2025新高考数学核心母题400道（教师版）

A.{-1,-2}B.{-1,2}C.{-2,4}D.{4}母题2.已知集合A={x∣y=lg(x-x2({,B={x∣x2-cx<0,c>0{,若A∩B=A,则实数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,则c≥1,母题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(的子集个数为:C+C+C+C=8.母题4.已知集合A={x∣x2-3x-4>0{,B={x∣-1≤x≤3},则(∁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},则(∁RA(∩B={x∣-1≤x≤3}=[-1,3[,母题5.已知集合M={x∣3x2-5x-2≤0{,N=[m,m+1[,若M∪N=M,则m的取值范围是()B.(-≤x≤2r{(,由M∪N=M可得N⊆M,则,解得-≤m≤1,母题6.已知集合A={x∣-1≤x<2},B={x∣x<a},若A∩B≠⌀,则实数a的取值范围为()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.母题7.设θ∈(0,π(,则“θ<”是“sinθ<”的()【解析】θ∈(0,π(时,若θ<,则sinθ<,充分性成立;若sinθ<,则0<θ<或<θ<π,必要性不成立;所以“θ<”是“sinθ<”的充分不必要条件.D.既不充分也不必要条件【解析】条件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.则¬q是¬p的必要不充分条件.母题10.设命题p:|4x-3|≤1,命题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.且等号不能同时成立.解得0≤a≤.母题11.已知方程x2+(m-3(x+m=0,在下列条件下,求m得范围:【解析】对于方程x2+(m-3(x+m=0,令f(x(=x2+(m-3(x+m,(1)方程有两个正根,等价于△=(m-3(2-4m=(m-1((m-9(≥0,且,求得0<m≤1;(2)方程有两个负根等价于△=(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.母题12.对于任意实数x,不等式(a-2(x2-2(a-2(x-4<0恒成立,则实数a的取值范围是()则有-2.[2-4×a-2.×-4<0,解得-2<a<2.母题13.若关于x的不等式2x2-8x-4+a≤0在1≤x≤3内有解,则实数a的取值范围是【解析】原不等式2x2-8x-4+a≤0在1≤x≤3内有解等价于a≤-2x2+8x+4在1≤x≤3内有解,设函数f(x(=-2x2+8x+4,x∈[1,3[,所以原问题等价于a≤f(x(max又当x=2时,f(x(max=12,所以a≤12.x+2y=1,则x+y的取值范围是()x+2y≥22x⋅2y=22x+y,所以2x+y≤,即x+y≤-2,当且仅当2x=2y=,即x=y=-1时取“=”,所以x+y的取值范围是(-∞,-2].母题15.已知正数x,y满足xy=x+y+3,(1)求x+y的最小值;∴x+y≥2、xy,则xy≤2,∴x+y+3=xy≤2,即(x+y(2-4(x+y(-12≥0,化简可得,(x+y+2((x+y-6(≥0,∴x+y≥6,当且仅当x=y=3时取等号,可得xy=x+y+3≥2xy+3,即xy-2xy-3≥0,可以变形为(xy-3((xy+1(≥0,当且仅当x=y=3时取等号,母题16.已知实数x>0,y>0,x+3y=1,则+的最小值为【解析】+=(x+3y(+=4++≥4+2、3母题17.设x>0，y>0，+2y=2，则x+的最小值为()A.B.2A.B.22C.+2D.3因为x>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，则x+y的最大值为()3+33A.3+33A.C.B.D.【解析】令t=x+y，则x=t-y，方程x2+xy+3y2=3可化为(t-y)2+(t-y(y+3y2-3=0，整理得3y2-ty+t2-3=0，则满足Δ=(-t)2-12(t2-3(≥0，解得t2≤,所以-6≤t≤6,即x+y≤6,所以x+y的最大值为6 2+xy-y2=1，得(2x-y)(x+y)=1，设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=,显然f(x(=g(x(,是同一函数.