新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)思想02 分類(lèi)與整合思想（練）（解析版）

第三篇思想方法篇思想02分類(lèi)與整合思想（練）一、單選題1．（2023·吉林·統(tǒng)考二模）已知點(diǎn)A，B，C為橢圓D的三個(gè)頂點(diǎn)，若SKIPIF1<0是正三角形，則D的離心率是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】首先由題得到SKIPIF1<0，結(jié)合SKIPIF1<0，即可求得SKIPIF1<0.【詳解】無(wú)論橢圓焦點(diǎn)位于SKIPIF1<0軸或SKIPIF1<0軸，根據(jù)點(diǎn)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0為橢圓SKIPIF1<0的三個(gè)頂點(diǎn),若SKIPIF1<0是正三角形,則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即有SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故選：C.2．（2023春·湖南常德·高一漢壽縣第一中學(xué)?？奸_(kāi)學(xué)考試）SKIPIF1<0表示不超過(guò)x的最大整數(shù)，已知函數(shù)SKIPIF1<0，有下列結(jié)論：①SKIPIF1<0的定義域?yàn)镾KIPIF1<0；②SKIPIF1<0的值域?yàn)镾KIPIF1<0；③SKIPIF1<0是偶函數(shù)；④SKIPIF1<0不是周期函數(shù)；⑤SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0．其中正確的結(jié)論個(gè)數(shù)是（

）A．3 B．2 C．1 D．0【答案】A【分析】直接根據(jù)解析式可知①正確；通過(guò)特殊值可知②和③不正確；根據(jù)周期函數(shù)的定義可知④正確；根據(jù)函數(shù)SKIPIF1<0的單調(diào)性可以判斷，可知⑤正確.【詳解】對(duì)于①，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故①正確；對(duì)于②，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故②錯(cuò)誤；對(duì)于③，SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的圖象不關(guān)于y軸對(duì)稱(chēng)，則SKIPIF1<0不是偶函數(shù)，故③錯(cuò)誤；對(duì)于④，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0表示x的小數(shù)部分，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0在SKIPIF1<0上是周期變化，當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．SKIPIF1<0是減函數(shù),SKIPIF1<0在R上不是周期函數(shù)，故④正確；對(duì)于⑤，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，表示x的小數(shù)部分，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0是減函數(shù)．故SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0，故⑤正確．故①④⑤正確.故選:A．3．（2023秋·天津·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有四個(gè)不同的零點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則下列結(jié)論中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】作出函數(shù)SKIPIF1<0圖象，根據(jù)函數(shù)圖象得出4個(gè)零點(diǎn)的關(guān)系及范圍，進(jìn)而得出結(jié)論.【詳解】函數(shù)SKIPIF1<0的四個(gè)不同的零點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，就是函數(shù)SKIPIF1<0與SKIPIF1<0兩個(gè)圖象四個(gè)交點(diǎn)的橫坐標(biāo)，作出函數(shù)SKIPIF1<0的圖象，對(duì)于A，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，結(jié)合圖象可知SKIPIF1<0，故A錯(cuò)誤；結(jié)合圖象可知SKIPIF1<0，解得SKIPIF1<0，故B正確；又SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故C錯(cuò)誤；根據(jù)二次函數(shù)的性質(zhì)和圖象得出SKIPIF1<0，所以SKIPIF1<0，故D錯(cuò)誤；故選：B4．（2023·河南·校聯(lián)考模擬預(yù)測(cè)）已知向量SKIPIF1<0的夾角為SKIPIF1<0，且SKIPIF1<0是函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn).若SKIPIF1<0，則SKIPIF1<0（

