第第頁第05講4.5.1函數(shù)的零點(diǎn)與方程的解課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①了解函數(shù)的零點(diǎn)與方程的解的關(guān)系，并能結(jié)合函數(shù)的圖象判定函數(shù)的零點(diǎn)。②能根據(jù)函數(shù)零點(diǎn)存在性定理對函數(shù)零點(diǎn)存在進(jìn)行判定，同時能處理與函數(shù)零點(diǎn)問題相結(jié)合的求參數(shù)及綜合類的問題。通過本節(jié)課的學(xué)習(xí)，要求能判定函數(shù)零點(diǎn)的存在，同時能解決與函數(shù)零點(diǎn)相結(jié)合的綜合問題知識點(diǎn)01：函數(shù)零點(diǎn)的概念1、函數(shù)零點(diǎn)的概念對于一般函數(shù)SKIPIF1<0，我們把使SKIPIF1<0的實(shí)數(shù)SKIPIF1<0叫做函數(shù)SKIPIF1<0的零點(diǎn).幾何定義:函數(shù)SKIPIF1<0的零點(diǎn)就是方程SKIPIF1<0的實(shí)數(shù)解,也就是函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸的公共點(diǎn)的橫坐標(biāo).

這樣：方程SKIPIF1<0有實(shí)數(shù)解SKIPIF1<0函數(shù)SKIPIF1<0有零點(diǎn)SKIPIF1<0函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸有公共點(diǎn)2、已學(xué)基本初等函數(shù)的零點(diǎn)①一次函數(shù)SKIPIF1<0只有一個零點(diǎn)SKIPIF1<0；②反比例函數(shù)SKIPIF1<0沒有零點(diǎn)；③指數(shù)函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）沒有零點(diǎn)；④對數(shù)函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）只有一個零點(diǎn)1；⑤冪函數(shù)SKIPIF1<0當(dāng)SKIPIF1<0時，有一個零點(diǎn)0；當(dāng)SKIPIF1<0時，無零點(diǎn)。知識點(diǎn)02：函數(shù)零點(diǎn)存在定理及其應(yīng)用1、函數(shù)零點(diǎn)存在定理如果函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的圖象是一條連續(xù)不斷的曲線，且有SKIPIF1<0，那么函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)至少有一個零點(diǎn)，即存在SKIPIF1<0，使得SKIPIF1<0，這個SKIPIF1<0也就是方程SKIPIF1<0的解.說明：定理要求具備兩個條件:①函數(shù)在區(qū)間SKIPIF1<0上的圖象是連續(xù)不斷的;②SKIPIF1<0.兩個條件缺一不可.2、函數(shù)零點(diǎn)的求法①代數(shù)法：根據(jù)零點(diǎn)定義，求出方程SKIPIF1<0的實(shí)數(shù)解;②數(shù)形結(jié)合法：作出函數(shù)圖象，利用函數(shù)性質(zhì)求解【即學(xué)即練1】（2023春·四川廣安·高一?？茧A段練習(xí)）函數(shù)SKIPIF1<0的零點(diǎn)為．【答案】2【詳解】令SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0.故答案為：SKIPIF1<03、函數(shù)零點(diǎn)個數(shù)的判斷①利用代數(shù)法，求出所有零點(diǎn)；②數(shù)形結(jié)合，通過作圖，找出圖象與SKIPIF1<0軸交點(diǎn)的個數(shù)；③數(shù)形結(jié)合，通過分離，將原函數(shù)拆分成兩個函數(shù)，找到兩個函數(shù)圖象交點(diǎn)的個數(shù)；④函數(shù)零點(diǎn)唯一：函數(shù)存在零點(diǎn)+函數(shù)單調(diào).知識點(diǎn)03：二次函數(shù)的零點(diǎn)問題一元二次方程SKIPIF1<0的實(shí)數(shù)根也稱為函數(shù)SKIPIF1<0的零點(diǎn).當(dāng)SKIPIF1<0時，一元二次方程SKIPIF1<0的實(shí)數(shù)根、二次函數(shù)SKIPIF1<0的零點(diǎn)之間的關(guān)系如下表所示：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0的實(shí)數(shù)根SKIPIF1<0（其中SKIPIF1<0）SKIPIF1<0方程無實(shí)數(shù)根SKIPIF1<0的圖象SKIPIF1<0的零點(diǎn)SKIPIF1<0SKIPIF1<0函數(shù)無零點(diǎn)【即學(xué)即練2】（2023·高一課時練習(xí)）若函數(shù)SKIPIF1<0的一個零點(diǎn)是1，則它的另一個零點(diǎn)是．【答案】3【詳解】由SKIPIF1<0，所以令SKIPIF1<0或SKIPIF1<0，故另一個零點(diǎn)為3故答案為：3題型01求函數(shù)的零點(diǎn)【典例1】函數(shù)SKIPIF1<0的零點(diǎn)為．【答案】4【詳解】依題意有SKIPIF1<0，所以SKIPIF1<0.故答案為：4.【典例2】已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)為.【答案】SKIPIF1<0和SKIPIF1<0【詳解】當(dāng)SKIPIF1<0時，令SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0內(nèi)有且僅有一個零點(diǎn)2；綜上所述：函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0和SKIPIF1<0.故答案為：SKIPIF1<0和SKIPIF1<0.【變式1】函數(shù)SKIPIF1<0的零點(diǎn)是【答案】SKIPIF1<0/SKIPIF1<0【詳解】令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<0.【變式2】求下列函數(shù)的零點(diǎn)．(1)SKIPIF1<0；(2)SKIPIF1<0.【答案】(1)9(2)答案見解析【詳解】（1）由SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0.（2）由SKIPIF1<0，得SKIPIF1<0，①當(dāng)SKIPIF1<0，SKIPIF1<0時，函數(shù)有唯一零點(diǎn)SKIPIF1<0；②當(dāng)SKIPIF1<0，即SKIPIF1<0時，函數(shù)有兩個零點(diǎn)SKIPIF1<0和SKIPIF1<0.題型02函數(shù)零點(diǎn)個數(shù)的判斷【典例1】函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)為（）A．1B．2C．1或2D．0【答案】C【詳解】由SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時，函數(shù)的零點(diǎn)個數(shù)為SKIPIF1<0；當(dāng)SKIPIF1<0時，函數(shù)的零點(diǎn)個數(shù)為SKIPIF1<0.所以該函數(shù)的零點(diǎn)個數(shù)是1或2.故選：C【典例2】方程SKIPIF1<0的實(shí)數(shù)解的個數(shù)是（

