解答題 10類導(dǎo)數(shù)答題模板學(xué)生版

11函數(shù)是高中數(shù)學(xué)主干知識(shí),單調(diào)性是函數(shù)的重要性質(zhì),用導(dǎo)數(shù)研究函數(shù)單調(diào)性是導(dǎo)數(shù)的一個(gè)主要應(yīng)導(dǎo)函數(shù)與原函數(shù)的關(guān)系，f/(x)>0,f(x)單調(diào)遞增，f/(x)<0,f(x)單調(diào)遞減1.(2024·全國(guó)·高考真題)已知函數(shù)f(x(=a(x-1(-lnx+1.(1)求f(x(的單調(diào)區(qū)間；2.(2023·全國(guó)·高考真題)已知函數(shù)f(x(=a(ex+a(-x.(1)討論f(x(的單調(diào)性；22(1)討論f(x)的單調(diào)性；(1)求函數(shù)f(x(的單調(diào)區(qū)間；5.(2024·江西新余·模擬預(yù)測(cè))已知函數(shù)f(x)=-alnx+(2a+1)x-x2.在處的切線方程.(2)討論f(x)的單調(diào)性.336.(2024·廣東佛山·一模)已知函數(shù)f(x(=e2x-2(a+1(ex+2ax+2a+1(a>0(.(1)求函數(shù)f(x(在x=0處的切線方程；(2)討論函數(shù)f(x(的單調(diào)性；(1)討論f(x)的單調(diào)性；44導(dǎo)函數(shù)與原函數(shù)的關(guān)系，f/(x)>0,f(x)單調(diào)遞增，f/(x)<0,f(x)單調(diào)遞減1.(2021·全國(guó)·高考真題)已知函數(shù)f(x)=x3-x2+ax+1.(1)討論f(x(的單調(diào)性；2.(2024·青海海西·模擬預(yù)測(cè))已知函數(shù)f(x(=x3-x2+ax.(1)討論函數(shù)f(x(的單調(diào)性；3.(2024·山東威?！ひ荒?已知函數(shù)f(x(=ln(ax(-x2+ax(a≠0(.(1)討論f(x(的單調(diào)性；55(1)證明曲線y=f(x(在x=1處的切線過原點(diǎn)；(2)討論f(x(的單調(diào)性；5.函數(shù)f(x)=(x2+ax(ex(a∈R).(1)求f(x(的單調(diào)區(qū)間；6.(2024·山西呂梁·三模)已知函數(shù)f(x(=x2-27.已知函數(shù)f(x(=xlnx-x-a，g(x(=x2+lnx-ax.66定義1:若函數(shù)f(x)的導(dǎo)函數(shù)f/(x)在點(diǎn)x=x0處可導(dǎo),則稱f(x)在點(diǎn)x=x0的導(dǎo)數(shù)為f(x)在點(diǎn)x=x0的二階導(dǎo)數(shù),記作fⅡ(x0(,同時(shí)稱f(x)在點(diǎn)x=x0為二階可導(dǎo).定義2:若f(x)在區(qū)間M上每一點(diǎn)都二階可導(dǎo),則得到一個(gè)定義在M上的二階可導(dǎo)函數(shù),記作fⅡ若f(x)在x=x0附近有連續(xù)的導(dǎo)函數(shù)fⅡ(x),且f/(x0(=0,fⅡ(x0(≠0(1)若fⅡ(x0(<0,則f(x)在點(diǎn)x0處取極大值;(2)若fⅡ(x0(>0,則f(x)在點(diǎn)x0處取極小值(1)討論f(x(的單調(diào)性；772.已知函數(shù)f(x(=x2-axlnx-1(a∈R(.3.已知函數(shù)f(x(=(ex-a-a(lnx+x+4.已知函數(shù)f(x(滿足f(x(=ex-x2.(1)討論f(x(的單調(diào)性；5.已知函數(shù)f(x(=x(alnx-x-1(，其中a∈R.6.