新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第12講 導(dǎo)數(shù)的綜合應(yīng)用（解析版）

第12講導(dǎo)數(shù)的綜合應(yīng)用學(xué)校____________姓名____________班級(jí)____________一、知識(shí)梳理1、不等式恒成立(1)分離變量.構(gòu)造函數(shù)，直接把問(wèn)題轉(zhuǎn)化為函數(shù)的最值問(wèn)題.(2)a≥f(x)恒成立?a≥f(x)max；a≤f(x)恒成立?a≤f(x)min；a≥f(x)能成立?a≥f(x)min；a≤f(x)能成立?a≤f(x)max.分類(lèi)討論求參數(shù)：根據(jù)不等式恒成立求參數(shù)范圍的關(guān)鍵是將恒成立問(wèn)題轉(zhuǎn)化為最值問(wèn)題，此類(lèi)問(wèn)題關(guān)鍵是對(duì)參數(shù)分類(lèi)討論，在參數(shù)的每一段上求函數(shù)的最值，并判斷是否滿(mǎn)足題意，若不滿(mǎn)足題意，只需找一個(gè)值或一段內(nèi)的函數(shù)值不滿(mǎn)足題意即可.雙變量恒成立含參不等式能成立問(wèn)題(有解問(wèn)題)可轉(zhuǎn)化為恒成立問(wèn)題解決，常見(jiàn)的轉(zhuǎn)化有：(1)?x1∈M，?x2∈N，f(x1)>g(x2)?f(x)min>g(x)min.(2)?x1∈M，?x2∈N，f(x1)>g(x2)?f(x)min>g(x)max.(3)?x1∈M，?x2∈N，f(x1)>g(x2)?f(x)max>g(x)min.(4)?x1∈M，?x2∈N，f(x1)>g(x2)?f(x)max>g(x)max.2、利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)利用導(dǎo)數(shù)求函數(shù)的零點(diǎn)常用方法(1)構(gòu)造函數(shù)g(x)，利用導(dǎo)數(shù)研究g(x)的性質(zhì)，結(jié)合g(x)的圖像，判斷函數(shù)零點(diǎn)的個(gè)數(shù).(2)利用零點(diǎn)存在定理，先判斷函數(shù)在某區(qū)間有零點(diǎn)，再結(jié)合圖像與性質(zhì)確定函數(shù)有多少個(gè)零點(diǎn).3、構(gòu)造函數(shù)證明不等式(1)五個(gè)常見(jiàn)變形：xex＝ex＋lnx，eq\f(ex,x)＝ex－lnx，eq\f(x,ex)＝elnx－x，x＋lnx＝lnxex，x－lnx＝lneq\f(ex,x).(2)三種基本模式①積型：aea≤blnbeq\o(→,\s\up17(三種同構(gòu)方式))eq\b\lc\{(\a\vs4\al\co1(同左：aea≤（lnb）elnb……f（x）＝xex，,同右：ealnea≤blnb……f（x）＝xlnx，,取對(duì)：a＋lna≤lnb＋ln（lnb）……f（x）＝x＋lnx，))②商型：eq\f(ea,a)＜eq\f(b,lnb)eq\o(→,\s\up17(三種同構(gòu)方式))eq\b\lc\{(\a\vs4\al\co1(同左：\f(ea,a)＜\f(elnb,lnb)……f（x）＝\f(ex,x)，,同右：\f(ea,lnea)＜\f(b,lnb)……f（x）＝\f(x,lnx)，,取對(duì)：a－lna＜lnb－ln（lnb）……f（x）＝x－lnx，))③和差型：ea±a＞b±lnbeq\o(→,\s\up17(兩種同構(gòu)方式))eq\b\lc\{(\a\vs4\al\co1(同左：ea±a＞elnb±lnb……f（x）＝ex±x，,同右：ea±lnea＞b±lnb……f（x）＝x±lnx.))考點(diǎn)和典型例題1、不等式恒成立【典例1-1】（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，若存在SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0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<0【答案】B【詳解】???1,??2∈??，使得SKIPIF1<0成立，等價(jià)于SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞增，所以當(dāng)x＝－1時(shí)，SKIPIF1<0取得最小值SKIPIF1<0；當(dāng)x＝－1時(shí)SKIPIF1<0取得最大值為SKIPIF1<0，所以SKIPIF1<0，即實(shí)數(shù)a的取值范圍是SKIPIF1<0故選：B．【典例1-2】（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，若對(duì)任意兩個(gè)不等的正實(shí)數(shù)SKIPIF1<0都有SKIPIF1<0成立，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】對(duì)任意兩個(gè)不等的正實(shí)數(shù)SKIPIF1<0，都有SKIPIF1<0恒成立，即為SKIPIF1<0時(shí)，SKIPIF1<0恒成立.所以SKIPIF1<0在SKIPIF1<0上恒成立，則SKIPIF1<0而SKIPIF1<0，則SKIPIF1<0．故選：A．【典例1-3】（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，若關(guān)于x的不等式SKIPIF1<0恒成立，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】解：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，顯然SKIPIF1<0為增函數(shù)，則原命題等價(jià)于SKIPIF1<0SKIPIF1<0SKIPIF1<0，又令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0恒成立，所以SKIPIF1<0，所以SKIPIF1<0，即得SKIPIF1<0．