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数学参考答案第1页（共5页）数学参考答案及评分标2023年高三年级第三次适应性检测准一、单项选择题：本题共8小题，每小题5分，共40分。1--8：CBADBCAD二、多项选择题：本题共4小题，每小题5分，共20分。9．BD10．

AC11．BCD12．ABD三、填空题：本题共4个小题，每小题5分，共20分。13．22143yx；14．3π4；15．28；16．1．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）解：(1)因为CcaB

ctan)2(sin2，所以CCcaBccossin)2(sin2··················1分所以CCACBCsin)sinsin2(cossinsin2，因为)π,0(C，0sinC，所以CACBsinsin2cos

sin2···························2分因为πCBA，所以CCBCBCCBCBsinsincos2cossin2sin)sin(2cossin2，所以CCBsinsi

ncos2，···········································································4分所以21cosB，又因为)π,0(B，所以3πB··

··············································5分（2）设xa，则xc3，由条件知)(21BCBABD·····································6分所以13413)cos2(41)(41||22222

xBaccaBCBABD························7分所以2x···································································································8分

所以22227cos2xBaccab，72b··················································9分所以周长728724xcbal·······························

·····················10分18．（12分）解：（1）取11CB的中点1O，连结11OA，因为ABAC，ABCCBA111为三棱台，4211BAAB，所以2π,21111111CABCABA，所以1111CBOA，211OA·············

········2分因为平面11BCCB平面ABC，平面//ABC平面111CBA，所以平面11BCCB平面111CBA，平面11BBCC平面111CBA11CB，所以11OA平面11BCCB······

·········································································3分由条件知梯形11BBCC的面积231)2422(21S

·································4分数学参考答案第2页（共5页）所以四棱锥111BBCCA的体积222331V·····························

············5分（2）取BC的中点O，连结AO，因为ACAB，所以AOBC，因为等腰梯形11BCCB中，OO,1分别为上下底面BCCB,11的中点所以1OOBC，因为平面11BCCB平面ABC，平面11BBCC平面ABCBC，所以1OO平面

ABC，以O为原点，分别以1,,OAOBOO所在直线为x轴，y轴，z轴建系如图，············6分则(22,0,0),(0,22,0),(0,22,0),ABC)1,2,0(1B，令1BBmBE)10(m，则),222

,0(mmE···········································7分设平面ACE的法向量为(,,)nxyz，因为(22,22,0)CA，),224,0(mmCE，则22220n

CAxy，0)224(mzymCEn，令mx，可得)224,,(mmmn··················································

········8分因为1OO平面ABC，所以)1,0,0(1OO为平面ABC的一个法向量···················9分记二面角EACB的平面角为，1027)224(2224cos

22mmm·······10分即0262mm，解得21m或32m（舍）···········································11分所以存在点E为BB1的中点，使得二面角EACB的余弦值为1027··········

······12分19．（12分）解：（1）因为nnnnaS3)1(1，所以nnnaS9122，nnnaS931212·········4分两式相减得，2212293nnnnaaa，所以2122293nnnnbaa，·············6分（2）

因为123,,aaa成等差数列，所以3122aaa，又因为321aa，9321aaa，所以0,3,6321aaa···························8分由nnnnaS3)1(1

可得，12113)1(nnnnaS，(1)n两式相减得，1121(1)(1)23nnnnnnaaa···········································9分当

22nk，2k且Nk时，22223kka，此时2122212221232333kkkkkkSa，2k································10分所以，2116,19,2nnnSn··············

··························································12分1OABC1A1B1COExyz数学参考答案第3页（共5页）20．（12分）解：（1）由题知：||4||PAPB,即32

4||||PAPB·······························2分所以曲线C是以)0,3(),0,3(BA为焦点，长轴长42a的椭圆························3分所以曲线C的方程为1422yx··········

···························································4分（2）设),(00yxQ，因为点),(00yxQ在椭圆C上，所以142020yx······················5分因为直

线txl:与圆Q相切于点M，所以||||0txQM··································6分2020)3(||yxQB····································

···········································7分假设存在，则||||)3(||02020txmQMmyxQB，即2022020)()3(txmyx，即2

202202200202332tmtxmxmyxx，因为142020yx，所以412020xy，所以04)322()43(2202202tmxtmxm恒成立·················

····················9分所以0432m，03222tm，0422tm············································10分解得334t，23m，所以存在23m及定直线334:xl

，满足条件············································12分21．（12分）解：（1）由题知：X的可能取值为1,0,1；················································

······1分321(1)(1)(1)4312PX；32325(0)(1)(1)434312PX；321(1)432PX··································································

················4分所以X的分布列为：X101P11251212所以1515()1011212212EX······································

··················6分（2）假设答题n轮，记nP表示“没有出现连续三轮每轮得1分”的概率，数学参考答案第4页（共5页）由题知：341234171131,1,1(),13()28216PPPP···········

····················7分经分析知：nP第n轮第1n轮第2n轮112nP没有得1分214nP得1分没有得1分318nP得1分得1分没有得1分所以123111(4)248nnnnPPPPn··································

·························9分解得111,,248abc··············································································10分

因为123111248nnnnPPPP，112111248nnnnPPPP，所以112111248nnnnnPPPPP12312111111()224848nnnnnPPPPP

31016nP所以1(4)nnPPn，且1234PPPP，可得12345PPPPP所以答题轮数越多（轮数不少于3），出现“连续三轮每轮得1分”的概率越大·······12分22．（12分）解:（1

）由题得：sin()exxcfx，cossin()exxxcfx·······························2分因为(0)10fc，得1c····················

···················································3分（2）因为sin1()1xbxfxa，所以sin10xabx······························

·······4分令()sin1xFxabx，()lncosxFxaabx，①若01a，则π(π)10Fa，不合题意··················································5分②若1a，1°当0b时，则()1sin0xFx

abx，适合········································6分2°当0lnba时，则()lncoslnln0xxFxaabxaabab，数学参考答案第5页（共5页）所以，()F

x在[0,π]单调递增，()(0)0FxF，适合····························7分3°当lnba，令()()gxFx，则2()(ln)sin0xgxaabx，所以()Fx在[0,π]上单

调递增，π(0)ln0,(π)ln0FabFaab，所以存在0(0,π)x使得0()0Fx，且()Fx在0(0,)x上单调递减，所以，当0(0,)xx时，()(0)0FxF；不合题意综上，lnba且1a·························

························································9分所以eelnabaa，令()eln(1)haaaa···············

····························10分则ee()1ahaaa，令()0ha，解得ea，当1ea时，()0ha，()ha在(1,e)上单调递减；当ea时，()0ha，()ha在

(e,)上单调递增；所以()(e)elnee0hah，所以eab···················································12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com