M17 LE3 Sem1 09-10 Solns - HIGH SCHOOL MATH PROBLEMS AND...

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Unformatted text preview: HIGH SCHOOL MATH PROBLEMS AND SOLUTIONS I. 1. We have 1 2 1 > 1 2 2 , but 1 is not greater than 2. 2. TRUE 3. TRUE 4. f is one-to-one, so f- 1 exists. However, ( f- 1 ◦ f )(1) 6 = 1. 5. log 20 , log 40 , log 80 is an arithmetic progression with a common difference of log 2. II. 1. f ( x ) = 6 x 4- x 3 + 5 x 2- x- 1. (a) ± 1 , ± 1 2 , ± 1 3 , ± 1 6 (b) It is given that i is a zero of f ; hence, its conjugate- i is also a zero of f . By synthetic division, we obtain the following: i | 6- 1 5- 1- 1 6 i- i- 6 1- i 1- i | 6- 1 + 6 i- i- 1- i- 6 i i i 6- 1- 1 ⇒ f ( x ) = ( x- i )( x + i )(6 x 2- x- 1) = ( x- i )( x + i )(3 x + 1)(2 x- 1) The zeroes of f are i,- i,- 1 3 , 1 2 . 2. h ( x ) = x 3 + 1 x 3- 1 x = y 3 + 1 y 3- 1 x ( y 3- 1) = y 3 + 1 xy 3- x = y 3 + 1 xy 3- y 3 = x + 1 y 3 ( x- 1) = x + 1 y 3 = x + 1 x- 1 y = 3 r x + 1 x- 1 = h- 1 ( x ) dom h- 1 = ran h = R \{ 1 } 3. (log 3 √ 5 25) log 6 3- (log . 25 √ 3) log 9 1 16 = log 5 25 log 5 3 √ 5 log 6 3- log √ 3 log 0 . 25...
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M17 LE3 Sem1 09-10 Solns - HIGH SCHOOL MATH PROBLEMS AND...

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