22(Guna π = ) 7 [4 marks]Answer:19. Table 1 shows the values of two variables, x and y, of a fuction. Rajah 1 menunjukkan nilai-nilai dua pemboleh ubah x dan y bagi suatu fungsi. x -6 -4 -2 0 2 4 6 y 7 32 47 52 47 32 7 Table 1 (a) By using scales of 2 cm to 2 units on the x-axis and 2cm to 10 units on the y-axis, label both axes. Dengan menggunakan skala 2 cm kepada 2 unit pada paksi-x dan 2 cm kepada 10 unit pada paksi-y, labelkan kedua-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: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 3 y 75 -15 -35 -15 15 25 -15 Table 2 11
  12. 12. (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, labelkan kedua-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 12
  13. 13. SMK SULTAN ABDUL SAMAD, PETALING JAYA SKEMA PEMARKAHAN MATEMATIK TINGKATAN 4 PEPERIKSAAN PERTENGAHAN TAHUN 20121 s t=6 45° 22 s = 6 + t -------------- (1) = × 2 × × 14 360° 7 s + t = 4 ------------- (2) = 11 cm 1(6 + t) + t = 4.........................(1) Length of arc HI (1) 6+t+t=4 45° 22 2t = 2 = × 2 × × 28 360° 7 t = 1.......................... (1) = 22 cm s = 6 + ( 1).................(1) Perimeter s = 5..........................(1) = 11 + 22 + 14 × 4..................(1) ∴ s = 5, t = 1 = 89 cm .......................(1) (b) Area of sector OFG2 4(x2 + 3) 16x = 0 45° 22 = 360° × 7 × 142 4x2 16x + 12 = 0...........................(1) 4(x2 4x + 3) = 0 = 77 cm2 4(x 3)(x 1) = 0..........................(1) Area of sector OHI @ (1) (x – 3)(x – 1) = 0 45° 22 2 @ = 360° × 7 × 28 x = 3 or x = 1..............................(2) = 308 cm2 Area of sector OGJ3 4t2 + 2 45° 22 2t = 3 = 360° × 7 × 142 6t = 4t2 + 2 = 77 cm2 4t2 6t + 2 = 0 Area of the shaded region 2t2 3t + 1 = 0.........................(1) = 77 + 308 – 77..................(1) (t 1)(2t 1) = 0.....................(1) (1) = 308 cm2................................ (t 1) = 0 or (2t 1) = 0 1 6 (a) Length of arc AB t = 1 or t = 2 .............................(2) 45° 22 = × 2 × × 14 360° 74 3x 2y 12 = 0 ------------ (1) = 11 cm 7x + 6y + 4 = 0 ------------ (2) Length of arc CDE @ (1) (1) × 3, 9x 6y 36 = 0 135° 22 = 360° × 2 × 7 × 28 9x 6y = 36 ---- (3) @ (1) (2) × 1, 7x + 6y + 4 = 0 = 66 cm Perimeter 7x + 6y = 4 ------------- (4) = 11 + 66 + 14 + 28.....................(1) (3) + (4) = 119 cm .....................(1) 9x + 7x = 36 + ( 4).............................(1) 16x = 32 (b) Area of sector AOB x = 2......................................(1) 45° 22 = 360° × 7 × 142 3(2) 2y 12 = 0...............(1) 2y = 6 12 = 77 cm2 Area of sector OCDE @ 2y = 6 135° 22 y = 3.......................(1) = 360° × 7 × 282 (1) ∴ x = 2, y = 3 @ = 924 cm25 (a) Length of arc FG Area of semicircle OFE 1 22 = 2 × 7 × 142 13
  14. 14. = 308 cm2 10 (a) Y ∩ Z = {q} .........................(1) Area of the shaded region (b) (i) X ∪ Y ∪ Z = {b, d, e, g, h, i, j, k, l, = 77 + 924 – 308....................(1) q, r, s, t, v, x, y}.......................(1) ................................(1) = 693 cm2 n(X ∪ Y ∪ Z) = 16 ...................(1) (ii Z = {b, d, e, g, h, j, r, s, t, v, x}(1) n(Z ) = 11 ..........................(1)7 (a) Some right prisms have cross-sections 11 (a) Some quadratic equations have in the form of trapezium................(1) negative roots.. ......................(1) (b) 1 (b) Premise 2: The area of triangle STU is 2 × 45 × 84 79 is not divisible by 4.. ...................(2) that is 1 890 cm2.......................(2) (c) 3(2)n + n ........................