）A．0B．1C．2D．3【答案】B【詳解】在同一直角坐標(biāo)系中畫出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，由圖象可知：兩個函數(shù)圖象只有一個交點(diǎn)，故方程SKIPIF1<0的實(shí)數(shù)解的個數(shù)為1，故選：B

【典例3】已知SKIPIF1<0，方程SKIPIF1<0的實(shí)根個數(shù)為．【答案】2【詳解】由SKIPIF1<0，則SKIPIF1<0，則令SKIPIF1<0，SKIPIF1<0，分別作出它們的圖象如下圖所示，

由圖可知，有兩個交點(diǎn)，所以方程SKIPIF1<0的實(shí)根個數(shù)為2．故答案為：2．【典例4】若函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個數(shù)是.【答案】2【詳解】作出SKIPIF1<0與SKIPIF1<0的函數(shù)圖像如圖：

由圖像可知兩函數(shù)圖像有SKIPIF1<0個交點(diǎn)，所以函數(shù)SKIPIF1<0有兩個零點(diǎn).故答案為：SKIPIF1<0【變式1】函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)是（

）A．0B．1C．2D．無數(shù)個【答案】C【詳解】令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的零點(diǎn)為1和SKIPIF1<0，故有兩個零點(diǎn)，故選：C【變式2】已知函數(shù)SKIPIF1<0.(1)作出函數(shù)SKIPIF1<0的圖象；(2)就a的取值范圍討論函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)．【答案】(1)作圖見解析(2)答案見解析【詳解】（1）先作出SKIPIF1<0的圖象，然后將其在x軸下方的部分翻折到x軸上方，原x軸上及其上方的圖象及翻折上來的圖象便是所要作的圖象．

.

