新高考數(shù)學(xué)一輪復(fù)習(xí) 導(dǎo)數(shù)專項(xiàng)重點(diǎn)難點(diǎn)突破專題18 構(gòu)造函數(shù)法解決導(dǎo)數(shù)問題(原卷版)

專題18構(gòu)造函數(shù)法解決導(dǎo)數(shù)問題1．以抽象函數(shù)為背景、題設(shè)條件或所求結(jié)論中具有“f(x)±g(x)，f(x)g(x)，eq\f(fx,gx)”等特征式、旨在考查導(dǎo)數(shù)運(yùn)算法則的逆向、變形應(yīng)用能力的客觀題，是高考試卷中的一位“常客”，常以壓軸題的形式出現(xiàn)，解答這類問題的有效策略是將前述式子的外形結(jié)構(gòu)特征與導(dǎo)數(shù)運(yùn)算法則結(jié)合起來，合理構(gòu)造出相關(guān)的可導(dǎo)函數(shù)，然后利用該函數(shù)的性質(zhì)解決問題．2．(1)當(dāng)題設(shè)條件中存在或通過變形出現(xiàn)特征式“f′(x)±g′(x)”時(shí)，不妨聯(lián)想、逆用“f′(x)±g′(x)＝[f(x)±g(x)]′”．構(gòu)造可導(dǎo)函數(shù)y＝f(x)±g(x)，然后利用該函數(shù)的性質(zhì)巧妙地解決問題．(2)當(dāng)題設(shè)條件中存在或通過變形出現(xiàn)特征式“f′(x)g(x)＋f(x)g′(x)”時(shí)，可聯(lián)想、逆用“f′(x)g(x)＋f(x)g′(x)＝[f(x)g(x)]′”，構(gòu)造可導(dǎo)函數(shù)y＝f(x)g(x)，然后利用該函數(shù)的性質(zhì)巧妙地解決問題．(3)當(dāng)題設(shè)條件中存在或通過變形出現(xiàn)特征式“f′(x)g(x)－f(x)g′(x)”時(shí)，可聯(lián)想、逆用“eq\f(f′xgx－fxg′x,[gx]2)＝eq\b\lc\[\rc\](\a\vs4\al\co1(\f(fx,gx)))′”，構(gòu)造可導(dǎo)函數(shù)y＝eq\f(fx,gx)，再利用該函數(shù)的性質(zhì)巧妙地解決問題．3．構(gòu)造函數(shù)解決導(dǎo)數(shù)問題常用模型(1)條件：f′(x)>a(a≠0)：構(gòu)造函數(shù)：h(x)＝f(x)－ax.(2)條件：f′(x)±g′(x)>0：構(gòu)造函數(shù)：h(x)＝f(x)±g(x)．(3)條件：f′(x)＋f(x)>0：構(gòu)造函數(shù)：h(x)＝exf(x)．(4)條件：f′(x)－f(x)>0：構(gòu)造函數(shù)：h(x)＝eq\f(fx,ex).(5)條件：xf′(x)＋f(x)>0：構(gòu)造函數(shù)：h(x)＝xf(x)．(6)條件：xf′(x)－f(x)>0：構(gòu)造函數(shù)：h(x)＝eq\f(fx,x).題型一構(gòu)造y＝f(x)±g(x)型可導(dǎo)函數(shù)1．設(shè)奇函數(shù)f(x)是R上的可導(dǎo)函數(shù)，當(dāng)x>0時(shí)有f′(x)＋cosx<0，則當(dāng)x≤0時(shí)，有()A．f(x)＋sinx≥f(0)B．f(x)＋sinx≤f(0)C．f(x)－sinx≥f(0) D．f(x)－sinx≤f(0)2．設(shè)定義在R上的函數(shù)f(x)滿足f(0)＝－1，其導(dǎo)函數(shù)f′(x)滿足f′(x)>k>1，則下列結(jié)論一定錯(cuò)誤的是()A．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,k)))<eq\f(1,k)B．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,k)))>eq\f(1,k－1)C．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,k－1)))<eq\f(1,k－1) D．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,k－1)))>eq\f(1,k－1)3．已知定義域?yàn)镽的函數(shù)f(x)的圖象經(jīng)過點(diǎn)(1,1)，且對于任意x∈R，都有f′(x)＋2>0，則不等式f(log2|3x－1|)<3－logeq\r(2)|3x－1|的解集為()A．(－∞，0)∪(0,1)B．(0，＋∞)C．(－1,0)∪(0,3) D．(－∞，1)4．設(shè)定義在R上的函數(shù)f(x)滿足f(1)＝2，f′(x)<1，則不等式f(x2)>x2＋1的解集為________．5．定義在R上的函數(shù)f(x)，滿足f(1)＝1，且對任意x∈R都有f′(x)＜eq\f(1,2)，則不等式f(lgx)＞eq\f(lgx＋1,2)的解集為__________．題型二構(gòu)造f(x)·g(x)型可導(dǎo)函數(shù)1．設(shè)函數(shù)f(x)，g(x)分別是定義在R上的奇函數(shù)和偶函數(shù)，當(dāng)x<0時(shí)，f′(x)g(x)＋f(x)g′(x)>0，且g(3)＝0，則不等式f(x)g(x)>0的解集是()A．(－3,0)∪(3，＋∞) B．(－3,0)∪(0,3)C．(－∞，－3)∪(3，＋∞) D．(－∞，－3)∪(0,3)2．設(shè)y＝f(x)是(0，＋∞)上的可導(dǎo)函數(shù)，f(1)＝2，(x－1)[2f(x)＋xf′(x)]>0(x≠1)恒成立．若曲線f(x)在點(diǎn)(1,2)處的切線為y＝g(x)，且g(a)＝2018，則a等于()A．－501B．－502C．－503 D．－5043．設(shè)定義在R上的函數(shù)f(x)滿足f′(x)＋f(x)＝3x2e－x，且f(0)＝0，則下列結(jié)論正確的是()A．f(x)在R上單調(diào)遞減B．f(x)在R上單調(diào)遞增C．f(x)在R上有最大值D．f(x)在R上有最小值4．已知f(x)是定義在R上的增函數(shù)，其導(dǎo)函數(shù)為f′(x)，且滿足eq\f(fx,f′x)＋x<1，則下列結(jié)論正確的是()A．