2025新高考重难点之构造函数 解析版

1.对于不等式f/(x(>k(k≠0(，构造函数g(x(=f(x(-kx+b2.对于不等式xf/(x(+f(x(>0，构造函数g(x(=xf(x(3.对于不等式xf/(x(-f(x(>0，构造函数g(x(=(x≠0(4.对于不等式xf/(x(+nf(x(>0，构造函数g(x(=xnf(x)5.对于不等式xf/(x(-nf(x(>0，构造函数g(x(=6.对于不等式f/(x(-f(x(>0，构造函数g(x(=7.对于不等式f/(x(+f(x(>0，构造函数g(x(=exf(x)8.对于不等式f/(x(+kf(x(>0，构造函数g(x(=ekxf(x)9.对于不等式f/(x(sinx+f(x(cosx>0，构造函数g(x(=sinxf(x)10.对于不等式f/(x(sinx-f(x(cosx>0，构造函数g(x(=11.对于不等式f/(x(cosx-f(x(sinx>0，构造函数g(x(=cosxf(x)12.对于不等式f/(x(cosx+f(x(sinx>0，构造函数g(x(=1.(23-24高三下·重庆)已知函数f(x(的定义域为(-∞,0(，f(-1(=-1，其导函数f/(x(满足xf/(x(-2f(x(>0，则不等式f(x+2025(+(x+2025(2<0的解集为()A.(-2026,0(B.(-2026,-2025(C.(-∞,-2026(D.(-∞,-2025(11【详解】根据题意可令g(x(=(x<0(⇒g/(x(=所以g(x(=在(-∞,0(上单调递减，则原不等式等价于<-1，由g(x+2025(=<-1=g(-1(⇒0>x+2025>-1，解之得x∈(-2026,-2025(.2.(2021·安徽高三月考(理))设函数f(x(是定义在(0,+∞(上的可导函数，其导函数为f'(x(，且有2f(x(>xf'(x(，则不等式4f(x-2021(>(x-2021(2f(2(的解集为()A.(2021,2023(B.(0,2022(C.(0,2020(D.(2022,+∞(构造函数利用它的导数确定单调性后可得不等式的解集.由条件'=<0，∴在(0,+∞(上单调递减，形式(一般需要适当变形).3.(2022·四川省眉山第一中学模拟预测(理))已知可导函数f(x)的定义域为(0,+∞)，满足xf/(x)-2f(x)<0，且f(2)=4，则不等式f(x)>x2的解集是.(0,2)调性即可求解.因为x>0，xf/(x)-2f(x)<0，所以g/(x)<0，可得g(x)在(0,+∞)上单调递减，224.(23-24高三上·云南昆明)已知定义域为R的函数f(x(，对任意的x∈R都有f/(x(>2x，且f(1(=2，则不等式f(2x(-4x2-1>0的解集为()A.(0,+∞(B.,+∞(C.(1,+∞(D.(2,+∞(【分析】令g(x)=f(x)-x2，由题意可得出g(x)在R上单调递增，所以不等式f(2x(-4x2-1>0可变形为得g(2x)>g(1)，由单调性解不等式即可得出答案.【详解】令g(x)=f(x)-x2，则g/(x)=f/(x)-2x>0，则g(x)在R上单调递增，g(1)=f(1)-1=1，由f(2x(-4x2-1>0可得f(2x(-(2x(2>1，即g(2x)>g(1)，1.(22-23高三下·广东)已知f(x)是定义在R上的偶函数，当x>0时，有xf/(x)+2f(x)<0恒成立，则()A.4f(1)>fB.<C.9f>4f(-D.9f(-1)<f(-【详解】设g(x)=x2f(x(，则g/(x)=2xf(x(+x2f/(x(=x[2f(x(+xf/(x([，因为当x>0时，有xf/(x)+2f(所以g(x)在(0,+∞(单调递减；又f(x)是定义在R上的偶函数，则g(-x)=x2f(-x(=x2f(x(=g(x(，故g(x)为偶函数，则g>g(1(⇒f>f(1(⇒4f(1)<f，A选项错误；g(2)>g(3(⇒4f(2(>9f(3(⇒>，B选项错误；公众号：慧博高中数学最新试题公众号：慧博高中数学最新试题g<g=g(-⇒f<f(-⇒9f<4f(-，C选项错误；=g(1(<g((=g(-(⇒f(-1(<(⇒9f<f(-选项正确；2.