新高考數(shù)學(xué)一輪復(fù)習(xí) 專項(xiàng)分層精練第11課 函數(shù)與方程（解析版）

第11課函數(shù)與方程（分層專項(xiàng)精練）【一層練基礎(chǔ)】一、單選題1．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）.A．1 B．2 C．3 D．4【答案】C【分析】通過(guò)解法方程SKIPIF1<0來(lái)求得SKIPIF1<0的零點(diǎn)個(gè)數(shù).【詳解】由SKIPIF1<0可得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，或SKIPIF1<0（舍去），當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0.故SKIPIF1<0是SKIPIF1<0的零點(diǎn)，SKIPIF1<0是SKIPIF1<0的零點(diǎn)，SKIPIF1<0是SKIPIF1<0的零點(diǎn).綜上所述，SKIPIF1<0共有SKIPIF1<0個(gè)零點(diǎn).故選：C2．（2012秋·上?！じ呷y(tǒng)考期中）已知SKIPIF1<0是函數(shù)SKIPIF1<0的零點(diǎn)，若SKIPIF1<0，則SKIPIF1<0的值滿足A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0的符號(hào)不確定【答案】B【分析】數(shù)形結(jié)合分析即可【詳解】不妨設(shè)SKIPIF1<0,則SKIPIF1<0,作出SKIPIF1<0圖像如下:則可以得到SKIPIF1<0點(diǎn)的橫坐標(biāo)即為SKIPIF1<0的零點(diǎn)SKIPIF1<0，此時(shí)SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0故選：B3．（2023春·廣東深圳·高一?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0有4個(gè)零點(diǎn)，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【分析】在同一坐標(biāo)系中作出SKIPIF1<0的圖象，根據(jù)SKIPIF1<0有4個(gè)零點(diǎn)求解.【詳解】解：令SKIPIF1<0，得SKIPIF1<0，在同一坐標(biāo)系中作出SKIPIF1<0的圖象，如圖所示：由圖象知：若SKIPIF1<0有4個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是SKIPIF1<0，故選：A二、多選題4．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0(即SKIPIF1<0，SKIPIF1<0)則（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是偶函數(shù) B．SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)C．設(shè)SKIPIF1<0最小值為SKIPIF1<0，則SKIPIF1<0 D．方程SKIPIF1<0可能有2個(gè)解【答案】ABD【分析】結(jié)合奇偶函數(shù)的定義和二次函數(shù)的性質(zhì)逐一判斷選項(xiàng)即可.【詳解】SKIPIF1<0：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，故正確；SKIPIF1<0：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，開(kāi)口向上，此時(shí)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，開(kāi)口向上，此時(shí)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，綜上，SKIPIF1<0在SKIPIF1<0上是增函數(shù)，故SKIPIF1<0正確；SKIPIF1<0：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)椴荒艽_定SKIPIF1<0的大小，所以最小值SKIPIF1<0無(wú)法判斷，故SKIPIF1<0錯(cuò)誤；SKIPIF1<0：令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0有2個(gè)解，故SKIPIF1<0正確.故選：ABD5．（2022秋·湖北省直轄縣級(jí)單位·高一?？茧A段練習(xí)）若函數(shù)SKIPIF1<0的圖像在R上連續(xù)不斷，且滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列說(shuō)法錯(cuò)誤的是（

）A．SKIPIF1<0在區(qū)間(0，1)上一定有零點(diǎn)，在區(qū)間(1，2)上一定沒(méi)有零點(diǎn)B．SKIPIF1<0在區(qū)間(0，1)上一定沒(méi)有零點(diǎn)，在區(qū)間(1，2)上一定有零點(diǎn)C．SKIPIF1<0在區(qū)間(0，1)上一定有零點(diǎn)，在區(qū)間(1，2)上可能有零點(diǎn)D．SKIPIF1<0在區(qū)間(0，1)上可能有零點(diǎn)，在區(qū)間(1，2)上一定有零點(diǎn)【答案】ABD【解析】根據(jù)SKIPIF1<0的圖像在SKIPIF1<0上連續(xù)不斷，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，結(jié)合零點(diǎn)存在定理，判斷出在區(qū)間SKIPIF1<0和SKIPIF1<0上零點(diǎn)存在的情況，得到答案．【詳解】由題知SKIPIF1<0，所以根據(jù)函數(shù)零點(diǎn)存在定理可得SKIPIF1<0在區(qū)間SKIPIF1<0上一定有零點(diǎn)，又SKIPIF1<0，無(wú)法判斷SKIPIF1<0在區(qū)間SKIPIF1<0上是否有零點(diǎn)，在區(qū)間(1，2)上可能有零點(diǎn)．故選：SKIPIF1<0．三、填空題6．（2019·浙江·高三專題練習(xí)）已知函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【分析】將函數(shù)的零點(diǎn)轉(zhuǎn)化為SKIPIF1<0有一個(gè)零點(diǎn)，SKIPIF1<0有兩個(gè)零點(diǎn)，結(jié)合零點(diǎn)分布分析運(yùn)算．