新高考數(shù)學(xué)一輪復(fù)習(xí) 函數(shù)專項重難點突破專題05 分段函數(shù)(解析版)

專題05分段函數(shù)真題再現(xiàn)1．（2023·北京·統(tǒng)考高考真題）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0，給出下列四個結(jié)論：①SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；②當(dāng)SKIPIF1<0時，SKIPIF1<0存在最大值；③設(shè)SKIPIF1<0，則SKIPIF1<0；④設(shè)SKIPIF1<0．若SKIPIF1<0存在最小值，則a的取值范圍是SKIPIF1<0．其中所有正確結(jié)論的序號是____________．【解析】依題意，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，易知其圖像為一條端點取不到值的單調(diào)遞增的射線；當(dāng)SKIPIF1<0時，SKIPIF1<0，易知其圖像是，圓心為SKIPIF1<0，半徑為SKIPIF1<0的圓在SKIPIF1<0軸上方的圖像（即半圓）；當(dāng)SKIPIF1<0時，SKIPIF1<0，易知其圖像是一條端點取不到值的單調(diào)遞減的曲線；對于①，取SKIPIF1<0，則SKIPIF1<0的圖像如下，

顯然，當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故①錯誤；對于②，當(dāng)SKIPIF1<0時，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0顯然取得最大值SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，綜上：SKIPIF1<0取得最大值SKIPIF1<0，故②正確；對于③，結(jié)合圖像，易知在SKIPIF1<0，SKIPIF1<0且接近于SKIPIF1<0處，SKIPIF1<0的距離最小，

當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0且接近于SKIPIF1<0處，SKIPIF1<0，此時，SKIPIF1<0，故③正確；對于④，取SKIPIF1<0，則SKIPIF1<0的圖像如下，

因為SKIPIF1<0，結(jié)合圖像可知，要使SKIPIF1<0取得最小值，則點SKIPIF1<0在SKIPIF1<0上，點SKIPIF1<0在SKIPIF1<0，同時SKIPIF1<0的最小值為點SKIPIF1<0到SKIPIF1<0的距離減去半圓的半徑SKIPIF1<0，此時，因為SKIPIF1<0的斜率為SKIPIF1<0，則SKIPIF1<0，故直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上，滿足SKIPIF1<0取得最小值，即SKIPIF1<0也滿足SKIPIF1<0存在最小值，故SKIPIF1<0的取值范圍不僅僅是SKIPIF1<0，故④錯誤.故答案為：②③.2．（2022·浙江·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0則SKIPIF1<0________；若當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0的最大值是_________．【解析】由已知SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0等價于SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0.3．（2022·北京·統(tǒng)考高考真題）設(shè)函數(shù)SKIPIF1<0若SKIPIF1<0存在最小值，則a的一個取值為________；a的最大值為___________．【解析】若SKIPIF1<0時，SKIPIF1<0，∴SKIPIF1<0；若SKIPIF1<0時，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0沒有最小值，不符合題目要求；若SKIPIF1<0時，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0∴SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0，綜上可得SKIPIF1<0；故答案為：0（答案不唯一），14．（2021·浙江·統(tǒng)考高考真題）已知SKIPIF1<0，函數(shù)SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0___________.【解析】SKIPIF1<0，故SKIPIF1<0，故答案為：2.考點一分段函數(shù)函數(shù)值(解析式)一、單選題1．已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．5B．3 C．2 D．1【解析】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B2．已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．-1 D．2【解析】由條件可得SKIPIF1<0，則SKIPIF1<0.故選：C.3．已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．4 B．8 C．16 D．32【解析】SKIPIF1<0，故選：C．4．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（）.A．SKIPIF1<01 B．SKIPIF1<02 C．SKIPIF1<03 D．SKIPIF1<04【解析】由函數(shù)SKIPIF1<0，可作圖如下：

