2017 LCHL Paper 1 Table of Contents

Transcription

1

2 2017 PAPER 1

3 INSTRUCTIONS There are two sections in this examination paper. Section A Concepts and Skills 150 marks 6 questions Section B Contexts and Applications 150 marks 3 questions Answer all nine questions. Write your answers in the spaces provided in this booklet. You may lose marks if you do not do so. There is space for extra work at the back of the booklet. You may ask the superintendent for more paper. Label any extra work clearly with the question number and part. The superintendent will give you a copy of the Formulae and Tables booklet. You must return it at the end of the examination. You are not allowed to bring your own copy into the examination. You will lose marks if you do not show all necessary work. You may lose marks if you do not include appropriate units of measurement, where relevant. You may lose marks if you do not give your answers in simplest form, where relevant. Write the make and model of your calculator(s) here:

4 2017 LCHL Paper 1 Table of Contents Section A Question 1 (a) (b) Question 1 (c) Question 2 (a) Question 2 (b) Question 2 (b) (ALTERNATE VERSION) Question 3 (a) Question 3 (b) Question 4 (a) Question 4 (b) Question 5 (a) Question 5 (b) Question 5 (c) Question 6 (a) Question 6 (b) Section B Question 7 (a) Question 7 (b) (c) Question 7 (d) Question 7 (e) Question 7 (f) Question 7 (g) Question 8 (a) Question 8 (b) (i) Question 8 (b) (ii) Question 8 (b) (iii) Question 8 (b) (iv) Question 8 (b) (v) Question 8 (b) (vi) Question 9 (a) Question 9 (b) (i) Question 9 (b) (ii) Question 9 (c)

5 Section A Concepts and Skills 150 Marks Answer all six questions from this section.

6 2017 LCHL Paper 1 Question 1 (a) 5+10 Marks Write the function f x = 2x 2 7x 10, where x R, in the form a x + h 2 + k, where a, h, and k Q. f x = 2x 2 7x 10 f x = 2 x x 5 f x = 2 x x f x = 2 x x f x = 2 x 7 4 f x = 2 x Complete the Square LCHL Paper 1 Question 1 (b) Hence, write the minimum point of f f x = a x + h 2 + k Min/Max = h, k f x = 2 x 7 4 Minimum 7 4,

7 2017 LCHL Paper 1 Question 1 (c) (i) 5+5 Marks Explain why f must have two real roots. Curve is shaped with a minimum value that is below the x axis. Therefore the curve crosses the x axis twice and has 2 real roots. Real Roots b 2 4ac 0 a = 2 b = 7 c = 10 b 2 4ac (ii) Hence, write the minimum point of f x = b ± b2 4ac 2a f x = 2x 2 7x 10 2x 2 7x 10 = 0 a = 2 b = 7 c = 10 = 7 ± ± = 4 = 7 ± 129 4

8 2017 LCHL Paper 1 Question 2 (a) 15 Marks z = 3 + i, where i 2 = 1. Use De Moivre s Theorem to write in the form z 4, where a + b ci, where a, b, and c Z. We can sketch 3 + i to find the modulus, r, and argument, θ and then write in polar form, z = r(cos θ + i sin θ) z = 3 + i z = r cos θ + i sin θ z = r cos θ + i sin θ z = 2 cos 5π 6 + i sin 5π 6 Im r = z = a 2 + b 2 Application of De Moivre s Theorem z = i r α θ 3 Re = = = 2 tan = opposite adjacent tan α = 1 3 α = tan 1 1 α = 30 3 θ = π π 6 θ = 5π 6 r cos θ + i sin θ n = r n cos nθ + i sin nθ z = 2 cos 5π 6 + i sin 5π 6 z 4 = 2 4 cos 4 5π + i sin 4 5π 6 6 z 4 = 16 cos 10π 10π + i sin 3 3 z 4 = i z 4 = 8 8 3i α = π 6

