2025年新高考數(shù)學(xué)名校選填壓軸好題匯編（四）（解析版）

1.(廣東省五校2023-2024學(xué)年高三10月聯(lián)考(二)數(shù)學(xué)試題)設(shè)函數(shù)f(x(=若方程f(x(A.(-∞,-1(B.(-1,0(C.(0,1(D.(1,+∞(令g(x(=f(x(-x=方程f(x(=x+b有3個(gè)不同的實(shí)根等價(jià)于g(x(與y=b’(x(=-1=，(x(<0；∴g(x(在(0,1(上單調(diào)遞增，在(1,+∞(上單調(diào)遞減，∴g(x(max=g(1(=-1；A.B.C.D.D1=D1所以A1E⊥平面DCC1D1，所以A1E⊥EQ，所以A1E=2×所以EQ=7-3=2，3.(廣東省七校2024屆高三第二次聯(lián)考數(shù)學(xué)公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)試卷)已知a>0，f(x(=(aex-ln(x+b(，當(dāng)x>0時(shí)，公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)f(x(≥0，則a(1-b(3的最大值為()A.B.C.D.x-在(0,+∞(為增函數(shù)，由y=aex與y=圖象知，y=aex-在(0,+∞(有唯一的零點(diǎn)x0，顯然f(x(=(aex-(ln(x+b(的定義域?yàn)閧x|-b〈x且x〈0{，且f(1-b(=0x-,y=ln(x+b(均在(-b,+∞(單調(diào)遞增，-b,+∞(時(shí)，ln(x+b(>0，因?yàn)閒(x(≥0，所以aex->0，所以當(dāng)x=1-b時(shí)，aex-=0,ln(x+b(=0， 故g(x(≤g(2(=，所以a(1-b(3=≤,當(dāng)且僅當(dāng)1-b=2即b=-1時(shí)等號(hào)成立；D.A.2B.5D.4C的漸近線的方程為y=±x，y=-1xrx=1x0-yy=-1xrx=1x0-y0y-y0=2(x-x0(,(y=-0+y0不妨取Ax0-y0,-x0+y0(，同理可得Bx0+y0,x0+y0(，則|OA|=x0-y0(2+(-x0+y0(2、52x、525=4x05=4x-y公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)+,1),，若方程[f(x)]2-af(()A.B.或1C.1D.或2由f(x(<0得x+2<0，即x<-2時(shí)，f(x)單調(diào)遞減，由f(x(>0得x+2>0，即-2<x≤0時(shí)，f(x)單調(diào)遞增，當(dāng)x=-2時(shí)，f(x)取得極小值f(-2)=-，f(0)=1，作出f(x)的圖象如圖：由圖象可知當(dāng)0<f(x(≤1時(shí)，有三個(gè)不同的x與f(x)對(duì)應(yīng)，設(shè)t=f(x(，方程[f(x)]2-af(x)+=設(shè)=t2-at+PQ|的最大值和最小值分別為m和n，則m+n的值為()02+max=|PO1|max+r=37+r，此時(shí)點(diǎn)P(2,-3(，|PQ|min=|PO1|min-r=27-r，此時(shí)P(-2,3(，即m=37+r,n=7-r，所以m+n=27A.B.-C.2D.-2所以f(x(=asinx+cosx，由函數(shù)f(x(圖象的對(duì)稱軸方程為x=kπ+(k∈Z(，得f(2kπ+-x(=f(x((k∈Z(，、、A.[1,3[B.[1,2[C.[3,2[A.B.4C.D.3-4kx-4=0，則x1+x2=4k,x1x2=-4，故y1y2=?=1，又|MF|=y1+=y1+1，|NF|=y2+=y2+1，則2|MF|+|NF|=2(y1+1(+(y2+1(=2y1+y2+≥22y1×y2+=，A.(1+x((1+x2((1+x3(?(1+x10(B.(1+x((1+2x((1+3x(?(1+10x(C.(1+x(2(1+x2(2(1+x3(2(1+x4(2?(1+x10(2D.(1+x(2(1+x+x2(2(1+x+x2+x3(2?(1+x+x2+?+x10(2總計(jì)m=66種.對(duì)C，(1+x(2(1+x2(2(1+x3(2(1+x4(2?(1+x10(2=(1+x((1+x2((1+x3((1+x4(?