高考數(shù)學(xué)壓軸難題歸納總結(jié)培優(yōu)專題2.12 交點(diǎn)零點(diǎn)有沒(méi)有極最符號(hào)異與否 (含解析)

高考資源網(wǎng)（），您身邊的高考專家全品高考網(wǎng)歡迎廣大教師踴躍來(lái)稿，稿酬豐厚。高考資源網(wǎng)（），您身邊的高考專家歡迎廣大教師踴躍來(lái)稿，稿酬豐厚。高考數(shù)學(xué)壓軸難題歸納總結(jié)提高培優(yōu)專題2.12交點(diǎn)零點(diǎn)有沒(méi)有極最符號(hào)異與否【題型綜述】導(dǎo)數(shù)研究函數(shù)圖象交點(diǎn)及零點(diǎn)問(wèn)題

利用導(dǎo)數(shù)來(lái)探討函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象的交點(diǎn)問(wèn)題，有以下幾個(gè)步驟：①構(gòu)造函數(shù)SKIPIF1<0；②求導(dǎo)SKIPIF1<0；③研究函數(shù)SKIPIF1<0的單調(diào)性和極值（必要時(shí)要研究函數(shù)圖象端點(diǎn)的極限情況）；④畫出函數(shù)SKIPIF1<0的草圖，觀察與SKIPIF1<0軸的交點(diǎn)情況，列不等式；⑤解不等式得解.探討函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，往往從函數(shù)的單調(diào)性和極值入手解決問(wèn)題，結(jié)合零點(diǎn)存在性定理求解.【典例指引】例1．已知函數(shù)SKIPIF1<0，SKIPIF1<0．（I）若曲線SKIPIF1<0在點(diǎn)（1，SKIPIF1<0）處的切線與直線SKIPIF1<0垂直，求a的值；（II）當(dāng)SKIPIF1<0時(shí)，試問(wèn)曲線SKIPIF1<0與直線SKIPIF1<0是否有公共點(diǎn)？如果有，求出所有公共點(diǎn)；若沒(méi)有，請(qǐng)說(shuō)明理由．【思路引導(dǎo)】（1）根據(jù)導(dǎo)數(shù)的幾何意義得到SKIPIF1<0，即SKIPIF1<0；（2）構(gòu)造函數(shù)SKIPIF1<0，研究這個(gè)函數(shù)的單調(diào)性，它和軸的交點(diǎn)個(gè)數(shù)即可得到SKIPIF1<0在（0，1）SKIPIF1<0（SKIPIF1<0）恒負(fù)，SKIPIF1<0，故只有一個(gè)公共點(diǎn)．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在（SKIPIF1<0）單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在（0，1）單調(diào)遞增．又SKIPIF1<0，所以SKIPIF1<0在（0，1）SKIPIF1<0（SKIPIF1<0）恒負(fù)因此，曲線SKIPIF1<0與直線SKIPIF1<0僅有一個(gè)公共點(diǎn)，公共點(diǎn)為（1，-1）．例2．已知函數(shù)f(x)=lnx，h(x)=ax(a為實(shí)數(shù))（1）函數(shù)f(x)的圖象與h(x)的圖象沒(méi)有公共點(diǎn)，求實(shí)數(shù)a的取值范圍；（2）是否存在實(shí)數(shù)m，使得對(duì)任意的SKIPIF1<0都有函數(shù)SKIPIF1<0的圖象在函數(shù)SKIPIF1<0圖象的下方？若存在，請(qǐng)求出整數(shù)m的最大值；若不存在，說(shuō)明理由（SKIPIF1<0）【思路引導(dǎo)】（Ⅰ）函數(shù)SKIPIF1<0與SKIPIF1<0無(wú)公共點(diǎn)轉(zhuǎn)化為方程SKIPIF1<0在SKIPIF1<0無(wú)解，令SKIPIF1<0，得出SKIPIF1<0是唯一的極大值點(diǎn)，進(jìn)而得到SKIPIF1<0，即可求解實(shí)數(shù)SKIPIF1<0取值范圍；（Ⅱ）由不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，即SKIPIF1<0對(duì)SKIPIF1<0恒成立，令SKIPIF1<0，則SKIPIF1<0，再令SKIPIF1<0，轉(zhuǎn)化為利用導(dǎo)數(shù)得到函數(shù)的單調(diào)性和極值，即可得出結(jié)論.當(dāng)且僅當(dāng)SKIPIF1<0故實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0∴存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，………9分∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，則SKIPIF1<0取到最小值SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增SKIPIF1<0，∴存在實(shí)數(shù)SKIPIF1<0滿足題意，且最大整數(shù)SKIPIF1<0的值為SKIPIF1<0.例3．已知二次函數(shù)f(x)的最小值為－4，且關(guān)于x的不等式f(x)≤0的解集為{x|－1≤x≤3，x∈R}．