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答案第1页,共9页2023-2024学年度上学期辽西联合校高三期中考试(数学参考答案,提示及评分细则)参考答案:1.A【详解】由5,3BCU,而{1,3}A,所以5,3,1ABCU.故选:A2.C【详解】命题“3,51xxxN”的否定是“3,51x
xxN”.故选:C3.C【详解】由E为AC边上的点,且3AEEC,得111111424224EDECCDACCBACCAABABAC
.故选:C4.C【详解】由21xm,即121xm,解得2112mxm,因为“12x”是“21xm”充分不必要条件,所以1,2真包含于21,12mm,所以122211mm(等号不能
同时取得),解得112m,所以实数m的取值范围为1,12.故选:C5.A【详解】∵33log5log31a,0.50.52log10logb,0.200551c,∴acb.故选:A.6.A{#
{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第2页,共9页【详解】解:因为关于x的不等式200axbxca的解集为3,1,所以0a,且31ba,31ca
,所以2ba,3ca,所以20cxbxa化为23210xx,解得113x.故选:A.7.B【详解】因为函数fx为奇函数,且在区间0,上是增函数,且102f,所以函数在(,0)上单调递增且11022ff
,所以当12x或102x时,()0fx,当102x或12x时,()0fx,由()()0fxfxx,即2()0fxx,故0fxx由0fxx可得0()0xfx或0()0xfx,所以102x或102x
,故选:B8.D【详解】设()1()exfxgx,()()1fxfx,即()()10fxfx,()()1()0exfxfxgx,()gx在R上单调递减,又(0)202
3f,不等式0()1(0)1e()e20222022(0)1eexxxfxffxf,即()(0)gxg,0x,原不等式的解集为(,0).故选:D9.ABD【详解】设izab,则izab,由26izz,可
得23i6izazb,所以36,1,ab解得2,1ab,因此2iz,A正确;{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第3页,共9页2i12i2ii
12i12i5z,为纯虚数,B正确;5z,C错误;2iz,其在复平面内对应的点为()2,1-,在第四象限,D正确.故选:ABD.10.ABD【详解】对A,与向量a共线且方向相同单位向量为222,21255,515aarr,故A正确.对B,因为
2,1ar,3,1b,故1,2ab,故1220abarrr,故abarrr成立,故B正确.对C,向量a在向量b上的投影数量是22231110231abb,故C错误.对D,22,
123,14,3abrr,故222435ab,故D正确.故选:ABD.11.BC【详解】因为0,0,2xyxy,所以22xyxy,故1xy,当且仅当xy时,取得等号,所以xy的最大值为1,故A正确;当13,22xy时,2219524
42xy,故B错误;因为2222224xyxyxyxy,所以2xy,当且仅当xy时,取得等号,即xy有最大值为2,故C错误;因为2221112xyxyxyxyxy,当且仅当xy时,取得等号,所以2
11xy有最大值为1,故D正确;故选:BC.12.AD{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第4页,共9页【详解】由函数的图象可得2A,由12πππ431
2,求得2.再根据五点法作图可得π22ππ3k,即π2π,Z3kk,又π2,求得π3,∴函数π2sin23fxx,5ππ2sin2122f,是最值,故A成立;2π5ππ2sin2sin
3333f,不等于零,故B不成立;将函数π3sin2cos22sin26yxxx的图象向左平移π2个单位得到函数ππ5πsin2sin2266yxx
的图象,故C不成立;当π,02x时,22,333πππx,π2ππ2sin2sin3233f,522122ππsinf,函数fx在π,02
上的图象如下,由图可知,2,3m时,函数fx与直线ym有两个交点,故方程fxm在π,02上有两个不相等的实数根时,m的取值范围是2,3,故D成立.故选:AD.13.12/0
.5【详解】因为12f,所以11(2)2fff,故答案为:12.14.1,11,3{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第5页,共9页【详解】由223010x
xx,解得131xx,所以函数的定义域为1,11,3故答案为:1,11,315.9316【详解】由题意可知,1137137137113713132221313222aaaSaTbbbb,所以7131377132931
3316aSbT.故答案为:9316.16.[8,0]【详解】命题“0Rx,20020axax”是假命题,命题的否定:“Rx,220axax”是真命题,即220axax恒成立,当0a时
,显然成立;当0a时,则2080aaa,解得:80a综上,实数a的取值范围是[8,0],故答案为:[8,0].17.(1)27nan;(2)26nSnn,nS的最小值为9.【详解】(1)设等差数列n
a的公差为d,由15a,3530aa,得3(52)(54)0dd,........................................................................................
.................................2分解得2d,.............................................................................................
........................................................3分于是1(1)27naandn,.所以数列na的通项公式是27nan..............
.................................5分(2)由(1)知,215(27)622nnaannnnnS,................................
