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- 1. ppr maths nbk NAMA : …………………………………………TINGKATAN : ……………… 1449/2 Matematik Kertas 2 Okt 2006 2 ½ jam SEKTOR SEKOLAH BERASRAMA PENUH BAHAGIAN SEKOLAH KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2006 MATEMATIK Kertas 2 Dua jam tiga puluh minit JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1. Kertas soalan ini mengandungi dua Pemeriksa bahagian : Bahagian A dan Bahagian Markah Markah Bahagian Soalan B. Penuh Diperoleh 2. Jawab semua soalan Bahagian A dan 1 4 empat soalan daripada Bahagian B. 2 4 3. Jawapan hendaklah ditulis dengan jelas 3 4 dalam ruang yang disediakan dalam 4 5 kertas soalan. 5 5 4. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh A 6 4 membantu anda untuk mendapatkan 7 7 markah. 8 5 5. Rajah yang mengiringi soalan tidak 9 3 dilukis mengikut skala kecuali dinyatakan. 10 5 6. Satu senarai rumus disediakan di 11 6 halaman 2 dan 3. 12 12 7. Anda dibenarkan menggunakan 13 12 kalkulator saintifik yang tidak boleh diprogram. B 14 12 15 12 16 12 Jumlah Kertas soalan ini mengandungi 22 halaman bercetak 1449/2 @ 2006 Hak Cipta SBP
- 2. SULIT 2 1449/2 The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. RELATIONS 1 am × an = am + n 2 am ÷ an = am - n 3 (am)n = am n 1 ⎛ d − b⎞ 4 A-1 = ⎜ ⎟ ad − bc ⎜ − c a ⎟ ⎝ ⎠ n ( A) 5 P(A) = n (S ) 6 P(A’) = 1 – P(A) ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 7 Distance = 8 Midpoint ⎛ x + x 2 y1 + y 2 ⎞ (x, y) = ⎜ 1 , ⎟ ⎝ 2 2 ⎠ distance traveled 9 Average speed = time taken sum of data 10 Mean = number of data Sum of (midpoint of interval × frequency) 11 Mean = Sum of frequencies 12 Pythagoras Theorem c2 = a2 + b2 y −y 13 m= 2 1 x −x 2 1 y − int ercept 14 m= x − int ercept 1449/2 SULIT
- 3. SULIT 3 1449/2 SHAPE AND SPACE 1 1 Area of trapezium = × sum of parallel sides × height 2 2 Circumference of circle = πd = 2πr 3 Area of circle = πr2 4 Curved surface area of cylinder = 2πrh 5 Surface area of sphere = 4πr2 6 Volume of right prism = cross sectional area × length 7 Volume of cylinder = πr2h 1 2 8 Volume of cone = πr h 3 4 3 9 Volume of sphere = πr 3 1 10 Volume of right pyramid = × base area × height 3 11 Sum of interior angles of a polygon = (n – 2) × 180° arc length angle subtended at centre 12 = circumference of circle 360° area of sector angle subtended at centre 13 = area of circle 360° PA' 14 Scale factor, k = PA 15 Area of image = k2 × area of object. 1449/2 SULIT
- 4. SULIT 4 1449/2 For Section A Examiner’s Use [52 marks] Answer all questions in this section. 2p2 −5 1 1 Solve the quadratic equation = . 3p 2 [4 marks] Answer: 2 Calculate the value of p and of q that satisfy the following simultaneous linear equations: 3p – 4q = –2 1 p + 2q = 13 2 [4 marks] Answer: 1449/2 SULIT
- 5. SULIT 5 1449/2 For 3 Diagram 1 shows a right prism. Right angled triangle PQR is the uniform cross- Examiner’s Use section of the prism. Q 18 cm R 9 cm P 12 cm T S U DIAGRAM 1 (a) Name the angle between the plane STP and the plane STQR, (b) Calculate the angle between the plane STP and the plane STQR. [4 marks] Answer: (a) (b) [Lihat sebelah 1449/2 SULIT
- 6. SULIT 6 1449/2 4 A straight line ST is parallel to line y = –2x + 5 and passes through point (4, –2). Find (a) the gradient of the straight line ST, (b) the equation of the straight line ST and hence, state its y-intercept. . [5 marks] Answer: (a) (b) 5 A bag contains 50 pens. 15 of them are blue and the rest are red and green. A pen is chosen at random from the bag. (a) Find the probability that a blue pen is chosen. 1 (b) The probability of choosing a green pen is . How many green pens 5 are there in the bag? ( c) Another 10 green pens are put into the bag. Find the probability that a green pen is chosen from the bag. [5 marks] Answer: (a) (b) ( c) 1449/2 SULIT