母题24.函数f(x(=+(x-1(0的定义域为.2>0,∴函数f(x(的定义域为{x∣-3<x<2且x≠1}.【答案】{x∣-3<x<2且x≠1}.母题25.已知函数y=x2+2ax+1的2+2ax+1的定义域为R,故△=4a2-4≤0,解得:-1≤a≤1,母题26.已知函数f(x(的定义域为(-1,1),则函数g(x(=+f(x-1(的定义域为()解得:0<x<2,(1)已知函数y=f(x(是一次函数,且[f(x([2-3f(x(=4x2-10x+4,求f(x(；(2)已知f(x(是二次函数且满足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)已知对任意实数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(是定义在R上的奇函数,当x<0时,f(x(=x(2-x(,求函数f(x(的解析式.设f(x(=ax+b(a≠0(,因为[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则{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)设二次函数的解析式为f(x(=ax2+bx+c(a≠0(,由f(0(=1,可得c=1,故f(x(=ax2+bx+1,因为f(x+1(-f(x(=2x,则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)设t=2x+1,则x=,则f(t(=4×2+8×+3=t2+2t,所以f(x(=x2+2x.(4)因为f(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(,则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(.当x>0时,-x<0,∴f(-x(=-x(2+x(,∴-f(x(=f(-x(=-x(2+x(,解得f(x(=x(2+x(母题28.已知函数f(x(=、x2-4x-5,则该函数的单调递增区间为2-4x-5≥0,解得x≤-1或x≥5,()函数t=x2-4x-5的图象是开口向上的抛物线,对称轴方程为x=2,∴函数f(x(=、x2-4x-5的单调递增区间为[5,+∞).母题29.已知f(x(是奇函数,且当x<0时,f(x(=-eax.若f(ln2(=8,则a=()A.3B.-C.-3D.-【解析】∵f(x(是奇函数,且当x<0时,f(x(=-eax.若f(ln2(=8,∴f(-ln2(=-f(ln2(=-8,则-e-aln2=-8,得e-aln2=8,得ln8=-aln2,即3ln2=-aln2,得-a=3,得a=-3,母题30.若函数f(x(=sinx⋅ln(ax+、1+9x2(的图象关于y轴对称,则实数a的值为()则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(,∴ln(-ax+1+9x2(+ln(ax+1+9x2(=0,∴(ax+1+9x2((-ax+1+9x2(=1,题型6单调性+奇偶性解不等式母题31.已知偶函数f(x(在区间[0,+∞)上单调递增,则满足f(2x-1(<f(2(的x的取值范围是()【解析】根据题意,f(x(是偶函数,则f(2x-1(=f(|2x-1|(,(<f(2(⇒|2x-1|(<f(2(⇒|2x-1|<2,即-2<2x-1<2,解可得-<x<;母题32.已知函数f(x(=(x-1((ax+b(为偶函数,且在(0,+∞(上单调递减,则f(3-x(<0【解析】∵f(x(=ax2+(b-a(x-b为偶函数,所以b-a=0,即b=a,∴f(x(=ax2-a,∴f(3-x(=a(3-x(2-a<0,可化为(3-x(2-1>0,即x2-6x+8>0,解得x<2或x>4母题33.已知函数f(x(=x3+2x+sinx,x∈[-1,1[,则不等式f(x-1(+f(2x-1(>0的解集【解析】函数f(x(=x3+2x+sinx,x∈[-1,1[是奇函数,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的解集为,1.母题34.