2025新高考數(shù)學(xué)核心母題400道（教師版）

A.{-1,-2}B.{-1,2}C.{-2,4}D.{4}母題2.已知集合A={x∣y=lg(x-x2({,B={x∣x2-cx<0,c>0{,若A∩B=A,則實(shí)數(shù)c的【解析】A={x∣y=lg(x-x2({={x∣x-x2>0{={x∣0<x<1},B={x∣x2-cx<0,c>0{={x∣0<x<c},若A∩B=A,則c≥1,母題3.已知全集U=A,A={1,2,3,4},B={x∣(x+1((x-3(>0,x∈Z{A.2B.4UB={x∣-1≤x≤3,x∈A}={1,2,3};UB(={1,2,3};UB(的子集個(gè)數(shù)為:C+C+C+C=8.母題4.已知集合A={x∣x2-3x-4>0{,B={x∣-1≤x≤3},則(?RA(∩B=()A.(-1,3)B.[-1,3[C.[-1,4[D.(-1,4)【解析】A={x∣x2-3x-4>0{={x∣x>4或x<-1},B={x∣-1≤x≤3},RA={x∣-1≤x≤4},則(?RA(∩B={x∣-1≤x≤3}=[-1,3[,母題5.已知集合M={x∣3x2-5x-2≤0{,N=[m,m+1[,若M∪N=M,則m的取值范圍是()B.(-≤x≤2r{(,由M∪N=M可得N?M,則,解得-≤m≤1,母題6.已知集合A={x∣-1≤x<2},B={x∣x<a},若A∩B≠?,則實(shí)數(shù)a的取值范圍為()A.-1<a≤2B.a>-1C.a>-2D.a≥2【解析】∵A={x∣-1≤x<2},B={x∣x<a},且A∩B≠?,∴a>-1.母題7.設(shè)θ∈(0,π(,則“θ<”是“sinθ<”的()【解析】θ∈(0,π(時(shí),若θ<,則sinθ<,充分性成立;若sinθ<,則0<θ<或<θ<π,必要性不成立;所以“θ<”是“sinθ<”的充分不必要條件.D.既不充分也不必要條件【解析】條件p:|x+1|>2,解得x>1,或x<-3.一p:-3≤x≤1.2<x<3.?q:x≤2,或x≥3.則?q是?p的必要不充分條件.母題10.設(shè)命題p:|4x-3|≤1,命題q:x2-(2a+1(x+a(a+1(≤0,若?p是?q的必要不充分2-(2a+1(x+a(a+1(≤0,解得:a≤x≤a+1.且等號(hào)不能同時(shí)成立.解得0≤a≤.母題11.已知方程x2+(m-3(x+m=0,在下列條件下,求m得范圍:【解析】對(duì)于方程x2+(m-3(x+m=0,令f(x(=x2+(m-3(x+m,(1)方程有兩個(gè)正根,等價(jià)于△=(m-3(2-4m=(m-1((m-9(≥0,且,求得0<m≤1;(2)方程有兩個(gè)負(fù)根等價(jià)于△=(m-3(2-4m=(m-1((m-9(≥0且,求得m≥9；求得<m≤1;<m≤1;求得0<m<;求得<m<0.母題12.對(duì)于任意實(shí)數(shù)x,不等式(a-2(x2-2(a-2(x-4<0恒成立,則實(shí)數(shù)a的取值范圍是()則有-2.[2-4×a-2.×-4<0,解得-2<a<2.母題13.若關(guān)于x的不等式2x2-8x-4+a≤0在1≤x≤3內(nèi)有解,則實(shí)數(shù)a的取值范圍是【解析】原不等式2x2-8x-4+a≤0在1≤x≤3內(nèi)有解等價(jià)于a≤-2x2+8x+4在1≤x≤3內(nèi)有解,設(shè)函數(shù)f(x(=-2x2+8x+4,x∈[1,3[,所以原問(wèn)題等價(jià)于a≤f(x(max又當(dāng)x=2時(shí),f(x(max=12,所以a≤12.x+2y=1,則x+y的取值范圍是()x+2y≥22x?2y=22x+y,所以2x+y≤,即x+y≤-2,當(dāng)且僅當(dāng)2x=2y=,即x=y=-1時(shí)取“=”,所以x+y的取值范圍是(-∞,-2].母題15.已知正數(shù)x,y滿足xy=x+y+3,(1)求x+y的最小值;∴x+y≥2、xy,則xy≤2,∴x+y+3=xy≤2,即(x+y(2-4(x+y(-12≥0,化簡(jiǎn)可得,(x+y+2((x+y-6(≥0,∴x+y≥6,當(dāng)且僅當(dāng)x=y=3時(shí)取等號(hào),可得xy=x+y+3≥2xy+3,即xy-2xy-3≥0,可以變形為(xy-3((xy+1(≥0,當(dāng)且僅當(dāng)x=y=3時(shí)取等號(hào),母題16.已知實(shí)數(shù)x>0,y>0,x+3y=1,則+的最小值為【解析】+=(x+3y(+=4++≥4+2、3母題17.設(shè)x>0，y>0，+2y=2，則x+的最小值為()A.B.2A.B.22C.+2D.3因?yàn)閤>0，y>0，所以x+=(x++y(=+xy++1=+xy+≥+2xy?=+2×O2=+2.2A.4B.42C.42+1D.22+1【解析】=+=+=1++≥1+2?=2、2+1，A.24B.25C.6+42D.62-3+=+(x+y(=13++≥13+2×=25，2+xy+3y2=3，則x+y的最大值為()3+33A.3+33A.C.B.D.【解析】令t=x+y，則x=t-y，方程x2+xy+3y2=3可化為(t-y)2+(t-y(y+3y2-3=0，整理得3y2-ty+t2-3=0，則滿足Δ=(-t)2-12(t2-3(≥0，解得t2≤,所以-6≤t≤6,即x+y≤6,所以x+y的最大值為6 2+xy-y2=1，得(2x-y)(x+y)=1，設(shè)2x-y=t,x+y=，其中t≠0. A.6B.4C.22D.2(1)f(x(=x2+2x-1,g(x(=t2+2t-1;(3)f(x(=x?x+1,g(x(=x2+x;f(x(=|3-x|+1=,顯然f(x(=g(x(,是同一函數(shù).母題24.函數(shù)f(x(=+(x-1(0的定義域?yàn)?2>0,∴函數(shù)f(x(的定義域?yàn)閧x∣-3<x<2且x≠1}.【答案】{x∣-3<x<2且x≠1}.母題25.已知函數(shù)y=x2+2ax+1的2+2ax+1的定義域?yàn)镽,故△=4a2-4≤0,解得:-1≤a≤1,母題26.