新高考數(shù)學(xué)一輪復(fù)習(xí) 講與練第6練 函數(shù)的圖像（解析版）

試卷第=page11頁，共=sectionpages33頁第6練函數(shù)的圖像學(xué)校____________姓名____________班級(jí)____________一、單選題1．設(shè)SKIPIF1<0，定義符號(hào)函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的圖像大致是（

）A． B． C． D．【答案】C【詳解】由函數(shù)SKIPIF1<0，故C選項(xiàng)正確．故選：C2．已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則函數(shù)SKIPIF1<0的解析式可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】A的函數(shù)即為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故排除A由圖象可知SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，則SKIPIF1<0為奇函數(shù)，排除B，C.故選：D.3．已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的大致圖象為（

）A． B．C． D．【答案】D【詳解】由題可知：函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故該函數(shù)為奇函數(shù)，排除A,C又SKIPIF1<0，所以排除B,故選：D4．函數(shù)SKIPIF1<0的圖象如圖所示，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由圖象可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故選：C5．已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的零點(diǎn)分別是a，b，c，則a，b，c的大小順序是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由已知條件得SKIPIF1<0的零點(diǎn)可以看成SKIPIF1<0與SKIPIF1<0的交點(diǎn)的橫坐標(biāo)，SKIPIF1<0的零點(diǎn)可以看成SKIPIF1<0與SKIPIF1<0的交點(diǎn)的橫坐標(biāo)，SKIPIF1<0的零點(diǎn)可以看成SKIPIF1<0與SKIPIF1<0的交點(diǎn)的橫坐標(biāo)，在同一坐標(biāo)系分別畫出SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的函數(shù)圖象，如下圖所示,可知SKIPIF1<0，故選：SKIPIF1<0.6．我國著名數(shù)學(xué)家華羅庚曾說：“數(shù)缺形時(shí)少直觀，形缺數(shù)時(shí)難入微，數(shù)形結(jié)合百般好，隔裂分家萬事休.”在數(shù)學(xué)的學(xué)習(xí)和研究中，常用函數(shù)的圖象來研究函數(shù)的性質(zhì)，也常用函數(shù)的解析式來研究函數(shù)圖象的特征.我們從這個(gè)商標(biāo)中抽象出一個(gè)函數(shù)的圖象如圖，其對(duì)應(yīng)的函數(shù)解析式可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】A：函數(shù)的定義域?yàn)镾KIPIF1<0，不符合；B：由SKIPIF1<0，不符合；C：由SKIPIF1<0，不符合；D：SKIPIF1<0且定義域?yàn)镾KIPIF1<0，SKIPIF1<0為偶函數(shù)，在SKIPIF1<0上SKIPIF1<0單調(diào)遞增，SKIPIF1<0上SKIPIF1<0單調(diào)遞減，結(jié)合偶函數(shù)的對(duì)稱性知：SKIPIF1<0上遞減，SKIPIF1<0上遞增，符合.故選：D7．已知某函數(shù)的圖象如圖所示，則該函數(shù)的解析式可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由圖象可知，SKIPIF1<0，對(duì)選項(xiàng)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)沒有意義，故排除；由圖象可知，SKIPIF1<0，對(duì)SKIPIF1<0：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，滿足圖象要求；對(duì)SKIPIF1<0：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不滿足圖象要求；故選：SKIPIF1<0.8．我國著名數(shù)學(xué)家華羅庚曾說：“數(shù)缺形時(shí)少直觀，形缺數(shù)時(shí)難入微，數(shù)形結(jié)合百般好，隔裂分家萬事休．”在數(shù)學(xué)的學(xué)習(xí)和研究中，常用函數(shù)的圖像來研究函數(shù)的性質(zhì)，也常用函數(shù)的解析式來分析函數(shù)的圖像的特征，如函數(shù)SKIPIF1<0的圖像大致是（

）A． B．C． D．【答案】D【詳解】由定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，排除A、C；而SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上不遞減，排除B.故選：D二、多選題9．已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的解析式可以是（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【詳解】對(duì)于A，SKIPIF1<0，為偶函數(shù)，則A不符合題意；對(duì)于B，畫出函數(shù)SKIPIF1<0的圖象，如圖，由圖可知，B符合題意；對(duì)于C，畫出函數(shù)SKIPIF1<0的圖象，如圖，由圖可知，C符合題意；對(duì)于D，畫出函數(shù)f(x)=ln由圖可知，D符合題意；故選：BCD．10．已知函數(shù)SKIPIF1<0SKIPIF1<0則下列結(jié)論正確的有（

）A．SKIPIF1<0N*B．SKIPIF1<0恒成立C．關(guān)于x的方程SKIPIF1<0R)有三個(gè)不同的實(shí)根，則SKIPIF1<0D．關(guān)于x的方程SKIPIF1<0N*)的所有根之和為SKIPIF1<0【答案】AC【詳解】由題知SKIPIF1<0，故A正確；由上可知，要使SKIPIF1<0恒成立，只需滿足SKIPIF1<0時(shí)，SKIPIF1<0成立，即SKIPIF1<0，即SKIPIF1<0成立，令SKIPIF1<0，則SKIPIF1<0得SKIPIF1<0，易知當(dāng)SKIPIF1<0時(shí)有極大值SKIPIF1<0，故B不正確；作函數(shù)圖象，由圖可知，要使方程SKIPIF1<0R)有三個(gè)不同的實(shí)根，則SKIPIF1<0，即SKIPIF1<0，故C正確；由SKIPIF1<0可知，函數(shù)在SKIPIF1<0上的函數(shù)圖象可以由SKIPIF1<0上的圖象向右平移一個(gè)單位長度，在將所有點(diǎn)的橫坐標(biāo)不變，縱坐標(biāo)變?yōu)樵瓉淼腟KIPIF1<0倍得到，由于SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，故SKIPIF1<0的兩根之和為SKIPIF1<0，同理，SKIPIF1<0的兩根之和為SKIPIF1<0，…，SKIPIF1<0的兩根之和為SKIPIF1<0，故所有根之和為SKIPIF1<0，故D錯(cuò)誤.故選：AC11．關(guān)于SKIPIF1<0的函數(shù)SKIPIF1<0有4個(gè)零點(diǎn)，則整數(shù)SKIPIF1<0的可能取值為（

