高三數(shù)學(xué)本試卷滿分150分,考試時間120分鐘.注意事項:2.回答選擇題時,選出每小題答案后,用鉛筆把答題卡上對應(yīng)題目的答案標(biāo)號涂黑。如需改動,用橡皮擦干凈后,再選涂其他答案標(biāo)號。回答非選擇題時,將答案寫在答題卡上,寫在本試卷上無效。3.考試結(jié)束后,將本試卷和答題卡一并交回。一、選擇題:本題共8小題,每小題5分,共40分.在每小題給出的四個選項中,只有一項是符合題目要求的.31.已知集合A=IXly=+4i,B==2·則AnB-A·x≥-4B.txlx≥-4且±1fC·xlx>0fD.txl-4≤≤-3或x>0f2i+a2.已知為純虛數(shù),其中i為虛數(shù)單位,則實數(shù)a=1-iA.-2B.2C.1D.-13.記s,為等差數(shù)列n的前n項和,若n的公差為d,S4=7s,則8=A.7dB.8dC.9dD.10d4.已知u為銳角C。s+6=3,則sin2+3=4444A.B.-C.D.-99995·已知個等腰梯形的下底邊長是上底邊長的3倍兩腰與下底邊所成角為3面積為8·若該等腰梯形是個圓臺的軸截面,則該圓臺的側(cè)面積為A.16mB.26mC.16D.32m6.為了研究變量x對變量Y的影響,對變量x和變量Y的觀測數(shù)據(jù)(i,yi)(i=1,2,…,5)進(jìn)行研究,計算得到55Y=bx+e,i-1i-1E(e)=0,D(e)=o,55Y=bx+e,i-1i-1E(e)=0,D(e)=o,則參數(shù)b的最小二乘估計為D.-0.3A.0.08B.-0.08CD.-0.34=324=32A.-1B.1C.D.-數(shù)學(xué)第1頁(共4頁)8.若函數(shù)y=f(x)與Y=f(-x)在區(qū)間[m,n]上的單調(diào)性相同,則把區(qū)間[m,n]n叫做y=f(x)的"平凡區(qū)間";若函數(shù)Y=f(x)與y=f(-x)在區(qū)間[m,n]上的單調(diào)性相反,則把區(qū)間[m,n]叫做y=f(x)的"非平凡區(qū)間".下列函數(shù)既有"平凡區(qū)間",又有"非平凡區(qū)間"的是xlxl,x≤-1或x≥1,2A.f(x)=(x-)2B.f(x)=2x<x<xlxl,x≤12xx>xlxl,x≤12xx>二、選擇題:本題共3小題,每小題6分,共18分.在每小題給出的選項中,有多項符合題目要求.全部選對的得6分,部分選對的得部分分,有選錯的得0分.A.AC=3B.△ABC的外接圓周長為5mC.CD=1D.DB·CB=16F2(,0),則A.曲線C是中心對稱圖形B.1OPI≥1C.滿足IPF1I=3,IPF2=1的點P有2個D.滿足PFPF2的點P有8個11.已知函數(shù)f(x)=lnx+a,則x+1A.函數(shù)f(x)僅有一個零點B.若函數(shù)f(x)在點(1,0)處與x軸相切,則a=2三、填空題:本題共3小題,每小題5分,共15分.12.在三棱錐P-ABC中,PA平面ABC,ABlCB,PC=13,則三棱錐PABC外接球的半徑為·13.已知函數(shù)f(x)=xe",g(x)=f(x)+f(2-x),則g(x)在點(1,g(1))處的切線方程為·14.互素是指兩個自然數(shù)a和b的最大公因數(shù)為1.歐拉函數(shù)p(n)表示不大于n(neN")且與n互素的正整數(shù)個數(shù),若數(shù)列ta,滿足a,=p(2"),且數(shù)列的前n項和為S,,則滿足s,<2025的n的最大值為·四、解答題:本題共5小題,共77分.解答應(yīng)寫出文字說明、證明過程或演算步驟.15.(13分)現(xiàn)需要對某人工智能芯片進(jìn)行性能測試,規(guī)則如下:首次測試(測試I)通過率為P(0<P<1),未通過測試I的芯片進(jìn)入第二次測試(測試I),通過率為q(0<q<1),未通過則報廢.通過任意一次測試即為合格芯片.(1)已知p=0.8,q=0.4,若某批次生產(chǎn)了10萬枚芯片,預(yù)估合格芯片的數(shù)量;(2)已知一枚芯片合格,求其是通過測試I的概率(結(jié)果用P,q表示).數(shù)學(xué)第2頁(共4頁)數(shù)學(xué)第3頁(共4頁)數(shù)學(xué)第4頁(共4頁)(1)當(dāng)a=0時,求f(x)的最小值;(2)若.2(<2)是f()的兩個極值點,且≥2求n的最大值·17.(15分)如圖,在棱長為5的正方體ABCD-AB,CD,中,點E在線段CDi上,滿足DiE=入D,C1(0<入<1),A,D與平面ACE交于點F.(1)若入=,求線段EF的長度;2(2)已知四邊形ACEF的周長為8+2·①求入的值;②求二面角E-AC-B的余弦值.18.(17分)已知圓心在x軸上移動的圓經(jīng)過點A(-4,0),且與x軸,Y軸分別交于B(x,0),C(0,y)兩個動點(B,C可以重合).(1)求點M(x,Y)的軌跡E的方程;(2)過點D(1,0)的兩條直線l,,l2相互垂直,直線l,與E交于G,H兩點,直線l2與E交于J,K兩點,線段GH,JK的中點分別為P,Q.①求四邊形GJHK面積的最小值;②判斷直線PQ是否過定點,若是,求出該定點;若不是,請說明理由.19.(17分)將正整數(shù)1,2,3,…,n的任意一種排列得到的有限數(shù)列記作n,若對ykeN*,k≤n,均有kk,則稱該數(shù)列為"n元全錯位數(shù)列",記"n元全錯位數(shù)列"的個數(shù)為b,如正整數(shù)1,2,3所對應(yīng)的"3元全錯位數(shù)列"有2,3,1和3,1,2,得b3=2.(1)求b,,b2;(3)求證:bn=n!·(n≥2).(2)求證:bn+(n+1)b(3)求證:bn=n!·(n≥2).ii(i+1)!高三數(shù)學(xué)參考答案123456789DBCCACBBABDABCBCD pp+(1p)q15.