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2007 HCI Prelims Paper 2 Solution

# 2007 HCI Prelims Paper 2 Solution - 2007 HCI H2 Math PAPER...

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2007 HCI H2 Math PAPER 2 SECTION A (PURE MATH) 1. From question, dt dx = rate of pop n – rate of pop n , therefore, dx r k dt x = . When 20, 0 dx x dt = = 0 20 r k = 20 r k = Thus, 20 20 1 dx k k k dt x x = = . (Shown) Solving DE: 20 20 dx x x k d dt x x = = x k dt 20 1 20 dx k dx x + = 20ln 20 x x kt + = c + When 104, 5 x t = = , 104 --- (1) 20ln84 5 k c + = + + When , 121 --- (2) 121, 6 x t = = 20ln101 6 k c + = (2) – (1) 20.6861 20.7 k = and finally 89.2 c When solve 7, t = 20ln 20 7 233.988 x x k c + = + = for x 139 . (Using GC). Possible Assumptions: - Number of BooBoo birds found in the island in 2005 and 2006 is close or equal to the actual number of BooBoo birds on the island. - There are no external factors like human interferences on the island in 2007 that may affect the population in a very different way that is not within the prediction of the model. A B C D 1 1 y x = y =1 0 x y Using graphs, 1 1 x 1 0 x 2(a) 4 1 3 z i + = 0 implies 2 4 3 1 3 2 i z i e π = − + = . Thus 2 2 4 3 2 i k z e π + π = and 3 , 2 , 1 , 0 , 2 2 6 1 4 1 = = = + k e z z k i k π . Suppose A = , B = , C = and D = . 0 z 1 z 2 z 3 z Observe that Lengths OA = OB = OC = OD . Also, the arguments of the roots differ by 90 ° the diagonals intersect at 90 ° . And deduce that ABCD forms a square: 2(b) Sub a = − into equation and get 2 2 2 2 2 4 1 1 2 b z b ± = = ± . To have all roots real, z 2 has to be real 1 b 0 b 1 ---(1) Further, we solve for z : 1 1 z b = ± ± . For real , z 1 1 must be . b ± 0 Since 1 1 is always > 0 for b b + 1, we need only worry about solving for b when 1 1 0 b 1 1 (see diagram). Thus 0 --- (2) b Combine (1) AND (2) (ans) 0 b

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2(b) ALTERNATIVE SOLUTION Sketch the graph of 4 2 y z z = . Turning points as intersection points or roots are in the form of a repeated root. If all roots are real of are real, the graph 4 2 2 z z b + = 0 4 2 y z = z and the line b y = must have 4 intersection points (taking into account the repeated root). (ans) ) 1 0 b − ≤ − 0 b 1 3(a) 2 2 2 2 3 1 4 1 7 1 2 1 2 2 1 2 2 1 x x dx dx dx x x x x x x + = + + + + + + 2 2 7 1 4 16 1 2 1 4 7 4 2 1 1 7 1 ln 2 1 2 2 2[( ) ] ( ) 1 7 4 ln 2 1 . tan 2 4 7 1 (4 1) ln 2 1 7 tan 2 7 x x dx x x x x c x x x c = + + + + + = + + + + = + + + 3(b) 3 2 2 0 a x a dx 3 2 2 2 2 0 3 3 3 2 2 0 3 ( ) ( ) 3 3 22 3 a a a a a a x a dx x a dx x x a x a x a = + = − + = 3(c) 2 2 1 1 ( ln ) = (ln ) e e 2 x x dx x x dx ( ) ( ) ( ) 3 3 2 1 1 3 2 1 3 3 3 3 1 1 1 = ln 2 ln 3 3 2 (ln ) 3 3 2 1 ln (5 2) 3 3 3 3 27 e e e e e x x x x dx x e x x dx e x x x dx x = = = 1 e y x 1 0