已知定义在R上的函数f(x(满足f(x(=f(x+5(,当x∈[-2,0)时,f(x(=-(x+2(2,f(x(=x,则f(1(+f(2(+⋯+f(2021(=()【解析】由f(x(=f(x+5(可知f(x(周期为5,3)时,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,∴每个周期: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.母题35.已知定义在R上的奇函数f(x(满足f(x+8(=f(x(,关于x=2对称且在区间[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】因为f(x(满足f(x+8(=f(x(,所以f(x(的周期为8,则f(-25(=f(-1(,f(80(=f(0(,f(11(=f(3(,又因为f(x(为R上的奇函数,且关于x=2对称,所以f(3(=f(1(,则f(x(在[-2,0[上也是单调递增,所以f(x(在[-2,2[上单调递增,故f(-1(<f(0(<f(1(,所以f(-25(<f(80(<f(11(.母题36.已知函数f(x((x∈R(满足f(-x(=2-f(x(,若函数与y=f(x(图象的交xi+yi(=()A.0B.mC.2mD.4m【解析】函数f(x((x∈R(满足f(-x(=2-f(x(,即为f(x(+f(-x(=2,,2-y1(也为交点,(x2,2-y2(也为交点,...=m.母题37.设函数f(x(是定义在R上的偶函数,对任意x∈R,都有f(x+2(=f(x-2(,且当x-2,0[时,f(x(=((x-1,若在区间(-2,6]内关于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(,时,-x∈[-2,0[,此时f(-x(=((-x-1=f(x(,即f(x(=2x-1,分别作出函数f(x((图中黑色曲线)和y=loga(x+2((图中红色曲线)图象如图:由在区间(-2,6]内关于x的方程f(x(-loga(x+2(=0(a>1(有3个不同的实数根,可得函数f(x(和y=loga(x+2(图象有3个交点,故有,求得<a<2,令4x+2=t,由-3≤x≤-1,可得-10≤t≤-2,当x=1时,y=,当x=2时,y=-1+,【解析】y=x2-2x+3=(x-1(2+2,母题41.函数y=的值域为()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∴函数的定义域为R当y≠2时,Δ=(2y-4(2-4(y-2((3y+7(≥0即:(2y+9((y-2(≤0母题43.函数y=x+4+、9-x2的值域+4[=3cosθ+4+3sinθ=3、2sin(θ++4.∵0≤θ≤π,∴≤θ+≤,∴-≤sin(θ+≤1,+4[.母题44.函数y=、x2+9+x2-10x+29的最小值为.【解析】y=(x-0(2+(3-0(2+(x-5(2+(0+2(2可以看作是x轴上的动点P(x,0(到两x=3时ymin==5√2.ymin=5/2.母题45.若不等式m+、-x2-2x≤x+1对x∈[-2,0[恒成立,则实数m的取值范围是 【解析】不等式即为、-x2-2x≤x+1-m.令f(x(=-x2-2x=-(x+1(2+1,g(x(=x+1-m.件.所以实数m的取值范围是m≤-、2.【答案】m≤-、2.336【解析】依题意f/(x(=(x2-2x-3(=(x-3((x+1(,故函数在区间小值为f(3(=f(x(=的最大值是因为f(-1(=e,f(1(=所以最大值e.故答案为: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.母题49.计算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,从而a>b.c3=(3=25a3=(23=16从而c>a.【答案】c>a>b.母题51.若a>b>1,0<c<1,则()A.ac<bcB.abc>bacC.alogbc<blogacD.logac<logbc因为a>b>1,所以ac>bc,故A错误；因为0<c<1,所以-1<c-1<0,所以函数y=xc-1在(0,+∞(上单调递减,因为a>b>1,所以ac-1<bc-1,即abc>bac,故B正确；因为0<c<1,所以y=logcx在(0因为a>b>1,所以logca<logcb<0,即<<0,所以0>logac>logbc,故D错误;因为0<-logac<-logbc,所以-blogac<-alogbc,所以alogbc<blogac,故C正确.