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F4 Final Sbp 2007 Maths P2
 

F4 Final Sbp 2007 Maths P2

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    F4 Final Sbp 2007 Maths P2 F4 Final Sbp 2007 Maths P2 Document Transcript

    • ppr maths nbk SULIT 1449/2 1449/2 NAMA : Matematik Kertas 2 TINGKATAN : Oktober 2007 1 2 jam 2 SEKTOR SEKOLAH BERASRAMA PENUH BAHAGIAN SEKOLAH KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN DIAGNOSTIK TINGKATAN EMPAT 2007 Pemeriksa MATEMATIK Markah Markah Bahagian Soalan Kertas 2 Penuh Diperoleh Dua jam tiga puluh minit 1 4 2 4 3 5 JANGAN BUKA KERTAS SOALAN INI 4 4 SEHINGGA DIBERITAHU 5 3 A 1. Kertas soalan ini mengandungi dua bahagian : 6 4 Bahagian A dan Bahagian B. Jawab semua soalan daripada Bahagian A dan empat soalan 7 6 dalam Bahagian B. 8 6 2. Jawapan hendaklah ditulis dengan jelas dalam 9 4 ruang yang disediakan dalam kertas soalan. 10 7 Tunjukkan langkah-langkah penting. Ini boleh membantu anda untuk mendapatkan markah. 11 5 12 12 3. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. 13 12 B 14 12 4. Satu senarai rumus disediakan di halaman 2 & 3. 15 12 16 12 5. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram. Jumlah Kertas soalan ini mengandungi 24 halaman bercetak. 1449/2 © 2007 Hak Cipta Sektor SBP [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk Kegunaan Pemeriksa MATHEMATICAL FORMULAE The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. RELATIONS 1 am x an = a m+ n 2 am ÷ an = a m – n 3 ( am )n = a mn 1 ⎛ d − b⎞ 4 A-1 = ⎜ ⎜−c a ⎟⎟ ad − bc ⎝ ⎠ n( A) 5 P(A)= n( S ) 6 P ( A′ ) = 1 − P(A) 7 Distance = ( x1 − x2 )2 + ( y1 − y2 ) 2 ⎛ x + x 2 y1 + y 2 ⎞ 8 Midpoint, ( x, y ) = ⎜ 1 , ⎟ ⎝ 2 2 ⎠ 9 Average speed = distance travelled time taken 10 Mean = sum of data number of data 11 Mean = sum of (class mark × frequency) sum of frequencies 12 Pythagoras Theorem c2 = a2 + b2 y 2 − y1 13 m= x 2 − x1 y -intercept 14 m= x-intercept 1449/2 2 SULIT
    • SULIT 1449/2 Untuk Kegunaan Pemeriksa SHAPES 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 360o area of sector angle subtended at centre 13 = area of circle 360o PA' 14 Scale factor , k = PA 15 Area of image = k 2 × area of object 1449/2 3 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk Kegunaan Section A Pemeriksa [52 marks] Answer all questions in this section. 3 3− x 1 Solve the quadratic equation x= 4 x−3 [4 marks] Answer : 2 Calculate the value of p and of q that satisfy the following simultaneous linear equations: 2 p − 3q = 6 4 p + 9q = −3 [4 marks] Answer : 1449/2 4 SULIT
    • SULIT 1449/2 Untuk 3 (a) State whether the following statement is true or false. Kegunaan Pemeriksa 6 ÷ 2 = 3 and 3 2 = 6 (b) Write down two implications based on the sentence below. 4m > 20 if and only if m > 5 (c) Given the number sequence 5, 18, 39, 68, … which follows the pattern 5 = 1 + 4(12) 18 = 2 + 4(22) 39 = 3 + 4(32) 68 = 4 + 4(42) …………….. …………….. Form a general conclusion by using the induction method for the numerical sequence above. [5 marks] Answer : (a) (b) Implication 1 : _________________________________________________ Implication 2 : _________________________________________________ (c) 1449/2 5 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk Kegunaan 4 Diagram 1 shows a combined solid consists of a cuboid JKLMPQRS and half cylinder Pemeriksa JMNTSP. J 8 cm K N 7 cm M L 16 cm S R T P Q DIAGRAM 1 22 Using π = , calculate the volume, in cm3, of the solid 7 [4 marks] Answers: 1449/2 6 SULIT