已知函數(shù)f(x(的定義域?yàn)?-1,1),則函數(shù)g(x(=+f(x-1(的定義域?yàn)?)解得:0<x<2,(1)已知函數(shù)y=f(x(是一次函數(shù),且[f(x([2-3f(x(=4x2-10x+4,求f(x(；(2)已知f(x(是二次函數(shù)且滿足f(0(=1,f(x+1(-f(x(=2x.求f(x(；(3)已知f(2x+1(=4x2+8x+3,求f(x(；(4)已知f(x+=x2+-3,求f(x(；(5)已知f(x(-2=3x+2,求f(x(;(6)已知對(duì)任意實(shí)數(shù)x、y都有f(x+y(-2f(y(=x2+2xy-y2+3x-3y,求f(x(；(7)=,求f(x(的解析式.(8)若f(x(是定義在R上的奇函數(shù),當(dāng)x<0時(shí),f(x(=x(2-x(,求函數(shù)f(x(的解析式.設(shè)f(x(=ax+b(a≠0(,因?yàn)閇f(x([2-3f(x(=4x2-10x+4,所以(ax+b(2-3(ax+b(=4x2-10x+4,故a2x2+(2ab-3a(x+b2-3b=4x2-10x+4,a2=4則{2ab-3a=-10,,解得a=-2,b=4或a=2,b=-1,b2-3ba2=4所以f(x(=-2x+4或f(x(=2x-1.(2)設(shè)二次函數(shù)的解析式為f(x(=ax2+bx+c(a≠0(,由f(0(=1,可得c=1,故f(x(=ax2+bx+1,因?yàn)閒(x+1(-f(x(=2x,則a(x+1(2+b(x+1(+1-(ax2+bx+1(=2x,即2ax+a+b=2x,所以f(x(=x2-x+1.(3)設(shè)t=2x+1,則x=,則f(t(=4×2+8×+3=t2+2t,所以f(x(=x2+2x.(4)因?yàn)閒(x+=x2+-3=(x+2-5,所以f(x(=x2-5(x≤-2或x≥2(.(5)由f(x(-2=3x+2①,由①②可得,f(x(=-x--2.(6)令x=y=0,可得f(0(=2f(0(,則f(0(=0,令y=0,可得f(x(=f(0(+x2+3x,即f(x(=x2+3x.所以f(t(=,故f(x(=(x≠-1(.當(dāng)x>0時(shí),-x<0,∴f(-x(=-x(2+x(,∴-f(x(=f(-x(=-x(2+x(,解得f(x(=x(2+x(母題28.已知函數(shù)f(x(=、x2-4x-5,則該函數(shù)的單調(diào)遞增區(qū)間為2-4x-5≥0,解得x≤-1或x≥5,()函數(shù)t=x2-4x-5的圖象是開(kāi)口向上的拋物線,對(duì)稱軸方程為x=2,∴函數(shù)f(x(=、x2-4x-5的單調(diào)遞增區(qū)間為[5,+∞).母題29.已知f(x(是奇函數(shù),且當(dāng)x<0時(shí),f(x(=-eax.若f(ln2(=8,則a=()A.3B.-C.-3D.-【解析】∵f(x(是奇函數(shù),且當(dāng)x<0時(shí),f(x(=-eax.若f(ln2(=8,∴f(-ln2(=-f(ln2(=-8,則-e-aln2=-8,得e-aln2=8,得ln8=-aln2,即3ln2=-aln2,得-a=3,得a=-3,母題30.若函數(shù)f(x(=sinx?ln(ax+、1+9x2(的圖象關(guān)于y軸對(duì)稱,則實(shí)數(shù)a的值為()則f(-x(=f(x(恒成立,∴sin(-x(?ln(-ax+1+9x2(=sinx?ln(ax+1+9x2(,∴-sinx?ln(-ax+1+9x2(=sinx?ln(ax+1+9x2(,∴l(xiāng)n(-ax+1+9x2(+ln(ax+1+9x2(=0,∴(ax+1+9x2((-ax+1+9x2(=1,題型6單調(diào)性+奇偶性解不等式母題31.已知偶函數(shù)f(x(在區(qū)間[0,+∞)上單調(diào)遞增,則滿足f(2x-1(<f(2(的x的取值范圍是()【解析】根據(jù)題意,f(x(是偶函數(shù),則f(2x-1(=f(|2x-1|(,(<f(2(?|2x-1|(<f(2(?|2x-1|<2,即-2<2x-1<2,解可得-<x<;母題32.已知函數(shù)f(x(=(x-1((ax+b(為偶函數(shù),且在(0,+∞(上單調(diào)遞減,則f(3-x(<0【解析】∵f(x(=ax2+(b-a(x-b為偶函數(shù),所以b-a=0,即b=a,∴f(x(=ax2-a,∴f(3-x(=a(3-x(2-a<0,可化為(3-x(2-1>0,即x2-6x+8>0,解得x<2或x>4母題33.已知函數(shù)f(x(=x3+2x+sinx,x∈[-1,1[,則不等式f(x-1(+f(2x-1(>0的解集【解析】函數(shù)f(x(=x3+2x+sinx,x∈[-1,1[是奇函數(shù),且f/(x(=3x2+2+cosx>0在x∈[-1,1[上恒成立,由f(x-1(+f(2x-1(>0,得f(x-1(>-f(2x-1(=f(1-2x(,r-1≤x-1≤1|-1≤1-2x≤1,解得:<x|(x-1>1-2x∴不等式f(x-1(+f(2x-1(>0的解集為,1.母題34.已知定義在R上的函數(shù)f(x(滿足f(x(=f(x+5(,當(dāng)x∈[-2,0)時(shí),f(x(=-(x+2(2,f(x(=x,則f(1(+f(2(+?+f(2021(=()【解析】由f(x(=f(x+5(可知f(x(周期為5,3)時(shí),f(x(=x,可知f(-2(=0,f(-1(=-1,f(0(=0,f(1(=1,f(2(=2,∴f(-2(+f(-1(+f(0(+f(1(+f(2(=2,∴每個(gè)周期:f(x(+f(x+1(+f(x+2(+f(x+3(+f(x+4(=2,∴f(1(+f(2(+?+f(2021(=f(1(+2×404=809.母題35.