TJC H2 Maths P2 Solutions

TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination 2014 Higher 2 MATHEMATICS 9740/02 Paper 2 15 September 2014 Additional Materials: Answer Paper 3 hours List of Formulae (MF 15) READ THESE INSTRUCTIONS FIRST Write your Civics Group and Name on all the work that you hand in. Write in dark blue 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 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 6 printed pages. © TJC/Prelim Exam 2014/MA H2 9740 [ Turn over

TJC/Prelim Exam/2014/MA H2 9740/02 2 Section A: Pure Mathematics [40 marks] 1 It is given that 1f( ) 3 15 xx  . Find the roots  and  of the equation f( )x x , where   . [2] A sequence of real numbers 123,,, . . .x xx satisfies the recurrence relation given by 1 fnnx x  for 1n , and 1x is a value that lies between  and . The diagram shows how the graphs of f( )y x and y x are used to obtain 2x from 1x . (i) Copy the above diagram, and illustrate how 3x can be obtained. [1] (ii) Describe the behaviour of the sequence in this case. [1] [Solution] Using GC, 0.928  and 2.64  (3sf) (i) (ii) The sequence decreases and converges to  (or 0.928 ).  fy x y x yx 1x  1f x 2x  fy x y x yx 1x  1f x 2x3x Need to label 3x clearly on the x-axis

TJC/Prelim Exam/2014/MA H2 9740/02 3 2 A circle C1 has equation given by 22 43 0xx y   . Another curve C2 with equation   22 ax b y ca   is obtained from circle C1 after going through the following successive transformations:  Stretching with scale factor 3 parallel to the y-axis followed by  Translation of 1 unit in the negative y-axis direction. (i) Find the values of a, b and c. [3] (ii) Sketch, on the same diagram, the two curves, C1 and C2, indicating clearly the points of intersection of C1 and C2. [3] (iii)  1,x k and  2 ,x k are points on C1 and C2 respectively such that 2123 xx , where k is a real constant. State the range of values of k. [1] [Solution] (i) 22 43 0xx y   2 221xy  After 1st transformation:  2 2 21 3 yx   After 2nd transformation:   22 121 1 9xy     22 92 1 9xy   a = 9, b = 2 , c = 1 (ii) (iii) From diagram, range of k is  0.25, 0.5 . y x 2 3 1 1 2 y = 0.5 4 y = 1 C1:  2 221xy  C2:   22 121 1 9xy   y = 0.25 1.13, 0.5  2.87, 0.5 1.03, 0.25  2.97, 0.25

TJC/Prelim Exam/2014/MA H2 9740/02 4 3 A curve has equation f( )y x , where  2 for 2,f 2 h( ) for 2. ax bx c xx x xx       The curve has a turning point at  3,5 . The only asymptotes are 1yx and 2x . (a) Find the values of a, b and c. [3] (b) It is given that the tangent to the curve f( )yx  at the point 70, 5   is parallel to the line 230yx . Sketch the graph of  f'yx , indicating clearly all asymptotes and axial intercepts. [3] (c) The function g is such that h( ) d g( )  x xx k for 2x  , where k is an arbitrary constant. By sketching the graph of  hyx for 22 x  , express  2 2 hd x x   in the form g( ) g( )p uq v , where ,,,pquv are constants to be determined. [3] [Solution] (a) Given  2 f for 2 2 ax bx cxx x   Since oblique asymptote is 1y x  , let 2 = 122 ax bx c d xxx    y x y = 1 – x x = 2 7 5  3, 5 f( )y x

TJC/Prelim Exam/2014/MA H2 9740/02 5   2 12 = 22 x xdax bx c xx    Compare coefficients of x2 : 1a Compare coefficient of x: 1b Since f35 93 5132 ab c c   (b) Given that the tangent to the given curve at the point 70, 5   is parallel to the line 230yx , this means that  f' 0 1 . 5 . (c)      hh hy x y x y x      y x y = – 1 x = – 1.5 3 'fyx Replace x by –x  Reflect y = h(x) in y-axis Replace x by |x|  Mirror y = h(x), x ≥ 0 in the y-axis

TJC/Prelim Exam/2014/MA H2 9740/02 6   02 2 2 hd 2 h dx xx x    [No labelling of asymptotes required for graph of  fy x ]   2g 0 2g 2  h(x), 2<x<2 Sharp as it mirror in y-axis

TJC/Prelim Exam/2014/MA H2 9740/02 7 4 The parametric equations of a curve C are given by sin cos x and sec 2y where 5 12 2    . (i) Sketch the curve C, indicating the coordinates of the endpoints clearly. [2] (ii) Show that the area of the region bounded by the curve C, x-axis and the lines 2 2x and 1x can be written as 2 5 12 1 dsin cos    . [3] (iii) Prove that sin cos 2 sin 4      . [1] Hence show that the exact area of the region is   2 ln 2 1 3 22   . [3] [Solution] (i) (ii) Required area =  1 2 52 122 ds e c 2 c o s s i n dyx        2 22 5 12 cos sin dcos sin       x y

TJC/Prelim Exam/2014/MA H2 9740/02 8   2 5 12 2 5 12 cos sin dcos sin cos sin 1 d [Shown]sin cos              (iii) RHS 2 sin 4   2s i n c o s c o s s i n44 sin cos LHS        (iv) Hence required area 2 5 12 1 dsin cos     2 5 12 2 5 12 1 d 2s i n 4 2 cos ec d24               2 5 12 2 ln cos ec cot24 4            2 ln cos ec cot ln cos ec cot24 4 6 6                             232 122 ln 2 1 ln 2 3 ln22 21 21          2 ln 2 1 2 32 

TJC/Prelim Exam/2014/MA H2 9740/02 9 5 (i) Show that 2ii1e 2 c o s e 2    . [2] (ii) Solve the equation  6 610 zz  , giving the complex number z i n t h e f o r m o f eir  , where 0r  and   . [4] (iii) Two of the roots are such that  Im 0z  . One of these roots is 6i 1 3 e3z   . State the other root 2z . [1] (iv) Given that  is the complex number such that  1 5arg 6  z , find the minimum value of 2 z . [4] [Solution] (i) RHS 2i 2c os e2  2c os cos i sin22 2         22cos i2sin cos22 2   1c o s i s i n   (Using Double Angle formulae) i1e  = LHS (ii)  6 610 zz  6 1 1z z   6 1 11z   2 61 1e , 0 ,1 ,2 , 3 ki kz      