(1+x10((1+x((1+x2((1+x3((1+x4(?(1+x10(x93422對(duì)D，(1+x(2(1+x+x2(2(1+x+x2+x3(2?(1+x+x2+?+x10(2=(1+x((1+x+x2((1+x+x2+x3(?(1+x+x2+?+x10(×(1+x((1+x+x2((1+x+x2+x3(?(1+x+x2+?+x10(，故D錯(cuò)誤.)D.)D.A.4412.(湖南省長(zhǎng)沙市長(zhǎng)郡中學(xué)2024屆高三月考(二)數(shù)學(xué)試卷)設(shè)函數(shù)f(x(=(x2+ax+b(lnx，若f(x(≥0，A.-2B.-1C.2D.1要使f(x(≥0，則二次函數(shù)y=x2+ax+b，在0<x<1上y<0，在x>1上y>0，A.1+2B.2+2C.5FF2=2=D.614.(湖北省云學(xué)部分重點(diǎn)高中聯(lián)盟2023-2024學(xué)年高三10月聯(lián)考數(shù)學(xué)試卷)已知函數(shù)f(x(=cosx-axA.(-∞,-B.(-C.,+∞(D.-,+∞(【解析】因?yàn)閒(x(=cosx-ax，則f’(x(=-sinx-a，由題意可得f’(x(=-sinx-a≥0對(duì)任意x∈0,|恒成立，即-sinx≥a對(duì)任意x∈0,|恒成立，所以實(shí)數(shù)a的取值范圍是(-∞,-.15.(湖北省云學(xué)部分重點(diǎn)高中聯(lián)盟2023-2024學(xué)年高三10月聯(lián)考數(shù)學(xué)試卷)已知函數(shù)f(x(=lnx-a(x-有兩個(gè)極值點(diǎn)x1,x2，則f(x1+x2(的取值范圍是()A.(0,ln2-【解析】由題意可知：f(x(的定義域?yàn)?0,+∞(，且f’(x(=-a(1+=-，2-x+a=0有兩個(gè)不相等的正根x1,x2，2-x+1=0有兩個(gè)不相等的正根x1,x2，Δ=-4>012=1>0則x1+x2=>0，解得0<2=1>0可得f(x1+x2(=f=ln-a-a(=a2-lna-1，令g(a(=a2-lna-1,0<a<，則g’(a(=2a-=<0，即g(a(的值域?yàn)?ln2-,+∞(，所以f(x1+x2(的取16.(湖北省武漢外國(guó)語(yǔ)學(xué)校2023-2024學(xué)年高三10月月考數(shù)學(xué)試題)已知a∈R，設(shè)函數(shù)f(x)=≤1,若關(guān)于x的不等式f(x)≥0在R上恒成立，則a的取值范圍為(1)當(dāng)0≤a≤1時(shí)，f(x)=x2-2ax+2a=(x-a)2+2a-a2≥2a-a2=a(2-a)>0，17.(湖北省武漢外國(guó)語(yǔ)學(xué)校2023-2024學(xué)年高三10月月考數(shù)學(xué)試題)已知函數(shù)f(x(=f(-x(,x∈R，f(5.5(=1，函數(shù)g(x(=(x-1(?f(x(，若g(x+1(為偶函數(shù)，則g(-0.5(的值為()A.3B.2.5C.2由g(x(=(x-1(?f(x(，可得(x-1(?f(x(=(1-x(?f(2-x(，又因?yàn)閒(x(=f(-x(，所以f(x(是定義在R上的偶函數(shù)，所以f(x(=-f(2-x(=-f(x-2)，所以f(x-4(=f[(x-2)-2]=-f(x-2)=-[f(x)]=f(x)，即f(x-4)=f(x)，所以函數(shù)f(x(是周期為4的周期函數(shù)，所以f(5.5)=f(1.5+4)=f(1.5)=f(-2.5)=f(2.5)=1，則g(-0.5)=g(2.5)=(2.5-1)f(2.5)=1.5.18.(湖北省新八校協(xié)作體2023-2024學(xué)年高三10月聯(lián)考數(shù)學(xué)試題)已知函數(shù)f(x(的定義域?yàn)镽，y=f(x(+2ex是偶函數(shù)，y=f(x(-4e-x是奇函數(shù)，則f(x(的最小值為()【解析】因?yàn)閥=f(x(+2ex是偶函數(shù)，所以f(x(+2ex=f(-x(+2e-x，即f(x)-f(-x)=2e-x-2ex①,又因?yàn)閥=f(x(-4e-x是奇函數(shù)，所以f(-x)-4e-x=-f(x)+4ex，即f(x)+f(-x)=4ex+4e-x②,聯(lián)立①②可得f(x)=ex+3e-x，A.