2012_VJC_H2_MA_Prelim_P2_Solutions

Victoria Junior College H2 Mathematics (9740) – 2012 Preliminary Examinations Paper 2 This list of common mistakes and comments are compiled to maximise your learning through the mistakes made. Do not miss the opportunity. Go through all the details carefully so that you will not make the same mistakes in the A level exams. 1. (i) The locus of point P is the half-line from (1, 2) (excluding it), inclined at an angle of  with the positive real axis. (ii) 2 6 2i 2w   (iii) 2 6 2i 2w   2. (i)   1 1 1 11r r r r      1 1 1 1 1 122 11 1 1 1212 1 2 2 n n n r r r rr r r r r n n                    1 .3 1.. n 1 1 1(1 ) 1 1 n nn n             1 11 12 1 nn rr rr u u r rr         1u 0 2 u u   1u 1 ... nnuu  0 1( 1) 1 1 11 ( 1) 1 (since 1 as given)1 1( 1) 2 (shown)1 n n nn n u n n u n u n n n                  Locus of P meets locus of Q more than once if 1122 55sin sin 0.412 0.412          2 2 5 |w – 6 – 2i| = 2 (ii)  4 (1, 2) /6  (1, 6) |z – 1 – 6i|min |z – 1 – 6i|min 4sin 26 4sin 3 23       (1, 2)

    1 1 2 1 Let 1. When 1, 2; when 1, 12( 1) ( 1)( 2) 12 1 1122 1 1(1 1) 15( 1) 1 12 13( 1) 12 n r n k n k k r r k r n k n r rr k kk k kk nn n nn n                                         3. i) Since replacing x with –x results in the same equation, the graph is symmetrical about the y-axis. (ii)   22 2 2 2 Differentiate with respect to : dd.2 2dd d 2 1 2d d2 (shown)d 2 1 y y y y e x y x yyex xx y exx yx xe      (iii) The line xk is vertical, so at x = k, d d y x is undefined. 2 112 1 0 ln . 22 yey     When 11ln ,22y     11 22 2 2 2 ln 11ln22 1 1 1 1ln 1 ln 22 2 2 2 ek k      Hence, 1 ln 2 2k  . (iv) When d1, 0, 2. d yxy x   Equation of tangent :  0 2 1yx   2 2.yx   Solving equations of the tangent and the curve simultaneously, we get 4 4 2 2 2.xe x x    From GC, 1x (rejected as this gives the previous point) or 2.73x (3 s.f.) (iii) Alternative    1 1 1 2( 1) 12 n r r rr      = 4 6 8 ... 2 n    1 1 1 ... 2.3 3.4 ( 1) nn         1 2 11 1 2 1.2 n r rr r    15 ( 1) 1 12 13 ( 1) 12 nn n nn n           

4. (i) 2 4 6 solve simultaneously using GC, 6 x y z x y z         1 2 1 2 5 1 0 xz yz zz    So the equation of the line l is 51 1 1 , . 02                  r