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2019—2020学年度下学期沈阳市郊联体期中考试高一试题数学答案一.选择题1.A2.B3.B4.D5.B6.C7.A8.A9.D10C11.B12.C13.6−14.115.1216.1718,3tan()sin(2)cos()217()9cos()tan(3)2tan(
sin)(sin)sin6(sin)(tan)sin()cos2xxxfxxxxxxxxx−−−=+−+−−==
−−−=−=、(1)(注:结果正确得满分,结果错误,诱导公式每正确一个得1分)(2)55cos7131312sin91312()13f=−=−
=−,得为第三象限角∴102221
211222221211221212118(32)912474(2)(2)4476(32)(2)6aeeeeeebeeeeeeabeeeee=−+=−++==+=++==−+
+=−、(1)221227282712cos,10277,0,180,=120eeeababababab++=−−===−
则∴1219A=1T=T==2322()2sin(2)2(,2)3fxxP
=+−、(1)由题意可知2,,,代入43+=+2,2,0532626()2sin(2)66272122366kkZkkZfxxxx
=+==++可得由,得∵,∴∴(2)∵∴maxmin82()2
1062672()112662xxfxxxfx+===+===−
当即时,当即时,(注:没写出1x相应的的值,每个扣分)221sin1cos20cossin1sin1cos(1sin)(1cos)cossin(1sin)(1sin)(1cos)(1cos)1sin1cosc
ossin3cossin2−+++−−+=++−−+−+=+、(1)由原式1=sin11cossincos5
51241+2sincos2sincos625257sincos12sincos25aa−++=+=
==−−=−=平方得,即∵,∴2273sin2sincoscos122221tantan2cossin
2sincos(cossin)10sincoscossin247168255125−−+++−==−+=−−=
(2)()()1221()3cos3sin23sin()2323BC=4T2=4T=8==4284()23sin()43fxxxxfxx=+=+
=+、(1)由已知可得
因为正三角形的高为,则即,,0522,243210288,33102()88733()=23kxkkZkxkkZyfxkkkZfx
−+++−++=−++(2)令即的单调增区间为,(3)由(1)有0000083sin()=4354sin()84351
02(,)(,)3343223cos()435xxxxx++=−+−+=
即由,得则0009(1)23sin()44323sin()104342437623()2555fxxx+=+
+=++=+=
1223122()2cossin()2cos(sincos)62231113cossincossin2cos2sin
(2)222262131()sin()4322fxabxxxxxxxxxxxf==−=−=−=−−=−−−=−=
、42,+61221()+712222()12sin()62()0212sin(),0,226sin(),0,6kxkkZxkZkfxkZgxmxygxmxxyxx−==−
=−−−+=−+=−=−(2)令,即,∴的对称中心为(,)(3)在,上恰有两个零点即有两个不等实根即与119225,0,sin,,666111122221+1221+1122mytxxyttmmm−=+
=−=−−+有两个不同交点令则由正弦函数的图象可知即实数的取值范围是,