新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練8.9 冪函數(shù)（基礎(chǔ)版）（解析版）

8.9冪函數(shù)（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一冪函數(shù)的三要素【例1-1】（2022·四川?。﹥绾瘮?shù)y＝SKIPIF1<0（m∈Z）的圖象如圖所示，則實(shí)數(shù)m的值為________.【答案】1【解析】有圖象可知：該冪函數(shù)在SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0,故SKIPIF1<0可取SKIPIF1<0，又因?yàn)樵摵瘮?shù)為偶函數(shù)，所以SKIPIF1<0為偶數(shù)，故SKIPIF1<0故答案為：SKIPIF1<0【例1-2】（2022課時練習(xí)）（1）函數(shù)SKIPIF1<0的定義域是________，值域是________；（2）函數(shù)SKIPIF1<0的定義域是________，值域是________；（3）函數(shù)SKIPIF1<0的定義域是________，值域是________；（4）函數(shù)SKIPIF1<0的定義域是________，值域是________．【答案】（1）SKIPIF1<0

SKIPIF1<0

（2）SKIPIF1<0

SKIPIF1<0

（3）SKIPIF1<0

SKIPIF1<0

（4）SKIPIF1<0

SKIPIF1<0【解析】（1）冪函數(shù)SKIPIF1<0圖像如圖所示，定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0,（2）冪函數(shù)SKIPIF1<0圖像如圖所示，定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0,（3）冪函數(shù)SKIPIF1<0圖像如圖所示，定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0,（4）冪函數(shù)SKIPIF1<0圖像如圖所示，定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0,故答案為：（1）SKIPIF1<0;SKIPIF1<0,（2）SKIPIF1<0;SKIPIF1<0,（3）SKIPIF1<0;SKIPIF1<0,（4）SKIPIF1<0;SKIPIF1<0.【一隅三反】1．（2022·云南師大附中高三階段練習(xí)）已知SKIPIF1<0為冪函數(shù)，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0為冪函數(shù)，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0.故選：B2．（2022·全國·模擬預(yù)測（文））設(shè)SKIPIF1<0，則“函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0”是“函數(shù)SKIPIF1<0在SKIPIF1<0上遞減”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，而SKIPIF1<0在SKIPIF1<0上遞減，函數(shù)SKIPIF1<0的圖象不一定經(jīng)過點(diǎn)SKIPIF1<0，如：SKIPIF1<0.所以“函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0”是“函數(shù)SKIPIF1<0在SKIPIF1<0上遞減”的充分不必要條件.故選：A.3．（2022·河北·邢臺市第二中學(xué)高三階段練習(xí)）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0，若SKIPIF1<0的最小值為SKIPIF1<0，則實(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【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時，等號成立；即當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，要使得函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，則滿足SKIPIF1<0，解得SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：A.4．（2022·河北）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則實(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】D【解析】函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，其函數(shù)值集合為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0的取值集合為SKIPIF1<0，SKIPIF1<0的值域SKIPIF1<0，不符合題意，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，其函數(shù)值集合為SKIPIF1<0，因函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：D考點(diǎn)二冪函數(shù)的性質(zhì)【例2-1】（2022·黑龍江·雞東縣第二中學(xué)二模）當(dāng)SKIPIF1<0時，冪函數(shù)SKIPIF1<0為減函數(shù)，則實(shí)數(shù)m的值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)楹瘮?shù)SKIPIF1<0既是冪函數(shù)又是SKIPIF1<0的減函數(shù)，所以SKIPIF1<0解得：SKIPIF1<0.故選：A.【例2-2】（2022·廣西）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0．又因?yàn)橹笖?shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0．故選:D．【例2-3】（2022·云南）已知冪函數(shù)SKIPIF1<0的圖象關(guān)于y軸對稱，且在SKIPIF1<0上單調(diào)遞減，則滿足SKIPIF1<0的a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，解得SKIPIF1<0．又SKIPIF1<0，故m＝1或2．當(dāng)m＝1時，SKIPIF1<0的圖象關(guān)于y軸對稱，滿足題意；當(dāng)m＝2時，SKIPIF1<0的圖象不關(guān)于y軸對稱，舍去，故m＝1．不等式化為SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．故應(yīng)選：D．【一隅三反】1．（2022·黑龍江·嫩江市高級中學(xué)高三開學(xué)考試）下列關(guān)于幕函數(shù)SKIPIF1<0的命題中正確的有（