    • SULIT 1449/2 Untuk 5 The Venn diagram in the answer space shows sets P, Q and R such that the universal set, Kegunaan Pemeriksa ξ = P ∪ Q ∪ R. On the diagrams in the answer space, shade (a) P′ ∩ R′ (b) Q ∩ R ∪ P [3 marks] Answer : (a) Q R P Q (b) R P 1449/2 7 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk 6 Diagram 2 shows a cuboid with horizontal rectangular base, ABCD. Kegunaan Pemeriksa E H F G 6 cm D C 8 cm A 10 cm B DIAGRAM 2 Calculate the angle between the plane ABE and the base ABCD. [4 marks] Answer : 1449/2 8 SULIT
    • SULIT 1449/2 7 In Diagram 3, O is the origin. KL, PQ and RS are straight lines. PQ is parallel to RS. Untuk The equation of the straight line KL is y = −2 x + 6 . Kegunaan Pemeriksa y K R P • x O S Q(−3, −3) • L DIAGRAM 3 Find the (a) coordinate of P, (b) gradient of the straight line PQ, (c) equation of the straight line RS. [6 marks] Answer: (a) (b) (c) 1449/2 9 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk 8 (a) Identify the antecedent and consequent in the following implication. Kegunaan Pemeriksa ‘ If a triangle has two equal sides, then it is an isosceles triangle.’ (b) State the converse of the following implication. If x > 5 , then x2 > 25 (c) Complete the premise in the following argument. Premise 1 : If x is an angle in a semicircle, then x = 90° Premise 2 : _________________________________________________ Conclusion : x = 90° (d) Complete the following argument Premise 1 : _________________________________________________ Premise 2 : M ∩ N ≠ M Conclusion : M ⊄ N [6 marks] Answer: (a) Antecedent : ………………………………………………………………… Consequent : …………………………………………………………………. (b) ………………………………………………………………………………… (c) Premise 2 : …………………………………………………………………… (d) Premise 1 : ……………………………………………………………………. 1449/2 10 SULIT
    • SULIT 1449/2 9 Ten cards are placed in an empty box. Untuk Kegunaan Pemeriksa E X C E L L E N C E (a) If a card is selected at random from the box, state the probability that the card marked C is selected. (b) A number of cards marked C are added into the box. If a card is selected at random 3 from the box, the probability of selecting a card marked C is . 7 Find the number of cards that has been added into the box. [4 marks] Answer : (a) (b) 1449/2 11 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk 10 In Diagram 4, ABOD is a square and PQR is a semicircle with centre O. Kegunaan Pemeriksa A B P Q D 14 cm O R DIAGRAM 4 22 OR = 7 cm. Using π = , calculate 7 (a) the perimeter, in cm, of the whole diagram, (b) the area, in cm2, of the shaded region. [7 marks] Answer : (a) (b) 1449/2 12 SULIT
    • SULIT 1449/2 Untuk 11 Diagram 5 shows a right pyramid with rectangular base PQRS . Kegunaan Pemeriksa The apex V is 5 cm vertically above the point T. V S R T 6 cm P 8 cm Q DIAGRAM 5 Calculate the angle between the line VP and the base PQRS. [5 marks] Answer: 1449/2 13 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk Kegunaan Section B Pemeriksa [48 marks] Answer any four questions in this section. 12 Table 1 in the answer space shows the distribution of the ages of 200 participants in a big walk event. (a) Using the data in Table 1, complete the table provided in the answer space. [4 marks] (b) By using a scale of 2 cm to 5 years on the x-axis and 2 cm to 20 participants on the y-axis , draw an ogive for the data. [5 marks] (c) Based on your ogive in (b), (i) find the inter quartile, (ii) explain briefly the meaning of the third quartile. [3 marks] Answer: (a) Age (years) Frequency Cumulative Upper Frequency Boundary 15 − 19 10 20 − 24 20 25 − 29 50 30 − 34 60 35 − 39 36 40 − 44 18 45 − 49 6 TABLE 1 (b) Refer graph on page 15. (c) (i) (ii) 1449/2 14 SULIT
