吉林省吉林市普通中学2024-2025学年高三上学期二模数学答案

令ab，则f，A选项正确；吉林地区普通高中—学年度高三年级第二次模拟考试1(1)020242025令ab1，则f(1)0；令a1,bx，则f(x)f(x)，B选项不正确；数学学科参考答案111令ax,b，则f(1)xf()f(x)0，一、单选题：本大题共8题，每小题5分，共40分。xxx11当x1时，01，f(x)0，f()0，即当0x1时，f(x)0.12345678xx又f(x)是奇函数，当1x0时，f(x)0；当x1时，f(x)0.CBACBCADf(0)f(1)f(1)0，C选项正确，D选项正确.8.教学提示lnt（法二）当x0时，f(x)0.令t[x](tN*)，则k.tf(ab)f(a)f(b)当x0时，.lntabab原不等式的解集对应区间的长度为1，不等式k的正整数解有且只有一个.fxt根据等式特点，构造函数()，logc|x|(c0且c1)lnxx易知f(x)在(0,e)上单调递增，(e,)上单调递减，x即f(x)xlogc|x|(c0且c1).ln3ln2又f(3)，f(2)f(4)，32又当x1时，f(x)0，c1，即f(x)xlogc|x|(c1).ln2ln3k.230,x0,综上易得选项正确，选项错误二、多选题：本大题共题，每小题分，共分。f(x)ACDB.3618xlogc|x|,x0.91011三、填空题：本大题共3小题，每小题5分，共15分。ACBCDACD32612.113.14.(2分)；[0,]（3分）10.教学提示334D.易证sinxx.14.教学提示（法一）由CP2PE得，点P轨迹是以A为球心，1为半径的球面，11f(x)sinxsin2xsinnx2n2n又点P在平面SAB内，点P在以A为圆心，1为半径，为圆3112sinxsin2xsinnx心角的圆弧上，因此点P的轨迹长度为.2n3112x2xnx建系如图，设P(cos,0,sin)([0,])，则AB(2,0,0),CP(cos1,3,sin).2n3nxABCP2(cos1)cos111.教学提示cosAB,CP.2(cos1)23sin252cos（法一）已知f(ab)af(b)bf(a)，ABCP令ab0，则f(0)0；第1页共6页5t2令t52cos,t[3,6],cos，32即ABC的面积S为.·························································································6分432t3tt6ππcosAB,CP[0,].（Ⅱ）因为CD为角C平分线，C，所以ACDBCD.2t2436在中，，6ABCSABCSACDSBCD故直线CP与直线AB所成角的余弦值的取值范围为[0,].41π1π1π所以absinaCDsinbCDsin，··························································8分（法二）设直线CP与直线AB所成角为，232626取AB的中点MCPM,PMA，,12232226由CD，得abab(ab)，所以ab(ab).·························10分根据三余弦定理可知，coscos1cos2248886CMtan，易知P从点M运动至N处，tan逐渐减小，则cos逐渐增大，ab6ab21PM11因为a0，b0，所以由基本不等式ab，得(ab)()，262由图可知，从点运动至处逐渐增大，PMNcos2266则P在点M处时，cos取得最小值，此时cos0，所以ab，当且仅当ab时取等号.33则在点处时，取得最大值，此时236，PNcoscoscos1cos222426所以ab的最小值为.·······················································································13分36故直线CP与直线AB所成角的余弦值的取值范围为[0,].416．【解析】（Ⅰ）f(x)的定义域为{x|x1}.············································································2分四、解答题：本大题共5小题，共77分。xex15．【解析】f(x).····································································································4分(x1)2（Ⅰ）由sin2AsinAsinBcos2Bcos2C(1sin2B)(1sin2C)sin2Csin2B，令f(x)0，得x1或1x0，得sin2Asin2Bsin2CsinAsinB.222f(x)的单调递减区间为(,1)，(1,0).·····························································7分由正弦定理得abcab.···················································································2分222abcab1exex所以cosC，（Ⅱ）x(1,),m(x1)，x10m.································9分2ab2ab2x1(x1)2π因为C(0,π)，所以C.·····················································································4分3exex(x1)设，，分g(x)2x(1,)g(x)3.···················································112222在ABC中，c3，ab6，由余弦定理cabab(ab)3ab，(x1)(x1)22令g(x)0，则x1.得(3)(6)3ab，解得ab1.当1x1时，g(x)0，g(x)在(1,1)上单调递减；1π133所以Sabsin1.23224第2页共6页当时，，在上单调递增分取，则，.分x1g(x)0g(x)(1,).······················································13y21z2hn2(0,1,h)········································································13eee则g(x)g(1)，m，即m的取值范围是(，).····································15分设平面BB1C1C与平面PBB1夹角为，min44417.【解析】|n1n2||h1|101则cos|cosn1,n2|，解得h2或.（Ⅰ）直三棱柱ABCABC中，AA平面ABC，2101111111|n1||n2|21h21又A1B1平面A1B1C1，AA1A1B1.·····································································3分所以正四棱锥PABBA的高为2或.·····································································15分112【此处也可直接证明：正四棱锥，.】（建系方式二）以为原点，，，所在直线PABB1A1中A1B1AA1AACAA1AB分别为x轴、y轴、z轴建立如图所示的空间直角坐标系，又A1B1A1C1，AA1A1C1A1,AA1,A1C1平面ACC1A1，此时可得平面的一个法向量为，BB1C1Cn1(1,0,1)A1B1平面ACC1A1.····························································································5分平面的一个法向量为.PBB1n2(1,0,h)又A1B1平面PA1B1，（法二）取BB1中点M,CC1中点N，连接PM，MN.平面PA1B1平面ACC1A1.·················································································7分正四棱锥，PABB1A1（Ⅱ）（法一）以为原点，，，所在直线分别为x轴、y轴、z轴建立如图所示的空间AAA1CAAB⊥PBPB1，PMBB1.直角坐标系，则,,.B(0,0,2)B1(2,0,2)C(0,2,0)又直三棱柱中,四边形是矩形，ABCA1B1C1BB1C1C设正四棱锥的高为，则，PABB1A1hP(1,h,1)，分别为，的中点，MNBB1CC1，，，分BB1(2,0,0)BC(0,2,2)BP(1,h,1)·························································9⊥MN//B1C1，B1C1BB1，MNBB1，设平面的一个法向量为.BB1C1Cn1(x1,y1,z1)∠PMN或其补角即为平面BB1C1C与平面PBB1的夹角，BBn02x10x10则11，BCn02y12z10y1z110101即cosPMN或.···············································································11分1010取，则，.分z11y11n1(0,1,1)·······································································112设正四棱锥PABB1A1的高为h,则PMh1,设平面的一个法向量为，PBB1n2(x2,y2,z2)A1B1A1C1且A1B1A1C12,BBn02x20x20则12，xhyz0xhyz.BPn20222222B1C1MN22第3页共6页2作PH⊥平面AA1C1C，垂足为H，连接NH，MQ2MP21h，则NH2h，PN(2h)21.在MNQ中，由余弦定理得，NQ2MN2MQ22MNMQcosNMQ,在PMN中，由余弦定理得，PN2PM2MN22PMMNcosPMN,即（22h）2844h222221h2cosNMQ,············································13分即((2h)21)2(h21)2(22)22h2122cosPMN,·············