(-1,e(B.(-∞,-1(∪[e，+∞(C.[-1,1)當(dāng)x<0時(shí)，g(x(=-=-+=，-k|；-k|=ex，x=1=0且y’|x=1=1，所以函數(shù)y=lnx在x=1的切線方程為y=x-1，又由函數(shù)y=-lnx，可得y’=-，可得y|x=1=0且y’|x=1=-1，所以函數(shù)y=-lnx在x=1的切線方程為y=-x+1，所以函數(shù)y=|lnx|與y=|x-1|只有一個(gè)公共點(diǎn)，公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)當(dāng)-1≤k<1時(shí)，g(x(=f(x(-||恰有2個(gè)零點(diǎn)；合求解.x和lnx相關(guān)的常見同構(gòu)模型a≤blnb?ealnea≤blnb，構(gòu)造函數(shù)f(x(=xlnx或g(x(=xex；?構(gòu)造函數(shù)f(x(=或g(x(=；a±a>b±lnb?ea±lnea>b±lnb，構(gòu)造函數(shù)f(x(=x±lnx或g(x(=ex±x.y=f(x(相切于(x1,f(x1((,(x2,f(x2((兩點(diǎn)，其中x1<x2．若當(dāng)x∈(0,x1(時(shí)，f(x(>k，則函數(shù)f(x(-kx在(0,+∞(上的極大值點(diǎn)個(gè)數(shù)為()A.0B.1C.2D.3,x2(,x5(,x3(,x4(設(shè)g(x(=f(x(-kx，則g(x(=f(x(-k，∴g(x(在(0,x1(,(x3,x2(,(x4,x5(上單調(diào)遞增，在(x1,x3(,(x2,x4(,(x5,+∞(上單調(diào)遞減，∴g(x(=f(x(-kx有x=x1，x=x2和x=x5三個(gè)極大值點(diǎn).6)的圖象向右平移個(gè)單位長(zhǎng)度得到函數(shù)g(x)的圖象，若g(xA.1B.2C.3D.4由題意得g(x)的圖象向左移個(gè)單位長(zhǎng)度得到函數(shù)f(x)的圖象，1為奇函數(shù)，f(x+2(為偶函數(shù)，則f(1(+f(2(+?+f(16(=()因?yàn)閒(x+2(為偶函數(shù)，所以f(x+2(=f(2-x(，則f(x+4(=f(-x(，所以f(x(+f(x+4(=2，f(x+4(+f(x+8(=2，所以f(x(=f(x+8(，故f(x(的周期為8，因?yàn)閒(1(+f(5(=2,f(2(+f(6(=2,f(3(+f(7(=2,f(4(+f(8(=2，所以f(1(+f(2(+?+f(16(=2[f(1(+f(2(+?+f(8([=16，均為R，若f(x(=f(-x(+2x,f(x(的圖象關(guān)于直線x=1對(duì)稱，且f(2(=0，則f()A.10B.20C.-10D.-20所以f’(x)=-f’(2-x)，又f(x)=f(-x)+2x，則f’(x)=-f’(-x)+2，所以-f’(2-x)=-f’(-x)+2，從而f’(2+x)=f’(x)-2，于是fI(1)+fI(3)+fI(5)+?+fI(19)=0-2-4-?-18=-90，fI(2)+fI(4)+?+fI(20)=-1-3-?-19=-100，所以=-190，又由f(x)=f(2-x)及f(x(=f(-x(+2x得f(2-x)=f(-x)+2x，從而f(2+x)=f(x)-2x，而f(2)=0所以f(20)=f(18)-36=f(16)-32-36=?=f(2)-4-8-?-32-36=-180，所以=-180-=10，故EM=EN=，易得Rt△MEO≌Rt△NEO，F(xiàn)(x(=ex+2f(x+2(是偶函數(shù)，其函數(shù)圖象為不間斷曲線且(x-2([fI(x(+f(x([>0，則不等式xf(lnx(<e3f(3(的解集為()【解析】由(x-2([f’(x(+f(x([>0,得x[f’(x+2(+f(x+2([>0,則當(dāng)x>0時(shí),得f’(x+2(+f(x+2(>0,F’(x(=ex+2f(x+2(+ex+2f’(x+2(=ex+2[f(x+2(+f’(x+2([，則當(dāng)x>0時(shí),F’(x(>0,得函數(shù)F(x(在(0,+∞(上單調(diào)遞增, 因?yàn)閤f(lnx(<e3f(3(,所以F(lnx-2(<F(1(,由于F(x(=ex+2f(x+2(是偶函數(shù),則F(|lnx-2|(<F(1(,而函數(shù)F(x(在(0,+∞(上單調(diào)遞增,得|lnx-2|<1,得-1<lnx-2<1,()A.x+y≤1B.x+y≥-2C.x2+y2≤2D.x2+y2≥1、、因?yàn)閤2+y2-xy=1變形可得(x-2+y2=1，設(shè)x-=cosθ,y=sinθ,所以x=cosθ+sinθ,、、27.