新高考數(shù)學(xué)一輪復(fù)習(xí)講練測(cè)專題2.3二次函數(shù)與一元二次方程、不等式（練）解析版

專題2.3二次函數(shù)與一元二次方程、不等式練基礎(chǔ)練基礎(chǔ)1．（浙江高考真題）已知a，b，c∈R，函數(shù)f(x)＝ax2＋bx＋c.若f(0)＝f（4）>f（1），則（）A．a(chǎn)>0，4a＋b＝0 B．a(chǎn)<0，4a＋b＝0C．a(chǎn)>0，2a＋b＝0 D．a(chǎn)<0，2a＋b＝0【答案】A【解析】由已知得f(x)的圖象的對(duì)稱軸為x＝2且f(x)先減后增，可得選項(xiàng).【詳解】由f(0)＝f（4），得f(x)＝ax2＋bx＋c圖象的對(duì)稱軸為x＝－SKIPIF1<0＝2，∴4a＋b＝0，又f(0)>f（1），f（4）>f（1），∴f(x)先減后增，于是a>0，故選：A.2．（2021·全國(guó)高三專題練習(xí)（文））已知函數(shù)SKIPIF1<0，則錯(cuò)誤的是（）A．SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱 B．方程SKIPIF1<0的解的個(gè)數(shù)為2C．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．SKIPIF1<0的最小值為SKIPIF1<0【答案】B【解析】結(jié)合函數(shù)的奇偶性求出函數(shù)的對(duì)稱軸，判斷SKIPIF1<0，令SKIPIF1<0，求出方程的解的個(gè)數(shù)，判斷B，令SKIPIF1<0，SKIPIF1<0，從而判斷C，D即可．【詳解】SKIPIF1<0定義域?yàn)镾KIPIF1<0，顯然關(guān)于原點(diǎn)對(duì)稱，又SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，關(guān)于SKIPIF1<0軸對(duì)稱，故選項(xiàng)A正確.令SKIPIF1<0即SKIPIF1<0，解得：SKIPIF1<0，1，SKIPIF1<0，函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，故B錯(cuò)誤；令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，SKIPIF1<0都為遞增函數(shù)，故SKIPIF1<0在SKIPIF1<0遞增，故C正確；由SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0，故SKIPIF1<0的最小值是SKIPIF1<0，故D正確．故選：B．3．（2021·北京高三其他模擬）設(shè)SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（）A．充分而不必要條件 B．必要而不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】分別解出兩個(gè)不等式的解集，比較集合的關(guān)系，從而得到兩命題的邏輯關(guān)系.【詳解】SKIPIF1<0SKIPIF1<0；SKIPIF1<0SKIPIF1<0；易知集合SKIPIF1<0是SKIPIF1<0的真子集，故是充分不必要條件.故選：A.4．（2021·全國(guó)高三月考）已知函數(shù)SKIPIF1<0，則“SKIPIF1<0”是“方程SKIPIF1<0有兩個(gè)不同實(shí)數(shù)解且方程SKIPIF1<0恰有兩個(gè)不同實(shí)數(shù)解”的（）A．充要條件 B．充分不必要條件C．必要不充分條件 D．既不充分也不必要條件【答案】C【解析】根據(jù)二次函數(shù)的圖象與性質(zhì)，求得SKIPIF1<0，反之若SKIPIF1<0有兩個(gè)正根SKIPIF1<0，當(dāng)SKIPIF1<0，得到方程SKIPIF1<0恰有四個(gè)不同實(shí)數(shù)解，結(jié)合充分條件、必要條件的判定方法，即可求解.【詳解】由SKIPIF1<0表示開(kāi)口向下的拋物線，對(duì)稱軸的方程為SKIPIF1<0，要使得方程SKIPIF1<0有兩個(gè)不同實(shí)數(shù)，只需SKIPIF1<0，要使得方程SKIPIF1<0恰有兩個(gè)不同實(shí)數(shù)解，設(shè)兩解分別為SKIPIF1<0，且SKIPIF1<0，則滿足SKIPIF1<0，因?yàn)镾KIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以必要性成立；反之，設(shè)SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0有兩個(gè)正根，且滿足SKIPIF1<0，若SKIPIF1<0，此時(shí)方程SKIPIF1<0恰有四個(gè)不同實(shí)數(shù)解，所以充分性不成立.所以“SKIPIF1<0”是“方程SKIPIF1<0有兩個(gè)不同實(shí)數(shù)解且方程SKIPIF1<0恰有兩個(gè)不同實(shí)數(shù)解”的必要不充分條件.故選：C.5．（2021·全國(guó)高三專題練習(xí)）若當(dāng)x∈(1，2)時(shí)，函數(shù)y=(x-1)2的圖象始終在函數(shù)y=logax的圖象的下方，則實(shí)數(shù)a的取值范圍是___________.【答案】1<a≤2.【解析】在同一個(gè)坐標(biāo)系中畫(huà)出兩個(gè)函數(shù)的圖象，結(jié)合圖形，列出不等式組，求得結(jié)果.【詳解】如圖，在同一平面直角坐標(biāo)系中畫(huà)出函數(shù)y=(x-1)2和y=logax的圖象.由于當(dāng)x∈(1，2)時(shí)，函數(shù)y=(x-1)2的圖象恒在函數(shù)y=logax的圖象的下方，則SKIPIF1<0，解得1<a≤2.故答案為：1<a≤2.6.（2020·山東省微山縣第一中學(xué)高一月考）若不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是_________.【答案】SKIPIF1<0【解析】∵不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，∴函數(shù)SKIPIF1<0的圖象始終在SKIPIF1<0軸下方，∴SKIPIF1<0，解得SKIPIF1<0，故答案為：SKIPIF1<0．