(1)求函數(shù)f(x)的解析式；(2)求函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)．【思路引導(dǎo)】（1）根據(jù)SKIPIF1<0是二次函數(shù)，且關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0，設(shè)出函數(shù)解析式，利用函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，可求函數(shù)SKIPIF1<0的解析式；（2）求導(dǎo)數(shù)，確定函數(shù)的單調(diào)性，可得當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，結(jié)合單調(diào)性由此可得結(jié)論．(2)∵SKIPIF1<0，∴SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0變化時(shí)，SKIPIF1<0，SKIPIF1<0的取值變化情況如下：SKIPIF1<0SKIPIF1<01SKIPIF1<03SKIPIF1<0SKIPIF1<0＋0－0＋SKIPIF1<0遞增極大值遞減極小值遞增當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，又因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，因而SKIPIF1<0在SKIPIF1<0上只有1個(gè)零點(diǎn)，故SKIPIF1<0在SKIPIF1<0上僅有1個(gè)零點(diǎn)．點(diǎn)睛：本題主要考查二次函數(shù)與一元二次不等式的關(guān)系，即一元二次不等式的解集區(qū)間的端點(diǎn)值即為對(duì)應(yīng)二次函數(shù)的零點(diǎn)，同時(shí)用導(dǎo)數(shù)研究函數(shù)圖象的意識(shí)、考查數(shù)形結(jié)合思想，利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性，根據(jù)零點(diǎn)存在性定理與單調(diào)性相結(jié)合可得零點(diǎn)個(gè)數(shù)．例4．已知函數(shù)SKIPIF1<0，SKIPIF1<0．（Ⅰ）求證：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；（Ⅱ）若函數(shù)SKIPIF1<0在（1，+∞）上有唯一零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍．【思路引導(dǎo)】（Ⅰ）求導(dǎo)SKIPIF1<0，得SKIPIF1<0，分析單調(diào)性得當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0即得證；（Ⅱ）SKIPIF1<0對(duì)t進(jìn)行討論①SKIPIF1<0，SKIPIF1<0在[1，+∞)上是增函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在(1，+∞)上沒(méi)有零點(diǎn)，②若SKIPIF1<0，SKIPIF1<0在[1，+∞)上是減函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在(1，+∞)上沒(méi)有零點(diǎn)，③若0<t<1時(shí)分析單調(diào)性借助于第一問(wèn)，找到SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即SKIPIF1<0成立；取SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，說(shuō)明存在SKIPIF1<0，使得SKIPIF1<0，即存在唯一零點(diǎn)．（Ⅱ）SKIPIF1<0①若SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在[1，+∞)上是增函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在(1，+∞)上沒(méi)有零點(diǎn)，所以SKIPIF1<0不滿足條件．②若SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在[1，+∞)上是減函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在(1，+∞)上沒(méi)有零點(diǎn)，所以SKIPIF1<0不滿足條件．點(diǎn)睛：本題考查了利用導(dǎo)數(shù)研究函數(shù)單調(diào)性，最值；考查了分類討論的思想；處理0<t<1時(shí)，注意前后問(wèn)間的聯(lián)系，找到SKIPIF1<0，使得SKIPIF1<0，根據(jù)單調(diào)性說(shuō)明唯一存在，這是本題的難點(diǎn)所在；【同步訓(xùn)練】1．已知函數(shù)SKIPIF1<0．