新高考數(shù)學(xué)二輪考點培優(yōu)專題（精講+精練）5 嵌套函數(shù)的零點問題（含解析）

素養(yǎng)拓展05嵌套函數(shù)的零點問題（精講+精練）一、知識點梳理一、知識點梳理1.嵌套函數(shù)形式：形如f2.解決嵌套函數(shù)零點個數(shù)的一般步驟(1)換元解套，轉(zhuǎn)化為t＝g(x)與y＝f(t)的零點．(2)依次解方程，令f(t)＝0，求t，代入t＝g(x)求出x的值或判斷圖象交點個數(shù)．注：抓住兩點：(1)轉(zhuǎn)化換元；(2)充分利用函數(shù)的圖象與性質(zhì)．二、題型精講精練二、題型精講精練【典例1】已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點個數(shù)是（

）A．4 B．5 C．6 D．7分析：令SKIPIF1<0→SKIPIF1<0→作函數(shù)SKIPIF1<0與SKIPIF1<0圖象→兩個交點的橫坐標(biāo)為SKIPIF1<0→SKIPIF1<0、SKIPIF1<0判斷SKIPIF1<0的零點個數(shù).【解析】令SKIPIF1<0，則SKIPIF1<0，作出SKIPIF1<0的圖象和直線SKIPIF1<0，由圖象可得有兩個交點，設(shè)橫坐標(biāo)為SKIPIF1<0，∴SKIPIF1<0.當(dāng)SKIPIF1<0時，有SKIPIF1<0，即有一解；當(dāng)SKIPIF1<0時，有三個解∴綜上，SKIPIF1<0共有4個解，即有4個零點，故選A【題型訓(xùn)練】一、單選題1．（2023春·高三平湖市當(dāng)湖高級中學(xué)校聯(lián)考期中）已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0零點個數(shù)最多是（

）A．10 B．12 C．14 D．16【答案】B【分析】畫出SKIPIF1<0的圖像，設(shè)SKIPIF1<0，首先討論SKIPIF1<0的根的情況，再分析SKIPIF1<0根的情況即可分析出SKIPIF1<0根的情況，即可得出答案．【詳解】畫出SKIPIF1<0的圖像，如圖所示，由SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，設(shè)SKIPIF1<0，由圖像可知SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0的圖像，如圖所示，由圖像可知，SKIPIF1<0，①當(dāng)SKIPIF1<0時，即SKIPIF1<0，沒有根；②當(dāng)SKIPIF1<0時，即SKIPIF1<0，此時有3個根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，即SKIPIF1<0，有3個根，當(dāng)SKIPIF1<0時，即SKIPIF1<0，有4個根，當(dāng)SKIPIF1<0時，即SKIPIF1<0，有4個根，故SKIPIF1<0時，SKIPIF1<0有11個根；③當(dāng)SKIPIF1<0時，SKIPIF1<0，此時有三個根，SKIPIF1<0，當(dāng)SKIPIF1<0時，即SKIPIF1<0，有4個根，當(dāng)SKIPIF1<0時，即SKIPIF1<0，有4個根，當(dāng)SKIPIF1<0時，即SKIPIF1<0，有4個根，故SKIPIF1<0時，SKIPIF1<0有12個根；綜上所述，SKIPIF1<0最多有12個根，故選：B．2．（2023春·廣東揭陽·高三校聯(lián)考階段練習(xí)）函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點個數(shù)為（

）A．2 B．3 C．4 D．5【答案】A【分析】令SKIPIF1<0，結(jié)合題意得到SKIPIF1<0的兩根為SKIPIF1<0，SKIPIF1<0，然后根據(jù)函數(shù)SKIPIF1<0的單調(diào)性和最值進(jìn)而求解.【詳解】令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0（舍去）；當(dāng)SKIPIF1<0時，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0的兩根為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，若SKIPIF1<0，易知方程無解，若SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0（舍去），此時方程有唯一的解；當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0，此時方程有唯一的解，綜上所述可知函數(shù)SKIPIF1<0的零點個數(shù)為SKIPIF1<0個，故選：A.3．（2023秋·福建廈門·高三統(tǒng)考期末）已知函數(shù)SKIPIF1<0，則方程SKIPIF1<0的實數(shù)解的個數(shù)至多是（

）A．5 B．6 C．7 D．8【答案】B【分析】根據(jù)復(fù)合方程問題，換元SKIPIF1<0，作函數(shù)圖象分別看內(nèi)外層分別討論方程SKIPIF1<0根的個數(shù)情況，即可得答案.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0化為SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，如圖為函數(shù)SKIPIF1<0的大致圖象：

