RVHS P2 MS

RV 2012 Yr 6 H2 MA Prelim Paper 2 Question 1 [7 Marks] 32 32 2 2 2 d (1 ) 0 (1)d ddLet dd Subs. into (1): d 0d d ( 1) (shown)d yx x y xx uyu xy x y xx u y y x y xx u ux x u xx                     3 3 3 2 3 3 3 3 3 1 dd1 ln | 1| 3 1 e where e e1 e 1 1 (shown) and f 3 x C x x u x xu xuC u A A uA Ay x x x             (i) 1A (ii) 1A

2 Question 2 [10 Marks] (a) 2 dtan e sec e d xx yyy x   2 2 2 2 2 1 1 de (1 e ) 1cot cot sec d1 tan cot d (cos ec 1) d cot e tan (e ) xx xx x y y y y y yy yy y y C C                   (b) 2 1exy  When 0x , ey and when 1x , 2ey . 2 1exy   2 ln 1xy Required volume 22π(1) e J where 2e e π (ln 1)dJ y y 2e e π (ln 1)d J yy      222 eee ee e π (ln ) 1dy y y y      2e e 2 2 2 π (ln ) 2 π e ln e eln e 2e 2e π e y y y      Required volume   22 2 π(1) e πe πe =πe e 1 J   Note that ln y d ln 1 d lny y y y y y y c    

3 Question 3 [11 Marks] ai aii aiii b 2 2 2 2 2 2 1 1 ( 1) 14 AByx x y x y         A: Scaling parallel to the y-axis with a scale factor of 1 2 . B: Translation of 1 unit in the positive direction of the y- axis. Alternatively 2 2 2 2 2 2 ( 2) (2 2)1 1 14 4 4 ABy y yx x x         Note that 2 2 (2 2) 14 yx  is equivalent to

4 22 ( 1) 1xy   . A: Translation of 2 units in the positive direction of the y-axis. B: Scaling parallel to the y-axis with a scale factor of 1 2 . Note: Scale factors must be positive numbers. Question 4 [11 Marks] i 3 (0) 8 8e iz  3 (0 2 ) (2 )8e 8ei k i kz  2 32e , 0, 1 ki zk    ii 2 3 1 2e i z  , 2 2z  , 2 3 3 2e i z  Note that the points 1Z , 2Z and 3Z lie on the circumference of a circle with centre O and radius 2. The points are equally spaced out separated by an angle of 2 3 radians. 1Z and 3Z represent complex numbers which are conjugates to each other. Thus 1Z and 3Z lie on the same vertical line. iii Since 1zz satisfy the equation 23| | | |z z z z   with 1 2 1 3| | | | 12 2 3z z z z     , thus locus of P passes through the point 1Z . Note 1 2 1 3| | | | 12z z z z    can be easily found using GC. Re Im Z1 Z2 Z3 O

5 iv Locus of P: 23| | | |z z z z   , is the perpendicular bisector of the line joining 2Z and 3Z . It passes through the origin and 1Z . Locus of Q: 1| | 2 3wz , is a circle with centre at 1Z and radius 23 . It passes through 2Z and 3Z . v Let V be the point of intersection of the loci of P and Q in the first quadrant. 2 12 3Z OZ  1 2 3VOZ    2 3 1 2e 1 3 i zi   