(多選題)(廣東省五校2023-2024學(xué)年高三10月聯(lián)考(二)數(shù)學(xué)試題)若正實(shí)數(shù)x，y滿足xex-1=y(1+lny(，則下列不等式中可能成立的是()A.1<x<yB.1<y<xC.x<y<1D.y<x<1x-1=y(1+lny(，所以xex-1=(1+lny(e(1+lny(-1，1+lny>0，令f(x(=xex-1，x∈(0,+∞(，則f’(x(=(x+1(ex-1>0，所以f(x(=xex-1在(0,+∞(上單調(diào)遞增，由f(x(=f(1+lny(，可得x=1+lny，令g(x(=lnx+1-x，則g’(x(=-1=，所以當(dāng)0<x<1時(shí)g’(x(>0，當(dāng)x>1時(shí)g’(x(<0，所以g(x(在(0,1(上單調(diào)遞增，在(1,+∞(上單調(diào)遞減，公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)當(dāng)y≠1時(shí)1<x<y或x<y<1，結(jié)合y=lnx+1與y=x的圖象也可得到所以1<x<y或x<y<1.D1D.若P是棱A1B1的中點(diǎn)，則三棱錐P-CEF的外接球的表面積是41πMN∩DN=N，EF∩CE=E，MN,DN?平面MNDB，EF,CE?平面CEF∴P點(diǎn)的軌跡為線段NM，又MN=2選項(xiàng)正∴三棱錐P-DEF的體積為定值，∴B選項(xiàng)正確；4，且AA1⊥平面A1B1C1D1，則A1P=，則FE⊥PE，∴G到E，F(xiàn)，P的距離相等，又GH⊥平面PEF，∴三棱錐P-CEF的外接球的球心O在GH上，設(shè)三棱錐P-CEF的外接球的半徑為R，則PO=CO=R，點(diǎn)M．N為拋物線C的準(zhǔn)線與y軸的交點(diǎn)．則以下結(jié)論正確的是()B.直線PN的傾斜角α≥設(shè)AB:y=kx+1，與C的方程聯(lián)立得x2-4kx-4=0，設(shè)A(x1,y1(,B(x2,y2(，，F(xiàn)·F=-4，所以A錯(cuò)誤；x記直線AB的斜率為k，令f(x)=x2，則f’(x)=x，則k1=f’(x1(=x1,k2=f’(x2(=x2， x又直線AB過(guò)點(diǎn)F(0,1)，故直線AB的方程為x-y+1=0,C正確；MA:y-y1=(x-x1(，又y1=，所以MA:y=x-，同理MB:y=x-，聯(lián)立解得M,，即M(2k,-1)，又F(0,1)，故選:BCD.是一條連續(xù)不斷的曲線，記f(x(的導(dǎo)函數(shù)為f’(x(，則()A.存在f(x(和實(shí)數(shù)t，使得f’(x(=tf(x(B.不存在f(x(和實(shí)數(shù)t，滿足f(x(+f(t(=f(2x(C.存在f(x(和實(shí)數(shù)t，滿足f(xt(=tf(x(D.若存在實(shí)數(shù)t滿足f’(x(=f(x+t(，則f(x(只能是指數(shù)函數(shù)【解析】令f(x(=ax，則存在實(shí)數(shù)lna使得f’(x(=axlna，A正確；存在t=2,f(x(=logax,f(x(+f(t(=logax+loga2=loga2x=f(2x(，故B錯(cuò)誤；令f(x(=logax，則f(xt(=logaxt=tlogax=tf(x(，C正確；若f(x(=sinx，f’(x(=cosx=sin(x+，故D錯(cuò)誤. ()C.yAyB=xPD.cos∠AFP=設(shè)N(xN,yN(且xN≠1，則P(2xN-1,2yN(，代入得x+y=，以PF為直徑的圓N的方程可寫為(x-xP((x-1(+y(y-yP(=0，令x=0，可得xP+y(y-yP(=0，即y2-yPy+xP=0，=y+1?y+1=yy+(y+y(+1=1-4xP，|PF|=(xP-1(2+y=1-4xP，=，故D正確.f(1((為曲線y=f(x(的對(duì)稱中心C.