已知定義在R上的奇函數(shù)f(x(滿足f(x+8(=f(x(,關(guān)于x=2對(duì)稱且在區(qū)間[0,2[A.f(-25(<f(11(<f(80(B.f(80(<f(11(<f(-25(C.f(11(<f(80(<f(-25(D.f(-25(<f(80(<f(11(【解析】因?yàn)閒(x(滿足f(x+8(=f(x(,所以f(x(的周期為8,則f(-25(=f(-1(,f(80(=f(0(,f(11(=f(3(,又因?yàn)閒(x(為R上的奇函數(shù),且關(guān)于x=2對(duì)稱,所以f(3(=f(1(,則f(x(在[-2,0[上也是單調(diào)遞增,所以f(x(在[-2,2[上單調(diào)遞增,故f(-1(<f(0(<f(1(,所以f(-25(<f(80(<f(11(.母題36.已知函數(shù)f(x((x∈R(滿足f(-x(=2-f(x(,若函數(shù)與y=f(x(圖象的交xi+yi(=()A.0B.mC.2mD.4m【解析】函數(shù)f(x((x∈R(滿足f(-x(=2-f(x(,即為f(x(+f(-x(=2,,2-y1(也為交點(diǎn),(x2,2-y2(也為交點(diǎn),...=m.母題37.設(shè)函數(shù)f(x(是定義在R上的偶函數(shù),對(duì)任意x∈R,都有f(x+2(=f(x-2(,且當(dāng)x-2,0[時(shí),f(x(=((x-1,若在區(qū)間(-2,6]內(nèi)關(guān)于x的方程f(x(-loga(x+2(=都有f(x+2(=f(x-2(,∴f(x-2(=f(x+2(=f(2-x(,即f(x(=f(x+4(,時(shí),-x∈[-2,0[,此時(shí)f(-x(=((-x-1=f(x(,即f(x(=2x-1,分別作出函數(shù)f(x((圖中黑色曲線)和y=loga(x+2((圖中紅色曲線)圖象如圖:由在區(qū)間(-2,6]內(nèi)關(guān)于x的方程f(x(-loga(x+2(=0(a>1(有3個(gè)不同的實(shí)數(shù)根,可得函數(shù)f(x(和y=loga(x+2(圖象有3個(gè)交點(diǎn),故有,求得<a<2,令4x+2=t,由-3≤x≤-1,可得-10≤t≤-2,當(dāng)x=1時(shí),y=,當(dāng)x=2時(shí),y=-1+,【解析】y=x2-2x+3=(x-1(2+2,母題41.函數(shù)y=的值域?yàn)?)A.B.D.2y+2xy+3y=2x2+4x-7∴(y-2(x2+(2y-4(x+3y+7=02+2x+3=(x+1(2+2>0∴函數(shù)的定義域?yàn)镽當(dāng)y≠2時(shí),Δ=(2y-4(2-4(y-2((3y+7(≥0即:(2y+9((y-2(≤0母題43.函數(shù)y=x+4+、9-x2的值域+4[=3cosθ+4+3sinθ=3、2sin(θ++4.∵0≤θ≤π,∴≤θ+≤,∴-≤sin(θ+≤1,+4[.母題44.函數(shù)y=、x2+9+x2-10x+29的最小值為.【解析】y=(x-0(2+(3-0(2+(x-5(2+(0+2(2可以看作是x軸上的動(dòng)點(diǎn)P(x,0(到兩x=3時(shí)ymin==5√2.ymin=5/2.母題45.若不等式m+、-x2-2x≤x+1對(duì)x∈[-2,0[恒成立,則實(shí)數(shù)m的取值范圍是 【解析】不等式即為、-x2-2x≤x+1-m.令f(x(=-x2-2x=-(x+1(2+1,g(x(=x+1-m.件.所以實(shí)數(shù)m的取值范圍是m≤-、2.【答案】m≤-、2.336【解析】依題意f/(x(=(x2-2x-3(=(x-3((x+1(,故函數(shù)在區(qū)間小值為f(3(=f(x(=的最大值是因?yàn)閒(-1(=e,f(1(=所以最大值e.故答案為:e【解析】(1(0.027--2+(2、2-1(0=-49+-1=-45(2)原式=+2×2-=2+2+3(=2-2-3=-3.母題49.計(jì)算log525+log336-log34-+lg5+lg3log32=N=log525+log336-log34-+lg5+lg3log32=log552+(log34+log39(-log34-+lg5+lg3×log32=2+log332-1+lg5+lg3×=2+2-1+lg(5×2(=2+2-1+1=4.2,b==2,從而a>b.c3=(3=25a3=(23=16從而c>a.【答案】c>a>b.母題51.若a>b>1,0<c<1,則()A.ac<bcB.abc>bacC.alogbc<blogacD.logac<logbc因?yàn)閍>b>1,所以ac>bc,故A錯(cuò)誤；因?yàn)?<c<1,所以-1<c-1<0,所以函數(shù)y=xc-1在(0,+∞(上單調(diào)遞減,因?yàn)閍>b>1,所以ac-1<bc-1,即abc>bac,故B正確；因?yàn)?<c<1,所以y=logcx在(0因?yàn)閍>b>1,所以logca<logcb<0,即<<0,所以0>logac>logbc,故D錯(cuò)誤;因?yàn)?<-logac<-logbc,所以-blogac<-alogbc,所以alogbc<blogac,故C正確.A.b<c<aB.a<b<cC.a<c<bD.c<a<b20.2<log21=0,∴a<0,0<c<1,∴a<c<b,x=3y=5z,則()A.2x<3y<5zB.5z<2x<3yC.3y<5z<2xD.3y<2x<5z令2x=3y=5z=k>1.lgk>0.3=69>3>lg2>lg55>0.∴3y<2x<5z.令2x=3y=5z=k>1.lgk>0.>1,可得2x>3y,>1.可得5z>2x.綜上可得:5z>2x>3y.母題54.已知函數(shù)f(x(的定義域是(0,+∞(,且滿足f(xy(=f(x(+f(y(,f((=1,如果對(duì)于0<x<y,都有f(x(>f(y(,則不等式f(-x(+f(3-x(≥-2的解集為f(xy(=f(x(+f(y(∴令x=y=1得f(1(=f(1(+f(1(,再令x=2,y=,∴f(2(=-1令x=y=2,∴令x=y=2得f(4(=f(2(+f(2(=-2,∵對(duì)于0<x<y,都有f(x(>f(y(.∵f(-x(+f(3-x(≥-2.