新高考數(shù)學(xué)二輪復(fù)習專題2-3 零點與復(fù)合嵌套函數(shù)（原卷版）

專題2-3零點與復(fù)合嵌套函數(shù)目錄TOC\o"1-1"\h\u題型01零點基礎(chǔ)：二分法 1題型02根的分布 2題型03根的分布：指數(shù)函數(shù)二次型 3題型04零點：切線法 3題型05抽象函數(shù)型零點 4題型06分段含參討論型 5題型07參數(shù)分界型討論 5題型08分離參數(shù)型水平線法求零點 6題型09對數(shù)絕對值水平線法 7題型10指數(shù)函數(shù)“一點一線”性質(zhì)型 8題型11零點：中心對稱性質(zhì)型 10題型12零點：軸對稱性質(zhì)型 10題型13嵌套型零點：內(nèi)外自復(fù)合型 11題型14嵌套型零點：內(nèi)外雙函數(shù)復(fù)合型 12題型15嵌套型零點：二次型因式分解 13題型16嵌套型零點：二次型根的分布 14題型17嵌套型零點：放大型函數(shù) 14高考練場 15題型01零點基礎(chǔ)：二分法【解題攻略】用二分法求函數(shù)零點近似值的步驟給定精確度SKIPIF1<0，用二分法求函數(shù)SKIPIF1<0零點SKIPIF1<0的近似值的一般步驟如下：①確定零點SKIPIF1<0的初始區(qū)間SKIPIF1<0，驗證SKIPIF1<0．②求區(qū)間SKIPIF1<0的中點c．③計算SKIPIF1<0，并進一步確定零點所在的區(qū)間：a．若SKIPIF1<0（此時SKIPIF1<0），則c就是函數(shù)的零點．b．若SKIPIF1<0（此時SKIPIF1<0），則令bSKIPIF1<0．c．若SKIPIF1<0（此時SKIPIF1<0，則令aSKIPIF1<0．④判斷是否達到精確度SKIPIF1<0：若SKIPIF1<0SKIPIF1<0，則得到零點近似值a（或b）；否則重復(fù)步驟②～④．【典例1-1】（2022·高三課時練習）已知函數(shù)SKIPIF1<0滿足：對任意SKIPIF1<0，都有SKIPIF1<0，且SKIPIF1<0.在用二分法尋找零點的過程中，依次確定了零點所在區(qū)間為SKIPIF1<0，又SKIPIF1<0，則函數(shù)SKIPIF1<0的零點為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2023·全國·高三專題練習）若函數(shù)SKIPIF1<0的一個正數(shù)零點附近的函數(shù)值用二分法計算，其參考數(shù)據(jù)如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0那么方程SKIPIF1<0的一個近似根（精確度SKIPIF1<0）可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2021秋·湖南·高三校聯(lián)考階段練習）已知函數(shù)SKIPIF1<0的一個零點SKIPIF1<0，用二分法求精確度為0.01的SKIPIF1<0的近似值時，判斷各區(qū)間中點的函數(shù)值的符號最多需要的次數(shù)為（

）A．6 B．7 C．8 D．9【變式1-2】（2021·江蘇南通·高三海安高級中學(xué)?？迹┖瘮?shù)SKIPIF1<0的零點與SKIPIF1<0的零點之差的絕對值不超過SKIPIF1<0，則SKIPIF1<0的解析式可能是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2020秋·湖南邵陽·高三湖南省邵東市第一中學(xué)?？茧A段練習）已知圖像連續(xù)不斷的函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有唯一零點，如果用“二分法”求這個零點（精確度0.0001）的近似值，那么將區(qū)間SKIPIF1<0等分的次數(shù)至少是（