7.（2021·全國(guó)高三專題練習(xí)）已知當(dāng)SKIPIF1<0時(shí)，不等式9x－m·3x＋m＋1>0恒成立，則實(shí)數(shù)m的取值范圍是________．【答案】SKIPIF1<0【解析】先換元3x＝t，SKIPIF1<0，使f（t）＝t2－mt＋m＋1>0在SKIPIF1<0上恒成立，再利用二次函數(shù)圖象特征列限定條件，計(jì)算求得結(jié)果即可.【詳解】令3x＝t，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則f（t）＝t2－mt＋m＋1>0在SKIPIF1<0上恒成立，即函數(shù)在SKIPIF1<0的圖象在x軸的上方，而判別式SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.8．（2021·浙江高一期末）已知函數(shù)SKIPIF1<0，若任意SKIPIF1<0、SKIPIF1<0且SKIPIF1<0，都有SKIPIF1<0，則實(shí)數(shù)a的取值范圍是___________．【答案】SKIPIF1<0【解析】本題首先可令SKIPIF1<0，將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，然后令SKIPIF1<0，通過(guò)函數(shù)單調(diào)性的定義得出函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，最后分為SKIPIF1<0、SKIPIF1<0兩種情況進(jìn)行討論，結(jié)合二次函數(shù)性質(zhì)即可得出結(jié)果.【詳解】因?yàn)槿我釹KIPIF1<0、SKIPIF1<0且SKIPIF1<0，都有SKIPIF1<0，所以令SKIPIF1<0，SKIPIF1<0即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，若SKIPIF1<0，則SKIPIF1<0，顯然不成立；若SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，綜合所述，實(shí)數(shù)a的取值范圍是SKIPIF1<0，故答案為：SKIPIF1<0.9.（2021·四川成都市·高三三模（理））已知函數(shù)SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)_______．【答案】SKIPIF1<0【解析】由SKIPIF1<0得，SKIPIF1<0，把SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，利用二次函數(shù)求最值.【詳解】SKIPIF1<0的圖像如圖示：不妨令SKIPIF1<0，由圖像可知，SKIPIF1<0，SKIPIF1<0由SKIPIF1<0，由SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．故答案為：SKIPIF1<0.10．（2021·浙江高一期末）已知函數(shù)SKIPIF1<0．（Ⅰ）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，求實(shí)數(shù)SKIPIF1<0的取值范圍；（Ⅱ）SKIPIF1<0，SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】（Ⅰ）SKIPIF1<0；（Ⅱ）SKIPIF1<0【解析】（Ⅰ）由題意討論SKIPIF1<0，SKIPIF1<0與SKIPIF1<0三種情況，求出函數(shù)的對(duì)稱軸，結(jié)合區(qū)間，列不等式求解；（Ⅱ）利用參變分離法得SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，根據(jù)單調(diào)性，求解出最值，即可得SKIPIF1<0的取值范圍.【詳解】（Ⅰ）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在區(qū)間SKIPIF1<0上單調(diào)遞減，符合題意；當(dāng)SKIPIF1<0時(shí)，對(duì)稱軸為SKIPIF1<0，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，符合題意，綜上，SKIPIF1<0的取值范圍為SKIPIF1<0.（Ⅱ）SKIPIF1<0，SKIPIF1<0恒成立，即SKIPIF1<0，SKIPIF1<0恒成立，令SKIPIF1<0，可知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0的取值范圍為SKIPIF1<0練提升TIDHNEG練提升TIDHNEG1．（2020·山東省高三二模）已知函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則實(shí)數(shù)m的范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0，（1）SKIPIF1<0，SKIPIF1<0恒成立等價(jià)于SKIPIF1<0或SKIPIF1<0恒成立，即SKIPIF1<0或SKIPIF1<0（不合題意，舍去）恒成立；即SKIPIF1<0，解得SKIPIF1<0，（2）SKIPIF1<0恒成立，符合題意；（3）SKIPIF1<0，SKIPIF1<0恒成立等價(jià)于SKIPIF1<0（不合題意，舍去）或SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0，解得SKIPIF1<0.綜上所述，SKIPIF1<0，故選：A.2．（2021·浙江高三二模）已知SKIPIF1<0，對(duì)任意的SKIPIF1<0，SKIPIF1<0．方程SKIPIF1<0在SKIPIF1<0上有解，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】對(duì)任意的SKIPIF1<0，SKIPIF1<0．方程SKIPIF1<0在SKIPIF1<0上有解，不妨取取SKIPIF1<0，SKIPIF1<0，方程有解SKIPIF1<0只能取4，則排除其他答案.