SAJC P2 MS

2 1 (i) 1 1 ( 1) 1 ! ( 1)! ( 1)! ( 1)! kk k k k k    Series 21 2 21 2 ( 1)! 11 ! ( 1)! 11 2! 3!11 3! 4! 11 (2 )! (2 1)! 11 (2 1)! (2 2)! 11 2! (2 2)! 11 2 (2 2)! n r n r r r rr nn nn n n                      (ii) 2 2 1 23 21 3 22 ( 2)! ( 1)! 1 1 2 1 ( 1)! 2 (2 2)! (2 1)! 6 nn rr n r rr r rn                2 (i) Distance travelled at the (n+1)th bounce = 10+2(10) 1 2 3 3 3 3 3 ...4 4 4 4 n                        = 10 + 20 33144 31 4 n       = 10 + 60 31 4 n 

3 = 70 - 60 3 4 n   (ii) Distance travelled at the (n+1)th bounce 70 - 60 3 4 n   So the largest value of is 4. In other words, at the (4+1)th = 5th bounce, the distance travelled is less than 55m, while at the 6th bounce, the distance travelled is more than 55m. Therefore, when the ball has travelled exactly 55m, it has only bounced 5 times. (iii) Total distance travelled at the maximum height after the 4th bounce (at the exact moment when the floor is raised) Height after 4th bounce, h = 10 4 3 4   = 3.1640625 m New height on platform, h’ = 3.1640625 - 2 = 1.1640625 m Distance travelled on the raised floor = h’ +2(h’) 1 2 3 3 3 3 ...4 4 4                   = h’ + 2h’ 3 4 31 4      = h’ + 2h’ (3) = 7h’= 8.1484375 m

4 Total distance travelled = Dist. travelled before floor is raised + dist. travelled after floor is raised = 3 (a) (b) Using long division 2 22 44 1( 2) ( 2) xx xx - =--- So A=1 and B = -4 Series of transformations: 1. Translate the graph of 2 1y x= by 2 units in the positive x direction 2. Stretch the resulting graph parallel to the y-axis with a scale factor of -4. 3. Translate the resulting graph by 1 unit in the positive y direction. x 4 2 -3 -2

5 4 2 22 10 (10 ) 1 10 1 5 ( 5) 1 5 ( 5)ln10 5 ( 5) ln 1010 , where 10 kt D dx kx xdt dx k dtxx dx k dtx x kt Cx x kt Dx x Be B ex              Where t = 0, x = 2 B = 1 4 10 10 10 10 1 10 4 4 (10 ) 10 4 kt kt kt kt x ex x e x ex e     (i) (ii) k < 0 A vaccination had been found to eliminate the virus OR The people infected by the virus had passed away OR any other reasonable answer. 5 zw- = ie - ie  = (cos α+isin α)- (cos+isin ) = (cos α- cos ) + i (sin α-sin ) = 2sin( )sin( ) (2sin( )cos( )2 2 2 2 i           ) 1 2 3 −5 5 10 x y 2 Asymptote: y = 10 k = 1 k = 0 k = - 1

6 = 2i sin( ) 2  [i sin( ) 2  + cos( ) 2  ] = 2i sin(