223321189-Tjc-h2-Math-p1-Solution.pdf - 2011 Preliminary...

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TJC/MA9740/P1/Preliminary Exam 2011/Solutions 1 2011 Preliminary Exam H2 Maths 9740 Paper 1 Solutions 1 Method of Difference 1 1 1 1 1 2 ( 1) n n r r r r u u r 2 1 u u 3 2 u u 4 3 1 n n u u u u 1 ( 1)( 2) n u u n n 2 2 2 ( 1) n u n n n n Thus ( 1) n u n n [Alternative Method] Summing up to n terms, we obtain 2 1 1 2 1 ( 3) 3 2 ( 1)( 2) Thus ( 1) n n n u u n n n n u n n n n u n n 2 Inequalities To find intersection point, 2 ( 2 ) x a x a   3 a x From the graph, for 2 2 x a x a , 3 a x Replace x by – x and let a = 2 in the above inequality, ( ) 2(2) 2( ) 2 x x becomes 4 2 2 x x Thus 2 2 3 3 x x   1 2 2 2 n n 3 a 2 a x y 2 a 2 a 2 y x a 2 y x a O a
TJC/MA9740/P1/Preliminary Exam 2011/Solutions 2 3 Functions & Transformations (i) Since R f = (0, 2] ± = D g , gf exists. Method 1 (mapping method) ( 1, 1) (0, 2] ( , ln 2] = Range of gf Method 2 (graphical method) gf( x ) = 2 2 1 ln 2 1 ln 2 ln(1 ) 2 x x , 1 < x < 1 Range of gf = ( , ln 2] (ii) A translation of 1 unit along the positive x -axis followed by a scaling of factor 1 2 along the x -axis. OR A scaling of factor 1 2 along the x -axis followed by a translation of 1 2 unit along the positive x -axis. 1 1 y x 2 y = f( x ) f g y y = gf( x )
TJC/MA9740/P1/Preliminary Exam 2011/Solutions 3 4 Mathematical Induction (i) 2 3 4 1 1 2(1) 2 1 1 2 2 3(2) 3 2 1 3 3 4(3) 4 S S S (ii) 1 n n S n , n 2 (iii) Let P n be the statement 1 n n S n where n Z , n 2. When n = 2, LHS = 2 1 2 S from (i) RHS = 2 1 1 2 2 P 2 is true. Assume P k be true for some k Z , k 2, i.e. 1 k k S k . When n = k + 1, 1 1 2 2 1 1 1 ( 1) ( 1) ( 1) k k k r r S r r r r k k 2 1 1 = ( 1) ( 1)( 1) 1 ( 1) ( 1) 1 ( 1) 1 1 k k k k k k k k k k k k k k k Thus P k +1 is true. Since P 2 is true, and P k +1 is true if P k is true, by the method of mathematical induction, P n is true for all n Z , n 2. (iv) As n , 1 1 1 n n S n n 1 . Thus 2 1 1 r r r converges to 1.
TJC/MA9740/P1/Preliminary Exam 2011/Solutions 4 5 Complex Numbers (i) 3 1 2 z 1 1 5 arg tan 6 3 z 1 1 5 1 4 i 4 6 4 3 i 2 z e   = 1 5 4 i 2 6 2e k 5 1 1 i 24 2 4 4 2 e k z , k = 2, 1, 0, 1 = 1 19 i 4 24 2 e , 1 7 i 4 24 2 e , 1 5 i 4 24 2 e , 1 17 i 4 24 2 e (ii) * 10 2i w z

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