SRJC P2

1 [TURN OVER ] SERANGOON JUNIOR COLLEGE 2012 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9740/2 Thursday 23 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 7 printed pages (inclusive of this page) and 1 blank page.

2 Section A: Pure Mathematics [40 marks] 1 A curve is defined by the parametric equations (3 ), = 2 (1+ )x ku u y k u , where k is a positive constant and u is a variable not equal to 3 2 . (i) Find d d y x in terms of u. [2] (ii) Given that x increases at the rate of 3 1 units per second, find, in terms of k, the rate of change of xy when y = 4k. [3] Solution (i) 2 (3 ) = 2 (1+ ) = 3 2 2 x ku u y k u ku ku k ku     d 32d x k kuu  d 2d y ku  d d d d d d 12 32 2 32 y y u x u x k k ku u      (ii) Given d1 d3 x t  , 4yk 1, 2u x k   d d d d d d 21 ()3 2(1) 3 2 3 y y x t x t    d d d()d d d 21(2 )( ) (4 )( )33 8 3 yxxy x yt t t kk k   

3 [TURN OVER ] 2 (a) (i) Using the substitution x = 3cos t + 1, show that 331 2 2 1 9 ( 1) dxx   = 93 2 3 4   . [4] (ii) The curve C has equation 29 ( 1)yx   . The region enclosed by the curve, the horizontal line y = 5, and the vertical lines x = 1 and 331 2x is denoted by S as shown in the diagram below. By considering the graph of 29 ( 1) 5yx    and using the results in (i), find the volume of the solid generated when the region S is rotated through 2 radians about the horizontal line y = 5, leaving your answer in the form 23 8 a b  where a and b are integers to be determined. [3] (b) The curve is defined parametrically by cos , sin ttx e t y e t , 0 2t  . Find the area of the region R enclosed by the curve and the axes as shown below. [3] Solution (i) x = 3cos t + 1 d d x t = – 3sin t When x = 1, 1 = 3 cos t + 1 cos t = 0 x y 1 331 2 5y S 29 ( 1)yx   R x y

4 t = 2  When x = 331 2 , 331 2 = 3 cos t + 1 cos t = 3 2 t = 6  331 2 2 1 9 ( 1) dxx   =   26 2 9 (3cos 1) 1 ( 3sin ) dttt       = 6 2 2 9 9cos ( 3sin ) dttt    = 6 2 2 3 1 cos ( 3sin ) dttt    = 6 2 2 9 sin dt t    = 2 6 1 cos 29 dt 2 t     = 2 6 9 sin 2 22 tt    = sin 29 sin 6 2 2 2 6 2                = 93 2 3 4    (Shown) (ii) Require volume = 331 2 22 1 ( 9 ( 1) 5) dxx     = 331 2 22 1 9 ( 1) 25 10 9 ( 1) dx x x        = 3 3 3 311 22 22 11 34 ( 1) d 10 9 ( 1) dx x x x       = 33 1+3 2 1 ( 1) 9 334 10 3 2 3 4 xx           

5 [TURN OVER ] = 227 3 45 351 3 15 84      = 2291 3 158   units3 291 and 15ab   (b) When y = 0, 0 sintet sin t = 0 ( 0te  ) t = 0 When x = 0, 0 costet cos t = 0 ( 0te  ) t = 2  sinty e t d sin cosd tty e t e tt  Required area = 1 0 d y xy = 2 0 ( cos ) ( sin cos ) dt t te t e t e t t   = 5.54 units2 3 (a) Relative to the origin O, the position vectors of A,B and C are a, b and c respectively where 32c = a b . Given that a is a unit vector, 3b and the angle between a and b is 2 3  , (i) find the exact value of  such that a is perpendicular to 34 b a c , [3] (ii) show that the exact area of triangle OBC is 93 4 units2. [2] (b) The diagram shows a rectangular box with unit vectors i, j, k as shown. The lengths of OA, OC and OD are 20 units, 12 units and 15 units respectively. The point P lies on BD and divides BD in the ratio 3: 2 . 1y R x y

6 A line has Cartesian equation 16 332 xy z   . (i) Find the vector equation of line PE. [3] (ii) The line PE intersects at Q, find the position vector of Q . [3] Solution (ai) Since angle between a and b is 2 3  , 2cos 3   ab ab 1 21 3 2   ab ( ) b ab   22 3 4 3 2 0 3 4 3 2 0 9 4 3 3 02 17 12 a b a a b a b a a (a b)                   (ii) 2 3 3 3sin( ) 33 2 2b a b a           1Area 3 22 1 322 1 32 93 4 b a b b a b b ba          C B O O j D E G A F i k

7 [TURN OVER ] (bi) 20 0 0 OA   , 0 12 0 OC   , 0 0 15 OD   Line : 13 16 3 6 232 31 r,xy z                            20 0 8 40 1 1 1(2 3 ) 2 12 3 0 4.8 245 5 5 0 15 9 45 OP OB OD                                            20 8 12 10 60 4.8 4.8 4 515 9 6 5 PE OE OP                                             Therefore line PE is r = 20 10 0 4 , 15 5                     . (ii) Since the line PE intersects , 20 10 1 3 0 4 6 2 15 5 3 1                                          10 3 21 4 2 6 5 18           Using GC, 3, 3  10 12 0 OQ    4 (i) The polynomial P(z) has real coefficients. The equation P(z) = 0 has a root ire , where r > 0 and 0 <  <  . Show that a quadratic factor of P(z) is 22 cos2 rrzz   . [3] (ii) Solve the equation 5 243,z  expressing the solutions in the form ire , where r > 0 and .     [2] (iii) Hence express 5 243z  as the product of one linear factor and two quadratic factors with real c