導(dǎo)數(shù)必考經(jīng)典壓軸解答題匯編（學(xué)生版）

活求解.其中，對于不等式證明中極值點(diǎn)偏移、隱零點(diǎn)問題和不等式的放縮應(yīng)用這三類問題是目前題的熱點(diǎn)方向.①求出函數(shù)y=f(x)在x=x0處的導(dǎo)數(shù)，即曲線y=f(x)在點(diǎn)(x0,f(x0))處切線的斜率；11②在已知切點(diǎn)坐標(biāo)和切線斜率的條件下，求得切線方程為y=y0+f'(x0)(x-x0).①設(shè)出切點(diǎn)坐標(biāo)T(x0,f(x0))(不出現(xiàn)y0)；②利用切點(diǎn)坐標(biāo)寫出切線方程：y=f(x0)+f'(x0)(x-x0)；③將已知條件代入②中的切線方程求解.1.運(yùn)用導(dǎo)數(shù)求函數(shù)f(x)極值的一般步驟：(1)確定函數(shù)f(x)的定義域；(2)求導(dǎo)數(shù)f'(x)；(4)列表檢驗(yàn)f'(x)在f'(x)=0的根x0左右兩側(cè)值的符號；(1)利用導(dǎo)數(shù)求函數(shù)f(x)在[a,b]上的最值的一般步驟：②求函數(shù)在區(qū)間端點(diǎn)處的函數(shù)值f(a)，f(b)；和22利用導(dǎo)數(shù)研究含參函數(shù)的零點(diǎn)(方程的根)主要有(1)利用導(dǎo)數(shù)研究函數(shù)f(x)的最值，轉(zhuǎn)化為f(x)圖象與x軸的交點(diǎn)問題，主要是應(yīng)用分類討論思想解決.(1)一般地，要證f(x)>g(x)在區(qū)間(a，b)上成立，需構(gòu)造輔助函數(shù)F(x)=f(x)-g(x)，通過分析F(x)在端點(diǎn)處的函數(shù)值來證明不等式．若F(a)=0，只需證明F極值點(diǎn)偏移的定義：對于函數(shù)y=f(x)在區(qū)間(a,b)內(nèi)只有一個極值點(diǎn)x0，方程f(x)的解分別為x1、x2，且a<x1<x2<b.(1)若≠x0，則稱函數(shù)y=f(x)在區(qū)間(x1,x2)上極值點(diǎn)x0偏移；若>x0，則函數(shù)y=f(x)在區(qū)間(x1,x2)上極值點(diǎn)x0左偏，簡稱極值點(diǎn)x0左偏；若<x0，則函數(shù)y=f(x)在區(qū)間(x1,x2)上極值點(diǎn)x0右偏，簡稱極值點(diǎn)x0右偏.33(1)函數(shù)f(x)存在兩個零點(diǎn)x1，x2且x1≠x2，求證：x1+x2>2x0(x0為函數(shù)f(x)的極值點(diǎn))；(2)函數(shù)f(x)中存在x1，x2且x1≠x2，滿足f(x1)=f(x2)，求證：x1+x2>2x0(x0為函數(shù)f(x)的極值點(diǎn))；f'①對結(jié)論x1+x2>2x0型，構(gòu)造函數(shù)F(x)=f(x)-f(2x0-x).②對結(jié)論x1x2>x型，方法一是構(gòu)造函數(shù)通過研究F(x)的單調(diào)性獲得不(2)(比值代換法)：通過代數(shù)變形將所證的雙變量不等式通過代換化為單變量的函數(shù)不等式，利用函數(shù)單調(diào)性證明.1.(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；(2)證明：f(x)>0.44-1,m(可以作3條直線與曲線y=f(x(相切，求m的取值范圍.3.(2024·湖北黃岡·一模)已知函數(shù)f(x(=2alnx+x2-(a+3(x,(a∈R((1)若曲線y=f(x(在點(diǎn)(1,f(1((處的切線方程為f(x(=-x+b，求a和b的值；(2)討論f(x(的單調(diào)性.55ax+5.(2024·浙江金華·一模)已知函數(shù)f(x(=x2-alnx+(1-a(x，(a>0(.若f(x(≥-的取值范圍.66(1)求函數(shù)y=f(x(的單調(diào)區(qū)間；(2)求不等式f(x(<2x的解集.7.(2024·廣東·模擬預(yù)測)已知函數(shù)f(x(=x3+(a-3(x2-ax+4.(2)討論f(x(的單調(diào)性.77≥0；8810.(2024·廣東肇慶·一模)已知函數(shù)f(x(=+ax+(2)若f(x(存在極大值，求a的取值范圍.11.