數(shù)學(xué)-【八省聯(lián)考】河南、山西、陜西、內(nèi)蒙古、四川、云南、寧夏、青海八省2025年高考綜合改革適應(yīng)性演練聯(lián)考試題和答案

1.已知集合A={-1,0,1},B={0,1,4},則A∩B=()A.{0}B.{1}C.{0,1}D.{-1,0,1,4}3.|2-4i|=()A.2A.2B.1C.0D.-12A.y=±xB.y=±2xC.y=±3xD.y=±4x7.在△ABC中,BC=8,AC=10,cos∠BAC=,則△ABC的面積為()8.已知函數(shù)f(x(=x|x-a|-2a2.若當(dāng)x>2時(shí),f(x(>0,則a的取值范圍是()A.(-∞,1]B.[-2,1[C.[-1,2[D.[-1,+∞)9.已知F(2,0(是拋物線(xiàn)C:y2=2px的焦點(diǎn),M是C上的點(diǎn),O為坐標(biāo)原點(diǎn),則()A.p=4B.|MF|≥|OF|C.以M為圓心且過(guò)F的圓與C的準(zhǔn)線(xiàn)相切D.當(dāng)∠OFM=120°時(shí),△OFM的面積為2310.在人工神經(jīng)網(wǎng)絡(luò)中,單個(gè)神經(jīng)元輸入與輸出的函數(shù)關(guān)系可以稱(chēng)為激勵(lì)函數(shù).雙曲正切函數(shù)是一種激勵(lì)函數(shù).定義雙曲正弦函數(shù)sinhx=雙曲余弦函數(shù)coshx=雙曲正切函數(shù)tanhx=,則()C.雙曲正切函數(shù)是增函數(shù)D.tanh(x+y(=A.B.C.D.12.已知函數(shù)f(x(=ax(a>0,a≠1(,若f(ln2(f(ln4(=8,則a=.14.已知曲線(xiàn)C:y=x3-兩條直線(xiàn)l1,l2均過(guò)坐標(biāo)原點(diǎn)O,l1和C交于M,N兩點(diǎn),l2和C交于P,Q兩點(diǎn).若△OPM的面積為2,則△MNQ的面積為.15.為考察某種藥物A對(duì)預(yù)防疾病B的效果,進(jìn)行了動(dòng)物(單位:St(1)求s,t;(2)記未服用藥物A的動(dòng)物患疾病B的概率為p,給出p的估計(jì)值；P(χ2≥k(k證明:數(shù)列為等比數(shù)列；(2)求{an{的通項(xiàng)公式；17.已知函數(shù)f(x(=alnx+(1)設(shè)a=1,b=-2,求曲線(xiàn)y=f(x(的斜率為2的切線(xiàn)方程；(3)設(shè)M是坐標(biāo)平面上的動(dòng)點(diǎn),且線(xiàn)段F1M的垂直平分線(xiàn)與C恰有一個(gè)公共點(diǎn),證明M的軌跡為圓,并求該19.在平面四邊形ABCD中,AB=AC=CD=1,∠ADC=30°,∠DAB=120°,將△ACD沿AC翻折至△ACP,其中P為動(dòng)點(diǎn).(1)設(shè)PC⊥AB,三棱錐P-ABC的各個(gè)頂點(diǎn)都在球O的球面上.①證明:平面PAC⊥平面ABC;(2)求二面角A-CP-B的余弦值的最小值.3.|2-4i|=()A.2|2-4i|=A.2B.1C.0D.-125.雙曲線(xiàn)x2-=1的漸近線(xiàn)方程為()A.y=±xB.y=±2xC.y=±3xD.y=±4x【解析】圓錐的底面半徑r=1,母線(xiàn)長(zhǎng)l=2,高h(yuǎn)=l2-r2=3,體積V=πr2h=33π.【解析】注意到sin∠BAC=sinA===,8.已知函數(shù)f(x(=x|x-a|-2a2.若當(dāng)x>2時(shí),f(x(>0,則a的取值范圍是()【解析】根據(jù)題意,f(2(≥0,于是2|2-a|-2a2≥0?-2≤a≤1,此時(shí)f(x(=x(x-a(-2a2,x=2在其對(duì)稱(chēng)軸x=右側(cè),符合題意,A.p=4B.|MF|≥|OF|綜上所述,正確的選項(xiàng)是ABC.定義雙曲正弦函數(shù)sinhx=雙曲余弦函數(shù)coshx=雙曲正切函數(shù)tanhx=,則()C.