    • SULIT 1449/2 Graph for Question 12 Untuk Kegunaan Pemeriksa 1449/2 15 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk Kegunaan 13 (a) Diagram 6 shows sectors OTQR, OABC and OCP with the same centre O. Pemeriksa TAOP and OCR are straight lines. Q R B C P T A O DIAGRAM 6 21 OR = 2OC , OC = cm and ∠COP = 60°. 2 22 Using π = , calculate 7 (i) the perimeter, in cm, of the whole diagram, [3 marks] (ii) the area, in cm2, of the shaded region. [4 marks] (b) Diagram 7 shows some number cards. 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 DIAGRAM 7 A card is picked at random. State the probability of choosing a card containing (i) digit 5, (ii) a multiple of 5, (iii) a number such that when divided by 5, the remainder is 1. [5 marks] 1449/2 16 SULIT
    • SULIT 1449/2 Untuk Answer: Kegunaan Pemeriksa 13 (a) (i) (ii) (b) (i) (ii) (iii) 1449/2 17 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk 14 The data in Diagram 8 show the marks of 40 students for the Mathematics monthly test. Kegunaan Pemeriksa 46 53 44 60 42 38 31 55 35 37 54 32 46 56 40 60 52 40 34 45 52 35 50 36 47 38 40 48 45 42 53 44 50 44 58 51 36 48 56 32 DIAGRAM 8 (a) Using the data in Diagram 8 and a class interval of 5 marks, complete Table 2 in the answer space. [4 marks] (b) Based on Table 2 in (a), (i) state the modal class, (ii) calculate the mean mark of the Mathematics monthly test and give your answer correct to 2 decimal places. [4 marks] (c) For this part of the question, use the graph paper provided on page 19. By using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a histogram for the data. [4 marks] 1449/2 18 SULIT
    • SULIT 1449/2 Answer: Untuk Kegunaan 14 (a) Pemeriksa Class interval Midpoint Frequency 31 – 35 36 - 40 TABLE 2 (b) (i) (ii) (c) Refer to the graph on page 20. 1449/2 19 [Lihat sebelah SULIT
    • SULIT 1449/2 Graph for Question 14 Untuk Kegunaan Pemeriksa 1449/2 20 SULIT
    • SULIT 1449/2 15 (a) Diagram 9 is a Venn diagram showing the number of students in a class who play at least Untuk one of the three games. Given the universal set, ξ = F ∪ R ∪ H. Kegunaan Pemeriksa F x 7 5 R 2x H 10 + x 18 – 2x 10 DIAGRAM 9 Given that set F = { students who play football } set R = { students who play rugby } and set H = { students who play hockey}. (i) Find the number of students who play hockey. (ii) If the number of students who play football is 30, find (a) the number of students who play rugby, (b) the number of students who play one game only. [6 marks] Answer: (a) (i) (ii) (a) (b) 1449/2 21 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk 15 (b) In Diagram 10 , TQRS, NR and OQ are straight lines drawn on a Cartesan plane. O is the Kegunaan Pemeriksa origin and OQ is parallel to NR. The gradient of NR = 2. y S •R • Q(p, 3) T • • x N(-6, 0) 0 DIAGRAM 10 Find (i) the value of p, (ii) the coordinate of R, (iii) the equation of the straight line ST. [6 marks] Answer: (b) (i) (ii) (iii) 1449/2 22 SULIT
    • SULIT 1449/2 2w 2 − 3 16 (a) Solve the equation = 1. Untuk 5w Kegunaan Pemeriksa [4 marks] (b) Calculate the values of x and y that satisfy the following simultaneous linear equations: 2x + y = 3 4x - 3y = 11 [4 marks] (c) (i) Form a true compound statement by combining the two statements given below. (a) 52 = 10 1 (b) = 0.25 4 (ii) Form a general conclusion by induction based on the numerical sequence below. 2, 9, 16, 23 … 2 = 2 + 7 (0) 9 = 2 + 7 (1) 16 = 2 + 7 (2) 23 = 2 + 7 (3) ………………. (iii) Complete the following sentence using a suitable quantifier to make it a true statement. ‘………………. prime numbers are odd numbers.’ [4 marks] 1449/2 23 [Lihat sebelah SULIT
    • SULIT 1449/2 Untuk Answer: Kegunaan Pemeriksa 16 (a) (b) (c) (i) (ii) (iii) 1449/2 24 SULIT