(2024·陜西榆林·模擬預(yù)測)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.9912.(2024·河南·二模)已知函數(shù)f(x(=x2+2(a-3(x+2alnx(a∈R(在定義域內(nèi)有兩個極值點(diǎn)x1,x2.(2)證明：f(x1(+f(x2(>-10.(1)討論函數(shù)f(x)的單調(diào)性；(1)討論函數(shù)f(x(的單調(diào)性；15.(2024·四川·一模)設(shè)f(x(=ex-x-ax(2)討論f(x(的零點(diǎn)數(shù)量.16.(2024·甘肅白銀·一模)已知函數(shù)f(x(=tx2-2lnx-1.(1)若曲線y=f(x(在x=2處的切線的斜率為3，求t.(2)已知f(x(恰有兩個零點(diǎn)x1,x2(x1<x2(.17.(2024·廣東廣州·模擬預(yù)測)已知函數(shù)f(x(=ex-kx2-x.18.(2024·四川·一模)已知函數(shù)f(x(=xlnx-ax2+1.(1)若f(x(在(0,+∞(上單調(diào)遞減，求a的取值范圍；19.(2024·山西·模擬預(yù)測)已知函數(shù)=lnx+x2-x+22(x1<x2(是函數(shù)f(x)的兩個極值點(diǎn)，求證：f(x1(-f(x2(<(a-((x1-x2(.x+(+2，記f/(x(是f(x(的導(dǎo)函數(shù).證明：當(dāng)x>1時(shí)，(x-1(e-x+xln(1+>lnx?ln(x+1(.21.(2024·河南·模擬預(yù)測)已知函數(shù)f(x(=ex-2elnx+ax22.(2024·福建·三模)函數(shù)f(x(=(1-x(eax-x-1，其中a為整數(shù).(1)求函數(shù)y=f(x)的單調(diào)遞減區(qū)間；(1)若曲線y=f(x(在點(diǎn)(1,f(1((處的切線為x+y+b=0，求實(shí)數(shù)b的值；25.(2024·四川樂山·三模)已知函數(shù)=ax+lnx,g-x-1(+1-x(1)討論f(x)的單調(diào)性；f(x(≤g(x(成立，求a的最小值.28.(2024·遼寧·模擬預(yù)測)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.(1)若f(x(在(0,2[上單調(diào)遞增，求a的取值范圍；+x2=4且0<x1<2，比較f(x1(與f(x2(的大小，并說明理由30.(2024·河南商丘·模擬預(yù)測)已知函數(shù)f(x(的定義域?yàn)?0,+∞(，其導(dǎo)函數(shù)f/(x(=2x+f(1(=1-2a.，滿足0<x1<x2，且f/(x1(=f/(x2(=0，求2f(x1(-f(x2(的取值范圍.，滿足f(x1(=f(x2(=0.+x2<π.32.(2024·安徽阜陽·一模)已知函數(shù)f(x(=3(1)討論f(x(的單調(diào)性.,x2是函數(shù)f(x(的兩個零點(diǎn)(x1<x2(.f/(x(是f(x(的導(dǎo)函數(shù)．證明：f/[λx1+(1-λ(x2[<0.(1)討論f(x)的單調(diào)性；，且f(x1(=f(x2(=22是f(x)1+x2>4a.(1)討論函數(shù)f(x(的單調(diào)性；(2)若關(guān)于x的方程f(x(=xex-lnx+有兩個不相等的實(shí)數(shù)根x1、x2，37.(2024·江蘇南通·三模)已知函數(shù)f(x(=(1+x)k-kx-1(k>1).(1)若x>-1，求f(x(的最小值；-n≥2-38.(24-25高三上·河北滄州·階段練習(xí))已知函數(shù)f(x(=lnx的圖象與函數(shù)g(x(的圖象關(guān)于直線y=-x+1對稱.f(x(-g(x(>0；(3)若圓M:(x-1)2+y2=r2(r>0(與曲線y=|f(x(|相交于A,B兩點(diǎn)，證明：∠AMB為銳角.當(dāng)0<x<1時(shí)，f(x(>求實(shí)數(shù)a的取值范圍；n=f(an+1(.證明：≤an≤>0；(3)設(shè)集合P={anan=,n∈N*對于正整數(shù)m，集合Qm={x|m<x<2m{，記P∩k=(k+1(ak+1-ak，稱g(x(=b0+b1x+b2x2+?