∴f(x(+f(x-3(≥f(4(,∴f[x(x-3([≥f(4(,解得-1≤x<0x+2x-1|x+2的圖象為y=|log2x-1|=圖象為:母題56.下列函數(shù)中,其圖象與函數(shù)y=lnx的圖象關(guān)于直線x=1對(duì)稱的是()A.y=ln(1-x(B.y=ln(2-x(C.y=ln(1+x(D.y=ln(2+x(則:函數(shù)y=lnx的圖象與y=ln(-x(的圖象關(guān)于y軸對(duì)稱.由于函數(shù)y=lnx的圖象關(guān)于直線x=1對(duì)稱.則:把函數(shù)y=ln(-x(的圖象向右平移2個(gè)單位即可得到:y=ln(2-x(.即所求得解析式為:y=ln(2-x(.題型22“知式選圖”母題57.函數(shù)f(x(=在[-π,π[的A.B.C.D.AB.C.AB.C.由y=f(x(=在[-6,6[,知f(-x(==-f(x(,又f(4(=因此排除A,D.D.③④②①A.①④②③B.①④③②C.③②④①母題60.已知定義在R上的奇函數(shù)f(x(滿足f(x-4(=-f(x(,且在區(qū)間[0,2[上是增函數(shù),若方程f(x(=m(m>0(在區(qū)間[-8,8[上有四個(gè)不同的根,則x1+x2+x3+x4=f(x-4(=-f(x(,可得周期T=8,且關(guān)于直線x=2對(duì)稱,根據(jù)圖象,可得x1+x2=-12,x3+x4=4,1+x2+x3+x4=-8,=f(x(的圖象與y=log3|x|的圖象的交點(diǎn)個(gè)數(shù)為-1,1[時(shí),f(x(=x2,∴f(3(=f(1(=1,=y=log33=1,作出函數(shù)f(x(和y=log3|x|的圖象如圖:母題62.已知函數(shù)f(x(=cosx+ex-2(x<0(與g(x(=cosx+ln(x+m(圖象上存在關(guān)于y軸【解析】由題意知,方程f(-x(-g(x(=0在(0,+∞(上有解,即e-x-ln(x+m(-2=0在(0,+∞(上有解,即函數(shù)y=e-x與y=ln(x+m(+2在(0,+∞(上有交點(diǎn),則lnm<1-2或m≤0,母題63.函數(shù)f(x(=1-xlog2x的零點(diǎn)所在區(qū)間是()【解析】∵函數(shù)f(x(=1-xlog2x,f(1(=1-0=1>0,f(2(=1-2=-1<0,根據(jù)函數(shù)零點(diǎn)的判定定理可得函數(shù)f(x(=1-xlog2x的零點(diǎn)所在母題64.若a<b<c,則函數(shù)f(x(=(x-a((x-b(+(x-b(?(x-c(+(x-c((x-a(的兩個(gè)零【解析】∵a<b<c,∴f(a(=(a-b((a-c(>0,f(b(=(b-c((b-a(<0,f(c(=(c-a((c-b(>0,母題65.已知函數(shù)f(x(是定義在R上的奇函數(shù),當(dāng)x>0時(shí),f(x(=lnx-x+2,試判斷f(x(【解析】函數(shù)f(x(是定義在R上的奇函數(shù),所以f(0(=0.x=0是函數(shù)的零點(diǎn),當(dāng)x>0時(shí),f(x(=lnx-x+2,x=1時(shí),函數(shù)取得最大值,f(1(=0-1+2=1>0,x=e2時(shí),f(e2(=2-e2+2<0,所以f(x(的零點(diǎn)個(gè)數(shù)5個(gè).母題66.已知函數(shù)f(x(=與g(x(=1-sinπx,則函數(shù)F(x(=f(x(-g(x(在區(qū)間母題67.已知函數(shù)f(x(=<0(a>0且a≠1(在R上單調(diào)遞減,且關(guān)于x函數(shù)f(x(=<0(a>0且α≠1(f(x(|=2-x有且僅有一個(gè)解,當(dāng)3a>2即a>時(shí),聯(lián)立|x2+(4a-3(x+3a|=2-x,則△=(4a-2(2-4(3a-2(=0,母題68.已知函數(shù)f(x(=x≤1,若關(guān)于x的方程f(x(=-x+a(a∈R(恰有兩個(gè)互A.B.C.D.關(guān)于x的方程f(x(=-x+a(a∈R(恰有兩個(gè)互異的實(shí)數(shù)解,即為y=f(x(和y=-的圖象有兩個(gè)交點(diǎn),考慮直線與在x>1相切,可得x2=1,由△=a2-1=0,解得a=1(-1舍去母題69.已知函數(shù)f(x(=g(x(=f(x(-kx(k∈R(由0<x<π,xsinx=x,即為sinx=1,解得x=;當(dāng)x≥π,x=kx,(k>0(,最多一解,解得0<k≤;又0<x<π時(shí),xsinx=kx,即為sinx=k有兩解,綜上可得0<k≤.母題70.已知函數(shù)f(x(=則函數(shù)g(x(=f(f(x((-的零點(diǎn)個(gè)數(shù)是()A.4B.3C.2D.1當(dāng)x≤0時(shí),由f(x(=得x+1=,即x=-1=-,當(dāng)x>0時(shí),由f(x(=得log2x=,即x==、2,由g(x(=f(f(x((-=0得f(f(x((=,則f(x(=-或f(x(=、2,若f(x(=-,此時(shí)方程f(x(=-有兩個(gè)交點(diǎn),若f(x(=2,此時(shí)方程f(x(=2只有一個(gè)交點(diǎn),則數(shù)g(x(=f(f(x((-的零點(diǎn)個(gè)數(shù)是3個(gè),.【解析】方程f2(x(-f(x(=0可解出f(x(=0或f(x(=1方程f2(x(-f(x(=0的不相等的實(shí)根個(gè)數(shù)即兩個(gè)函數(shù)f(x(=0或f(x(=1的所有不相等的根的個(gè)數(shù)的和,方程的根的個(gè)數(shù)與兩個(gè)函數(shù)y=0,y=1的圖象與函數(shù)f(xy=0的圖象與函數(shù)f(x(的圖象的交點(diǎn)個(gè)數(shù)有三個(gè),故方程f2(x(-f(x(=0有7個(gè)解,故答母題72.