【詳解】根據(jù)題意得：SKIPIF1<0有一個(gè)零點(diǎn)，SKIPIF1<0有兩個(gè)零點(diǎn)若SKIPIF1<0有一個(gè)零點(diǎn)，則SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有兩個(gè)零點(diǎn)則可得SKIPIF1<0，得SKIPIF1<0故答案為：SKIPIF1<0．【二層練綜合】一、單選題1．（2022·全國(guó)·高三專題練習(xí)）求下列函數(shù)的零點(diǎn)，可以采用二分法的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0不是單調(diào)函數(shù)，SKIPIF1<0，不能用二分法求零點(diǎn)；SKIPIF1<0是單調(diào)函數(shù)，SKIPIF1<0，能用二分法求零點(diǎn)；SKIPIF1<0不是單調(diào)函數(shù)，SKIPIF1<0，不能用二分法求零點(diǎn)；SKIPIF1<0不是單調(diào)函數(shù)，SKIPIF1<0，不能用二分法求零點(diǎn).故選：B2．（2023秋·高一課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是（

）A．2 B．3 C．4 D．5【答案】D【分析】令SKIPIF1<0，根據(jù)SKIPIF1<0分別求出函數(shù)SKIPIF1<0的零點(diǎn)或零點(diǎn)所在區(qū)間，再作出函數(shù)SKIPIF1<0的圖象，根據(jù)數(shù)形結(jié)合即可求出函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)；【詳解】令SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由于SKIPIF1<0，由零點(diǎn)存在定理可知，存在SKIPIF1<0，使得SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0．作出函數(shù)SKIPIF1<0，直線SKIPIF1<0的圖象如下圖所示：

由圖象可知，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)；直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)；直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有且只有一個(gè)交點(diǎn)．綜上所述，函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為5．故選：D．3．（2023·全國(guó)·高二專題練習(xí)）函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)有極值，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由可導(dǎo)函數(shù)在開(kāi)區(qū)間內(nèi)有極值的充要條件即可作答.【詳解】由SKIPIF1<0得，SKIPIF1<0，因函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)有極值，則SKIPIF1<0時(shí)，SKIPIF1<0有解，即在SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與直線y=a有公共點(diǎn)，而SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，則SKIPIF1<0，顯然在SKIPIF1<0零點(diǎn)左右兩側(cè)SKIPIF1<0異號(hào)，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C【點(diǎn)睛】結(jié)論點(diǎn)睛：可導(dǎo)函數(shù)y＝f(x)在點(diǎn)x0處取得極值的充要條件是f′(x0)＝0，且在x0左側(cè)與右側(cè)f′(x)的符號(hào)不同．二、多選題4．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0若關(guān)于x的方程SKIPIF1<0有5個(gè)不同的實(shí)根，則實(shí)數(shù)a的取值可以為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【分析】換元，將原方程根的個(gè)數(shù)問(wèn)題轉(zhuǎn)化二次函數(shù)零點(diǎn)的分布問(wèn)題，結(jié)合圖象可解.【詳解】令SKIPIF1<0，記SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0，則由SKIPIF1<0的圖象可知：方程SKIPIF1<0有5個(gè)不同的實(shí)根SKIPIF1<0SKIPIF1<0與SKIPIF1<0的圖象共有5個(gè)交點(diǎn)SKIPIF1<0，且SKIPIF1<0（不妨設(shè)SKIPIF1<0）.則SKIPIF1<0解得SKIPIF1<0.故選：BCD5．（2023·廣東廣州·廣州市培正中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0恰有兩個(gè)不同解SKIPIF1<0，則SKIPIF1<0的取值可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．2【答案】BC【分析】利用函數(shù)的單調(diào)性以及已知條件得到SKIPIF1<0，代入SKIPIF1<0，令SKIPIF1<0，求導(dǎo)，利用導(dǎo)函數(shù)的單調(diào)性分析原函數(shù)的單調(diào)性，即可求出取值范圍.【詳解】因?yàn)镾KIPIF1<0的兩根為SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上恒成立，從而SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0，故選：BC．【點(diǎn)睛】關(guān)鍵點(diǎn)睛：本題考查利用導(dǎo)數(shù)解決函數(shù)的范圍問(wèn)題.構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)求取值范圍是解決本題的關(guān)鍵.三、填空題6．（2023·全國(guó)·高一專題練習(xí)）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則x、y、z由小到大的順序是.【答案】SKIPIF1<0【分析】把給定的三個(gè)等式作等價(jià)變形，比較函數(shù)SKIPIF1<0的圖象與曲線SKIPIF1<0交點(diǎn)的橫坐標(biāo)大小作答.