）A．3 B．4 C．5 D．6【答案】A【分析】由題知SKIPIF1<0或SKIPIF1<0.，再根據(jù)向量垂直的數(shù)量積表示，數(shù)量積的運(yùn)算律分別討論求解即可.【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0的兩個(gè)零點(diǎn)分別為2，3，所以SKIPIF1<0或SKIPIF1<0.又SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0（舍去）；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，滿(mǎn)足SKIPIF1<0.綜上，SKIPIF1<0故選：A5．（2023秋·云南德宏·高三統(tǒng)考期末）已知函數(shù)SKIPIF1<0的周期為2，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．如果SKIPIF1<0，那么SKIPIF1<0的零點(diǎn)個(gè)數(shù)是（

）A．3 B．4C．5 D．6【答案】C【分析】先將問(wèn)題SKIPIF1<0的零點(diǎn)問(wèn)題轉(zhuǎn)化為函數(shù)SKIPIF1<0與SKIPIF1<0的交點(diǎn)，分析出SKIPIF1<0的值域，由此判斷出零點(diǎn)個(gè)數(shù)．【詳解】函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為函數(shù)SKIPIF1<0與SKIPIF1<0的圖象的交點(diǎn)的個(gè)數(shù)，因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象沒(méi)有交點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．又函數(shù)SKIPIF1<0的周期為2，所以SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象沒(méi)有交點(diǎn)，作函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的圖象，觀察圖象可得兩函數(shù)圖象有5個(gè)交點(diǎn),所以函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為5.故選：C.6．（2022秋·福建廈門(mén)·高三廈門(mén)一中?？计谥校╇p曲線(xiàn)SKIPIF1<0：SKIPIF1<0的右焦點(diǎn)和虛軸上的一個(gè)端點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0為雙曲線(xiàn)SKIPIF1<0左支上一點(diǎn)，若SKIPIF1<0周長(zhǎng)的最小值為SKIPIF1<0，則雙曲線(xiàn)SKIPIF1<0的離心率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由題意求得SKIPIF1<0，SKIPIF1<0的坐標(biāo)，設(shè)出SKIPIF1<0，運(yùn)用雙曲線(xiàn)的定義可得SKIPIF1<0，則SKIPIF1<0的周長(zhǎng)為SKIPIF1<0，運(yùn)用三點(diǎn)共線(xiàn)取得最小值，可得SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的關(guān)系，結(jié)合離心率公式，計(jì)算即可得到所求值.【詳解】解：由題意可得SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，由雙曲線(xiàn)的定義可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的周長(zhǎng)為SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0共線(xiàn)，取得最小值，且為SKIPIF1<0，由題意可得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故選：B.7．（2023·河北·高三河北衡水中學(xué)?？茧A段練習(xí)）若平面向量SKIPIF1<0滿(mǎn)足SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)題意分析可得點(diǎn)SKIPIF1<0的軌跡為以SKIPIF1<0為焦點(diǎn)的橢圓，結(jié)合橢圓的定義分析求解.【詳解】∵SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，即SKIPIF1<0，故點(diǎn)SKIPIF1<0的軌跡為以SKIPIF1<0為焦點(diǎn)的橢圓，∴SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)點(diǎn)SKIPIF1<0為SKIPIF1<0的延長(zhǎng)線(xiàn)與橢圓的交點(diǎn)SKIPIF1<0時(shí)等號(hào)成立，SKIPIF1<0，當(dāng)且僅當(dāng)點(diǎn)SKIPIF1<0為SKIPIF1<0的延長(zhǎng)線(xiàn)與橢圓的交點(diǎn)SKIPIF1<0時(shí)等號(hào)成立，即SKIPIF1<0，故SKIPIF1<0.故選：D.8．（2023·安徽合肥·統(tǒng)考一模）已知線(xiàn)段PQ的中點(diǎn)為等邊三角形ABC的頂點(diǎn)A，且SKIPIF1<0，當(dāng)PQ繞點(diǎn)A轉(zhuǎn)動(dòng)時(shí)，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】以SKIPIF1<0點(diǎn)為原點(diǎn)，建立直角坐標(biāo)系，可知SKIPIF1<0兩點(diǎn)都是圓SKIPIF1<0上的動(dòng)點(diǎn)，當(dāng)直線(xiàn)SKIPIF1<0斜率不存在時(shí)，可得SKIPIF1<0，直線(xiàn)SKIPIF1<0斜率存在時(shí)，可得到SKIPIF1<0或SKIPIF1<0，再討論SKIPIF1<0與SKIPIF1<0的大小關(guān)系，即可求解.