）A．冪函數(shù)圖象都通過點(diǎn)SKIPIF1<0B．當(dāng)冪指數(shù)SKIPIF1<0時，冪函數(shù)SKIPIF1<0的圖象都經(jīng)過第一、三象限C．當(dāng)冪指數(shù)SKIPIF1<0時，冪函數(shù)SKIPIF1<0是增函數(shù)D．若SKIPIF1<0，則函數(shù)圖象不通過點(diǎn)SKIPIF1<0【答案】B【解析】對于A，當(dāng)SKIPIF1<0時，冪函數(shù)圖象不通過點(diǎn)SKIPIF1<0，A錯誤；對于B，冪指數(shù)SKIPIF1<0時，冪函數(shù)分別為SKIPIF1<0，三者皆為奇函數(shù)，圖象都經(jīng)過第一、三象限，故B正確；對于C，當(dāng)SKIPIF1<0時，冪函數(shù)SKIPIF1<0在SKIPIF1<0上皆單調(diào)遞減，C錯誤；對于D，若SKIPIF1<0，則函數(shù)圖象不通過點(diǎn)SKIPIF1<0，通過SKIPIF1<0點(diǎn)，D錯誤，故選：B2．（2023·全國·高三專題練習(xí)）冪函數(shù)SKIPIF1<0在xSKIPIF1<0（0，+∞）上是減函數(shù)，則m＝（

）A．﹣1 B．2 C．﹣1或2 D．1【答案】A【解析】∵冪函數(shù)SKIPIF1<0，∴m2﹣m﹣1＝1，解得m＝2，或m＝﹣1；又xSKIPIF1<0（0，+∞）時f（x）為減函數(shù)，∴當(dāng)m＝2時，m2+m﹣3＝3，冪函數(shù)為y＝x3，不滿足題意；當(dāng)m＝﹣1時，m2+m﹣3＝﹣3，冪函數(shù)為SKIPIF1<0，滿足題意；綜上，SKIPIF1<0．故選：A．3．（2022·全國·高三專題練習(xí)）“冪函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)”是“函數(shù)SKIPIF1<0為奇函數(shù)”的（

）條件A．充分不必要 B．必要不充分C．充分必要 D．既不充分也不必要【答案】A【解析】要使函數(shù)SKIPIF1<0是冪函數(shù)，且在SKIPIF1<0上為增函數(shù)，則SKIPIF1<0，解得：SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0為奇函數(shù)，即充分性成立；“函數(shù)SKIPIF1<0為奇函數(shù)”，則SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，故必要性不成立，故選：A．4．（2022·全國課時練習(xí)）如圖所示是函數(shù)SKIPIF1<0(SKIPIF1<0且互質(zhì))的圖象，則（