【詳解】SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.要對(duì)任意的SKIPIF1<0，SKIPIF1<0．方程SKIPIF1<0在SKIPIF1<0上都有解,取SKIPIF1<0，SKIPIF1<0，此時(shí)，任意SKIPIF1<0，都有SKIPIF1<0，其他SKIPIF1<0的取值，方程均無(wú)解，則SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D.3.（2020·浙江省高三二模）已知函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)三個(gè)象限，則實(shí)數(shù)a的取值范圍是________.【答案】SKIPIF1<0或SKIPIF1<0.【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)圖象經(jīng)過(guò)第三象限，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)圖象恒經(jīng)過(guò)第一象限，當(dāng)SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0時(shí)，函數(shù)圖像經(jīng)過(guò)第一、四象限，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)圖象恒經(jīng)過(guò)第一象限，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，函數(shù)圖像經(jīng)過(guò)第一、四象限，綜上所述：SKIPIF1<0或SKIPIF1<0.4．（2020·陜西省西安中學(xué)高三其他（理））記SKIPIF1<0函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是_________.【答案】SKIPIF1<0【解析】令SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即1是函數(shù)SKIPIF1<0的零點(diǎn)，因?yàn)楹瘮?shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，所以根據(jù)題意，若函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)，則二次函數(shù)SKIPIF1<0沒(méi)有零點(diǎn)，SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<05．（2021·浙江高三專題練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的最大值是___________.【答案】SKIPIF1<0【解析】根據(jù)函數(shù)SKIPIF1<0，分SKIPIF1<0，SKIPIF1<0和SKIPIF1<0三種情況討論，分別求得其最大值，即可求解.【詳解】由題意，函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)镾KIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)镾KIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0知，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，綜上可得，SKIPIF1<0的最大值是SKIPIF1<0.故答案為：SKIPIF1<06.（2021·浙江高三期末）已知函數(shù)SKIPIF1<0，若對(duì)于任意SKIPIF1<0，均有SKIPIF1<0，則SKIPIF1<0的最大值是___________.【答案】SKIPIF1<0【解析】首先討論SKIPIF1<0、SKIPIF1<0時(shí)SKIPIF1<0的最值情況，由不等式恒成立求SKIPIF1<0的范圍，再討論SKIPIF1<0并結(jié)合SKIPIF1<0的單調(diào)情況求SKIPIF1<0的范圍，最后取它們的并集即可知SKIPIF1<0的最大值.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有SKIPIF1<0；SKIPIF1<0有SKIPIF1<0；由SKIPIF1<0有SKIPIF1<0，有SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有SKIPIF1<0；SKIPIF1<0有SKIPIF1<0；由SKIPIF1<0有SKIPIF1<0，有SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0：SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞減，SKIPIF1<0上遞增；SKIPIF1<0：SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增；SKIPIF1<0：SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增，SKIPIF1<0上遞增；∴綜上，SKIPIF1<0在SKIPIF1<0上先減后增，則SKIPIF1<0，可得SKIPIF1<0∴SKIPIF1<0恒成立，即SKIPIF1<0的最大值是-1.故答案為：SKIPIF1<0.7．（2020·武漢外國(guó)語(yǔ)學(xué)校（武漢實(shí)驗(yàn)外國(guó)語(yǔ)學(xué)校）高一期中）已知函數(shù)SKIPIF1<0，且SKIPIF1<0的解集為SKIPIF1<0．（1）求SKIPIF1<0的解析式；（2）設(shè)SKIPIF1<0，在定義域范圍內(nèi)若對(duì)于任意的SKIPIF1<0，使得SKIPIF1<0恒成立，求M的最小值．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0．【解析】（1）代入方程的根，求得參數(shù)值.

