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1 Leaving Certificate Examination, 213 Sample paper prepared by Mathematics Project Maths - Phase 2 Paper 1 Higher Level Saturday 18 May Paper written by J.P.F. Charpin and S. King 3 marks 1

2 Sample Instructions There are three sections in this examination paper: Section A Concepts and Skills 1 marks 4 questions Section B Contexts and Applications 1 marks 2 questions Section C Functions and Calculus (old syllabus) 1 marks 2 questions Answer questions as follows: In Section A, answer all four questions. In Section B, answer Question 5 and Question 6. In Section C, answer Question 7 and Question 8. Write your answers in the spaces provided in this booklet. There is space for extra work at the back of the booklet. You may also 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 booklet of Formulae and Tables. You must return it at the end of the examination. You are not allowed to bring your own copy into the examination. Marks will be lost if all necessary work is not clearly shown. Answers should include the appropriate units of measurement, where relevant. Answers should be given in simplest form, where relevant. Write the make and model of your calculator(s) here: 2

3 Section A Concepts and Skills Answer all four questions from this section. Question 1 1 Marks (a) If (x-2) is a factor in the cubic function f : x x 3 +x 2 +ax+1, find the value of a. (b) Find the other two roots of the function. x 3 +x 2-11x +1 (x-2) x 3 +2x 2 x 2 +3x 5 +3x 2 11x 3x 2 +6x 5x +1 5x 1 The two other roots are f(2) = a(2)+1 = 8+4+2a+1 = a(2)+1 = 2a+22 = a = 11 x = (1)( 5) = (1) 2 or x = (1)( 5) = (1) 2 (c) Using parts (a) and (b) or otherwise, solve the equation (Hint substitute x = y 2 ) y 6 +y 4 11y 2 +1 = (25 Marks) y 2 = 2 y 2 = 1.19 y 2 = 4.19 The third case is impossible, The four solutions are y = y = y = y =

4 Question 2 (a) Write the complex number z = 1+ 3i in polar form. z = 1+ 3i = = 1+3 = 4 = 2 z = 1+ ( ) 1 3 ( 3i = i = 2 cos π 3 +isin π ) 3 (25 Marks) (b) Calculate z 2 and z 3 using de Moivre s theorem. Place the three points corresponding to z, z 2 and z 3 on the Argand diagram. ( z 2 = 2 2 cos π 3 +isin π ) ( 2 = 4 cos 2π 3 3 +isin 2π ) 3 ( z 3 = 2 3 cos π 3 +isin π 3 = 8(cosπ +isinπ) = 8 3) 8 z z Im z 2 2 (c) What geometrical transformation is observed on the Argand diagram when a complex number is multiplied by z? Question 3 Rotation of angle π/3 Enlargement by factor 2 (a) Solve the following inequalities (i) x 4 x 1 > 3. Re (25 Marks) 4

5 Multiply both sides by (x+1) 2 (ii) x 4 > 3 (x 4)(x 1) > 3(x 1)2 x 1 x 2 5x+4 > 3x 2 6x+3 > 2x 2 x 1 > (2x+1)(x 1) The two roots of the polynomial on the right hand side of the inequality are x = 1/2 and x = 1. It is very easy to see that for example is a solution so the answer is ( ) x 1 > < x < 1 ( ) x 1 > ( ) x ( ) 1 1 ln > ln xln > ln 1 xln2 > ln5 5 xln2 < ln5 x < ln5 ln2 (b) Prove by induction that for any value of n N the number 1 n ( 1) n is a multiple of 11. For n= 1 ( 1) = 1 1 = This is a multiple of 11 Assume 1 n ( 1) n is a multiple of 11. It may be written 1 n ( 1) n = 11m = 1 n = 11m+( 1) n. At level n+1, 1 n+1 ( 1) n+1 = 1 1 n +( 1) n = 1 (11m+( 1) n )+( 1) n = 11m+1 ( 1) n +( 1) n = 11m+11 ( 1) n Both terms are multiple of 11, so if 1 n ( 1) n is a multiple of 11, 1 n+1 ( 1) n+1 is a multiple of 11. If the property is true at level n, it is also true at level n+1. Since the property is valid at level n =, it is valid for n N. Question 4 (a) Solve the system of equations (25 Marks) { log2 x+log 2 y = log 2 3 log 2 (4x+4) = log 2 y

