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 ) = iftan A (sec A − sin A tan A )
n π
A ≠ . 2
英文原题
Find the range of f A A A A A A A
A A A A
( ) sin ( cos cos sin sin cos )
tan (sec sin tan )
= + + +
−
3 3 2 4 2 2 2
if
A n
≠
π
2 .
解析
英文解析
- 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 ))
if A 6 = .nπ
Answer: (3 , 4) . We factor the numerator and write the denominator in term of fractions to get 2
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 Acos 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