新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)專(zhuān)題01 不等式綜合問(wèn)題（練）（解析版）

第一篇熱點(diǎn)、難點(diǎn)突破篇專(zhuān)題01不等式綜合問(wèn)題（練）【對(duì)點(diǎn)演練】一、單選題1．（2022·遼寧·朝陽(yáng)市第一高級(jí)中學(xué)高三階段練習(xí)）已知命題SKIPIF1<0：SKIPIF1<0，SKIPIF1<0是假命題，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】D【分析】根據(jù)一元二次不等式恒成立求解實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】由題意得SKIPIF1<0是真命題，即SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0符合題意；當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，且SKIPIF1<0，解得SKIPIF1<0.綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選:D.2．（2022·全國(guó)·高三專(zhuān)題練習(xí)）不等式SKIPIF1<0的解集為SKIPIF1<0,則函數(shù)SKIPIF1<0的圖像大致為（

）A． B．C． D．【答案】C【分析】根據(jù)題意，可得方程SKIPIF1<0的兩個(gè)根為SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0，結(jié)合二次方程根與系數(shù)的關(guān)系得到SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的關(guān)系，再結(jié)合二次函數(shù)的性質(zhì)判斷即可．【詳解】根據(jù)題意，SKIPIF1<0的解集為SKIPIF1<0，則方程SKIPIF1<0的兩個(gè)根為SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0.則有SKIPIF1<0，變形可得SKIPIF1<0，故函數(shù)SKIPIF1<0是開(kāi)口向下的二次函數(shù)，且與SKIPIF1<0軸的交點(diǎn)坐標(biāo)為SKIPIF1<0和SKIPIF1<0.對(duì)照四個(gè)選項(xiàng)，只有C符合.故選：C．3．（2022·重慶市云陽(yáng)縣高陽(yáng)中學(xué)高三階段練習(xí)（理））關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立的一個(gè)充分不必要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)二次不等式恒成立得SKIPIF1<0，再根據(jù)充分不必要條件的概念求解即可.【詳解】解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，該不等式成立；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，該不等式成立；綜上，得當(dāng)SKIPIF1<0時(shí)，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，所以，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立的一個(gè)充分不必要條件是SKIPIF1<0．故選：D．4．（2022·寧夏·銀川一中高三階段練習(xí)（理））《忠經(jīng)·廣至理章第十二》中有言“不私，而天下自公”，在實(shí)際生活中，新時(shí)代的青年不僅要有自己“不私”的覺(jué)悟，也要有識(shí)破“詐公”的智慧.某金店用一桿不準(zhǔn)確的天平（兩邊臂不等長(zhǎng)）稱(chēng)黃金，顧客要購(gòu)買(mǎi)SKIPIF1<0黃金，售貨員先將SKIPIF1<0的砝碼放在左盤(pán)，將黃金放于右盤(pán)使之平衡后給顧客；然后又將SKIPIF1<0的砝碼放入右盤(pán)，將另一黃金放于左盤(pán)使之平衡后又給顧客，則顧客實(shí)際所得黃金（

）A．大于SKIPIF1<0 B．小于SKIPIF1<0 C．等于SKIPIF1<0 D．以上都有可能【答案】A【分析】根據(jù)已知條件，結(jié)合基本不等式的公式，即可求解．【詳解】由于天平兩臂不相等，故可設(shè)天平左臂長(zhǎng)為SKIPIF1<0，右臂長(zhǎng)為SKIPIF1<0（不妨設(shè)SKIPIF1<0，第一次稱(chēng)出的黃金重為SKIPIF1<0，第二次稱(chēng)出的黃金重為SKIPIF1<0，由杠桿平衡定理可得，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故顧客實(shí)際所得黃金大于SKIPIF1<0．故選：SKIPIF1<0．5．（2022·湖北·高三階段練習(xí)）已知隨機(jī)變量SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．9 B．8 C．SKIPIF1<0 D．6【答案】B【分析】由正態(tài)曲線(xiàn)的對(duì)稱(chēng)軸得出SKIPIF1<0，再由基本不等式得出最小值.【詳解】由隨機(jī)變量SKIPIF1<0，則正態(tài)分布的曲線(xiàn)的對(duì)稱(chēng)軸為SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，故最小值為SKIPIF1<0.故選：B二、多選題6．（2020·山東·青島二中高三期中）設(shè)SKIPIF1<0，SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【分析】結(jié)合已知條件，可得到SKIPIF1<0，對(duì)于選項(xiàng)A：對(duì)SKIPIF1<0兩邊同時(shí)平方，并利用不等式性質(zhì)即可判斷；對(duì)于B：利用不等式性質(zhì)即可判斷；對(duì)于CD：結(jié)合均值不等式即可判斷.【詳解】由SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，對(duì)于A：由SKIPIF1<0兩邊平方并整理得，SKIPIF1<0，故A錯(cuò)誤；對(duì)于B：SKIPIF1<0，故B正確；對(duì)于C：由選項(xiàng)B知，SKIPIF1<0，又SKIPIF1<0，故C正確；對(duì)于D：因?yàn)镾KIPIF1<0，又SKIPIF1<0，SKIPIF1<0，故D正確.