2007 SRJC Prelim Paper 1 Solutions

2007 SRJC Prelim Paper 1 Solutions - 1 2007 JC2 H2 MATH...

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1 [Turn Over 2007 JC2 H2 MATH PRELIM PAPER 1 1 Solve the inequality 2 29 2 28 x x xx  . Hence solve the inequality   2 2 42 1 2 9 1 x x . <Solution> 0 2 ( 2)( 4) 1 0 ( 2)( 4) x x x x x    + + –2 –1 4 2 1 or 4        2 2 1 2 9 1 x x         2 2 2 2 2 2 22 2 2 2 8 x x Thus, 2 2 2 1 or 2 4 (rejected) 2 2 or 2 x        [2] [3]
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2 2 Show that 22 2(2 1) ( 3 1)( n n n n n   can be written in the form 11 3 1 1 n n n n   . Hence, or otherwise, find 3 21 ( 3 N n n n n n n   . Use the above result to show that the sum to infinity 2 2 2 2 2 2 2 2 1 1 1 1 ......... 1 .2 2 .3 3 .4 4 .5 is less than 1. <Solution> 1 1 1 3 1 2 2 n n n n = ) 1 )( 1 3 ( ) 1 3 ( 1 2 2 2 2 n n n n n n n n = ) 1 )( 1 3 ( ) 1 2 ( 2 2 2 n n n n n shown. N n n n n n n 3 2 2 ) 1 )( 1 3 ( 1 2 = N n n n n n 3 2 2 1 1 1 3 1 2 1 1 1 1 2 1 11 5 19 11 29 19 41 ................ ( 3( 1) 1 ( ( 3 1 1 N N N N N N N N   = 1 1 1 1 5 6 2 1 2 2 N N N N [3] [1] [2]
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3 [Turn Over = ) 1 )( 1 ( 1 5 3 2 2 2 N N N N N For n ≥ 3, 1 3 2 2 n n n and 1 ) 1 ( 2 2 n n n Thus, 2 2 ) 1 ( 1 n n ) 1 )( 1 3 ( 1 2 2 n n n n < ) 1 )( 1 3 ( ) 1 2 ( 2 2 n n n n n < 5 3 2 2 2 2 11 ...... 3 4 4 5  < 5 3 2 2 2 2 2 2 1 1 1 ...... 1 2 2 3 3 4  < 36 1 4 1 5 3 < 1 ( Shown)
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4 3 Sketch the locus of arg(1 2 ) 4 z  in an Argand diagram. Find the least value of zc where 1 ci  . Find also the range of values of k , where k such that there is exactly one complex number z , that satisfies both z c k  and arg(1 2 ) 4 z . <Solution> 1 arg(1 2 ) arg( 2)( ) 4 2 4 1 arg( 2) arg( ) 24 13 arg( ) 2 4 4 zz z z    From the Argand diagram, 22 11 1 44 5 d min 1 1 2 (or 1.77) 4 To have exactly one complex number z that satisfy both, z c k and arg(1 2 ) 4 z , from the Argand diagram, 12 d or d kk  . Now   2 2 3 2 2 13 d 1 (or 1.80) 2 i.e. 5 13 2 or 42 [3] [4] Locus of arg(1 2 ) 4 z 3 4 1 2 1 2 (–1,–1) d 1 d 2
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5 [Turn Over 4 (a) A sequence {} n x of positive numbers is defined as 2 1 2 23 n n n x x x for n + . The sequence n x converges to a number as n tends to infinity. Find the value of exactly. <Solution> If } { n x converges to as n tends to infinity , then x n+1 = x n = as n to infinity.
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2007 SRJC Prelim Paper 1 Solutions - 1 2007 JC2 H2 MATH...

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