（Ⅰ）若SKIPIF1<0在SKIPIF1<0處取極值，求SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；（Ⅱ）當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0有唯一的零點(diǎn)SKIPIF1<0，求證：SKIPIF1<0【思路引導(dǎo)】本題考查導(dǎo)數(shù)的幾何意義及導(dǎo)數(shù)在研究函數(shù)單調(diào)性、極值中的應(yīng)用．（Ⅰ）根據(jù)函數(shù)在SKIPIF1<0處取極值可得SKIPIF1<0，然后根據(jù)導(dǎo)數(shù)的幾何意義求得切線方程即可．（Ⅱ）由（Ⅰ）知SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．結(jié)合函數(shù)的單調(diào)性和函數(shù)值可得SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，設(shè)為SKIPIF1<0，證明SKIPIF1<0即可得結(jié)論．（Ⅱ）由（Ⅰ）知SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0由SKIPIF1<0，可得SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．又SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；又SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，設(shè)為SKIPIF1<0，從而可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0有唯一零點(diǎn)SKIPIF1<0，故SKIPIF1<0且SKIPIF1<02．已知函數(shù)SKIPIF1<0SKIPIF1<0．（1）當(dāng)SKIPIF1<0時(shí)，若函數(shù)SKIPIF1<0恰有一個(gè)零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍；（2）當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，對(duì)任意SKIPIF1<0，有SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【思路引導(dǎo)】（1）討論SKIPIF1<0、SKIPIF1<0兩種情況，分別利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，結(jié)合函數(shù)的單調(diào)性，利用零點(diǎn)存在定理可得函數(shù)SKIPIF1<0恰有一個(gè)零點(diǎn)時(shí)實(shí)數(shù)SKIPIF1<0的取值范圍；（2）對(duì)任意SKIPIF1<0，有SKIPIF1<0成立，等價(jià)于SKIPIF1<0，利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，分別求出最大值與最小值，解不等式即可的結(jié)果．②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．要使函數(shù)SKIPIF1<0有一個(gè)零點(diǎn)，則SKIPIF1<0即SKIPIF1<0．綜上所述，若函數(shù)SKIPIF1<0恰有一個(gè)零點(diǎn)，則SKIPIF1<0或SKIPIF1<0．（2）因?yàn)閷?duì)任意SKIPIF1<0，有SKIPIF1<0成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，所以SKIPIF1<0．從而SKIPIF1<0SKIPIF1<0．所以SKIPIF1<0即SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0，則SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．又SKIPIF1<0，所以SKIPIF1<0，即為SKIPIF1<0，解得SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0的取值范圍為SKIPIF1<0．3．已知函數(shù)SKIPIF1<0(I)若函數(shù)SKIPIF1<0處取得極值，求實(shí)數(shù)SKIPIF1<0的值；并求此時(shí)SKIPIF1<0上的最大值；(Ⅱ)若函數(shù)SKIPIF1<0不存在零點(diǎn)，求實(shí)數(shù)a的取值范圍；【思路引導(dǎo)】（1）根據(jù)函數(shù)的極值的概念得到SKIPIF1<0，SKIPIF1<0，根據(jù)函數(shù)的單調(diào)性得到函數(shù)的最值．（2）研究函數(shù)的單調(diào)性，找函數(shù)和軸的交點(diǎn)，使得函數(shù)和軸沒(méi)有交點(diǎn)即可；分SKIPIF1<0和SKIPIF1<0，兩種情況進(jìn)行討論．