9 2017 LCHL Paper 1 Question 2 (b) 10 Marks The complex number w is such that w = 3 and w makes an angle of 30 with the positive sense of the real axis. If t = zw, write t in its simplest form. Division/ Multiplication in Polar Form z 1 = r 1 cos θ 1 + i sin θ 1 z 2 = r 2 cos θ 2 + i sin θ 2 w = 3 arg w = 30 z = 3 + i z 1 z 2 = r 1 r 2 cos θ 1 + θ 2 + i sin θ 1 + θ 2 z 1 = r 1 cos θ z 2 r 1 θ 2 + i sin θ 1 θ 2 2 = π 6 z = 2 cos 5π 6 + i sin 5π 6 Express in Polar Form Polar Form z = r(cos θ + i sin θ) t = zw t = 2 cos 5π 6 + i sin 5π 6 3 cos π 6 + i sin π 6 w = 3 cos π 6 + i sin π 6 t = 2 3 cos 5π 6 + π 6 + i sin 5π 6 + π 6 t = 6 cos π + i sin π t = i t = 6

10 2017 LCHL Paper 1 Question 2 (b) 10 Marks The complex number w is such that w = 3 and w makes an angle of 30 with the positive sense of the real axis. If t = zw, write t in its simplest form. Alternate Method Let w = a + bi w = 3 a + bi = 3 a 2 + b 2 = 3 a 2 + b 2 = 9 tan 30 = b a 1 3 = b a a = 3b a 2 + b 2 = 9 3b 2 + b 2 = 9 3b 2 + b 2 = 9 4b 2 = 9 b 2 = 9 4 b = 3 2 Positive only as we are dealing with positive sense of real axis. a = 3b a = a = w = i z = t = zw 3 + i t = 3 + i i t = i + i i t = i i i2 t = t = 12 2 t = 6 TOO LONG.but works

11 2017 LCHL Paper 1 Question 3 (a) 10 Marks Differentiate 1 3 x2 x + 3 from 1 st principles with respect to x. f x = 1 3 x2 x + 3 By Rule (no marks for this) 2 3 x 1 Differentiation from 1 st Principles f x = f x + h = f x + h f x = f x + h f x h = lim h 0 f x + h f x h = f x + h = 1 3 x + h 2 x + h + 3 f x + h f x = 1 3 x + h 2 x + h x2 x + 3 = 1 3 x2 + 2xh + h 2 x h x2 + x 3 = 1 3 x xh h2 x h x2 + x 3 = 2 3 xh h2 h f x + h f x = 2 h 3 x h 1 f x + h f x lim = 2 n h 3 x 1

12 2017 LCHL Paper 1 Question 3 (b) 5 Marks f x = ln 3x and g x = x + 5, where x R. Find the value of the derivative of f g x at x = 1 4. Give your answer correct to 3 decimal places. We first need to find the f g x We do this by subbing g x in for x in the f x function. Then find the derivative of this. f x = ln 3x g x = x + 5 f g x = ln 3 x = ln 3 x x = ln 3x x = ln 3x x + 77 Notation f g x = f g x y is the same as f x dy dx the same as f x differentiate derive get the derivative of f g x = ln 3x x f g x = 3x 2 6x x x + 30 = 3x x f 1 4 = f 1 4 = f x = ln u f x = 1 du u dx x = 1 4 Only 5 marks for this whole question. Differentiate anything correctly and you got 3.

13 2017 LCHL Paper 1 Question 4 (a) 15 Marks The amount of a substance remaining in a solution reduces exponentially over time. An experiment measures the percentage of the substance remaining in the solution. The percentage is measured at the same time each day. The data collected over the first 4 days are given in the table below. Based on the data in the table, estimate which is the first day on which the percentage of the substance in the solution will be less than 0.01%. General Term of a Geometric Series T n = ar n 1 r = T 2 T 1 a = 0.95 r = T 2 T 1 r = r = 9 20 T n = ar n 1 T n = n log n n 1 < n 1 n 1 log < < log n 1 > log log 9 20 n > log log 9 20 n > Day Percentage of substance (%) For what value of n is the percentage of the substance less than < log Get the log of both sides. Rule of Logs log a x q = q log a x Note: Change the direction of the inequality as log 9 is negative. 20 During the 12 th Day the substance in the solution will be less than 0.01%.

