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

試卷第=page11頁(yè)，共=sectionpages33頁(yè)第7練函數(shù)與方程學(xué)校____________姓名____________班級(jí)____________一、單選題1．設(shè)函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】易知SKIPIF1<0在R上單調(diào)遞增且連續(xù)．由于SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0．故選：B2．已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0零點(diǎn)個(gè)數(shù)為（

）A．0 B．1 C．2 D．3【答案】A【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以不存在零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，也不存在零點(diǎn)，所以函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為0.故選：A.3．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，即為函數(shù)SKIPIF1<0與直線SKIPIF1<0有兩個(gè)交點(diǎn)，函數(shù)SKIPIF1<0圖象如圖所示：所以SKIPIF1<0，故選：D.4．已知函數(shù)SKIPIF1<0，且f（x）在[0，SKIPIF1<0]有且僅有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．[SKIPIF1<0，SKIPIF1<0） B．[SKIPIF1<0，SKIPIF1<0） C．[SKIPIF1<0，SKIPIF1<0） D．[SKIPIF1<0，SKIPIF1<0）【答案】D【詳解】因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上有且只有3個(gè)零點(diǎn)，由余弦函數(shù)性質(zhì)可知SKIPIF1<0，解得SKIPIF1<0.故選：D．5．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SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的大小順序?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0、SKIPIF1<0均為SKIPIF1<0上的增函數(shù)，故函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0、SKIPIF1<0在SKIPIF1<0上均為增函數(shù)，故函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，因此，SKIPIF1<0.故選：A.6．已知直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象恰有SKIPIF1<0個(gè)公共點(diǎn)，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】A【詳解】根據(jù)題意，函數(shù)SKIPIF1<0，作出SKIPIF1<0的圖象：當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0和函數(shù)SKIPIF1<0的圖象只有一個(gè)交點(diǎn)；當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0和函數(shù)SKIPIF1<0的圖象只有一個(gè)交點(diǎn)，直線SKIPIF1<0和函數(shù)SKIPIF1<0的圖象有2個(gè)交點(diǎn)，即方程SKIPIF1<0在SKIPIF1<0上有2個(gè)實(shí)數(shù)根，SKIPIF1<0，則有SKIPIF1<0，解可得SKIPIF1<0，即SKIPIF1<0的取值范圍為SKIPIF1<0，SKIPIF1<0；故答案為：SKIPIF1<0，SKIPIF1<0.7．設(shè)函數(shù)SKIPIF1<0有5個(gè)不同的零點(diǎn)，則正實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】易知函數(shù)SKIPIF1<0、SKIPIF1<0在SKIPIF1<0上為增函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0無(wú)限接近0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上存在一點(diǎn)SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上有且只有一個(gè)零點(diǎn)；所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有4個(gè)零點(diǎn)，令SKIPIF1<0，即SKIPIF1<0Z，解得SKIPIF1<0SKIPIF1<0Z，由題可得SKIPIF1<0區(qū)間內(nèi)的4個(gè)零點(diǎn)分別是SKIPIF1<0，所以SKIPIF1<0即在SKIPIF1<0之間，即SKIPIF1<0，解得SKIPIF1<0故選：A8．已知函數(shù)SKIPIF1<0若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0僅有1個(gè)零點(diǎn)，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0R，有SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0同號(hào)，所以SKIPIF1<0在R上單調(diào)遞增，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，故SKIPIF1<0；因?yàn)镾KIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0沒(méi)有公共點(diǎn)，若函數(shù)SKIPIF1<0僅有一個(gè)零點(diǎn)，所以函數(shù)SKIPIF1<0與SKIPIF1<0圖象僅有一個(gè)交點(diǎn)，則SKIPIF1<0與SKIPIF1<0有且僅有1個(gè)公共點(diǎn)，且為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0處的切線的斜率k大于等于1，而SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，綜上，SKIPIF1<0的取值范圍為SKIPIF1<0.故選：C.二、多選題9．下列函數(shù)有兩個(gè)零點(diǎn)的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【詳解】解：對(duì)于A：令SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故A正確；對(duì)于B：令SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0時(shí)SKIPIF1<0，即函數(shù)在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即函數(shù)在SKIPIF1<0上單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)函數(shù)取得極小值即最小值，SKIPIF1<0，即SKIPIF1<0在定義域上只有一個(gè)零點(diǎn)，綜上可得函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0和SKIPIF1<0，故B正確；對(duì)于C：令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故C錯(cuò)誤；對(duì)于D：因?yàn)镾KIPIF1<0，所以函數(shù)的定義域?yàn)镾KIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，解SKIPIF1<0得SKIPIF1<0；解SKIPIF1<0即SKIPIF1<0，即SKIPIF1<0（舍去），所以SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0和SKIPIF1<0，故D正確；故選：ABD10．已知函數(shù)f（x）是定義在R上的偶函數(shù)，且當(dāng)x>0時(shí)，f（x）SKIPIF1<0，m∈R，那么函數(shù)g（x）=f（x）﹣2在定義域內(nèi)的零點(diǎn)個(gè)數(shù)可能是（