由方程SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.SKIPIF1<0.故選：B.5．已知函數(shù)SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】當(dāng)SKIPIF1<0時，由SKIPIF1<0①，得SKIPIF1<0②，①②聯(lián)立，可得SKIPIF1<0,得SKIPIF1<0③把①代入③可得SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，故選：C.6．已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．4 B．5 C．6 D．7【解析】由題意可得SKIPIF1<0SKIPIF1<0SKIPIF1<0，故選：D.7．已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．1 B．e C．SKIPIF1<0 D．SKIPIF1<0【解析】因為SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0.故選：D8．若函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由于SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，故選C.二、多選題9．函數(shù)f(x)的圖象如圖所示，則f(x)的解析式是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】結(jié)合圖象可知，當(dāng)x≤0時，設(shè)SKIPIF1<0，將SKIPIF1<0代入函數(shù)，得SKIPIF1<0，SKIPIF1<0，同理，當(dāng)x>0時，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：AC三、填空題10．已知函數(shù)SKIPIF1<0，則SKIPIF1<0___________．【解析】由SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<011．已知函數(shù)SKIPIF1<0，則SKIPIF1<0___________．【解析】因為SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0.12．已知函數(shù)SKIPIF1<0，則SKIPIF1<0_________．【解析】SKIPIF1<0，所以SKIPIF1<013．設(shè)函數(shù)SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的解析式為____________．【解析】因為函數(shù)解析式為SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.14．設(shè)SKIPIF1<0定義在SKIPIF1<0上且SKIPIF1<0，則SKIPIF1<0______．【解析】因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，同理可得SKIPIF1<0.考點二分段函數(shù)定義域和值域一、單選題1．設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0則SKIPIF1<0的值域是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】當(dāng)SKIPIF1<0,即SKIPIF1<0，SKIPIF1<0時,SKIPIF1<0或SKIPIF1<0,SKIPIF1<0,因為SKIPIF1<0，所以SKIPIF1<0，因此這個區(qū)間的值域為SKIPIF1<0.當(dāng)SKIPIF1<0時,即SKIPIF1<0，得SKIPIF1<0,SKIPIF1<0其最小值為SKIPIF1<0，其最大值為SKIPIF1<0，因此這區(qū)間的值域為SKIPIF1<0.綜上，函數(shù)值域為:SKIPIF1<0.故選：D2．若定義運算SKIPIF1<0，則函數(shù)SKIPIF1<0的值域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】若SKIPIF1<0，即SKIPIF1<0時SKIPIF1<0；若SKIPIF1<0，即SKIPIF1<0時SKIPIF1<0；綜上，SKIPIF1<0值域為SKIPIF1<0.故選：A二、多選題3．已知函數(shù)SKIPIF1<0，關(guān)于函數(shù)SKIPIF1<0的結(jié)論正確的是（

）A．SKIPIF1<0的定義域是R B．SKIPIF1<0的值域是SKIPIF1<0C．若SKIPIF1<0，則x的值為SKIPIF1<0 D．SKIPIF1<0【解析】A：函數(shù)的定義域為SKIPIF1<0，所以本選項不正確；B：當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，所以有SKIPIF1<0，綜上所述：SKIPIF1<0的值域是SKIPIF1<0，所以本選項正確；C：當(dāng)SKIPIF1<0時，SKIPIF1<0，不符合SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，或SKIPIF1<0不符合SKIPIF1<0，綜上所述：當(dāng)SKIPIF1<0時，x的值為SKIPIF1<0，所以本選項正確；D：SKIPIF1<0，所以本選項正確，故選：BCD三、填空題4．函數(shù)SKIPIF1<0的定義域是________.【解析】解：因為SKIPIF1<0，所以SKIPIF1<0定義域為[0，1]∪(1，2)∪[2，＋∞)＝[0，＋∞).5．已知SKIPIF1<0，則SKIPIF1<0的值域是______；【解析】當(dāng)SKIPIF1<0時,根據(jù)指數(shù)函數(shù)的圖象與性質(zhì)知SKIPIF1<0，當(dāng)SKIPIF1<0時,SKIPIF1<0.綜上:SKIPIF1<0的值域為SKIPIF1<0.6．函數(shù)SKIPIF1<0的值域是______．【解析】當(dāng)SKIPIF1<0時，滿足SKIPIF1<0；當(dāng)SKIPIF1<0時，由SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0．7．已知函數(shù)SKIPIF1<0SKIPIF1<0的最大值為m，SKIPIF1<0的最小值為n，則SKIPIF1<0______．【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，所以此時SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以此時SKIPIF1<0，綜上所述，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.8．已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最小值為_________．【解析】SKIPIF1<0的圖象如圖所示，故SKIPIF1<0的最小值為1，故答案為：1.四、雙空題9．函數(shù)y＝SKIPIF1<0的定義域為________，值域為________.【解析】定義域為各段的并集，即（－∞，0）∪（0，＋∞）.因為x>0，所以x2>0，由于值域為各段的并集，所以函數(shù)的值域為{－2}∪（0，＋∞）.10．如圖為一分段函數(shù)的圖象，則該函數(shù)的定義域為________，值域為________．【解析】由圖象可知，第一段的定義域為[－1，0)，值域為[0，1)；第二段的定義域為[0，2]，值域為[－1，0]．所以該分段函數(shù)的定義域為[－1，2]，值域為[－1，1)．故答案為：[－1，2]；[－1，1)五、解答題11．已知函數(shù)SKIPIF1<0的圖象如圖所示，其中SKIPIF1<0軸的左側(cè)為一條線段，右側(cè)為某拋物線的一段．(1)寫出函數(shù)SKIPIF1<0的定義域和值域；(2)求SKIPIF1<0的值．【解析】（1）由圖可知，函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，值域為SKIPIF1<0.（2）當(dāng)SKIPIF1<0時，設(shè)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時，可設(shè)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以，SKIPIF1<0，則SKIPIF1<0，因此，SKIPIF1<0.考點三分段函數(shù)單調(diào)性一、單選題1．定義運算：SKIPIF1<0，例如：SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞減，當(dāng)SKIPIF1<0的遞增區(qū)間為SKIPIF1<0，故選：A2．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0和SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<0【解析】SKIPIF1<0，則由二次函數(shù)的性質(zhì)知，當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0和SKIPIF1<0.故選：B3．若函數(shù)SKIPIF1<0是SKIPIF1<0上的單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】函數(shù)SKIPIF1<0,當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)圖像的對稱軸為SKIPIF1<0，函數(shù)不是單調(diào)函數(shù)，不滿足題意，排除B、C；當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)圖像的對稱軸為SKIPIF1<0，函數(shù)不是單調(diào)函數(shù)，排除D.故選：A．二、多選題4．已知函數(shù)SKIPIF1<0在R上單調(diào)遞增，則實數(shù)a的取值可以是（