（2）SKIPIF1<0，由于SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是增函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0存在零點(diǎn)．②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．在SKIPIF1<0上SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0時(shí)SKIPIF1<0取最小值．函數(shù)SKIPIF1<0不存在零點(diǎn)，等價(jià)于SKIPIF1<0，解得SKIPIF1<0．綜上所述：所求的實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．點(diǎn)睛：這個(gè)題目考查的是另用導(dǎo)數(shù)研究函數(shù)的極值和最值問(wèn)題，函數(shù)的零點(diǎn)問(wèn)題；對(duì)于函數(shù)有解求參的問(wèn)題，常用的方法是，轉(zhuǎn)化為函數(shù)圖像和軸的交點(diǎn)問(wèn)題，或者轉(zhuǎn)化為兩個(gè)函數(shù)圖像的交點(diǎn)問(wèn)題，還可以轉(zhuǎn)化為方程的根的問(wèn)題．4．已知函數(shù)SKIPIF1<0，其中SKIPIF1<0是自然數(shù)的底數(shù)，SKIPIF1<0．（Ⅰ）求實(shí)數(shù)SKIPIF1<0的單調(diào)區(qū)間．（Ⅱ）當(dāng)SKIPIF1<0時(shí)，試確定函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，并說(shuō)明理由．【思路引導(dǎo)】（Ⅰ）SKIPIF1<0，令SKIPIF1<0，解出SKIPIF1<0，SKIPIF1<0，解出SKIPIF1<0，即可得SKIPIF1<0的單調(diào)區(qū)間（Ⅱ）SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，現(xiàn)考慮函數(shù)SKIPIF1<0的零點(diǎn)，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，考慮函數(shù)SKIPIF1<0與SKIPIF1<0的交點(diǎn)，兩者相切SKIPIF1<0，解得SKIPIF1<0，此時(shí)SKIPIF1<0，所以SKIPIF1<0，故函數(shù)SKIPIF1<0與SKIPIF1<0無(wú)交點(diǎn)，即可得結(jié)果．點(diǎn)睛：本題考查了利用導(dǎo)數(shù)研究函數(shù)單調(diào)區(qū)間，研究函數(shù)零點(diǎn)問(wèn)題，第二問(wèn)中對(duì)SKIPIF1<0進(jìn)行這樣處理，很容易確定一個(gè)零點(diǎn)0，考慮函數(shù)SKIPIF1<0的零點(diǎn)時(shí)使用換元法，簡(jiǎn)化函數(shù)式，很容易利用初等函數(shù)即可解決．5．已知函數(shù)SKIPIF1<0，SKIPIF1<0．（Ⅰ）求曲線SKIPIF1<0在SKIPIF1<0處的切線方程．（Ⅱ）求SKIPIF1<0的單調(diào)區(qū)間．（Ⅲ）設(shè)SKIPIF1<0，其中SKIPIF1<0，證明：函數(shù)SKIPIF1<0僅有一個(gè)零點(diǎn)．【思路引導(dǎo)】（Ⅰ）求導(dǎo)SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0可得SKIPIF1<0在SKIPIF1<0處的切線方程（Ⅱ）令SKIPIF1<0，解出SKIPIF1<0，令SKIPIF1<0，解出SKIPIF1<0，可得SKIPIF1<0的單調(diào)區(qū)間．（Ⅲ）SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0極大值SKIPIF1<0，SKIPIF1<0極小值SKIPIF1<0可得SKIPIF1<0在SKIPIF1<0無(wú)零點(diǎn)，在SKIPIF1<0有一個(gè)零點(diǎn)，所以SKIPIF1<0有且僅有一個(gè)零點(diǎn)．點(diǎn)睛：本題考查了利用導(dǎo)數(shù)求函數(shù)在某點(diǎn)處的切線，考查了函數(shù)的單調(diào)區(qū)間，考查了利用導(dǎo)數(shù)研究零點(diǎn)問(wèn)題，注意SKIPIF1<0處理時(shí)采用因式分解很容易得出SKIPIF1<0的根，考查了學(xué)生推理運(yùn)算的能力，屬于中檔題．6．設(shè)函數(shù)SKIPIF1<0（Ⅰ）當(dāng)SKIPIF1<0（SKIPIF1<0為自然對(duì)數(shù)的底數(shù)）時(shí)，求SKIPIF1<0的極小值；（Ⅱ）若函數(shù)SKIPIF1<0存在唯一零點(diǎn)，求SKIPIF1<0的取值范圍．【思路引導(dǎo)】（1）先求導(dǎo)數(shù)，再求導(dǎo)函數(shù)零點(diǎn)，列表分析導(dǎo)函數(shù)符號(hào)變化規(guī)律，進(jìn)而確定極值（2）先化簡(jiǎn)SKIPIF1<0，再利用參變分離法得SKIPIF1<0，利用導(dǎo)數(shù)研究函數(shù)SKIPIF1<0，由圖像可得存在唯一零點(diǎn)時(shí)SKIPIF1<0的取值范圍試題解析：（1）由題設(shè)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0．