(2) (c) Premise 1: (d) Implication 1: If 4xx + 2x + 10 is a quadratic equation n n If 2 > 9, then n > 0.......................(1) in x, then the value of x is 2...............(2) Implication 2:8 (a) n n If n > 0, then 2 > 9......................(1) 12 (a) 1 1 −2 1 10 > 5 or 5 = 25.........................(2) (2) (b) Conclusion: The right prism does not have a cross- sectional area of 70 cm2 and a height of 15 cm.............................(2) (c) All the numbers of 4, 7, 12, 19, ... can (2) be written in the form n2 + 3, n = 1, 2, 3, 4, ... ..........................(2) (c) 13 2(x2 + 4) 5 = 2x 2 2(x + 4) = 10x 2x2 + 8 = 10x 2x2 10x + 8 = 0 (2) x2 5x + 4 = 0.............................(1) (x 4)(x 1) = 0.........................(1)9 ξ = {25, 26, 27, 28, 29, 30, 31, 32, 33, 34, (x 4) = 0 or (x 1) = 0 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45} x = 4 or x = 1.................................(2) X = {30, 36, 42} Y = {25} 14 (a) YZ = 8 × 7................................(1) Z = {25, 26, 27, 28, 29, 34, 35, 36, 37, 38, = 56..........................(1) 39, 43, 44, 45} ∴q = 56 (a) (i) X = {30, 36, 42}.......................(1) (b) Gradient of XY (ii) Y = {25}........................(1) (−7) = −(−9) ................(2) (b) (i) Y ∩ Z = {25}...............(1) n(Y ∩ Z) = 1 .................(1) 7 = − .......................(1) (ii) X ∪ Z = {25, 26, 27, 28, 29, 30, 9 34, 35, 36, 37, 38, 39, 42, 43, 44, 45} 15 (a) x-intercept of PQ (X ∪ Z) = {31, 32, 33, 40, = -4 ..............(1) 41}.........................(1) (b) Gradient of QR n[(X ∪ Z)] = 5..............(1) −3 − (−8) = 0 − (−4) ...(1) 14
  15. 15. 5 = 4 ...........(1) 5 = 4 .............(1)16 (a) Length of AC = 4 – 1..............(1) =3 AC = CB ∴ p = −3 ..........(1) (b) intercept-y −intercept-x 5 = −4 ..............(1) 5 intercept-y = −(−4 × 4) = 5...(1)17 Volume of cylinder ........................ = 924 cm3 (1) Volume of hemisphere 20. Skala .............................(2) 16 ......................(1) Plot ..............................(2) = 1621 cm3 Sambung titik...................(2) Volume of remaining solid 16 = 924 − 16 ......................(1) 21 5 ........................(1) = 90721 cm318. Volume (1) (1) 22 1 22 = 7 × 132 × 14 + 3 × 7 × 62 × 7 22 1 22 = 7 × 169 × 14 + 3 × 7 × 36 × 7 = 7 436 + 264........................(1) ...........................(1) = 7 700 cm319 . Skala .............................(2) Plot ..............................(2) Sambung titik...................(2) 15
  16. 16. SULIT NAMA : TINGKATAN : SMK SULTAN ABDUL SAMAD PETALING JAYAPEPERIKSAAN AKHIR TAHUN 2012 1449/2TINGKATAN 4MATHEMATICSKertas 2Mei2 jam Dua jam tiga puluh minit JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU Untuk Kegunaan Pemeriksa Kod Pemeriksa :1. Kertas soalan ini mengandungi hanya SATU Bahagian Soalan Markah Markah Penuh Diperolehi bahagian: Bahagian A 1 4 2 42. Jawab semua soalan. 3 4 4 53. Tulis jawapan pada ruang yang disediakan 5 6 dan tunjukkan kerja mengira anda untuk 6 6 7 5 membantu mendapatkan markah. 8 6 9 64. Satu senarai formula disediakan. 10 6 A 11 7 12 65. Anda dibenarkan menggunakan kalkulator 13 4 sainstifik 14 5 15 4 16 4 17 4 18 4 19 5 20 5 Jumlah 100 Kertas soalan ini mengandungi 11 halaman bercetak Disediakan oleh: Disemak Oleh: Disahkan Oleh: 16 ..................................... .......................................... ..................................... (Pn Nor Mala Bt Mahadi) (Cik Rafeidah Bt Yaacob (Pn. Naik Soo Fong) Ketua Panitia Matematik Guru Matematik Ting. 5 GKMP Sains Dan Math
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