由圖可得，當(dāng)SKIPIF1<0時，SKIPIF1<0有兩個根SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，此時方程SKIPIF1<0最多有5個根；當(dāng)SKIPIF1<0時，SKIPIF1<0有三個根SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，此時方程SKIPIF1<0最多有6個根；當(dāng)SKIPIF1<0時，SKIPIF1<0有兩個根SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，此時方程SKIPIF1<0有4個根；當(dāng)SKIPIF1<0時，SKIPIF1<0有一個根SKIPIF1<0，即SKIPIF1<0，此時方程SKIPIF1<0有2個根；綜上，方程SKIPIF1<0的實數(shù)解的個數(shù)至多是6個.故選：B.4．（2023·全國·高三期末）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0的所有實根之和為4，則實數(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】由題對SKIPIF1<0取特殊值，利用數(shù)形結(jié)合，排除不合題意的選項即得.【詳解】令SKIPIF1<0，當(dāng)SKIPIF1<0時，方程為SKIPIF1<0，即SKIPIF1<0，作出函數(shù)SKIPIF1<0及SKIPIF1<0的圖象，由圖象可知方程的根為SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象，結(jié)合圖象可得所有根的和為5，不合題意，故BD錯誤；當(dāng)SKIPIF1<0時，方程為SKIPIF1<0，即SKIPIF1<0，由圖象可知方程的根SKIPIF1<0，即SKIPIF1<0，結(jié)合函數(shù)SKIPIF1<0的圖象，可得方程有四個根，所有根的和為4，滿足題意，故A錯誤.故選：C.5．（2023秋·河南信陽·高三信陽高中?？计谀┮阎瘮?shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點個數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】確定函數(shù)SKIPIF1<0的值域，利用換元法令SKIPIF1<0，則SKIPIF1<0，則將函數(shù)SKIPIF1<0的零點問題轉(zhuǎn)化為函數(shù)SKIPIF1<0的圖象的交點問題，作函數(shù)SKIPIF1<0圖象，確定其交點以及其橫坐標(biāo)范圍，再結(jié)合SKIPIF1<0的圖象，即可確定SKIPIF1<0的零點個數(shù).【詳解】已知SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，作出其圖象如圖示：可知SKIPIF1<0值域為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，則函數(shù)SKIPIF1<0的零點問題即為函數(shù)SKIPIF1<0的圖象的交點問題，而SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖示：可知：SKIPIF1<0的圖象有兩個交點，橫坐標(biāo)分別在SKIPIF1<0之間，不妨設(shè)交點橫坐標(biāo)為SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0圖象和直線SKIPIF1<0可知，二者有兩個交點，即此時SKIPIF1<0有兩個零點；當(dāng)SKIPIF1<0時，由SKIPIF1<0圖象和直線SKIPIF1<0可知，二者有3個交點，即此時SKIPIF1<0有3個零點，故函數(shù)SKIPIF1<0的零點個數(shù)是5，故選：B.【點睛】本題考查了復(fù)合函數(shù)的零點個數(shù)的確定問題，綜合性較強，涉及到函數(shù)的值域以及分段函數(shù)的性質(zhì)的應(yīng)用和數(shù)形結(jié)合的思想方法，解答的關(guān)鍵是采用換元法將函數(shù)的零點問題轉(zhuǎn)化為函數(shù)圖象的交點問題.6．（2023春·江西吉安·高三吉安一中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰有兩個零點，則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】設(shè)SKIPIF1<0，進(jìn)而考慮SKIPIF1<0與SKIPIF1<0的交點，分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0五種情況討論求解即可.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，我們先來考慮SKIPIF1<0與SKIPIF1<0的交點，令SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0只有1個交點，交點橫坐標(biāo)SKIPIF1<0，此時SKIPIF1<0有1個零點；當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0只有2個交點，交點橫坐標(biāo)SKIPIF1<0，此時SKIPIF1<0有3個零點.