【答案】(1)88000 pp+(1p)q【解析】(1)設(shè)事件A：芯片合格，記X為該生產(chǎn)批次合格芯片的數(shù)量，則每個芯片通過測試的概率為，··········································································································5分則E(X)=105×0.88=88000，·································································································6分 （備注：推導(dǎo)出P(A)給3分；推導(dǎo)出E(X)給3分；下結(jié)論給1分）PC|B)=p+(1p)q，································································10分P(AB)=P(B)P(A|B)=p.1=p，·························································································12分(AEQ\*jc3\*hps23\o\al(\s\up9(B),A)（備注：設(shè)事件給（備注：設(shè)事件給1分；推導(dǎo)出P(A)給2分；推導(dǎo)出P(AB)給2分；推導(dǎo)出P(B|A)給1分,過程酌情給分)16.【答案】(1)1+ln2（6分29分）【解析】（1）當(dāng)a=0時，f(x)=lnx+，定義域為(0,+∞)，·········································所以f(x)=xEQ\*jc3\*hps21\o\al(\s\up7(一),x2)2，··············································································································2分((備注：f(x)求導(dǎo)正確給2分；判斷出f(x)的單調(diào)性給3分；求出f(x)的最小值給1分）（2）由題意知，函數(shù)f(x)的定義域為(0,+∞)，求導(dǎo)得f(x)=一ax22x一2，·································7分因為x1,x2（x1<x2）是f(x)的兩個極值點，〔a2lx1x2解得0<a<.······················································································································10分x1（備注：（備注：f(x)求導(dǎo)正確給1分；列出根與系數(shù)的關(guān)系給2分；化簡出+=一2給3分；推導(dǎo)出0<a≤給給2分；下結(jié)論給1分，過程酌情給分）17.【答案】（1）EF=（4分2）①λ=（5分）②-（6分）則則因為平面ABCD//平面A1B1C1D1，又平面ACE∩平面ABCD=AC，平面ACE∩平面A1B1C1D1=EF，所以EF//AC，····················································································································2分連接A1C1（圖略又因為AC//A1C1，所以EF//A1C1.又λ=，即E為D1C1的中點，所以EF為△D1A1C1的中位線，所以F為D1A1的中點，則EF=.······················································································4分(備注：推導(dǎo)出EF//AC給2分；得出EF=給2分）（2）①由（1）知，EF//AC，D1E=D1F，所以C1E=A1F，所以CE=AF，所以四邊形ACEF為等腰梯形，E=A1F=5(1-λ)，又0<λ<1，所以λ=.······································································································9分②以D為原點，DA,DC,DD1所在直線分別為x軸、y軸、z軸，建立空間直角坐標(biāo)系（圖略則E(0,3,5)，A(5,0,0)，C(0,5,0)，·······················································································10分則AC=(-5,5,0)，AE=(-5,3,5).設(shè)平面ACE的法向量為n=92==92==m.n（備注：①判斷出四邊形ACEF為等腰梯形給1分；寫出EF，CE給2分；求出λ=給2分②建系并寫出點坐標(biāo)給1分；求出平面ACE的法向量給2分；寫出平面ABC的法向量給1分；求出二面角EE-AC-B的余弦值給2分，過程酌情給分）18.【答案】(1)y2=4x（3分）(2)①32（7分）②直線PQ過定點，定點坐標(biāo)為(3,0)（7分）(2,2(2,2x-4(備注：列出關(guān)系式給2分；推導(dǎo)出點M的軌跡方程給1分）(2,x-4(備注：列出關(guān)系式給2分；推導(dǎo)出點M的軌跡方程給1分）(2,2（2）①因為直線l1,l2的斜率一定存在且不為0，故設(shè)l1：y=k(x-1)，l2：y=-(x-1)，G(x1,y1)，H(x2,y2)，y1y2=-4.·········································································6分y1y2=-4.·········································································6分2,同理JK=4(k2+1)，··········································································································8分1(1)(1)所以S四邊形GJHK=2GH.JK=8|(k2+k2+2,≥8(|2k2.k2+2,=1(1)(1)當(dāng)且僅當(dāng)k=±1時，四邊形GJHK的面積最小，最小值為32.·························································10分當(dāng)直線PQ斜率存在時，設(shè)直線lPQ：y=mx+n，聯(lián)立方程得分又，得(m+n)k2-2k+2m=0，···································································14分同理可得分所以是方程(m+n)x2-2x+2m