THUSSAT2023年3月诊断性测试数学答案

中学生标准学术能力诊断性测试2023年3月测试数学参考答案一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．12345678ADBDCCAB二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分．9101112ACBCDABDAD三、填空题：本题共4小题，每小题5分，共20分．13．52−14．(−−,1)15．10816．1四、解答题：本题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．17．（10分）23（1）(391=2nnana+++)()nn+1()332(nana++)nn+1()=(nn++21)22()(nana++32)nn+11()即=·····································································2分(nn++21)223()(12+)a11(na+2)1又=，所以数列n是首项为，公比为的等比数列232(11+)(n+1)3n2(na+2)n1(n+1)从而2=，则an=··················································5分(n+1)3(n+23)n2(n+1)nnn++11+1（2）a==···························································6分n(nn++2)32nnn3323n+1+++Sn31323n23n+1123n+1设T=+++，则T=+++n31323n3n33233n+1第1页共7页两式相减得：n−1111−2211nn+1293+1T=+++−=+−n121nn++11n3333331−33n−12111525nn++=+−−=−1nn++11················································9分363623352n+552n+5从而T=−，故S−·····················································10分n443nn443n18．（12分）（1）由已知及正弦定理得：2sincossin2sinBAAC−=又在ABC中，sinsinsincoscossinCABABAB=+=+()·························2分−=+2sincossin2sincos2cossinBAAABAB即2sincos=sinABA−1又sin0A，=−cosB········································································4分222又0B，B=，即角B的大小为··············································5分3313（2）SacBac==sin·····································································6分ABC24BD是ABC的角平分线，而SSSABCABDBCD=+311ac=ABBDsin60+BDBCsin6042233ac即acBDac=+()，BD=·················································8分44ac+BD=2，ac=2(a+c)acac+2，ac4ac，即ac16···············································10分13当且仅当ac==4时取等号，则S=acsinB16=43ABC24即的面积的最小值为43·······························································12分第2页共7页19．（12分）1（1）在BE上取一点N，使得BNNE=，连接FN，NM21BD=6，=B=NBD1，NE=2，ED=36AF1BNAF1=，==FE2NEFE2则FNAB···························································································2分FN面ABC，AB面，FN面BNCM1==，NMBC·······························································4分NDMD5NM面，BC面，NM面FNNMN=，面FNM面FM面FNM，FM面··························································5分（2）AEBD⊥，CEBD⊥2所以二面角ABDC−−的平面角为=AEC············································6分3又AECEE=，⊥BD面AECBD面ABD，面⊥面面ABD面AECAE=，过点C作CHAE⊥，则CH⊥面3则CHCECE==sin322336CE=−=333322，CH=32=·······························8分()2236即到面的距离为2553656MDCD=，M到面的距离为=······························9分662431计算EM：cos==CDB33353在DME中，DM=，DE=32第3页共7页253+−322EM1512==EM···············································11分3532232564534EM与面ABD所成角的正弦值为=·········································12分51342（其他方法酌情给分）20．（12分）（1）列联表如下：感兴趣不感兴趣合计男生12416女生9514合计21930··········································2分3012549−()2K2=0.40822.0721614219所以没有85%的把握认为学生对“数学建模”选修课的兴趣度与性别有关············5分（2）由题意可知X的取值可能为0，1，2，33C55则PX(===0)3···········································································6分C94212CC4510PX(===1)3············································································7分C92121CC455PX(=2)=3=············································································8分C9143C41PX(=3)=3=················································································9分C921故X的分布列为X012351051P42211421第4页共7页510514EX()=+++=0123···············································12分42211421321．（12分）（1）因为双曲线C以2xy=50为渐近线设双曲线方程为(2525xyxy+−=)()，即45xy22−=·························1分yx22−=1F(0,3)，0，即：−−549−−=9，−=9，即=−20·····················································3分5420yx22所以双曲线的方程为：−=1···························································4分45（2）设直线ly:3kx=+，Px(y11,)，Qx(y22,)225420yx−=2+−=53420(kxx)2ykx=+3化简得：(5430250kxkx22−++=)···························································6分30kxx+=−1254k2−此方程的两根为xx,，则1225xx=1254k2−29kk22−−5422900k100()PQ=1+k22−2=101+k(5kk22−−4)54k−(54)244k2+20(k+1)=101+k2=22··············································8分(54k2−)54k−15k−12PQ中点M坐标为−22,·······················································9分5kk−−45412115kPQ中垂线方程为：yx+=−+225454kkk−−−27−27令x=0，y=，T0,254k2−54k−第5页共7页271515k2+则TF=+=3····························································11分5454kk22−−1515k2+TF54k2−3==········································································12分PQ201(k2+)454k2−22．（12分）xxee1−−xx（1）令fx()==2x0···································································1分(ex)ex1，即函数fx()的单调递增区间为(−,1)，函数的单调递减区间为(1,+)··········································2分1111当x=时，112，切点为,f==1222e22ee21又112，fx()在处的切线方程为：f==122ee211111yxyx−=−=+·················································4分2e2e2e4e2ex2xxxln1ln1++（2）+kx(ln1)，=kkexeeexxxln1+xln1x+exx(1,+)，0，k···············································6分eln1x+ln1x+eln1x+x由（1）可知fx()=在上单调递减，下证：xx+ln1ex即证：xx−ln1在x(1,+)恒成立11x−令gxxx()=−ln，则gx()=10−=xxgx()在上单调递增又x1，g(x)g(1)=1−ln1=1································