- 7. SULIT 7 1449/2 6 Diagram 2(a) shows a container in the form of a right pyramid fully filled with For water. Diagram 2(b) shows an empty cylindrical container. The height of the Examiner’s pyramid is 15 cm. Use A 12 cm B 14 cm 8 cm D C E DIAGRAM 2(a) DIAGRAM 2(b) All the water from the right pyramid container is poured into the cylindrical container. 22 By using π = , calculate the height, in cm, of the water level in the cylinder. 7 [4 marks] Answer: [Lihat sebelah 1449/2 SULIT
- 8. SULIT 8 1449/2 For 7 In Diagram 3, OPQS is a quadrant with the centre O and OSQR is a semicircle Examiner’s Use with the centre S. Q R S 60° T O P DIAGRAM 3 22 Given that OP = 14 cm. Using π = , calculate 7 (a) the area, in cm2, of the shaded region, (b) the perimeter, in cm, of the whole diagram. [6 marks] Answer: (a) (b) 1449/2 SULIT
- 9. SULIT 9 1449/2 8 (a) Determine whether the following is a statement or not. Give a reason to For your answer. Examiner’s Use 64 = 42 (b) Rewrite the following statement by inserting the word ‘not’ into the original statement. State the truth value of your new statement. 3 is the factor of 15 ( c) Construct a true statement using a suitable quantifier for the given object and the property. Object : Triangles Property : Have a right angle [5 marks] Answer: (a) …………………………………………………………………………….. (b) …………………………………………………………………………….. ( c) ……………………………………………………………………………. [Lihat sebelah 1449/2 SULIT
- 10. SULIT 10 1449/2 For 9 The Venn diagram in the answer space shows set P, set Q and set R with the Examiner’s Use universal set ξ = P ∪ Q ∪ R. On the diagrams provided in the answer space, shade (a) the set Q ∩ (P ∪ R), (b) the set Q ∪ R’ ∩ P. . [3 marks] Answer: (a) P (b) P Q Q R R 10 In Diagram 4 , AE is a tangent to the circle centre O, at A. CDE is a straight line. A B z° x° O 70° 20° E y° D C DIAGRAM 4 Find the value of (a) x (b) y ( c) z [5 marks] Answer: (a) (b) ( c) 1449/2 SULIT
- 11. SULIT 11 1449/2 11 (a) Diagram 5 shows a unit circle. y For Examiner’s 1 Use v° –1 O 1 x (– 0.42, – 0.91) –1 DIAGRAM 5 Find the value of (i) sin v° (ii) cos v° + sin 270° [3 marks] (b) Diagram 6 shows a rhombus PQRS. R 26 cm x° T Q S P DIAGRAM 6 5 It is given that TSQ is a straight line and tan x° = − . 12 Find (i) the length of RP (ii) sin x° [3 marks] Answer: (a) (i) (ii) (b) (i) (ii) [Lihat sebelah 1449/2 SULIT
- 12. SULIT 12 1449/2 For Section B Examiner’s Use [48 marks] Answer four questions in this section. 12 (a) In Diagram 7, a straight line RS is parallel to the straight line PQ. The equation of PQ is 2y = x – 4. y S(4, 5) R Q(8, m) O P x DIAGRAM 7 Find (i) the y-intercept of PQ, (ii) the value of m, (iii) the equation of RS. [6 marks] Answer: (a) (i) (ii) (iii) 1449/2 SULIT
- 13. SULIT 13 1449/2 12 (b) In Diagram 8, EFGH is a parallelogram and O is the origin. The gradient For Examiner’s of EF is 2. Use y F(4, 10) E O G x H DIAGRAM 8 Find (i) the equation of line EF, (ii) the coordinates of H, (iii) the gradient of HF. [6 marks] Answer: (b) (i) (ii) (iii) [Lihat sebelah 1449/2 SULIT
- 14. SULIT 14 1449/2 For 13 (a) Diagram 9 shows a right prism with a horizontal rectangular base PQRS. Examiner’s Use W E V T 5 cm U 7 cm S P F 12 cm R Q 8 cm DIAGRAM 9 Given that E and F are midpoints of WV and SR respectively. (i) Find the length of PF, (ii) Calculate the angle between the line PE and the plane PQRS, (iii) Name the angle between the plane PQVE and the plane PQUT. [6 marks] Answer: (a) (i) (ii) (iii) 1449/2 SULIT