6 { log2 x+log 2 y = log 2 3 log 2 (4x+4) = log 2 y +4 { 2 log 2 x+log 2 = y = 2 log log 2 (4x+4) = 2 log 2 y+4 { xy = 3 = 4x+4 = 16y { xy = 3 = x+1 = 4y { y(4y 1) = 3 = x = 4y 1 { 4y = 2 +y 3 = x = 4y 1 { (4y +3)(y 1) = = x = 4y 1 { y = 1 = x = 4(1) 1 = 3 The first solution has two solutions. y = 3/4 is impossible as log 2 y must be defined. (b) Thetwopoints(,4)and(2,1)areonacurve oftheformy = ae bx where aandbareconstants. Find these two values and give your answers in the form (p,qlnr) where p Z, q Z, r N. At point (,4) At point (2,1) 4 = a e b() = a 1 = a e b(1) = 4e 2b = e 2b = 1 4 ( ) 1 = 2b = ln = ln4 = ln2 2 = 2ln2 4 = b = ln2 The points are on the curve y = 4e xln2 6

7 Section B Answer Question 5 and Question 6. Question 5 Contexts and Applications 1 Marks (5 Marks) An engineering company has to build a new concrete dam. Concrete is a mixture of water, cement and sand or gravel. When these ingredients are mixed, a chemical reaction occurs which creates a lot of heat. To prevent this heat from damaging the structure of the dam, a network of pipes is installed inside the dam to cool it down. The average temperature in the dam is calculated using Newton s law of cooling; T = T a +(T T a )e kt wheretisthetime in months, T a istheambient orsurroundingtemperature, T isthetemperature at t = and k is a positive constant. (a) Engineerswanttoestimatehowitwilltakethedamtocooldownwithoutwaterflowingthrough the pipes. They think that a couple of months after completion, the average temperature in the dam will be T = 6 C (this corresponds to time t=). They estimate that two months later, this average temperature will be T = C. (i) Using these two temperatures measured by the engineers and taking an average ambient temperature of 1 C, identify or calculate T a and k (k should be calculated to 4 decimal places). T a may be identified from the text The temperature may be written as T a = 1 T = 1+(6 1)e kt = 1+5e kt The temperature at t = 2 is T = so = 1+5e 2k = 5e 2k = = e 2k = ( ) = 2k = ln 5 = k = 1 ( ) ln

8 (ii) What will be the average temperature in the dam after 5 years (to one decimal place)? 5 years = 5 12=6 months T = 1+5e.25(6) 21.2 (iii) How long will it take the dam to reach 12 C (express your answer in years)? What do you think? 12 = 1+5e.25t = 5e.25t = 2 = e.25t = 2 5 ( ) 2 =.25t = ln 5 = t = 1 ( ) 2.25 ln It takes the dam 1288 months to cool down. This corresponds to 17 years and 4 months. This is way too long. (b) To reduce the cooling down time, engineers decide that the dam will not be built as a single concrete block and that cold water will be flown through the pipe system. In this situation, the average temperature in the dam is (i) Calculate the value of a for which at t=2. T = 1+5e.25t(2t a) 1+5e.25t(2t a) = 1+5e.25t 1+5e.25(2)(2(2) a) = 1+5e.25(2) = 5e.5(4 a) = 5e.5 =.5(4 a) =.5 = 4 a = 1 = a = 3 8