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,则()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.综上可得:5z>2x>3y.母题54.已知函数f(x(的定义域是(0,+∞(,且满足f(xy(=f(x(+f(y(,f((=1,如果对于0<x<y,都有f(x(>f(y(,则不等式f(-x(+f(3-x(≥-2的解集为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,∵对于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的图象为y=|log2x-1|=图象为:母题56.下列函数中,其图象与函数y=lnx的图象关于直线x=1对称的是()A.y=ln(1-x(B.y=ln(2-x(C.y=ln(1+x(D.y=ln(2+x(则:函数y=lnx的图象与y=ln(-x(的图象关于y轴对称.由于函数y=lnx的图象关于直线x=1对称.则:把函数y=ln(-x(的图象向右平移2个单位即可得到:y=ln(2-x(.即所求得解析式为:y=ln(2-x(.题型22“知式选图”母题57.函数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母题60.已知定义在R上的奇函数f(x(满足f(x-4(=-f(x(,且在区间[0,2[上是增函数,若方程f(x(=m(m>0(在区间[-8,8[上有四个不同的根,则x1+x2+x3+x4=f(x-4(=-f(x(,可得周期T=8,且关于直线x=2对称,根据图象,可得x1+x2=-12,x3+x4=4,1+x2+x3+x4=-8,=f(x(的图象与y=log3|x|的图象的交点个数为-1,1[时,f(x(=x2,∴f(3(=f(1(=1,=y=log33=1,作出函数f(x(和y=log3|x|的图象如图:母题62.已知函数f(x(=cosx+ex-2(x<0(与g(x(=cosx+ln(x+m(图象上存在关于y轴【解析】由题意知,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,即函数y=e-x与y=ln(x+m(+2在(0,+∞(上有交点,则lnm<1-2或m≤0,母题63.函数f(x(=1-xlog2x的零点所在区间是()【解析】∵函数f(x(=1-xlog2x,f(1(=1-0=1>0,f(2(=1-2=-1<0,根据函数零点的判定定理可得函数f(x(=1-xlog2x的零点所在母题64.若a<b<c,则函数f(x(=(x-a((x-b(+(x-b(⋅(x-c(+(x-c((x-a(的两个零【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,母题65.已知函数f(x(是定义在R上的奇函数,当x>0时,f(x(=lnx-x+2,试判断f(x(【解析】函数f(x(是定义在R上的奇函数,所以f(0(=0.x=0是函数的零点,当x>0时,f(x(=lnx-x+2,x=1时,函数取得最大值,f(1(=0-1+2=1>0,x=e2时,f(e2(=2-e2+2<0,所以f(x(的零点个数5个.母题66.已知函数f(x(=与g(x(=1-sinπx,则函数F(x(=f(x(-g(x(在区间母题67.已知函数f(x(=<0(a>0且a≠1(在R上单调递减,且关于x函数f(x(=<0(a>0且α≠1(f(x(|=2-x有且仅有一个解,当3a>2即a>时,联立|x2+(4a-3(x+3a|=2-x,则△=(4a-2(2-4(3a-2(=0,母题68.已知函数f(x(=x≤1,若关于x的方程f(x(=-x+a(a∈R(恰有两个互A.B.C.D.关于x的方程f(x(=-x+a(a∈R(恰有两个互异的实数解,即为y=f(x(和y=-的图象有两个交点,考虑直线与在x>1相切,可得x2=1,由△=a2-1=0,解得a=1(-1舍去母题69.已知函数f(x(=g(x(=f(x(-kx(k∈R(由0<x<π,xsinx=x,即为sinx=1,解得x=;当x≥π,x=kx,(k>0(,最多一解,解得0<k≤;又0<x<π时,xsinx=kx,即为sinx=k有两解,综上可得0<k≤.母题70.