【詳解】以SKIPIF1<0點(diǎn)為原點(diǎn)，以與SKIPIF1<0平行的直線(xiàn)為SKIPIF1<0軸，與SKIPIF1<0垂直的直線(xiàn)為SKIPIF1<0軸，建立平面直角坐標(biāo)系，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，易知SKIPIF1<0兩點(diǎn)都是圓SKIPIF1<0上的動(dòng)點(diǎn)，當(dāng)直線(xiàn)SKIPIF1<0斜率不存在時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0當(dāng)直線(xiàn)SKIPIF1<0斜率不存在時(shí)，可設(shè)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0；同理，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，綜上所述，SKIPIF1<0的取值范圍是SKIPIF1<0，故答案選：D.二、多選題9．（2023秋·河南鄭州·高一統(tǒng)考期末）已知實(shí)數(shù)a，b滿(mǎn)足等式SKIPIF1<0，下列式子可以成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【分析】根據(jù)指數(shù)函數(shù)圖象分析判斷.【詳解】設(shè)SKIPIF1<0，分別作出SKIPIF1<0的函數(shù)圖象，如圖所示：當(dāng)SKIPIF1<0，則SKIPIF1<0，A成立；當(dāng)SKIPIF1<0，則SKIPIF1<0，B成立，C不成立；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，D成立.故選：ABD.10．（2023·廣東深圳·統(tǒng)考一模）已知拋物線(xiàn)C：SKIPIF1<0的準(zhǔn)線(xiàn)為SKIPIF1<0，直線(xiàn)SKIPIF1<0與C相交于A、B兩點(diǎn)，M為AB的中點(diǎn)，則（

）A．當(dāng)SKIPIF1<0時(shí)，以AB為直徑的圓與SKIPIF1<0相交B．當(dāng)SKIPIF1<0時(shí)，以AB為直徑的圓經(jīng)過(guò)原點(diǎn)OC．當(dāng)SKIPIF1<0時(shí)，點(diǎn)M到SKIPIF1<0的距離的最小值為2D．當(dāng)SKIPIF1<0時(shí)，點(diǎn)M到SKIPIF1<0的距離無(wú)最小值【答案】BC【分析】將直線(xiàn)SKIPIF1<0代入SKIPIF1<0，結(jié)合韋達(dá)定理求得SKIPIF1<0坐標(biāo)、點(diǎn)SKIPIF1<0到準(zhǔn)線(xiàn)SKIPIF1<0的距離SKIPIF1<0及SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可判斷A；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可判斷B；當(dāng)SKIPIF1<0時(shí)，得SKIPIF1<0的關(guān)系式，代入SKIPIF1<0表達(dá)式，利用基本不等式可判斷C；當(dāng)SKIPIF1<0時(shí)，得SKIPIF1<0的關(guān)系式，代入SKIPIF1<0表達(dá)式，利用對(duì)勾函數(shù)的性質(zhì)可判斷D．【詳解】拋物線(xiàn)SKIPIF1<0，準(zhǔn)線(xiàn)SKIPIF1<0方程是SKIPIF1<0，直線(xiàn)SKIPIF1<0代入SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，點(diǎn)SKIPIF1<0到準(zhǔn)線(xiàn)SKIPIF1<0的距離SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，點(diǎn)SKIPIF1<0到準(zhǔn)線(xiàn)SKIPIF1<0的距離SKIPIF1<0，則以AB為直徑的圓與SKIPIF1<0相切，故A錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，則以AB為直徑的圓經(jīng)過(guò)原點(diǎn)O，故B正確；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故C正確；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，由對(duì)勾函數(shù)的性質(zhì)得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最小值SKIPIF1<0，故D錯(cuò)誤．故選：BC．11．（2023·遼寧·校聯(lián)考模擬預(yù)測(cè)）已知F是拋物線(xiàn)SKIPIF1<0的焦點(diǎn)，點(diǎn)SKIPIF1<0在拋物線(xiàn)W上，過(guò)點(diǎn)F的兩條互相垂直的直線(xiàn)SKIPIF1<0，SKIPIF1<0分別與拋物線(xiàn)W交于B，C和D，E，過(guò)點(diǎn)A分別作SKIPIF1<0，SKIPIF1<0的垂線(xiàn)，垂足分別為M，N，則（

）A．四邊形SKIPIF1<0面積的最大值為2B．四邊形SKIPIF1<0周長(zhǎng)的最大值為SKIPIF1<0C．SKIPIF1<0為定值SKIPIF1<0D．四邊形SKIPIF1<0面積的最小值為32【答案】ACD【分析】根據(jù)給定條件，求出拋物線(xiàn)SKIPIF1<0的方程，確定四邊形SKIPIF1<0形狀，利用勾股定理及均值不等式計(jì)算判斷A，B；設(shè)出直線(xiàn)SKIPIF1<0的方程，與拋物線(xiàn)方程聯(lián)立，求出弦SKIPIF1<0長(zhǎng)即可計(jì)算推理判斷C，D作答.【詳解】因?yàn)辄c(diǎn)SKIPIF1<0在拋物線(xiàn)SKIPIF1<0上，所以SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，拋物線(xiàn)SKIPIF1<0的焦點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以四邊形SKIPIF1<0面積的最大值為2，故A正確.