）A．SKIPIF1<0是奇數(shù)且SKIPIF1<0 B．SKIPIF1<0是偶數(shù)，SKIPIF1<0是奇數(shù)，且SKIPIF1<0C．SKIPIF1<0是偶數(shù)，SKIPIF1<0是奇數(shù)，且SKIPIF1<0 D．SKIPIF1<0是偶數(shù)，且SKIPIF1<0【答案】C【解析】SKIPIF1<0函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對稱，故SKIPIF1<0為奇數(shù)，SKIPIF1<0為偶數(shù)，在第一象限內(nèi)，函數(shù)是凸函數(shù)，故SKIPIF1<0，故選：C.5．（2022·黑龍江·雙鴨山一中高三開學(xué)考試）已知SKIPIF1<0，則a，b，c大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0.所以SKIPIF1<0.因?yàn)镾KIPIF1<0.所以SKIPIF1<0.所以SKIPIF1<0.故選：A.考點(diǎn)三二次函數(shù)根的分布【例3-1】（2022·全國·高一課時練習(xí)）關(guān)于x的方程SKIPIF1<0恰有一根在區(qū)間SKIPIF1<0內(nèi)，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】方程SKIPIF1<0對應(yīng)的二次函數(shù)設(shè)為：SKIPIF1<0因?yàn)榉匠蘏KIPIF1<0恰有一根屬于SKIPIF1<0，則需要滿足：①SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0；②函數(shù)SKIPIF1<0剛好經(jīng)過點(diǎn)SKIPIF1<0或者SKIPIF1<0，另一個零點(diǎn)屬于SKIPIF1<0，把點(diǎn)SKIPIF1<0代入SKIPIF1<0，解得：SKIPIF1<0，此時方程為SKIPIF1<0，兩根為SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，不合題意，舍去把點(diǎn)SKIPIF1<0代入SKIPIF1<0，解得：SKIPIF1<0，此時方程為SKIPIF1<0，兩根為SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，故符合題意；③函數(shù)與x軸只有一個交點(diǎn)，橫坐標(biāo)屬于SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時，方程SKIPIF1<0的根為SKIPIF1<0，不合題意；若SKIPIF1<0，方程SKIPIF1<0的根為SKIPIF1<0，符合題意綜上：實(shí)數(shù)m的取值范圍為SKIPIF1<0故選：D【例3-2】（2022·湖北·華中師大一附中）關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個不相等的實(shí)數(shù)根SKIPIF1<0，且SKIPIF1<0，那么SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】當(dāng)SKIPIF1<0時，SKIPIF1<0即為SKIPIF1<0，不符合題意；故SKIPIF1<0，SKIPIF1<0即為SKIPIF1<0，令SKIPIF1<0，由于關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個不相等的實(shí)數(shù)根SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0與x軸有兩個交點(diǎn)，且分布在1的兩側(cè)，故SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，故選：D【一隅三反】1．（2022·江蘇）已知方程SKIPIF1<0有兩個不相等的實(shí)數(shù)根，且兩個實(shí)數(shù)根都大于2，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】令SKIPIF1<0由題可知：SKIPIF1<0則SKIPIF1<0，即SKIPIF1<0故選：C2．（2022·廣西·高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個不同的實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【解析】當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0是增函數(shù)，函數(shù)值集合是SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0是減函數(shù)，函數(shù)值集合是SKIPIF1<0，關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個不同的實(shí)根，即函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有兩個交點(diǎn)，在坐標(biāo)系內(nèi)作出直線SKIPIF1<0和函數(shù)SKIPIF1<0的圖象，如圖，觀察圖象知，當(dāng)SKIPIF1<0時，直線SKIPIF1<0和函數(shù)SKIPIF1<0的圖象有兩個交點(diǎn)，即方程SKIPIF1<0有兩個不同的實(shí)根，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：A3（2022·江蘇）方程SKIPIF1<0的兩根都大于SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是_____．【答案】SKIPIF1<0【解析】由題意，方程SKIPIF1<0的兩根都大于SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0.4．（2022貴州）方程SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有兩個不同的根，SKIPIF1<0的取值范圍為__．【答案】SKIPIF1<0【解析】令SKIPIF1<0，圖象恒過點(diǎn)SKIPIF1<0，SKIPIF1<0方程SKIPIF1<00在區(qū)間SKIPIF1<0內(nèi)有兩個不同的根，SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<08.9冪函數(shù)（精練）（基礎(chǔ)版）題組一題組一冪函數(shù)的三要素1．（2023·全國·高三專題練習(xí)）現(xiàn)有下列函數(shù)：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0；⑦SKIPIF1<0，其中冪函數(shù)的個數(shù)為（

）A．1 B．2 C．3 D．4【答案】B【解析】冪函數(shù)滿足SKIPIF1<0形式，故SKIPIF1<0，SKIPIF1<0滿足條件，共2個故選：B2．（2022·全國·高三專題練習(xí)）已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值等于（