（2）使不等式恒成立，根據(jù)函數(shù)單調(diào)性求得函數(shù)的最值，從而求得參數(shù)的值.【詳解】解：（1）由題意SKIPIF1<0

解得SKIPIF1<0SKIPIF1<0（2）由題意SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0當(dāng)SKIPIF1<0令SKIPIF1<0，當(dāng)SKIPIF1<0，當(dāng)SKIPIF1<0取等號(hào)，當(dāng)SKIPIF1<0當(dāng)SKIPIF1<0取等號(hào)，SKIPIF1<0SKIPIF1<0綜上，SKIPIF1<0SKIPIF1<0SKIPIF1<08．（2021·浙江高一期末）設(shè)函數(shù)SKIPIF1<0．（1）若SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0，求SKIPIF1<0的取值范圍；（2）若SKIPIF1<0在區(qū)間SKIPIF1<0上有零點(diǎn)，求SKIPIF1<0的最小值．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0.【解析】（1）對(duì)實(shí)數(shù)SKIPIF1<0的取值進(jìn)行分類討論，分析函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性，求得SKIPIF1<0，再由SKIPIF1<0可求得實(shí)數(shù)SKIPIF1<0的取值范圍；（2）設(shè)函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0、SKIPIF1<0，由韋達(dá)定理化簡(jiǎn)SKIPIF1<0，設(shè)SKIPIF1<0，由SKIPIF1<0結(jié)合不等式的基本性質(zhì)求出SKIPIF1<0的最小值，即為所求.【詳解】（1）二次函數(shù)SKIPIF1<0的圖象開(kāi)口向上，對(duì)稱軸為直線SKIPIF1<0.①當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0.綜上所述，SKIPIF1<0.所以，當(dāng)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0；（2）設(shè)函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)為SKIPIF1<0、SKIPIF1<0，由韋達(dá)定理可得SKIPIF1<0，所以，SKIPIF1<0SKIPIF1<0，設(shè)SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以，SKIPIF1<0.此時(shí)，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0.所以，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0取最小值SKIPIF1<0.9．（2020·全國(guó)高一單元測(cè)試）已知函數(shù)f（x）=9x﹣aSKIPIF1<03x+1+a2（x∈[0，1]，a∈R），記f（x）的最大值為g（a）．（Ⅰ）求g（a）解析式；（Ⅱ）若對(duì)于任意t∈[﹣2，2]，任意a∈R，不等式g（a）≥﹣m2+tm恒成立，求實(shí)數(shù)m的范圍．【答案】（Ⅰ）g（a）=SKIPIF1<0；（Ⅱ）m≤﹣SKIPIF1<0或m≥SKIPIF1<0．【解析】（Ⅰ）令u=3x∈[1，3]，得到f（x）=h（u）=u2﹣3au+a2，分類討論即可求出，（Ⅱ）先求出g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0，再根據(jù)題意可得﹣m2+tm≤﹣SKIPIF1<0，利用函數(shù)的單調(diào)性即可求出．【詳解】解：（Ⅰ）令u=3x∈[1，3]，則f（x）=h（u）=u2﹣3au+a2．當(dāng)SKIPIF1<0≤2，即a≤SKIPIF1<0時(shí)，g（a）=h（u）min=h（3）=a2﹣9a+9；當(dāng)SKIPIF1<0，即a＞SKIPIF1<0時(shí)，g（a）=h（u）min=h（1）=a2﹣3a+1；故g（a）=SKIPIF1<0；（Ⅱ）當(dāng)a≤SKIPIF1<0時(shí)，g（a）=a2﹣9a+9，g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0；當(dāng)aSKIPIF1<0時(shí)，g（a）=a2﹣3a+1，g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0；因此g（a）min=g（SKIPIF1<0）=﹣SKIPIF1<0；對(duì)于任意任意a∈R，不等式g（a）≥﹣m2+tm恒成立等價(jià)于﹣m2+tm≤﹣SKIPIF1<0．