已知函数f(x(=则函数g(x(=f(f(x((-的零点个数是()A.4B.3C.2D.1当x≤0时,由f(x(=得x+1=,即x=-1=-,当x>0时,由f(x(=得log2x=,即x==、2,由g(x(=f(f(x((-=0得f(f(x((=,则f(x(=-或f(x(=、2,若f(x(=-,此时方程f(x(=-有两个交点,若f(x(=2,此时方程f(x(=2只有一个交点,则数g(x(=f(f(x((-的零点个数是3个,.【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的不相等的实根个数即两个函数f(x(=0或f(x(=1的所有不相等的根的个数的和,方程的根的个数与两个函数y=0,y=1的图象与函数f(xy=0的图象与函数f(x(的图象的交点个数有三个,故方程f2(x(-f(x(=0有7个解,故答母题72.已知函数f(x(=|x|-1,关于x的方程f2(x(-|f(x(|+k=0,下列四个结论中正确的∴当a=-1时,f(x(=a有且只有一个当a>-1时,f(x(=a有两个不同的解,∵令f(x(=t,则方程f2(x(-|f(x(|+k=0可化为k=|t|-t2,作函数k=|t|-t2的图象如右,当时,k=|t|-t2有两个不同的解,故方程方程f2(x(-|f(x(|+k=0当0<k<时,k=|t|-t2有4个不同的解,且-1<t<1,故方程方程f2(x(-|f(x(|+k=0当k=0时,k=|t|-t2有三个不同的解,分别为-1,0,1；故方程方程f2(x(-|f(x(|+k=0有5个不同的解,则③正确;当k<0时,k=|t|-t2有两个不同的解,且t<-1或t>1,故方程方程f2(x(-|f(x(|+k=0有2个不同的解,则①正确;母题73.函数f(x(=则函数y=2[f(x([2-3f(x(+1的零点个数为()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(=,函数f(x(=当f(x(=1时,方程有2个根,x=e,x=0;当f(x(=时,方程有2个根,x=1舍去,x=e,母题74.已知函数f(x(=1,x≤0,若方程f2(x(+bf(x(+2=0有8个相异实根,则A.(-4,-2)B.(-4,-22(C.(-3,-2)D.(-3,-2、2(【解析】令f(x(=t,则方程f2(x(+bf(x(+2=0⇔方程t2+bt+2=0.如图是函数f(x(=1,x≤0,的图象,根据图象可得:方程f2(x(+bf(x(+2=0有8个相异实根⇔方程t2+bt+2=0.有两个不等实数解t1,t2,则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.了某地区新冠肺炎累计确诊病例数I(t((t的单位:天)的Logistic模型:I(t(=其中K为最大确诊病例数.当I(t*(=0.95K时,标志着已初步遏制疫情,则t*【解析】由已知可得=0.95K,解得e-0.23(t-52(=,两边取对数有-0.23(t-52(=-ln19≈-3,解得t≈65,可以用指数模型:I(t(=ert描述累计感染病例数I(t(随时间t(单位:天)的变化规律,指数增长率r与R0,T近似满足R0=1+rT.有学者基于已有数据估计出R0=3.28,T=6.据()【解析】因为R0=3.28,T=6,且R0=1+rT,设累计感染病例数增加3倍需要的时间约为t则I(t(=2I(0(,即ert=2,即e0.38t=2,两边取自然对数可得,lne0.38t=ln2,(PB=y(.令P与Q同时分别从A,C出发,那么,定义x为y的纳皮尔对数【解析】设P运动到第一个三等分点的时间为t1,此时Q运动的距离为x1,P运动到中点的时间为t2,此时Q运动的距离为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(母题80.已知函数f(x(=ln(2x-3(+axe-x,若f/(2(=1,则a=.22(=2+ae-2-2ae-2=2-ae-2=1,母题81.已知函数f(x(的导函数为f/(x(,f(x(=2x2-3xf/(1(+lnx,则f(1(=.【答案】-【解析】∵f(x(=2x2-3xf/(1(+lnx,x(=4x-3f/(1(+,将x=1代入,得f/(1(=4-3f/(1(+1,得f/(1(=.∴f(1(=2-=-.母题82.已知曲线y=-3lnx的一条切线的斜率为-,则切点的横坐标为令-=-,解得x=2或x=-3(舍去)。母题83.已知曲线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,母题84.