）A．SKIPIF1<0 B．4 C．8 D．SKIPIF1<0【答案】D【解析】設(shè)冪函數(shù)SKIPIF1<0，冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0．故選：D.3．（2022福建）下列冪函數(shù)中，定義域?yàn)镽的冪函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】ASKIPIF1<0，則需要滿足SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故A不符合題意；BSKIPIF1<0，則需要滿足SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故B不符合題意；CSKIPIF1<0，則需要滿足SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故C不符合題意；DSKIPIF1<0，故函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故D正確；故選：D.4．（2022·全國·高一專題練習(xí)）已知函數(shù)SKIPIF1<0是冪函數(shù)，則SKIPIF1<0的值為_____．【答案】8【解析】依題意得，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0故答案為：85．（2022·上海）函數(shù)SKIPIF1<0的定義域?yàn)開_________．【答案】SKIPIF1<0【解析】函數(shù)解析式為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.因此，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.題組二題組二冪函數(shù)的性質(zhì)1．（2023·全國·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則a、b、c的大小關(guān)系為（）A．a(chǎn)＜b＜c B．c＜a＜b C．b＜a＜c D．c＜b＜a【答案】C【解析】函數(shù)SKIPIF1<0是定義域R上的單調(diào)減函數(shù)，且SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，于是得SKIPIF1<0，即SKIPIF1<0，所以a、b、c的大小關(guān)系為SKIPIF1<0．故選：C2．（2022·全國·高三專題練習(xí)）冪函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對稱，且在SKIPIF1<0上是增函數(shù)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，由冪函數(shù)性質(zhì)得，在SKIPIF1<0上是減函數(shù)；所以當(dāng)SKIPIF1<0時，SKIPIF1<0，由冪函數(shù)性質(zhì)得，在SKIPIF1<0上是常函數(shù)；所以當(dāng)SKIPIF1<0時，SKIPIF1<0，由冪函數(shù)性質(zhì)得，圖象關(guān)于y軸對稱,在SKIPIF1<0上是增函數(shù)；所以當(dāng)SKIPIF1<0時，SKIPIF1<0，由冪函數(shù)性質(zhì)得，圖象關(guān)于y軸對稱,在SKIPIF1<0上是增函數(shù)；故選：D.3．（2022·全國·高三專題練習(xí)）已知冪函數(shù)SKIPIF1<0為偶函數(shù)，則實(shí)數(shù)SKIPIF1<0的值為（

）A．3 B．2 C．1 D．1或2【答案】C【解析】SKIPIF1<0冪函數(shù)SKIPIF1<0為偶函數(shù)，SKIPIF1<0，且SKIPIF1<0為偶數(shù)，則實(shí)數(shù)SKIPIF1<0，故選：C4．（2021·新疆維吾爾自治區(qū)喀什第二中學(xué)高三階段練習(xí)）下列函數(shù)中，不是奇函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】對于A、D：由冪函數(shù)SKIPIF1<0定義域?yàn)镽，當(dāng)α為奇數(shù)，SKIPIF1<0是奇函數(shù).故A、D為奇函數(shù)；對于B：SKIPIF1<0為奇函數(shù)；對于C：SKIPIF1<0為偶函數(shù).故選：C5．（2021·全國·高三專題練習(xí)）已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍為(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意可知，SKIPIF1<0，解得，SKIPIF1<0，故SKIPIF1<0，易知，SKIPIF1<0為偶函數(shù)且在SKIPIF1<0上單調(diào)遞減，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得，SKIPIF1<0或SKIPIF1<0.故SKIPIF1<0的取值范圍為SKIPIF1<0.故選：C.6．（2022·黑龍江）已知SKIPIF1<0是冪函數(shù)，且在SKIPIF1<0上單調(diào)遞增，則滿足SKIPIF1<0的實(shí)數(shù)SKIPIF1<0的范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，易知SKIPIF1<0是偶函數(shù)，所以由SKIPIF1<0得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故選：D.7．（2022·河北·青龍滿族自治縣實(shí)驗(yàn)中學(xué)高三開學(xué)考試）“當(dāng)SKIPIF1<0時，冪函數(shù)SKIPIF1<0為減函數(shù)”是“SKIPIF1<0或2”的（