已知函數(shù)f(x(=ex+a-ln(x+1(-a(a∈R(.88(2)構(gòu)造新的函數(shù)h(x(；(3)利用導(dǎo)數(shù)研究h(x(的單調(diào)性或1.(2024·全國(guó)·高考真題)已知函數(shù)f(x(=a(x-1(-lnx+1.(1)求f(x(的單調(diào)區(qū)間；992.(2023·天津·高考真題)已知函數(shù)f(x(=((ln(x+1(.(1)求曲線y=f(x(在x=2處的切線斜率；3.(2021·全國(guó)·高考真題)設(shè)函數(shù)f(x(=ln(a-x(，已知x=0是函數(shù)y=xf(x(的極值點(diǎn).4.(2024·陜西榆林·模擬預(yù)測(cè))已知函數(shù)f(x(=ax-ln(x+1(+1.(2)求f(x(的極值；(3)當(dāng)a≤2時(shí)，證明：當(dāng)-1<x<0時(shí)，f(x(>ex.5.(2024·浙江寧波·一模)已知函數(shù)f(x(=-axsinx.(1)判斷f(x(的奇偶性；6.(2024·廣東·二模)已知函數(shù)f(x)=ex-1-xlnx.(1)求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程；(3)求證：f(x)≥x-lnx+e-1.(1)f(x(的值域?yàn)閇m,M[①?x∈D,g(a(≤f(x(，則只需要g(a(≤f(x(min=m?x∈D,g(a(<f(x(，則只需要g(a(<f(x(min=m②?x∈D,g(a(≥f(x(，則只需要g(a(≥f(x(max=M?x∈D,g(a(>f(x(，則只需要g(a(>f(x(max=M(2)若f(x(的值域?yàn)?m,M(①?x∈D,g(a(≤f(x(，則只需要g(a(≤m?x∈D,g(a(<f(x(，則只需要g(a(≤m(注意與(1)中對(duì)應(yīng)情況進(jìn)行對(duì)比)②?x∈D,g(a(≥f(x(，則只需要g(a(≥M?x∈D,g(a(>f(x(，則只需要g(a(≥M(注意與(1)中對(duì)應(yīng)情況進(jìn)行對(duì)比)點(diǎn).(1)若f(x(的值域?yàn)閇m,M[①?x∈D,g(a(≤f(x(，則只需要g(a(≤f(x(max=M?x∈D,g(a(<f(x(，則只需要g(a(<f(x(max=M②?x∈D,g(a(≥f(x(，則只需要g(a(≥f(x(min=m?x∈D,g(a(>f(x(，則只需要g(a(>f(x(min=m(2)若f(x(的值域?yàn)?m,M(①?x∈D,g(a(≤f(x(，則只需要g(a(<M(注意與(1)中對(duì)應(yīng)情況進(jìn)行對(duì)比)?x∈D,g(a(<f(x(，則只需要g(a(<M②?x∈D,g(a(≥f(x(，則只需要g(a(>m(注意與(1)中對(duì)應(yīng)情況進(jìn)行對(duì)比)?x∈D,g(a(>f(x(，則只需要g(a(>m點(diǎn).1.(2024·全國(guó)·高考真題)已知函數(shù)f(x(=(1-ax(ln(1+x(-x.4.(2024·遼寧·模擬預(yù)測(cè))已知函數(shù)f(x(=(ax-1(ex+1+3(a≠0(.(1)求f(x(的極值；(2)設(shè)a=1，若關(guān)于x的不等式f(x(≤(b-1(ex+1-x在區(qū)間[-1,+∞(內(nèi)有解，求b的取值范圍.(2)若存在x06.(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)f(x(=x2-2alnx-2(a∈R).(1)討論f(x(的單調(diào)性；(2)若不等式f(x(≤2(lnx(2+x2-2x在區(qū)間(1,+∞)上有解，求實(shí)數(shù)a的取值范圍.(1)求函數(shù)f(x)的單調(diào)區(qū)間；8.