）A．0 B．1 C．2 D．3【解析】由題意可知，SKIPIF1<0在SKIPIF1<0上遞增，則SKIPIF1<0，即SKIPIF1<0.SKIPIF1<0在SKIPIF1<0上遞增，則SKIPIF1<0.又SKIPIF1<0，則SKIPIF1<0.綜上，SKIPIF1<0，根據(jù)選項只有CD符合.故選：CD三、填空題5．設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的遞增區(qū)間為________.【解析】由題知SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖所示.∴函數(shù)SKIPIF1<0的遞增區(qū)間為SKIPIF1<0，SKIPIF1<0.6．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為__________.【解析】由函數(shù)SKIPIF1<0，根據(jù)一次函數(shù)的性質(zhì)，可得SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，則函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0.7．己知函數(shù)SKIPIF1<0滿足對任意SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0成立，則實數(shù)a的取值范圍是__________．【解析】因為對任意SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.所以SKIPIF1<0，解得SKIPIF1<0.8．已知函數(shù)SKIPIF1<0，設(shè)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的取值范圍是____________.【解析】因為函數(shù)f(x)在區(qū)間SKIPIF1<0，SKIPIF1<0上都是單調(diào)遞增函數(shù)，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，滿足SKIPIF1<0，必有SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，令SKIPIF1<0，在SKIPIF1<0上遞增，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.四、雙空題9．函數(shù)SKIPIF1<0的單調(diào)性為______；奇偶性為______．【解析】SKIPIF1<0在SKIPIF1<0時嚴(yán)格單調(diào)遞增，SKIPIF1<0在SKIPIF1<0時，嚴(yán)格單調(diào)遞增，且SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0是嚴(yán)格單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0時嚴(yán)格單調(diào)遞增，又SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上是嚴(yán)格增函數(shù)，易知SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，所以對定義域內(nèi)任意的SKIPIF1<0都有SKIPIF1<0，因此SKIPIF1<0是奇函數(shù)，故答案為：嚴(yán)格增；奇函數(shù)．五、解答題10．已知函數(shù)SKIPIF1<0(1)求SKIPIF1<0，SKIPIF1<0的值；(2)若SKIPIF1<0，求實數(shù)a的值；(3)直接寫出SKIPIF1<0的單調(diào)區(qū)間．【解析】（1）根據(jù)分段函數(shù)解析式可得SKIPIF1<0，易知SKIPIF1<0；所以SKIPIF1<0

