新高考數(shù)學(xué)二輪復(fù)習(xí)導(dǎo)數(shù)培優(yōu)專(zhuān)題02 利用導(dǎo)數(shù)求函數(shù)單調(diào)區(qū)間與單調(diào)性（含解析）

專(zhuān)題02利用導(dǎo)數(shù)求函數(shù)單調(diào)區(qū)間與單調(diào)性專(zhuān)項(xiàng)突破一利用導(dǎo)數(shù)判斷或證明函數(shù)單調(diào)性一、多選題1．若函數(shù)f（x）的導(dǎo)函數(shù)在定義域內(nèi)單調(diào)遞增，則f（x）的解析式可以是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】A：由SKIPIF1<0，令SKIPIF1<0，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0是實(shí)數(shù)集上的增函數(shù)，符合題意；B：由SKIPIF1<0，因?yàn)橐淮魏瘮?shù)SKIPIF1<0是實(shí)數(shù)集上的增函數(shù)，所以符合題意；C：由SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0是周期函數(shù)，所以函數(shù)SKIPIF1<0不是實(shí)數(shù)集上的增函數(shù)，因此不符合題意；D：由SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，因此不符合題意，故選：AB二、解答題2．已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0至少有兩個(gè)零點(diǎn)，求a的取值范圍．【解析】(1)由SKIPIF1<0，在SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，在SKIPIF1<0上SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增，SKIPIF1<0上遞減.(2)由（1）知：SKIPIF1<0極小值為SKIPIF1<0，極大值為SKIPIF1<0，要使SKIPIF1<0至少有兩個(gè)零點(diǎn)，則SKIPIF1<0，可得SKIPIF1<0.3．設(shè)函數(shù)SKIPIF1<0.(1)若曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處與直線(xiàn)SKIPIF1<0相切，求a，b的值；(2)討論函數(shù)SKIPIF1<0的單調(diào)性.【解析】(1)由題意知，SKIPIF1<0，又SKIPIF1<0即SKIPIF1<0，解得SKIPIF1<0；(2)已知SKIPIF1<0，令SKIPIF1<0，知SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.4．已知函數(shù)SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，求證：SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.【解析】證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，且SKIPIF1<0（1）SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0,SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0,SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.5．已知函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0單調(diào)遞減，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0單調(diào)遞減，若SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0單調(diào)遞增．綜上可得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)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<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增．6．已知SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0.(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0恒成立，求實(shí)數(shù)a的取值范圍.【解析】(1)SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，①SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；②SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；③SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增(2)SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞減，則只需滿(mǎn)足SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，則SKIPIF1<0，整理可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則可得SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，則SKIPIF1<0，故SKIPIF1<0時(shí)，SKIPIF1<0恒成立，綜上，SKIPIF1<0；7．已知函數(shù)SKIPIF1<0.(1)當(dāng)a=2時(shí),求曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程；(2)設(shè)函數(shù)SKIPIF1<0,討論SKIPIF1<0的單調(diào)性并判斷有無(wú)極值,有極值時(shí)求出極值.【解析】（1）由題意SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因此，曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程是SKIPIF1<0，即SKIPIF1<0.（2）因?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增.所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取到極大值，極大值是SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取到極小值，極小值是SKIPIF1<0.（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0無(wú)極大值也無(wú)極小值.（3）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增.所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取到極大值，極大值是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取到極小值，極小值是SKIPIF1<0.綜上所述：當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，函數(shù)既有極大值，又有極小值，極大值是SKIPIF1<0，極小值是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，無(wú)極值；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，函數(shù)既有極大值，又有極小值，極大值是SKIPIF1<0，極小值是SKIPIF1<0.專(zhuān)項(xiàng)突破二利用導(dǎo)數(shù)求函數(shù)單調(diào)區(qū)間(不含參)一、單選題1．函數(shù)SKIPIF1<0的單調(diào)減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】SKIPIF1<0,由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.故選：B2．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由題得函數(shù)的定義域?yàn)镾KIPIF1<0.SKIPIF1<0，令SKIPIF1<0.所以函數(shù)的單調(diào)遞減區(qū)間為SKIPIF1<0.故選：A3．已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以由SKIPIF1<0得SKIPIF1<0，故選：C．4．已知函數(shù)f(x)滿(mǎn)足SKIPIF1<0，則f(x)的單調(diào)遞減區(qū)間為（

