新高考數(shù)學(xué)一輪復(fù)習(xí)提升訓(xùn)練3.5 冪函數(shù)與一元二次函數(shù)（精講）（解析版）

3.5冪函數(shù)與一元二次函數(shù)（精講）（提升版）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)及性質(zhì)【例1-1】（2022·全國(guó)·高三專(zhuān)題練習(xí)）冪函數(shù)SKIPIF1<0是偶函數(shù)，且在（0，+∞）上是減函數(shù)，則m的值為（

）A．﹣6 B．1 C．6 D．1或﹣6【答案】B【解析】∵冪函數(shù)SKIPIF1<0是偶函數(shù)，且在（0，+∞）上是減函數(shù)，∴SKIPIF1<0，且SKIPIF1<0為偶數(shù)SKIPIF1<0或SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0滿(mǎn)足條件；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，舍去因此：m＝1故選：B【例1-2】（2022·全國(guó)·高三專(zhuān)題練習(xí)）冪函數(shù)SKIPIF1<0是偶函數(shù)，在SKIPIF1<0上是減函數(shù)，則整數(shù)SKIPIF1<0的值為（

）A．0 B．1 C．0或1 D．2【答案】A【解析】因?yàn)閮绾瘮?shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，滿(mǎn)足題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0定義域?yàn)镾KIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，不滿(mǎn)足題意，舍去；綜上，SKIPIF1<0.故選：A【一隅三反】1．（2022·全國(guó)·高三專(zhuān)題練習(xí)）（多選）已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0，則下列說(shuō)法正確的有（

）A．函數(shù)是偶函數(shù) B．函數(shù)是增函數(shù)C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0【答案】BCD【解析】因?yàn)閮绾瘮?shù)SKIPIF1<0的圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，其定義域?yàn)镾KIPIF1<0，不關(guān)于原點(diǎn)對(duì)稱(chēng)，所以該函數(shù)是非奇非偶函數(shù)，故A錯(cuò)；又SKIPIF1<0，所以SKIPIF1<0是增函數(shù)，故B正確；因此當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故C正確；當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，故D正確.故選：BCD.2．（2022·全國(guó)·高三專(zhuān)題練習(xí)）（多選）已知函數(shù)SKIPIF1<0是冪函數(shù)，對(duì)任意SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，滿(mǎn)足SKIPIF1<0.若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0的值為負(fù)值，則下列結(jié)論可能成立的有（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】BC【解析】由于函數(shù)SKIPIF1<0為冪函數(shù)，故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.由于“對(duì)任意SKIPIF1<0，且SKIPIF1<0，滿(mǎn)足SKIPIF1<0”知，函數(shù)在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0.易見(jiàn)SKIPIF1<0，故函數(shù)SKIPIF1<0是單調(diào)遞增的奇函數(shù).由于SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，此時(shí)，若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0知，SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.綜上可知，SKIPIF1<0，且SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.故選：BC.3．（2022·全國(guó)·高三專(zhuān)題練習(xí)（理））已知冪函數(shù)SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，與SKIPIF1<0軸及SKIPIF1<0軸均無(wú)交點(diǎn)，則由SKIPIF1<0的值構(gòu)成的集合是__________．【答案】SKIPIF1<0【解析】由冪函數(shù)SKIPIF1<0與SKIPIF1<0軸及SKIPIF1<0軸均無(wú)交點(diǎn)，得SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，即函數(shù)為偶函數(shù)，故SKIPIF1<0為偶數(shù)，所以SKIPIF1<0，故答案為：SKIPIF1<0.4．（2022·上海·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0為冪函數(shù)，且為奇函數(shù)，則實(shí)數(shù)a的值_____．【答案】1【解析】因?yàn)楹瘮?shù)SKIPIF1<0為冪函數(shù)，所以SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為偶函數(shù)，不符合題意，所以舍去；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為奇函數(shù)，符合題意.故答案為：1考點(diǎn)二一元二次函數(shù)【例2-1】（2021·重慶市清華中學(xué)校高三階段練習(xí)）若函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0，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】C【解析】函數(shù)SKIPIF1<0的圖象如圖所示，因?yàn)镾KIPIF1<0當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)的定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0．故選：C．【例2-2】（2022·寧夏·平羅中學(xué)模擬預(yù)測(cè)（理））已知SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)的概率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題設(shè)SKIPIF1<0對(duì)稱(chēng)軸為SKIPIF1<0，而SKIPIF1<0，函數(shù)開(kāi)口向上，所以SKIPIF1<0的增區(qū)間為SKIPIF1<0，故在SKIPIF1<0上是增函數(shù)有SKIPIF1<0，綜上，SKIPIF1<0對(duì)應(yīng)可行域如下陰影部分：所以陰影部分面積為SKIPIF1<0，而SKIPIF1<0的面積為1，故在SKIPIF1<0上是增函數(shù)的概率為SKIPIF1<0.故選：D【例2-3】（2022·全國(guó)·高三專(zhuān)題練習(xí)）（多選）若函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0，則正整數(shù)a的值可能是（

）A．2 B．3 C．4 D．5【答案】BC【解析】函數(shù)SKIPIF1<0的圖象如圖所示：因?yàn)楹瘮?shù)在SKIPIF1<0上的值域?yàn)镾KIPIF1<0，結(jié)合圖象可得SKIPIF1<0，結(jié)合a是正整數(shù)，所以BC正確.故選：BC.【一隅三反】1．（2022·全國(guó)·高三專(zhuān)題練習(xí)）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列，則二次函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸的交點(diǎn)個(gè)數(shù)為（

