HMMT 二月 1998 · ADV 赛 · 第 4 题
HMMT February 1998 — ADV Round — Problem 4
题目详情
- Find the range of f ( A ) = if tan A (sec A − sin A tan A ) n π A ≠ . 2
解析
- Find the range of ( ) 2 2 2 4 2 (sin A ) 3 cos A + cos A + 3 sin A + (sin A )(cos A ) f ( A ) = (tan A ) (sec A − (sin A )(tan A )) nπ if A 6 = . 2 Answer: (3 , 4) . We factor the numerator and write the denominator in term of fractions to get 2 2 2 2 2 2 (sin A )(3 + cos A )(sin A + cos A ) (sin A )(3 + cos A )(sin A + cos A ) ( ) = . ( ) 2 2 (sin A )(1 − sin A ) sin A 1 sin A − 2 cos A cos A cos A cos A 2 2 2 2 2 Because sin A + cos A = 1, 1 − sin A = cos A, so the expression is simply equal to 3 + cos A . The 2 nπ 2 range of cos A is (0 , 1) (0 and 1 are not included because A 6 = , so the range of 3 + cos A is (3 , 4). 2