新高考數(shù)學(xué)二輪復(fù)習(xí)專題培優(yōu)練習(xí)專題03 分段函數(shù)（解析版）

專題03分段函數(shù)一、單選題1.（2024屆肅省蘭州市第五十中學(xué)高三上學(xué)期開學(xué)考試）函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．4 B．2 C．8 D．6【答案】B【解析】因為SKIPIF1<0，所以SKIPIF1<0.故選B2．（2024屆遼寧省六校高三上學(xué)期期初考試）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】依題意SKIPIF1<0，所以可能有以下兩種情形：情形一：若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0(不符題意，舍去).情形二：若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0,解得SKIPIF1<0.綜上有SKIPIF1<0.故SKIPIF1<0.故選A.3．（2024屆湖南省株洲市第三中學(xué)高三上學(xué)期8月月考）已知SKIPIF1<0，且SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)，由函數(shù)解析式可得函數(shù)在R上單調(diào)遞增不滿足題意，故SKIPIF1<0在R上單調(diào)遞減，所以SKIPIF1<0，解得SKIPIF1<0.故選D.4．（2024屆吉林省通化市輝南縣高三上學(xué)期月考）已知函數(shù)SKIPIF1<0有最大值，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因為當(dāng)SKIPIF1<0時，SKIPIF1<0，要使SKIPIF1<0有最大值，則SKIPIF1<0時，函數(shù)值的范圍不超過SKIPIF1<0可得SKIPIF1<0解得SKIPIF1<0.故選A5．（2024屆內(nèi)蒙古包頭市高三上學(xué)期調(diào)研）設(shè)函數(shù)SKIPIF1<0則滿足SKIPIF1<0的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0不成立；當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0，與SKIPIF1<0矛盾，舍去，當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0．綜上，滿足SKIPIF1<0的SKIPIF1<0的取值范圍是SKIPIF1<0．故選B．6．（2024屆百師聯(lián)盟高三上學(xué)期聯(lián)考）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有5個不同的實根，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】

當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，做出SKIPIF1<0的圖像，如圖所示，SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0共5個不等實根，由圖可知SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0有兩個根，若使SKIPIF1<0與SKIPIF1<0共5個不等實根，只需滿足SKIPIF1<0.故選D.7．（2023屆河南省開封市通許縣高三沖刺）已知SKIPIF1<0若函數(shù)SKIPIF1<0有兩個零點，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減.所以SKIPIF1<0時，SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減.所以SKIPIF1<0時，SKIPIF1<0.畫出函數(shù)SKIPIF1<0的圖象如圖所示：因為函數(shù)SKIPIF1<0有兩個零點，所以SKIPIF1<0與SKIPIF1<0的圖象有兩個交點，由圖可知SKIPIF1<0或SKIPIF1<0.所以SKIPIF1<0的取值范圍為SKIPIF1<0.故選C.8．（2024屆重慶市南開中學(xué)高三上學(xué)期第一次質(zhì)量檢測）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【答案】A【解析】

由圖可知當(dāng)且僅當(dāng)SKIPIF1<0時，方程SKIPIF1<0有四個不同的根，且SKIPIF1<0，由題：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0設(shè)SKIPIF1<0則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0故SKIPIF1<0在SKIPIF1<0遞增，在SKIPIF1<0遞減，SKIPIF1<0.故選A.9．（2024屆黑龍江省佳木斯市高三上學(xué)期第二次調(diào)研）已知函數(shù)SKIPIF1<0是定義域上的單調(diào)減函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意可得二次函數(shù)對稱軸為SKIPIF1<0，由于整個函數(shù)單調(diào)遞減，則有SKIPIF1<0，解之得SKIPIF1<0.故選A10．（2024屆遼寧省沈陽市第二十中學(xué)高三上學(xué)期第一次模擬）函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上的所有零點之和為（

）A．SKIPIF1<0 B．32 C．16 D．8【答案】D【解析】∵函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，∴SKIPIF1<0.又∵函數(shù)SKIPIF1<0，∴SKIPIF1<0∴函數(shù)SKIPIF1<0是偶函數(shù)，∴函數(shù)SKIPIF1<0的零點都是以相反數(shù)的形式成對出現(xiàn)的.∴函數(shù)SKIPIF1<0在SKIPIF1<0上所有的零點的和為0，∴函數(shù)SKIPIF1<0在SKIPIF1<0上所有的零點的和，即函數(shù)SKIPIF1<0在SKIPIF1<0上所有的零點之和.即方程SKIPIF1<0在SKIPIF1<0上的所有實數(shù)解之和.由SKIPIF1<0時，SKIPIF1<0，故有SKIPIF1<0，∴函數(shù)SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時，SKIPIF1<0.又∵當(dāng)SKIPIF1<0時，SKIPIF1<0，如圖：

