PAPER 1 Question-Answer Book INSTRUCTIONS A1(33) A2(33) B(34) Total. Name: MATHEMATICS Compulsory Part 1. Class: ( ) Marker s Use Only 1 /3 2 /3 3 /3

Transcription

1 MATHEMATICS Compulsory Part 1 F5 Final Exam 016 Name: Class: ( ) PAPER 1 Question-Answer Book Question No. Marker s Use Only Marks Time allowed: hours 15 mins This paper must be answered in English. 1 /3 /3 3 /3 INSTRUCTIONS 1. Write your Name, Class and Class No. in the space provided on Page 1.. This paper consists of THREE sections, A(1), A() and B. A(1) and A() carries 33 marks and Section B carries 34 marks. 3. Attempt ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer Book. Do not write in the margins. 4. Unless otherwise specified, all working must be clearly shown. 5. Unless otherwise specified, numerical answers should be either exact or correct to 3 significant figures. 6. The diagrams in this paper are not necessarily drawn to scale. 7. Graph paper and supplementary answer sheets will be supplied on request. Write your name, class and class no. and put them INSIDE this book. 4 /4 5 /4 6 /4 7 /4 8 /4 9 /4 10 /6 11 /6 1 /6 13 /7 14 /8 15 /6 16 /6 17 /6 18 /8 19 /8 A1(33) A(33) B(34) Total F5 Final Exam 016 1

2 F5 Final Exam 016

3 SECTION A(1) (33 marks) Answer ALL questions in this section and write your answers in the spaces provided. 1. Simplify 7 8 (x y ) x 4 y 6 and express the answer with positive indices. (3 marks). Make x the subject of the formula Ax = ( 3x + B) C. (3 marks) F5 Final Exam 016 3

4 F5 Final Exam 016 4

5 Solve the equation = 1 x 1 x +. (3 marks) 4. Factorize (a) 5m 10n, (b) m + mn 8n, (c) m + mn 8n 5m + 10n. (4 marks) F5 Final Exam 016 5

6 5. Given that log = a and log 3 = b, express each of the following in terms of a and b. (a) log 4, (b) log 15, (c) log 18. (4 marks) F5 Final Exam 016 6

7 6. It is given that f (x) is the sum of two parts, one part varies as x and the other part varies as Suppose that f ( 3) = 48 and f ( 9) = 198. (a) Find f (x). x. (b) Solve the equation f ( x) = 90. (4 marks) F5 Final Exam 016 7

8 7. Bobby is going to arrange 7 different magazines on a bookshelf. (a) How many ways are there to arrange the magazines? (b) If two soccer magazines should be put together, how many ways are there to arrange the magazines? (4 marks) 8. Simplify 5i 3 i and express the answer in standard form. (4 marks) F5 Final Exam 016 8

9 9. The following stem-and-leaf diagram shows the weights of 0 boys. Stem (10 kg) Leaf (1 kg) (a) Find the median weight. (1 mark) (b) Find the inter-quartile range of the weight of the boys. (1 mark) (c) Construct a box-and-whisker diagram to represent the data in the graph paper. ( marks) F5 Final Exam 016 9

10 Section A() (33 marks) Answer ALL questions in this section and write your answers in the spaces provided. 10. Consider a quadratic equation ax bx + c = 0. (a) Find the discriminant of the equation. (1 mark) (b) If a, b, c form a geometric sequence, (i) show that the equation has one double root; (ii) express the root of the equation in terms of a and b. (3 marks) (c) Using the result of (b)(ii), solve 5x 10x + 5 = 0. ( marks) F5 Final Exam

11 4 11. In Figure 1, L 1 : y = x + 8 and L : x + y = 7 cuts the x-axis at A and B respectively. 3 Two lines intersect at P. (a) Find the coordinates of P. ( marks) (b) What are the coordinates of A and B? ( marks) (c) Find the area of PAB. ( marks) Figure 1 F5 Final Exam

12 a 1. Figure shows the graph of y =. x + 1 Figure (a) Find, in terms of a, the y-intercept. (3 marks) (b) If the graph passes through (1, 1), find the value of a. (3 marks) F5 Final Exam 016 1

13 13. In Figure 3, BD is an angle bisector of ADC. AD is a diameter of the circle. If DAC = 4, (a) find (b) find ADC, (3 marks) CAB. (4 marks) Figure 3 F5 Final Exam

