．20.【详解】?1?sin
B?C??A
?cos2A?sin2?2cos2A?122
A1?cosA?cos2?2cos2A?1??2cos2A?1
2211?
3?2?1?1??1；?299
2(2)由cosA?
1122，可得sinA?1??，39322222由余弦定理可得a?b?c?2bccosA?b?c?即有bc≤
3293
a?，当且仅当b?c?，取得等号．442
1192232．bcsinA????
22434224
bc?2bc?bc?bc，333
则?ABC面积为即有b?c?
332时，?ABC的面积取得最大值．242?f?x??sin2x?3cos2x
21.解：（1）?f?x??2sinxcosx?23cosx?3???
?f?x??2sin?2x??由3??
?A??
?sin????1
?23?
A???A??
f???2，?2sin?2????2
43??4??
?
?2?A????解得A?，所以B?C?2323333cosC4
?sinB?
?2??33?sin??C??cosC
4?3??sin
2?2?33cosC?cossinC?cosC334?3133cosC?sinC?cosC22413?sinC?cosC24?tanC?
32
??
（2）由（1）知f?x??2sin?2x?
?????
x?m,?，??3??2?
?2x?
???4??
??2m?,3?33??
??f?x?????3,2???????2m??
332??解得??m?
312?即m的最小值为?
3
31
a1?，得a1?1，22
3
当n?2时，Sn?Sn?1?an??an?an?1?得an?3an?1，2
22.（1）当n?1时a1?所以an?3n-12nn
?n，an?2?an?13
（2）由（1）得：bn?
12n?2?......?n①333112n
得Tn?2?3?......?n?1②33332111n
两式相减得：Tn??2?......?n?n?1，33333
又Tn?
1?1?1???2n3?3n?
故Tn??n?1，1331?3所以Tn?
33?2n3?,?T?.n44?3n4
河北省鸡泽一中2019-2020学年高一下学期开学考试数学试题 PDF版含答案
