新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第7講 函數(shù)與方程（解析版）

第7講函數(shù)與方程學(xué)校____________姓名____________班級(jí)____________一、知識(shí)梳理1.函數(shù)的零點(diǎn)(1)概念：一般地，如果函數(shù)y＝f(x)在實(shí)數(shù)α處的函數(shù)值等于零，即f(α)＝0，則稱α為函數(shù)y＝f(x)的零點(diǎn).(2)函數(shù)的零點(diǎn)、函數(shù)的圖像與x軸的交點(diǎn)、對(duì)應(yīng)方程的根的關(guān)系：2.函數(shù)零點(diǎn)存在定理如果函數(shù)y＝f(x)在區(qū)間[a，b]上的圖像是連續(xù)不斷的，并且f(a)·f(b)<0(即在區(qū)間兩個(gè)端點(diǎn)處的函數(shù)值異號(hào))，則函數(shù)y＝f(x)在區(qū)間(a，b)中至少有一個(gè)零點(diǎn)，即?x0∈(a，b)，f(x0)＝0.考點(diǎn)和典型例題1、函數(shù)零點(diǎn)所在區(qū)間的判斷【典例1-1】（2022·天津紅橋·一模）函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】函數(shù)SKIPIF1<0是SKIPIF1<0上的連續(xù)增函數(shù),SKIPIF1<0,可得SKIPIF1<0,所以函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是SKIPIF1<0.故選：C【典例1-2】（2021·山西·太原五中高三階段練習(xí)（文））利用二分法求方程SKIPIF1<0的近似解，可以取的一個(gè)區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：設(shè)SKIPIF1<0，SKIPIF1<0當(dāng)連續(xù)函數(shù)SKIPIF1<0滿足SKIPIF1<0（a）SKIPIF1<0（b）SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上有零點(diǎn)，即方程SKIPIF1<0在區(qū)間SKIPIF1<0上有解，又SKIPIF1<0（2）SKIPIF1<0，SKIPIF1<0（3）SKIPIF1<0，故SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<0，故方程SKIPIF1<0在區(qū)間SKIPIF1<0上有解，即利用二分法求方程SKIPIF1<0的近似解，可以取的一個(gè)區(qū)間是SKIPIF1<0.故選：C．【典例1-3】（2019·全國·高三專題練習(xí)）若SKIPIF1<0的一個(gè)正數(shù)零點(diǎn)附近的函數(shù)值用二分法逐次計(jì)算，數(shù)據(jù)如下表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0那么方程SKIPIF1<0的一個(gè)近似根（精確到0.1）為（

）A．1.2 B．1.3 C．1.4 D．1.5【答案】C【詳解】根據(jù)二分法，結(jié)合表中數(shù)據(jù)，由于SKIPIF1<0所以方程SKIPIF1<0的一個(gè)近似根所在區(qū)間為SKIPIF1<0所以符合條件的解為1.4故選:C.【典例1-4】（2022·天津·靜海一中高三階段練習(xí)）已知函數(shù)SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)，且當(dāng)時(shí)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0零點(diǎn)個(gè)數(shù)就是SKIPIF1<0圖象交點(diǎn)個(gè)數(shù)，作出SKIPIF1<0圖象，如圖：由圖可得有SKIPIF1<0個(gè)交點(diǎn)，故SKIPIF1<0有SKIPIF1<0個(gè)零點(diǎn)．故選：B．【典例1-5】（2022·河南河南·三模（理））函數(shù)SKIPIF1<0的所有零點(diǎn)之和為（

）A．0 B．2 C．4 D．6【答案】B【詳解】令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0圖象關(guān)于SKIPIF1<0對(duì)稱，在SKIPIF1<0上遞減.SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，圖象關(guān)于原點(diǎn)對(duì)稱，所以SKIPIF1<0圖象關(guān)于SKIPIF1<0對(duì)稱，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0與SKIPIF1<0有兩個(gè)交點(diǎn)，兩個(gè)交點(diǎn)關(guān)于SKIPIF1<0對(duì)稱，所以函數(shù)SKIPIF1<0的所有零點(diǎn)之和為SKIPIF1<0.故選：B2、圖像零點(diǎn)個(gè)數(shù)的判定【典例2-1】（2022·北京·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）A．SKIPIF1<0個(gè) B．SKIPIF1<0個(gè) C．SKIPIF1<0個(gè) D．SKIPIF1<0個(gè)【答案】C【詳解】由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，或SKIPIF1<0，或SKIPIF1<0則SKIPIF1<0的零點(diǎn)個(gè)數(shù)為3故選：C【典例2-2】（2022·安徽·巢湖市第一中學(xué)高三期中（文））已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）A．3 B．4 C．5 D．6【答案】B【詳解】令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0且遞增，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0且遞減，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0且遞增，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0且遞增，此時(shí)SKIPIF1<0，所以，SKIPIF1<0的零點(diǎn)等價(jià)于SKIPIF1<0與SKIPIF1<0交點(diǎn)橫坐標(biāo)SKIPIF1<0對(duì)應(yīng)的SKIPIF1<0值，如下圖示：由圖知：SKIPIF1<0與SKIPIF1<0有兩個(gè)交點(diǎn)，橫坐標(biāo)SKIPIF1<0、SKIPIF1<0：當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，在SKIPIF1<0、SKIPIF1<0、SKIPIF1<0上各有一個(gè)解；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，在SKIPIF1<0有一個(gè)解.綜上，SKIPIF1<0的零點(diǎn)共有4個(gè).故選：B【典例2-3】（2016·天津市紅橋區(qū)教師發(fā)展中心高三學(xué)業(yè)考試）函數(shù)SKIPIF1<0．若在SKIPIF1<0內(nèi)恰有一個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：當(dāng)SKIPIF1<0時(shí)，函數(shù)為常函數(shù)，沒有零點(diǎn)，不滿足題意，所以SKIPIF1<0為一次函數(shù)，因?yàn)镾KIPIF1<0在SKIPIF1<0內(nèi)恰有一個(gè)零點(diǎn)，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C【典例2-4】（2022·湖南衡陽·二模）已知定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0恒有SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，已知SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上的零點(diǎn)個(gè)數(shù)為（

