YJC P2 MS

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 1 of 10 Qn Solution 1 (i) Let  be the acute angle between the 2 planes. 12 21 11 3cos 666                   3   (ii) 41 12 21 OB                     for some R 4 12 2         Since B is on 1 , (4 ) 2( 1 2 ) (2 ) 2         62  1 3  11 3 11 51 533 55 3 OB            Thus the coordinates of B is 11 5 5,,3 3 3   . (iii) Normal of 3 1 2 3 1 2 1 3 3 1 1 1 3 1                                            Cartesian equation of 3 : 1 4 1 1 1 1 1 2 1                               r

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 2 of 10 7x y z    7x y z   From G.C. the position vector of the point of intersection of the three planes is 10 3 53 2     2 OPQR forms a rectangle with length OR which is 3 times the length of OP. (a) Let the position vector of complex number z2 be OS 3z x xi   22 2 2 3 8 6z x xi x x i    2 2 8 6 xOS x     3 3 3 3 86 z iz x xi i x xi x xi        8 6 xOQ x   8 6 xOS x xOQx     Since x is a constant and OS xOQ ,  S is collinear with the origin and the point Q. Im(z) Re(z) –3 Q R P O

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 3 of 10 (b) Area of region P =   2 11113 tan 3 tan 4.1723   3 (i)   2f 4 4x x x          22 2 2 2 2 4 2 4 4 x x           Since horizontal line y = 0 cuts the graph  fyx twice, f is not a one-one function.  f does not have an inverse. (ii) Largest value of k = 2 Let   2 22 4 4yx     22 4 4xy     Since 2x  , 22 4 4xy     1 2 2f : 2 4 4 , 4 4x x x          (iii)  1f ,2R      g ,1D   Given 1 2  ,    1 gf , 2 ,1RD        1gf  exists   1gf ln 1 2 ,R       22 , 4 4 Im(z) Re(z) 3 3 –3 –3 P O

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/2 Pg 4 of 10 4 (i) 2 3 4 2 3 4 5 7 9,,333u u u (ii) 21 3 n n nu  (iii) Let Pn be the statement “ 21 ,3 n n nun  ” When n = 1, LHS = 1 1u  , RHS = 21 13   Since LHS = RHS, P1 is true. Assume Pk is true for some values of k  ie 21 3 k k ku  When n = k + 1 Want to show: 1 1 23 3 k k ku      1 1 1 1 LHS 1 323 1 2 13233 23 3 RHS k k kk k kk k u u k k              1P is truek By Mathematical Induction,