）A．2 B．4 C．6 D．8【答案】BC【詳解】解：由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0不是方程的根．當(dāng)x>0時(shí)，f（x）SKIPIF1<0，當(dāng)0<x≤2時(shí)，令3x﹣x2=2，解得x=1或2共有兩個(gè)解；當(dāng)x>2時(shí)，令SKIPIF1<0，即（m﹣2）x=2m，當(dāng)m=2時(shí)，方程無(wú)解，當(dāng)m>2時(shí)，方程有解xSKIPIF1<02，符合題意，當(dāng)m<2時(shí)，xSKIPIF1<02，不符合題意，方程無(wú)解．所以當(dāng)x>0時(shí)，f（x）=2有2個(gè)或3個(gè)根，而函數(shù)f（x）是定義在R上的偶函數(shù)，所以函數(shù)g（x）=f（x）﹣2在定義域內(nèi)的零點(diǎn)個(gè)數(shù)可能是4或6，故選：BC.11．已知函數(shù)SKIPIF1<0，ω>0．若函數(shù)SKIPIF1<0在SKIPIF1<0上恰有2個(gè)零點(diǎn)，則ω的可能值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【詳解】SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0上恰好有2個(gè)零點(diǎn)，∴SKIPIF1<0，則SKIPIF1<0，故B、C、D中的對(duì)應(yīng)值在SKIPIF1<0內(nèi).故選：BCD12．已知函數(shù)SKIPIF1<0(SKIPIF1<0為正整數(shù))，則下列判斷正確的有（

）A．對(duì)于任意的正整數(shù)SKIPIF1<0，SKIPIF1<0為奇函數(shù)B．存在正整數(shù)SKIPIF1<0，SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對(duì)稱C．當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0有四個(gè)零點(diǎn)D．當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0有兩個(gè)零點(diǎn)【答案】BD【詳解】當(dāng)SKIPIF1<0為偶數(shù)時(shí)，可得SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0為偶函數(shù)，所以函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，所以A不正確，B正確；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0僅有2個(gè)零點(diǎn)，所以C不正確；令SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0僅有2個(gè)零點(diǎn)，所以D正確.故選：BD.三、填空題13．函數(shù)SKIPIF1<0的圖象在區(qū)間（0,2）上連續(xù)不斷，能說(shuō)明“若SKIPIF1<0在區(qū)間（0,2）上存在零點(diǎn)，則SKIPIF1<0”為假命題的一個(gè)函數(shù)SKIPIF1<0的解析式可以為SKIPIF1<0=___________.【答案】SKIPIF1<0（答案不唯一）【詳解】函數(shù)SKIPIF1<0的圖象在區(qū)間（0，2）上連續(xù)不斷，且“若SKIPIF1<0在區(qū)間（0,2）上存在零點(diǎn)，則SKIPIF1<0”為假命題，可知函數(shù)SKIPIF1<0滿足在（0,2）上存在零點(diǎn)，且SKIPIF1<0，所以滿足題意的函數(shù)解析式可以為SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）.14．函數(shù)SKIPIF1<0有三個(gè)不同的零點(diǎn)，則實(shí)數(shù)t的范圍是__________.【答案】SKIPIF1<0【詳解】作出函數(shù)SKIPIF1<0的圖象和直線SKIPIF1<0，如圖，由圖象可得SKIPIF1<0時(shí)，直線與函數(shù)圖象有三個(gè)交點(diǎn)，即函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)．SKIPIF1<0．故答案為：SKIPIF1<0．15．已知函數(shù)SKIPIF1<0.若函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【詳解】畫(huà)出SKIPIF1<0的圖象如下圖所示，SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，由圖可知，SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<016．已知SKIPIF1<0是定義在R上的奇函數(shù)，且SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．設(shè)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)m的取值范圍是__________．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0是偶函數(shù)，所以有SKIPIF1<0，所以函數(shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，由SKIPIF1<0SKIPIF1<0，而SKIPIF1<0是定義在R上的奇函數(shù)，所以有SKIPIF1<0，因此有SKIPIF1<0，因此SKIPIF1<0，所以SKIPIF1<0，因此函數(shù)SKIPIF1<0的周期為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此有：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF