东北三省精准教学联盟2024-2025学年高三下学期3月联考数学答案

高三数学参考答案1234567891011CDDBCBCDBDABDACD12.160(5分）5π13.6（5分）14.35π（5分）π315．【答案】（1）（6分）（2）（7分）32A【解析】（）由正弦定理，得，分1sinABBsinsincos····················································12又B(0,π)，所以sinB0，A所以sinAcos，····································································································2分2AAA所以2sincoscos，·························································································4分222AπA因为，所以，所以，分A(0,π)0,cos0·························································5222A1Aππ所以sin，解得，即A.·········································································6分22263AA1（备注：推导出sinAcos给2分；推导出sin给3分，得出角A的大小给1分）222（2）因为D为BC的中点，1所以AD()ABAC,······························································································8分27又AD，b2，2第1页7122两边平方得到(c42c)，整理可得c2c30，···········································10分44解得c1或c3（舍去）·······················································································11分13所以ABC的面积SbcsinA.·······································································13分221（备注：写出中线定理AD()ABAC给2分；求出c=1给3分；得出三角形面积给2分）2116.【答案】（1）2xy20（6分）（2）,（9分）e2x11【解析】（1）当a1时，f(x)(x1)lnx，f(x)lnxlnx1，……………………2分xxf(1)0，f(1)2，…………………………………………………………………………………4分∴曲线yf()x在(1,f(1))处的切线方程为yf(1)f(1)(x1)，整理得，y2(x1)，∴曲线yf()x在(1,f(1))处的切线方程为2xy20．………………………………………6分（备注：求导给2分；求出斜率和切点给2分；求出切线方程给2分）xaa（2）f(x)lnxlnx1,x＞0，xxf()x是增函数，即f(x)0在（0，+∞）上恒成立，…………………………………………8分方法一：即在（，）上恒成立，所以，axxlnx0+∞axxlnxmax设g(x)xxlnx，x＞0，则g(x)2lnx，x＞0，…………………………………………11分1当x0,时，g(x)0，g()x单调递增，e21当x,时，g(x)0，g()x单调递减，2e111∴当时，取得极大值，也是最大值，∵，………………………………分x2g()xg2214eee1∴a的取值范围是,．………………………………………………………………………15分2e（备注：得出f'(x)≥0在（0，+∞）恒成立给2分；分离参数给1分，得出g()x的最大值给5分；下结论给1分，共4页过程酌情给分）aa方法二：即lnx10在（0，+∞）上恒成立，所以lnx10，xxmina1axa设h(x)lnx1，x＞0，则h()x，x＞0，…………………………………9分xxx2x2①若a0，则h(x)0，h()x在(0,)上单调递增，当x趋近于0时，h()x趋近于，即f(x)0不恒成立，所以f()x在(0,)上不单调递增，与题意不符，舍去．…………………………………………11分②若a0，则当x(0,a)时，h(x)0，h()x单调递减，当x(,)a时，h(x)0，h()x单调递增，则当xa时，h()x取得极小值，也是最小值，∴h(a)lna20，1解得a，…………………………………………………………………………………………14分e21∴a的取值范围是,．……………………………………………………………………分215e（备注：得出f'(x)≥0在（0，+∞）恒成立给2分；构造h()x给1分；令h(x)≥0求a的取值范围给5分，未讨min论a的正负扣2分；下结论给1分，过程酌情给分）2517.【答案】（1）证明见解析（6分）（2）（9分）5【解析】（1）证明：因为ABC为正三角形，且DEF,,分别是各边的中点，所以ADE,,CDFBEF均为正三角形.分别取的中点，DE,,EFFDABC1,,11则AA1DE,,BB1EFCC1DF，AA1BB1CC1,···········································1分又因为平面底面，平面底面ADEDEFADEDEFDE,所以平面，同理可得平面分AA1DEFBB1DEF,················································3第2页所以分AA1BB1,·······································································································4所以四边形为平行四边形，所以，AA1B1BABA1B1因为AB平面DEF，AB平面DEF，所以AB平面DEF.·····································5分11同理可得CB平面DEF，又ABBCB,AB平面ABC，BC平面ABC，所以平面ABC∥平面DEF.·······················································································6分（备注：做出辅助线给1分；证明AB∥平面DEF给4分；证明平面ABC∥平面DEF给1分，过程酌情给分）（2）以A为坐标原点，分别以AE，AF，AA为x,,yz轴的正方向,建立空间直角坐标系，则1111313311AC0,0,,,,,DE,0,0,,0,0，2442221313所以分AC,,0,AD,0,,·······························································8442213nAC0xy0，144设平面CDA的法向量为n(,,)xyz，则···························10分113nAD0xz0，122令，得，所以，分x3y3,z3n1(3,3,3)···················································11易知平面的一个法向量为，分EDAn2(0,1,0)·································································12nn35所以12，分cosn1,n2·····························································14|n1||n2|151525所以二面角CDAE的正弦值为.······································································15分5（备注：建系给1分；写出点的坐标给1分；写出平面EDA的法向量给1分；计算出平面CDA的法向量给3分；求出二面角的正弦值给3分，过程酌情给分）12【答案】（），（分）（）（分）18.1p(2,0)0p(2,1)42p(Y1|X2)593（3）分布列见解析（8分）共4页【解析】（1）X2，Y0的情况有，甲抢到2题并答对2题，乙未抢到题，不符合题意；甲抢到2题并答对2题，乙抢到题并答对1题答错1题，不符合题意.2所以，分p(2,0)0·····································································································2X2，Y1的情况有，甲抢到2题并答对2题，乙抢到1题并答错1题，321221所以2分p(2,1)C3.·········································································42339（）备注：求得p(2,0)给2分；求得p(2,1)给2分322121（），分2p(X2)C3······································································6236192故p(Y1|X2).··················································································9分136（备注：求得p(X2)给2分；求得p(Y1/X2)给3分）（3）X0表示：甲抢到2题并答对1题答错1题，或甲抢到0题，3312117故21，p(X0)C3C2233224已知X0，则Y的可能取值有3,1,1,3，3312238p(Y3|X0)，·····································································11分763243311231p(Y3|X0)，·······································································12分763243231121211212C3CC3223323334p(Y1|X0)，··························14分7