）A．4個(gè) B．5個(gè) C．3個(gè)或4個(gè) D．4個(gè)或5個(gè)【答案】D【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0的周期為2，又因?yàn)镾KIPIF1<0為奇函數(shù)，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0單調(diào)遞減得函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，得SKIPIF1<0，作出函數(shù)圖象如圖所示，由圖象可知當(dāng)SKIPIF1<0過點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)在SKIPIF1<0上只有3個(gè)零點(diǎn).當(dāng)SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)有5個(gè)零點(diǎn).當(dāng)SKIPIF1<0時(shí)，有4個(gè)零點(diǎn).當(dāng)SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)有5個(gè)零點(diǎn).當(dāng)SKIPIF1<0時(shí)，有4個(gè)零點(diǎn).當(dāng)SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)在SKIPIF1<0上只有3個(gè)零點(diǎn).當(dāng)SKIPIF1<0時(shí)，有4個(gè)零點(diǎn).所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上有4個(gè)或5個(gè)零點(diǎn).故選：D【典例2-5】（2022·寧夏銀川·一模（理））設(shè)函數(shù)SKIPIF1<0，已知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0在SKIPIF1<0上的零點(diǎn)最多有（

）A．2個(gè) B．3個(gè) C．4個(gè) D．5個(gè)【答案】A【詳解】由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，取SKIPIF1<0，可得SKIPIF1<0.若SKIPIF1<0在SKIPIF1<0上單詞遞增，則SKIPIF1<0，解得SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0所以函數(shù)SKIPIF1<0在SKIPIF1<0上的零點(diǎn)最多有2個(gè).所以SKIPIF1<0在SKIPIF1<0上的零點(diǎn)最多有2個(gè).故選：A3、圖像零點(diǎn)的綜合應(yīng)用【典例3-1】（2022·安徽·模擬預(yù)測(cè)（文））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【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【典例3-2】（2022·黑龍江·哈師大附中三模（文））已知有且只有一個(gè)實(shí)數(shù)x滿足SKIPIF1<0，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】D【詳解】SKIPIF1<0顯然不是SKIPIF1<0的根.所以SKIPIF1<0因此只有一個(gè)實(shí)數(shù)x滿足SKIPIF1<0等價(jià)于方程SKIPIF1<0只有一個(gè)實(shí)數(shù)根.令SKIPIF1<0，令SKIPIF1<0，故可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0圖像如圖：故SKIPIF1<0.故選：D【典例3-3】（2022·河南安陽·模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），若函數(shù)SKIPIF1<0的零點(diǎn)有5個(gè)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【詳解】解：依題意函數(shù)SKIPIF1<0的零點(diǎn)即為方程SKIPIF1<0的根，①當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0的函數(shù)圖象如下所示：所以SKIPIF1<0有兩個(gè)根SKIPIF1<0，SKIPIF1<0（SKIPIF1<0，SKIPIF1<0），而SKIPIF1<0對(duì)應(yīng)2個(gè)根，所以需要SKIPIF1<0對(duì)應(yīng)3個(gè)根，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0的函數(shù)圖象如下所示：所以SKIPIF1<0有兩個(gè)根SKIPIF1<0，SKIPIF1<0（SKIPIF1<0，SKIPIF1<0），而SKIPIF1<0對(duì)應(yīng)2個(gè)根，SKIPIF1<0對(duì)應(yīng)2個(gè)根，即共四個(gè)根，所以不滿足題意；③當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0的函數(shù)圖象如下所示：所以SKIPIF1<0有三個(gè)根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所對(duì)應(yīng)2、2、1個(gè)根，即共5個(gè)根，所以滿足題意；④當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0的函數(shù)圖象如下所示：所以SKIPIF1<0有三個(gè)根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，（SKIPIF1<0，SKIPIF1<0，SKIPIF1<0），而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0分別對(duì)應(yīng)2、2、0個(gè)根，即共四個(gè)根，所以不滿足題意；綜上可得實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0或SKIPIF1<0；故選：D【典例3-4】（2022·福建三明·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，即SKIPIF1<0有兩根，又SKIPIF1<0，故可轉(zhuǎn)換為SKIPIF1<0有兩根，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞減；在SKIPIF1<0上SKIPIF1<0，SKIPIF