（2）由圖象易知，函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)就是函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0的交點(diǎn)的個數(shù)．SKIPIF1<0.當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)為0；當(dāng)SKIPIF1<0與SKIPIF1<0時，函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)為2；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)為4；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的零點(diǎn)的個數(shù)為3.【變式3】若SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0至多有個零點(diǎn).【答案】4【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0或SKIPIF1<0.綜上所述，SKIPIF1<0的零點(diǎn)可能是SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.所以，SKIPIF1<0的零點(diǎn)至多有4個.故答案為：4.【變式4】函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象的交點(diǎn)個數(shù)為個．【答案】2【詳解】在同一坐標(biāo)系中作出兩個函數(shù)的圖象，如圖，它們交點(diǎn)個數(shù)為2．

故答案為：2.題型03判斷函數(shù)零點(diǎn)所在的區(qū)間【典例1】若SKIPIF1<0是方程SKIPIF1<0的解，則SKIPIF1<0（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】因?yàn)楹瘮?shù)SKIPIF1<0在定義上單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間是SKIPIF1<0，即SKIPIF1<0．故選：C.【典例2】設(shè)SKIPIF1<0為方程SKIPIF1<0的解，若SKIPIF1<0，則SKIPIF1<0的值為.【答案】SKIPIF1<0【詳解】由題意可知SKIPIF1<0是方程SKIPIF1<0的解，所以SKIPIF1<0，令SKIPIF1<0，由SKIPIF1<0，所以SKIPIF1<0，再根據(jù)SKIPIF1<0，可得SKIPIF1<0，故答案為:SKIPIF1<0.【變式1】函數(shù)SKIPIF1<0的零點(diǎn)所在區(qū)間是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【詳解】易知函數(shù)定義域?yàn)镾KIPIF1<0，且函數(shù)SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0上沒有零點(diǎn)；SKIPIF1<0，SKIPIF1<0，由零點(diǎn)存在定理可知SKIPIF1<0，所以零點(diǎn)所在區(qū)間是SKIPIF1<0.故選：D【變式2】函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則k的值為（

）A．1B．2C．0D．3【答案】A【詳解】解：因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0，故選：A題型04已知零點(diǎn)個數(shù)求參數(shù)的取值范圍【典例1】若方程SKIPIF1<0有兩個不同的實(shí)數(shù)根，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】令SKIPIF1<0，由于當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖所示，則當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有兩個交點(diǎn)，即方程SKIPIF1<0有兩個不同的實(shí)數(shù)根，SKIPIF1<0的取值范圍是SKIPIF1<0．故選：C．【典例2】設(shè)SKIPIF1<0表示m，n中的較小數(shù)．若函數(shù)SKIPIF1<0至少有3個零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【詳解】由題意可得SKIPIF1<0有解，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時，必有SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時，必有SKIPIF1<0，不等式組無解，綜上所述，SKIPIF1<0，∴SKIPIF1<0的取值范圍為SKIPIF1<0.故選：A【典例3】若函數(shù)SKIPIF1<0有2個零點(diǎn)，求實(shí)數(shù)a的取值范圍．【答案】SKIPIF1<0【詳解】令SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，畫出SKIPIF1<0的大致圖象如下：由圖象可知：當(dāng)SKIPIF1<0或SKIPIF1<0時，直線SKIPIF1<0與SKIPIF1<0的圖象有兩個交點(diǎn)，符合題意，故a的取值范圍為SKIPIF1<0，