∴函數(shù)SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0；函數(shù)SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0；函數(shù)SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時，SKIPIF1<0，即方程SKIPIF1<0在SKIPIF1<0上的有一個實數(shù)解SKIPIF1<0，即SKIPIF1<0有一個零點SKIPIF1<0；綜上，函數(shù)SKIPIF1<0在SKIPIF1<0上的所有零點之和為8.故選D.11.（2024屆黑龍江省哈爾濱市第三中學(xué)校高三上學(xué)期第一次驗收）已知函數(shù)SKIPIF1<0的最大值為1，則實數(shù)SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】A【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時取等號，依題意，SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，若SKIPIF1<0，則當(dāng)SKIPIF1<0時，SKIPIF1<0，解得SKIPIF1<0，符合題意，若SKIPIF1<0，則當(dāng)SKIPIF1<0時，SKIPIF1<0，解得SKIPIF1<0，矛盾，所以實數(shù)SKIPIF1<0的值為SKIPIF1<0.故選A12.（2024屆江西省宜春市宜豐中學(xué)高三上學(xué)期開學(xué)考試）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有四個不同的實根SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】對于SKIPIF1<0，可知其對稱軸為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；作出函數(shù)SKIPIF1<0的圖象，如圖所示，

若方程SKIPIF1<0有四個不同的實根SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0有四個不同的交點，交點橫坐標(biāo)依次為SKIPIF1<0，對于SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0；對于SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0；所以SKIPIF1<0，由對勾函數(shù)可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選B.二、多選題13．（2024屆廣東省潮州市潮安區(qū)鳳塘中學(xué)高三上學(xué)期統(tǒng)測）已知函數(shù)SKIPIF1<0，則下列結(jié)論中正確的是（

）A．函數(shù)SKIPIF1<0有且僅有一個零點0 B．SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減【答案】BC【解析】由函數(shù)SKIPIF1<0，可得SKIPIF1<0有兩個零點0、1，故A錯誤；由于SKIPIF1<0，故B正確；當(dāng)SKIPIF1<0時SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C正確；當(dāng)SKIPIF1<0時SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0上單調(diào)遞增，故D錯誤．故選BC．14．（2023屆山西省三晉名校聯(lián)盟高三下學(xué)期4月測試）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的最小值為SKIPIF1<0B．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增C．若SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的最大值為SKIPIF1<0D．SKIPIF1<0有三個零點【答案】BD【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，故SKIPIF1<0的最小值為SKIPIF1<0，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0和SKIPIF1<0，故A錯誤，B正確；若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，根據(jù)分段函數(shù)不難判斷出SKIPIF1<0，故SKIPIF1<0的最大值為SKIPIF1<0，故C錯誤；根據(jù)題意，函數(shù)SKIPIF1<0在SKIPIF1<0上有一個零點，函數(shù)SKIPIF1<0在SKIPIF1<0上有兩個零點SKIPIF1<0和SKIPIF1<0，故D正確，故選BD.15．（2024屆福建省泉州市高三高中畢業(yè)班質(zhì)量監(jiān)測）已知函數(shù)SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0為增函數(shù)C．SKIPIF1<0的值域為SKIPIF1<0 D．方程SKIPIF1<0最多有兩個解【答案】ACD【解析】對于A，顯然SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，A正確；對于B，顯然SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，B錯誤；對于C，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，因此SKIPIF1<0的值域為SKIPIF1<0，C正確；對于D，如圖，當(dāng)SKIPIF1<0時，方程SKIPIF1<0無解；當(dāng)SKIPIF1<0時，方程SKIPIF1<0有兩個解；

當(dāng)SKIPIF1<0時，方程SKIPIF1<0有一個解，因此方程SKIPIF1<0最多有兩個解，D正確.故選ACD16．（2023屆遼寧省凌源市高三下學(xué)期開學(xué)抽測）已知函數(shù)SKIPIF1<0則下列說法正確的是（

）A．SKIPIF1<0B．當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0值域為SKIPIF1<0C．當(dāng)SKIPIF1<0時，方程SKIPIF1<0恰有6個實根D．若SKIPIF1<0恒成立，則SKIPIF1<0．【答案】ABD【解析】依題意，根據(jù)分段函數(shù)SKIPIF1<0可得圖象如圖所示：因為SKIPIF1<0，故A正確；由題知函數(shù)SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0，在SKIPIF1<0上函數(shù)SKIPIF1<0值域為SKIPIF1<0，故當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0值域為SKIPIF1<0，故B正確；當(dāng)SKIPIF1<0時有5個實數(shù)根，當(dāng)SKIPIF1<0時有7個實數(shù)根，故C錯誤；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的圖象與SKIPIF1<0的圖象交于SKIPIF1<0點，結(jié)合圖象SKIPIF1<0，即SKIPIF1<0，故D正確．故選ABD.17.（2024屆安徽省六校教育研究會高三上學(xué)期入學(xué)素質(zhì)測試）高斯是德國著名的數(shù)學(xué)家，近代數(shù)學(xué)奠基者之一，享有“數(shù)學(xué)王子”的稱號.設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)，SKIPIF1<0也被稱為“高斯函數(shù)”，例如：SKIPIF1<0，SKIPIF1<0.已知函數(shù)SKIPIF1<0，下列說法中正確的是（