- 15. SULIT 15 1449/2 13 (b ) In Diagram 10, MP and LQ are two vertical flagpoles standing on horizontal ground KLM . For Examiner’s Use Q P 5m K 70° 400 M hm L DIAGRAM 10 Given that the angle of depression of P from Q is 38°, MP = 5 m, LQ = h m and KM = 10 m. Calculate (i) the angle of elevation of P from K, (ii) the value of h. [6 marks] Answer: (b) (i) (ii) [Lihat sebelah 1449/2 SULIT
- 16. SULIT 16 1449/2 For Examiner’s 14 (a) (i) Fill the blanks with ‘and’ or ‘or’ in order to form a true statement. Use 1 1 (a) sin 30˚ = cos 30˚ = 2 2 (b) All the multiples of 3 are multiples of 6 all multiples of 6 are multiples of 3. (ii) Complete the argument below. Premise 1 : If a ∈ A, then a ∈ A ∪ B. Premise 2 : __________________________________ Conclusion : 5 ∉ A (iii) Make a general conclusion by induction for the sequence 13, 28, 49, 76,… which follows the following pattern: 13 = 3 (2)2 + 1 28 = 3 (3)2 + 1 49 = 3 (4)2 + 1 76 = 3 (5)2 + 1 ……………… ……………… [6 marks] Answer: (a) (i) (a) ………………………………………………………………. (b) ………………………………………………………………. (ii) .......................................................................................................... (iii) .......................................................................................................... 1449/2 SULIT
- 17. SULIT 17 1449/2 For Examiner’s 14 (b) Diagram 11 is an incomplete Venn diagram. A group of 45 students Use involved in their favourite games. B = Badminton T = Tennis 2 3 5 6 11 S = Squasy DIAGRAM 11 Given that ξ = B ∪ S ∪ T and the number of students who involve in badminton only is the same as the students who involve in tennis only. Find the number of students who involve in (i) all the three games, (ii) neither badminton nor tennis, (iii) two games only. (iv) tennis only. [6 marks] Answer: (b) (i) (ii) (iii) (iv) [Lihat sebelah 1449/2 SULIT
- 18. SULIT 18 1449/2 For 15 The daily distances, in km, travelled in 40 days by a salesman are shown in Examiner’s Diagram 12. Use 42 62 51 37 70 14 12 93 53 45 54 27 70 76 91 72 68 58 57 65 21 18 22 87 64 63 57 59 70 19 41 55 36 56 50 57 73 75 13 39 DIAGRAM 12 (a) Using data in Diagram 12 and a class interval of 10 km, complete Table 1 in the answer space. [4 marks] Answer: (a) Distance (km) Midpoint Frequency 11 – 20 21 – 30 31 – 40 TABLE 1 (b) Determine the range of the given data. [1 mark] ( c) Based on your table in (a), calculate the estimated mean distance travelled. [3 marks] Answer: (b) ( c) (i) (ii) (d) For this part of the question, use the graph paper on page 19. By using a scale of 2 cm to 10 km on the x-axis and a scale of 2 cm to 1 day on the y-axis, draw a frequency polygon for the data above. [4 marks] Answer: (d) Refer graph on page 19. 1449/2 SULIT
- 19. SULIT 19 1449/2 Graph for Question 15(d) 1449/2 SULIT
- 20. SULIT 20 1449/2 For 16 The histogram in Diagram 13 shows the marks achieved in a test by a group of Examiner’s students in a particular school. Use Frequency 20 16 12 8 4 29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5 Marks DIAGRAM 13 (a) Find the total number of students who sat for the test. [1 mark] (b) Find the modal class. [1 mark] ( c) Based on the histogram, complete the Table 3 below. [3 marks] Answer: (a) (b) ( c) Marks Upper boundary Cumulative frequency 20 – 29 29.5 0 30 – 39 39.5 4 TABLE 3 (d) For this part of the question, use the graph paper on page 22. By using the scale of 2 cm to 10 marks on the x-axis, and 2 cm to 10 students on the y-axis, draw an ogive for the data. [4 marks] ( e) From the ogive, find (i) the median, (ii) the number of students who passed the test if the passing mark is 45%. [3 marks] 1449/2 SULIT
- 21. SULIT 21 1449/2 For Answer: Examiner’s (d) Refer graph on page 22. Use ( e) (i) (ii) 1449/2 SULIT
- 22. SULIT 22 1449/2 Graph for Question 16(d) END OF QUESTION PAPER 1449/2 SULIT

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