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2007 SAJC Prelim Paper 1 solutions

# 2007 SAJC Prelim Paper 1 solutions - 2007 H2 Maths Prelim...

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1 2007 H2 Maths Prelim Exam(Paper 1) Solutions 1 x -coordinates of point of intersections = 1 1 , 2 6 Graphs of 2 9 y x and 4 | | 2 y x : For 2 9 4 | | 2 x x , the solution set is 2 1 or 6 1 : { x x x }. 2(i) a = 0 2 2 1 f : 2 , , 0 2 2 since 0 f ( ) 2 , 2 x x x x y x x y x x x x     (ii) Since f g ( ,2] ( ,3) R D     , gf exists. 3(i) 2 2 1 1 1 1 2 x C Bx x A x x ) 1 )( ( ) 1 ( 2 2 x C Bx x A Subts. x = -1, A = 1 Comparing coefficients of 1 : 1 : 0 2 C x B x 1 , 1 , 1 C B A (ii)   1 1 1 1 1 1 1 1 1 2 f 1 2 1 2 2 x x x x x x x x x 1 1 1 2 3 2 x x x x x 2 3 2 3 1 1 x x x x x x x 2 2 (Shown)

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2 4 (i) Let n P be the statement 2 2 1 1 ln 1 ln 2 n r n n r where 2 n . When n = 2, 2 1 3 ln 1 ln 2 4 LHS , 2 1 3 ln ln 2(2) 4 RHS . 2 P is true. Assume k P is true for some positive integer k 2 , i.e. 2 2 1 1 ln 1 ln 2 k r k k r Want to prove that 1 k P is true based on k P is true, i.e. 1 2 2 1 2 ln 1 ln 2( 1) k r k r k 1 2 2 2 2 2 2 2 2 1 ln 1 1 1 ln ln 1 2 ( 1) 1 1 ln 1 2 ( 1) 1 2 1 1 ln 2 ( 1) 1 2 ln 2 ( 1) 2 ln 2( 1) k r LHS r k k k k k k k k k k k k k k k k k RHS k So 1 k P is true. Since 1 P is true and 1 k P is true based on k P is true, n P is true for all 2 n , by Mathematical Induction.
3 (ii) 2 2 1 2 2 2 2 2 1 ln 1 1 1 ln 1 ln 1 2 1 1 ln ln 2(2 ) 2 2 1 2 ln 4 1 2 1 ln 2( 1) 2 1 ln 2 2 n r n n n r r r r r n n n n n n n n n n n n  

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2007 SAJC Prelim Paper 1 solutions - 2007 H2 Maths Prelim...

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