雙曲正切函數(shù)是增函數(shù)D.tanh(x+y(=【解析】對(duì)于選項(xiàng)A,由于y=ex和y=-e-x均為增函數(shù),選項(xiàng)正確;對(duì)于選項(xiàng)C,有tanhx=選項(xiàng)正確;==tanh(x+y(,選項(xiàng)正確;綜上所述,正確的選項(xiàng)是ACD.A.B.C.D.12.已知函數(shù)f(x(=ax(a>0,a≠1(,若f(ln2(f(ln4(=8,則a=.【解析】由f(ln2(f(ln4(=8,可得aln2?aln414.已知曲線(xiàn)C:y=x3-兩條直線(xiàn)l1,l2均過(guò)坐標(biāo)原點(diǎn)O,l1和C交于M,N兩點(diǎn),l2和C交于P,Q兩點(diǎn).若△OPM的面積為2,則△MNQ的面積為.15.為考察某種藥物A對(duì)預(yù)防疾病B的效果,進(jìn)行了動(dòng)物(單位:St(1)求s,t;(2)記未服用藥物A的動(dòng)物患疾病B的概率為p,給出p的估計(jì)值；P(χ2≥k(k證明:數(shù)列為等比數(shù)列；(2)求{an{的通項(xiàng)公式；n(1)略；(2)an=1(nn-2(*單調(diào)遞增,n-2>0恒成立,單調(diào)遞減,:{bn{關(guān)于n∈N*單調(diào)遞增,且bn<1。:bn<bn+1<1.17.已知函數(shù)f(x(=alnx+-x.(1)設(shè)a=1,b=-2,求曲線(xiàn)y=f(x(的斜率為2的切線(xiàn)方程；(1)y=2x-5;(2){a-1},其中a>2.【解析】1)f(x(=lnx--x,fI(x(=+-1=2→(3x+2((x-1(=0,“x>0,:x=1,故切點(diǎn)為(1,-3),k=2,切線(xiàn)方程為y=2x-5.fI(x(=-1,“x=1為f(x(的極小值點(diǎn),:fI(1(=a-b-1=a=b+1①當(dāng)b≤0時(shí),x-b>0,令fI(x(=0→x=1此時(shí)當(dāng)0<x<1時(shí),fI(x(>0,f(x(單調(diào)遞增；當(dāng)x>1時(shí),fI(x(<0,f(x(單調(diào)遞減,f(x(在x=1取得極大值,舍去.②當(dāng)b=1時(shí),fI(x(=-≤0,f(x(在(0,+∞(上單調(diào)遞減,f(x(不存在極值,③當(dāng)0<b<1時(shí),當(dāng)0<x<b時(shí),fI(x(<0,f(x(單調(diào)遞減；當(dāng)b<x<1時(shí),fI(x(>0,f(x(單調(diào)遞增；當(dāng)x>1時(shí),fI(x(<0,f(x(單調(diào)遞減此時(shí),f(x(在x=1取得極大值,舍去④當(dāng)b>1時(shí),當(dāng)0<x<1時(shí),fI(x(<0,f(x(單調(diào)遞減；當(dāng)1<x<b時(shí),f/(x(>0,f(x(單調(diào)遞增；當(dāng)x>b時(shí),f/(x(<0,f(x(單調(diào)遞減此時(shí)f(x(在x=1取得極小值,符合-1(2+y2=16.FM的垂直平分線(xiàn)為:y=-x-+=x+2+4y2=12Ly0y0y0Ly=-x+?3x2+4|Γ-2+4y2=12Ly0y0y0L2-xx+?3(x-1(2-36-=0?y+(2x-14(y+x-18x-32x0-15=0?y+(2x-14(y+(x0+1(2(x0+3((x0-5(=0?y+(2x-14(y+(x+2x0+1((x-2x0-15(=0(y+x+2x0+1((y+x-2x0-15(=0∵x+y+2x0+1=(x0+1(2+y>0,∴x+y-2x0-15=0∴M的軌跡方程為(x-1(2+y2=16,它為一個(gè)圓其中P為動(dòng)點(diǎn).(1)設(shè)PC⊥AB,三棱錐P-ABC的各個(gè)頂點(diǎn)都在球O的球面上.①證明:平面PAC⊥平面ABC;(2)求二面角A-CP-B的余弦值的最小值.①AB⊥AC,AB⊥PC,于是AB⊥PAC,從而面PAC⊥面ABC;三棱錐P-ABC的外接球球心O在底面ABC上的投影為△ABC的外