故選：BCD.7．（2022·江蘇江蘇·高三階段練習(xí)）已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【分析】由題可得SKIPIF1<0，進(jìn)而可得SKIPIF1<0可判斷A，根據(jù)特值可判斷B，根據(jù)基本不等式可判斷C，利用二次函數(shù)的性質(zhì)可判斷D.【詳解】由SKIPIF1<0，可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故A正確；取SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，故B錯(cuò)誤；因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0，故C正確；由SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故D正確.故選：ACD.8．（2022·河北·開(kāi)灤第一中學(xué)高三階段練習(xí)）若SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，其中SKIPIF1<0，SKIPIF1<0是整數(shù)，則SKIPIF1<0的可能取值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【分析】對(duì)SKIPIF1<0分類(lèi)討論，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，由一次函數(shù)的圖象知不存在；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，利用數(shù)形結(jié)合的思想可得出SKIPIF1<0的整數(shù)解.【詳解】當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，此時(shí)SKIPIF1<0不存在；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，可設(shè)SKIPIF1<0，SKIPIF1<0，作出SKIPIF1<0的圖象如下，由題意可知SKIPIF1<0，再由SKIPIF1<0，SKIPIF1<0是整數(shù)可得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0所以SKIPIF1<0的可能取值為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0故選：BCD三、填空題9．（2022·廣西南寧·模擬預(yù)測(cè)（文））若直線(xiàn)SKIPIF1<0平分圓SKIPIF1<0的周長(zhǎng)，則ab的最大值為_(kāi)_______【答案】SKIPIF1<0【分析】因?yàn)橹本€(xiàn)平分圓，則直線(xiàn)過(guò)圓心，再利用基本不等式求出ab的最大值.【詳解】由題意得，直線(xiàn)SKIPIF1<0過(guò)圓心SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，（當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，取“=”），又SKIPIF1<0，所以ab的最大值為SKIPIF1<0.故答案為：SKIPIF1<0.10．（2022·河南安陽(yáng)·高三階段練習(xí)（文））已知點(diǎn)P（m，n）是函數(shù)SKIPIF1<0圖象上的點(diǎn)，當(dāng)SKIPIF1<0時(shí)，2m＋n的最小值為_(kāi)_____．【答案】SKIPIF1<0【分析】根據(jù)基本不等式即可求解最小值.【詳解】P（m，n）是函數(shù)SKIPIF1<0圖象上的點(diǎn)，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0【沖刺提升】一、單選題1．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知m，n，s，t為正數(shù)，SKIPIF1<0，SKIPIF1<0，其中m，n是常數(shù)，且SKIPIF1<0的最小值是SKIPIF1<0，點(diǎn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的一條弦AB的中點(diǎn)，則弦AB所在直線(xiàn)方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由已知得SKIPIF1<0，化簡(jiǎn)后利用基本不等式可求出其最小值，再結(jié)合其最小值為SKIPIF1<0和SKIPIF1<0可求出SKIPIF1<0，從而可得點(diǎn)SKIPIF1<0的坐標(biāo)，再利用點(diǎn)差法可求出直線(xiàn)AB的斜率，從而可求出直線(xiàn)方程.【詳解】因?yàn)閙，n，s，t為正數(shù)，SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0，又SKIPIF1<0，又SKIPIF1<0為正數(shù)，所以解得SKIPIF1<0，即SKIPIF1<0，設(shè)弦兩端點(diǎn)分別為SKIPIF1<0，則SKIPIF1<0，兩式相減得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以直線(xiàn)的斜率為SKIPIF1<0，所以直線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0.經(jīng)檢驗(yàn)直線(xiàn)SKIPIF1<0與橢圓SKIPIF1<0有兩個(gè)交點(diǎn)，所以直線(xiàn)方程為SKIPIF1<0，故選：D2．（2022·全國(guó)·高考真題）已知正四棱錐的側(cè)棱長(zhǎng)為l，其各頂點(diǎn)都在同一球面上.若該球的體積為SKIPIF1<0，且SKIPIF1<0，則該正四棱錐體積的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】設(shè)正四棱錐的高為SKIPIF1<0，由球的截面性質(zhì)列方程求出正四棱錐的底面邊長(zhǎng)與高的關(guān)系，由此確定正四棱錐體積的取值范圍.