已知函數(shù)f(x(=|x|-1,關(guān)于x的方程f2(x(-|f(x(|+k=0,下列四個(gè)結(jié)論中正確的∴當(dāng)a=-1時(shí),f(x(=a有且只有一個(gè)當(dāng)a>-1時(shí),f(x(=a有兩個(gè)不同的解,∵令f(x(=t,則方程f2(x(-|f(x(|+k=0可化為k=|t|-t2,作函數(shù)k=|t|-t2的圖象如右,當(dāng)時(shí),k=|t|-t2有兩個(gè)不同的解,故方程方程f2(x(-|f(x(|+k=0當(dāng)0<k<時(shí),k=|t|-t2有4個(gè)不同的解,且-1<t<1,故方程方程f2(x(-|f(x(|+k=0當(dāng)k=0時(shí),k=|t|-t2有三個(gè)不同的解,分別為-1,0,1；故方程方程f2(x(-|f(x(|+k=0有5個(gè)不同的解,則③正確;當(dāng)k<0時(shí),k=|t|-t2有兩個(gè)不同的解,且t<-1或t>1,故方程方程f2(x(-|f(x(|+k=0有2個(gè)不同的解,則①正確;母題73.函數(shù)f(x(=則函數(shù)y=2[f(x([2-3f(x(+1的零點(diǎn)個(gè)數(shù)為()A.1B.2C.3D.4【解析】由y=2[f(x([2-3f(x(+1=0得[f(x(-1[[2f(x(-1[=0,即f(x(=1或f(x(=,函數(shù)f(x(=當(dāng)f(x(=1時(shí),方程有2個(gè)根,x=e,x=0;當(dāng)f(x(=時(shí),方程有2個(gè)根,x=1舍去,x=e,母題74.已知函數(shù)f(x(=1,x≤0,若方程f2(x(+bf(x(+2=0有8個(gè)相異實(shí)根,則A.(-4,-2)B.(-4,-22(C.(-3,-2)D.(-3,-2、2(【解析】令f(x(=t,則方程f2(x(+bf(x(+2=0?方程t2+bt+2=0.如圖是函數(shù)f(x(=1,x≤0,的圖象,根據(jù)圖象可得:方程f2(x(+bf(x(+2=0有8個(gè)相異實(shí)根?方程t2+bt+2=0.有兩個(gè)不等實(shí)數(shù)解t1,t2,則x+x+x+x+x等于①若x=1,f(x(=1,故12+b+=0,b=-解=1得:x=0或x=2;解得:x=-1或x=∴x+x+x+x+x=12+02+22+(-1(2+32=15.了某地區(qū)新冠肺炎累計(jì)確診病例數(shù)I(t((t的單位:天)的Logistic模型:I(t(=其中K為最大確診病例數(shù).當(dāng)I(t*(=0.95K時(shí),標(biāo)志著已初步遏制疫情,則t*【解析】由已知可得=0.95K,解得e-0.23(t-52(=,兩邊取對(duì)數(shù)有-0.23(t-52(=-ln19≈-3,解得t≈65,可以用指數(shù)模型:I(t(=ert描述累計(jì)感染病例數(shù)I(t(隨時(shí)間t(單位:天)的變化規(guī)律,指數(shù)增長(zhǎng)率r與R0,T近似滿足R0=1+rT.有學(xué)者基于已有數(shù)據(jù)估計(jì)出R0=3.28,T=6.據(jù)()【解析】因?yàn)镽0=3.28,T=6,且R0=1+rT,設(shè)累計(jì)感染病例數(shù)增加3倍需要的時(shí)間約為t則I(t(=2I(0(,即ert=2,即e0.38t=2,兩邊取自然對(duì)數(shù)可得,lne0.38t=ln2,(PB=y(.令P與Q同時(shí)分別從A,C出發(fā),那么,定義x為y的納皮爾對(duì)數(shù)【解析】設(shè)P運(yùn)動(dòng)到第一個(gè)三等分點(diǎn)的時(shí)間為t1,此時(shí)Q運(yùn)動(dòng)的距離為x1,P運(yùn)動(dòng)到中點(diǎn)的時(shí)間為t2,此時(shí)Q運(yùn)動(dòng)的距離為x2,1=107log2=107log,(1)y=x7+x6-3x5;(2)y=x+x-1;(3)y=(3x2+2((x-5(;(6)y=(x+1((x+2((x+3(.【解析】(1)“y=x7+x6-3x5,:y/=7x6+6x5-15x4;(2(“y=x+x-1,:y/=1-x-2;(3(“y=(3x2+2((x-5(,:y/=(3x2+2(/(x-5(+(3x2+2((x-5(/=6x(x-5(+(3x2+2(=9x2-30x+2;(4)“y=,(sinx(/.x-sinx.(x(/:y/=x2=xcosx-sinxx2;(5)“y=,(6)“y=(x+1((x+2((x+3(,:y/=(x+1(/(x+2((x+3(+(x+1((x+2(/(x+3(+(x+1((x+2((x+3(/=(x+2((x+3(+(x+1((x+3(+(x+1((x+2(=(x2+5x+6(+(x2+4x+3(+(x2+3x+2(母題80.已知函數(shù)f(x(=ln(2x-3(+axe-x,若f/(2(=1,則a=.22(=2+ae-2-2ae-2=2-ae-2=1,母題81.已知函數(shù)f(x(的導(dǎo)函數(shù)為f/(x(,f(x(=2x2-3xf/(1(+lnx,則f(1(=.【答案】-【解析】∵f(x(=2x2-3xf/(1(+lnx,x(=4x-3f/(1(+,將x=1代入,得f/(1(=4-3f/(1(+1,得f/(1(=.∴f(1(=2-=-.母題82.已知曲線y=-3lnx的一條切線的斜率為-,則切點(diǎn)的橫坐標(biāo)為令-=-,解得x=2或x=-3(舍去)。母題83.已知曲線y=aex+xlnxA.a=e-1,b=1B.a=e-1,b=-1C.a=e,b=-1D.a=e,b=1x+xlnx,∴y/=aex+lnx+1,母題84.已知函數(shù)f(x(在R上滿足f(x(=2f(2-x(-x2+8x-8,則曲線y=f(x(在點(diǎn)A.y=-2x+3B.y=xC.y=3x-2D.y=2x-1【解析】∵f(x(=2f(2-x(-x2+8x-8,∴f(2-x(=2f(x(-(2-x(2+8(2-x(-8.∴f(2-x(=2f(x(-x2+4x-4+16-8x-8.