SRJC P1

1 [TURN OVER ] SERANGOON JUNIOR COLLEGE 2012 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9740/1 Wednesday 15 Aug 2012 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 no blank page.

2 Answer all questions [100 marks]. 1 The graph of 2 1y x  for 0x , is shown in the diagram below. Region R is bounded by the x-axis, the y-axis, the line 1x and the curve 2 1y x  . The area of region R may be approximated by the total area, A, of n rectangles, each of width 1 n , as shown in the diagram. (i) Show that 1 0 2n r A nr     . [2] (ii) By considering the exact area of region R, show that 1 0 1 ln 2 n r nr    . [2] Solution (i) Total area of all the n rectangles, 112 1 2 2 2 2 1 0 1 1 1 n n n n A n            1 2 2 2 2 0 1 1 2 1 n n n n n n n n             2 2 2 2 0 1 2 1n n n n n         1 0 2 () n r shownnr     (ii) 1 1 0 0 2 d 2ln 1 2ln 21 xxx     Area of the n rectangles > Area of region R 3 n 2 n 1 1n n  2n n  3n n  x y 2 1y x  0 1 n

3 [TURN OVER ]  1 0 2 2ln 2 n r nr     1 0 1 ln 2 n r nr    (Shown) 2 A sequence of real numbers x1, x2, x3, … satisfies the recurrence relation 1 2 3 nn nxx  . Given that 1 2 3x  , write down 234,,x x x in the form of  ! n na b  , where a and b are positive integers. [2] Hence make a conjecture for xn and prove the conjecture by Mathematical Induction. [4] Solution 1 2 3x  , 21 1 2 2 33xx  32 2 2 8 39xx  43 3 2 40 3 27xx    1 11 1 1 !2 1.2 2! 3 3 3 3x       2 2 2 2 2 1 !2 1.2.3 3! 3 3 3 3x       3 3 3 3 3 1 !8 1.2.3.4 4! 9 3 3 3x       4 4 4 4 4 1 !40 1.2.3.4.5 5! 27 3 3 3x      1!Conjecture: 3 n n nx  Let nP be the statement  1! 3 n n nx  for all n  .  1 21,LHS given 3nx     1 1 1 ! 2RHS LHS33    1P is true. Assume kP is true for some k  i.e.  1! 3 k k kx 

4 To show 1kP  is true i.e.   1 1 2! 3 k k kx    1 2LHS 3 kk kxx   1!2 33 k kk      1 2! RHS3k k   1kP  is true if kP is true. Since 1P is true and 1kP  is true if kP is true, nP is true for all n  . 3 A geometric series, G, has common ratio r, 1r , and an arithmetic series, A, has a non-zero first term a. The first three terms of G are equal to the seventh, third and first term of A respectively. (i) Show that 22 3 1 0rr   . [3] (ii) Deduce that G is convergent. [1] (iii) Find the sum to infinity of the even-numbered terms of G in terms of a. [3] Solution (i) Let b be the first term of the G and d and b be the common difference of the AP.   2 6 2 1 b a d br a d br a      2 22br br d  43b br d     2 3 gives, 22 b br br br     212 r r r   22 3 1 0rr   (ii)   2 1 1 0rr   1 or 1 (rejected 1)2rr   Since 1 1, is convergent.2rG   (iii) From (1) , 4ba Sum to infinity of the even-numbered terms 22 14 2 1 11 2 abr r  

5 [TURN OVER ] 8 3 a 4 Given that ln 1yx , show that 22 2 d 2 d d2ln ddd y y yy y x xx   . [2] (i) By further differentiation of this result, or otherwise, find the Maclaurin’s series of y up to and including the term in 3x . [3] (ii) Deduce the series expansion of 1e 1 x y x    up to and including the term in 2x . [2] Solution ln 1yx 1 21 d 1 (1 ) ( 1)d2 y xyx    1 d 1 1 d2 1 y yx x   d2ln d yyy x  2 2 d d 1 d d2ln 2 d d dd y y y yy x y x xx             22 2 d 2 d d2ln ddd y y yy y x xx   (i) 32 32 d d 1 d2ln 2 ddd y y yy yxxx     222 2 2 2 4 d d d 1 d d 2d d ddd y y y y y y x x x x y x                       When 0x , ey de d2 y x  2 2 d 0 d y x  3 3 de 8d y x  Maclaurin’s series of y is

6 23eee ( ) (0)2 2! 8 3! xxyx                 3eee 2 48y x x    (ii) ln 1yx 1 xye  1 2 21 d e e d 2 16 x x ye xx       1 2 1 ee e + 8 x x x      5 Paul, a life guard standing at point A along a straight stretch of the beach, looks through his binoculars and sees a boy clinging on to his overturned canoe and struggling to keep afloat at point B in the sea. P is the point on the straight stretch of the beach nearest to B such that BP = 1 km and PA = 2 km. To reach the boy, Paul first runs to Q and then swims in a straight line to B. When Paul runs, he covers 1 km in 4 minutes. When he swims, he covers 1 km in 10 minutes. (i) If PQ = x km, 02 x , show that the time T minutes taken by Paul to reach B is given by 28 4 10 1T x x    . [1] (ii) Find the exact value of x such that he would take the shortest time to reach the boy. [4] (iii) Hence, find the shortest time he would take to reach the boy, leaving your answers in exact form. [2] Solution (i) Total time, T = run swimTT  = 24 (2 ) 10 1xx     28 4 10 1xx    (ii)     1 2 2d1 4 10 1 2d2 T xxx     B x km 1 km 2 km P A Q

7 [TURN OVER ] 2 104 1 x x    For shortest time, d 0d T x  2 104 1 x x   22 1 5 xx       224 1 5 xx   221 4x 2 4 21x 2 since 0 21 xx   Method 1 x 2 21    2 21 2 21    d d T x − 0 + sketch Method 2 2 2 2 22 210 1 10 d 21 d1 xxx T x xx         22 32 2 32 2 10(1 ) 10 1 10 0 is a minimum 1 xx x T x       Hence, when x = 2 21 , Paul would take the shortest time. (iii)When x = 2 21 , 248 4 10 1 2121 T    