2007 SRJC Prelim Paper 1 Solutions

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

This preview shows pages 1–6. Sign up to view the full content.

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]

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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]
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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 22

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online