∴當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，結(jié)合SKIPIF1<0的圖象（如圖），可知當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)．所以，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)．點(diǎn)睛：利用函數(shù)零點(diǎn)的情況求參數(shù)值或取值范圍的方法(1)利用零點(diǎn)存在的判定定理構(gòu)建不等式求解．(2)分離參數(shù)后轉(zhuǎn)化為函數(shù)的值域(最值)問(wèn)題求解．(3)轉(zhuǎn)化為兩熟悉的函數(shù)圖象的上、下關(guān)系問(wèn)題，從而構(gòu)建不等式求解．7．已知函數(shù)SKIPIF1<0．（1）若SKIPIF1<0，求函數(shù)SKIPIF1<0的極值；（2）若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍．【思路引導(dǎo)】（1）函數(shù)求導(dǎo)得SKIPIF1<0，討論導(dǎo)數(shù)的單調(diào)性即可得極值；（2）函數(shù)求導(dǎo)得SKIPIF1<0，討論SKIPIF1<0，SKIPIF1<0，SKIPIF1<0和SKIPIF1<0時(shí)函數(shù)的單調(diào)性及最值即可下結(jié)論．（2）SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，易知函數(shù)SKIPIF1<0只有一個(gè)零點(diǎn)，不符合題意；當(dāng)SKIPIF1<0時(shí)，在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0→SKIPIF1<0，SKIPIF1<0→SKIPIF1<0，所以函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)．當(dāng)SKIPIF1<0時(shí)，在SKIPIF1<0和SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0，函數(shù)SKIPIF1<0至多有一個(gè)零點(diǎn)，不符合題意．當(dāng)SKIPIF1<0時(shí)，在SKIPIF1<0和SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞增；在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0，函數(shù)SKIPIF1<0至多有一個(gè)零點(diǎn)，不符合題意．綜上：實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．點(diǎn)睛：根據(jù)函數(shù)零點(diǎn)求參數(shù)取值，也是高考經(jīng)常涉及的重點(diǎn)問(wèn)題，（1）利用零點(diǎn)存在的判定定理構(gòu)建不等式求解；（2）分離參數(shù)后轉(zhuǎn)化為函數(shù)的值域（最值）問(wèn)題求解，如果涉及由幾個(gè)零點(diǎn)時(shí)，還需考慮函數(shù)的圖象與參數(shù)的交點(diǎn)個(gè)數(shù)；（3）轉(zhuǎn)化為兩熟悉的函數(shù)圖象的上、下關(guān)系問(wèn)題，從而構(gòu)建不等式求解．8．已知SKIPIF1<0，SKIPIF1<0．（1）求函數(shù)SKIPIF1<0的增區(qū)間；（2）若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍，并說(shuō)明理由；（3）設(shè)正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足當(dāng)SKIPIF1<0時(shí)，求證：對(duì)任意的兩個(gè)正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0總有SKIPIF1<0．(參考求導(dǎo)公式：SKIPIF1<0)【思路引導(dǎo)】(1)求導(dǎo)SKIPIF1<0，對(duì)SKIPIF1<0進(jìn)行分類討論，可得函數(shù)SKIPIF1<0的增區(qū)間；（2）由（1）知：若SKIPIF1<0函數(shù)在SKIPIF1<0的上為增函數(shù)，函數(shù)SKIPIF1<0有至多有一個(gè)零點(diǎn)，不合題意．