14 2017 LCHL Paper 1 Question 4 (b) 10 Marks A square has sides of length 2 cm. The midpoints of the sides of this square are joined to form another square. This process is continued. The first three squares in the process are shown below. Find the sum of the perimeters of the squares if this process is continued indefinitely. Give your answer in the form a + b c cm, where a, b, and c N. x Find the perimeters of the 1 st three squares. Square 1 l = 2 T 1 = 4 2 = 8 Sum of Perimeters r = T 2 T 1 a = 8 Sum to Infinity of a Geometric Series S = a 1 r given r < 1 = 8 Square 2 l = 2 T 2 = 4 2 = 4 2 Square 3 l = 1 r = T 2 T 1 r = r = = = = x 1 Pythagoras c 2 = a 2 + b 2 1 T 2 = 4 1 = 4 x 2 = x 2 = 2 x = 2

15 2017 LCHL Paper 1 Question 5 (a) 15 Marks The function f is such that f x = 2x 3 + 5x 2 4x 3, where x R. Show that x = 3 is a root of f x and find the other two roots. f x = 2x 3 + 5x 2 4x 3 f 3 = = = 0 Therefore x = 3 is a root of f x 2x 2 x 1 x + 3 2x 3 + 5x 2 4x 3 2x 3 + 6x 2 x 2 4x 3 + x 2 + 3x x x 3 0 If x = a is a root then x a is a factor. No remainder in division may be stated as reason for x = 3 as root. 2x 2 x 1 = 0 2x + 1 x 1 = 0 2x = 1 x = 1 x = 1 2

16 2017 LCHL Paper 1 Question 5 (b) 5 Marks Find the co-ordinates of the local maximum point and the local minimum point of the function f. f x = 2x 3 + 5x 2 4x 3 Max/Min f x = 0 and solve for x f x = 6x x 4 6x x 4 = 0 3x 2 + 5x 2 = 0 3x 1 x + 2 = 0 Calculate the y co-ordinate by subbing x = 1 into f x. 3 f 1 3 = = , f 2 = = 9 3x = 1 x = 1 3 x = 2 2,9

17 2017 LCHL Paper 1 Question 5 (c) 5 Marks f x + a, where a is a constant, has only one real root. Find the range of possible values of a. f x = 2x 3 + 5x 2 4x 3 f x + a = 2x 3 + 5x 2 4x 3 + a Adding a shifts all points on the curve up a units. For f x to have one real root then the minimum point on the curve would have to be above the x-axis. 1 3, a >

18 2017 LCHL Paper 1 Question 6 (a) 15 Marks The graph of the function g x = e x, x R, 0 x 1, is shown on the diagram below. On the same diagram, draw the graph of h x = e x, x R, in the domain 0 x 1. Fill in an input-output table. x h x = e x y 0 e e e e e e h x = e x

19 2017 LCHL Paper 1 Question 6 (b) 10 Marks Find the area enclosed by g x = e x, h x = e x and the line x = Give your answer correct to 4 decimal places. x = 0.75 We can find the required area by integrating g x from x = 0 to x = and subtracting the integral of h x from x = 0 to x = A = න 0.75 e x dx න 0.75 e x dx 0 0 = e x e x = e 0.75 e 0 e 0.75 e 0 = e e = e e = units 2 h x = e x

20 Section B Contexts and Applications 150 Marks Answer all three questions from this section.

21 2017 LCHL Paper 1 Question 7 (a) 10 Marks Sometimes it is possible to predict the future population in a city using a function. The population in Sapphire City, over time, can be predicted using the following function: p t = Se 0.1t The population in Avalon, over time, can be predicted using the following function: q t = 3.9e kt In the functions above, t is time, in years; t = 0 is the beginning of 2010; and both S and k are constants. The population in Sapphire City at the beginning of 2010 is people. Find the value of S. Sapphire City p t = Se 0.1t 10 6 At the beginning of 2010, t = 0 and the population p 0 = 1, 100, 000 p t = Se 0.1t 10 6 Se = 1,100,000 Se 0 = 1,100, S 0 = 1.1 S = 1.1

22 2017 LCHL Paper 1 Question 7 (b) 10+5 Marks Find the predicted population in Sapphire City at the beginning of Sapphire City p t = 1.1e 0.1t 10 6 S = 1.1 At the beginning of 2015, t = 5 and we have to find p 5. p t = 1.1e 0.1t 10 6 p 5 = 1.1e p 5 = 1.1e p 5 = p 5 1,813,593 (c) Find the predicted change in the population in Sapphire City during Sapphire City p t = 1.1e 0.1t 10 6 S = 1.1 At the end of 2015, t = 6 and we have to find p 6. p t = 1.1e 0.1t 10 6 p 6 = 1.1e p 6 = 1.1e p 6 = p 6 2,004,331 Change in population = P 6 P 5 = = 190,738