令h（t）=mt﹣m2，由于h（t）是關(guān)于t的一次函數(shù)，故對(duì)于任意t∈[﹣2，2]都有h（t）≤﹣SKIPIF1<0等價(jià)于SKIPIF1<0，即SKIPIF1<0，解得m≤﹣SKIPIF1<0或m≥SKIPIF1<0．10．（2021·全國(guó)高一課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0，在區(qū)間SKIPIF1<0上有最大值16，最小值SKIPIF1<0.設(shè)SKIPIF1<0．（1）求SKIPIF1<0的解析式；（2）若不等式SKIPIF1<0在SKIPIF1<0上恒成立，求實(shí)數(shù)k的取值范圍；【答案】（1）SKIPIF1<0SKIPIF1<0；（2）SKIPIF1<0．【解析】（1）由二次函數(shù)的性質(zhì)知SKIPIF1<0在SKIPIF1<0上為減函數(shù)，在SKIPIF1<0上為增函數(shù)，結(jié)合其區(qū)間的最值，列方程組求SKIPIF1<0，即可寫(xiě)出SKIPIF1<0解析式；（2）由題設(shè)得SKIPIF1<0在SKIPIF1<0上恒成立，即k只需小于等于右邊函數(shù)式的最小值即可.【詳解】（1）∵SKIPIF1<0(SKIPIF1<0)，即SKIPIF1<0在SKIPIF1<0上為減函數(shù)，在SKIPIF1<0上為增函數(shù)．又在SKIPIF1<0上有最大值16，最小值0，∴SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0；（2）∵SKIPIF1<0∴SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上為減函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最小值為1，∴SKIPIF1<0，即SKIPIF1<0.練真題TIDHNEG練真題TIDHNEG1．（浙江省高考真題）若函數(shù)SKIPIF1<0在區(qū)間[0,1]上的最大值是M,最小值是m,則SKIPIF1<0的值（）A．與a有關(guān)，且與b有關(guān) B．與a有關(guān)，但與b無(wú)關(guān)C．與a無(wú)關(guān)，且與b無(wú)關(guān) D．與a無(wú)關(guān)，但與b有關(guān)【答案】B【解析】因?yàn)樽钪翟赟KIPIF1<0中取，所以最值之差一定與SKIPIF1<0無(wú)關(guān)，選B．2.（2018·浙江高考真題）已知λ∈R，函數(shù)f(x)=x?4,x≥λx2?4x+3,x<λ，當(dāng)λ=2時(shí)，不等式f(x)<0的解集是___________．若函數(shù)f(x【答案】(1,4)(1,3]∪(4,+∞)【解析】由題意得x≥2x?4<0或x<2x2?4x+3<0，所以2≤x<4或1<x<2，即1<x<4，不等式f當(dāng)λ>4時(shí)，f(x)=x?4>0，此時(shí)f(x)=x2?4x+3=0,x=1,3，即在(?∞,λ)上有兩個(gè)零點(diǎn)；當(dāng)λ≤4時(shí)，f(x)=x?4=0,x=4，由f(x)=x2?4x+3在(?∞,λ)上只能有一個(gè)零點(diǎn)得3.（北京高考真題）已知SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0的取值范圍是_____.【答案】SKIPIF1<0【解析】試題分析：SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，取最大值1；當(dāng)SKIPIF1<0時(shí)，取最小值SKIPIF1<0.因此SKIPIF1<0的取值范圍為SKIPIF1<0.4.（2018·天津高考真題（理））已知SKIPIF1<0，函數(shù)SKIPIF1<0若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有2個(gè)互異的實(shí)數(shù)解，則SKIPIF1<0的取值范圍是______________.【答案】SKIPIF1<0【解析】分析：由題意分類討論SKIPIF1<0和SKIPIF1<0兩種情況，然后繪制函數(shù)圖像，數(shù)形結(jié)合即可求得最終結(jié)果.詳解：分類討論：當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0即SKIPIF1<0，整理可得：SKIPIF1<0，很明顯SKIPIF1<0不是方程的實(shí)數(shù)解，則SKIPIF1<0，當(dāng)SKIPIF1<0