9 (ii) For what range of values of t will the following inequality be true, where a is the value calculated in the previous question? 1+5e.25t(2t a) > 1+5e.25t. 1+5e.25t(2t 3) > 1+5e.25t = 5e.25t(2t 3) > 5e.25t This is only valid because t >. = e.25t(2t 3) > 5e.25t =.25t(2t 3) >.25t =.25t(2t 3) <.25t = 2t 3 < 1 = 2t < 4 = t < 2 (iii) Interpret this result in terms of temperature, in particular evaluate the validity of the model for the first few months. The temperature with the second method is higher for the first two months. This is very unlikely so the model is probably not valid at the start. (iv) Calculate the average temperature in the dam after 2 years (to one decimal place) when the new construction method is used. T = 1+5e.25(24)(48 3) 13.6 C (v) Conclude. This is a more realistic figure, typically much shorter than the temperature calculated with the previous method. The model seems to be more appropriate Question 6 (5 Marks) Stephen wants to start saving money to pay for the study of his new born child. He has two options: He could invest in nine year Irish bonds which pay 4% at maturity. So after nine years, you receive 14% of your initial investment. An investment account with a fixed interest rate of 4% per year. (a) Stephen would like to compare the two possible investments. (i) Calculate the AER of the nine year bond. 1.4 = (1+r) 9 = 1+r = (1.4) 1/9 = r = (1.4) 1/9 1 =.38 = 3.8% The AER is 3.8% (ii) Which investment should Stephen choose? Stephen should choose the investment account. 9

10 (b) Stephen chooses the investment account. He pays e2 into the account every year. (i) Calculate the amount on the account at the end of the first year? At the end of the first year, the investment is worth 2 (1+.4) =e28 (ii) Show how to use the sum of a geometric series to calculate the value of Stephen s investment after 18 years. Money deposited on the account at the beginning of year 1 is worth 2(1+.4) 18 at the end. Money deposited on the account at the beginning of year 2 is worth 2(1+.4) 17 at the end.... Money deposited on the account at the beginning of year 18 is worth 2(1+.4) at the end. At the end of the 18 years, the investment is worth I = 2 (1+.4) (1+.4) (1+.4) = 2 ( ) = ( ) This is the sum of a geometric series. (iii) Calculate the value of the investment after 18 years to the nearest euro. The investment is worth =e53,342.4 (c) Stephen actually wants to pay a fixed sum of money every month. Interests are compounded monthly and the AER is still 4%. (i) Find the monthly interest rate with the rate in % correct to 3 decimal places. Using the same method as before 1.4 = (1+r) 12 = 1+r = 1.4 1/12 = r = 1.4 1/12 1 =.327 =.327% (ii) Correct to 2 decimal places, how much should Stephen pay every month reach e28 at the end of the first year? At the end of the year, the investment is worthe28 28 = A(1.327) = A = =e (iii) If he pays this amount of money monthly for 18 years in total, calculate the value of the investment to the nearest euro. 18 years corresponds to 18 12=216 months. The value of the investment after 216 months is I = = e

11 (iv) Compare this value with the result of the previous section. Where does the difference come from? The two values are nearly equal. The difference is due to rounding up errors (d) After 18 years, Stephen adds some money to the fund and tops it up toe6.. This money will be used to finance the studies of Stephen s child. Assuming his child studies for 4 years and receives equal payments every month from Stephen investment account, (i) How many months are there before the study time of Stephen s child finishes? There are 4 12=48 months before the end of the study time. (ii) If the interest rate on the account is 4% AER, which is actually compounded monthly, using the monthly rate calculated in the previous section, calculate the value of each monthly payment. Using equal repayments at equal intervals, the sum is A = 6.327( ) =e ( ) 1 (iii) How much money should Stephen add to the e6 starting fund to guarantee his child a monthly income ofe15 for the duration of his degree? Using equal repayments at equal intervals, the sum is 15 = P.327( ) ( ) 1 Stephen should add an extra e = P = 15 ( ) 1.327( ) = P e