已知函数f(x(在R上满足f(x(=2f(2-x(-x2+8x-8,则曲线y=f(x(在点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.将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∴函数y=f(x(在(1,f(1((处即y=2x-1.【答案】x-y-1=0故切线方程为y-0=(x-1(⇒x-y-1=0.故答案为:x-y-1=0母题86.若直线y=kx+b是曲线y=lnx+2的切线,也是曲线y=ln(x+1(的切线,则b=()A.1B.C.1-ln2D.1-2ln2【解析】设y=kx+b与y=lnx+2和y=ln(x+1(的切点分别为(x1,kx1+b(、(x2,kx2+b(由导数的几何意义可得,得x1=x2+1,kx1+b=lnx1+22+bkx1+b=lnx1+2联立上述式子解得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当a≤0时,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当a>ln2时,f(x(>0,f(x(<0,③当a<ln2时,f(x(>0,f(x(<0,母题88.函数f(x(=x-sin2x+asinx在R上单调递增,则a的取值范围为【解析】函数f(x(=x-sin2x+asinx的导数为f(x(=1-cos2x+acosx,由题意可得f(x(≥0恒成立,即为1-cos2x+acosx≥0,即有-cos2x+acosx≥0,设t=cosx(-1≤t≤1(,即有5-4t2+3at≥0,当0<t≤1时,3a≥4t-,可得3a≥-1,即a≥-；当-1≤t<0时,3a≤4t-,x(=2x-kex,∵f(x(在(0,+∞(上单调递减,∴f/(x(=2x-kex≤0在(0,+∞(上恒成立,即k≥,母题90.对任意x∈R,函数y=f(x(的导数都存在,若f(x(+f/(x(>0恒成立,且a>0,则下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(.母题91.已知奇函数f(x(的导函数为faf(a(≥2f(2-a)+af(a-2(,则实数a的取值范围是()A.(-∞,-1(B.[-1,1[∵f(x(为奇函数,即f(-x(=-f(x(,∴g(-x(=-xf(-x(=xf(x(=g(x(,即函数g(x(为偶函数“x∈(0,+∞(时,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母题92.若定义在R上的函数f(x(满足f(x(+f/(x(<1,f(0(=4,则不等式ex[f(x(-1[>3(e【解析】设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(在定义域上单调递减,“exf(x(>ex+3,:g(x(>3,又“g(0(=e0f(0(-e0=4-1=3,:g(x(<g(0(,:x<0母题93.已知函数f(x(=x3+ax2+bx+a2在x=1处有极值10,则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当3时,f/(x(=3(x-1(2≥0,:在x=1处不存在极值；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,若函数f(x(=xlnx-aex有两个极值点,只需0<a<,母题95.已知函数f(x(=ax2+bx+clnx(a>0(在x=1和x=2处取得极值,且极大值为-A.0C.2ln2-4D.4ln2-41(=2a+b+c=0①则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,此时为增函数,由fl(x(<0得1<x<2,此时f(x(为减函数,则当x=1时,f(x(取得极大值,极大值为-,又f(4(=8-12+2ln4=4ln2-4>-,母题96.若函数f(x(=-x3+ax2+x-1有且只有一个零点,则实数a的取值范围为()【解析】函数f(x(=-x3+ax2+x-1有且只有一个零点,等价于关于x的方程ax2=x3-x设函数h(x(=x-,则hl(x(=1+设g(x(=x3+x-2,hl(x(=3x2+1>0,h(x(为增函数,母题97.