(2024·廣東·模擬預(yù)測(cè))已知函數(shù)f(x(=x-1-alnx,a∈R.(1)判斷函數(shù)f(x(的單調(diào)性；9.(2024·河南鄭州·模擬預(yù)測(cè))已知函數(shù)f(x)=xlnx-ax2，g(x)=ax2-ax+1，h(x)=f(x)+g(x).10.(2024·貴州安順·二模)已知函數(shù)f(x(=ex-1-k(x-1(,k∈R.(1)討論f(x(的單調(diào)性；1.(2022·全國(guó)·高考真題)已知函數(shù)=ax-lnx.2.(2022·全國(guó)·高考真題)已知函數(shù)f(x(=ln(1+x(+axe-x(1)當(dāng)a=1時(shí)，求曲線y=f(x(在點(diǎn)(0,f(0((處的切線方程；(2)若f(x(在區(qū)間(-1,0(,(0,+∞(各恰有一個(gè)零點(diǎn)，求a的取值范圍.3.(2021·全國(guó)·高考真題)已知函數(shù)f(x)=(x-1)ex-ax2+b.(1)討論f(x)的單調(diào)性；①<a≤,b>2a；②0<a<,b≤2a.(2)若曲線y=f(x(與直線y=1有且僅有兩個(gè)交點(diǎn)，求a的取值范圍.5.(2022·全國(guó)·高考真題)已知函數(shù)f(x)=ex-ax和g(x)=ax-lnx有相同的最小值.若a>e，則0<b-f-1(；若0<a<e,x1<x2<x3，則7.(2024·河北邯鄲·模擬預(yù)測(cè))已知函數(shù)f(x(=(lnx+x((ex-((a∈8.(2024·四川·一模)設(shè)f(x(=ex-x-ax(2)討論f(x(的零點(diǎn)數(shù)量.若f有兩個(gè)不同的實(shí)數(shù)根，求實(shí)數(shù)m的取值范圍.10.(2023·江蘇南通·一模)已知函數(shù)f(x(=和g(x(=在同一處取得相同的最大值.(1)求實(shí)數(shù)a；(2)設(shè)直線y=b與兩條曲線y=f(x(和y=g(x(共有四個(gè)不同的交點(diǎn)，其橫坐標(biāo)分別為x1,x2,x3,x4(x1<x2<x3<x4)，證明：x1x4=x2x3.1.(2024·廣東佛山·二模)已知f(x(=-e2x+4ex-ax-5.(2)若f(x(有兩個(gè)極值點(diǎn)x1，x2，證明：f(x1(+f(x2(+x1+x2<0.2.已知f(x(=(x+1(ekx,k≠0.(1)若k=1，求f(x(在(0,f(0((處的切線方程；(2)設(shè)g(x(=f/(x(，求g(x(的單調(diào)區(qū)間；f(m+n(+1>f(m(+f(n(.3.(2024·四川德陽·二模)已知函數(shù)f(x(=lnx+x2-2ax,a∈R，(2)若函數(shù)f(x(有兩個(gè)極值點(diǎn)x1,x2(x1<x2(，求2f(x1(-f(x2(的最小值.4.(2024·河北保定·二模)已知函數(shù)f(x)=ax-xlnx,f/(x)為其導(dǎo)函數(shù).(1)若f(x)≤1恒成立，求a的取(2)若存在兩個(gè)不同的正數(shù)x1,x2，使得f(x1(=f(x2(，證明：f/((>0.5.(2024·山西·模擬預(yù)測(cè))已知函數(shù)=lnx+x2-x+22(x1<x2(是函數(shù)f(x)的兩個(gè)極值點(diǎn)，求證：f(x1(-f(x2(<(a-((x1-x2(.6.已知函數(shù)f(x(=ln(x+1(-x2-ax-1(a∈R(.f(x1(-f(x2(≥M，求M的最大值；零點(diǎn)問題是高考的熱點(diǎn)問題，隱零點(diǎn)的代換與估計(jì)問題是函數(shù)第1步:用零點(diǎn)存在性定理判定導(dǎo)函數(shù)零點(diǎn)的存在性,列出零點(diǎn)方程fl(x0(=0,并結(jié)合f(x)的第3步:將零點(diǎn)方程fl(x0(=0適當(dāng)變形,整體代入f(x)最值式子進(jìn)行化簡(jiǎn):從而得到f(x)最值式的估計(jì).f(x(=xex?f(-lnx(=-?x2ex+lnx=01.