【典例4】已知函數(shù)SKIPIF1<0是偶函數(shù).當(dāng)SKIPIF1<0時，SKIPIF1<0.(1)求函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，求實(shí)數(shù)a的取值范圍；(3)已知SKIPIF1<0，試討論SKIPIF1<0的零點(diǎn)個數(shù)，并求對應(yīng)的m的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0(3)答案見解析【詳解】（1）設(shè)SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0∵SKIPIF1<0為偶函數(shù)∴SKIPIF1<0綜上，有SKIPIF1<0（2）由（1）作出SKIPIF1<0的圖像如圖：因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上具有單調(diào)性，由圖可得SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0或SKIPIF1<0.（3）由（1）作出SKIPIF1<0的圖像如圖：由圖像可知：當(dāng)SKIPIF1<0時，SKIPIF1<0有兩個零點(diǎn)；當(dāng)SKIPIF1<0時，SKIPIF1<0有四個零點(diǎn)；當(dāng)SKIPIF1<0時，SKIPIF1<0有六個零點(diǎn)；當(dāng)SKIPIF1<0時，SKIPIF1<0有三個零點(diǎn)；當(dāng)SKIPIF1<0時，SKIPIF1<0沒有零點(diǎn).【變式1】設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0若SKIPIF1<0恰有一個零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【詳解】畫出函數(shù)SKIPIF1<0的圖象如下圖所示：函數(shù)SKIPIF1<0可由SKIPIF1<0分段平移得到，易知當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0恰有一個零點(diǎn)，滿足題意；當(dāng)SKIPIF1<0時，代表圖象往上平移，顯然沒有零點(diǎn)，不符合題意；當(dāng)SKIPIF1<0時，圖象往下平移，當(dāng)SKIPIF1<0時，函數(shù)有兩個零點(diǎn)；當(dāng)SKIPIF1<0時，SKIPIF1<0恰有一個零點(diǎn)，滿足題意，即SKIPIF1<0；綜上可得SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D【變式2】（多選）已知函數(shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0恰有兩個互異的實(shí)數(shù)解，則實(shí)數(shù)a的值可以是（

）A．0B．1C．SKIPIF1<0D．2【答案】BCD【詳解】函數(shù)SKIPIF1<0的圖象，如圖所示：由題意知，直線SKIPIF1<0與SKIPIF1<0的圖象有2個交點(diǎn)．當(dāng)直線SKIPIF1<0過點(diǎn)SKIPIF1<0時，SKIPIF1<0，當(dāng)直線SKIPIF1<0過點(diǎn)SKIPIF1<0時，SKIPIF1<0．結(jié)合圖象如圖可知，當(dāng)SKIPIF1<0時，直線SKIPIF1<0與SKIPIF1<0的圖象有2個交點(diǎn)，如圖所示：又當(dāng)直線SKIPIF1<0與曲線SKIPIF1<0相切在第一象限時，直線SKIPIF1<0與SKIPIF1<0的圖象也有2個交點(diǎn)，如圖所示：SKIPIF1<0，化簡可得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，又由圖可知SKIPIF1<0，所以SKIPIF1<0，此時切點(diǎn)的橫坐標(biāo)為2符合.綜上，實(shí)數(shù)a的取值范圍是SKIPIF1<0.故選：BCD．【變式3】若函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)有且只有一個零點(diǎn)，則SKIPIF1<0的取值集合是.【答案】SKIPIF1<0【詳解】

由已知得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.由二次函數(shù)圖象及函數(shù)零點(diǎn)存在定理可知，該函數(shù)在SKIPIF1<0內(nèi)只有一個零點(diǎn)，只需SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.【變式4】已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)．(1)求SKIPIF1<0的值；(2)畫出SKIPIF1<0的圖象，并指出其單調(diào)減區(qū)間；(3)若關(guān)于SKIPIF1<0的方程SKIPIF1<0有2個不相等的實(shí)數(shù)根，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)作圖見解析；答案見解析(3)SKIPIF1<0【詳解】（1）因?yàn)镾KIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0；（2）由（1）得SKIPIF1<0，列表：SKIPIF1<0…SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<001234…SKIPIF1<0…521252125…描點(diǎn)連線得圖象如圖所示：由圖象可得單調(diào)減區(qū)間為SKIPIF1<0和SKIPIF1<0；（3）因?yàn)殛P(guān)于SKIPIF1<0的方程SKIPIF1<0有2個不相等的實(shí)數(shù)根，所以SKIPIF1<0與SKIPIF1<0的圖象有兩個不同的交點(diǎn)，由（2）中的圖可知SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.題型05已知零點(diǎn)所在區(qū)間求參數(shù)的取值范圍【典例1】函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【詳解】由零點(diǎn)存在定理可知，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在零點(diǎn)，顯然函數(shù)為增函數(shù)，只需滿足SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D【典例2】（多選）函數(shù)SKIPIF1<0的一個零點(diǎn)在區(qū)間SKIPIF1<0內(nèi)，則實(shí)數(shù)a的可能取值是（

