【文档说明】江苏省扬州市高邮市2022-2023学年高一上学期10月月考试题 数学(答案).docx,共(5)页,185.215 KB,由envi的店铺上传
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高一10月阶段测试答案1.B2.D3.D4.A5.B6.B7.D8.C9.BC10.AD11.ABD12.BCD13.3或-214.+815.03−m,03−mm(0,3−均可16.925,4917.解(1)因为122+
−=mxmRxB当1=m时31−=xxB,又因为43−=xxA,所以43−=xxBA.............................................................2分因为31−=xxxBC
R或..............................................................................................................3分所以4313)
(−−=xxxBCAR或..................................................................................5分(2)=BA时,当B=时,122
+−mm,解得3−m,..................................................................................6分当B时,−++
−312122mmm或−+−42122mmm,解得623−−mm或.........................8分综上,实数m的取值范围是62−mmm或.....................................
.......................................10分18.(1)41=+−xx,142)(2122=−+=+−−xxxx................................2分62)(12
2121=++=+−−xxxx又6021212121=++−−xxxx且.......................4分367614212122==++−−xxxx.............................................
.............6分(2)3log9log9424===bb,.............................................8分bab+=+===35log3log3log3)53(log3log15log27
log27log2222322215...........................12分19.解:(1)因为p为真命题,所以方程0622=+−axx有解,即04362−=a..................2分所以33
−a即33−=aaA................................................................................4分(2)因为Ax是Bx的必要不充分条件,所以,AB且AB
...............................................6分ⅰ)当B=时,123−−mm,解得:21m.................................................................
.........8分ⅱ)当B时,−−−−−两等号不同时取31323123mmmm解得:2131−m...................................................
11分综上:31−m..................................................................................................................................1
2分20.解:(1))3817()428(xttxtty++−++=txt81242−−+=...................................................2分因为)3(+=xt,所以381453
812423+−−=+−−++=xxxxxy....................................................5分(2)因为38145+−−=xxy48]381)3[(++++−=xx..
.....................................................................7分又因为(20,0x,所以0381,03++xx,所以18381)3(2381)3(=++++++xxxx(当且仅当63813=+=+xx
x即时取“=”)........9分所以304818=+−y.................................................................................................
.............................11分即当6=x万元时,y取最大值30万元。...............................................................................................12分
21.解:(1)因为不等式01222−+bxax的解集为)1,3(−−,所以-3和-1是方程01222=−+bxax的两个实数根,所以−=−−−=−+−aaba12)1)(3(2)1()3(0,解得a=﹣4,b=﹣8.......................
.................4分(2)3−=ab时,不等式为),(012)3(22Rbaxaax−−+,即0)6)(2(−+axx,.......5分当0=a时,解不等式得2−x;....................
..................................6分当0a时,不等式化为0)6)(2(−+axx,解不等式得2−x或ax6;..................7分当0a时,不等式化为0)6)(2(−+axx,若3−=a,不等式化为0)2(2+x,不等式无解;
......................................8分若03−a,解不等式得26−xa;...............................................9分若3−a,解不等式得ax62−;..................
................................10分综上知,3−a时,不等式的解集为(﹣2,a6);3−=a时,不等式的解集为;03−a时,不等式的解集为(a6,﹣2);0=a时,不等式的解集为(﹣∞,﹣2);0a时,不等式的解集为(﹣∞,﹣
2)∪(a6,+∞)...................................12分22.解:(1)11==yx时,,abba−−==++1,12即恒成立即恒成立02)1(,0),5,2(2++−xaa
xyx,2)1(−−xxax恒成立)5,2(x,)1(2−−xxxa恒成立,最大值))1(2(−−xxxa.................................2分,2−=xt令则)3,0(t,则223322132123)
1)(2()1(22−=+++=++=++=−−ttttttttxxx当且仅当22==ttt即,此时22+=x时取”“=,所以实数a的取值范围时),223(+−...4分(2)11==yx时,,abba−−==++1,12即0,1,2−−ya恒
成立,即恒成立02)1(2++−xaax,02)(2+−−xaxx恒成立...6分+−++−0202222xxx41714171+−x,所以实数x的取值范围是)4171,4171(+−...........8分(3)对00,ybRx时,
恒成立,08,02−=aba,则82ba...................................10分1282282822=+=++bbbbbbba,当且仅当bb28=且82ba=,即2,4==ab时取等号所以ba2+最小值是1..
.........................................................................................................................12分获得更
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