RVHS P1 MS

RV 2012 Yr 6 H2 MA Prelim Paper 1 Question 1 [7 Marks] Let 4 3 2f ( ) 2 14z z z z az b     Consider 4 3 22 14 0z z z az b     ---- (1) Sub 1 2iz into (1), using GC ( 7 24i) 2( 11 2i) 14( 3 4i) (1 2i) 0 ab            ( 7 22 42 ) ( 24 4 56 2 )i 0a b a           ( 27 ) (36 2 )i 0 0ia b a       Comparing the real and imaginary coefficients 27 0 ----(2) 36 2 0 ----(3) ab a       Solving (2) and (3), 18a and 45b Therefore 4 3 2f ( ) 2 14 18 45z z z z z     Using GC to solve 4 3 2f ( ) 2 14 18 45 0z z z z z      , i2z  , i2z  , 3z , 3z Replace z with iz in (1), we obtain 4 3 22 14 18 45 0z iz z iz     i 1 2iz , i 1 2iz , i 3iz , i 3iz i2z  , i2z  , 3z , 3z Alternatively, since all the coefficients of the polynomial f ( )z are real  1 2iz and 1 2iz are roots of f ( ) 0z      2(1 2i) (1 2i) 2 5z z z z       is a quadratic factor of f ( )z . Let 2z qz r be the other quadratic factor of f ( )z . 4 3 22 14z z z az b     22 25z z z qz r        Comparing coefficient of 3z : 22 q    0q Comparing coefficient of 2z : 14 5r  9r Therefore f ( )z 22 2 5 9z z z        4 3 2f ( ) 2 14 18 45z z z z z     18a and 45b f ( ) 0z   1 2iz , 1 2iz , 3iz , 3iz

2 Question 2 [8 Marks] 9 ( 3)( 3) 0 x x xx x    3 0 or 3xx    4 34 3 3422 3 2 2 9 d 99 dd 9ln | | 9ln | |22 999ln 3 9ln 8 9ln 4 9ln 32 2 2 18ln 3 1 9ln(4 ) 2 n n n xx x x x x xxx xxxx nn nn I                                              As 0n , ln(4 )n  I  Question 3 [9 Marks] i OAQB is a parallelogram OA BQ OA OQ OB   11 02 22 OQ          2 2 4 OQ     ii  a c c is the projection vector of a onto b . iii |a| < |b| 22 ( 1) 4 1 4 4pp      2 20 ( 1)( 2) 0 12 pp pp p          But p > 0, therefore 02 p . 3 –3 0

3 iv 2 1 2 1 3 1 1 2 . 1 2 1 . 3 3 3 0 0 2 2 2 2 4 0                                                                                  Thus ab and ab are perpendicular. Note: ab and ab are the diagonals of the parallelogram with a and b as the adjacent sides. The parallelogram with a and b as the adjacent sides must be a rhombus. | a b| is the area of a rhombus formed by the vectors a and b. OR | a b| is the area of the rhombus OAQB. Question 4 [9 Marks]   21cos xy  --- (1)            2 1 1 1cos2d d x xx y --- (2) Squaring both sides, we get   2 212 1 cos4 d d x x x y       yx yx 4d d)1( 2 2  