故選：B【典例1-4】（2022·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)實(shí)數(shù)SKIPIF1<0，若不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，則t的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0對(duì)SKIPIF1<0恒成立，即SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0，故SKIPIF1<0，問(wèn)題轉(zhuǎn)化為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0遞增，在SKIPIF1<0遞減，故SKIPIF1<0（e）SKIPIF1<0，故SKIPIF1<0.故選：B．【典例1-5】（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，則實(shí)數(shù)a的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0與1的大小不定，但當(dāng)實(shí)數(shù)a最小時(shí)，只需考慮其為負(fù)數(shù)的情況，此時(shí)SKIPIF1<0因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，兩邊取對(duì)數(shù)得：SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0故a的最小值是SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，從四個(gè)選項(xiàng)均為負(fù)，考慮SKIPIF1<0，此時(shí)有SKIPIF1<0，SKIPIF1<0兩邊取對(duì)數(shù)得：SKIPIF1<0，所以SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，無(wú)最大值，此時(shí)無(wú)解，綜上：故a的最小值是SKIPIF1<0．故選：C2、利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)【典例2-1】（2022·河南·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象恰有3個(gè)交點(diǎn)，則實(shí)數(shù)k的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因?yàn)楹瘮?shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象恰有3個(gè)交點(diǎn)，所以SKIPIF1<0有3個(gè)根.經(jīng)驗(yàn)證：x=1為其中一個(gè)根.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0可化為SKIPIF1<0，及SKIPIF1<0i.SKIPIF1<0或SKIPIF1<0時(shí)，方程有且僅有一個(gè)根x=-1；ii.SKIPIF1<0且SKIPIF1<0時(shí)，方程SKIPIF1<0有兩個(gè)根，SKIPIF1<0或x=-1.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0可化為SKIPIF1<0.令SKIPIF1<0，（x>0）.則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單減.因?yàn)镾KIPIF1<0，所以SKIPIF1<0有且只有1個(gè)根x=1.所以需要SKIPIF1<0有兩個(gè)根SKIPIF1<0或x=-1，SKIPIF1<0才有3個(gè)根，此時(shí)SKIPIF1<0且SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有且僅有一個(gè)根x=-1，所以只需SKIPIF1<0在SKIPIF1<0有2個(gè)根.此時(shí)SKIPIF1<0.在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單減；在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單增.且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以只需SKIPIF1<0，即SKIPIF1<0，亦即SKIPIF1<0.記SKIPIF1<0.則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.所以SKIPIF1<0，即SKIPIF1<0（當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)）.所以要使SKIPIF1<0成立，只需SKIPIF1<0，解得：SKIPIF1<0.所以SKIPIF1<0且SKIPIF1<0.綜上所述：實(shí)數(shù)k的取值范圍是SKIPIF1<0.故選：B【典例2-2】（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)不同的交點(diǎn)，則實(shí)數(shù)m取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由題意得：SKIPIF1<0，則SKIPIF1<0，問(wèn)題轉(zhuǎn)化為y＝m和SKIPIF1<0有2個(gè)交點(diǎn)，而SKIPIF1<0，在SKIPIF1<0和SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞增，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞減，當(dāng)x趨于正無(wú)窮大時(shí)，SKIPIF1<0無(wú)限接近于0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象，如圖所示：觀察圖象得：函數(shù)SKIPIF1<0和SKIPIF1<0的圖象有2個(gè)不同的交點(diǎn)時(shí)，實(shí)數(shù)SKIPIF1<0.故選：D．【典例2-3】（2022·陜西·寶雞中學(xué)模擬預(yù)測(cè)）已知曲線(xiàn)SKIPIF1<0與SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個(gè)公共點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】曲線(xiàn)SKIPIF1<0與SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個(gè)公共點(diǎn)，即SKIPIF1<0在區(qū)間SKIPIF1<0上有兩根，設(shè)SKIPIF1<0，則SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減.又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在區(qū)間SKIPIF1<0上有兩根則SKIPIF1<0故選：A【典例2-4】（2022·江西·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0）有三個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是（

