EJC H2 MATH P2

EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2018 General Certificate of Education Advanced Level Higher 2 MATHEMATICS Paper 2 [100 marks] 9758/02 21 September 2018 3 hours Additional Materials: Answer Paper List of Formulae (MF26) READ THESE INSTRUCTIONS FIRST Write your name, civics group and question number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphi ng calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 7 printed pages.

Section A: Pure Mathematics [40 marks] 1 (i) Given that f is a continuous function, explain, with the aid of a sketch, why the value of 11 2 2lim f f ... f n nn n nn n n→∞  ++     ++ +         is 2 1 f( ) dxx [2] (ii) Hence evaluate 1 1lim →∞ = + n n r n nn r exactly. [3] Suggested Solution (i) As shown in the diagram, the area under the curve y = f( x) from x =1 to x =2 can be approximated by the total areas of the n rectangles with width 1 n and heights given by 123f1 , f1 , f1 , , f1 . n nnn n    +++ +       L As n increases, the approximations will get better and approach the exact area as a limit, i.e. 2 1 11 2 2lim f f ... f f ( ) d n nn n xxnn n n→∞  ++     ++ + =          [Shown] (ii) 2 11 11lim f ( ) d where f ( ) n n r n xx xnn r x→∞ = ==+  [] 2 2 1 1 1 d l n| | l n2 l n1 l n2xxx== = − = … x y O 1+ y = f(x) 1+ 1+ 1+ 2 1

2 The diagrams below show the graphs of f( )yx= and f' ( )yx= . On separate diagrams, sketch the graphs of: (i) f( 2 ) 1yx=+ [2] (ii) 1 f( )y x= ′ [3] (iii) f( )yx= [3] showing clearly, in each case, the intersection(s) with the axes, the coordinates of the turning point(s) and the equation(s) of the asymptotes. Suggested Solution y x O 1 5 (8, 3− ) 3− y x O 1 2 y=2 (5, 3) 1

(i) (ii) 1 f( )y x= ′ (iii) y = f(x) 3 (a) A retirement savings account pays a compound interest of 0.2% per month on the amount of money in the account at the end of each month. A one-time principal amount of $P is deposited y x O 1 y=3 (5/2, 4) 2 1 3− y x O 1 5 18, 3  −   5x = y x O 1 2 2 (5, 3) 1

to open the account and a monthly pay-out of $x is withdrawn from the account at the beginning of each month, starting from the month that the account is opened. (i) Show that the amount in the account at the end of n months after the interest has been added is given by (1.002 ) 501 (1.002 1) nnPx −− . [4] (ii) Suppose a fixed monthly pay-out of $2,000 is to be sustained for at least 25 years, find the minimum principal amount required correct to the nearest dollar. [2] (iii) If a principal amount of $600,000 is placed in the account, find the number of years for which a monthly pay-out of $2,000 pe r month can be sustained, leaving your answer correct to the nearest whole number. [2] (b) A different retirement savi ngs account provides an increas ing amount of monthly pay-out over a period of 25 years. The pay-out in the first month is $ a. The pay-out for each subsequent month is an increment of $c from the pay-out of the previous month. The pay-out in the final mont h is $4,000, and the total pay-out at the end of 25 complete years is $751,500. Find the month in which the pay-out is $2,000. [5] Suggested Solution (a)(i) Month Balance at the end of the month 1 ( )(1.002) (1.002) (1.002)Px P x−= − 2 ( (1.002) (1.002) )(1.002)Pxx −− 22(1.002) (1.002) (1.002)Pxx=− − 3 22( (1.002) (1.002) (1.002))(1.002)Pxx −− 332(1.002) (1.002) (1.002) (1.002)Pxxx=− − − n 1(1.002) (1.002) (1.002) (1.002)nnnPxx x −=− − − − L Balance after n months 1(1.002) (1.002) (1.002) (1.002)nnnPxx x −=− − − − L (1.002)(1.002 1)(1.002) 1.002 1 n n xP −=− − (1.002) 501 (1.002 1)nnPx=− − (shown) (a)(ii) 300 300(1.002) 501(2000)(1.002 1) 0P −− ≥ 300 300 501(2000)(1.002 1) 1.002P −≥= 451761.1356