...................................7分{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第6页,共9页显然2(3)99nnS
,当且仅当3n时取等号,.................................................................................8分所以2
6nSnn,nS的最小值为9...............................................................................................................10
分18.(1)25ABxx,R{|2ABxxð或2}x(2)2m或12m【详解】(1)∵25Axx,121Bxmxm,∴当3m时,则27Bxx,所以25ABxx,...........
................................................2分R{|2Axxð或5}x,..........................................
..................................................................................3分又27Bxx,所以R{|2ABxxð或2}x......................
......................................................................................5分(2)∵ABA,∴BA,∴当B
时,则有121mm,即2m,满足题意;................................................................7分当B时,则有121mm,即2m,.............................
.........................................................8分可得12215mm,解得:12m......................................
...................................................................10分综上所述,m的范围为2m或12m.........
..................................................................................12分19.(1)减区间为(0,2),增区间为(2,),极小值为2ln2,无极大值(2)1a【详解】(1)函数fx的定
义域为0,,.......................................................................................1分当1a时,212ln
2fxxxx求导得21fxxx,整理得:21xxfxx.....................................................................2分由()0fx¢>得2x
;由0fx得02x从而,函数fx减区间为(0,2),增区间为(2,)...............................................................................
4分所以函数fx极小值为22ln2f,无极大值.................................................................................6分(2)由已知1,x时,0fx恒成立,即
20xax恒成立,即2axx恒成立,则min2axx.............................................................................
..........................8分令函数21gxxxx,由2210gxx知gx在1,单调递增,.................................9
分{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第7页,共9页从而min11agxg....................
...................................................................................................10分经检验知,当1a时,函数fx不是常函数,所以a的取值范围是1a
...................................12分20.(1)324ac(2)12【详解】(1)若选①2222abc,由余弦定理可得2222cosbacacB,∴1cosacB
,...................................................................................................................................2分又1sin3
B,∴2cos1si23n2BB,......................................................................................4分∴324ac..................
......................................................................................................
.................6分若选②1ABBC,则coscos1ABBCBacB,.......................................................................
.......................2分又1sin3B,∴2cos1si23n2BB,............................................................................
..........4分∴324ac............................................................................................................
..............................6分(2)由正弦定理2sinsinsinabcRABC(R为ABC外接圆半径),可得22sin2sin4sinsinacRARCRA
C,又∵2sinsin3AC,324ac.........................................................................................................
8分∴2322443R,解得34R.................................................................................
.......................10分∴3112sin2432bRB.......................................................................................................
........12分21.(1)12;(2)43310.【详解】(1)2()2cos123cossincos23sin22sin26fxxxxxxx,...
........4分由于直线3x是函数()2sin26fxx的一条对称轴,2sin136,2()362kkZ,...........
......................................................................5分31()22kkZ,又01,1133k,kZ,从而0
k,12................................6分(2)()2sin6fxx,由题意可得12()2sin236gxx,{#{QQAB
IYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}答案第8页,共9页则1()2cos2gxx,∵622cos365g,3cos65
,..........................................7分又0,2,2663,4sin65,..................................
................................8分sinsinsincoscossin666666...........................
.....................................10分4331433525210...........................................................
......................................................................12分22.(1)1y(2)见解析(3)362e,0e
【详解】(1)解:若1a,则exfxx,e1xfx,由01,00ff,曲线yfx在点(0,)0f处的切线方程为1y;.............
.....................................................................2分(2)解:函数fx的定义域为,,e1xfxa,当0a时
,0fx,所以fx在,上单调递减;....................................................................4分当
>0a时,当1lnxa时,0fx,当1lnxa时,()0fx¢>所以fx在1(,ln)a上单调递减,fx在1(ln,)a上单调递增;......................................................
6分综上,当0a时,fx在,上单调递减;当>0a时,fx在1(,ln)a上单调递减,在1(ln,)a上单调递增;......................................................7分(3)
解:22e3e3xxgxaxxxax,函数gx有两个零点,即方程2e30xax有两个不相等的实数根,也即方程23exxa有两个不相等的实数根,.................................
...............................................................8分即直线ya与函数23exxy的图像有两个交点,令23exxhx,则21323eexxxxxxhx
,当3x或1x时,0hx,当13x时,0hx,所以函数hx在,1和3,上递减,在1,3上递增,{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAA
BAANABAA=}#}答案第9页,共9页所以12ehxh极小值,363ehxh极大值,当>3x时,23>0exxhx,且2362e>eh.................................
...........................................................10分所以,函数hx的图像大致如图,则a的取值范围是362e,0e..................................
.................................................................12分{#{QQABIYaQggggQAIAAAhCUwFiCgAQkBCCAKoORBAAsAABAANABAA=}#}获得更多资源请扫码加入享学资源网微信公众号ww
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