對于任意x∈R，f(x)<0B．對于任意x∈R，f(x)>0C．當(dāng)且僅當(dāng)x∈(－∞，1)時(shí)，f(x)<0D．當(dāng)且僅當(dāng)x∈(1，＋∞)時(shí)，f(x)>05．若定義在R上的函數(shù)f(x)滿足f′(x)＋f(x)>2，f(0)＝5，則不等式f(x)<eq\f(3,ex)＋2的解集為________．6．設(shè)函數(shù)f(x)在R上的導(dǎo)函數(shù)為f′(x)，且2f(x)＋xf′(x)＞x2，則下列不等式在R上恒成立的是()A．f(x)＞0B．f(x)＜0C．f(x)＞xD．f(x)＜x7．已知定義在R上的函數(shù)f(x)滿足f(x)＋2f′(x)＞0恒成立，且f(2)＝eq\f(1,e)(e為自然對數(shù)的底數(shù))，則不等式exf(x)－eSKIPIF1<0＞0的解集為________．題型三構(gòu)造eq\f(fx,gx)型可導(dǎo)函數(shù)1．設(shè)f′(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù)，f(－1)＝0,當(dāng)x＞0時(shí)，xf′(x)－f(x)＜0，則使得f(x)＞0成立的x的取值范圍是()A．(－∞，－1)∪(0,1) B．(－1,0)∪(1，＋∞)C．(－∞，－1)∪(－1,0) D．(0,1)∪(1，＋∞)2．已知f(x)為定義在(0，＋∞)上的可導(dǎo)函數(shù)，且f(x)>xf′(x)，則不等式x2feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,x)))－f(x)<0的解集為________．3．已知f(x)為R上的可導(dǎo)函數(shù)，且?x∈R，均有f(x)＞f′(x)，則有()A．e2019f(－2019)＜f(0)，f(2019)＞e2019f(0)B．e2019f(－2019)＜f(0)，f(2019)＜e2019f(0)C．e2019f(－2019)＞f(0)，f(2019)＞e2019f(0)D．e2019f(－2019)＞f(0)，f(2019)＜e2019f(0)4．已知定義在R上函數(shù)f(x)，g(x)滿足：對任意x∈R，都有f(x)>0，g(x)>0，且f′(x)g(x)－f(x)g′(x)<0.若a，b∈R＋且a≠b，則有()A．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))geq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))>f(eq\r(ab))g(eq\r(ab))B．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))geq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))<f(eq\r(ab))g(eq\r(ab))C．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))g(eq\r(ab))>geq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))f(eq\r(ab))D．feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))g(eq\r(ab))<geq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))f(eq\r(ab))5．設(shè)f′(x)是函數(shù)f(x)(x∈R)的導(dǎo)函數(shù)，且滿足xf′(x)－2f(x)>0，若在△ABC中，角C為鈍角，則()A．f(sinA)·sin2B>f(sinB)·sin2AB．f(sinA)·sin2B<f(sinB)·sin2AC．f(cosA)·sin2B>f(sinB)·cos2AD．f(cosA)·sin2B<f(sinB)·cos2A6．定義在R上的函數(shù)f(x)滿足：f′(x)＞f(x)恒成立，若x1＜x2，則ex1f(x2)與ex2f(x1)的大小關(guān)系為()A．ex11f(x2)＞ex2f(x1)B．ex1f(x2)＜ex2f(x1)C．ex1f(x2)＝ex2f(x1)D．ex1f(x2)與ex2f(x1)的大小關(guān)系不確定專項(xiàng)突破練構(gòu)造函數(shù)法解決導(dǎo)數(shù)問題一、單選題1．已知SKIPIF1<0是定義在R上的偶函數(shù)，SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的圖象是連續(xù)不斷的一條曲線，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．SKIPIF1<0是定義在R上的函數(shù)，SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，已知SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0成立，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．已知函數(shù)SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對稱，且當(dāng)SKIPIF1<0，SKIPIF1<0成立，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且滿足SKIPIF1<0（SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)），則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．