，即SKIPIF1<0.（2）①當(dāng)SKIPIF1<0時，SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0（舍）.②當(dāng)SKIPIF1<0時，SKIPIF1<0，解得SKIPIF1<0（舍）.綜上可得SKIPIF1<0.即實數(shù)a的值為SKIPIF1<0（3）畫出函數(shù)圖象如下所示：所以，單調(diào)遞增區(qū)間SKIPIF1<0，單調(diào)遞減區(qū)間SKIPIF1<0，SKIPIF1<0考點四分段函數(shù)求參一、單選題1．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．1 D．2【解析】由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，舍去；若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，符合題意；故選：C.2．設(shè)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．1 D．2【解析】依題意，SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，故選：C3．已知函數(shù)SKIPIF1<0SKIPIF1<0的值域為SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由題意知當(dāng)SKIPIF1<0時，SKIPIF1<0，由于函數(shù)SKIPIF1<0的值域為SKIPIF1<0，故SKIPIF1<0時，SKIPIF1<0的取值范圍應(yīng)包含SKIPIF1<0，故此時SKIPIF1<0，且SKIPIF1<0，故SKIPIF1<0，故選：C.4．已知函數(shù)SKIPIF1<0的最大值為0，則實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】若SKIPIF1<0，SKIPIF1<0即當(dāng)SKIPIF1<0時SKIPIF1<0，∴SKIPIF1<0的最大值為0，滿足題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，不滿足題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，當(dāng)SKIPIF1<0時等號成立，滿足題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時等號成立，滿足題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，不滿足題意；所以SKIPIF1<0；故選：A.5．若函數(shù)SKIPIF1<0，在R上單調(diào)遞增，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0是單調(diào)遞增函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0當(dāng)SKIPIF1<0時，對勾函數(shù)取值要大于或等于指數(shù)式的值，所以SKIPIF1<0，解之得：SKIPIF1<0，綜上所述：實數(shù)a的取值范圍是SKIPIF1<0，故選：B6．已知函數(shù)SKIPIF1<0，滿足對任意的實數(shù)SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，都有SKIPIF1<0，則實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】對任意的實數(shù)SKIPIF1<0，都有SKIPIF1<0,即SKIPIF1<0成立，可得函數(shù)圖像上任意兩點連線的斜率小于0，說明函數(shù)是減函數(shù)；可得：SKIPIF1<0，解得SKIPIF1<0，故選：C7．已知函數(shù)SKIPIF1<0在SKIPIF1<0是減函數(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【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，所以，SKIPIF1<0在SKIPIF1<0上為減函數(shù)，因為SKIPIF1<0在SKIPIF1<0是減函數(shù)，且函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)，只需SKIPIF1<0，解得SKIPIF1<0.故選：B.8．函數(shù)SKIPIF1<0，若SKIPIF1<0互不相等，且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】函數(shù)SKIPIF1<0的圖像如圖所示：設(shè)SKIPIF1<0，由函數(shù)圖像數(shù)形結(jié)合可知：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0．故選：C．二、填空題9．已知SKIPIF1<0，函數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0________．【解析】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.10．已知函數(shù)SKIPIF1<0．若SKIPIF1<0，則實數(shù)m的取值范圍是______．【解析】作出函數(shù)SKIPIF1<0的圖象，如圖所示，

