SRJC H2 MATHS P1 Soln

1 SERANGOON JUNIOR COLLEGE 2014 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9740/1 Wednesday 20 Aug 2014 Additional materials: Writing paper List of Formulae (MF15) TIME : 3 hours READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the 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 a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not al lowed 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. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. Total marks for this paper is 100 marks. This question paper consists of 6 printed pages (inclusive of this page) and 2 blank pages.

2 Answer all questions [100 marks]. 1 A cubic polynomial has turning points at A(–1, 13) and B(2, –14). (i) Find the equation of this polynomial. [3] (ii) Hence find the coordinates of the point C on the graph of this polynomial such that AC is parallel to the x-axis. [2] Suggested Solutions Let 32y ax bx cx d    So, 2d 32d y ax bx cx    Using A(–1, 13), 13a b c d     --- (1) 3 2 0a b c   --- (2) Using B(2, –14), 8 4 2 14a b c d    --- (3) 12 4 0a b c   --- (4) Using GC, 2, 3, 12, 6a b c d    Hence, 322 3 12 6y x x x    Since AC is parallel to the x-axis, y-coordinate at C is 13. So, 3213 2 3 12 6x x x    Solving, we get 3.5 or 1 (N.A.)xx  Therefore, the coordinates of C is (3.5, 13). 2 SRBank introduces the UniSave Bank Account to encourage young parents to save up for their child’s university education. This account consists of two independent components – Ordinary Account and Birthday Account. The bank will provide an interest rate of 1% of the total amount in the Ordinary Account at the end of every year. As a bonus, the bank will deposit an amount equivalent to ten times the age of the child into the Birthday Account on the child’s birthday each year, with the last deposit on the 18th birthday. Mr and Mrs Hon intend to save up for their child’s university education by depositing a fixed amount of $3000 into the UniSave’s Ordinary Account at the start of every year from the year their child turns one year old. (i) Show that the total amount in the UniSave Bank Account at the end of the year when their child is n years old, where 19,n is given by $  303000 1.01 301290n  . [3] (ii) Given that 19,n find the least n such that the total amount in the UniSave Bank account will exceed $70000. [2]

3 Suggested Solution (i) End of Year Total Amount in Ordinary Account Total Amount in Birthday Account 1 $  3000 1.01 $10 2 $    3000 3000 1.01 1.01 = $     23000 1.01 3000 1.01 $  10 20 3 $       23000 3000 1.01 3000 1.01 1.01  = $       233000 1.01 3000 1.01 3000 1.01 $  10 20 30 … … … 17 $   2 173000 1.01 1.01 ... 1.01   $  10 ... 170 18 $   2 183000 1.01 1.01 ... 1.01   $  10 ... 180 19 $   2 193000 1.01 1.01 ... 1.01   $  10 ... 180 Total amount at the end of Year n (where n 19 ) = $ 23000(1.01 1.01 ... 1.01 ) 10 20 ... 180n       = $     1.01 1.01 1 183000 10 1801.01 1 2 n    = $  303000 1.01 1 1710n = $  303000 1.01 301290n  (ii)  303000 1.01 301290 70000n  1.01 1.22538n  20.43n Least n = 21 3 A sequence of real numbers 1 2 3, , ,u u u … satisfies the recurrence relation 1 3 n n n uu u    , n  . Given that 1 1u  , and by considering 2 1 nu  for 1,2,3n , make a suitable conjecture for nu in the form of n k ab , where ,,a b k  . [2] Prove the conjecture by Mathematical Induction. [4]

4 Solution       1 1 1 21 2 12 32 3 23 221 1 1 3 1 1 1 2 2 1 1 3 13 1 3 4 4 1 1 2 24 1 1 3113 13 34 13 u u uu uu uu uu                               Conjecture: 2 31 n nu   Let Pn be the statement “ 2 31 n nu   , n  ”. When n = 1, LHS = 1 1 u  RHS = 1 2 1 LHS31  Hence 1P is true. Assume Pk is true for some k  i.e 2 31 k ku   To show that 1Pk is true, i.e 1 1 2 31 k ku    LHS = 1ku  3 k k u u  = 2 31 2 331 k k      = 2 3131 2 31331 kk k k       =   2 2 3 3 1k =   2 3 3 1k  = 1 2 31k  = RHS Therefore, 1Pk is true when Pk is true. Since 1P is true and 1Pk is true when Pk is true, by Mathematical Induction, Pn is

5 true for all n  . 4 (a) A graph with equation f ( )yx undergoes in succession, the following transformations: A : A translation of 3 units in the direction of the negative x-axis B : A reflection about the y-axis C : A scaling parallel to the x-axis by a factor of 2 The equation of the resulting curve is given by 4 2 18 xy x   . Find the equation f ( )yx . [3] (b) The graphs of y = g( )x and y = g( )x are shown below. Graph of y = |g(x)| Graph of y = g( )x Sketch the graph of y = g(x), showing clearly any equations of asymptote and intercepts with the axes. [3] Solution 4 2 18 xy x   'C  2 4 2 4 18 2 9 xxy xx   2 29 xy x   'B  22 2 9 2 9 xxy xx      2 29 xy x   'A  ( 3) 2 1 2( 3) 9 2 3 xxy xx       x y x = 3 x = −3 −1 1 y = 2 y = − −1 1 x x = 3 x = −3 y 2 −

6 y = g(x) 5 The curve C has equation given by 2 1, , 1.1 xxy x x x    (i) Without using a calculator, find the set of values that y can take. [3] (ii) Sketch the graph of C, indicating clearly the equations of any asymptotes and the coordinates of any turning points of the curve. [2] (iii) Show that  1, 1 lies on 1, where .y kx k k    Hence, find the range of values of k where   2 11 xxkx x  has two real roots. [2] Suggested Solution (i) Consider any horizontal line ,y k k . Consider the intersection of the graphs 2 1 1 xxy x   and yk , i.e. 2 1 1 xx x k   2 ( 1)1x xx k   2 (1 ) (1 ) 0x x k k   For the equation to have real solutions, we need Discriminant 0 2 4(1)((1 ) 1 ) 0kk    2 2 3 0kk   ( 1)( 03)kk    We have 3 or 1 kk   The set of values that y can take is  : 3 or 1yy y   2 1 1 11 xxy x x x     x y x = 3 x = −3 −1 1 y = 2 2

7 (ii) When 1x , ( 1) 1 1y k k     (Shown) To find the range of values of k where   2 11 xxkx x  has 2 real roots is the