）A．0B．1C．2D．3【答案】BC【詳解】因?yàn)楹瘮?shù)SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞增，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由函數(shù)SKIPIF1<0的一個零點(diǎn)在區(qū)間SKIPIF1<0內(nèi)，得SKIPIF1<0，解得SKIPIF1<0,故選：BC【典例3】設(shè)SKIPIF1<0為實(shí)數(shù)，函數(shù)SKIPIF1<0在SKIPIF1<0上有零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0在SKIPIF1<0單調(diào)遞增，且有零點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<0【變式1】若函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)恰有一個零點(diǎn)，則a的取值范圍（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0只有一個零點(diǎn)SKIPIF1<0，不符合題意，當(dāng)SKIPIF1<0時，若SKIPIF1<0，即SKIPIF1<0，函數(shù)SKIPIF1<0只有一個零點(diǎn)SKIPIF1<0，不符合題意，因函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)恰有一個零點(diǎn)，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以a的取值范圍是SKIPIF1<0.故選：A【變式2】）若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有解，則實(shí)數(shù)SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】方程SKIPIF1<0在SKIPIF1<0上有解，等價于函數(shù)SKIPIF1<0與SKIPIF1<0在SKIPIF1<0有交點(diǎn)，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0【變式3】若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】由題意得：SKIPIF1<0為連續(xù)函數(shù)，且在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以只需SKIPIF1<0或SKIPIF1<0，解得：SKIPIF1<0，故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0題型06二次函數(shù)的零點(diǎn)問題【典例1】已知函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0開口向上，對稱軸為SKIPIF1<0，要想滿足SKIPIF1<0，則要SKIPIF1<0，解得：SKIPIF1<0.故選：B【典例2】方程SKIPIF1<0的一根大于1，一根小于1，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】∵方程SKIPIF1<0的一根大于1，另一根小于1，令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.【典例3】（1）判斷二次函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)是否存在零點(diǎn)；（2）若二次函數(shù)SKIPIF1<0SKIPIF1<0的兩個零點(diǎn)均為正數(shù)，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】（1）存在；（2）SKIPIF1<0【詳解】（1）由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，因?yàn)镾KIPIF1<0，所以二次函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)存在零點(diǎn).（2）因?yàn)槎魏瘮?shù)SKIPIF1<0SKIPIF1<0的兩個零點(diǎn)均為正數(shù)，所以二次SKIPIF1<0SKIPIF1<0有兩個正實(shí)數(shù)根．設(shè)為SKIPIF1<0，由一元二次方程的根與系數(shù)的關(guān)系得SKIPIF1<0，解得SKIPIF1<0.即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.【變式1】已知關(guān)于SKIPIF1<0的方程SKIPIF1<0，SKIPIF1<0存在兩個不同的實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【詳解】由題意可得，SKIPIF1<0即SKIPIF1<0在SKIPIF1<0時有2個不同的解，設(shè)SKIPIF1<0，根據(jù)雙勾函數(shù)的性質(zhì)可知，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，要使SKIPIF1<0在SKIPIF1<0時有2個不同的解，則SKIPIF1<0，故選:D.【變式2】若關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0內(nèi)有解，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【詳解】由題意SKIPIF1<0在SKIPIF1<0內(nèi)有解，SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0．故選：A．【變式3】已知函數(shù)SKIPIF1<0SKIPIF1<0.