）條件A．既不充分也不必要 B．必要不充分C．充分不必要 D．充要【答案】C【解析】當(dāng)SKIPIF1<0時，冪函數(shù)SKIPIF1<0為減函數(shù)，所以有SKIPIF1<0，所以冪函數(shù)SKIPIF1<0為減函數(shù)”是“SKIPIF1<0或2”的充分不必要條件，故選：C8．（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0與SKIPIF1<0均單調(diào)遞減的一個充分不必要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】函數(shù)SKIPIF1<0單調(diào)遞減可得SKIPIF1<0及SKIPIF1<0；函數(shù)SKIPIF1<0單調(diào)遞減可得SKIPIF1<0，解得SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0均單調(diào)遞減，可得SKIPIF1<0，由題可得所求區(qū)間真包含于SKIPIF1<0，結(jié)合選項(xiàng)，函數(shù)SKIPIF1<0與SKIPIF1<0均單調(diào)遞減的一個充分不必要條件是C.故選：C.9．（2022·全國·模擬預(yù)測）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，e是自然對數(shù)的底數(shù)，則a，b，c的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，綜上，SKIPIF1<0，故選：B.10．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，若當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，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】C【解析】由題意，SKIPIF1<0，即SKIPIF1<0為奇函數(shù)，同時也為增函數(shù)，∵SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0恒成立，SKIPIF1<0，若不等式恒成立，只需SKIPIF1<0，令SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．故選：C11．（2023·全國·高三專題練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0在R上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，∵SKIPIF1<0，∴SKIPIF1<0．故選：D．12．（2022·遼寧·黑山縣黑山中學(xué)高三階段練習(xí)）下列命題中正確的是（

）A．當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的圖像是一條直線；B．冪函數(shù)的圖像都經(jīng)過SKIPIF1<0和SKIPIF1<0點(diǎn)；C．冪函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0；D．冪函數(shù)的圖像不可能出現(xiàn)在第四象限．【答案】D【解析】對于A，SKIPIF1<0時，函數(shù)SKIPIF1<0的圖像是一條直線除去SKIPIF1<0點(diǎn)，故SKIPIF1<0錯誤；對于B，冪函數(shù)的圖像都經(jīng)過SKIPIF1<0點(diǎn)，當(dāng)指數(shù)大于SKIPIF1<0時，都經(jīng)過SKIPIF1<0點(diǎn)，當(dāng)指數(shù)小于SKIPIF1<0時，不經(jīng)過SKIPIF1<0點(diǎn)，故B錯誤；對于C，函數(shù)SKIPIF1<0，故定義域?yàn)镾KIPIF1<0，故錯誤；對于D，由冪函數(shù)的性質(zhì)，冪函數(shù)的圖像一定過第一象限，不可能出現(xiàn)在第四象限，故正確.故選：D.13．（2022·全國·課時練習(xí)）（多選）下列結(jié)論中正確的是（

）A．冪函數(shù)的圖像都經(jīng)過點(diǎn)SKIPIF1<0，SKIPIF1<0B．冪函數(shù)的圖像不經(jīng)過第四象限C．當(dāng)指數(shù)SKIPIF1<0取1，3，SKIPIF1<0時，冪函數(shù)SKIPIF1<0是增函數(shù)D．當(dāng)SKIPIF1<0時，冪函數(shù)SKIPIF1<0在其整個定義域上是減函數(shù)【答案】BC【解析】A選項(xiàng)，當(dāng)指數(shù)SKIPIF1<0時，冪函數(shù)SKIPIF1<0的圖像不經(jīng)過原點(diǎn)，故A錯誤；B選項(xiàng)，所有的冪函數(shù)在區(qū)間SKIPIF1<0上都有定義且SKIPIF1<0，所以冪函數(shù)的圖像不可能經(jīng)過第四象限，故B正確；C選項(xiàng)，當(dāng)α為1，3，SKIPIF1<0時，SKIPIF1<0是增函數(shù)，顯然C正確；D選項(xiàng)，當(dāng)SKIPIF1<0時，SKIPIF1<0在區(qū)間SKIPIF1<0和SKIPIF1<0上是減函數(shù)，但在整個定義域上不是減函數(shù)，故D錯誤.故選：BC14．（2022·廣東）（多選）已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，則（

