SAJC H2 MATHS P2 Solutions

2017 Prelim Paper 2 Solution 1 Method 1 i3zw  3 3i ii zwz    --- (1) Substitute (1) into 2 6 3i 0zw    2 (3i i) 6 3i 0zz      2 i 6 0zz      2 i i 4(1)(6) i 1 24 22z        i 5i 2   2iz or 3iz  3i (2i)iw 3i ( 3i)iw   23 i 3 3i  Method 2 i3zw  3izw …. (1) Substitute (1) into 2 6 3i 0zw      2 3 i 6 3i 0ww      29 6 i 6 3i 0w w w       2 (1 6i) 15 3i 0ww       2 (1 6i) 15 3i 0ww       2 (1 6i) 15 3i 0ww      2(1 6i) (1 6i) 4(1)( 15 3i) 2w       

1 6i 1 12i 36 60 12i 2        1 6i 25 2     2 3iw or 3 3iw   3 (2 3i)i 2iz    3 ( 3 3i)i 3iz     Method 3 i 3 i 3i wz wz      2 ( i 3i) 6 3i 0zz       2 i 6 0zz   Let iz a b where ,ab  2( i) i( i) 6 0a b a b      22 2 i i 6 0a b ab a b       22 6 (2 )i 0a b b ab a      By comparing the real and imaginary parts, 22 60a b b    … (1) 20ab a … (2) From (2), 0a or 1 2b When 0a , 2 60bb   ( 2)( 3) 0bb   2b or 3b Hence 2i, i(2i) 3i 2 3izw     or 3i, i( 3i) 3i 3 3izw      When 1 2b , 2 2511 4 2 4 6a    

There is no real solution for a. 2(i)                    2 2 2 3 3 2 3 1 11 2 1 3 1 1 1 11 2 2 3 3 3 n n n n n n n n n n n n n n n n n nn n nn                  (ii) 3 2 3 2 2 26 32 2 3 12 11 2 3 1 1 2 3 231 2 3 4 2 3 1 2 3 4 5 ... 231 21 2 3 1 +11 n r n r n r r rr r rr r r r n n n n n n                                 2 3 2 1 3 12 1 2 2 1 3 2 12 21 423 1 n n n nn nn                (iii)    2 2 10 1 2 3 n r r r r r     Let 22r p r p    

    2 22 2 3 4 23 33 22 26 11 26 2 6 2 6 4 2 4 233 2 3 3 4 5 4 2 6 2 3 pn p n p n pp p p p p p pp pp p p p p nn nn                                   3(i) 2 2 2 1f : 1 1Let 1 1 1 11 x x y x x y x y       Since 1x , 11x y   1 1f1 x x   = 1 x x  . From graph of f,  fR 0,   1fD x   0, . x =1 y =0

(ii) (iii) Since 11ff ( ) f f ( )x x x  have the same rule, we investigate the domain  1ff 1,D    1ff 0,D   Taking the intersection of these domains, Range of values is 1x . 4 (i) Equation of plane is 1 1 2 3 2 0 , , 2 0 1                               r A normal vector to plane is 1 2 2 2 0 1 0 1 4                           Hence vector equation of the plane is x =1 y =1 y = f(x) =1 y =0 x =0

2 1 2 1 3 1 4 2 4 2 13 4                           r r (ii) 52 : 2 1 , 24 ACl s s                  r Thus 52 2 1 for some 24 OC s s                  . Since C lies on the plane: 5 2 2 2 1 1 3 2 4 4 2( 5+2 )+(2+ ) 4(2 4 ) 3 3 21 s s s s s                                Thus 325 21 3731 2 1321 7 10 342 21 OC             (iii) Using mid-point theorem '2 37 5 3921 13 2 1277 10 2 6 OA OC OA                            B is the point of intersection of l1 and  .

'' 39 11 12 37 62 461 97 20 BA OA OB                     2 39 461: 12 9 ,7 6 20 l t t                   r or 2 1 46 : 3 9 , 2 20 l t t                  r 5(i) The height of triangle ADG is tan aa t  . Hence 122 aAH a a tt      . 2 tan (2 1)BH BE EH a a a t       Area 1 ( )( )2S AH BC  12 2 (2 1)2 aS a t t    2 12 (2 1)S a t t    2 144S a t t    (ii) 2 2 d1 4d S att   When d 0d S t  , 2 1 4t 

 1 2t  Reject 1tan 2t   as  is acute 2 2 23 d2 d S att   When 1 2t  , 2 22 32 d2 16 0d 1 2 S aat      . Hence the minimum value of S occurs when 1 2t  . Minimum   22 4 2 2 8S a a    . (iii) To sketch the graph of 2 14 4 tan tanSa      6 (a) Since adjacent balls do not sum up to two, balls numbered ‘1’ needs be separated. Number of ways of arranging the other balls with no restriction = 6! 2   0  121tan ,82 a  S 

Slotting in the balls numbered ‘1’, permutation is done as balls are of different colour = 7 3 3!C  No of ways 7 3 3!6! 151200 C   (b) Method 1 Table of 5 Table of 6 Case 1 – 2 friends are seated together at table of 5 No. of ways to select 3 other friends and arrange them at the table of 5 = 9 3 (4 1)!C  No. of ways to arrange the 2 friends = 2! No. of ways to sit the remaining friends at the table of 6 = (6-1)! = 5! = 120 Total no. of ways = 9 3 (4 1)! 2! 5!C     =120960 Case 2 – 2 friends are seated together at table of 6 Table of 5 Table of 6 No. of ways to select 4 other friends and arrange them at the table of 6 = 9 4 (5 1)!C  = 3024 No. of ways to sit the 2 friends at the table of 6 = 2! 2 friends X X X (6-1)! (5-1)! X 2 friends X X X

No. of ways to sit the remaining friends at the table of 5 = (5-1)! = 4! = 24 Total no. of ways = 9 4 (5 1)! 2! 4!C     = 145152 No of ways to arrange 11 friends without restrictions = 11 5 (5 1)! (6 1)!C     = 1330560 Total no. of ways of arranging 11 people such that 2 particular friends are not seated together = 1330560 – 120960 – 145152 = 1064448 Method 2 Alternative Method Case 1: Two particular friends seated at table of 5 No of ways 9 3C 2! 3 2 1209 0 5! 6       9 3C : Selection of friends to be seated at table of 5. This automatically selects friends to be seated at table of 6. (3-1)!: Arranging the 3 other friends in table of 5. 3 2P : Slotting in the 2 particular friends 5!: Arranging the 6 other friends in table of 6. Case 2: Two particular friends seated at table of 6 No of ways 9 4C 4! 3! 4 21772 3 8      9 4C : Selection of friends to be seated at table of 5. This automatically selects friends to be seated at table of 6. (5-1)!: Arranging the 5 friends in table of 5. 4!: Arranging the 5 friends in table of 6. 4 2P : Slotting in the 2 particular friends