Maths Revision Book - Mathematics C Semester 2 Revision...

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Mathematics C Semester 2 Revision Book SOLUTIONS 2008
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UNSW Foundation Year UNSW Global Pty Limited UNSW Sydney NSW 2052 Copyright © 2008 UNSW Global Pty Limited, trading as UNSW Foundation Year All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner.
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1 MATHEMATICS C SOLUTIONS TO REVISION EXERCISES #1 9 and 16 3 9 16 3 3 9 16 6 3 4 3 5 10 ) 3 2 )( 3 2 5 ( . 1 = = = = + = y x y x ********************************************************************************* 3 1 3 10 3 or 3 10 15 26 12 36 ) 2 )( 13 ( ) 6 )( 6 ( 6 13 2 + 6 G.P. a form 13 , 6 , + 2 Then . be number Let the . 2 2 2 = = + + = + + + + = + + + + = + + + x x x x x x x x x x x x x x x x x x ********************************************************************************* 3. f(x) x 2 12 (a) f(x) x (b) (c) y 33 x ********************************************************************************* units. 5 16 9 ) 2 6 ( ) 1 2 ( ) ( ) ( ) a ( . 4 2 2 2 1 2 2 1 2 = + = + + = + = y y x x d AB 0 10 3 4 4 4 6 3 3 4 ) 1 ( 3 4 2 1 2 2 6 ) ( : AB line of Equation ) b ( 1 1 1 2 1 2 = + + = = + = + = = = y x x y x y x x m y y x x y y m AB 3 4 = AB to parallel line of Gradient ) c ( = AB m 0 13 3 4 16 4 3 3 ) 4 ( 1 ) ( is C(4,1) through line of Equation 3 4 1 1 = = = = y x x y x y x x m y y *********************************************************************************
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2 5. ) 3 ( ) 3 )( 3 ( lim 3 9 lim ) a ( 3 2 3 + = x x x x x x x 6 3 3 ) 3 ( lim 3 = + = + = x x 12 4 ) 2 ( 2 ) 2 ( ) 4 2 ( lim ) 2 ( ) 4 2 )( 2 ( lim 2 8 lim ) b ( 2 2 2 2 2 2 2 = + = + = + + + = + + x x x x x x x x x x x ********************************************************************************* 6. nU () = 99 nA = =− = 33 99 33 66 ********************************************************************************* 2 1 3 4 5 2 3 4 4 2 4 4 4 2 3 3 1 3 3 1 3 3 1 3 ) c ( 2 3 4 1 2 7 1 ) b ( ) a ( . 7 2 5 2 7 = + = + = + + = = = = = = = x dx dy x x x x y x x x dx dy x x x x x x x x x dx dy x x x y x x x y *********************************************************************************
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3 ). 2 , 3 ( is point The 2 1 ) 3 ( 2 ) 3 ( find to curve the of equation the into 3 substitute parabola, on the lies point the Since 3 6 2 4 2 2 4 4 is tangent the of gradient the line, the to parallel is tangent the As 4 gradient 2 2 0 7 4 : Line 1 2 : Curve (c) 4 gradients, equal have lines Parallel 4 1 4 gradient 0 7 4 : Line (b) 2 2 (a) . 8 2 2 = + = = = = = + = = + = = + + = = = = = + + = y y y x x x x dx dy m x dx dy y x x x y m m y x x dx dy l ********************************************************************************* 2 2 3 1 2 3 1 3 3 ) 3 ( ) 3 ( 9 27 . 9 + + + = = = x x x x x x Since the bases are equal, equate indices: 2 2 2 3 = + = x x x ********************************************************************************* . 4 3 reject 3 or 4 0 ) 3 )( 4 ( 0 12 1 3 4 1 3 log 4 log ) 1 ( log 3 log 4 log log . 10 2 = = = = = + = = = = x x x x x x x x x x x x x x b b b b b b *********************************************************************************
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4 SOLUTIONS TO REVISION EXERCISES #2. 2 5 3 3 or 2 45 3 2 36 9 3 0 9 3 27 9 3 0 9 4 18 9 ) 3 2 )( 3 2 ( ) 3 )( 6 ( . 1 2 2 2 2 ± ± = + ± = = = = + + + = + + x x x x x x x x x x x x ********************************************************************************* below graph see 1 3 0 ) 1 )( 3 ( 0 3 2 3 2 ) c ( 6 4 4 6 6 2 4 (b) 4 20 5 3 24 4 2 3 24 ) 2 ( 2 6] [ 2 4 3 2 ) a ( . 2 2 2 + + + < < < < < < + + × + x x x x x x x x x x x x x x x x x x x f(x) x -3 1 ********************************************************************************* n n n n a n S d n a T n d a n n 32 4 128 160. 4 4 4 68 ) 192 68 ( 2 32 4 ) 1 ( 68 192 ) ( 2 ) 1 ( : find To . 192 , 4 , 68 with A.P. an is This 192. + . . . + 72 + 68 find To . 3 = = = + = + = + = + = + = = = = = A A A *********************************************************************************
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5 () . 4 and 0 then , 2 4 2 If . 2 4 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 1 2 . 4 2 2 = = = + = + + + = + + + = + b a b a ********************************************************************************* 58 1 0 8 0 81 0 8 45 4 9 22 .
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This note was uploaded on 04/07/2012 for the course COMMERCE 101 taught by Professor Louis during the Three '07 term at University of New South Wales.

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Maths Revision Book - Mathematics C Semester 2 Revision...

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