）A．4 B．6 C．7 D．10題型02根的分布【解題攻略】根的分布1.基礎(chǔ)分布：0分布特征：（1）、兩正根；（2）、兩負跟；（3）、一正一負兩根。方法：判別式+韋達定理區(qū)間分布與K分布特征：（1）、根比某個常數(shù)K大或者??；（2）、根在某個區(qū)間（a，b）內(nèi)（外）方法：借助復(fù)合條件的大致圖像，從以下四點入手開口方向；判別式；對稱軸位置；（4）根的分布區(qū)間端點對應(yīng)的函數(shù)值正負【典例1-1】（2023上·甘肅武威·高三統(tǒng)考開學(xué)考試）關(guān)于SKIPIF1<0的一元二次方程SKIPIF1<0有兩個不相等的正實數(shù)根，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0且SKIPIF1<0【典例1-2】（2023·高三課時練習）關(guān)于x的方程SKIPIF1<0至少有一個負根的充要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2022上·江蘇揚州·高三統(tǒng)考階段練習）已知一元二次方程SKIPIF1<0的兩根都在SKIPIF1<0內(nèi)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2022上·廣東廣州·高三廣州市第二中學(xué)?？茧A段練習）已知關(guān)于SKIPIF1<0的方程SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有實根，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2022上·遼寧沈陽·高三沈陽市外國語學(xué)校校考階段練習）一元二次方程SKIPIF1<0有一個正根和一個負根的一個充要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型03根的分布：指數(shù)函數(shù)二次型【解題攻略】指數(shù)型根的分布換元，令SKIPIF1<0，有指數(shù)函數(shù)性質(zhì)知，t的最大范圍為正。注意題中對方程根的正負范圍，對應(yīng)的t的取值范圍根據(jù)換元后新“根”的范圍，用一元二次型“根的分布”求解。特殊的函數(shù)式子，可以分離參數(shù)，轉(zhuǎn)化為“水平線型”求解?！镜淅?-1】（2021上·上海浦東新·高三上海市建平中學(xué)?？计谥校╆P(guān)于SKIPIF1<0的方程SKIPIF1<0恰有兩個根為SKIPIF1<0、SKIPIF1<0，且SKIPIF1<0、SKIPIF1<0分別滿足SKIPIF1<0和SKIPIF1<0，則SKIPIF1<0【典例1-2】．（2021·高三課時練習）設(shè)a為實數(shù)，若關(guān)于x的方程SKIPIF1<0有實數(shù)解，則a的取值范圍是.【變式1-1】（2021·山西臨汾·統(tǒng)考二模）已知函數(shù)SKIPIF1<0．若存在SKIPIF1<0，使得SKIPIF1<0，則m的取值范圍是．【變式1-2】（2021上·四川遂寧·高三階段）已知方程SKIPIF1<0有兩個不相等實根，則SKIPIF1<0的取值范圍為．【變式1-3】（2022下·浙江舟山·高三舟山中學(xué)校考開學(xué)考試）關(guān)于x的方程k?4x﹣k?2x+1+6(k﹣5)＝0在區(qū)間[﹣1，1]上有解,則實數(shù)k的取值范圍是.題型04零點：切線法【解題攻略】切線法求零點或者零點個數(shù)：適用于小題。大題則過程證明不嚴謹，容易丟過程分。數(shù)形結(jié)合，或者求導(dǎo)“畫圖”，求導(dǎo)畫圖，有時候需要判斷“上凸或者下凹”特殊的函數(shù)，需要通過“虛設(shè)零點”求得?！镜淅?-1】（2020上·湖北武漢·高三校聯(lián)考）已知函數(shù)SKIPIF1<0有三個零點SKIPIF1<0SKIPIF1<0SKIPIF1<0，則SKIPIF1<0（

）A．7 B．8 C．15 D．16【典例1-2】（2020上·河南·高三校聯(lián)考階段練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上有SKIPIF1<0個不同的零點，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2021·湖南長沙·高三長郡中學(xué)階段練習）函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0為偶函數(shù)，當SKIPIF1<0時，SKIPIF1<0，若函數(shù)SKIPIF1<0恰有一個零點，則實數(shù)SKIPIF1<0的取值集合是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2024·安徽黃山·屯溪一中?？寄M預(yù)測）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有三個零點，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2020·天津武清·天津市武清區(qū)楊村第一中學(xué)?？寄M預(yù)測）已知函數(shù)，若函數(shù)SKIPIF1<0有三個零點，則實數(shù)SKIPIF1<0的取值范圍是．題型05抽象函數(shù)型零點【解題攻略】抽象型函數(shù)判斷函數(shù)圖像定義域判斷。函數(shù)奇偶性判斷。函數(shù)簡單性判斷。函數(shù)值正負判斷利用極限，判斷無窮遠處的值與“比值”【典例1-1】（安徽省示范高中培優(yōu)聯(lián)盟2022-2023學(xué)年高三上學(xué)期11月冬季聯(lián)考數(shù)學(xué)試題）已知定義域為SKIPIF1<0的偶函數(shù)SKIPIF1<0的圖象是連續(xù)不斷的曲線，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0在區(qū)間SKIPIF1<0上的零點個數(shù)為（