如SKIPIF1<0，則SKIPIF1<0，又因為SKIPIF1<0，結(jié)合圖象可知：SKIPIF1<0，所以實數(shù)m的取值范圍是SKIPIF1<0，11．設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，則實數(shù)a＝_____．【解析】當(dāng)SKIPIF1<0時，則SKIPIF1<0；當(dāng)SKIPIF1<0時，則SKIPIF1<0；綜上所述：SKIPIF1<0在定義域內(nèi)恒成立，令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，則SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去）；綜上所述：SKIPIF1<0或SKIPIF1<0.12．已知函數(shù)SKIPIF1<0且SKIPIF1<0，則正數(shù)SKIPIF1<0的值為______.【解析】當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0單調(diào)遞增，有SKIPIF1<0，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0單調(diào)遞增，有SKIPIF1<0，因為SKIPIF1<0，所以有SKIPIF1<0，故答案為：SKIPIF1<013．設(shè)函數(shù)SKIPIF1<0存在最小值，則SKIPIF1<0的取值范圍是________.【解析】①當(dāng)SKIPIF1<0時，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因此SKIPIF1<0不存在最小值；②當(dāng)SKIPIF1<0時，SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，故函數(shù)SKIPIF1<0存在最小值；③當(dāng)SKIPIF1<0時，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0不存在最小值，故SKIPIF1<0，解得SKIPIF1<0.此時SKIPIF1<0滿足題設(shè)；④當(dāng)SKIPIF1<0時，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0.因為SKIPIF1<0，所以SKIPIF1<0，因此SKIPIF1<0不存在最小值.綜上，SKIPIF1<0的取值范圍是SKIPIF1<0.14．已知SKIPIF1<0，若存在三個不同實數(shù)SKIPIF1<0使得SKIPIF1<0，則SKIPIF1<0的取值范圍是______．【解析】作出函數(shù)SKIPIF1<0的圖像如下圖所示：設(shè)SKIPIF1<0，由圖像可知SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0.SKIPIF1<0.15．設(shè)SKIPIF1<0且SKIPIF1<0，已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0是遞增數(shù)列，則a的取值范圍是____．【解析】因為SKIPIF1<0是遞增數(shù)列，所以SKIPIF1<0解得SKIPIF1<0，故答案為:SKIPIF1<0.16．已知SKIPIF1<0，其中SKIPIF1<0且SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0___________．【解析】因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0考點五解分段函數(shù)不等式一、單選題1．已知SKIPIF1<0,則使SKIPIF1<0成立的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】（方法1）當(dāng)SKIPIF1<0時SKIPIF1<0，不等式SKIPIF1<0可化為SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，不等式SKIPIF1<0可化為SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.綜上，使不等式SKIPIF1<0成立的SKIPIF1<0的取值范圍是SKIPIF1<0.故選:A.（方法2）函數(shù)SKIPIF1<0的圖象如圖所示，虛線表示SKIPIF1<0，函數(shù)SKIPIF1<0圖象在虛線SKIPIF1<0及以上的部分中SKIPIF1<0的取值范圍即不等式SKIPIF1<0的解集.由圖可知，SKIPIF1<0的取值范圍就是點的橫坐標(biāo)與點SKIPIF1<0的橫坐標(biāo)之間的范圍.在SKIPIF1<0中，令SKIPIF1<0，得SKIPIF1<0，所以點SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0.在SKIPIF1<0中，令SKIPIF1<0，得SKIPIF1<0（舍去）或SKIPIF1<0，所以點SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0，所以使不等式SKIPIF1<0成立的SKIPIF1<0的取值范圍是SKIPIF1<0.故選：A.2．已知函數(shù)SKIPIF1<0，則滿足不等式SKIPIF1<0的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】畫出SKIPIF1<0的圖象，如下：顯然要滿足SKIPIF1<0，則要SKIPIF1<0，且SKIPIF1<0，解得：SKIPIF1<0.故選：C3．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】因為SKIPIF1<0.①當(dāng)SKIPIF1<0時，SKIPIF1<0.②當(dāng)SKIPIF1<0時，SKIPIF1<0.③當(dāng)SKIPIF1<0時，SKIPIF1<0.綜上所述：SKIPIF1<0.故選：D.4．已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0即為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0無解，當(dāng)SKIPIF1<0時，SKIPIF1<0即為SKIPIF1<0，在同一平面直角坐標(biāo)系下畫出SKIPIF1<0和SKIPIF1<0的大致圖像如圖，由圖可得當(dāng)且僅當(dāng)SKIPIF1<0時，SKIPIF1<0，綜上所述，SKIPIF1<0的解為SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0，綜上所述，不等式SKIPIF1<0的解集是SKIPIF1<0.故選：D.5．若SKIPIF1<0，且SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0遞增，所以SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，可得當(dāng)SKIPIF1<0時，SKIPIF1<0的解集為SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0的解集為SKIPIF1<0，不滿足題意，舍去因為關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，滿足SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，不滿足SKIPIF1<0綜上可得:SKIPIF1<0的取值范圍是SKIPIF1<0故選:B.6．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【解析】作出函數(shù)SKIPIF1<0的圖象如圖，因為SKIPIF1<0，若SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0；綜上，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.所以實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選：A.二、多選題7．已知SKIPIF1<0是定義在SKIPIF1<0的奇函數(shù)，且SKIPIF1<0時，SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0時，SKIPIF1<0B．SKIPIF1<0有3個零點C．SKIPIF1<0增區(qū)間為SKIPIF1<0D．SKIPIF1<0的解集為SKIPIF1<0【解析】由SKIPIF1<0是定義在SKIPIF1<0的奇函數(shù)知SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，A錯誤；由上可知SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故B正確；由SKIPIF1<0的解析式知SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上均單調(diào)遞增，但在SKIPIF1<0上不具有單調(diào)性，如SKIPIF1<0，但SKIPIF1<0，故C錯誤；由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，解得