已知f（x）為定義在R上的可導(dǎo)函數(shù)，SKIPIF1<0為其導(dǎo)函數(shù)，且SKIPIF1<0恒成立，其中e是自然對數(shù)的底數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08．已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，其導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0，則下列式子一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<09．已知函數(shù)SKIPIF1<0為SKIPIF1<0上的可導(dǎo)函數(shù)，其導(dǎo)函數(shù)為SKIPIF1<0，且滿足SKIPIF1<0恒成立，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0，SKIPIF1<0為其導(dǎo)函數(shù)，滿足①SKIPIF1<0，②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.若不等式SKIPIF1<0有實(shí)數(shù)解，則其解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<011．已知定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，e為自然對數(shù)的底數(shù)，若關(guān)于x的不等式SKIPIF1<0恒成立，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．已知函數(shù)SKIPIF1<0為定義域在R上的偶函數(shù)，且當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<013．已知函數(shù)SKIPIF1<0，若SKIPIF1<0且SKIPIF1<0，則有（

）A．SKIPIF1<0可能是奇函數(shù)，也可能是偶函數(shù) B．SKIPIF1<0C．SKIPIF1<0時(shí)，SKIPIF1<0 D．SKIPIF1<014．定義在R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，則不等式SKIPIF1<0（其中e為自然對數(shù)的底數(shù)）的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<015．設(shè)函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，有SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<016．已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足：SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題17．設(shè)SKIPIF1<0，SKIPIF1<0是定義在R上的恒大于零的可導(dǎo)函數(shù)，且滿足SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<018．已知定義在R上的函數(shù)SKIPIF1<0圖像連續(xù)，滿足SKIPIF1<0，且SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則不等式SKIPIF1<0中的x可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<019．定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且SKIPIF1<0恒成立，則必有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<020．已知SKIPIF1<0是SKIPIF1<0上的可導(dǎo)函數(shù)，且SKIPIF1<0對于任意SKIPIF1<0恒成立，則下列不等關(guān)系正確的是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0三、填空題21．已知SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，SKIPIF1<0是在SKIPIF1<0上無零點(diǎn)的偶函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則使得SKIPIF1<0的解集是________22．已知函數(shù)SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，SKIPIF1<0，對SKIPIF1<0，SKIPIF1<0成立，則SKIPIF1<0的解集為_________．23．已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且對任意SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是___________.24．定義在SKIPIF1<0上的函數(shù)滿足SKIPIF1<0，且對任意SKIPIF1<0都有SKIPIF1<0，則不等式SKIPIF1<0的解集為__________.25．若SKIPIF1<0為定義在SKIPIF1<0上的連續(xù)不斷的函數(shù)，滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0的取值范圍___________.26．已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為___________.27．已知定義在SKIPIF1<0的函數(shù)SKIPIF1<0滿足SKIPIF1<0，則不等式SKIPIF1<0的解集為___________.28．若定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0