當(dāng)a<0時(shí)，x=a是f(x(的極大值點(diǎn)D.當(dāng)a>1時(shí)，f(x(有三個(gè)零點(diǎn)即存在這樣的a,b使得f(x)=f(2b-x)，3-3ax2+1=2(2b-x)3-3a(2b-x)2+1，因?yàn)榈仁接疫?(2b-x(3展開式含有x3的項(xiàng)為2C(2b(0(-x(3=-2x3，且f(x)+f(2-x)=2x3-3ax2+1+2(2-x)3-3a(2-x)2+1=(12-6a)x2+(12a-24)x+18-12a，f’所有f(x)在x=a處取到極大值，x=a是f(x(的極大值點(diǎn)f’f’B12B3=3，?,Bn-1Bn=n(n≥2)，OC1=1，C1C2=C2C3=?=Cn-1Cn=(n≥2)，四邊形OB1D1C1，D3C3B.長(zhǎng)方形OBnDnCn的面積為xDn=1+2+3+?+n=,yDn=1+(n-1(=xDn+==(yDn(2，所以C正確.2+22+?+n2=，B.水面EFGH所在四邊形的面積為定值f(x(=g(x(，g(x(=f(x(，若f(2x(=2f(x(g(x(，則()A.g(x(的圖象關(guān)于直線x=0對(duì)稱B.g(2x(=g2(x(+f2(x(C.g(0(=0或1D.g2(x(-f2(x(=1【解析】對(duì)于A，由f(x(為R上的奇函數(shù)，則f(0(=0，f(x(+f(-x(=0，則f’(x(-f’(-x(=0，所以g(x(-g(-x(=0，即g(x(為偶函數(shù)，因此關(guān)于直線x=0對(duì)稱，故A正確；對(duì)于B，由f(2x(=2f(x(g(x(，則兩邊同時(shí)求導(dǎo)得：2f’(2x(=2f’(x(g(x(+2f(x(g’(x(，即g(2x(=g2(x(+f2(x(，故B正確；由g(x(f(x(-f(x(g(x(=0，則2g(x(g’(x(-2f(x(f’(x(=0，即[g2(x([’-[f2(x([’=0，即[g2(x(-f2(x([’=則g2(x(-f2(x(=C(C為常數(shù))，設(shè)h(x(=g2(x(-f2(x(=C(C為常數(shù))，對(duì)于C，由g(2x(=g2(x(+f2當(dāng)g(0(=0，則h(0(=g2(0(-f2(0(=0，則h(x(=g2(x(-f2(x(=0，即f(x(=±g(x(，又g(x(為偶函數(shù)，則f(x(即是奇函數(shù)也是偶函數(shù)，與f(x(在R上單調(diào)遞增矛盾，對(duì)于D，當(dāng)g(0(=1時(shí)，則h(0(=g2(0(-f2(0(=1，則h(x(=g2(x(-f2(x(=1，即g2(x(-f2(x(=1，故D正36.(多選題)(湖南省長(zhǎng)沙市長(zhǎng)郡中學(xué)2024屆高三月考(二)數(shù)學(xué)試卷)已知函數(shù)f(x)=sinωx+acosωx(xA.a>0A.a>0設(shè)F(x(=f(x-=2sin2(x-+=2sin2x，F(xiàn)(-x(=2sin[2(-x)]=-2sin2x=-F(x(，所以F(x(為奇函數(shù)，即函數(shù)f(x-為奇函數(shù)，所以B不正確.M在y軸上的射影為點(diǎn)N.若|MN|=|NF|，則()A.l的斜率為31>0,y2<0，,Nx-(，、x=或x=、(y=2px(y=3p(y=-3p、即A,-3p,N、A.a>0,c<0B.+<-2C.a+c>0D.<-12+c2>-2ac，且ac<0，所以+<-2，故B正確；對(duì)于D：因?yàn)?a+2c(+(a+b(=(2a+b+c(+c=c<0，可得a+2c<-(a+b(，又因?yàn)?a+b+c=(a+b(+(a+c(=0，可得a+b=-(a+c(>0，所以<-1，故D正確；A.sinα+sinβ>1B.tanα?tanβ<1C.cosα+cosβ<、2D.tan(α-β(>tan可得：0<β<α<，且α+β>，<α<，對(duì)于選項(xiàng)A：因?yàn)棣?gt;-α,且0<-α<，則sinβ>sin=cosα,可得sinα+sinβ>sinα+cosα=、2sin，因?yàn)?lt;α<，則<α+<，可得<sin<1，所以sinα+sinβ>、2sin>1，故A正確；對(duì)于選項(xiàng)B：因?yàn)閠an(α+β(=-<0，且tanα,tanβ>0，即tanα+tanβ>0，則1-tanα?tanβ<0，即tanα?tanβ>1，故B錯(cuò)誤；對(duì)于選項(xiàng)C：因?yàn)棣?gt;-α,則cosβ<cos=sinα,則cosα+cosβ<cosα+sinα=、2si由選項(xiàng)A可知：<sin<1，對(duì)于選項(xiàng)D：因?yàn)閠an(α-β(= tan1-tan2又因?yàn)?