）A．0 B．1 C．2 D．1或2【答案】D【解析】由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列，可得SKIPIF1<0，所以SKIPIF1<0，所以二次函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸交點(diǎn)的個(gè)數(shù)為1或2.故選：D.2．（2022·天津·南開(kāi)中學(xué)二模）已知函數(shù)SKIPIF1<0是R上的單調(diào)函數(shù)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】當(dāng)函數(shù)SKIPIF1<0是R上的單調(diào)遞減函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，因?yàn)镾KIPIF1<0且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0不可能是增函數(shù)，所以函數(shù)SKIPIF1<0在R上不可能是增函數(shù)，綜上：實(shí)數(shù)a的取值范圍為SKIPIF1<0，故選：B3（2022·重慶·模擬預(yù)測(cè)）已知二次函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)都在區(qū)間SKIPIF1<0內(nèi)，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】二次函數(shù)SKIPIF1<0，對(duì)稱(chēng)軸為SKIPIF1<0，開(kāi)口向上，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，要使二次函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)都在區(qū)間SKIPIF1<0內(nèi)，需SKIPIF1<0，解得SKIPIF1<0故實(shí)數(shù)a的取值范圍是SKIPIF1<0故選：C4．（2022·全國(guó)·高三專(zhuān)題練習(xí)（理））若集合SKIPIF1<0中有且只有一個(gè)元素，則正實(shí)數(shù)SKIPIF1<0的取值范圍是___________【答案】SKIPIF1<0【解析】由題意，不等式SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0是一個(gè)二次函數(shù)，圖象是確定的一條拋物線(xiàn)，而SKIPIF1<0一次函數(shù)，圖象是過(guò)一定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)，作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，如圖所示，其中SKIPIF1<0，又因?yàn)镾KIPIF1<0，結(jié)合圖象，要使得集合SKIPIF1<0中有且只有一個(gè)元素，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.即正實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.考點(diǎn)三一元二次函數(shù)與其他知識(shí)綜合【例3】（2022·山東濟(jì)寧·三模）已知二次函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】若SKIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，不合乎題意，因?yàn)槎魏瘮?shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故選：B.【一隅三反】1．（2021·廣東·湛江二十一中）若函數(shù)SKIPIF1<0有最大值，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】令SKIPIF1<0，要使函數(shù)SKIPIF1<0有最大值，則內(nèi)層函數(shù)SKIPIF1<0要有最小正值，且外層函數(shù)SKIPIF1<0為減函數(shù)，可知0＜a＜1．要使內(nèi)層函數(shù)SKIPIF1<0要有最小正值，則SKIPIF1<0，解得SKIPIF1<0．綜合得a的取值范圍為SKIPIF1<0．故選：B.2．（2022·黑龍江）若關(guān)于SKIPIF1<0的方程SKIPIF1<0有解，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【解析】方程SKIPIF1<0有解，SKIPIF1<0有解，令SKIPIF1<0，則可化為SKIPIF1<0有正根，則SKIPIF1<0在SKIPIF1<0有解，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以SKIPIF1<0，故選：SKIPIF1<0．3．（2022·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，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】A【解析】因?yàn)楹瘮?shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，可得真數(shù)部分SKIPIF1<0取到所有的正數(shù)，即函數(shù)SKIPIF1<0取到所有的正數(shù)，所以SKIPIF1<0是函數(shù)SKIPIF1<0的值域的子集，所以SKIPIF1<0解得：SKIPIF1<0或SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是：SKIPIF1<0.故選：A.考點(diǎn)四圖像問(wèn)題【例4-1】（2022·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）與函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）在同一個(gè)坐標(biāo)系內(nèi)的圖象可能是（

）A． B．C． D．【答案】C【解析】?jī)蓚€(gè)函數(shù)分別為指數(shù)函數(shù)和二次函數(shù)，其中二次函數(shù)圖象過(guò)點(diǎn)（0，－1），故排除A，D；二次函數(shù)圖象的對(duì)稱(chēng)軸為直線(xiàn)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，指數(shù)函數(shù)遞減，SKIPIF1<0，C符合題意；當(dāng)SKIPIF1<0時(shí)，指數(shù)函數(shù)遞增，SKIPIF1<0，B不符合題意．故選：C．【例4-2】（陜西省部分地市學(xué)校2022屆高三下學(xué)期高考全真模擬考試?yán)砜茢?shù)學(xué)試題）函數(shù)SKIPIF1<0的圖象大致是（

）A． B．C． D．【答案】C【解析】由題意，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱(chēng)，且滿(mǎn)足SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，其圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，排除B選項(xiàng)；當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；排除A選項(xiàng)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以排除D選項(xiàng)，選項(xiàng)C符合.故選：C.【一隅三反】1．（2021·山東·新泰市第一中學(xué)高三階段練習(xí)）若不等式SKIPIF1<0的解集為SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象可以為（

）A． B．C． D．【答案】C【解析】由題可得SKIPIF1<0和SKIPIF1<0是方程SKIPIF1<0的兩個(gè)根，且SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，則函數(shù)圖象開(kāi)口向下，與SKIPIF1<