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怀仁市2020—2021学年度下学期一模高三教学质量调研测试理科数学答案一.选择题:123456789101112ACBBCADACCBD二.填空题:13.1514.215.313216.1,23三.解答题:17.(12分)1解:选择①:由,cos
213cosBB得BBBcos21sin23cos21即21cos21sin23BB所以216sinB因为,,0B所以,6566B故66
B所以3B.....................................6分选择②:由正弦定理BbACcAasinsinsinsin,可化为acbca222由余弦定理得212cos222acbcaB因为,0B所以3
B,....................6分选择③:由正弦定理得ABCAbccossinsin3cos3,又BACBABABBAABAcoscossincoscossincossincossintantan
,由BAAbctantancos3得ABCBACcossinsin3coscossin因为0sinC所以3tanB因为,0B所以3B。....................................
............................6分2在ABC中,由1及42332sinsinsin,32CcAaBbb,CcAasin4,sin4,....................................................
........................................................8分所以AAAACAcacos34sin832sin8sin4sin8sin42Asin74.....
...............................................................................................10分因为320A且为锐角,所以存在角A使得,2A所以ca2的最大
值为74........................................................................................................................
....................12分18.(12分)详解:(1)列联表如下:喜欢不喜欢合计男生51520女生101020合计152540072.2667.220202515)1015105(402k......
..............................................................4分所以,有85%的把握认为该校学生是否喜欢健身操与性别有关。...............................5分(2)记事件A为“从样本中随
机抽取男生,女生各1人,其中恰有1人喜欢健身操”则21)(12012011011511015CCCCCCAP....................................................................
.................7分(3)由题意知的可能取值为0,1,2832143)0(xP2121412143)1(xP812141)2(xP所以X的发布列为X012P832
18143812211830)(XE...............................................................................
...12分19.(本大题12分详解(1)证明:因为,2,,//ADABDCADCDAB所以22,22BCBD又因为,4CD所以222BCBDCD,所以BDBC,取,OBD的中点为连接,,OSOA因为6SDSBSA所以SOASOBBDSO,所以,OASO
因为,OOBOA所以ABCDSO平面所以,SOBC又因为,OBDSO所以.SBDBC平面.................................................
...............5分(2)如图,以A为原点,分别以ABAD,和垂直平面ABCD的方向为zyx,,轴的正方向,建立空间直角坐标系xyzA,则2,1,1,0,0,2,0,4,2,0,2,0,0,0,0SDCBA设10
CSCM,则2,34,2M,由(1)得平面SBD的一个法向量为0,2,2BC设zyxn,,为平面ABM的一个法向量,2,34,2020AMAB,,,由00AMnABn,得
0234202zyxy不妨设2,0,2n......................................................................................................8分设平面SBD与
平面ABM所成的二面角为所以2958445224224cos222整理得(舍去)或,解得21310162.................................................10分所以3110949925,32,3
,35AMAM.................................................12分20.(本大题12分)详解(1)根据题意知121caac,解的1,2ca由此可得,32b故椭圆的标准方程为13422yx
............................................................................................4分(2)由(1)知,)01(2,F,直线BA的斜率不可能为0,因此设直线
BA的方程为)0(mtmyx,与椭圆C联立,得关于y的一元二次方程0123643222tmtyym设,,),,(2211yxByxA则),(11yxA根据韦达定理有436221mmtyy,431232221mty
y①...........................................7分而AB所在的直线经过点)01(2,F,因此0111221121122yyxyxyxyxy等价于0122121yytymy将①式代入,得
,06112322mtttm化简得,4t因此直线BA恒过定点04,。.....................................................................................
.......12分21,(本大题12分)详解:(1)因为xaexfxln)(,所以),0(),1(ln)(xxxaexfx。令,1ln)(xxxk,则,1-x)(2xxk当,0)()1,0(xkx时,函数)(
xk单调递减;当,0)(),1(xkx时,函数)(xk单调递增。所以,01)1()(kxk又因为0,0xea,所以)(,0)(xfxf在定义域,0上单调递增.......................5分(2)由0)()(g0)(xfxxh得
即,即;xaeaxxxlnln2即:xaeaxxxlnln)(,xaeaexxxlnln所以,)ln(lnxxaeaexx即,)ln(lnxxaeaexx对任意)1,0(x恒成立,设xxxHln)(,则2ln1xxxH)(所以,当时
,)1,0(x0)(xH,函数)(xH单调递增,且当),1(x时,0)(xH,时,)1,0(x0)(xH。若xaex1,则)(0)(xHaeHx,若,10xae因为)()(xHaeHx,且)(xH在10,上单调
递增,所以xaex,综上可知,xaex对任意)1,0(x恒成立,即xexa对任意)1,0(x恒成立。................................................
.............................................................................................10分设)1,0(,)(xexxGx,则)1,0(,01)(
xexxGx所以)(xG在),(10单调递增,所以aeGxG1)1()(,即a的取值范围为,1e。............................................
............................12分22.(本大题10分)解:(1)把参数方程14341tytx(t为参数)消去参数t得310xy由2C的极坐标方程为4sin,两边同乘以,得24sin,
将cossinxy且222yx代入,得曲线2C的直角坐标方程为2240xyy;.......................................................................
..5分(2)直线1C的标准参数方程1,2312xtyt(t为参数)把直线1C的参数方程1,2312xtyt(t为参数)代入曲线2C的普通方程2240xyy中,整理得2330tt,012
3tt,123tt<0,利用参数的几何意义知:1212121212121212211115413ttttttttPAPBtttttttt....................................................
..............................................................................................10分23.(本大题10分)详解:(1)由题可知
3,4332,82,43)(xxxxxxxf当2x时,1043)(xxf当32x时,1058)(,xxf当3x时,543)(xxf所以函数)(xfy的值域为,5,若不等式mxf)(恒成立,则5m............
.....................................................................5分(2)由(1)知5cba证明:abba222bccb222aca2c22acbc
abcba222)(2222acbcabcbacba2223222222)(即2532222)()(cbacba325222cba当且仅当”号时取“cba......
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