）A．100 B．102 C．200 D．202【典例1-2】（山東省德州市2022屆高三三模數(shù)學(xué)試題）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，對于任意SKIPIF1<0，必有SKIPIF1<0，若函數(shù)SKIPIF1<0只有一個零點，則函數(shù)SKIPIF1<0有（

）A．最小值為SKIPIF1<0 B．最大值為SKIPIF1<0 C．最小值為4 D．最大值為4【變式1-1】已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0，則函數(shù)SKIPIF1<0的零點個數(shù)是（

）A．3 B．4 C．5 D．6【變式1-2】（2023秋·浙江杭州·高三杭州市長河高級中學(xué)校考）定義在R上的單調(diào)函數(shù)SKIPIF1<0滿足：SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上有零點，則a的取值范圍是題型06分段含參討論型【典例1-1】（湘豫名校聯(lián)考2022-2023學(xué)年高三上學(xué)期10月一輪復(fù)習診斷考試（一）數(shù)學(xué)（文科）試題）已知函數(shù)SKIPIF1<0有且僅有兩個零點，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【典例1-2】（2021·江蘇·高三專題練習）設(shè)SKIPIF1<0，e是自然對數(shù)的底數(shù)，函數(shù)SKIPIF1<0有零點，且所有零點的和不大于6，則a的取值范圍為．【變式1-1】已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0只有兩個零點，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有三個零點，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】已知函數(shù)SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0恰有2個零點”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件題型07參數(shù)分界型討論【解題攻略】參數(shù)在分段函數(shù)分界處，需要分類討論。分類討論討論點，首先是兩段函數(shù)的交點處，齊次是每段函數(shù)的各自“特色”處，如二次函數(shù)如果二次項有參，則“開口”即位討論點之一，要“多畫圖”，每一種情況，畫處各自“小圖”做對比【典例1-1】（2023·全國·高三專題練習）函數(shù)SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0的零點個數(shù)為；若SKIPIF1<0恰有4個零點，則SKIPIF1<0的取值范圍是.【典例1-2】.（2021秋·山東濟南·高三濟南外國語學(xué)校?？计谥校┮阎瘮?shù)SKIPIF1<0,如果函數(shù)SKIPIF1<0恰有兩個零點，那么實數(shù)SKIPIF1<0的取值范圍為．【變式1-1】（2022秋·天津河西·高三天津?qū)嶒炛袑W(xué)校考階段練習）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)恰有3個零點，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2023春·天津南開·高三南開大學(xué)附屬中學(xué)校考階段練習）已知SKIPIF1<0，函數(shù)SKIPIF1<0恰有3個零點，則m的取值范圍是（

）SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2023春·河南南陽·高三河南省桐柏縣第一高級中學(xué)校考階段練習）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0，若SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)恰有9個零點，則a的取值范圍是．題型08分離參數(shù)型水平線法求零點【解題攻略】分離參數(shù)水平線法求零點1.分離參數(shù)。2.構(gòu)造函數(shù)于水平線。3.構(gòu)造函數(shù)時，要注意函數(shù)是否有“水平漸近線”【典例1-1】（2021上·山東濰坊·高三統(tǒng)考）已知SKIPIF1<0，符號SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)，若函數(shù)SKIPIF1<0有且僅有3個零點，則SKIPIF1<0的取值范圍是SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2023·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0，若存在SKIPIF1<0，使得方程SKIPIF1<0有兩個不同的實數(shù)根且兩根之和為6，則實數(shù)SKIPIF1<0的取值范圍是．【變式1-1】（2022上·廣東汕頭·高三?？迹┮阎瘮?shù)SKIPIF1<0，令SKIPIF1<0，則下列說法正確的（