(2020·新Ⅰ卷·統(tǒng)考高考真題第21題)已知函數(shù)f(x)=aex-1-lnx+lna.(1)當(dāng)a=e時(shí)，求曲線y=f(x(在點(diǎn)(1,f(1((處的切線與兩坐標(biāo)軸圍成的三角形的面積；2.(2024·山東威?！ざ?已知函數(shù)f(x(=lnx-ax+1.(1)求f(x(的極值；(2)證明：lnx+x+1≤xex3.已知函數(shù)f(x)=ln(ax),a>0，若f(x)≤(x-1)ex-a,求a的取值范圍.4.(2024·陜西西安·模擬預(yù)測(cè))已知函數(shù)f(x)=ax-lnx-a，若f(x)的最小值為0，(2)若g(x)=xf(x)，證明:g(x)存在唯一的極大值點(diǎn)x0，且g(x0(<.5.(2024·黑龍江·模擬預(yù)測(cè))已知函數(shù)f(x)=xlnx+kx-3k，求：6.(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)f(x(=x2-alnx.(1)討論函數(shù)f(x(的單調(diào)性；(2)若函數(shù)f(x(的最小值為，不等式f(x(≥(x-1(2ex-e2+m在上恒成立，求實(shí)數(shù)m的取值范圍.7.(2024·陜西安康·模擬預(yù)測(cè))已知函數(shù)f(x(=，g(x(=lnx.(1)求f(x(的極值；證明：xg(x(+2>exf(x(-(1)求出函數(shù)f(x)的極值點(diǎn)x0；(2)構(gòu)造一元差函數(shù)F(x)=f(x0+x)-f(x0-x)；(3)確定函數(shù)F(x)的單調(diào)性；(4)結(jié)合F(0)=0，判斷F(x)的符號(hào)，從而確定f(x0+x)、f(x0-x)的大小關(guān)系.1.(2022·全國(guó)·統(tǒng)考高考真題)已知函數(shù)f(x(=-lnx+x-a.2.已知函數(shù)f(x(=lnx-ax+1,a∈R.3.(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)f(x(1)求f(x(的單調(diào)區(qū)間；4.設(shè)函數(shù)f(x(=lnx-ax(a∈R(.5.已知函數(shù)f(x(=2xlnx+(a∈R(有兩個(gè)零點(diǎn)x1,x2(x1<x2(.6.已知函數(shù)f(x(=lnx-ax2.a+eb<1.1.(2024·河南·模擬預(yù)測(cè))已知函數(shù)g(x(=lnx+mx+1.(1)當(dāng)m<0時(shí)，求g(x(的單調(diào)區(qū)間；(2)當(dāng)m=1時(shí)，設(shè)正項(xiàng)數(shù)列{xn{滿足：x1=1,xn+1=g(xn(，(1+X的數(shù)學(xué)期望E(X(取最大值時(shí)正整數(shù)n的值.的概率均為p(0<p<1(.(2)若這N件產(chǎn)品中恰好有M(0≤M≤N(件不合格，以(1)中確定的p0作為p的值，則當(dāng)M=45時(shí)，若以使得P(M=45(最大的N值作為N的估計(jì)值，求N的估計(jì)值.平面PAB⊥平面PCD．E,F分別是AB,CD的中點(diǎn)．AB=2BC=2.(2)求四棱錐P-ABCD體積的最大值；(3)求平面PEF與平面PBC的夾角余弦值的范圍.(2)已知正項(xiàng)數(shù)列{an{滿足an+1=an-(n∈N*