【詳解】∵球的體積為SKIPIF1<0，所以球的半徑SKIPIF1<0,[方法一]：導(dǎo)數(shù)法設(shè)正四棱錐的底面邊長(zhǎng)為SKIPIF1<0，高為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0,所以SKIPIF1<0，SKIPIF1<0所以正四棱錐的體積SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，正四棱錐的體積SKIPIF1<0取最大值，最大值為SKIPIF1<0，又SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0,所以正四棱錐的體積SKIPIF1<0的最小值為SKIPIF1<0，所以該正四棱錐體積的取值范圍是SKIPIF1<0.故選：C.[方法二]：基本不等式法由方法一故所以SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0取到SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，得SKIPIF1<0，則SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，球心在正四棱錐高線(xiàn)上，此時(shí)SKIPIF1<0，SKIPIF1<0，正四棱錐體積SKIPIF1<0，故該正四棱錐體積的取值范圍是SKIPIF1<0二、多選題17．（2020·海南·高考真題）已知a>0，b>0，且a+b=1，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【分析】根據(jù)SKIPIF1<0，結(jié)合基本不等式及二次函數(shù)知識(shí)進(jìn)行求解.【詳解】對(duì)于A，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，故A正確；對(duì)于B，SKIPIF1<0，所以SKIPIF1<0，故B正確；對(duì)于C，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，故C不正確；對(duì)于D，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，故D正確；故選：ABD【點(diǎn)睛】本題主要考查不等式的性質(zhì)，綜合了基本不等式，指數(shù)函數(shù)及對(duì)數(shù)函數(shù)的單調(diào)性，側(cè)重考查數(shù)學(xué)運(yùn)算的核心素養(yǎng).3．（2022·黑龍江·哈爾濱市第六中學(xué)校高三階段練習(xí)）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．若SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0C．SKIPIF1<0取到最大值時(shí)，SKIPIF1<0 D．設(shè)SKIPIF1<0，則數(shù)列SKIPIF1<0的最小項(xiàng)為SKIPIF1<0【答案】AD【分析】求得等差數(shù)列SKIPIF1<0的通項(xiàng)公式判斷選項(xiàng)A；求得SKIPIF1<0的最小值判斷選項(xiàng)B；求得SKIPIF1<0取到最大值時(shí)n的值判斷選項(xiàng)C；求得數(shù)列SKIPIF1<0的最小項(xiàng)判斷選項(xiàng)D.【詳解】由SKIPIF1<0，可得SKIPIF1<0，則等差數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，則選項(xiàng)A判斷正確；若SKIPIF1<0，則SKIPIF1<0則SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立）又SKIPIF1<0，則SKIPIF1<0的最小值為不為SKIPIF1<0.則選項(xiàng)B判斷錯(cuò)誤；等差數(shù)列SKIPIF1<0中，SKIPIF1<0則等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0取到最大值時(shí)，SKIPIF1<0或SKIPIF1<0.則選項(xiàng)C判斷錯(cuò)誤；設(shè)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0則SKIPIF1<0則數(shù)列SKIPIF1<0的最小項(xiàng)為SKIPIF1<0.則選項(xiàng)D判斷正確故選：AD4．（2022·廣東·廣州大學(xué)附屬中學(xué)高三階段練習(xí)）已知SKIPIF1<0的左，右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，長(zhǎng)軸長(zhǎng)為4，點(diǎn)SKIPIF1<0在橢圓C外，點(diǎn)Q在橢圓C上，則下列說(shuō)法中正確的有（

）A．橢圓C的離心率的取值范圍是SKIPIF1<0B．已知SKIPIF1<0，當(dāng)橢圓C的離心率為SKIPIF1<0時(shí)，SKIPIF1<0的最大值為3C．存在點(diǎn)Q使得SKIPIF1<0D．SKIPIF1<0的最小值為1【答案】ACD【分析】易得SKIPIF1<0，再根據(jù)點(diǎn)SKIPIF1<0在橢圓C外，可得SKIPIF1<0，從而可求得SKIPIF1<0的范圍，再根據(jù)離心率公式即可判斷A；根據(jù)離心率求出橢圓方程，設(shè)點(diǎn)SKIPIF1<0，根據(jù)兩點(diǎn)的距離公式結(jié)合橢圓的有界性即可判斷B；當(dāng)點(diǎn)Q位于橢圓的上下頂點(diǎn)時(shí)SKIPIF1<0取得最大值，結(jié)合余弦定理判斷SKIPIF1<0是否大于等于SKIPIF1<0即可判斷C；根據(jù)SKIPIF1<0結(jié)合基本不等式即可判斷D.【詳解】解：根據(jù)題意可知SKIPIF1<0，則橢圓方程為SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0在橢圓C外，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，則離心率SKIPIF1<0，故A正確；對(duì)于B，當(dāng)橢圓C的離心率為SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以橢圓方程為SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故B錯(cuò)誤；對(duì)于C，當(dāng)點(diǎn)Q位于橢圓的上下頂點(diǎn)時(shí)SKIPIF1<0取得最大值，此時(shí)SKIPIF1<0，SKIPIF1<0，即當(dāng)點(diǎn)Q位于橢圓的上下頂點(diǎn)時(shí)SKIPIF1<0為鈍角，所以存在點(diǎn)Q使得SKIPIF1<0為直角，所以存在點(diǎn)Q使得SKIPIF1<0，故C正確；對(duì)于D，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，取等號(hào)，所以SKIPIF1<0的最小值為1，故D正確.