(1)若該函數(shù)有兩個不相等的正零點(diǎn)，求SKIPIF1<0的取值范圍；(2)若該函數(shù)有兩個零點(diǎn)，一個大于1，另一個小于1，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0或SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0【詳解】（1）因?yàn)槎魏瘮?shù)SKIPIF1<0有兩個不相等的正零點(diǎn)，且對稱軸SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.（2）因?yàn)槎魏瘮?shù)SKIPIF1<0有兩個零點(diǎn)，一個大于1，另一個小于1，所以SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0.題型07函數(shù)與方程綜合【典例1】已知函數(shù)SKIPIF1<0，常數(shù)SKIPIF1<0.(1)若SKIPIF1<0是奇函數(shù)，求SKIPIF1<0的值；(2)若SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有且僅有一個零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）①若SKIPIF1<0有定義，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，此時SKIPIF1<0符合題意；②若SKIPIF1<0無定義，則SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，不關(guān)于原點(diǎn)對稱，故SKIPIF1<0不是奇函數(shù)，不符合題意.綜上，SKIPIF1<0.（2）當(dāng)SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0（舍），所以SKIPIF1<0，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有且僅有一個零點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0.【典例2】已知函數(shù)SKIPIF1<0(1)證明：函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；(2)討論關(guān)于x的方程SKIPIF1<0的實(shí)數(shù)解的個數(shù)．【答案】(1)證明見解析(2)答案見解析【詳解】（1）任取SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．（2）關(guān)于x的方程SKIPIF1<0的實(shí)數(shù)解的個數(shù)，等價于函數(shù)SKIPIF1<0與常函數(shù)SKIPIF1<0的交點(diǎn)個數(shù)，由（1）可得：SKIPIF1<0，令SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，結(jié)合（1）可得：函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，令SKIPIF1<0，且SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故函數(shù)SKIPIF1<0的圖像如圖所示：可得函數(shù)SKIPIF1<0的圖像如圖所示：對于函數(shù)SKIPIF1<0與常函數(shù)SKIPIF1<0的交點(diǎn)個數(shù)，則有：當(dāng)SKIPIF1<0時，交點(diǎn)個數(shù)為0個；當(dāng)SKIPIF1<0或SKIPIF1<0時，交點(diǎn)個數(shù)為2個；當(dāng)SKIPIF1<0時，交點(diǎn)個數(shù)為3個；當(dāng)SKIPIF1<0時，交點(diǎn)個數(shù)為4個．【變式1】已知函數(shù)SKIPIF1<0為奇函數(shù).(1)求實(shí)數(shù)a的值；(2)若方程SKIPIF1<0在區(qū)間SKIPIF1<0上無解，求實(shí)數(shù)m的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）若函數(shù)SKIPIF1<0為奇函數(shù)，即SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.（2）由（1）可得：SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，構(gòu)建SKIPIF1<0，對SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，由SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，故方程SKIPIF1<0無解，則實(shí)數(shù)m的取值范圍SKIPIF1<0.【變式2】已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0軸左側(cè)的圖象如圖所示．(1)求函數(shù)SKIPIF1<0的解析式；(2)若關(guān)于SKIPIF1<0的方程SKIPIF1<0有SKIPIF1<0個不相等的實(shí)數(shù)根，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）由圖象知：SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0上的偶函數(shù)，SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0；綜上所述：SKIPIF1<0；（2）SKIPIF1<0為偶函數(shù)，SKIPIF1<0圖象關(guān)于SKIPIF1<0軸對稱，可得SKIPIF1<0圖象如下圖所示，SKIPIF1<0有SKIPIF1<0個不相等的實(shí)數(shù)根，等價于SKIPIF1<0與SKIPIF1<0有SKIPIF1<0個不同的交點(diǎn)，由圖象可知：SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.A夯實(shí)基礎(chǔ)一、單選題1．已知函數(shù)SKIPIF1<0，則SKIPIF1<0的零點(diǎn)所在的區(qū)間為（