）A．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0B．當SKIPIF1<0時，SKIPIF1<0有3個零點C．當SKIPIF1<0時，SKIPIF1<0的所有零點之和為SKIPIF1<0D．當SKIPIF1<0時，SKIPIF1<0有1個零點【變式1-2】（2023上·山西朔州·高三懷仁市第一中學(xué)校?？迹┮阎瘮?shù)SKIPIF1<0，若SKIPIF1<0恰有3個零點SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2021上·河南新鄉(xiāng)·高三?？茧A段練習）若函數(shù)SKIPIF1<0有三個零點，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型09對數(shù)絕對值水平線法【解題攻略】對數(shù)絕對值對于SKIPIF1<0，SKIPIF1<0若有兩個零點，則滿足1.SKIPIF1<02.SKIPIF1<03.要注意上述結(jié)論在對稱軸作用下的“變與不變”【典例1-1】．（2021上·江蘇連云港·高三統(tǒng)考）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有4個不同的實根SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2020上·河南信陽·高三統(tǒng)考）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有四個不同的解SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2019·湖南·高三湖南師大附中?？茧A段練習）已知函數(shù)SKIPIF1<0是定義域為SKIPIF1<0的奇函數(shù)，且當SKIPIF1<0時，SKIPIF1<0，若函數(shù)SKIPIF1<0有六個零點，分別記為SKIPIF1<0，則SKIPIF1<0的取值范圍是.A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2023上·湖南長沙·高三周南中學(xué)校考開學(xué)考試）已知函數(shù)SKIPIF1<0，若SKIPIF1<0有四個解SKIPIF1<0，則SKIPIF1<0的取值范圍是．【變式1-3】．（2020上·河南鄭州·高三校聯(lián)考中）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有4個不同的實根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0題型10指數(shù)函數(shù)“一點一線”性質(zhì)型【解題攻略】指數(shù)函數(shù)，無論平移或者翻折，始終要注意函數(shù)的核心性質(zhì)“一點一線”是否變化。要把“一點一線”這個核心性質(zhì)提升到底數(shù)大于1或者小于1的分類討論相同地位SKIPIF1<0SKIPIF1<0SKIPIF1<0圖象性質(zhì)①定義域SKIPIF1<0，值域SKIPIF1<0②SKIPIF1<0，即時SKIPIF1<0，SKIPIF1<0，圖象都經(jīng)過SKIPIF1<0點③SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0等于底數(shù)SKIPIF1<0④在定義域上是單調(diào)減函數(shù)在定義域上是單調(diào)增函數(shù)⑤SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0⑥既不是奇函數(shù)，也不是偶函數(shù)【典例1-1】（2023上·云南昆明·高三昆明八中校考）已知SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2023上·安徽·高三池州市第一中學(xué)校聯(lián)考階段練習）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有四個不同的零點SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2023上·四川成都·高三四川省成都列五中學(xué)?？迹┤絷P(guān)于x的方程SKIPIF1<0有兩個不等的實數(shù)解，則實數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2023上·重慶沙坪壩·高三重慶南開中學(xué)?？茧A段練習）已知函數(shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0有四個不同的根SKIPIF1<0（SKIPIF1<0SKIPIF1<0），則SKIPIF1<0的最大值是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2023下·四川達州·高三校考階段練習）已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有三個不同的零點，則實數(shù)SKIPIF1<0的取值范圍是（