【詳解】依題意，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，因此，SKIPIF1<0成立的x值是函數(shù)SKIPIF1<0與SKIPIF1<0的圖象交點(diǎn)的橫坐標(biāo)SKIPIF1<0，SKIPIF1<0成立的y值是函數(shù)SKIPIF1<0與SKIPIF1<0的圖象交點(diǎn)的橫坐標(biāo)SKIPIF1<0，SKIPIF1<0成立的z值是函數(shù)SKIPIF1<0與SKIPIF1<0的圖象交點(diǎn)的橫坐標(biāo)SKIPIF1<0，在同一坐標(biāo)系內(nèi)作出函數(shù)SKIPIF1<0，SKIPIF1<0的圖象，如圖，觀察圖象得：SKIPIF1<0，即SKIPIF1<0，所以x、y、z由小到大的順序是SKIPIF1<0.故答案為：SKIPIF1<0【點(diǎn)睛】思路點(diǎn)睛：涉及某些由指數(shù)式、對(duì)數(shù)式給出的幾個(gè)數(shù)大小比較，可以把這幾個(gè)數(shù)視為對(duì)應(yīng)的指數(shù)、對(duì)數(shù)函數(shù)與另外某個(gè)函數(shù)圖象交點(diǎn)橫坐標(biāo)，利用圖象的直觀性質(zhì)解決.【三層練能力】一、單選題1．（2023春·陜西西安·高二西安市第三中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰有5個(gè)零點(diǎn)SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】將SKIPIF1<0看成整體解出SKIPIF1<0或SKIPIF1<0，作出SKIPIF1<0的大致圖象，將式子化為SKIPIF1<0，然后轉(zhuǎn)化為SKIPIF1<0的范圍進(jìn)行分類討論即可判斷.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減且恒負(fù)，在SKIPIF1<0上單調(diào)遞增且恒負(fù)，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出SKIPIF1<0的大致圖象如圖所示，函數(shù)SKIPIF1<0恰有5個(gè)零點(diǎn)SKIPIF1<0，等價(jià)于方程SKIPIF1<0有5個(gè)不同的實(shí)數(shù)根，解得：SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，該方程有5個(gè)根，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，綜上：SKIPIF1<0的取值范圍是：SKIPIF1<0.故選：B.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題的關(guān)鍵點(diǎn)是對(duì)SKIPIF1<0的理解，將SKIPIF1<0看成一個(gè)SKIPIF1<0，解出其值，然后通過(guò)圖象分析，轉(zhuǎn)化為直線SKIPIF1<0與圖象的交點(diǎn)情況.2．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．2 B．3 C．4 D．5【答案】B【分析】對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，再對(duì)k分類討論以確定函數(shù)的單調(diào)性，函數(shù)有唯一零點(diǎn)的條件，轉(zhuǎn)化為函數(shù)最值即可作答.【詳解】因SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0時(shí)，恒有SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)，SKIPIF1<0時(shí)，SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，從而SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，因函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，則SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，而SKIPIF1<0，于是得SKIPIF1<0的零點(diǎn)SKIPIF1<0，所以SKIPIF1<0.故選：B二、多選題3．（2023·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0分別是函數(shù)SKIPIF1<0和SKIPIF1<0的零點(diǎn)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0D．SKIPIF1<0【答案】BCD【分析】利用函數(shù)與方程思想，得到兩根滿足的方程關(guān)系，然后根據(jù)結(jié)構(gòu)構(gòu)造函數(shù)SKIPIF1<0，求導(dǎo)，研究單調(diào)性，得到SKIPIF1<0及SKIPIF1<0，結(jié)合指對(duì)互化即可判斷選項(xiàng)A、B、C，最后再通過(guò)對(duì)勾函數(shù)單調(diào)性求解范圍即可判斷選項(xiàng)D.【詳解】令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，記函數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故A錯(cuò)誤；又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B正確；所以SKIPIF1<0，故C正確；又SKIPIF1<0，所以SKIPIF1<0，結(jié)合SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0，故D正確；故選：BCD【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題考查函數(shù)的零點(diǎn)問(wèn)題，解題方法是把函數(shù)的零點(diǎn)轉(zhuǎn)化為方程的根，通過(guò)結(jié)構(gòu)構(gòu)造函數(shù)，利用函數(shù)單調(diào)性及指對(duì)互化找到根的關(guān)系得出結(jié)論.三、填空題4．（2023春·安徽滁州·高二?？计谀┮阎瘮?shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0恰有兩個(gè)不相等的實(shí)數(shù)根SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【分析】根據(jù)給定分段函數(shù)，求出函數(shù)SKIPIF1<0的解析式，確定給定方程有兩個(gè)不等實(shí)根的a的取值