）A．5 B．6 C．7 D．9【答案】ABC【詳解】由對(duì)勾函數(shù)得單調(diào)性可知，SKIPIF1<0的圖象大致如下：x>0時(shí)，有兩個(gè)零點(diǎn)，須滿足：k>0，且SKIPIF1<0；x<0時(shí)，有兩個(gè)零點(diǎn)，須滿足：k>0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，無零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，有一個(gè)零點(diǎn)，故不合題意；當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，故不可能有4個(gè)零點(diǎn)，綜上：實(shí)數(shù)k的取值范圍為[5，9)，故選：ABC.12．定義域和值域均為SKIPIF1<0（常數(shù)SKIPIF1<0）的函數(shù)SKIPIF1<0和SKIPIF1<0圖象如圖所示，給出下列四個(gè)命題，那么，其中正確命題是（

）A．方程SKIPIF1<0有且僅有三個(gè)解B．方程SKIPIF1<0有且僅有三個(gè)解C．方程SKIPIF1<0有且僅有九個(gè)解D．方程SKIPIF1<0有且僅有一個(gè)解【答案】AD【詳解】解：對(duì)于A中，設(shè)SKIPIF1<0，則由SKIPIF1<0，即SKIPIF1<0，由圖象知方程SKIPIF1<0有三個(gè)不同的解，設(shè)其解為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由于SKIPIF1<0是減函數(shù)，則直線SKIPIF1<0與函數(shù)SKIPIF1<0只有1個(gè)交點(diǎn)，所以方程SKIPIF1<0，SKIPIF1<0，SKIPIF1<0分別有且僅有一個(gè)解，所以SKIPIF1<0有三個(gè)解，故A正確；對(duì)于B中，設(shè)SKIPIF1<0，則由SKIPIF1<0，即SKIPIF1<0，由圖象可得SKIPIF1<0有且僅有一個(gè)解，設(shè)其解為b，可知SKIPIF1<0，則直線SKIPIF1<0與函數(shù)SKIPIF1<0只有2個(gè)交點(diǎn)，所以方程SKIPIF1<0只有兩個(gè)解，所以方程SKIPIF1<0有兩個(gè)解，故B錯(cuò)誤；對(duì)于C中，設(shè)SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0，方程SKIPIF1<0有三個(gè)不同的解，設(shè)其解為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則由函數(shù)SKIPIF1<0圖象，可知SKIPIF1<0，SKIPIF1<0，由圖可知，直線SKIPIF1<0和直線SKIPIF1<0分別與函數(shù)SKIPIF1<0有3個(gè)交點(diǎn)，直線SKIPIF1<0與函數(shù)SKIPIF1<0只有1個(gè)交點(diǎn)，所以SKIPIF1<0或SKIPIF1<0或SKIPIF1<0共有7個(gè)解，所以SKIPIF1<0共有七個(gè)解，故C錯(cuò)誤；對(duì)于D中，設(shè)SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0，由圖象可得SKIPIF1<0有且僅有一個(gè)解，設(shè)其解為b，可知SKIPIF1<0，因?yàn)镾KIPIF1<0是減函數(shù)，則直線SKIPIF1<0與函數(shù)SKIPIF1<0只有1個(gè)交點(diǎn)，所以方程SKIPIF1<0只有1解，所以方程SKIPIF1<0只有一個(gè)解，故D正確.故選：AD.三、填空題13．已知函數(shù)SKIPIF1<0，SKIPIF1<0，若方程SKIPIF1<0恰有兩個(gè)不同的實(shí)數(shù)根，則實(shí)數(shù)SKIPIF1<0的取值范圍是________【答案】SKIPIF1<0【詳解】由題意，作出如下函數(shù)圖象，由圖象可知：當(dāng)SKIPIF1<0過點(diǎn)SKIPIF1<0即SKIPIF1<0時(shí)，方程SKIPIF1<0有一個(gè)實(shí)數(shù)根；當(dāng)SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上相切時(shí)，SKIPIF1<0有一個(gè)實(shí)數(shù)根，即SKIPIF1<0，SKIPIF1<0，有切點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0；當(dāng)SKIPIF1<0與SKIPIF1<0平行即SKIPIF1<0時(shí)，方程SKIPIF1<0恰有兩個(gè)不同的實(shí)數(shù)根；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有一個(gè)實(shí)數(shù)根；綜上，當(dāng)SKIPIF1<0或SKIPIF1<0或SKIPIF1<0時(shí)，方程SKIPIF1<0有一個(gè)實(shí)數(shù)根；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0恰有三個(gè)不同的實(shí)數(shù)根；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0恰有兩個(gè)不同的實(shí)數(shù)根；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0無實(shí)數(shù)根.故答案為：SKIPIF1<014．函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍是______．【答案】SKIPIF1<0【詳解】設(shè)SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0有三個(gè)零點(diǎn)SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0的圖象與直線SKIPIF1<0交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0，且SKIPIF1<0，作出SKIPIF1<0的圖象如圖所示，由圖可知SKIPIF1<0，且SKIPIF1<0是方程SKIPIF1<0的兩個(gè)實(shí)根，所以SKIPIF1<0，因?yàn)镾KIPIF1<0滿足SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0