23 2017 LCHL Paper 1 Question 7 (d) 15 Marks The predicted population in Avalon at the beginning of 2011 is people. Write down and solve an equation in k to show that k = 0.05, correct to 2 decimal places. Avalon q t = 3.9e kt 10 6 At the beginning of 2011, t = 1 and we have to find q 1 = q t = 3.9e kt e k = e k = e k = e k = k = log e k = ln k = 0.05 Rule of Logs a x = y log a y = x Note: log e x = ln x

24 2017 LCHL Paper 1 Question 7 (e) 5 Marks Find the year during which the populations in both cities will be equal. Sapphire City Avalon p t = 1.1e 0.1t 10 6 q t = 3.9e 0.05t 10 6 k = e 0.1t 10 6 = 3.9e 0.05t e 0.1t = 3.9e 0.05t 1.1e 0.1t = 3.9 e 0.05t e 0.1t e 0.05t = e 0.15t = t = log e t = ln ln t = 0.15 t = 8.44 years The populations will be equal during 2018 Rule of Indices a p a q = a p+q

25 2017 LCHL Paper 1 Question 7 (f) 5 Marks Find the predicted average population in Avalon from the beginning of 2010 to the beginning of Avalon q t = 3.9e 0.05t 10 6 q t = 3.9e 0.05t 10 6 Average Value of a Function b 1 b a න a f x dx න න e 0.05t e 0.05t dt dt The constant can go to the front of the integral. = e 0.05t = e 0.05t 0 15 = e e = e = = ,743,694

26 2017 LCHL Paper 1 Question 7 (g) 5 Marks Use the function q t = 3.9e 0.05t 10 6 to find the predicted rate of change of the population in Avalon at the beginning of Avalon q t = 3.9e 0.05t 10 6 Rate of Change q t q t = 3.9e 0.05t 10 6 q t = 3.9e 0.05t q t = 0.195e 0.05t 10 6 q 8 = 0.195e q 8 = 0.195e q 8 = q 8 = 130,712 At the beginning of 2018, t = 8.

27 2017 LCHL Paper 1 Question 8 (a) 5 Marks When a loan of P is repaid in equal repayments of amount A, at the end of each of t equal periods of time, where i is the periodic compound interest rate (expressed as a decimal), the formula below can be used to find the amount of each repayment. A = P i 1 + i t 1 + i t 1 Show how this formula is derived. You may use the formula for the sum of a finite geometric series. A 1 + i + A 1 + i 2 + A 1 + i A 1 + i t Sum of net present values of the Repayments is equal to P Sum of a Geometric Series a 1 rn S n = 1 r a = A 1 + i r = i n = t P = P = A 1 + i A 1 + i i i 1 + i i t i t A 1 + i t i t P = i P = A 1 + i t 1 i 1 + i t t i 1 + i A = P 1 + i t 1

28 2017 LCHL Paper 1 Question 8 (b) (i) 10 Marks Alex has a credit card debt of One method of clearing this debt is to make a fixed repayment at the end of each month. The amount of this repayment is 2.5% of the original debt. What is the fixed monthly repayment, A, required to pay the debt of 5000? Simply find 0.025% of = 125

29 2017 LCHL Paper 1 Question 8 (b) (ii) 10 Marks The annual percentage rate (APR) charged on debt by the credit card company is 21.75%, fixed for the term of the debt. Find as a percentage, correct to 3 significant figures, the monthly interest rate that is equivalent to an APR of 21.75%. This is not as simple as just dividing by 12. Compound Interest Formula F = P 1 + i t We do not know what P and F are so we can choose any initial amount, P. The easiest is to let P = 1. By doing that we can say that after 1 year P will have earned 21.75% interest so F = There are 12 months in a year so t = 12 and we need to solve for i = i = 1 + i = i i = i = 1.65%

30 2017 LCHL Paper 1 Question 8 (b) (iii) 10 Marks Assume Alex pays the fixed monthly repayment, A, each month and does not have any further transactions on that card. Complete the table below to show how the balance of the debt of 5000 is reducing each month for the first three months, assuming an APR of 21.75%, charged and compounded monthly % of Balance

31 2017 LCHL Paper 1 Question 8 (b) (iv) 10 Marks Using the formula you derived on the previous page, or otherwise, find how long it would take to pay off a credit card debt of 5000, using the repayment method outlined at the beginning of part (b) above. Give your answer in months, correct to the nearest month. Amortisation Formula A = P P = 5000 A = 125 i = t =? t i 1 + i 1 + i t 1 i 1 + i t A = P 1 + i t 1 t = = t t t t 1 = t t 125 = t t t = t = t = t = log t = t 66 months Rule of Logs a x = y log a y = x