14 14. There are n rows of seats in a theatre. It is known that the first row has 15 seats and each row has seats more than the preceding row, as shown in the Figure 4. Figure 4 (a) Find the number of seats in the nth row in terms of n, where n > 1. (3 marks) (b) Find the total number of seats in the first n rows. (1 mark) (c) If the number of seats in the theatre is 79, find the number of rows of seats in the theatre. (4 marks) F5 Final Exam

15 Section B (34 marks) Answer ALL questions in this section and write your answers in the spaces provided. 15. A committee consists of 4 teachers and 6 parents. Four persons are selected randomly from the committee. (a) Find the probability that at least parents are selected. (3 marks) (b) Eric claims that the probability of selecting same numbers of teachers and parents is equal to 0.5. Do you agree? Explain your answer. (3 marks) F5 Final Exam

16 16. In a test, the mean of the distribution of the scores of a class of students is 61 marks. The standard scores of Albert and Mary are.6 and 1.4 respectively. Albert gets marks. A student claims that the range of the distribution is at most 59 marks. Is the claim correct? Explain your answer. (6 marks) F5 Final Exam

17 17. Solve the equation 6cos θ + sinθ 4 = 0 for 0 θ < 360. (Give the answers correct to the nearest degree.) (6 marks) F5 Final Exam

18 18. The total cost $C of running a camp is partly constant and partly varies directly as the number of students n who join the camp. If 60 students join the camp, the cost per student is $95. If 90 students join the camp, the cost per student is $85. (a) What is the total cost if 60 students join the camp? ( marks) (b) Set up an equation in relating C and n. (3 marks) (c) If 80 students join the camp and the school subsidizes $000 for the camp, how much should each student pay? (3 marks) F5 Final Exam

19 F5 Final Exam

20 19. Figure 5 shows a triangular board with AB = 8 cm, BC = 6 cm and CA = 9 cm. The board is supported by a rod AP, where AP = 5 cm. Figure 5 P, B and C lie on the ground and the bearing of C from P is S 10 E. (a) Find the area of ABC. (3 marks) (b) Find BPC. (3 marks) (c) Hence find the bearing of B from P. ( marks) F5 Final Exam 016 0

21 F5 Final Exam 016 1

22 END OF PAPER F5 Final Exam 016

23 Solution SECTION A(1) (33 marks) 7 8 (x y ) 1. Simplify and express the answer with positive indices. (3 marks) x 4 y (x y ) 4 6 x y x 14 y 16 = 4 x y 6 = x 14 4 y 16 ( 6) = x 10 y A. Make x the subject of the formula Ax = ( 3x + B) C. (3 marks) Ax = 3 Cx + BC Ax 3 Cx = BC x ( A 3C) = BC BC x = ( A 3C) A Solve the equation = 1 (3 marks) x 1 x = 1 x 1 x + 6(x + ) 10(x 1) = (x 1)(x + ) 6x x + 10 = x + x x x + 5x 4 = 0 (x 3)(x + 8) = 0 x = 8 or Factorize (a) 5m 10n, (b) m + mn 8n, (c) m mn 8n + 5m + 10n. (4 marks) (a) 5m 10n = 5( m n) (b) m + mn 8n = ( m + 4n)( m n) (c) m + mn 8n 5m + 10n = 8 m + mn n (5m 10n) = ( m + 4n)( m n) 5( m n) = ( m + 4n 5)( m n) A F5 Final Exam 016 3

24 5. Given that log = a and log 3 = b, express each of the following in terms of a and b. (a) log 4, (b) log 15, (c) log 18. (4 marks) (a) log 4 = log( 3 3) (b) = log 3 + log3 = 3 log + log3 = 3 a + b 30 log 15 = log 3 10 = log = log3 + log10 log = a +1 b 1 (c) log 18 = log18 1 = log(3 ) 1 = (log3 + log ) 1 = (b + a) 1 b + a = 6. It is given that f (x) is the sum of two parts, one part varies as x and the other part varies as Suppose that f ( 3) = 48 and f ( 9) = 198. (a) Find f (x). (b) Solve the equation f ( x) = 90. (4 marks) (a) Let f ( x) = hx + kx, where h and k are non-zero constants So, we have 3 h + 9k = 48 and 9 h + 81k = 198. Solving, we have h = 13 and k = 1 Thus, we have f ( x) = 13x + x (b) f ( x) = x + x = 90 x + 13x 90 = 0 ( x 5)( x + 18) = 0 x = 5 or 18 x. 7. Bobby is going to arrange 7 different magazines on a bookshelf. (a) (b) How many ways are there to arrange the magazines? If two soccer magazines should be put together, how many ways are there to arrange the magazines? (4 marks) (a) Number of ways = 7! = 5040 (b) Number of ways = (5 + 1)!! =1440 F5 Final Exam 016 4