由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以四邊形SKIPIF1<0周長(zhǎng)的最大值為SKIPIF1<0，故B不正確.設(shè)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0消x得SKIPIF1<0，方程SKIPIF1<0的判別式SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，同理得SKIPIF1<0，SKIPIF1<0，C正確.SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，此時(shí)SKIPIF1<0，故D正確.故選：ACD.12．（2022秋·福建廈門(mén)·高三廈門(mén)一中?？计谥校┮阎襟wSKIPIF1<0的棱長(zhǎng)為2（如圖所示），點(diǎn)SKIPIF1<0為線(xiàn)段SKIPIF1<0（含端點(diǎn)）上的動(dòng)點(diǎn)，由點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0確定的平面為SKIPIF1<0，則下列說(shuō)法正確的是（

）A．平面SKIPIF1<0截正方體的截面始終為四邊形B．點(diǎn)SKIPIF1<0運(yùn)動(dòng)過(guò)程中，三棱錐SKIPIF1<0的體積為定值C．平面SKIPIF1<0截正方體的截面面積的最大值為SKIPIF1<0D．三棱錐SKIPIF1<0的外接球表面積的取值范圍為SKIPIF1<0【答案】BCD【分析】根據(jù)線(xiàn)面平行的判定定理，運(yùn)動(dòng)變化思想，函數(shù)思想，即可分別求解.【詳解】對(duì)A選項(xiàng)，當(dāng)SKIPIF1<0與SKIPIF1<0點(diǎn)重合時(shí)，平面SKIPIF1<0截正方體的截面為SKIPIF1<0，錯(cuò)誤；對(duì)B選項(xiàng)，∵SKIPIF1<0，又SKIPIF1<0平面SKIPIF1<0，SKIPIF1<0平面SKIPIF1<0，∴SKIPIF1<0平面SKIPIF1<0，又點(diǎn)SKIPIF1<0為線(xiàn)段SKIPIF1<0（含端點(diǎn)）上的動(dòng)點(diǎn)，∴SKIPIF1<0到平面SKIPIF1<0的距離為定值，又SKIPIF1<0的面積也為定值，∴三棱錐SKIPIF1<0的體積為定值，正確；對(duì)C選項(xiàng)，當(dāng)SKIPIF1<0由SKIPIF1<0移動(dòng)到SKIPIF1<0的過(guò)程中，利用平面的基本性質(zhì)，延長(zhǎng)SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，連接SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，所以，從SKIPIF1<0到SKIPIF1<0之間，平面SKIPIF1<0截正方體的截面為SKIPIF1<0為等腰梯形，且SKIPIF1<0，當(dāng)SKIPIF1<0與SKIPIF1<0重合時(shí)，截面為矩形SKIPIF1<0，此時(shí)面積最大為SKIPIF1<0，正確；對(duì)D選項(xiàng)，如圖，分別取左右側(cè)面的中心SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0垂直于左右側(cè)面，根據(jù)對(duì)稱(chēng)性易知：三棱錐SKIPIF1<0的外接球的球心SKIPIF1<0在線(xiàn)段SKIPIF1<0上，設(shè)SKIPIF1<0到SKIPIF1<0的距離為SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，又易知SKIPIF1<0，外接球SKIPIF1<0的半徑SKIPIF1<0，在SKIPIF1<0與SKIPIF1<0中，由勾股定理可得：SKIPIF1<0，兩式相減得：SKIPIF1<0，∴SKIPIF1<0，令SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的對(duì)稱(chēng)軸為SKIPIF1<0，的開(kāi)口向上，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，最小值為SKIPIF1<0，最大值為SKIPIF1<0，即SKIPIF1<0，∴三棱錐SKIPIF1<0的外接球表面積SKIPIF1<0，正確.故選：BCD.三、填空題13．（2023·陜西咸陽(yáng)·?？家荒＃┮阎瘮?shù)SKIPIF1<0，則不等式SKIPIF1<0的解集為_(kāi)_____.【答案】SKIPIF1<0【分析】由題意結(jié)合函數(shù)的解析式分類(lèi)討論求解不等式的解集即可.【詳解】解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，綜上，不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<014．（2023秋·云南德宏·高三統(tǒng)考期末）已知橢圓C：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為橢圓的左右焦點(diǎn)．若點(diǎn)P是橢圓上的一個(gè)動(dòng)點(diǎn)，點(diǎn)A的坐標(biāo)為（2，1），則SKIPIF1<0的范圍為_(kāi)____．【答案】SKIPIF1<0【分析】利用橢圓定義可得SKIPIF1<0，再根據(jù)三角形三邊長(zhǎng)的關(guān)系可知，當(dāng)SKIPIF1<0共線(xiàn)時(shí)即可取得SKIPIF1<0最值.【詳解】由橢圓標(biāo)準(zhǔn)方程可知SKIPIF1<0，SKIPIF1<0又點(diǎn)P在橢圓上，根據(jù)橢圓定義可得SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0易知SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0三點(diǎn)共線(xiàn)時(shí)等號(hào)成立；又SKIPIF1<0，所以SKIPIF1<0；即SKIPIF1<0的范圍為SKIPIF1<0.