若函数f(x(=2x3-ax2+1(a∈R(在(0,+∞(内有且只有一个x(=2x(3x-a(,x∈(0,+∞(,①当a≤0时,f/(x(=2x(3x-a(>0,函数f(x(在(0,+∞(上单调递增,f(0(=1,②当a>0时,f/(x(=2x(3x-a(>0的解为x>,又f(x(只有一个零点,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.母题98.已知函数f(x(=ex-ax2-bx.当a>0,b=0时,讨论函数f(x(在区间(0,+∞(上零【解析】当x>0,a>0,b=0时,函数f(x(零点的个数即方程ex=ax2根的个数.x(=,当,+∞母题99.已知函数f(x(=lnx-.讨论f(x(的单调性,并证明f(x(有且仅有两个零点;又∵f(e(<0,f(e2(>0,f(e(⋅f(e2(<0,故f(x(在定义域内有且仅有两个零点;母题100.函数f(x(=2ex-a(x-1(2有且只有一个零点,则实数a的取值范围是()【解析】f(x(=2ex-a(x-1(2=0,x=1时不成立,g(x(单调递增;1<x<3时,g/(x(<0时,函数g(x(单调递减;母题101.已知函数f(x(=ex-m-xlnx,f(x(的导函数为f/(x(.若g(x(=f/(x(-m+1,讨论【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(,则xex=xem(lnx+m(=em+lnx(lnx+m(.令φ(x(=xex,则φ(x(=φ(m+lnx(,x(=(x+1(ex>0,∴当x>0时φ(x(=xex为增函数,∴x=m+lnx,即m=x-lnx(x>0(,令t(x(=x-lnx(x>0(,则t/(x(=1-.当0<x<1时,t/(x(<0,当x>1时,t/(x(>0,∴t(x(min=t(1(=1.当m=1时,m=x-lnx有一个解,即g当m>1时,m=x-lnx有两个解,即g(x(有两个零点.母题102.已知函数f(x(=x3-x2+x.当x∈[-2,4[时,求证: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,当x∈[-2,4[时恒成立,母题103.设函数f(x(=e2x-alnx.证明:当a>0时,f(x(≥2a+aln.f(x(=e2x-alnx的定义域为(0,+∞(,∴f/(x(=2e2x-.当a≤0时,f/(x(>0恒成立,故f/(x(没有零点,当a>0时,∵y=e2x为单调递增,y=-单调递增,∴f/(x(在(0,+∞(单调递增,又f/(a(>0,假设存在b满足0<b<时,且b<,f/(b(<0,故当a=0,所以f(x0(=+2ax0+aln≥2a+aln.故当a>0时,f(x(≥2a+aln.母题104.已知函数f(x(=ex2-xlnx.求证:当x>0时,f(x(<xex+.【解析】要证f(x(<xex+,只需证ex-lnx<ex+,即ex-ex<lnx+.令h(x(调递增,则h(x(min==0,所以lnx+≥0.再令φ(x(=ex-ex,则φ/(x(=e-exex≤0.因为h(x(与φ(x(不同时为0,所以ex-ex<lnx+,故原不等式成立.母题105.已知函数f(x(=aln(x-1(+,其中a为正实数.证明:当x>2时,f(x(<ex+(a-1(x-2a.x(<0,即lnx≤x-1,当且仅当x=1时取“=”.当x>2时,ln(x-1(<x-2,又a>0,∴aln(x-1(<a(x-2(.要证f(x(<ex+(a-1(x-2a,只需证aln(x-1(+-<ex+(a-1(x-2a,只需证a(x-2(+-<ex+(a-1(x-2a,x-x-->0对于任意的x>2恒成立.所以h(x(>h(2(=e2-4>0,所以当x>2时,f(x(<ex+(a-1(x-2a.A.4B.3C.2D.1【解析】当x≤0时,∀a>0,xa≤ex-1+x2+1恒成立；当x>0,a≤+x+,则0<x<1时,f(x(<0,f(x(递减;x>1时,f(x(>0,f(x(递增,则f(x(的最小值为f(1(=3,A.e+4B.e+3C.e+2D.e+1【解析】分类参数得a≤=+2x+对于任意x>0恒成立,当0<x<1时,g(x(<0,当x>1时,g(x(>0,所以g(x(min=g(1(=e+2+2=e+4,所以a≤e+4,()A.设f(x(=,f/(x(==,由x>0,可得0<x<时,f/(x(>0,f(x(递增;<x<3时,f/(x(<0,f(x(递减;x>3时,f/(x(>0,f(x(递增.且x>3时,f(x(<0,即有x=处,f(x(取得最大值,且母题109.