當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0只有3個交點，交點橫坐標(biāo)SKIPIF1<0，此時SKIPIF1<0有5個零點.若SKIPIF1<0與SKIPIF1<0相切時，設(shè)切點SKIPIF1<0，所以，切線斜率SKIPIF1<0，解得SKIPIF1<0，故當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0沒有交點，SKIPIF1<0沒有零點.當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0有2個交點，交點橫坐標(biāo)SKIPIF1<0，此時SKIPIF1<0有2個零點.故選：C【點睛】關(guān)鍵點點睛：本題解題的關(guān)鍵在于通過換元SKIPIF1<0，將問題轉(zhuǎn)化為直線SKIPIF1<0與SKIPIF1<0的交點個數(shù)，進(jìn)而數(shù)形結(jié)合，分類討論求解即可.7．（2023春·安徽滁州·高一?？奸_學(xué)考試）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有兩個零點，則函數(shù)SKIPIF1<0的零點個數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】作出函數(shù)SKIPIF1<0的圖象，根據(jù)題意利用圖象分析可得SKIPIF1<0，令SKIPIF1<0并將問題轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0交點橫坐標(biāo)t對應(yīng)x值的個數(shù)，結(jié)合數(shù)形結(jié)合法求零點個數(shù)即可.【詳解】當(dāng)SKIPIF1<0時，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0；當(dāng)SKIPIF1<0時，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.作出函數(shù)SKIPIF1<0的圖象如圖所示，令SKIPIF1<0，則SKIPIF1<0，若函數(shù)SKIPIF1<0有兩個零點，則函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有兩個交點，所以SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，由圖象可得SKIPIF1<0有SKIPIF1<0個實數(shù)根，SKIPIF1<0有SKIPIF1<0個實數(shù)根，故SKIPIF1<0的零點個數(shù)為SKIPIF1<0，故選：B．8．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0為自然對數(shù)的底數(shù)），則函數(shù)SKIPIF1<0的零點個數(shù)為（

）A．3 B．5 C．7 D．9【答案】C【分析】作出函數(shù)SKIPIF1<0的圖象，可設(shè)SKIPIF1<0，可得SKIPIF1<0，判斷SKIPIF1<0與SKIPIF1<0交點個數(shù)，進(jìn)而將SKIPIF1<0的零點個數(shù)問題轉(zhuǎn)化為函數(shù)SKIPIF1<0的圖象交點個數(shù)問題，數(shù)形結(jié)合，可得答案.【詳解】設(shè)SKIPIF1<0，令SKIPIF1<0可得：SKIPIF1<0,對于SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0處切線的斜率值為SKIPIF1<0，設(shè)SKIPIF1<0與SKIPIF1<0相切于點SKIPIF1<0，SKIPIF1<0切線斜率SKIPIF1<0，則切線方程為：SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0；由于SKIPIF1<0，故作出SKIPIF1<0與SKIPIF1<0圖象如下圖所示，SKIPIF1<0與SKIPIF1<0有四個不同交點，即SKIPIF1<0與SKIPIF1<0有四個不同交點，設(shè)三個交點為SKIPIF1<0，由圖象可知：SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖，由此可知SKIPIF1<0與SKIPIF1<0無交點，與SKIPIF1<0有三個不同交點，與SKIPIF1<0各有兩個不同交點，SKIPIF1<0的零點個數(shù)為7個，故選：C【點睛】方法點睛：解決此類復(fù)合函數(shù)的零點問題，常常采用換元的方法，將零點問題轉(zhuǎn)化為函數(shù)圖象的交點問題，數(shù)形結(jié)合，即可解決.9．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0恰有5個零點，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題意可先做出函數(shù)SKIPIF1<0的大致圖象，利用數(shù)形結(jié)合和分類討論，即可確定m的取值范圍.