<α-β<，則0<<，可得0<tan<1，即0<1-tan2<1，所以tan(α-β(=->tan，故D正確；3=-f(x(在x=x0處的切線與函數(shù)y=f(x(的圖象有且僅有兩個(gè)交點(diǎn)令f’(x(=6x2=0解得x=0，且f(0(=B選項(xiàng)，f(x)=2x3-3ax2+1,f’(x(=6x2-6ax=6x(x-a(，若f(x)=2x3-3ax2+1有三個(gè)不同的零點(diǎn)x1,x2,x3，則f(x(=2(x-x1((x-x2((x-x3(=2x3-2(x1+x2+x3(x2+(x1x2+x2x3+x1x3(x-2x1x2x3，通過(guò)對(duì)比系數(shù)可得-2x1x2x3=1?x1x2x3=-，正確.則f(x(=f(2b-x(，即2x3-3ax2+1=2(2b-x(3-3a(2b-x(2+1，f(x)=2x3-3ax2+1,f’(x(=6x2-6ax，則f(x0)=2x-3ax+1,f’(x0(=6x-6ax0，所以f(x(在x=x0處的切線方程為y-(2x-3ax+1(=(6x-6ax0((x-x0(，y=(6x-6ax0((x-x0(+(2x-3ax+1(，由{-3ax+1(，消去y得2x3-3ax2+1=2x-3ax+1+(6x-6ax0((x-x0(①,3-2x=2(x3-x(=2(x-x0((x2+xx0+x(，-3ax2+3ax=-3a(x2-x(=-3a(x-x0((x+x0(，所以①可化為2(x-x0((x2+xx0+x(-3a(x-x0((x+x0(-(6x-6ax0((x-x0(=0，提公因式x-x0得(x-x0([2(x2+xx0+x(-3a(x+x0(-(6x-6ax0([=0，化簡(jiǎn)得(x-x0([2x2+(2x0-3a(x-(4x-3ax0([=0，進(jìn)一步因式分解得(x-x0(2(2x+4x0-3a(=0，解得x1=x0,x2=，所以x1-x2=x0-==≠0，所以x1≠x2，f(x(在x=x0處的切線與函數(shù)y=f(x(的圖象有且僅有兩個(gè)交點(diǎn)，正確.f(x+為奇函數(shù)，f(x+π(為偶函數(shù)．當(dāng)x∈[0,π[時(shí)，f(x(=cosxA.f(x(在(3π,4π(上單調(diào)遞減B.f(C.點(diǎn)(-π,0(是函數(shù)f(x(的一個(gè)對(duì)稱中心D.方程f(x(+lgx=0有5個(gè)實(shí)數(shù)解因?yàn)閒(x+為奇函數(shù)，所以f(-x+(=-f(x+(，即f(-x(=-f(x+π(，因?yàn)閒(x+π(為偶函數(shù)，所以f(-x+π(=f(x+π(，所以f(-x+π)=-f(-x)，即f(x+π)=-f(x)?f(x+2π)=f(x)，則f(x)周期為2π,由f(-x+π(=f(x+π(得，f(x)的一條對(duì)稱軸為直線x=π,f(x(=cosx，對(duì)于A，f(x(在(π,2π(上單調(diào)遞增，所以f(x(在(3π,4π(上單42.(多選題)(湖北省新八校協(xié)作體2023-2024學(xué)年高三10月聯(lián)考數(shù)學(xué)試題)[x[表示不超過(guò)x的最大整A.若x∈(0,1(，則f(-x(+f(x(+B.f(x+y(<f(x(+f(y(C.設(shè)g(x(=f(2x(+f(g(k(=401D.所有滿足f(m(=f(n((m,n(的點(diǎn)(m,n(組成的區(qū)域的面積和為f(x(=0，f(-x(=-1，則f(-x(+=-1+<0=-f(x(+，故A正確；B選項(xiàng)，取x=2.5，y=3.5，則f(x+y(=f(6(=6>5=f(2.5(故B錯(cuò)誤；R，且滿足f(x(+f(y(=f(x+y(-2xy+1,f(1(=3，則下列結(jié)論正確的是()A.f(4(=21B.