新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第10練 導(dǎo)數(shù)與函數(shù)的單調(diào)性（解析版）

第10練導(dǎo)數(shù)與函數(shù)的單調(diào)性學(xué)校____________姓名____________班級____________一、單選題1．下列函數(shù)中，定義域是R且為增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】對于A，函數(shù)SKIPIF1<0的定義域是R，且SKIPIF1<0，SKIPIF1<0是R上的增函數(shù)，滿足題意；對于B，函數(shù)SKIPIF1<0是R上的減函數(shù)，SKIPIF1<0不滿足題意；對于C，函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，SKIPIF1<0不滿足題意；對于D，函數(shù)SKIPIF1<0在定義域R上不是單調(diào)函數(shù)，SKIPIF1<0不滿足題意．故選：A．2．函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0為偶函數(shù)，且在SKIPIF1<0上單調(diào)遞增B．SKIPIF1<0為偶函數(shù)，且在SKIPIF1<0上單調(diào)遞減C．SKIPIF1<0為奇函數(shù)，且在SKIPIF1<0上單調(diào)遞增D．SKIPIF1<0為奇函數(shù)，且在SKIPIF1<0上單調(diào)遞減【答案】A【詳解】函數(shù)SKIPIF1<0定義域為R，且SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)，故排除選項C，D；又當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故選項A正確，選項B錯誤，故選：A．3．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：因為函數(shù)SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，故選：C.4．函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象如圖所示，則函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】A【詳解】由SKIPIF1<0圖象知，當(dāng)SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，函數(shù)為增函數(shù)，當(dāng)SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，函數(shù)為減函數(shù)，對應(yīng)圖象為A.故選：A．5．若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實數(shù)a的取值范圍(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題可知，SKIPIF1<0恒成立，故SKIPIF1<0，即SKIPIF1<0．故選：A﹒6．設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，若對任意SKIPIF1<0恒成立，則下列選項正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：因為對任意SKIPIF1<0，SKIPIF1<0，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，即SKIPIF1<0的圖象增長得越來越慢，從圖象上來看函數(shù)是上凸遞增的，所以SKIPIF1<0，又SKIPIF1<0，表示點SKIPIF1<0與點SKIPIF1<0的連線的斜率，由圖可知SKIPIF1<0即SKIPIF1<0，故選：A7．若對任意的SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0成立，則實數(shù)m的最小值是（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0等價于SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減函數(shù)，又由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以實數(shù)SKIPIF1<0的最小值為SKIPIF1<0.故選：D.8．已知關(guān)于x的方程SKIPIF1<0有三個不同的實數(shù)根，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：令SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，又SKIPIF1<0，所以存在SKIPIF1<0，使得SKIPIF1<0，所以在SKIPIF1<0上函數(shù)SKIPIF1<0有唯一的零點，即方程SKIPIF1<0有唯一的解，又因為關(guān)于x的方程SKIPIF1<0有三個不同的實數(shù)根，所以當(dāng)SKIPIF1<0時，原方程要有兩個不同的實數(shù)根，當(dāng)SKIPIF1<0時，由SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的圖像有兩個交點，設(shè)SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，結(jié)合圖像可知，SKIPIF1<0，則SKIPIF1<0．故選：C．二、多選題9．已知函數(shù)SKIPIF1<0（e為自然對數(shù)的底數(shù)，SKIPIF1<0），則關(guān)于函數(shù)SKIPIF1<0，下列結(jié)論正確的是（

）A．有2個零點 B．有2個極值點 C．在SKIPIF1<0單調(diào)遞增 D．最小值為1【答案】BC【詳解】SKIPIF1<0定義域為R，SKIPIF1<0，令SKIPIF1<0得：SKIPIF1<0或1，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，如下表：SKIPIF1<0SKIPIF1<00SKIPIF1<01SKIPIF1<0SKIPIF1<0-0+0-SKIPIF1<0遞減極小值1遞增極大值SKIPIF1<0遞減從而判斷出函數(shù)有兩個極值點，在SKIPIF1<0上單調(diào)遞增，BC正確，由于SKIPIF1<0恒成立，所以函數(shù)SKIPIF1<0無零點，A錯誤，當(dāng)SKIPIF1<0時，SKIPIF1<0，故函數(shù)無最小值，D錯誤；.故選：BC10．函數(shù)SKIPIF1<0的導(dǎo)函數(shù)的圖象如圖所示，則下列說法正確的是（

）A．3是SKIPIF1<0的極小值點B．SKIPIF1<0是SKIPIF1<0的極小值點C．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減D．曲線SKIPIF1<0在SKIPIF1<0處的切線斜率小于零【答案】AD【詳解】A：由導(dǎo)函數(shù)的圖象可知：當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，所以3是SKIPIF1<0的極小值點，因此本選項說法正確；B：由導(dǎo)函數(shù)的圖象可知：當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0不是SKIPIF1<0的極小值點，因此本選項說法不正確；C：由導(dǎo)函數(shù)的圖象可知：當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，所以本選項說法不正確；D：：由導(dǎo)函數(shù)的圖象可知：SKIPIF1<0，所以本選項說法正確，故選：AD11．已知SKIPIF1<0，下列說法正確的是（

）A．SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0 B．SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0C．SKIPIF1<0的極大值為SKIPIF1<0 D．方程SKIPIF1<0有兩個不同的解【答案】BC【詳解】對于A，由SKIPIF1<0（SKIPIF1<0），得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，所以A錯誤，對于B，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，所以B正確，對于C，由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0取得極大值SKIPIF1<0，所以C正確，對于D，由C選項可知SKIPIF1<0的最大值為SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0與SKIPIF1<0的交點個數(shù)為1，所以SKIPIF1<0有1個解，所以D錯誤，故選：BC12．已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的極大值為SKIPIF1<0 B．SKIPIF1<0的極大值為SKIPIF1<0C．曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0 D．曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0【答案】BD【詳解】解：因為SKIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0或SKIPIF1<0時SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0的極大值為SKIPIF1<0，故A錯誤，B正確；因為SKIPIF1<0.所以曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，故C錯誤，D正確；故選：BD三、解答題13．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間.【詳解】由SKIPIF1<0，SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0單調(diào)遞增區(qū)間為SKIPIF1<0；單調(diào)遞減區(qū)間為SKIPIF1<0.14．已知SKIPIF1<0.（1）當(dāng)SKIPIF1<0時，討論SKIPIF1<0的單調(diào)區(qū)間；（2）若SKIPIF1<0在定義域R內(nèi)單調(diào)遞增，求a的取值范圍.【詳解】（1）當(dāng)SKIPIF1<0時，SKIPIF1<0則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0所以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（2）由題可知：SKIPIF1<0在定義域R內(nèi)單調(diào)遞增等價于SKIPIF1<0由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0則SKIPIF1<015．已知函數(shù)SKIPIF1<0，其中k∈R.當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；【詳解】由題設(shè)，SKIPIF1<0，

當(dāng)SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，故SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<