RI 9740/01 2013

1 | Page   Question 1 No. Suggested Solution Remarks for Student (i) When 3,  we have 24 , 22 6 , 543 9 . xz xy z xyz      From GC, the point of intersection is 38 119 25,,.363   (ii) When 0,  we have 24 , 22 6 , 54 9 . xz xy z xy      From GC, p, q and r have no point of intersection. Since no two of the planes are parallel, they form a triangular prism. It’s 3 marks, so obviously they are looking for more than “no solution”. The normal can be read off the Cartesian form of the equation. Eg. 24xz has normal 1 0. 2    Hence it is easy to check that no two of the planes are parallel. Question 2 No. Suggested Solution Remarks for Student 2 We have 2 2 2 .1 11 (1 ) ( 1) 0. xxyx y y x xx xy x y          If y is a value the expression can take, the final quadratic equation has a real solution for x. Therefore 2 2 (1 ) 4(1)( 1) 0 0 32 3 o r 3 63 . . . 23 yy yy yy           Set of values y can take is ,3 2 3(] [ 3 2 3 , ) .   While this is the fastest method, it can sometimes be hard to see how to use the discriminant in such questions. If you have difficulty, you may want to try the alternative method. Raffles Institution H2 Mathematics Solution for 2013 A-Level Paper 1

2 | Page   Alternative Method: We look for the range of the function 2 13f( ) 2 11 xxxx x x    by finding its maximum and minimum values. 2 3f( ) 1 0 (1 )x x  when 13x or 13 .x These give the stationary points (1 3, 3 2 3 ) and (1 3, 3 2 3 ). 3 6() , (1f )x x   so 1(1 3 ) 0 2 f 3     and 1(1 3 ) 0. 2 f 3     Thus (1 3, 3 2 3 ) is a maximum point and (1 3, 3 2 3 ) is a minimum point. Therefore the set of values f( )yx can take is ,3 2 3(] [ 3 2 3 , ) .   From here, it is best to use GC to check how the graph of f( )yx looks to figure out the range. Question 3 No. Suggested Solution Remarks for Student (i) 11 3 21 2 2 ( 21 ) xy xx   Simplifying the equation first will make it easier to find the asymptotes. Remember to follow the instructions of the question. It asks for the equations of asymptotes and coordinates of intercepts. x   y        2

3 | Page   (ii) 1 121 x x   when 2.x From the graph in (i), 1 121 x x   when 1 2x or 2.x Don’t waste time solving algebraically. Use the graph. Question 4 No. Suggested Solution Remarks for Student (i) 33 2 3(1 2i) 1 3(2i) 3(2i) (2i) 11 2iw        Alternative Method: 22 3 12 i (1 2i) 3 4i (1 2i)( 3 4i) 11 2i w w w           Binomial Theorem. Usually this more tedious method is to be avoided. But in this case 2w is need later, so there is no loss in time. (ii) Since 12 iw is a root, 32 ( 11 2i) 5( 3 4i) (1 17(1 2i) 0 (1 1 2i) 5(1