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0的大致圖象如圖所示．設(shè)SKIPIF1<0，則SKIPIF1<0，由圖可知當(dāng)SKIPIF1<0時，SKIPIF1<0有且只有1個實根，則SKIPIF1<0最多有3個不同的實根，不符合題意．當(dāng)SKIPIF1<0時，SKIPIF1<0的解是SKIPIF1<0，SKIPIF1<0．SKIPIF1<0有2個不同的實根，SKIPIF1<0有2個不同的實根，則SKIPIF1<0有4個不同的實根，不符合題意．當(dāng)SKIPIF1<0時，SKIPIF1<0有3個不同的實根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．SKIPIF1<0有2個不同的實根，SKIPIF1<0有2個不同的實根，SKIPIF1<0有3個不同的實根，則SKIPIF1<0有7個不同的實根，不符合題意．當(dāng)SKIPIF1<0時，SKIPIF1<0有2個不同的實根SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0．SKIPIF1<0有2個不同的實根，SKIPIF1<0有3個不同的實根，則SKIPIF1<0有5個不同的實根，符合題意．當(dāng)SKIPIF1<0時，SKIPIF1<0有2個不同的實根SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0有2個不同的實根，SKIPIF1<0，有2個不同的實根，則SKIPIF1<0有4個不同的實根，不符合題意．當(dāng)SKIPIF1<0時，SKIPIF1<0有且只有1個實根，則SKIPIF1<0最多有3個不同的實根，不符合題意，綜上，m的取值范圍是SKIPIF1<0．故選：C.【點睛】方法點睛：對于函數(shù)零點問題，若能夠畫圖時可作出函數(shù)圖像，利用數(shù)形結(jié)合與分類討論思想，即可求解.本題中，由圖看出，m的討論應(yīng)有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0這幾種情況,也是解題關(guān)鍵.二、填空題10．（2023秋·貴州畢節(jié)·高三統(tǒng)考期末）已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的所有零點之和為___________．【答案】SKIPIF1<0【分析】利用分段函數(shù)，分類討論，即可求出函數(shù)SKIPIF1<0的所有零點，從而得解．【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，①當(dāng)SKIPIF1<0時，SKIPIF1<0，得SKIPIF1<0；②當(dāng)SKIPIF1<0時，SKIPIF1<0，得SKIPIF1<0；綜上所述：若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0.故SKIPIF1<0或SKIPIF1<0，則有：①由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；②由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；綜上所述：函數(shù)SKIPIF1<0的所有零點為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，4.故所有零點的和為SKIPIF1<0．故答案為：SKIPIF1<0.【點睛】關(guān)鍵點點睛：根據(jù)題意分SKIPIF1<0和SKIPIF1<0兩種情況討論，運算求解,11．（2023·福建福州·高三福州三中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0則函數(shù)SKIPIF1<0的零點個數(shù)為___________．【答案】5【分析】方法一：令SKIPIF1<0，將問題轉(zhuǎn)化為SKIPIF1<0，根據(jù)圖象分析得SKIPIF1<0有兩個零點為SKIPIF1<0，SKIPIF1<0，從而考慮SKIPIF1<0與SKIPIF1<0根的個數(shù)即可求解；方法二：利用導(dǎo)函數(shù)以及零點的存在性定理討論SKIPIF1<0的根分別為SKIPIF1<0．SKIPIF1<0，從而用數(shù)形結(jié)合的方法確定SKIPIF1<0與SKIPIF1<0根的個數(shù)即可求解.【詳解】方法一：SKIPIF1<0大致圖象如下令SKIPIF1<0SKIPIF1<0所以SKIPIF1<0式方程的一個根SKIPIF1<0，再由圖可知SKIPIF1<0式方程的另一個根SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0的圖象有2個交點，所以SKIPIF1<0有2個實根，當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0的圖象有3個交點，所以SKIPIF1<0有3個實根，SKIPIF1<0共有5個零點.