(22-23高三下·广东东莞)已知函数f(x(的定义域为(-∞,0(,其导函数f/(x(满足xf/(x(-2f(x(>0,则不等式f(x+2023(-(x+2023(2f(-1(<0的解集为()A.(-2024,-2023)B.(-2024,0)C.(-∞,-2023)D.(-∞,-2024)由题可得当x∈(-∞,0(时，xf(x(-2f(x(>0，构造函数g(x)=，可判断g(x)在(-∞,0)上的单调性，由题意知，当x∈(-∞,0)时，xf/(x)-2f(x)>0，设设则所以g(x(在(-∞,0)上单调递减，不等式f(x+2023)-(x+2023)2f(-1)<0等价于解得-2024<x<-2023.3.(22-23高三上·山东泰安·阶段练习)已知f(x(是定义在R上的偶函数，f/(x(是f(x(的导函数，当x≥0时，f/(x(-2x>0，且f(1(=2，则f(x(>x2+1的解集是()A.(-1,0(∪(1,+∞)B.(-∞,-1(∪(1,+∞(C.(-1,0(∪(0,1(D.(-∞,-1(∪(0,1(【分析】根据f/(x(-2x>0构造函数g(x(=f(x(-x2，然后分析g(x(的奇偶性和单调性，再将问题转化为解不等式g(x(>g(1(，由此可得结果.【详解】构造函数g(x(=f(x(-x2，因为f(x(是R上的偶函数且y=x2也是R上的偶函数，所以g(x(是R上的偶函数，因为x≥0时，g/(x(=f/(x(-2x>0，所以g(x(在[0,+∞(上单调递增，44所以g(x(在(-∞,0(上单调递减，又因为f(x(>x2+1，所以g(x(>1且g(1(=f(1(-12=1，4.(2024·陕西商洛·模拟预测)已知函数f(x(=2xlnx-ax2，若对任意的x1,x2∈(0,+∞(，当x1>x2时，都有2x1+f(x2(>2x2+f(x1(，则实数a的取值范围为()A.,+∞(B.[1,+∞(C.,+∞(D.[2,+∞(公众号：慧博高中数学最新试题【分析】构造函数F(x(=f(x(-2x，求导，分离参数求最值即可.公众号：慧博高中数学最新试题+f(x2(>2x2+f(x1(等价于f(x1(-2x1<f(x2(-2x2，令F(x(=f(x(-2x,x∈(0,+∞(，根据题意对任意的x1,x2∈(0,+∞(，时，F(x1(<F(x2(，所以函数F(x(=f(x(-2x在(0,+∞(上单调递减，所以F/(x(=f/(x(-2=2lnx-2ax≤0在(0,+∞(上恒成立，即≤a在(0,+∞(上恒成立.令g(x(=,x∈(0,+∞(，则g/(x(=，所以当x∈(0,e(时，g/(x(>0，g(x(单调递增，,+∞(时，g/(x(<0,g(x(单调递减.所以g(x)max=g(e(=，所以a≥.(1)a≥f(x(恒成立⇔a≥f(x(max；(2)a≤f(x(恒成立⇔a≤f(x(min.5.(2023·广东佛山·校考模拟预测)已知f/(x(是函数y=f(x((x∈R(的导函数，对于任意的x∈R都有f/(x(+f(x(>1，且f(0(=2023，则不等式exf(x(>ex+2022的解集是()A.(2022,+∞(B.(-∞,0(∪(2023,+∞(C.(-∞,0(∪(0,+∞(D.(0,+∞(法二：构造辅助函数.令g(x(=exf(x(-ex，则g/(x(=ex(f(x(+f/(x(-1(>0，所以g(x(在R上单调递增，55又因为g(0(=f(0(-1=2022，所以exf(x(>ex+2022⇔g(x(>g(0(，所以x>0，A.e2f(-1(<1B.f(1(>e2C.f<eD.f(1(>ef2x-2f(x(e2xf/(x)-2x-2f(x(e2xf/(x)-2f(x(因为f/(x)-2f(x(<0在R上恒成立，所以g/(x(<0在R上恒成立，故g(x(在R上单调递减，所以g(-1(>g(0(，=e2f(-1(>=1，故A不正确；7.(22-23高三下·天津)已知可导函数f(x(的导函数为f/(x(,f(0(=2023，若对任意f(x(<f/(x(，则不等式f(x(<2023ex的解集为()A.(0,+∞(,+∞C.-D.(-∞,0(【分析】由题意f/(x(-f(x(>0【详解】由题意对任意的x∈R，都有f(x(<f/(x(，即f/(x(-f(x(>0，即g(x)=为R上的增函数，公众号：慧博高中数学最新试题公众号：慧博高中数学最新试题8.