）.A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型11零點：中心對稱性質(zhì)型【解題攻略】函數(shù)中心對稱：（1）若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的一個對稱中心為SKIPIF1<0（2）若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的一個對稱中心為SKIPIF1<0（3）若函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的一個對稱中心為SKIPIF1<0.【典例1-1】（2023·全國·高三專題練習）函數(shù)SKIPIF1<0在SKIPIF1<0上的所有零點之和等于.【典例1-2】（2021·全國·高三專題練習）已知函數(shù)SKIPIF1<0為定義在SKIPIF1<0上的奇函數(shù)，當SKIPIF1<0時，SKIPIF1<0，則關(guān)于SKIPIF1<0的函數(shù)SKIPIF1<0SKIPIF1<0(SKIPIF1<0)的所有零點之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2022上·甘肅張掖·高三階段練習）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，若SKIPIF1<0，則關(guān)于SKIPIF1<0的方程SKIPIF1<0的所有根之和為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2021下·江西宜春·高三階段性）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，若SKIPIF1<0，則關(guān)于SKIPIF1<0的方程SKIPIF1<0的所有根之和為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】．（2022上·吉林松原·高三統(tǒng)考）定義在R上的奇函數(shù)f（x），當x≥0時，f（x）=，則關(guān)于x的函數(shù)F（x）=f（x）﹣a（0＜a＜1）的所有零點之和為A．3a﹣1 B．1﹣3a C．3﹣a﹣1 D．1﹣3﹣a題型12零點：軸對稱性質(zhì)型【解題攻略】軸對稱性的常用結(jié)論如下：若函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的一條對稱軸為SKIPIF1<0若函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的一條對稱軸為SKIPIF1<0若函數(shù)SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0的一條對稱軸為SKIPIF1<0(4)f(a－x)＝f(b＋x)?f(x)的圖象關(guān)于直線x＝eq\f(a＋b,2)對稱；【典例1-1】（2020·廣東中山·校聯(lián)考模擬預(yù)測）定義域為SKIPIF1<0的函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有3個不同的實數(shù)解SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2020上·遼寧沈陽·高三校聯(lián)考）已知定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0，滿足SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的所有零點之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2018上·湖南衡陽·高三衡陽市一中?？迹┮阎瘮?shù)SKIPIF1<0，若SKIPIF1<0互不相等），則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2019上·天津南開·高三天津二十五中統(tǒng)考）已知三個函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的零點依次為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2018上·貴州貴陽·高三貴陽一中階段練習）已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，滿足SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上所有零點之和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型13嵌套型零點：內(nèi)外自復(fù)合型【解題攻略】對于嵌套型復(fù)合函數(shù)SKIPIF1<0的零點個數(shù)問題，求解思路如下：（1）確定內(nèi)層函數(shù)SKIPIF1<0和外層函數(shù)SKIPIF1<0；（2）確定外層函數(shù)SKIPIF1<0的零點SKIPIF1<0；（3）確定直線SKIPIF1<0與內(nèi)層函數(shù)SKIPIF1<0圖象的交點個數(shù)分別為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則函數(shù)SKIPIF1<0的零點個數(shù)為SKIPIF1<0.【典例1-1】（2015下·浙江嘉興·高三階段練習）已知函數(shù)SKIPIF1<0，則下列關(guān)于函數(shù)SKIPIF1<0的零點個數(shù)的判斷正確的是A．當SKIPIF1<0時，有3個零點；當SKIPIF1<0時，有4個零點B．當SKIPIF1<0時，有4個零點；當SKIPIF1<0時，有3個零點C．無論k為何值，均有3個零點D．無論k為何值，均有4個零點【典例1-2】（2022上·河北石家莊·高三統(tǒng)考）已知函數(shù)若函數(shù)的零點個數(shù)為A．3 B．4 C．5 D．6【變式1-1】（2021上·天津·高三天津?qū)嶒炛袑W(xué)?？计谥校┮阎猄KIPIF1<0為偶函數(shù)，當SKIPIF1<0時，SKIPIF1<0，若函數(shù)SKIPIF1<0恰有4個零點，則實數(shù)SKIPIF1<0的取值范圍（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2021上·安徽滁州·高三安徽省定遠中學(xué)校聯(lián)考）已知函數(shù)SKIPIF1<0，若SKIPIF1<0存在四個互不相等的實數(shù)根，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2019上·黑龍江哈爾濱·高三哈爾濱三中?？迹┮阎瘮?shù)SKIPIF1<0，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題型14嵌套型零點：內(nèi)外雙函數(shù)復(fù)合型【典例1-1】（2021上·陜西安康·高三統(tǒng)考階段練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若方程SKIPIF1<0有四個不等的實數(shù)根，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2023上·江西吉安·高三吉安一中?？迹┮阎瘮?shù)SKIPIF1<0，SKIPIF1<0，當SKIPIF1<0時，方程SKIPIF1<0根的個數(shù)為（

）.A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2021上·安徽池州·高三池州市第一中學(xué)?？迹┰O(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，若方程SKIPIF1<0有四個實數(shù)根，則實數(shù)t的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2020上·江蘇南京·高三南京市第五高級中學(xué)?？茧A段練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0有6個零點，則實數(shù)SKIPIF1<0的取值范圍為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2021·全國·高三專題練習）已知函數(shù)SKIPIF1<0SKIPIF1<0，則方程SKIPIF1<0的實數(shù)根的個數(shù)為（