已知函数f(x(=lnx,若对任意的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(单调递减,t(>0,g(t(单调递增,t(min>g(1(=0.母题110.已知函数f(x(=ax+lnx(a∈R(f(x1(<g(x2(,求f/(x(=a+,x>0⋯(2分(f/(x(>0,所以函数f(x(的单调增区间为(0,+∞(,⋯(4分(当a<0时,令f/(x(=0,得x=-.当x变化时,f/(x(与f(x(变化情况如下表:x-af/(x(+0-f(x(所以函数f(x(的单调增区间为(0,-,函数f(x(的单调减区间为,+∞(⋯(6分((2)由已知,转化为f(x(max<g(x(max⋯(8分(所以g(x(max=2⋯(9分((或者举出反例:存在f(e3(=ae3+3>2,故不符合题意.)⋯(10分(当a<0时,f(x(在(0,-上单调递增,在(-,+∞(上单调递减,故f(x(的极大值即为最大值,f(-=-1+ln(-=-1-ln(-a(,⋯(11分(所以2>-1-ln(-a(,解得a<-.母题111.已知函数f(x(=ex-mx2-2x.若x∈[0,+∞)时,f(x(>-1恒成立,求m的取值范围.>-1恒成立,x-2x-+1>mx2恒成立.当x=0时,对于任意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,+∞(单调递增,h/(x(>h/(0(=1>0.故m<-1.母题112.已知函数f(x(=x+-2.若关于x的不等式f(x(≥alnx+-2恒成立,求实数【解析】由题意,x+-alnx-≥0恒成立,令p(x(=x+-alnx-,则p/(x(=由题意需p(x0(≥0,即x0+-(x0-(lnx0- -e,e-|.母题113.已知函数f(x(=aex-1-lnx+lna(e是自然对数的底).解法一:∵f(x(=aex-1-lnx+lna,x(=aex-1-,且a>0.设g(x(=f/(x(,则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,且当x∈(0,x0(时,f/(x(<0,,+∞(时f/(x(>0,∴aex0-1=,∴lna+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,当0<a<1时,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等价于elna+x-1+lna+x-1≥lnx+x=elnx+lnx,令g(x(=ex+x,上述不等式等价于g(lna+x-1(≥g(lnx(,显然g(x(为单调增函数,∴又等价于lna+x-1≥lnx,即lna≥lnx-x+1,令h(x(=lnx-x+1,则h/(x(=-1=解法三:由(1)得f(x(=ex-1-lnxx-1-lnx≥1---(7分(对任意x0>0,g(x0(=aex-1-lnx0+lna在(0,+∞(上单调递增,所以,当a≥1时,f(x(=aex-1-lnx+lna≥ex-1-lna≥1,当0<a<1时,f(x(=aex-1-lnx+lna<ex-1-lnx即存在x=1使f(1(=a+lna<1,不合题意,母题114.已知函数f(x(=,若x1>x2>0,求证:[x1f(x1(-x2f(x2([(x+x(>2x2(x1-x2(.【解析】当x1>x2>0时,要证明[x1f(x1(-x2f(x2([(x+x(>2x2(x1-x2(,令>1,设u(t(=lnt-,则u/(t(=即[x1f(x1(-x2f(x2([(x+x(>2x2(x1-x2(.母题115.已知函数f(x(=x3+klnx(k∈R(,f/(x(为f(x(的导函数.=f(x(-f/(x(+-1的单调区间和极值；所以曲线y=f(x(在点(1,f(1((处的切线方程为y-1=5(x-1(,即y=5x-4.xx=1x(-0+g(x((3)证明:由f(x(=x3+klnx,得f/(x(=3x2+.1>x2,令=t(t>1(,则(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,当x>1时,h/(x(=1+-=(1-2>0,所以当t>1时,h(t(>h(1(,即t--2lnt>0,因为x2≥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】-当k为偶数时,A=当k为奇数时=-2.