MIPU3 H2 Mathematics Paper 2 2012 Prelim II Answer Key

PU3 H2 Mathematics Paper 2 2012 Prelim II 1(a)(i) 1(a)(ii) 4 2 5 1 4 2 5z     1(b)(i) 1(b)(ii) 12zi 2(i) x (3, 4) -1 5 x (1, 2) 2 3 1 y = -1 0

2(ii)       2 1 1 2 1 ln 1 2 ln 1 2 : ln 1 , 1 xye yx yx f x x x      2(iii) 2(iv) 2 2 1 1 0.797 0 x x ex ex x x     3(i)       2 2 2 2 1 2 1 2 2 1 22 dx tdt dy dt t dy tdx t dy dx tt         3(ii) 2 1 2 2 1 2 1 1 1 1t x t t yt             At 1,t 1 2 dy dx  Gradient of normal 2  1 2 2 25 yx yx     0 x = -1

3(iii)   211 22 32 12 2 4 3 5 0 1.158 1 2.158 t t t t t t t t            4(a)                1 2 11 62 6 6 4 4 1 2 1 4 1 1 2 1 12 1 2 11 n r nn rr nn n n rr rr n n n nn nn                   4(b)(i)         11 11 12 11 23 11 1 1 1 1 1 1 11 ... 1 nn rr nn n r r r r              4(b)(ii)    11 21 211rr r r r r    As 1 1, 1 1nn     1 2 21r rr     5(a) (i) Consider the green, yellow and purple tiles as 1 unit. Number of ways to arrange the green, yellow and purple tiles within the unit = 5! 3! Arranging the unit with the rest of the tiles = 7! 3!3! Thus number of possible arrangements for the tiles = 5! 7! 3! 3!3! =2800

(ii)Number of ways to arrange the 3 red, 3 green, 1 yellow and 1 purple tile = 8! 3!3! Number of ways to slot in the blue tiles = 9 3   Number of possible arrangements such that no blue tiles are placed next to another = 8! 3!3! 9 3   =94080 (iii) Number of possible arrangements such that a red tile at the beginning and another red tile at the end of the line = (3 3 3)! 3!3!  = 10080 5(b) (i) Number of ways =  10 1 ! 362880 (ii) Insertion method Number of ways   7 27 1 ! 2! 2! 60480C      (iii) Number of ways  9 1 ! 10 2! 806400     6(i) P(same colour) 45 22 9 2 4 9 CC C  Die is 1 Die is 1 Die is not 1 Die is not 1 Die is 1 5 6 Same colour Different colour Die is not 1 Die is not 1 Die is 1 1 6 4 9 5 9 1 6 1 6 1 6 5 6 5 6 5 6

P(wins exactly one DVD) 4 1 5 5 1 37 2 (0.228 (3 s.f.))9 6 9 6 6 162       6(ii) P(different colour|did not win any DVD) P(different colour and did not win any DVD) P(did not win any DVD) 5 5 5 25966 (0.510 (3 s.f.))4 5 5 5 5 49 9 6 9 6 6        6(iii) P(wins 2 DVDs| different colour) 1 1 1 6 6 36   6(iv) Let X be the number of DVDs John can win, out of 2 dice throws. Then X ~ B(2, 1 6 ). Mean number of DVDs = 112 63 7(i) 0.946r 7(ii) The scatter diagram shows that the relationship between t and x is non-linear. 7(iii) Fro