）A．5 B．6 C．7 D．8題型15嵌套型零點：二次型因式分解【典例1-1】（2020·山東德州·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有且只有兩個不同實數(shù)根，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2020下·江蘇無錫·高三?？迹┮阎瘮?shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0有5個不同的實數(shù)解，則實數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2022秋·天津河?xùn)|·高三?？茧A段練習）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰有4個不同的零點，則a的取值范圍為．【變式1-2】（2023春·上海寶山·高三校考）已知SKIPIF1<0滿足SKIPIF1<0，當SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0在SKIPIF1<0上恰有八個不同的零點，則實數(shù)SKIPIF1<0的取值范圍為.【變式1-3】（2019·浙江衢州·衢州二中?？级＃┰O(shè)SKIPIF1<0（其中SKIPIF1<0為自然對數(shù)的底數(shù)），SKIPIF1<0，若函數(shù)SKIPIF1<0恰有4個不同的零點，則實數(shù)SKIPIF1<0的取值范圍為.題型16嵌套型零點：二次型根的分布【典例1-1】（2023·遼寧沈陽·東北育才雙語學(xué)校?？家荒＃┮阎瘮?shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0有6個不同的實數(shù)根，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2022下·河南信陽·高三信陽高中?？茧A段練習）已知函數(shù)SKIPIF1<0若關(guān)于x的方程SKIPIF1<0有8個不同的實數(shù)根，則實數(shù)b的取值范圍是（）A．SKIPIF1<0

B．

SKIPIF1<0C．SKIPIF1<0 D．[﹣5，﹣4]【變式1-1】（2020下·河南·高撒統(tǒng)考）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個不相等的實數(shù)根，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2020·河北邯鄲·統(tǒng)考二模）已知SKIPIF1<0若函數(shù)SKIPIF1<0恰有5個零點，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2023上·江蘇常州·高三江蘇省前黃高級中學(xué)?？奸_學(xué)考試）已知函數(shù)SKIPIF1<0，關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有SKIPIF1<0個不同實數(shù)解，則SKIPIF1<0的取值范圍為．題型17嵌套型零點：放大型函數(shù)【解題攻略】（1）如果函數(shù)在SKIPIF1<0上滿足SKIPIF1<0，則此類函數(shù)在局部范圍上具有與周期函數(shù)相類似的性質(zhì)．（2）復(fù)雜函數(shù)的零點，可以轉(zhuǎn)化為熟悉函數(shù)圖像的交點來處理．滿足SKIPIF1<0形式，一般情況下，b多是0或者1.俗稱為“放大鏡函數(shù)”。【典例1-1】（寧夏石嘴山市平羅中學(xué)2022屆高三第四次模擬考試數(shù)學(xué)（理）試題）已知定義為R的奇函數(shù)SKIPIF1<0滿足：SKIPIF1<0，若方程SKIPIF1<0在SKIPIF1<0上恰有三個根，則實數(shù)k的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】.已知函數(shù)f(x)=1?2x?3,1≤x≤2A．關(guān)于x的方程f(x)?12nB．若函數(shù)y=f(x)?kx有4個零點，則實數(shù)k的取值范圍為1C．對于實數(shù)x∈[1,+∞)，不等式2xf(x)?3≤0恒成立D．當x∈[2n?1,2n【變式1-1】（河北省邯鄲市大名縣第一中學(xué)2020-2021學(xué)年高三下學(xué)期5月月考數(shù)學(xué)試題）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0滿足以下三點條件：①定義域為SKIPIF1<0；②對任意SKIPIF1<0，有SKIPIF1<0；③當SKIPIF1<0時，SKIPIF1<0則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上零點的個數(shù)為__________個.【變式1-2】（浙江省寧波市鎮(zhèn)海中學(xué)2020-2021學(xué)年高三下學(xué)期開學(xué)考試數(shù)學(xué)試題）定義在R上的奇函數(shù)SKIPIF1<0滿足，當SKIPIF1<0時，SKIPIF1<0，且SKIPIF1<0時，有SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上的零點個數(shù)為A．9 B．8 C．7 D．6【變式1-3】已知SKIPIF1<0，則函數(shù)SKIPIF1<0零點的個數(shù)為___________.高考練場1.（2021秋·吉林長春·高三?？茧A段練習）若函數(shù)SKIPIF1<0的零點與函數(shù)SKIPIF1<0的零點之差的絕對值不超過SKIPIF1<0，則SKIPIF1<0可以是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02.（2021上·湖南長沙·高三長沙一中?？奸_學(xué)考試）關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個不相等的實數(shù)根SKIPIF1<0且SKIPIF1<0，那么SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03.（2020·內(nèi)蒙古包頭·高三北重三中?？迹╆P(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個不相等的實數(shù)根，則實數(shù)SKIPIF1<0的取值范圍是.4.（2023春·新疆烏魯木齊·高三新疆師范大學(xué)附屬中學(xué)?？奸_學(xué)考試）若SKIPIF1<0若SKIPIF1<0有兩個零點，則