方程f(x(=x有整數(shù)解C.f(x+1(是偶函數(shù)D.f(x-1(是偶函數(shù)【解析】對(duì)于A，因?yàn)楹瘮?shù)f(x)的定義域?yàn)镽，且滿足f(取x=y=0，得：f(0)+f(0)=f(0)取x=y=1，得f(1)+f(1)=f(2)-2+1，則f(2)=7，取x=y=2，得f(2)+f(2)=f(4)-8+對(duì)于B，取y=1，得f(x)+f(1)=f(x+1)-2x+1，則f(x+1)-f(x)=2x+2，f(x)-f(x-1)=2(x-1)+2,f(x-1)-f(x-2)=2(x-2)+2,?,f(2)-f(1)=2+2，以上各式相加得f(x(-f(1(=+2(x-1(=x2-x+2(x-1(=x2+x-2，所以f(x(=x2-x+1,x≠1，而f(-x(+f(x(=2+2x2，故當(dāng)x<-1時(shí)，有f(x(=2+2x2-f(-x(=x2-x+1對(duì)于D,若f(x-1(是偶函數(shù)，則應(yīng)有f(0(=f(-2(，由f(x)+f(y)=f(x+y)-2xy+1,，f(0)=1，f(2)=7取x=2,y=-2，得f(2)+f(-2)=f(0)+8+1,所以f(-2)=3而f(0(=1≠f(-2(=3，故D錯(cuò)誤;44.(多選題)(河南省七校聯(lián)考2024屆高三第二次聯(lián)合教學(xué)質(zhì)量檢測(cè)數(shù)學(xué)試題)如圖，在長(zhǎng)方體ABCD-A.線段DP長(zhǎng)度的最小值為B.存在點(diǎn)P，使AP+PC=23C.存在點(diǎn)P，使AC⊥平面MNPD.以B為球心，為半徑的球體被平面ABC所以邊AB上高為BDsin∠ABD=2，A正確；sin∠ABA=，則cos∠ABC=cos(∠ABA+(=-sin∠ABA=-，,因此B錯(cuò)；所以∠DMQ+∠MDQ=∠DMQ+∠DNM=，所以DC⊥MN，又長(zhǎng)方體中BC與側(cè)面DCCD垂直，MN?側(cè)面DCCD，因此BC⊥MN，BC與CD是平面BCDA內(nèi)兩條相交直線，因此MN⊥平面BCDA，又AC?平面BCDA，所以MN⊥AC，PQ∩MN=Q，且PQ,MN?平面PMN，所以AC⊥平面PMN，C正確；由BB⊥平面ABCD，AC?平面ABCD得BB⊥AC，又正方形ABCD中，AC⊥BD，BD∩BB=B，BD,BB?平面BDDB，所以AC⊥平面BDDB，而BH?平面BDDB，所以AC⊥BH，AC∩BO=O，AC,BO?平面ABC，所以BH⊥平面ABC，平面ABC截球所得截面圓半徑為r，則r=45.(多選題)(河南省部分名校2023-2024學(xué)年高三階段性測(cè)試(二)數(shù)學(xué)試題)已知函數(shù)f(x)=sin2x+A.f(x)為奇函數(shù)B.f(x)的值域?yàn)镃.f(x)的圖象關(guān)于直線x=對(duì)稱D.f(x)以π為周期滿足f(-x(=-f(x(，所以函數(shù)是奇函數(shù)，故A正確；的值域是(-∞,-3[∪[3,+∞(，故B錯(cuò)誤；f-x(=sin(3π-2x(+=sin2x+=f(x(，所以函數(shù)f(x(關(guān)于x=對(duì)稱，故C正f(x+π(=sin(2x+2π(+=sin2x+=f(x(，所以函數(shù)f(x(的周期為π,故D正確.x-ax3+2ax2lnx≥0恒成立，則實(shí)數(shù)a的可能取值為()A.1B.C.eD.e2則又可化為-a(x-lnx2(≥0?-aln≥0，x2x3’x2x3’’則關(guān)于t的不等式t-alnt≥0在,+∞(恒成立，所以h(t)min=h(e)=e，公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)公眾號(hào)：慧博高中數(shù)學(xué)題庫(kù)A.f(x(的圖象無(wú)對(duì)稱中心B.f(x(+f=2C.f(x(的圖象與g(x(=--1的圖象關(guān)于原點(diǎn)對(duì)稱D.f(x(的圖象與h(x(=ex-1的圖象關(guān)于直線y=x對(duì)稱假設(shè)f(x(的圖象有對(duì)稱中心(x0,y0)，取P(x,y)，其中x>2x0，P關(guān)于點(diǎn)(x0,y0)的對(duì)稱點(diǎn)是Q(2x0-x,2y0-選項(xiàng)C，設(shè)P(x,y)是f(x)圖象關(guān)于原點(diǎn)對(duì)稱的圖象上任一點(diǎn)，它關(guān)于原點(diǎn)的對(duì)稱點(diǎn)為Q(-x,-y)在f(x)的因此-y=+1，即y=--1，所以f(x)的圖象上任一點(diǎn)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)在g(x)的圖象上，111ex-148.