）A．函數(shù)SKIPIF1<0為增函數(shù) B．函數(shù)SKIPIF1<0為偶函數(shù)C．當(dāng)SKIPIF1<0時，SKIPIF1<0 D．當(dāng)SKIPIF1<0時，SKIPIF1<0【答案】ACD【解析】設(shè)冪函數(shù)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故A正確，因?yàn)镾KIPIF1<0的定義域不關(guān)于原點(diǎn)對稱，所以函數(shù)SKIPIF1<0不是偶函數(shù)，故B錯誤，當(dāng)SKIPIF1<0時，SKIPIF1<0，故C正確，當(dāng)SKIPIF1<0時，SKIPIF1<0SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，D正確．故選：ACD.15．（2022·遼寧營口）已知冪函數(shù)SKIPIF1<0的圖像經(jīng)過點(diǎn)SKIPIF1<0，則下列命題正確的有（

）A．函數(shù)SKIPIF1<0為非奇非偶函數(shù) B．函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0C．SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0 D．若SKIPIF1<0，則SKIPIF1<0【答案】AC【解析】設(shè)冪函數(shù)SKIPIF1<0，SKIPIF1<0為實(shí)數(shù)，其圖像經(jīng)過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0為非奇非偶函數(shù)，故A正確，B錯誤.且SKIPIF1<0在SKIPIF1<0上為增函數(shù)，故C正確.因?yàn)楹瘮?shù)SKIPIF1<0是凸函數(shù)，所以對定義域內(nèi)任意SKIPIF1<0，都有SKIPIF1<0成立，故D錯誤.故選:AC.16．（2022·全國·高三專題練習(xí)）（多選）已知函數(shù)SKIPIF1<0，則下列結(jié)論中錯誤的是（

）A．SKIPIF1<0的值域?yàn)镾KIPIF1<0 B．SKIPIF1<0的圖象與直線SKIPIF1<0有兩個交點(diǎn)C．SKIPIF1<0是單調(diào)函數(shù) D．SKIPIF1<0是偶函數(shù)【答案】ACD【解析】函數(shù)SKIPIF1<0的圖象如圖所示，由圖可知SKIPIF1<0的值域?yàn)镾KIPIF1<0，結(jié)論A錯誤，結(jié)論C，D顯然錯誤，SKIPIF1<0的圖象與直線SKIPIF1<0有兩個交點(diǎn)，結(jié)論B正確．故選：ACD17．（2022·廣西北海）已知冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，函數(shù)SKIPIF1<0，對任意SKIPIF1<0，總存在SKIPIF1<0使得SKIPIF1<0，則SKIPIF1<0的取值范圍為__________.【答案】SKIPIF1<0【解析】因?yàn)楹瘮?shù)SKIPIF1<0是冪函數(shù)，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，根據(jù)題意有SKIPIF1<0，SKIPIF1<0的范圍為SKIPIF1<0.故答案為：SKIPIF1<0.18．（2022·福建·泉州科技中學(xué)）已知冪函數(shù)SKIPIF1<0為奇函數(shù)，且在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0_______．【答案】SKIPIF1<0【解析】因?yàn)閮绾瘮?shù)SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0或1或3，又因?yàn)閮绾瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，故答案為：SKIPIF1<0.19．（2021·全國·模擬預(yù)測）寫出一個同時具有下列性質(zhì)①②③的函數(shù)：SKIPIF1<0______．①SKIPIF1<0為奇函數(shù)；②SKIPIF