(22-23高三下·全国)定义域为R的可导函数f(x(的导函数为f/(x(，满足f/(x(-f(x(<0，且f(0(A.(0,+∞(B.(2,+∞(C.(-∞,0(D.(-∞,2(不等式的解集.【详解】由f/(x(-f(x(<0得：<0，即/<0，令h(x(=，则h/(x(<0，∴h(x(在R上单调递减，1.(2023·山东烟台·二模)已知函数f(x(的定义域为R，其导函数为f/(x(，且满足f/(x(+f(x(=e-x，f(0(=0，则不等式(e2x-1(f(x(<e-的解集为().A.(-1,B.,e(C.(-1,1(D.(-1,e(【分析】先由题中条件求出f(x(=xe-x，根据不等式可构造g(x(=xex-xe-x，利用g(x(为偶函数且在区间[0,+∞(上单调递增可解.【详解】由f/(x(+f(x(=e-x得exf/(x(+exf(x(=1，即[exf(x([/=1，可设exf(x(=x+m，所以f(x(=xe-x，(e2x-1(f(x(<e-可化为xe-x(e2x-1(<e-，即e-，设g(x(=xex-xe-x，因g(-x(=-xe-x+xex=g(x(，故g(x(为偶函数/(x(=ex+xex+xe-x-e-x，77x+xe-x≥0，ex-e-x≥0，故g/(x(=ex+xex+xe-x-e-x≥0，所以g(x(在区间[0,+∞(上单调递增，因g(1(=e-e-1，所以当x≥0时ge-的解集为[0,1(，又因g(x(为偶函数，故g(x(<e-的解集为(-1,1(.2.(2022·青海西宁·二模(理))已知定义在R上的可导函数f(x(的导函数为f/(x(，满足f/(x(<f(x(，且f(x+3(为偶函数，f(6(=1，则不等式f(x(>ex的解集为.【答案】(-∞,0(g(0(，而原不等式等价于g(x(>g(0(，即/(x(<f(x(，所以g/(x(<0，即g(x(在R上是减函数，因为f(x+3(为偶函数，所以f(x+3(图象关于y轴对称，而f(x+3(向右平移3个单位可得f(x(，所以f(x(对称轴为x=3，则f(0(=f(6(=1，所以g(0(==1，不等式f(x(>ex等价于g(x(=>1=g(0(，故x<0，所以不等式f(x(>ex的解集为(-∞,0(.故答案为：(-∞,0(3.(23-24高三下·广东佛山)已知函数f(x(的定义域为(0，+∞(，且f/(x(>-f(x(ln2恒成立，则不等A.(1,e2(B.(0,e2(C.(1,e3(D.(0,e3(【分析】根据不等式f/(x(>-f(x(ln2和<的结构特征构造函数F(x(=2xf(x((x>0(，研究其单调性即可求解.【详解】令F(x(=2xf(x((x>0(，因为f/(x(>-f(x(ln2即f(x(ln2+f/(x(>0，则F/(x(=2x[f(x(ln2+f/(x([>0(x>0(，所以F(x(在(0,+∞(上单调递增，故若F(lnx(<F(2(，即2lnxf(lnx(<22f(2(，即<，由定义域及单调性可得0<lnx<2=lne2，⇒1<所以不等式<的解集为(1,e2(.884.(23-24高三下·福建)设f(x)在R上存在导数f/(x)，满足f/(x)+f(x)>0，且有f(2)=2,ex-2f(x)>2A.(-∞,1)B.(-∞,2)C.(1,+∞)D.(2,+∞)【分析】根据题目中已知f/(x)+f(x)>0和f(2)=2，以及不等式ex-2f(x)>2结构特征可构造函数F(x(=exf(x(,x∈R，再由函数的特殊值F(2(和单调性性质即可求解.【详解】令F(x(=exf(x(,x∈R，则F/(x(=ex[f/(x(+f(x([>0,x∈R，所以函数F(x(在R上单调递增，又由题F(2(=e2f(2(=2e2，所以ex-2f(x)>2，即exf(x)>2e2，即F(x(>F(2(的解集为{x|x>2{，1.(22-23高三上·重庆沙坪坝)已知f/(x(是函数f(x(的导函数，f(x(-f(-x(=0，且对于任意的x∈(0,有f/(x(cosx>f(-x(sin(-x(．则下列不等式一定成立的是()A.f(-<f(-cosB.f(->f(-C.f(-1(<、2fcos1D.f>f(-正确.【详解】因为对于任意的x∈(0,有f/(x(cosx>f(-x(sin(-x(．