山东省青岛市四区统考2023-2024学年高三上学期期中高三数学答案

2023-2024学年度第一学期期中学业水平检测高三数学评分标准一、单项选择题：本题共8小题，每小题5分，共40分。1--8：DAACBCAB二、多项选择题：本题共4小题，每小题5分，共20分。（选不全得2分）9．BCD10．AC11．BCD12．ABD三、填空题：本题共4个小题，每小题5分，共20分。（16题第一个空2分，第二个空3分）13．ye3x；14．6；15．3；16．（1）14；（2）2．四、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CDPA，CDAD，PAADA，所以CD平面PAD····················································································3分又因为CD平面ABCD，所以平面ABCD平面PAD···································4分（2）记AD中点为F，因为PAPD，所以PFAD····································5分又因为平面ABCD平面PADAD，所以PF平面ABCD···························6分故以F为坐标原点，分别以FM,FD,FP所在射线为x轴,y轴,z轴，建立如图空间直角坐标系，所以P(0,0,2),M(2,0,0),D(0,2,0),C(2,2,0)，···············································7分设平面的法向量为，PDMn1(x1,y1,z1)zn1PD02y12z10则,，2x2z0Pn1PM011令，可得分z11n1(1,1,1)··················8设平面的法向量为PCDn2(x2,y2,z2)n2PC02x22y22z20FD则,，A2y2z0yn2PD022令，可得分z21n2(0,1,1)················9BC设平面PDM与平面PCD夹角为，Mx|nn|26则，12分cos|cosn1n2|·············································10|n1||n2|6318.（12分）解：（1）因为sin2Asin2B4sinAsinBcosC0，由正弦定理知：a2b24abcosC①···························································2分又由余弦定理知：c2a2b22abcosC②····················································3分c2由①②得：cosC·············································································4分6abc23ab1又因为c23ab，所以cosC··········································5分6ab6ab22π因为c(0，π)，所以C········································································6分3ππ5π2π5π14（2）因为T2()，所以，，27775因为N*，所以1或2···································································8分高三数学答案第1页（共4页）{#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}12π1若1，则f(x)sin(x)，因为f(C)，所以sin()，232π2ππ7π因为||，所以(,)，所以无解·············································10分236614π1若2，则f(x)sin(2x)，因为f(C)，所以sin()，232π4π5π11π4π7ππ因为||，所以(,)，所以，解得··········11分2366366此时f(x)sin(2x)在(，)上不单调，所以无解······························12分67219.（12分）1解：（1）由题f(x)a，x0································································1分x当a0时，f(x)0，f(x)在(0,)上单调递减；·······································3分1当a0时，由f(x)0解得x································································4分a11所以，当x(0,)时，f(x)0；当x(,)时，f(x)0；aa11所以，f(x)在(0,)上单调递减，在(,)上单调递增；·································6分aa1（2）由（1）知：当a0时，f(x)f()1lna·····································7分minaa2a2所以，存在a0，使1lnab成立，即存在a0，使1lnab成立···8分22a211a2令g(a)1lna，则g(a)a··············································9分2aa所以，g(a)在(0,1)上单调递增，在(1,)上单调递减，···································10分1所以g(a)g(1)···················································································11分21所以b的取值范围为(,]··········································································12分220.（12分）解：（1）因为APBD，PCBD，APPCP，所以BD平面APC········1分所以BDAC····························································································2分因为四边形ABCD是圆柱底面的内接四边形，且AC为其直径所以BEED，ABAD，BCCD，ABCADC90·····························4分又因为ACBCCD，所以AC2BC，1所以在RTABC中，sinBAC，所以BAC302所以BAD60，BAD是等边三角形························································5分（2）因为AC4，由（1）知，在RTABC中，AE3，EC1，所以CE:EA1:3······················································································6分高三数学答案第2页（共4页）{#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}因为PF:FA1:3，所以PC//EF又因为PC平面BFD，EF平面BFD，所以PC//平面BFD·····················7分所以点P到平面FBD的距离等于点C到平面FBD的距离···································9分因为CEPC，所以CEEF，又因为CEBD，EFBDE，所以CE平面BFD，···············································································10分所以点C到平面FBD的距离为CE1，点P到平面FBD的距离为1··················12分21.（12分）CDsinACD1解：（1）由题知，在△ACD中，由正弦定理得sinCAD····2分AD2π因为ADCD，所以ACDCAD，所以CAD·································3分6π所以DπACDCAD，所以ADCD········································4分2（2）在△ABC中，AC2··········································································5分AB2BC2AC23由余弦定理知：cosABC·······································6分2ABBC2所以AB2BC243ABBC，所以2ABBC43ABBC4解得：ABBC843，等号当仅当ABBC时取··························7分231所以SABBCsinB23····························································8分ABC25πABAC（3）在△ABC中，设BAC(0,)，由正弦定理知：6sinACBsinB5π5π所以AB4sin()，AE2sin()····················································9分66在△ADE中，由余弦定理知：DE2AD2AE22ADAEcosDAE5π5ππ所以DE234sin2()43sin()cos()666πππ34sin2()43sin()cos()666ππππ52cos(2)23sin(2)54sin(2)3336π54sin(2)54cos2(1,9]···········································11分2π所以DE29，等号当仅当BAC时取，所以DE的最大值等于3················12分222.（12分）解：（1）因为f(x)ex2ax1，································································1分设g(x)f(x)，则g(x)ex2a，·····························································2分高三数学答案第3页（共4页）{#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}当a0时，g