）.A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以由零點(diǎn)存在性定理知，SKIPIF1<0的零點(diǎn)所在的區(qū)間為SKIPIF1<0.故選：B.2．已知方程SKIPIF1<0的解在SKIPIF1<0內(nèi)，則SKIPIF1<0（

）A．3B．2C．1D．0【答案】C【詳解】令函數(shù)SKIPIF1<0，顯然函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0，SKIPIF1<0，因此函數(shù)SKIPIF1<0的零點(diǎn)SKIPIF1<0，所以方程SKIPIF1<0的解在SKIPIF1<0內(nèi)，即SKIPIF1<0.故選：C3．已知函數(shù)SKIPIF1<0的兩個零點(diǎn)都大于2，則實(shí)數(shù)m的取值范圍是（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】因?yàn)槎魏瘮?shù)圖象的開口向上，對稱軸SKIPIF1<0，函數(shù)SKIPIF1<0的兩個零點(diǎn)都大于2，所以SKIPIF1<0，解得SKIPIF1<0.故選：C4．已知二次函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)的零點(diǎn)情況是（

）A．有兩個零點(diǎn)B．有唯一零點(diǎn)C．沒有零點(diǎn)D．不確定【答案】C【詳解】因?yàn)楹瘮?shù)SKIPIF1<0開口向下，又SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)沒有零點(diǎn).故選：C5．已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的零點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【詳解】由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0與SKIPIF1<0圖象交點(diǎn)的橫坐標(biāo)，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0與SKIPIF1<0圖象交點(diǎn)的橫坐標(biāo)，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0與SKIPIF1<0圖象交點(diǎn)的橫坐標(biāo)，分別作出SKIPIF1<0和SKIPIF1<0的圖象，則由圖象可得SKIPIF1<0，

因?yàn)镾KIPIF1<0和SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，故選：B6．設(shè)實(shí)數(shù)a為常數(shù)，則函數(shù)SKIPIF1<0存在零點(diǎn)的充分必要條件是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0存在零點(diǎn)，等價于方程SKIPIF1<0在SKIPIF1<0上存在零點(diǎn)，注意到SKIPIF1<0的圖像開口向上，對稱軸為SKIPIF1<0，且SKIPIF1<0，故上述條件等價于SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.所以函數(shù)SKIPIF1<0存在零點(diǎn)的充分必要條件是SKIPIF1<0.故選：A.7．已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有五個零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0則SKIPIF1<0，此時SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0則SKIPIF1<0，此時SKIPIF1<0，則SKIPIF1<0，故問題轉(zhuǎn)為SKIPIF1<0，SKIPIF1<0共有四個零點(diǎn)，畫出函數(shù)圖象如下可知：則SKIPIF1<0，故選：D

8．享有“數(shù)學(xué)王子”稱號的德國數(shù)學(xué)家高斯，是近代數(shù)學(xué)奠基者之一，SKIPIF1<0被稱為“高斯函數(shù)”，其中SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)，例如：SKIPIF1<0，設(shè)SKIPIF1<0為函數(shù)SKIPIF1<0的零點(diǎn)，則SKIPIF1<0（

）A．3B．4C．5D．6【答案】B【詳解】因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，SKIPIF1<0，則存在唯一零點(diǎn)SKIPIF1<0，使得SKIPIF1<0，由高斯函數(shù)的定義可知，SKIPIF1<0.故選：B.二、多選題9．函數(shù)SKIPIF1<0的零點(diǎn)可以是（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】CD【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0的零點(diǎn)是SKIPIF1<0和SKIPIF1<0，故選：CD10．設(shè)SKIPIF1<0為定義在R上的奇函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0為常數(shù)），則（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．函數(shù)SKIPIF1<0僅有一個零點(diǎn)【答案】ABD【詳解】對A：因?yàn)镾KIPIF1<0是SKIPIF1<0上的奇函數(shù)，故SKIPIF1<0，解得SKIPIF1<0，故A正確；對BC：SKIPIF1<0，故B正確，C錯誤；對D：當(dāng)SKIPIF1<0時，SKIPIF1<0；因?yàn)镾KIPIF1<0是增函數(shù)，SKIPIF1<0也是增函數(shù)，故SKIPIF1<0