將f(2-x(代入f(x(=2f(2-x(-x2+8x-8得f(x(=4f(x(-2x2-8x+8-x2+8x-8.∴f(x(=x2,f/(x(=2x∴函數(shù)y=f(x(在(1,f(1((處即y=2x-1.【答案】x-y-1=0故切線方程為y-0=(x-1(?x-y-1=0.故答案為:x-y-1=0母題86.若直線y=kx+b是曲線y=lnx+2的切線,也是曲線y=ln(x+1(的切線,則b=()A.1B.C.1-ln2D.1-2ln2【解析】設(shè)y=kx+b與y=lnx+2和y=ln(x+1(的切點(diǎn)分別為(x1,kx1+b(、(x2,kx2+b(由導(dǎo)數(shù)的幾何意義可得,得x1=x2+1,kx1+b=lnx1+22+bkx1+b=lnx1+2聯(lián)立上述式子解得k=2,x1=,x2=-.代入kx1+b=lnx1+2,解得b=1-ln2.(1)f(x(=x-alnx;(2)g(x(=(x-a-1(ex-(x-a(2.令f(x(=0,得x=a,①當(dāng)a≤0時(shí),f(x(>0在(0,+∞(上恒成立,f(x(<0,g(x(=(x-a(ex-2(x-a(=(x-a((ex-2(,令g(x(=0,得x=a或x=ln2,①當(dāng)a>ln2時(shí),f(x(>0,f(x(<0,③當(dāng)a<ln2時(shí),f(x(>0,f(x(<0,母題88.函數(shù)f(x(=x-sin2x+asinx在R上單調(diào)遞增,則a的取值范圍為【解析】函數(shù)f(x(=x-sin2x+asinx的導(dǎo)數(shù)為f(x(=1-cos2x+acosx,由題意可得f(x(≥0恒成立,即為1-cos2x+acosx≥0,即有-cos2x+acosx≥0,設(shè)t=cosx(-1≤t≤1(,即有5-4t2+3at≥0,當(dāng)0<t≤1時(shí),3a≥4t-,可得3a≥-1,即a≥-；當(dāng)-1≤t<0時(shí),3a≤4t-,x(=2x-kex,∵f(x(在(0,+∞(上單調(diào)遞減,∴f/(x(=2x-kex≤0在(0,+∞(上恒成立,即k≥,母題90.對(duì)任意x∈R,函數(shù)y=f(x(的導(dǎo)數(shù)都存在,若f(x(+f/(x(>0恒成立,且a>0,則下A.f(a(<f(0(B.f(a(>f(0(C.ea?f(a(<f(0(D.ea?f(a(>f(0(x(=ex(f(x(+f/(x((>0,a?f(a(>f(0(.母題91.已知奇函數(shù)f(x(的導(dǎo)函數(shù)為faf(a(≥2f(2-a)+af(a-2(,則實(shí)數(shù)a的取值范圍是()A.(-∞,-1(B.[-1,1[∵f(x(為奇函數(shù),即f(-x(=-f(x(,∴g(-x(=-xf(-x(=xf(x(=g(x(,即函數(shù)g(x(為偶函數(shù)“x∈(0,+∞(時(shí),xf/(x(+f(x(>0.:g/(x(>0,“af(a(≥2f(2-a(+af(a-2(=(2-a(f(2-a(,:g(a(≥g(2-a(,:|a|≥|2-a|解可得,a≥1母題92.若定義在R上的函數(shù)f(x(滿足f(x(+f/(x(<1,f(0(=4,則不等式ex[f(x(-1[>3(e【解析】設(shè)g(x(=exf(x(-ex,(x∈R(,x(=exf(x(+exf/(x(-ex=ex[f(x(+f/(x(-1[,“f(x(+f/(x(<1,:f(x(+f/(x(-1<0,:g/(x(<0,:y=g(x(在定義域上單調(diào)遞減,“exf(x(>ex+3,:g(x(>3,又“g(0(=e0f(0(-e0=4-1=3,:g(x(<g(0(,:x<0母題93.已知函數(shù)f(x(=x3+ax2+bx+a2在x=1處有極值10,則f(2(等于()【解析】f/(x(=3x2+2ax+b,3+2a+b=0rb=-3-2a#/DEL/#ra=4ra=-3:3+2a+b=0rb=-3-2a#/DEL/#ra=4ra=-3①當(dāng)3時(shí),f/(x(=3(x-1(2≥0,:在x=1處不存在極值；a=4f/(x(=3x2+8x-11=(3x+11((a=4a=4:b=-11,:f(2(=8+16-22+16a=4【解析】f/(x(=lnx-aex+1,若函數(shù)f(x(=xlnx-aex有兩個(gè)極值點(diǎn),只需0<a<,母題95.已知函數(shù)f(x(=ax2+bx+clnx(a>0(在x=1和x=2處取得極值,且極大值為-A.0C.2ln2-4D.4ln2-41(=2a+b+c=0①則f(1(=a+b+cln1=a+b=-,③,由①②③得a=,b=-3,c=2,即f(x(=x2-3x+2lnx,由fl(x(>0得4≥x>2或0<x<1,此時(shí)為增函數(shù),由fl(x(<0得1<x<2,此時(shí)f(x(為減函數(shù),則當(dāng)x=1時(shí),f(x(取得極大值,極大值為-,又f(4(=8-12+2ln4=4ln2-4>-,母題96.若函數(shù)f(x(=-x3+ax2+x-1有且只有一個(gè)零點(diǎn),則實(shí)數(shù)a的取值范圍為()【解析】函數(shù)f(x(=-x3+ax2+x-1有且只有一個(gè)零點(diǎn),等價(jià)于關(guān)于x的方程ax2=x3-x設(shè)函數(shù)h(x(=x-,則hl(x(=1+設(shè)g(x(=x3+x-2,hl(x(=3x2+1>0,h(x(為增函數(shù),母題97.若函數(shù)f(x(=2x3-ax2+1(a∈R(在(0,+∞(內(nèi)有且只有一個(gè)x(=2x(3x-a(,x∈(0,+∞(,①當(dāng)a≤0時(shí),f/(x(=2x(3x-a(>0,函數(shù)f(x(在(0,+∞(上單調(diào)遞增,f(0(=1,②當(dāng)a>0時(shí),f/(x(=2x(3x-a(>0的解為x>,又f(x(只有一個(gè)零點(diǎn),f(x(=2x3-3x2+1,f/(x(=6x(x-1(,x∈[-1,1[,f(-1(=-4,f(0(=1,f(1(=0,∴f(x(min=f(-1(=-4,f(x(max=f(0(=1,f(x(max+f(x(min=-4+1=-3.