）A．(-,0) B．(1,+∞) C．(-,1) D．(0,+∞)【解析】由題設(shè)SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，而SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0遞增，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即SKIPIF1<0遞減，故SKIPIF1<0遞減區(qū)間為(-,0).故選：A二、多選題5．函數(shù)SKIPIF1<0的一個(gè)單調(diào)遞減區(qū)間是（

）A．（e，+∞） B．SKIPIF1<0 C．（0，SKIPIF1<0） D．（SKIPIF1<0，1）【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞減，所以AD選項(xiàng)符合題意.故選：AD三、填空題6．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是______．【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，因?yàn)槎x域?yàn)镾KIPIF1<0，所以單調(diào)遞增區(qū)間為SKIPIF1<0.7．函數(shù)SKIPIF1<0，SKIPIF1<0的增區(qū)間為_(kāi)__________.【解析】由已知得SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0，故答案為:SKIPIF1<0.四、解答題8．已知函數(shù)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)求曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程．【解析】(1)SKIPIF1<0，SKIPIF1<0，解SKIPIF1<0得SKIPIF1<0解SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0的單調(diào)減區(qū)間是SKIPIF1<0的單調(diào)增區(qū)間是SKIPIF1<0．(2)由（1）知SKIPIF1<0，而SKIPIF1<0，所以曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0．專(zhuān)項(xiàng)突破三利用導(dǎo)數(shù)求函數(shù)單調(diào)區(qū)間(含參)1．設(shè)函數(shù)SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間．【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．若SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．綜上所述，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．2．已知函數(shù)SKIPIF1<0.(1)求函數(shù)f(x)的單調(diào)區(qū)間；(2)若f(x)≥0對(duì)定義域內(nèi)的任意x恒成立，求實(shí)數(shù)a的取值范圍.【解析】(1)求導(dǎo)可得SKIPIF1<0①SKIPIF1<0時(shí)，令SKIPIF1<0可得SKIPIF1<0，由于SKIPIF1<0知SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；②SKIPIF1<0時(shí)，令SKIPIF1<0可得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，由于SKIPIF1<0知SKIPIF1<0或SKIPIF1<0；∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；③SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；④SKIPIF1<0時(shí)，令SKIPIF1<0可得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，由于SKIPIF1<0知SKIPIF1<0或SKIPIF1<0∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；(2)由（1）SKIPIF1<0時(shí)，SKIPIF1<0，（不符合，舍去）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故函數(shù)在SKIPIF1<0處取得最小值，所以函數(shù)SKIPIF1<0對(duì)定義域內(nèi)的任意x恒成立時(shí)，只需要SKIPIF1<0即可，∴SKIPIF1<0.綜上，SKIPIF1<0.3．設(shè)函數(shù)SKIPIF1<0其中SKIPIF1<0．（1）當(dāng)SKIPIF1<0時(shí)，求曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)斜率；（2）求函數(shù)SKIPIF1<0的單調(diào)區(qū)間．【解析】（1）由題設(shè)，SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，故點(diǎn)SKIPIF1<0處的切線(xiàn)斜率為1.（2）由題設(shè)，SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0單調(diào)遞減；∴SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0、SKIPIF1<0上遞減.4．已知函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性．【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0恒成立，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)；若SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，在SKIPIF1<0上為減函數(shù)，綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上為減函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上為增函數(shù)，在SKIPIF1<0上為減函數(shù)．5．已知函數(shù)SKIPIF1<0SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0恰有一個(gè)零點(diǎn)，求a的值．【解析】(1)SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0．因?yàn)镾KIPIF1<0，則SKIPIF1<0，即原方程有兩根設(shè)為SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0（舍去），SKIPIF1<0．則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0在SKIPIF1<0上是減函數(shù)，在SKIPIF1<0上是增函數(shù)．(2)由（1）可知SKIPIF1<0．①若SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減所以SKIPIF1<0至多有一解且SKIPIF1<0，則SKIPIF1<0，代入解得SKIPIF1<0．②若SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，結(jié)合①可得SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0存在一個(gè)零點(diǎn)．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0存在一個(gè)零點(diǎn)．因此SKIPIF1<0存在兩個(gè)零點(diǎn)，不合題意綜上所述：SKIPIF1<0．6．已知函數(shù)SKIPIF1<0SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0在SKIPIF1<0處的切線(xiàn)方程；(2)討論SKIPIF1<0的單調(diào)性.【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0處的切線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0；(2)SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在R上單調(diào)遞增；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在R上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．