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( ) 2  () 2i e  (Proved). * 2 () 22 ()2 2 ( )( * *) ( )( ) || | 2   sin( ) |2 4sin ( ), since | | 1 & | | 12 2(1 cos( )) i i z w z w z w z w zw ie ie                    Therefore, k = 2. Alternatively, Since arg( *z ) = -α and arg( *w )=-  , we have ( )( * *)z w z w = [2i sin( ) 2  () 2i e  ][2i sin( ) 2  () 2i e  ] =4i2 2( sin ( )) 2  =4 2sin ( ) 2  = 2(1 cos( )) Therefore, k = 2. (b) 2 3 1 i 320 1 i 3 z z     54 3 2(1 3 ) 0 (1 3 ) zi iz    54 2(1 3 )zi =0 54 2(1 3 )zi   = 4()4 32(2 ) i e   = 25 4() 3 since e 1iiee    = 25 7() 3i e  = 2 ()3i e  , correct to principal range.

7 Therefore ( 2 )55 32 , 2, 1,0,1, 2 ik z e k      6() 152 ki ze   , k = -2, -1,0,1,2 The roots are 11 5 7 13( ) ( ) ( ) ( ) ( )15 15 15 15 15 1 2 3 4 52 , 2 , 2 , 2 , 2 i i i i i z e z e z e z e z e           . Pi represent the complex no zi, where i= 1,2,3,4,5. 6 Three cases: Case 1: All three balls distinct (ABC) n(ABC) = 6 3C (6 types of balls available) Case 2: two identical (AAB) n(AAB) = 25 11CC .(2 types of balls available to choose the 2 identical balls from, then 5 types of balls available remaining to choose the last ball) Case 3: three identical (AAA). n(AAA) = 1 n(Total) = n(ABC) + n(AAB) + n(AAA) = 20+10+1=31 7 (i) It is necessary for the research company to collect a sample because it is impossible for the company to interview all eligible voters in the United States before the election as the population is too large (90 million) (ii) Quota Sampling. The method is not appropriate because it might be biased towards households that have eligible voters that are present only during the phone interviews. Moreover, choosing a particular telephone book means that the sample comes from a particular area/state of the United States, and the sample may not be representative of the Im(z) P4 P5 P3 Re(z) 0 P1 P2

8 sentiments of all eligible voters in the United States. (iii) One could use stratified sampling. This could be conducted by ensuring that the sample of 1000 voters be sampled in the following manner Educational Qualification No degree Bachelor Degree Master degree or higher Percentage / (Number) 58.61% 586 30.44% 304 10.95% 110 The research company could ensure that they sample 586 eligible voters that have no degrees, 304 eligible voters have Bachelor degrees and 110 eligible voters that have Master degrees or higher. The company can randomly choose the required number provided by the respective educational institutions (representing different regions/states in America) to ensure that each of the chosen interviewees within each stratum have an equal chance of being selected. 8 (i) By symmetry, 250 300 2752  P(X < 250) = 0.2 P 250 275Z    = 0.2 25 0.84162 29.7      (ii) P(X > c) = 0.01  P(X < c) = 0.99 c = 344 (correct to 3 s.f.) 9 (i) P(last ball is green) =

9 (ii) 10 20 2 ~ P ( ) 1P( 2) P( 0) 8 1 2! 8 0! 11 0.5 (-ve rejected)42 oX XX ee             Alternative: Using GC P(1  X < 4) = P(X = 1) + P(X = 2) + P(X = 3) or P( 3X  ) – P(X = 0) = 0.392 (3 s.f.) Let Y be the no. of defects in the carpet of area 500 m2. o 500~ P 0.520Y   o~ P (12.5)Y Since =12.5 > 10, ~ N(12.5,12.5)Y approximately. P(Y > 10) P(Y >10.5) = 0.714 (3 s.f.) 11 (i) Let X be the r.v. “ volume of content of randomly chosen milk carton” and µ be the mean volume of content in milk cartons. H0: µ = 2 H1: µ < 2 We carry out 1-tailed t-test at 10% significance level, since the population variance is unknown and sample size n is small. Under H0, T= ~ (9) / X t sn  From GC,  .cc 

10 1.9834 -value 0.00882 x p   Since p-value < 0.10, we reject H0 and conclude that at 10% significance level, there is sufficient evidence that the manufacture