）A．（0，SKIPIF1<0） B．（0，SKIPIF1<0） C．（0，1） D．（0，e）【答案】A【詳解】令SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，h(x)在（－∞，0）上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，h(x)在（0，＋∞）上單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0，所以g(x)在R上單調(diào)遞減，又SKIPIF1<0，g（0）＝SKIPIF1<0，所以存在SKIPIF1<0使得SKIPIF1<0，所以方程SKIPIF1<0有兩個(gè)異于SKIPIF1<0的實(shí)數(shù)根，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，k(x)在（－∞，1）上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，k(x)在（1，＋∞）上單調(diào)遞減，且SKIPIF1<0．所以SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0的部分圖象大致如圖所示，由圖知SKIPIF1<0，故選：A．【典例2-5】（2022·浙江·赫威斯育才高中模擬預(yù)測(cè)）已知SKIPIF1<0，函SKIPIF1<0，若函數(shù)SKIPIF1<0有三個(gè)不同的零點(diǎn)，SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖所示：由圖象知：當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有兩不等實(shí)根，當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有一個(gè)實(shí)根；令SKIPIF1<0，顯然SKIPIF1<0,所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恒成立，則SKIPIF1<0在SKIPIF1<0上遞增，且SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖所示：由圖象知：當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0在SKIPIF1<0恰有一個(gè)實(shí)根，即此時(shí)SKIPIF1<0有三個(gè)不同的零點(diǎn)，綜上，SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B3、構(gòu)造函數(shù)證明不等式【典例3-1】（2021·重慶合川·高二階段練習(xí)）已知函數(shù)SKIPIF1<0(1)當(dāng)SKIPIF1<0，證明：SKIPIF1<0；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上恰有一個(gè)極值，求a的值．【答案】(1)證明見(jiàn)解析；(2)SKIPIF1<0.【解析】(1)由題設(shè)SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，則SKIPIF1<0，得證.(2)由題設(shè)SKIPIF1<0在SKIPIF1<0有且僅有一個(gè)變號(hào)零點(diǎn)，所以SKIPIF1<0在SKIPIF1<0上有且僅有一個(gè)解，令SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，故SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0、SKIPIF1<0上遞增，在SKIPIF1<0上遞減，故極大值SKIPIF1<0，極小值SKIPIF1<0，SKIPIF1<0，要使SKIPIF1<0在SKIPIF1<0上與SKIPIF1<0有一個(gè)交點(diǎn)，則SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.經(jīng)驗(yàn)證，SKIPIF1<0或SKIPIF1<0時(shí)SKIPIF1<0對(duì)應(yīng)零點(diǎn)不變號(hào)，而SKIPIF1<0時(shí)SKIPIF1<0對(duì)應(yīng)零點(diǎn)為變號(hào)零點(diǎn)，所以SKIPIF1<0.【典例3-2】（2022·浙江·鎮(zhèn)海中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0(1)求證：函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)SKIPIF1<0；(2)若方程SKIPIF1<0有且僅有一個(gè)正數(shù)解SKIPIF1<0，求證：SKIPIF1<0．【解析】(1)解：由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因?yàn)镾KIPIF1<0，由零點(diǎn)存在定理可知，函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)SKIPIF1<0.(2)解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，又由當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有且僅有一個(gè)正數(shù)解SKIPIF1<0，現(xiàn)證不等式左側(cè)：SKIPIF1<0，要證SKIPIF1<0，只需證SKIPIF1<0在SKIPIF1<0上恒成立，只需證SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，解得