方法二：令SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0有且僅有一個零點SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0有且僅有一個零點SKIPIF1<0，其中SKIPIF1<0．SKIPIF1<0時，SKIPIF1<0時，SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0有且僅有一個零點SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時，結(jié)合函數(shù)圖象可知SKIPIF1<0無解，SKIPIF1<0有兩個根SKIPIF1<0因為SKIPIF1<0，所以由圖象可得SKIPIF1<0與SKIPIF1<0的圖象有2個交點，所以SKIPIF1<0有2個實根，當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0的圖象有3個交點，所以SKIPIF1<0有3個實根，SKIPIF1<0共有5個零點.故答案為:5.12．（2023秋·廣東深圳·高三深圳市高級中學(xué)?？茧A段練習(xí)）已知SKIPIF1<0，SKIPIF1<0為三次函數(shù)，其圖象如圖所示.若SKIPIF1<0有9個零點，則SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【分析】根據(jù)SKIPIF1<0的圖象判斷與SKIPIF1<0在不同m取值范圍下的交點個數(shù)，并確定交點橫坐標(biāo)的范圍，結(jié)合SKIPIF1<0解析式求橫坐標(biāo)關(guān)于m的表達(dá)式，結(jié)合題圖及SKIPIF1<0有9個零點，列不等式組求m范圍.【詳解】由題設(shè)SKIPIF1<0，其圖象如下，當(dāng)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0只有一個交點且SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0有兩個交點且SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0有三個交點且SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0有兩個交點且SKIPIF1<0；由題圖，要使SKIPIF1<0，SKIPIF1<0有9個零點，則SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0有SKIPIF1<0，根據(jù)SKIPIF1<0解析式：SKIPIF1<0，綜上，SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<0【點睛】關(guān)鍵點點睛：根據(jù)SKIPIF1<0、SKIPIF1<0的圖象及SKIPIF1<0零點個數(shù)，確定SKIPIF1<0時函數(shù)SKIPIF1<0對應(yīng)零點的范圍，進(jìn)而求各零點關(guān)于m的表達(dá)式，列不等式求參數(shù)范圍.13．（2023秋·江蘇泰州·高三統(tǒng)考期末）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有三個實數(shù)解，則實數(shù)SKIPIF1<0的取值集合為______.【答案】SKIPIF1<0【分析】當(dāng)SKIPIF1<0時，易知SKIPIF1<0無解；當(dāng)SKIPIF1<0時，設(shè)SKIPIF1<0，采用數(shù)形結(jié)合的方式可知SKIPIF1<0，可知SKIPIF1<0無解；當(dāng)SKIPIF1<0時，設(shè)SKIPIF1<0，采用數(shù)形結(jié)合的方式可知SKIPIF1<0，通過討論SKIPIF1<0的范圍可確定SKIPIF1<0或SKIPIF1<0的取值，由此可構(gòu)造方程求得SKIPIF1<0的值.【詳解】SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，此時SKIPIF1<0無解，不合題意；當(dāng)SKIPIF1<0時，設(shè)SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的大致圖象如下圖所示，則SKIPIF1<0對應(yīng)的兩根為SKIPIF1<0，此時SKIPIF1<0與SKIPIF1<0無解，即方程SKIPIF1<0無解，不合題意；當(dāng)SKIPIF1<0時，設(shè)SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的大致圖象如下圖所示，則SKIPIF1<0對應(yīng)的兩根為SKIPIF1<0，若SKIPIF1<0恰有三個實數(shù)解，則SKIPIF1<0和SKIPIF1<0與SKIPIF1<0共有SKIPIF1<0個不同的交點，①當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0有兩個不同交點，如圖所示，SKIPIF1<0與SKIPIF1<0有且僅有一個交點，則SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0；②當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0有兩個不同交點，SKIPIF1<0