(多選題)(河南省部分名校2023-2024學(xué)年高三10月月考數(shù)學(xué)試卷)記函數(shù)f(x(=ex-的零點(diǎn)為A.x0-lnx0=0B.x0∈,C.當(dāng)x>時(shí)，f(x(>x+1D.x0為函數(shù)g(x(=的極值點(diǎn)x0-=0?ex0=?x0=-lnx0?x0+lB選項(xiàng)，f(x(=ex+>0，則f(x(在(0,+∞(上單調(diào)遞增，即f(x(在(0,+∞(上有唯一零點(diǎn)x0.>0.又f>0.C選項(xiàng)，令h(x(=f(x(-x-1=ex-x--1，x∈(0,+∞(.則g/(x(=ex-1+>0，得g(x(在(0,+∞(上單調(diào)遞增，則當(dāng)時(shí)，h(x(=f(x(-x-1>h((>0?f(x(>x+1，故C正確；由A選項(xiàng)分析，-(1++x0+lnx0+1=-(1+得-數(shù)為f/(x(，且滿足f(x+y(=f(x(+f(y(+xy，f(1(=0,f/(1(=，則A.f(x(的圖像關(guān)于點(diǎn)(1,0(成中心對(duì)稱B.f/(2(=對(duì)于B，令y=1，則有f(x+1(=f(x(+f(1(+x=f(x(+x，兩邊同時(shí)求導(dǎo)，得f/(x+1(=f/(x(+1，對(duì)C：令y=1，則有f(x+1(=f(x(+f(1(+x，即f(x+1(-f(x(=x，則f(2024(=f(2024(-f(2023(+f(2023(-f(2022(+?-f(1(+f(1(對(duì)D：令y=1，則有f(x+1(=f(x(+f(1(+x，即f(x+1(=f(x(+x，則f/(x+1(=f/(x(+1，即f/(x+1(-f/(x(=1，又f/(1(=，故f/(k(=+k-1=k-50.(多選題)(河南省部分名校2024屆高三月考(一)數(shù)學(xué)試題)設(shè)函數(shù)f(x)的A.f(x+4π)=f(x)B.f(x)的圖象關(guān)于直線對(duì)稱C.f(x)在區(qū)間,2π(上為增函數(shù)D.方程f(x)-lgx=0僅有4個(gè)實(shí)數(shù)解因?yàn)閒(x-為奇函數(shù)，所以f(x)的圖象關(guān)于點(diǎn)中心因?yàn)閒(x+為偶函數(shù)，所以f(x)的圖象關(guān)于直線對(duì)稱.可畫出y=f(x)的部分圖象大致如下(圖中x軸上相鄰刻度間距離均為對(duì)于A，由圖可知f(x)的最小正周期為2π,所以f(x+4π)=f(x)，故A正確.數(shù)為f’(x)，若xf’(x)-1<0.f(e)=2，則關(guān)于x的不等式f(ex)<x+1的解集為.g(e)=f(e)-lne=1，因此f(ex)<x+1?f所以不等式f(ex)<x+1的解集為(1,+∞). . 可知f(x)為定義在R上的奇函數(shù)；因?yàn)閒(a2(+f(a3+a4(=0，則f(a3+a4(=-f(a2(=f(-a2(，n+3=an(n∈N*(可知：3為數(shù)列{an{的周期，則an+an+1+an+2=0，53.(廣東省七校2024屆高三第二次聯(lián)考數(shù)學(xué)試卷)函數(shù)f(x(=8ln(sinx(+sin22x在個(gè)數(shù)為個(gè).【解析】f(x(=8ln(sinx(+sin22x=8ln(sinx(+1-cos22x=8ln(sinx(+1-(1-2sin2x)2=8ln(sinx(+4sin2x-4sin4x，令t=sinx∈(0,1)，則f(t)=8lnt+4t2-4t4，所以f(t)在(0,1)上單調(diào)遞增，所以f(t)<f(1)=8ln1+4×12-4×14=0，所以函數(shù)f(x(=8ln(sinx(+sin22x在區(qū)間上的零點(diǎn)個(gè)數(shù)為0個(gè).f(x(>0,f(x(單調(diào)遞增，5+125+122