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

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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 0 - 6 i i i 6 - 1 - 1 0 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 0 . 25 3) log 9 1 16 = log 5 25 log 5 3 5 log 6 3 - log 3 log 0 . 25 log 1 16 log 9 = 2 1 / 3 log 6 3 - log 3 log 9 log 1 16 log 0 . 25 = 6 log 6 3 - log 3 3 log 3 9 log 2 1 16 log 2 0 . 25 = 3 - 1 / 2 2 - 4 - 2 = 3 - 1 2 = 5 2 4. (a) The sum of the first two terms of the arithmetic progression is 23. The
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