12 Section C Answer Question 7 and Question 8 Question 7 (a) Differentiate the following functions Functions and Calculus (old syllabus) ) f(x) = ln( x2 +1, g(x) = x+2 x 2 +3, h(x) = x 4 +2x 2 Using the chain rule and the properties of logaritms, ) f(x) = ln( x2 +1 = 1 2 ln( x 2 +1 ) = f (x) = 1 2x 2x 2 +1 = x x 2 +1 Use the quotient rule Use the chain rule (b) The family of functions f n is defined by For example, for n = 3 (i) Show that g (x) = (1)(x2 +3) 2x(x+2) (x 2 +3) 2 = x2 4x+3 (x 2 +3) 2 h (x) = 1 4x 3 4x 2 x4 +2x = 2 x3 x 2 x4 +2x 2 f n (x) = x3 +x 2 +(n 2)x+3 n+3 f (x) = x3 +x 2 2x+3 3 f 3 (x) = x3 +x 2 +x+3 6 and n N f 1 (x) = x3 +x 2 x Marks (5 Marks) Just Replace n with and 1 in the expression above (ii) Graphthetwofunctionsf (x)andf 1 (x)correspondington = andn = 1for 4 x 2. (iii) Estimate from the graph for what value of x is f (x) = f 1 (x) for 4 x 2. The two curves are equal for x = 1 and x =

13 (c) Show that all the functions f n have the same value at x = 1. f n (1) = (n 2)1+3 = n+3 n+3 n+3 = 1 (d) Calculate the equation of the tangent to any function f n at the point x = 1. Note that n is a parameter and should be treated as a constant. f n (x) = 3x2 +2x+(n 2) = f n n+3 (1) = 3(1)2 +2(1)+(n 2) = n+3 n+3 n+3 = 1 The equation of the tangent to all curves is y = 1+(x 1) = x (e) (i) Assuming the curve admits a local maximum or a minimum, calculate the value(s) of x at which this will happen. If the curve admits a local maximum or a minimum, f n(x) = f f1 f n (x) = 3x2 +2x+(n 2) = = 3x 2 +2x+(n 2) = n+3 Using the standard formula, there are two solutions x = n 6 or x = n

14 (ii) For what values of n can the curve have a local maximum or a minimum? These values only make sense if 28 12n is defined. This is the case when 28 12n > = 28 > 12n = n < A maximum or a minimum will occur for n =, n = 1 and n = 2. (iii) If possible, using your graph, estimate the value of the local maximum or minimum for f and f 1. Question 8 (a) Find f. Min.8, Max f 1. Min.7, Max 1. (b) Evaluate the following integrals (i) (ii) Use substitution ( ) 1 2x +3x2 dx ( ) 1 2x +3x2 dx = 1 2 lnx+x3 +C 4 x+e 2x x 2 +e 2xdx u = x 2 +e 2x = du = ( 2x+2e 2x) dx x+e 2x x 2 +e 2xdx = 1 du 2 u = 1 2 lnu = 1 2 ln( x 2 +e 2x) 4 x+e 2x x 2 +e 2xdx = 1 2 ln( x 2 +e 2x) 1 x 1+x2dx and (5 Marks) (Note: use radians if necessary) 1 x 1+x 2dx = 1 2 ln( 1+x 2) = 1 2 ln( 16+e 8) 1 1+x 2dx = 1 ln x 2dx = tan 1 x 1 = tan 1 1 = π

15 (iii) Use radians to calculate 2π cos 2 xdx = (c) (i) Show that 2π 2π cos2x+1 dx = cos 2 xdx 2π { sinx x π sinx = sinx π x 2π If x π, sinx so sinx = sinx. If π x 2π, sinx so sinx = sinx. (ii) Use radians to calculate the following integral 2π sinx dx = π 2π sinxdx (cos2x+1)dx = 1 ( ) sin2x 2π +1 = π 2 2 sinx dx 2π π sinxdx = cosx π +cosx 2π π =