+bnxn為f(x(的“伴生函數(shù)”.tt(<=mq-np，已知函數(shù)1,g-1+1+1+(?1+<e.一個常數(shù)k，使f(x1(-f(x2(=k(x1-x2(成立，則稱函數(shù)f(x(為極值可差比函數(shù)，常數(shù)k稱為該函數(shù)的極值差比系數(shù)．已知函數(shù)f(x(=x--alnx.<?<xn-1<?,xn-1,xn∈D，都有不等式|f(xi(-f稱函數(shù)y=f(x(,x∈D是“絕對差有界函數(shù)”f(x1(-f(x2(|≤k|x1-x2|恒45.(2024·海南省直轄縣級單位·模擬預(yù)測)已知函數(shù)f(x(=x-lnx-2.(1)求曲線y=f(x(在(e,e-3(處的切線方程；(2)若a≥0，g(x(=ax2-2(ax+1(-f(x(，討論函數(shù)g(x(的單調(diào)性.46.(2024·湖北·一模)已知f(x(=(ax2+x+1(ex.(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(-3,-1(內(nèi)存在極小值點(diǎn)，求a的取值范圍.(1)當(dāng)函數(shù)f(x(在點(diǎn)(2,f(2((處的切線m與直線l:3x-2y-1=0平行時(shí)，求切線m的方程；(2)若過點(diǎn)(3,m(可作曲線y=f(x(的三條不同的切線，求實(shí)數(shù)m的取值范圍.49.(2024·西藏拉薩·一模)已知函數(shù)f(x(=x2-(λ+3(x+λlnx.(1)若λ=-3，求f(x(的單調(diào)區(qū)間；50.(2024·廣東·模擬預(yù)測)已知函數(shù)f(x(=x-(1)判斷函數(shù)f(x(的單調(diào)性；2.7萬元，設(shè)該公司年內(nèi)共生產(chǎn)該品牌服裝x千件并全部銷售完，銷售收入為R(x)萬元，且R(x)=2(x,年利潤=年銷售收入-年總成本)(1)寫出年利潤W(萬元)關(guān)于年產(chǎn)量x(千件)的函數(shù)解析式；(2)求公司在這一品牌服裝的生產(chǎn)中所獲年利潤最大時(shí)的年產(chǎn)量.(2)若f(x)在區(qū)間(1,+∞)上有且只有一個極值點(diǎn)，求實(shí)數(shù)a的取值范圍.53.(2024·新疆·模擬預(yù)測)已知函數(shù)f(x(=(x-1(emx.(2)若不等式f(x(≥x2-x在[1,+∞(上恒成立，求實(shí)數(shù)m的取值范圍.證明：0<f(x(<55.(2024·四川內(nèi)江·一模)已知函數(shù)f(x(=a(x+a(-ln(x+1(，a∈(1)討論函數(shù)f(x(的單調(diào)性；56.(2024·河北邯鄲·模擬預(yù)測)已知函數(shù)f(x(=(lnx+x((ex-((a∈R(.(2)若f(x(有兩個不同的零點(diǎn)，求實(shí)數(shù)a的取值范圍.57.(2024·四川樂山·三模)已知函數(shù)f(x(=ax(2)若存在x059.(2024·吉林·模擬預(yù)測)已知函數(shù)f(x(=x(ex-a(-60.(2024·河南·三模)設(shè)函數(shù)f(x(的導(dǎo)函數(shù)為f/(x(,f/(x(的導(dǎo)函數(shù)為f″(x(,f″(x(的導(dǎo)函數(shù)為f川(x(.若,f(x0((為曲線y=f(x(的拐點(diǎn).(2)已知函數(shù)f(x(=ax5-5x3，若,f為曲線y=f(x(的一個拐點(diǎn)，求f(x(的單調(diào)區(qū)間與極值.61.(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=-x2+2lnx,g(x(=a(x2+2x(.(1)若曲線f(x(在點(diǎn)(1,-1(處的切線與曲線g(x(有且只有一個公共點(diǎn)，求實(shí)數(shù)a的值.(2)若方程g(x(-f(x(=1有兩個不相等的實(shí)數(shù)根x1,x2，(1)當(dāng)a>0時(shí)，試討論f(x)的單調(diào)性；(2)若函數(shù)f(x)有兩個不相等的零點(diǎn)x1，x2，(ii)證明：x1+x2>4.63.(202