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 2i) 17(1 2i 2) ( 2 )0 54)i 0 . . . ab ab ab a                     Comparing imaginary and real parts, we have 27a and 11 2 295.ba  It is possible to do (ii) and (iii) together. See below. (iii) Since a and b are real and 12 iw is a root, * 12 iw  is also a root. Therefore 32 2 27 5 17 295 [ (1 2i)][ (1 2i)]( ) (2 5 ) ( ) , zz z z z c z d zz c z d          so clearly 27c and 295 59.5d  Hence the roots of this equation are 12 i , 12 i and 59 .27 Alternative Method: Since a and b are real and 12 iw is a root, * 12 iw  is also a root. Therefore 32 2 32 5 17 [ (1 2i)][ (1 2i)]( ) (2 5 ) ( ) (2 ) ( 5 2) 5. az z z b z z cz d zz c z d cz c d z c d z d              Remember the trick to evaluate *() ( ) .zw zw This method deals with both (ii) and (iii) together, so we do not use the values of a and b from (ii).

4 | Page   Comparing coefficients of 2z and z gives 25 52 1 7 . cd cd   Thus 27,c 59,d  and 27,ac 52 9 5 .bd The roots of this equation are 12 i , 12 i and 59 .27 Question 5 No. Suggested Solution Remarks for Student (i) Graph of f( )yx for 64: x a a  We can assume 0a here as we are given .a x a Sketch the graph for 2aax first, then use the condition f( 3 ) f( ) ,x ax which says that f is periodic with period 3a, to fill in the rest. (ii) Since 31 22 ,aa a a   33 222 21 1 2 2 f( ) d 1 d . aa a a xx xx a      Using the substitution sin ,xa  we have 2 2 2 dsin cos and cos . d11 x ax a       Also, 26 1xa  and .3 3 2xa  Therefore  3 2 1 2 2 3 2 6 3 6 3 6 3 6 2 cos ( cos ) d1d 2 .21 cos d 1c o s 2 d si 2 22 n a a ax xa a a a a                               For substitutions, change x, dx and the limits. y   a  2a  3a  4a  5a  6a a 2a 3a 4a  x   1

5 | Page   Question 6 No. Suggested Solution Remarks for Student (i) Since a and b are not parallel, a vector equation for the plane OAB is ,, .   rab  As C lies on this plane, we have   cab for some real constants  and . (ii) 1 (4 3 )7ON  ac  Ratio Theorem (iii) The area of triangle ONC is 11 1 (4 3 )22 7 2 () 7 22 .77 OON C         ac a a c ab a c b      Similarly, the area of triangle OMC is ( 11 1 22 ) 2 1 .44 MO CO     bc ab b ab   Since , ba ab we have 2 74 8 .7    Note the use of properties of the cross product, in particular .aa 0 Note that  and  below can be taken out of the magnitude because we are told both are positive. Question 7 No. Suggested Solution Remarks for Student (i) The length, ,ka of the kth piece of string cut off forms a geometric progression with first term 128 and common ratio 2 ,3 so 1 2128 . 3 n npa    Hence 1 7 2ln ln 128 3 ln128 ( 1)(ln 2 ln 3) ln 2 ( 1)(ln 2 ln 3) ( 6 )l n2 ( 1 )l n3 , n p n n nn              giving 1, 6, 1AB C  and 1.D

6 | Page   (ii) The total length of string cut off cannot exceed 1 128 384.21 3 k k a      (iii) Let nS denote the total length of the first n pieces of string cut off. Then 2128 1 3 2384 1 .2 31 3 n n n S          When this exceeds 380 we have 22 951 380 133 96 21 39 6 1ln 96 11.3.2 3 8 ln 34 nn n n             Therefore 12 pieces must be cut off before the total length is greater than 380 cm. The question asks for working to justify your answer, so you cannot just read the answer from the GC. In particular, this means no “Table of Values”. Question 8 No. Suggested Solution Remarks for Student (i) We have i31i3 2 e ,   so i 3i i3 (1 i 3 ) 2e 2 e . wz er r         Therefore 2wr and 3arg .w    Alternative Method: ( 1i3 ) 1i3 2wz z r     arg arg (1 i 3) arg(1 i 3) arg 3wz z     Be sure to consider the quadrant when finding arguments. Both presentations are fine, but the first is more convenient if you need to add or subtract 2 to the argument.

7 | Page   (ii) Recall that the effect of multiplying a complex number z by ies  is to scale it by a factor of s and rotate it counter-clockwise about the origin by angle . Thus the locus of w is obtained from that of z by scaling it by a factor of 2 and rotating it clockwise by .3  (iii) 10 2a rg 10a rg 2arg 228 310 3 z zww         Therefore 8 2 ,3   giving .24   More complete solution: Arguments of complex numbers can be changed by multiples of 2 without changing the number, so in fact 10 2 2 23arg 8w kz        For .k  Therefore ,24 4 k   and fo