Minimum principal amount required is $451762. (a)(iii) (600000)(1.002) 501(2000)(1.002 1) 0nn−− ≥ (1.002) ( 402000) 501(2000) 0 1002000 167(1.002) 40200 67 167ln 67 457.11ln1.002 n n n −+ ≥ ≤= ≤= No. of years ≤ 457.11 38.09212 = No. of years for which the pay-out can be sustained is 38. (b) Month Payout 1 a 2 a+c 3 a+2c M M 300 4000 Let $a be the pay-out in the first month. 299 4000ac+= --- (1) ()300 2 299 7515002 ac+= 2 299 5010ac+= --- (2) Solving the simultaneous equations, 1010, 10ac== To find the month n with a pay-out of $2,000: (1010) ( 1)(10) 2000n+− = 100n = The pay-out is $2,000 in the 100th month. Alternative solution: Let $a be the pay-out in the first month. 299 4000ac+= --- (1) ()300 4000 751500 10102 aa+=  = . Hence from (1) 10c = Consider () ( )1010 1 10 2000 100nUn n=+ − =  =

The pay-out is $2,000 in the 100th month.

4 The function f is defined by () 2 4 for 1 3,f( ) for 3 4,4 x xx xx − ≤<=  ≤<− and it is given that () ( )f3 fx x−= for all real values of x. (i) State a reason why f does not have an inverse. [1] (ii) Sketch the graph of ()fyx= for 16−< <x . [3] (iii) Evaluate ()f2 0 1 7 . [2] The function g has domain [1, 4)and is defined by () 2 4 for 1 3,g( ) for 3 4.4 x xx xx − ≤<=  ≤<− (iv) By sketching g( )yx= and 1g( )y x−= on the same diagram, state the values of x such that () () 1ggx x−= . [3] The function h, is defined by () 4 1 for 0 1,h( ) for 1 3.1 x xx xx  − ≤<=  ≤≤− (v) Explain why 1hg− doesn’t exist. [1] (vi) Given that hg exist, define hg in similar form as function h. [2] (vii) Find the range of hg . [2] Suggested Solution (a)(i) Since () ( )f3 fx x−= , we can easily find 2 values of x (for example () ( )f3 f 0−= ) to show that it’s not 1-1, hence no inverse.

(ii) (iii) () () () ( )f 2017 f 2014 f 2011 f 1 3=== = = L (iv) From the graph, the values of x such that () () 1ggxx −= is 13 x≤≤ . (v) [ ) [ ]1 hgR1 , 4 0 , 3 D− =⊄= , 1hg− doesn’t exist (vi) () () 4 2 3 for 1 3,hg( ) for 3 4.14 x xx xx  − ≤<=  ≤<−− (vii) Range of hg [ ]0,16= y x y x Intersection between ()gy x= and () 1gyx −= () 1gy x−= ()gy x=

Section B: Probability and Statistics [60 marks] 5 Two fair 4-sided dice each has its faces labelled ‘1’, ‘2’, ‘3’ and ‘4’. The two dice are thrown and the absolute difference in score on their bottom faces is denoted by X. (i) Find P( )X x= for all possible values of x. [2] (ii) Find E( )X and Var( )X . [ 2 ] Suggested Solution (i) Consider the number of outcomes with the respective differences out of 16: 41P( 0) 16 4X == = 63P( 1) 16 8X == = 41P( 2) 16 4X == = 21P( 3) 16 8X == = (ii) () () () ()13 11E( ) 0 1 2 3 1.2548 4 8X    =+ ++=       () 22Var( ) E( ) E( )X XX=− () () () () 22 2 2 213 110123 ( 1 . 2 5 )48 4 8    =+ ++ −       0.9