7．設(shè)函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，求SKIPIF1<0的極值；(2)討論函數(shù)SKIPIF1<0的單調(diào)性.【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，所以當(dāng)SKIPIF1<0時(shí)，該函數(shù)有極小值SKIPIF1<0，無(wú)極大值.(2)由SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)在SKIPIF1<0時(shí)，單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，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，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，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，綜上所述：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增8．已知函數(shù)SKIPIF1<0(其中常數(shù)SKIPIF1<0)，討論SKIPIF1<0的單調(diào)性；【解析】SKIPIF1<0，記SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增．②當(dāng)SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有兩個(gè)實(shí)根SKIPIF1<0，SKIPIF1<0，注意到SKIPIF1<0，SKIPIF1<0且對(duì)稱(chēng)軸SKIPIF1<0，故SKIPIF1<0，所以當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減．綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減．專(zhuān)項(xiàng)突破四利用函數(shù)單調(diào)性比較大小一、單選題1．已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則以下不等式正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】令SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0單調(diào)遞減,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0故選:C2．設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】根據(jù)題意，SKIPIF1<0，則SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，所以SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0.故選：A3．已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】根據(jù)題意，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0；由SKIPIF1<0得SKIPIF1<0；則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，所以SKIPIF1<0，因此SKIPIF1<0．故選：D．4．已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0大?。?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，SKIPIF1<0單調(diào)遞增，又由SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B.5．已知SKIPIF1<0，則a、b、c的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由題可知SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，即c最大；對(duì)于a、b，構(gòu)造函數(shù)SKIPIF1<0，因?yàn)镾KIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；所以SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，綜上，SKIPIF1<0.故選：A6．若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0恒成立，故SKIPIF1<0單調(diào)遞增，由SKIPIF1<0可得：SKIPIF1<0，故SKIPIF1<0，A錯(cuò)誤，B正確；SKIPIF1<0可能比1大，可能等于1，也可能SKIPIF1<0，故不能確定SKIPIF1<0與0的大小關(guān)系，CD錯(cuò)誤.故選：B7．已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，于是當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0單調(diào)遞減，注意到SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0.故選：B.8．已知函數(shù)SKIPIF1<0為函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，滿(mǎn)足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下面大小關(guān)系正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】根據(jù)題意，SKIPIF1<0，變換可得：SKIPIF1<0SKIPIF1<0，解析可得，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，故選：A.9．已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0構(gòu)造函數(shù)SKIPIF1<0且SKIPIF1<0當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,此時(shí)SKIPIF1<0;當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,此時(shí)SKIPIF1<0.故SKIPIF1<0當(dāng)SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0單調(diào)遞增.故SKIPIF1<0故SKIPIF1<0，SKIPIF1<0又SKIPIF1<0即SKIPIF1<0，故SKIPIF1<0，故選:B10．若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】對(duì)于A,B,令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，且SKIPIF1<0故存在SKIPIF1<0，使得SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0遞增，由于SKIPIF1<0，此時(shí)SKIPIF1<0大小關(guān)系不確定，故A,B均不正確；對(duì)于C,D,設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0單調(diào)遞減，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，故C錯(cuò)誤，D正確，故選：D11．設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】∵SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單減，∴SKIPIF1<0，故SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單減，∴SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0.∴SKIPIF1<0，同理可得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，故選：A二、多選題12．下列命題為真命題的個(gè)數(shù)是（）A．SKIPIF1<0

B．SKIPIF1<0

C．SKIPIF1<0 D．SKIPIF1<0【解析】設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，對(duì)于A，由SK