25 8. Simplify 5i 3 i 5i 5i 3 + i = 3 i 3 i 3 + i 5i(3 + i) = 3 ( i) and express the answer in standard form. 15i + 5i = 9 i i = = + i A 9. The following stem-and-leaf diagram shows the weights of 0 boys. (4 marks) Stem (10 kg) Leaf (1 kg) (a) Find the median weight. (1 mark) (b) Find the inter-quartile range of the weight of the boys. (1 mark) (c) Construct a box-and-whisker diagram to represent the data in the graph paper. ( marks) (a) Median = kg = 43 kg (b) (c) Q1 = kg = 40 kg Q3 = kg = 55 kg Inter-quartile range = Q3 Q1 = (55 40) kg = 15 kg A F5 Final Exam 016 5

26 Section A() (33 marks) Answer ALL questions in this section and write your answers in the spaces provided. 10. Consider a quadratic equation ax bx + c = 0. (a) Find the discriminant of the equation. (1 mark) (b) If a, b, c form a geometric sequence, (i) show that the equation has one double root; (ii) express the root of the equation in terms of a and b. (3 marks) (c) Using the result of (b)(ii), solve 5x 10x + 5 = 0. ( marks) (a) = ( b) 4( a)( c) = 4b 4ac (b) (i) Let r be the common ratio. (ii) b = ar c = ar = 0 By the result of (a), = 4( ar) 4a( ar ) = 4a r 4a r The equation has one double root. The root of the equation b ± 0 = a b = a (c) Let a = 5, b = 10 and c = 5. b a = 10 = 5 c 5 = = a 10 5, 10, 5 form a geometric By (b)(ii), sequence. x = 10 = In Figure 1, L 1 : y = x + 8 and L : x + y = 7 cuts the x-axis at A and B respectively. 3 Two lines intersect at P. (a) Find the coordinates of P. ( marks) (b) What are the coordinates of A and B? ( marks) (c) Find the area of PAB. ( marks) Figure 1 F5 Final Exam 016 6

27 (a) 4 y = x (1) 3 x + y = 7...() Substituting (1) into (), 4 x + x + 8 = x + x + 16 = x = 11 3 x = 3 Substituting x = 3 into (1), 4 y = (3) = 1 The coordinates of P are (3, 1). (b) For L, when y = 0, = x x = 6 The coordinates of A are ( 6, 0). For L, when y = 0, x + (0) = 7 x = 7 The coordinates of B are (7, 0). (c) AB = [7 ( 6)] units = 33 units For the base AB, height = (1 0) units = 1 units 33 1 The area of PAB = sq. units = 198 sq. units A a 1. Figure shows the graph of y =. x + 1 Figure F5 Final Exam 016 7

28 (a) Find, in terms of a, the y-intercept. (3 marks) (b) If the graph passes through (1, 1), find the value of a. (3 marks) (a) When x = 0, a y = = a The y-intercept is a. (b) Substituting (1, 1) into the function, a 1 = 1+ 1 a = 1 A 13. In Figure 3, BD is an angle bisector of ADC. AD is a diameter of the circle. If DAC = 4, (a) find ADC, (3 marks) (b) find CAB. (4 marks) A Figure 3 (a) ACD = 90 ( in semicircle) In ACD, DAC + ADC + ACD = 180 ( sum of ) 4 + ADC + 90 = 180 ADC = 48 (b) BD is an angle bisector of ADC. 1 BDC = ADC = 4 CAB = BDC ( s in the same segment) = 4 A 14. There are n rows of seats in a theatre. It is known that the first row has 15 seats and each row has seats more than the preceding row, as shown in the Figure 4. Figure 4 (a) Find the number of seats in the nth row in terms of n, where n > 1. (3 marks) (b) Find the total number of seats in the first n rows. (1 mark) (c) If the number of seats in the theatre is 79, find the number of rows of seats in the theatre. (4 marks) (a) First term = 15 Common difference = T(n) = 15 + (n 1)() F5 Final Exam 016 8