母題98.已知函數(shù)f(x(=ex-ax2-bx.當(dāng)a>0,b=0時(shí),討論函數(shù)f(x(在區(qū)間(0,+∞(上零【解析】當(dāng)x>0,a>0,b=0時(shí),函數(shù)f(x(零點(diǎn)的個(gè)數(shù)即方程ex=ax2根的個(gè)數(shù).x(=,當(dāng),+∞母題99.已知函數(shù)f(x(=lnx-.討論f(x(的單調(diào)性,并證明f(x(有且僅有兩個(gè)零點(diǎn);又∵f(e(<0,f(e2(>0,f(e(?f(e2(<0,故f(x(在定義域內(nèi)有且僅有兩個(gè)零點(diǎn);母題100.函數(shù)f(x(=2ex-a(x-1(2有且只有一個(gè)零點(diǎn),則實(shí)數(shù)a的取值范圍是()【解析】f(x(=2ex-a(x-1(2=0,x=1時(shí)不成立,g(x(單調(diào)遞增;1<x<3時(shí),g/(x(<0時(shí),函數(shù)g(x(單調(diào)遞減;母題101.已知函數(shù)f(x(=ex-m-xlnx,f(x(的導(dǎo)函數(shù)為f/(x(.若g(x(=f/(x(-m+1,討論【解析】g(x(=f/(x(-m+1=ex-m-lnx-m(m>0(.令g(x(=0,得ex-m=lnx+m.x=em(lnx+m(,則xex=xem(lnx+m(=em+lnx(lnx+m(.令φ(x(=xex,則φ(x(=φ(m+lnx(,x(=(x+1(ex>0,∴當(dāng)x>0時(shí)φ(x(=xex為增函數(shù),∴x=m+lnx,即m=x-lnx(x>0(,令t(x(=x-lnx(x>0(,則t/(x(=1-.當(dāng)0<x<1時(shí),t/(x(<0,當(dāng)x>1時(shí),t/(x(>0,∴t(x(min=t(1(=1.當(dāng)m=1時(shí),m=x-lnx有一個(gè)解,即g當(dāng)m>1時(shí),m=x-lnx有兩個(gè)解,即g(x(有兩個(gè)零點(diǎn).母題102.已知函數(shù)f(x(=x3-x2+x.當(dāng)x∈[-2,4[時(shí),求證:x-6≤f(x(≤x.【解析】x-6≤f(x(≤x,即-6≤x3-x2≤0,x∈[-2,4[恒成立,令g(x(=x3-x2,x∈x(=x2-2x=x(x-(,令g/(x(>0得-2<x<0,或<x<4,g/(x(g(x(min=-6,g(0(=0,g(4(=0,故g(x(max=0,故-6≤g(x(≤0,當(dāng)x∈[-2,4[時(shí)恒成立,母題103.設(shè)函數(shù)f(x(=e2x-alnx.證明:當(dāng)a>0時(shí),f(x(≥2a+aln.f(x(=e2x-alnx的定義域?yàn)?0,+∞(,∴f/(x(=2e2x-.當(dāng)a≤0時(shí),f/(x(>0恒成立,故f/(x(沒(méi)有零點(diǎn),當(dāng)a>0時(shí),∵y=e2x為單調(diào)遞增,y=-單調(diào)遞增,∴f/(x(在(0,+∞(單調(diào)遞增,又f/(a(>0,假設(shè)存在b滿足0<b<時(shí),且b<,f/(b(<0,故當(dāng)a=0,所以f(x0(=+2ax0+aln≥2a+aln.故當(dāng)a>0時(shí),f(x(≥2a+aln.母題104.已知函數(shù)f(x(=ex2-xlnx.求證:當(dāng)x>0時(shí),f(x(<xex+.【解析】要證f(x(<xex+,只需證ex-lnx<ex+,即ex-ex<lnx+.令h(x(調(diào)遞增,則h(x(min==0,所以lnx+≥0.再令φ(x(=ex-ex,則φ/(x(=e-exex≤0.因?yàn)閔(x(與φ(x(不同時(shí)為0,所以ex-ex<lnx+,故原不等式成立.母題105.已知函數(shù)f(x(=aln(x-1(+,其中a為正實(shí)數(shù).證明:當(dāng)x>2時(shí),f(x(<ex+(a-1(x-2a.x(<0,即lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)取“=”.當(dāng)x>2時(shí),ln(x-1(<x-2,又a>0,∴aln(x-1(<a(x-2(.要證f(x(<ex+(a-1(x-2a,只需證aln(x-1(+-<ex+(a-1(x-2a,只需證a(x-2(+-<ex+(a-1(x-2a,x-x-->0對(duì)于任意的x>2恒成立.所以h(x(>h(2(=e2-4>0,所以當(dāng)x>2時(shí),f(x(<ex+(a-1(x-2a.A.4B.3C.2D.1【解析】當(dāng)x≤0時(shí),?a>0,xa≤ex-1+x2+1恒成立；當(dāng)x>0,a≤+x+,則0<x<1時(shí),f(x(<0,f(x(遞減;x>1時(shí),f(x(>0,f(x(遞增,則f(x(的最小值為f(1(=3,A.e+4B.e+3C.e+2D.e+1【解析】分類參數(shù)得a≤=+2x+對(duì)于任意x>0恒成立,當(dāng)0<x<1時(shí),g(x(<0,當(dāng)x>1時(shí),g(x(>0,所以g(x(min=g(1(=e+2+2=e+4,所以a≤e+4,()A.設(shè)f(x(=,f/(x(==,由x>0,可得0<x<時(shí),f/(x(>0,f(x(遞增;<x<3時(shí),f/(x(<0,f(x(遞減;x>3時(shí),f/(x(>0,f(x(遞增.且x>3時(shí),f(x(<0,即有x=處,f(x(取得最大值,且母題109.已知函數(shù)f(x(=lnx,若對(duì)任意的x1,x2∈(0,+∞(,都有[f(x1(-f(x2([(x-x(>A.