又f(x)-f(-x)=0，-sinx=sin(-x)，所以f/(x)cosx+f(x)sinx>0，x)cosx+f(x)sinx>0，所以g/(x)>0，所以g(x)在(0,上为增函数，99f<fcos，所以f(-(<f(-cos，故A正确；f<f，所以·2f(-(<f(1)，所以·2cos1f<f(f<f，所以f(<f(-，故D不正确；2.(2023秋·陕西西安)已知函数f(x(的定义域为(-,，其导函数是f/(x(.有f/(x(cosx+f(x(sinx<0，则关于x的不等式f(x(<2fcosx的解集为()A.,B.,C.(-,-D.(-,所以，函数g(x(在(-,上单调递减，3.(22-23高三上·全国·阶段练习)已知函数f(x)及其导函数f/(x)的定义域均为(-π,0(，f(-=-2，3f(x)cosx+f/(x)sinx>0，则不等式f(x)sin3x->0的解集为()A.(-,0(B.(-,0(.C.(-,-D.(-π,-【详解】令g(x(=f(x(sin3x-，则g/(x(=f/(x(sin3x+3f(x(sin2xcosx=sin2x(f/(x)sinx+3f(x(cosx(，故g(x(为(-π,0(上的增函数，故f(x)sin3x->0即为g(x(>0，而g(-=f(-sin3(--=0，故g(x(>0的解为x>-，即f(x+cos3x->0的解为(-,0(..A.f>、2fB.f>2cos1⋅f(1(C.f<、2cos1⋅f(1(D.f<f解：构造函数g(x(=f(x(cosx，则g/(x(=cosx⋅f/(x(-sinx⋅f(x(，/(x(=cosx⋅f/(x(-sinx⋅f(x(>0，由g<g，即fcos<fcos，即f<f，故A正确；由g<g(1(，即fcos<f(1(cos1，即f<f(1(cos1，故C正确；由g<g，即fcos<fcos，即f<f，即f<f，故错误的是D．故选D.的函数.如已知是xf/(x(-f(x(<0，可构造g(x(=，可得g/(x(=<0.5.(2020高三·全国·专题练习)已知偶函数y=f(x)对于任意的x∈0,满足f/(x).cosx+f(x).sinx>0(其中f/(x)是函数f(x)的导函数)，则下列不等式中不成立的是()A.、2f(-<fB.、2f(->fC.f(0)<、2f(-D.f<、3f公众号：慧博高中数学最新试题f<、2f<2f，从而可判断A,B,D，由g(-=、2f(->g(0)=f(0)可判断C选公众号：慧博高中数学最新试题项.【详解】“偶函数y=f(x)对于任意的x∈L|Γ0,满足f/且f/(x).cosx+f(x).sinx=f/(x).cosx-f(x).(cosx)/，:可构造函数g(x)=，则g/(x)=>0，:g(x)为偶函数且在0,上单调递增，:g(-=g=2f(g(-=g==2f:BD对，A错，对于C，g(-=g=、2f(->g(0)=f(0)，:C正确，故选:A.1.(21-22高三上·江西南昌·期末)设函数f/(x(是定义在(0，π(上的函数f(x(的导函数，有f/(x)cosxA.a>b>cB.b>c>aC.c>a>bD.c>b>a=-f=cosf，即可得出答案.则gl(x)=fl(x)cosx-f(x)sinx，又因为fl(x)cosx-f(x)sinx>0，所以gl(x)>0，所以g(x)在(0,π)上单调递增，b=f=cosf，c=-f=cosf，所以cosf<cosf<cosf，2.(2021·东莞市东华高级中学高二期末)已知函数y=f(x)为R上的偶函数，且对于任意的x∈0,满足f'(x)cosx+f(x)sinx<0，则下列不等式成立的是()A.、3f>fB.f(0)>、2f(-C.f<、2f(-D.-、3f(->f(-g(->g=g(-，结合条件分别判断四个选项即可.解：偶函数y=f(x)对于任意的x∈L|Γ0,满 又g'(x)=<0，故g(x)在区间0,上是减函数，所以g(0)>g>g=g(->g=g(-，即f(0)=>=、2f=·2f(-(，故B正确；>⇒·3f<f((，故A错误；>=⇒f>·2f(-(，故C错误；=>=⇒·3f(-<f(-(，故D错误；3.(2022·安徽·合肥一中模拟)已知函数y=f(x-1(图象关于点(1,0(对称，且当x>0时，f/(x(sinx+