32 2017 LCHL Paper 1 Question 8 (b) (v) 10 Marks Alex decides to borrow 5000 from the local Credit Union to pay off this credit card debt of The APR charge for the Credit Union loan is 8.5% fixed for the term of the loan. The loan is to be repaid in equal weekly repayments, at the end of each week, for 156 weeks. Find the amount of each weekly repayment. Convert the yearly interest rate to a weekly interest rate. This is not as simple as just dividing by 52. Compound Interest Formula F = P 1 + i t We do not know what P and F are so we can choose any initial amount, P. The easiest is to let P = 1. By doing that we can say that after 1 year P will have earned 8.5% interest so F = There are 52 weeks in a year so t = 52 and we need to solve for i = i = 1 + i = i i = i = 0.157% Amortisation Formula A = P P = 5000 A =? i = t = 156 i 1 + i t 1 + i t 1 i 1 + i t A = P 1 + i t A = A = A 36.16

33 2017 LCHL Paper 1 Question 8 (b) (vi) 5 Marks How much will Alex save by paying off the credit card debt using the loan from the Credit Union instead of paying the fixed repayment from part (b)(i) each month to the credit card company? Total amount using fixed payment of 125 for 66 months = 8250 Total amount using Credit Union loan for 156 weeks = Savings by using the Credit Union Loan

34 2017 LCHL Paper 1 Question 9 (a) 20 Marks The depth of water, in metres, at a certain point in a harbour varies with the tide and can be modelled by a function of the form f t = a + b cos ct where t is the time in hours from the first high tide on a particular Saturday and a, b, and c are constants. (Note: ct is expressed in radians.) On that Saturday, the following were noted: The depth of the water in the harbour at high tide was 5.5 m The depth of the water in the harbour at low tide was 1.7 m High tide occurred at 02:00 and again at 14:34. Use the information you are given to add, as accurately as you can, labelled and scaled axes to the diagram below to show the graph of f over a portion of that Saturday. The point P should represent the depth of the water in the harbour at high tide on that Saturday morning. Range and Period of Trigonometric Functions f x g x = a + b sin nx = a + b cos nx Range a + b, a b Period = 2π n f t High Tide = 5.5 Low Tide = 1.7 High Tide = 5.5 m Low Tide = 1.7 m P is on the vertical axis as t is measured in hours from the first high tide. At this time, t = 0, the height is 5.5 m Note: t is NOT the time on a clock. It is the time in hours, after the first high tide! t

35 2017 LCHL Paper 1 Question 9 (b) (i) 10 Marks Find the value of a and the value of b. High Tide = 5.5 a + b = 5.5 Low Tide = 1.7 a b = 1.7 f t a + b = 5.5 a b = 1.7 2a = 7.2 a = 3.6 = a + b cos ct a + b = b = 5.5 b = b = 1.9 Range and Period of Trigonometric Functions f x g x = a + b sin nx = a + b cos nx Range a + b, a b Period = 2π n f t 6 5 High Tide = Low Tide = Note: t is NOT the time on a clock. It is the time in hours, after the first high tide! t

36 2017 LCHL Paper 1 Question 9 (b) (ii) 5 Marks Show that c = 0.5, correct to 1 decimal place. 14: = 12 hours 34 mins There is hours between 60 two successive high tides. f t hours 34 mins f t = cos ct = 2π c c = 2π c = c 0.5 Range and Period of Trigonometric Functions f x g x = a + b sin nx = a + b cos nx Range a + b, a b Period = 2π n Note: t is NOT the time on a clock. It is the time in hours, after the first high tide! t

37 5 Marks 2017 LCHL Paper 1 Question 9 (c) Use the equation f t = a + b cos ct to find the times on that Saturday afternoon when the depth of the water in the harbour was exactly 5 2 m. Give each answer correct to the nearest minute. f t 6 5 f t = cos 0.5t Tide = cos 0.5t = cos 0.5t = cos 0.5t = 1.6 cos 0.5t = t = cos cos t = 0.5 t = hours 4 = 1 hour 8 mins Times 14: 34 1: 08 = 13: 26 14: : 08 = 15: Note: t is NOT the time on a clock. It is the time in hours, after the first high tide! t