29 = 13 + n There are 13 + n seats in the nth row. (b) The total number of seats in first n rows n = ( n) = n + 14n (b) n + 14n = 79 n + 14n 79 = 0 (n + 36)(n ) =0 n= or 36 (rejected) + The number of rows of seats in the theatre is. Section B (34 marks) Answer ALL questions in this section and write your answers in the spaces provided. 15. A committee consists of 4 teachers and 6 parents. Four persons are selected randomly from the committee. (a) Find the probability that at least parents are selected. (3 marks) (b) Eric claims that the probability of selecting same numbers of teachers and parents is equal to 0.5. Do you agree? Explain your answer. (3 marks) (a) P(at least parents) = 1 P(no parent) P(1 parent) C 4 C1 C3 = C C = = 4 (b) P(same numbers of teachers and parents) 6 4 C C C = 10 4 = 7 3 No. It is because the probability equals 7 3, not 0.5. F5 Final Exam 016 9

30 16. In a test, the mean of the distribution of the scores of a class of students is 61 marks. The standard scores of Albert and Mary are.6 and 1.4 respectively. Albert gets marks. A student claims that the range of the distribution is at most 59 marks. Is the claim correct? Explain your answer. (6 marks) Let σ marks be the standard deviation of the distribution 61 =.6 σ σ =15 The score of Mary = σ = (15) = 8 marks The difference of the score of Mary and the score of Albert = 8 = 60 marks > 59 marks A Note that the range of the distribution is at least the difference of the score of Mary and the score of Albert. Therefore, the range of the distribution exceeds 59 marks. Thus the claim is incorrect. 17. Solve the equation 6cos θ + sinθ 4 = 0 for 0 θ < 360. (Give the answers correct to the nearest degree.) (6 marks) 6cos θ + sinθ 4 = 0 6(1 sin θ ) + sinθ 4 = 0 6sin θ sinθ = 0 ( 3sinθ )(sinθ + 1) = 0 1 sinθ = or A 3 θ , , or = 4, 138, 10 or 330 (cor. to the nearest degree) A 18. The total cost $C of running a camp is partly constant and partly varies directly as the number of students n who join the camp. If 60 students join the camp, the cost per student is $95. If 90 students join the camp, the cost per student is $85. (a) What is the total cost if 60 students join the camp? ( marks) (b) Set up an equation in relating C and n. (3 marks) (c) If 80 students join the camp and the school subsidizes $000 for the camp, how much should each student pay? (3 marks) F5 Final Exam

31 (a) Total cost = $95 60 = $ 5700 A (b) For n = 60, C = 5700 (from (a)) 5700 = k (1) 1 k For n = 90, C = = = k () 1 k () (1): 1950 = 30k k = 65 Substituting k = 65 into (1), 5700 = k (65) k 1 = 1800 C = n (c) When n = 80, C = (80) = 7000 The total cost is $7000. The cost for each student = $ 80 = $ Figure 5 shows a triangular board with AB = 8 cm, BC = 6 cm and CA = 9 cm. The board is supported by a rod AP, where AP = 5 cm. Figure 5 P, B and C lie on the ground and the bearing of C from P is S 10 E. (a) Find the area of ABC. (3 marks) (b) Find BPC. (3 marks) (c) Hence find the bearing of B from P. ( marks) F5 Final Exam

32 (a) s = cm = = = 11.5 cm Area of ABC 11.5(11.5 8)(11.5 6)(11.5 9) cm 11.5(3.5)(5.5)(.5) cm = 3.5 cm (cor. to 3 sig. fig.) (b) Consider ACP, CP = AC AP (Pyth. theorem) = 9 5 cm = 56 cm Consider ABP, BP = AB AP (Pyth. theorem) = 8 5 cm = 39 cm Consider BPC, by the cosine formula, ( 39) + ( 56) 6 cos BPC = ( 39)( 56) 59 cos BPC = BPC = 50.9 (cor. to 3 sig. fig.) (c) The bearing of B from P is S( )W = S40.9 W. A END OF PAPER F5 Final Exam 016 3