-1B.0C.1D.2【解析】∵f(x(=lnx,∴f(x1(-f(x2(=lnx1-lnx2=ln,f(x1(-f(x2([(x-x(>k(x1x2+x(恒成立,且x1,x2∈(0,+∞(,1x2+x>0,x1+x2>0,令t=,g(t(=tlnt-lnt,(t>0且t≠1(,t(=lnt+1-,令g/(t(=0,得t=1.t(<0,g(t(單調(diào)遞減,t(>0,g(t(單調(diào)遞增,t(min>g(1(=0.母題110.已知函數(shù)f(x(=ax+lnx(a∈R(f(x1(<g(x2(,求f/(x(=a+,x>0?(2分(f/(x(>0,所以函數(shù)f(x(的單調(diào)增區(qū)間為(0,+∞(,?(4分(當(dāng)a<0時(shí),令f/(x(=0,得x=-.當(dāng)x變化時(shí),f/(x(與f(x(變化情況如下表:x-af/(x(+0-f(x(所以函數(shù)f(x(的單調(diào)增區(qū)間為(0,-,函數(shù)f(x(的單調(diào)減區(qū)間為,+∞(?(6分((2)由已知,轉(zhuǎn)化為f(x(max<g(x(max?(8分(所以g(x(max=2?(9分((或者舉出反例:存在f(e3(=ae3+3>2,故不符合題意.)?(10分(當(dāng)a<0時(shí),f(x(在(0,-上單調(diào)遞增,在(-,+∞(上單調(diào)遞減,故f(x(的極大值即為最大值,f(-=-1+ln(-=-1-ln(-a(,?(11分(所以2>-1-ln(-a(,解得a<-.母題111.已知函數(shù)f(x(=ex-mx2-2x.若x∈[0,+∞)時(shí),f(x(>-1恒成立,求m的取值范圍.>-1恒成立,x-2x-+1>mx2恒成立.當(dāng)x=0時(shí),對(duì)于任意m都成立,?(5分(令h(x(=(x-2(ex+2x+e-2,注意到h(1(=0,hx(=(x-1(ex+2,h/(x(=xex>0,0,+∞(單調(diào)遞增,h/(x(>h/(0(=1>0.故m<-1.母題112.已知函數(shù)f(x(=x+-2.若關(guān)于x的不等式f(x(≥alnx+-2恒成立,求實(shí)數(shù)【解析】由題意,x+-alnx-≥0恒成立,令p(x(=x+-alnx-,則p/(x(=由題意需p(x0(≥0,即x0+-(x0-(lnx0- -e,e-|.母題113.已知函數(shù)f(x(=aex-1-lnx+lna(e是自然對(duì)數(shù)的底).解法一:∵f(x(=aex-1-lnx+lna,x(=aex-1-,且a>0.設(shè)g(x(=f/(x(,則g/(x(=aex-1+∴f(x(min=f(1(=1,1(=-1-1((a-1(<0,0>0,使得f/(x0(=aex0-1-=0,且當(dāng)x∈(0,x0(時(shí),f/(x(<0,,+∞(時(shí)f/(x(>0,∴aex0-1=,∴l(xiāng)na+x0-1=-lnx0,因此f(x(min=f(x0(=aex0-1-lnx0+lna=+lna+x0-1+lna≥2lna-1+2?x0=2lna+1>1,當(dāng)0<a<1時(shí),f(1(=a+lna<a<1,∴f(1(<1,f(x(≥1不是恒成立.解法二:f(x(=aex-1-lnx+lna=elna+x-1-lnx+lna≥1等價(jià)于elna+x-1+lna+x-1≥lnx+x=elnx+lnx,令g(x(=ex+x,上述不等式等價(jià)于g(lna+x-1(≥g(lnx(,顯然g(x(為單調(diào)增函數(shù),∴又等價(jià)于lna+x-1≥lnx,即lna≥lnx-x+1,令h(x(=lnx-x+1,則h/(x(=-1=解法三:由(1)得f(x(=ex-1-lnxx-1-lnx≥1---(7分(對(duì)任意x0>0,g(x0(=aex-1-lnx0+lna在(0,+∞(上單調(diào)遞增,所以,當(dāng)a≥1時(shí),f(x(=aex-1-lnx+lna≥ex-1-lna≥1,當(dāng)0<a<1時(shí),f(x(=aex-1-lnx+lna<ex-1-lnx即存在x=1使f(1(=a+lna<1,不合題意,母題114.已知函數(shù)f(x(=,若x1>x2>0,求證:[x1f(x1(-x2f(x2([(x+x(>2x2(x1-x2(.【解析】當(dāng)x1>x2>0時(shí),要證明[x1f(x1(-x2f(x2([(x+x(>2x2(x1-x2(,令>1,設(shè)u(t(=lnt-,則u/(t(=即[x1f(x1(-x2f(x2([(x+x(>2x2(x1-x2(.母題115.已知函數(shù)f(x(=x3+klnx(k∈R(,f/(x(為f(x(的導(dǎo)函數(shù).=f(x(-f/(x(+-1的單調(diào)區(qū)間和極值；所以曲線y=f(x(在點(diǎn)(1,f(1((處的切線方程為y-1=5(x-1(,即y=5x-4.xx=1x(-0+g(x((3)證明:由f(x(=x3+klnx,得f/(x(=3x2+.1>x2,令=t(t>1(,則(x1-x2((f/(x1(+f/(x2((-2(f(x1(-f(x2((=(x1-x2((3x++3x+(x-x+klnx2+3x1x-2kln=x(t3-3t2+3t-1(+k(t--2lnt(①,當(dāng)x>1時(shí),h/(x(=1+-=(1-2>0,所以當(dāng)t>1時(shí),h(t(>h(1(,即t--2lnt>0,因?yàn)閤2≥1,t3-3t2+3t-1=(t-1(3>0,k≥-3,所以x(t3-3t2+3t-1(+k(t--2lnt(≥t3-3t2+3t-1-3(t--2lnt(=t3-3t2+6lnt+-1②,t(>g(1(,即t3-3t2+6lnt+>1,故t3-3t2+6lnt+-1>0③,由①②③可得(x1-x2((f/(x1(+f/(x2((-2(f(x1(-f(x2((>0,(α-=.(α-=sin2=1-cos2=1-(-2=【答案】-當(dāng)k為偶數(shù)時(shí),A=當(dāng)k為奇數(shù)時(shí)=-2.