TPJC_JC2_H2_Maths_2012_Year_End_Exam_Solutions_Paper_1 -...

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2012 JC2 Preliminary Examination H2 Mathematics Paper 1 Solution 1 2 2 2 1 1 3 x x x - + < + 2 2 2 1 3 x x x - + < + 2 2 0 x x - - < ( 1)( 2) 0 x x + - < 1 2 x - < < 2 (i) 1 1 2 d 1 e e d x x x x =- (ii) 1 1 3 2 1 1 2 1 1 1 1 1 e d e d 1 1 e e d 1 e e x x x x x x x x x x x x x x C x = - - � � � � = - - � � � � � � � � =- + + 3 Distance between (1, 0) and ( x , y ), ( ) ( ) 2 2 1 0 W x y = - + - Since ( x , y ) lies on the curve, then 2 e x y = Thus, ( ) ( ) 2 2 2 1 e x W x = - + ( ) 2 2 4 1 e x W x = - + Differentiating w.r.t. x , ( ) 4 d 2 2 1 4e d x W W x x = - + ( ) 4 d When 0, 1 2e 0 d x W x x = - + = From GC, x = - 0.14043 - 0.14043 - - 0.14043 - 0.14043 + 1
x d d W x - ve 0 +ve \ / Thus, W is minimum when x = - 0.14043 When x = - 0.14043, 2( 0.14043) e 0.75513 y - = = The coordinate of point P is ( - 0.140, 0.755) Alternative Method: Curve ( ) 2 4 1 e x y x = - + The x value of the minimum point of ( ) 2 4 1 e x y x = - + gives the minimum value of ( ) 2 4 1 e x x - + . When x = -0.1404361, 2( 0.1404361) e y - = = 0.75513. The coordinate of point P is ( - 0.140, 0.755) 4 Let n P be the statement ( ) ( ) 1 3 3 1 3 2 1 4 n r n r r n = = + - , n + Z When 1 n = , LHS = ( ) ( ) 1 1 1 3 1 3 3 r r r = = = RHS = ( ) ( ) 1 3 1 3 2 1 1 3 4 + - = Since LHS = RHS, 1 P is true. Assume that k P is true for some k + Z , i.e. ( ) ( ) 1 3 3 1 3 2 1 4 k r k r r k = = + - To show 1 k P + is true, i.e. ( ) ( ) ( ) 1 1 1 3 3 1 3 2 1 1 4 k r k r r k + + = = + + - 2
LHS = ( ) ( ) ( ) ( ) 1 1 1 1 3 3 1 3 k k r r k r r r r k + + = = = + + = ( ) ( ) ( ) 1 3 1 3 2 1 1 3 4 k k k k + + - + + = ( ) ( ) ( ) 1 3 4 1 3 2 1 1 3 4 3 k k k k + + - + +

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