若SKIPIF1<0可知SKIPIF1<0，要使得函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，則SKIPIF1<0SKIPIF1<0，以下證明SKIPIF1<0函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)即可（3）證明：不妨設(shè)SKIPIF1<0，以SKIPIF1<0為變量，令SKIPIF1<0，則可以證明SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增；因?yàn)镾KIPIF1<0所以SKIPIF1<0，這樣就證明了SKIPIF1<0SKIPIF1<0而SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0存在惟一零點(diǎn)；又SKIPIF1<0令SKIPIF1<0SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上遞增，所以的SKIPIF1<0SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0也存在惟一零點(diǎn)；綜上：SKIPIF1<0函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)方法2：(先證：SKIPIF1<0有SKIPIF1<0)SKIPIF1<0SKIPIF1<0而SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0也存在惟一零點(diǎn)；綜上：SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)．（3）證明：不妨設(shè)SKIPIF1<0，以SKIPIF1<0為變量令SKIPIF1<0，則SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0；即SKIPIF1<0在定義域內(nèi)遞增．又因?yàn)镾KIPIF1<0且SKIPIF1<0所以SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0；又因?yàn)镾KIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增；因?yàn)镾KIPIF1<0所以SKIPIF1<0即SKIPIF1<0【點(diǎn)睛】本題考查運(yùn)用導(dǎo)數(shù)知識(shí)研究函數(shù)的圖象與性質(zhì)、函數(shù)的應(yīng)用、不等式問(wèn)題、數(shù)學(xué)歸納法等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力、推理論證能力，考查數(shù)形結(jié)合思想、函數(shù)與方程思想、特殊與一般思想等．9．已知函數(shù)SKIPIF1<0，SKIPIF1<0．（1）當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程；（2）令SKIPIF1<0，討論函數(shù)SKIPIF1<0的零點(diǎn)的個(gè)數(shù)；（3）若SKIPIF1<0，正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，證明SKIPIF1<0【思路引導(dǎo)】（1）求出SKIPIF1<0的解析式，求出切點(diǎn)坐標(biāo)，再求出SKIPIF1<0，由出SKIPIF1<0的值，可得切線斜率，利用點(diǎn)斜式求出切線方程即可；（2）求導(dǎo)數(shù)，分三種情況討論，利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，分別結(jié)合函數(shù)單調(diào)性判斷出函數(shù)SKIPIF1<0的零點(diǎn)的個(gè)數(shù)；（3）SKIPIF1<0，化為SKIPIF1<0，設(shè)SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，然后結(jié)合函數(shù)單調(diào)性得到SKIPIF1<0，解不等式可得結(jié)論．（3）證明：當(dāng)所以即為：所以令所以所以所以因?yàn)椤痉椒c(diǎn)晴】本題主要考查利用導(dǎo)數(shù)求曲線切線以及利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性與最值，屬于難題．求曲線切線方程的一般步驟是：（1）求出SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù)，即SKIPIF1<0在點(diǎn)SKIPIF1<0SKIPIF1<0出的切線斜率（當(dāng)曲線SKIPIF1<0在SKIPIF1<0處的切線與SKIPIF1<0軸平行時(shí)，在處導(dǎo)數(shù)不存在，切線方程為SKIPIF1<0）；（2）由點(diǎn)斜式求得切線方程SKIPIF1<0．10．已知函數(shù)SKIPIF1<0（SKIPIF1<0）．