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 rk = Thus, 20 20 1 dx k kk dt x x = ⎝⎠ . (Shown) Solving DE: 20 20 dx x x kd dt x x ⎛⎞ =⇒= ⎜⎟ x k d t 20 1 20 dx k dx x += ∫∫ 20ln 20 x xk t +− = c + When 104, 5 x t , 104 --- (1) 20ln84 5 kc + + When , 121 --- (2) 121, 6 xt 20ln101 6 (2) – (1) 20.6861 20.7 k = and finally 89.2 c When solve 7, t = 20ln 20 7 233.988 xx 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 13 zi = 0 implies 2 4 3 2 i e π =− + = . Thus 2 2 4 3 2 ik ze 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 22 4 11 2 b zb ±− = =± − . To have all roots real, z 2 has to be real 1 b 0 b 1 ---(1) Further, we solve for z : ± . For real , z must be . b 0 Since is always > 0 for b b 1, we need only worry about solving for b when 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 zz =− . 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 42 2 b −+ = 0 4 2 y z z and the line b y = must have 4 intersection points (taking into account the repeated root). (ans) ) 10 b −≤−≤ 0 b ≤≤ 1 3(a) 222 23 1 41 7 1 21 2 2 xx dx dx dx x x ++ ∫∫∫ 2 2 7 1 6 1 4 7 4 17 1 ln 2 1 22 2 [ ( ) ] () 4 ln 2 1 . tan 24 7 1( 4 1 ) ln 2 1 7 tan 2 7 x xd x x x x xc x x =+ + + + + + + + 3(b) 3 0 a x ad x 3 0 3 33 0 3 22 3 aa a a x x x a d x ax a + ⎡⎤ + ⎢⎥ ⎣⎦ = ∫∫ 3(c) 1 1 (l n) = ( l n) ee 2 x xd x x x 2 1 1 3 2 1 3 3 1 1 1 =l n 2 l n 2 (ln ) ln (5 2) 33 3 3 2 7 e e e e e d x x e d x ex x x x ⎧⎫ ⎪⎪ = ⎨⎬ ⎩⎭ 1 e y x 1 0
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4(i) Sub equation of line into plane: 10 1 11 . 0 3 01 2 ⎡⎤ ⎛⎞ ⎛⎞⎛⎞ ⎜⎟ ⎜⎟⎜⎟ + = and get . ⎢⎥ ⎝⎠ ⎝⎠⎝⎠ ⎣⎦ λμ 2 32 λ= − μ 2 Sub 3 2 into Π : 3 (3 2 ) 1 1 3 1 0 1 ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ =−μ + μ = + μ− ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ r . (ans) OR if using 1 (3 ) 2 μ= −λ and substitute into 2 Π , 02 3 2 ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟ =+ β
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This note was uploaded on 08/29/2009 for the course MA 9740 taught by Professor Moe during the Summer '07 term at Singapore Management.

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

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