2023年高三年级第三次适应性检测数学参考答案及评分标准一、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。y2x24π13．1；14．；15．28；16．1．433四、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）sinC解：(1)因为2csinB(2ac)tanC，所以2csinB(2ac)··················1分cosC所以2sinCsinBcosC(2sinAsinC)sinC，因为C(0,π)，sinC0，所以2sinBcosC2sinAsinC···························2分因为ABCπ，所以2sinBcosC2sin(BC)sinC2sinBcosC2cosBsinCsinC，所以2cosBsinCsinC，···········································································4分1π所以cosB，又因为B(0,π)，所以B················································5分231（2）设ax，则c3x，由条件知BD(BABC)·····································6分21113所以|BD|2(BABC)2(a2c22accosB)x213························7分444所以x2···································································································8分所以b2a2c22accosB7x2，b27··················································9分所以周长labc4x27827····················································10分18．（12分）解：（1）取B1C1的中点O1，连结A1O1，因为ABAC，A1B1C1ABC为三棱台，AB2A1B14，π所以ABAC2,BAC，所以AOBC，AO2·····················2分11111112111111因为平面BCC1B1平面ABC，平面ABC//平面A1B1C1，所以平面BCC1B1平面A1B1C1，平面BCC1B1平面A1B1C1B1C1，所以A1O1平面BCC1B1···············································································3分1由条件知梯形BCCB的面积S(2242)132·································4分112数学参考答案第1页（共5页）1所以四棱锥A1BCC1B1的体积V3222·········································5分3z（2）取BC的中点O，连结AO，C1O1B1因为ABAC，所以AOBC，A1E因为等腰梯形BCC1B1中，O1,O分别为上下底面B1C1,BC的中点所以OO1BC，OBCy因为平面BCC1B1平面ABC，平面BCC1B1平面ABCBC，A所以OO1平面ABC，x以O为原点，分别以OA,OB,OO1所在直线为x轴，y轴，z轴建系如图，············6分则A(22,0,0),B(0,22,0),C(0,22,0),B1(0,2,1)，令BEmBB(0m1)，则E(0,222m,m)···········································7分1设平面ACE的法向量为n(x,y,z)，因为CA(22,22,0)，CE(0,422m,m)，则nCA22x22y0，nCE(422m)ymz0，令xm，可得n(m,m,422m)··························································8分因为OO1平面ABC，所以OO1(0,0,1)为平面ABC的一个法向量···················9分422m72记二面角EACB的平面角为，cos·······10分2m2(422m)21012即6m2m20，解得m或m（舍）···········································11分2372所以存在点E为BB的中点，使得二面角EACB的余弦值为················12分11019．（12分）nnn1n解：（1）因为S(1)a3，所以Sa9，Sa9·········4分nn12n2n12n12n322两式相减得，aaa9n，所以ba2a9n，·············6分2n2n12n3n2n12n3（2）因为a1,a2,a3成等差数列，所以2a2a1a3，又因为a1a23，a1a2a39，所以a16,a23,a30···························8分nnn1n1由Sn(1)an13可得，Sn1(1)an23，(n1)n1nn两式相减得，an1(1)an2(1)an123···········································9分2k2当n2k2，k2且kN时，a2k23，2k12k22k12k2此时S2k1a2k32333，k2································10分6,n1所以，········································································12分S2n1n19,n2数学参考答案第2页（共5页）20．（12分）解：（1）由题知：|PB|4|PA|,即|PB||PA|423·······························2分所以曲线C是以A(3,0),B(3,0)为焦点，长轴长2a4的椭圆························3分x2所以曲线C的方程为y21·····································································4分4x2（2）设Q(x,y)，因为点Q(x,y)在椭圆C上，所以0y21······················5分000040因为直线l:xt与圆Q相切于点M，所以|QM||x0t|··································6分22|QB|(x03)y0···············································································7分22假设存在，则|QB|(x03)y0m|QM|m|x0t|，22222222222即(x03)y0m(x0t)，即x023x03y0mx02mtx0mt，x2x2因为0y21，所以y210，40043所以(m2)x2(2m2t23)x4m2t20恒成立·····································9分4003所以m20，2m2t230，4m2t20············································10分4433解得t，m，32343所以存在m及定直线l:x，满足条件············································12分2321．（12分）解：（1）由题知：X的可能取值为1,0,1；······················································1分321P(X1)(1)(1)；431232325P(X0)(1)(1)；434312321P(X1)··················································································4分432所以X的分布列为：X101151P121221515所以E(X)101························································6分1212212（2）假设答题n轮，记Pn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3页（共5页）17113由题知：P1,P1,P1()3,P13()4·······························7分123284216经分析知：Pn第n轮第n1轮第n2轮1P没有得1分2n11P得1分没有得1分4n21P得1分得1分没有得1分8n3111所以PPPP(n4)···························································9分n2n14n28n3111解得a,b,c··············································································10分248111111因为PPPP，PPPP，n2n14n28n3n12n4n18n2111所以PPPPPn1n2n4n18n2111111(PPP)PP22n14n28n34n18n21P016n3所以Pn1Pn(n4)，且P1P2P3P4，可得P1P2P3P4P5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）sinxccosxsinxc解:（1）由题得：f(x)，f(x)·······························2分exex因为f(0)1c0，得c1·······································································3分bsinx1（2）因为f(x)1，所以axbsinx10·····································4分ax令F(x)axbsinx1，F(x)axlnabcosx，①若0a1，则F(π)aπ10，不合题意··················································5分②若a1，1°当b0时，则F(x)ax1bsinx0，适合········································6分2°当0blna时，则F(x)axlnabcosx