（1）判斷函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上零點(diǎn)的個(gè)數(shù)；（2）當(dāng)SKIPIF1<0時(shí)，若在SKIPIF1<0（SKIPIF1<0）上存在一點(diǎn)SKIPIF1<0，使得SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【思路引導(dǎo)】SKIPIF1<0令SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，記SKIPIF1<0，SKIPIF1<0，求得導(dǎo)數(shù)，利用函數(shù)單調(diào)性可以求得函數(shù)極值點(diǎn)以此判斷函數(shù)SKIPIF1<0在SKIPIF1<0上的零點(diǎn)個(gè)數(shù)；SKIPIF1<0本題不宜分離，因此作差構(gòu)造函數(shù)SKIPIF1<0，利用分類討論法求函數(shù)最小值，由于SKIPIF1<0，所以討論SKIPIF1<0與SKIPIF1<0的大小，分三種情況，當(dāng)SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，解對(duì)應(yīng)不等式即可．②當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0的最小值為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0．③當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，可得SKIPIF1<0的最小值為SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0不成立．綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．點(diǎn)睛：對(duì)于求不等式成立時(shí)的參數(shù)范圍問(wèn)題，在可能的情況下把參數(shù)分離出來(lái)，使不等式一端是含有參數(shù)的不等式，另一端是一個(gè)區(qū)間上具體的函數(shù)，這樣就把問(wèn)題轉(zhuǎn)化為一端是函數(shù)，另一端是參數(shù)的不等式，便于問(wèn)題的解決．但要注意分離參數(shù)法不是萬(wàn)能的，如果分離參數(shù)后，得出的函數(shù)解析式較為復(fù)雜，性質(zhì)很難研究，就不要使用分離參數(shù)法．11．已知函數(shù)SKIPIF1<0．（1）求SKIPIF1<0在SKIPIF1<0處的切線方程；（2）試判斷SKIPIF1<0在區(qū)間SKIPIF1<0上有沒(méi)有零點(diǎn)？若有則判斷零點(diǎn)的個(gè)數(shù)．【思路引導(dǎo)】（1）利用導(dǎo)數(shù)的幾何意義求切線方程．（2）利用導(dǎo)數(shù)求出函數(shù)的極大值和極小值，判斷極值與0的關(guān)系明確零點(diǎn)個(gè)數(shù)．試題解析：12．已知函數(shù)SKIPIF1<0，其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)．（1）當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極值；（2）當(dāng)SKIPIF1<0時(shí)，討論函數(shù)SKIPIF1<0的定義域內(nèi)的零點(diǎn)個(gè)數(shù)．【思路引導(dǎo)】（1）求出SKIPIF1<0，SKIPIF1<0求得SKIPIF1<0的范圍，可得函數(shù)SKIPIF1<0增區(qū)間，SKIPIF1<0求得SKIPIF1<0的范圍，可得函數(shù)SKIPIF1<0的減區(qū)間，根據(jù)單調(diào)性可得函數(shù)的極值；（2）利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，可證明函數(shù)SKIPIF1<0恒成立，即證明SKIPIF1<0在定義域內(nèi)無(wú)零點(diǎn)．試題解析：（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0單調(diào)增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0單調(diào)減，所以SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，極大值是SKIPIF1<0．（2）由已知SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，【方法點(diǎn)睛】本題主要考查利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性、函數(shù)的極值以及函數(shù)零點(diǎn)問(wèn)題，屬于難題．求函數(shù)SKIPIF1<0極值的步驟：(1)確定函數(shù)的定義域；(2)求導(dǎo)數(shù)SKIPIF1<0；(3)解方程SKIPIF1<0求出函數(shù)定義域內(nèi)的所有根；(4)列表檢查SKIPIF1<0在SKIPIF1<0的根SKIPIF1<0左右兩側(cè)值的符號(hào)，如果左正右負(fù)（左增右減），那么SKIPIF1<0在SKIPIF1<0處取極大值，如果左負(fù)右正（左減右增），那么SKIPIF1<0在SKIPIF1<0處取極小值．13．已知函數(shù)SKIPIF1<0．（1）討論SKIPIF1<