）A．SKIPIF1<0是周期函數(shù)B．SKIPIF1<0的值域是SKIPIF1<0C．SKIPIF1<0在SKIPIF1<0上是增函數(shù)D．若方程SKIPIF1<0有3個不同實根，則SKIPIF1<0【答案】AB【解析】由題意，列出部分定義域函數(shù)SKIPIF1<0，所以部分定義域的SKIPIF1<0，如圖：可得函數(shù)SKIPIF1<0是周期為1的函數(shù)，且值域為SKIPIF1<0，在SKIPIF1<0上單調(diào)遞減，

故選項A、B正確，C錯誤；對于選項D，若方程SKIPIF1<0有3個不同實根，則SKIPIF1<0的圖象與直線SKIPIF1<0有3個交點，又直線SKIPIF1<0恒過點SKIPIF1<0，結(jié)合圖象知，SKIPIF1<0或SKIPIF1<0，故選項D錯誤.故選AB三、填空題18．（2024屆寧夏銀川一中高三上學(xué)期月考）已知SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【解析】若SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，當(dāng)SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，當(dāng)SKIPIF1<0時，則SKIPIF1<0不符合要求，綜上可知：SKIPIF1<0的取值范圍為SKIPIF1<019．（2024屆湖南省長沙市高三上學(xué)期入學(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．【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上遞減，函數(shù)值集合為SKIPIF1<0，在SKIPIF1<0上遞增，函數(shù)值集合為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上遞減，函數(shù)值集合為SKIPIF1<0，在SKIPIF1<0上遞增，函數(shù)值集合為SKIPIF1<0，顯然函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象關(guān)于直線SKIPIF1<0對稱，如圖，方程SKIPIF1<0的四個解SKIPIF1<0是直線SKIPIF1<0與曲線SKIPIF1<0的四個交點橫坐標(biāo)為SKIPIF1<0，顯然SKIPIF1<0，不妨令SKIPIF1<0，則有SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，因此SKIPIF1<0，而對勾函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0.20．（2024屆北京市景山學(xué)校高三上學(xué)期開學(xué)考試）已知SKIPIF1<0，函數(shù)SKIPIF1<0，若存在三個互不相等的實數(shù)SKIPIF1<0，使得SKIPIF1<0成立，則a的取值范圍是.【答案】SKIPIF1<0【解析】若存在三個互不相等的實數(shù)SKIPIF1<0，使得SKIPIF1<0成立，則方程SKIPIF1<0存在三個不相等的實根，當(dāng)SKIPIF1<0時，SKIPIF1<0解得SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0有兩個不等的實根，即SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，在區(qū)間SKIPIF1<0單調(diào)遞減，則SKIPIF1<0，所以SKIPIF1<0.21．（2023屆四川省射洪中學(xué)校高三模擬預(yù)測）已知函數(shù)SKIPIF1<0，則下列命題中正確的有．①函數(shù)SKIPIF1<0有兩個極值點；②若關(guān)于x的方程SKIPIF1<0恰有1個解，則SKIPIF1<0；③函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0有且僅有一個交點；④若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0無最值．【答案】①③【解析】由函數(shù)SKIPIF1<0可得SKIPIF1<0，函數(shù)SKIPIF1<0的圖像如下圖所示：

對于①，由圖可知，SKIPIF1<0和SKIPIF1<0是函數(shù)SKIPIF1<0的兩個極值點，故①正確；對于②，若函數(shù)SKIPIF1<0恰有1個零點，即函數(shù)SKIPIF1<0與SKIPIF1<0的圖像僅有一個交點，可得SKIPIF1<0或SKIPIF1<0，故②不正確；對于③，因為函數(shù)SKIPIF1<0，SKIPIF1<0在點SKIPIF1<0處切線斜率SKIPIF1<0，在點SKIPIF1<0處的切線為SKIPIF1<0，函數(shù)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0處的切線斜率為SKIPIF1<0，在SKIPIF1<0處切線為SKIPIF1<0，如圖中虛線所示，易知當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0的圖像與直線SKIPIF1<0恰有一個交點；當(dāng)SKIPIF1<0，即SKIPIF1<0時，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，由二次函數(shù)的圖像及零點存在定理可知，方程SKIPIF1<0有且只有一個實數(shù)根；當(dāng)SKIPIF1<0，即SKIPIF1