高二数学参考答案第1页(共8页)2023—2024学年第二学期三市联合期末检测高二数学参考答案1.【答案】B【解析】将y=6.2代入方程^y=2.5x+1.2,解得x=2.2.【答案】D【解析】根据题意,分2步进行分析:①甲,乙必须相邻且甲在乙的右边,将甲乙看成一个整体,有1种顺序,②将甲乙整体与丙丁全排列,有A33=6种情况,则有1×6=6种排法.3.【答案】C【解析】P(1200<ξ<1600)=1-2P(ξ≤800)2=0.35.4.【答案】D【解析】由题意,n∈N*,在等比数列an中,a1=2,a1+a2+a3=6,设公比为q,∴a1+a1q+a1q2=6,即2+2q+2q2=6,解得q=-2或q=1,∴a6+a7+a8=a1+a2+a3q5=6q5,当q=1时,a6+a7+a8=6,当q=-2时,a6+a7+a8=-192.5.【答案】C【解析】由题意,可设圆的方程为(x-1)2+(y-1)2=r2,令y=0,得x2-2x+2-r2=0,x1+x2=2,x1x2=2-r2,|x1-x2|=(x1+x2)2-4x1x2=4r2-4=2,解得r2=2,∴圆的方程是(x-1)2+(y-1)2=2,即x2+y2-2x-2y=0.6.【答案】A【解析】如图,取BD的中点为O,由正方形ABCD的边长为2,则OB=OD=OA&#039;=OC=2,因此O为空间四边形A&#039;BCD的外接球球心,外接球半径R=2,设球心到平面α的距离为d,截面圆的半径为r,则有R2=r2+d2,即r=R2-d2,当ON⊥截面α时,d最大,此时截面面积最小,且ON=d,在△OND中,OD=2,DN=12,∠ODN=π4,由余弦定理可得,ON=DN2+OD2-2DN·OD·cosπ4=52,此时r=R2-d2=32,所以截面面积最小值为πr2=3π4.7.【答案】B{#{QQABBYSUogioAIIAAAgCAQ24CgAQkACAASgGxEAAoAAAABFABAA=}#}高二数学参考答案第2页(共8页)【解析】令t=12lnx,则x=e2t,所以不等式f12lnx<x等价转化为不等式ft<e2t,即fte2t<1,构造函数gt=fte2t,则g&#039;t=f&#039;t-2fte2t,由题意,g&#039;t=f&#039;t-2fte2t>0,所以gt为R上的增函数,又f(1012)=e2024,所以g1012=f1012e2024=1,所以gt=fte2t<1=g1012,解得t<1012,即12lnx<1012,所以0<x<e2024.8.【答案】C【解析】设Qy204,y0,Py214,y1,Ty224,y2,直线l的斜率kQP=y0-y1y20-y214=4y0+y1,直线l方程:y-y0=4y0+y1x-y204,化简得y=4x+y0y1y0+y1同理直线QT方程:y=4x+y0y2y0+y2,直线PT的方程:y=4x+y1y2y1+y2,点A(3,-2),B(3,-6)分别代入直线QP,QT方程,即-2=12+y0y1y0+y1,-6=12+y0y2y0+y2,􀮠􀮢􀮡􀪁􀪁􀪁􀪁消除y0,得y1y2=12,代入直线PT方程:y=4x+y1y2y1+y2,得y=4y1+y2(x+3),直线PT过定点(-3,0).9.【答案】AD【解析】由分布列的性质,可得a+b+16=1①,因为Eξ=53,所以1×a+2×b+3×16=53②,联立①②解得a=12,b=13,因为η=3ξ+1,所以Eη=3Eξ+1=6,Dη=9Dξ=9×59=5.10.【答案】ACD【解析】由fx=ax3+bx2+4,可得f&#039;x=3ax2+2bx,则f″x=6ax+2b,{#{QQABBYSUogioAIIAAAgCAQ24CgAQkACAASgGxEAAoAAAABFABAA=}#}高二数学参考答案第3页(共8页)因为点(1,2)是对称中心,结合题设中“拐点”的定义可知,f″1=6a+2b=0且f1=a+b+4=2,解得a=1,b=-3,所以A正确;由a=1,b=-3,可知fx=x3-3x2+4,则f&#039;x=3x2-6x,令f&#039;x=0,可得x=0或x=2,当x∈-∞,0时,f&#039;x>0,fx单调递增;当x∈0,2时,f&#039;x<0,fx单调递减;当x∈2,+∞时,f&#039;x>0,fx单调递增;又f0=4,f2=0,函数fx只有两个零点,所以B错误;因为1,2是函数fx的对称中心,所以f1+x+f(1-x)=2f1,令S=f12024+f22024+f32024+…+f20232024+f1+f20252024+f20262024…f40472024,可得S=f40472024+…+f20262024+f20252024+f1+f20232024+…+f12024,所以2S=[f12024+f40472024]+[f22024+f40462024]+…+[f40472024+f12024]=4×4047,所以S=2×4047,即f12024+f22024+f32024+…+f20232024+f1+f20252024+f20262024+…+f40472024=2×4047,所以C正确;拐点处的切线方程为y=-3x+5,所以-3×2+5<m<f2,故-1<m<0,D正确.11.【答案】ABD【解析】 a1=1,1an+1=1an+an,又a1=1>0,归纳可得an>0,∴an-an+1anan+1=an>0,∴an>an+1,数列an单调递减,故选项A正确;当n=1时,S1=1;当n≥2时,Sn=a1+a2+…+an<a1+a1+…+a1=n,故选项D正确; 1an+1=1an+an,∴1a2n+1=1an+an2=a2n+2+1a2n>1a2n+2,∴1a2n+1-1a2n>2,∴1a2n-1a21>2(n-1),∴1a2n>2n-1,∴an<12n-1,a2024<14047<163,故选项B正确;又1a2n+1=1an+an2=a2n+2+1a2n≤1a2n+3,∴1a2n+1-1a2n<3,∴1a22-1a21=3,1a23-1a22<3,…,1a2n-1a2n-1<3,∴1a2n-1a21<3(n-1),∴1a2n<3n-2,∴an>13n-2(n≥2);当n=1时,an=13n-2.综上,an≥13n-2.故选项C错误.{#{QQABBYSUogioAIIAAAgCAQ24CgAQ...