故選：ACD.三、填空題5．（2022·重慶市云陽(yáng)縣高陽(yáng)中學(xué)高三階段練習(xí)（理））已知SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)__________【答案】SKIPIF1<0【分析】由題知SKIPIF1<0，進(jìn)而令SKIPIF1<0，將SKIPIF1<0的最小值轉(zhuǎn)化為SKIPIF1<0的最小值，再根據(jù)基本不等式求解即可.【詳解】解：由SKIPIF1<0得SKIPIF1<0，所以，整理得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，此時(shí)SKIPIF1<0.故答案為：SKIPIF1<06．（2022·黑龍江·鐵人中學(xué)高三階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0對(duì)于SKIPIF1<0恒成立，且方程SKIPIF1<0有實(shí)根，則SKIPIF1<0的最小值為_(kāi)_____.【答案】SKIPIF1<0【分析】根據(jù)題意結(jié)合一元二次不等式在R上恒成立可得SKIPIF1<0，消b整理得SKIPIF1<0，注意到SKIPIF1<0，結(jié)合基本不等式求最值.【詳解】由題意可得：不等式SKIPIF1<0對(duì)于SKIPIF1<0恒成立，則SKIPIF1<0方程SKIPIF1<0有實(shí)根，則SKIPIF1<0∴SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0∵SKIPIF1<0，則SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立∴SKIPIF1<0，則SKIPIF1<0故答案為：SKIPIF1<0.7．（2021·天津·高考真題）若SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)___________．【答案】SKIPIF1<0【分析】?jī)纱卫没静坏仁郊纯汕蟪?【詳解】SKIPIF1<0SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0且SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.8．（2019·天津·高考真題（理））設(shè)SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_____.【答案】SKIPIF1<0【分析】把分子展開(kāi)化為SKIPIF1<0，再利用基本不等式求最值．【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)成立，故所求的最小值為SKIPIF1<0．9．（2022·河北·唐山市第十一中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，若對(duì)任意SKIPIF1<0恒有SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)__________.【答案】SKIPIF1<0【分析】根據(jù)分段函數(shù)的性質(zhì)，結(jié)合對(duì)任意SKIPIF1<0恒有SKIPIF1<0，求得實(shí)數(shù)SKIPIF1<0的取值范圍，在根據(jù)對(duì)數(shù)函數(shù)與指數(shù)函數(shù)的單調(diào)性得函數(shù)SKIPIF1<0的單調(diào)性，從而可求解SKIPIF1<0的最大值.【詳解】解：由已知函數(shù)SKIPIF1<0，則SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取到最小值SKIPIF1<0；SKIPIF1<0時(shí)，若對(duì)任意SKIPIF1<0恒有SKIPIF1<0，則此時(shí)SKIPIF1<0單調(diào)遞減，則對(duì)稱(chēng)軸SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0；結(jié)合可知對(duì)任意SKIPIF1<0恒有SKIPIF1<0，則有SKIPIF1<0，所以SKIPIF1<0.又SKIPIF1<0，其中SKIPIF1<0在SKIPIF1<0時(shí)是增函數(shù)，SKIPIF1<0在SKIPIF1<0時(shí)是減函數(shù)，故SKIPIF1<0在SKIPIF1<0時(shí)是增函數(shù)，故SKIPIF1<0.故答案為：SKIPIF1<0.10．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知P是曲線(xiàn)SKIPIF1<0上的一動(dòng)點(diǎn)，曲線(xiàn)C在P點(diǎn)處的切線(xiàn)的傾斜角為SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的取值范圍是___________【答案】SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義，求導(dǎo)表示出切線(xiàn)斜率，根據(jù)傾斜角與斜率的變化關(guān)系，將問(wèn)題等價(jià)轉(zhuǎn)化為含參不等式恒成立，利用參變分離以及基本不等式，可得答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)榍€(xiàn)在M處的切線(xiàn)的傾斜角SKIPIF1<0，所以SKIPIF1<0對(duì)于任意的SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，即SKIPIF1<0，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故SKIPIF1<0，所以a的取值范圍是SKIPIF1<