故答案為：SKIPIF1<015．（2022秋·上海青浦·高三統(tǒng)考階段練習(xí)）在平面直角坐標(biāo)系SKIPIF1<0中，若動(dòng)點(diǎn)SKIPIF1<0到兩直線(xiàn)SKIPIF1<0和SKIPIF1<0的距離之和為SKIPIF1<0，則SKIPIF1<0的最大值為_(kāi)__________.【答案】8【分析】由已知可知兩直線(xiàn)SKIPIF1<0，取SKIPIF1<0在SKIPIF1<0的右側(cè)時(shí)，分別過(guò)SKIPIF1<0作兩直線(xiàn)的垂線(xiàn)，結(jié)合幾何性質(zhì)確定SKIPIF1<0點(diǎn)軌跡，即可求得SKIPIF1<0的最大值,其他位置同理可得．【詳解】若動(dòng)點(diǎn)SKIPIF1<0到兩直線(xiàn)SKIPIF1<0和SKIPIF1<0的距離之和為SKIPIF1<0，SKIPIF1<0交點(diǎn)為SKIPIF1<0的斜率分別為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0的右側(cè)時(shí)，過(guò)SKIPIF1<0分別向SKIPIF1<0引垂線(xiàn)，垂足分別為SKIPIF1<0，那么SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0軸的平行線(xiàn)，與SKIPIF1<0交點(diǎn)為SKIPIF1<0如圖，則SKIPIF1<0，所以SKIPIF1<0，其它位置同理，那么點(diǎn)SKIPIF1<0軌跡為正方形SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0時(shí)，SKIPIF1<0取得最大值SKIPIF1<0，即SKIPIF1<0取得最大值8.故答案為：8．16．（2022·北京·統(tǒng)考模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0的極小值點(diǎn)為1，則實(shí)數(shù)a的取值范圍是__________，【答案】SKIPIF1<0【分析】令SKIPIF1<0得SKIPIF1<0，討論SKIPIF1<0與SKIPIF1<0的大小關(guān)系，確定1是否為極值點(diǎn)即可.【詳解】由SKIPIF1<0得，SKIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0有兩個(gè)不同的根，令SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0SKIPIF1<0單調(diào)遞增，①當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0的大致圖象如圖1：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0的極大值點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以1為SKIPIF1<0的極小值點(diǎn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0的極大值點(diǎn)，故SKIPIF1<0時(shí)滿(mǎn)足題意.②當(dāng)SKIPIF1<0時(shí),SKIPIF1<0是SKIPIF1<0的最大根，SKIPIF1<0的大致圖象如圖2：SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以1為SKIPIF1<0的極大值點(diǎn)，此時(shí)不滿(mǎn)足題意.③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的大致圖象如圖3圖4，SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以1為SKIPIF1<0的極大值點(diǎn)，此時(shí)不滿(mǎn)足題意.④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以1為SKIPIF1<0的極大值點(diǎn)，此時(shí)不滿(mǎn)足題意.綜上：SKIPIF1<0的取值范圍：SKIPIF1<0，故答案為：SKIPIF1<0【點(diǎn)睛】已知SKIPIF1<0為極值點(diǎn)求參數(shù)方法：由極值點(diǎn)定義知SKIPIF1<0，（1）若由SKIPIF1<0可以求得參數(shù)值，再證明SKIPIF1<0為函數(shù)的極值點(diǎn)；（2）若SKIPIF1<0恒成立，求不到參數(shù)值，則SKIPIF1<0為變號(hào)零點(diǎn)，通過(guò)含參討論確保SKIPIF1<0兩側(cè)的單調(diào)性不同求得參數(shù)值或范圍.（相似題2018新課標(biāo)3卷理21題）四、解答題17．（2022·北京·統(tǒng)考模擬預(yù)測(cè)）已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，在數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)求數(shù)列SKIPIF1<0，SKIPIF1<0的通項(xiàng)公式；(2)設(shè)SKIPIF1<0，SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和，求SKIPIF1<0的最值．【答案】(1)SKIPIF1<0，SKIPIF1<0(2)最小值為SKIPIF1<0，最大值為1【分析】（1）利用累加法和等差數(shù)列的通項(xiàng)公式可求SKIPIF1<0，由SKIPIF1<0及SKIPIF1<0可求SKIPIF1<0；（2）利用錯(cuò)位相減法求出SKIPIF1<0，分情況討論可得答案.【詳解】（1）由己知得，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0SKIPIF1<0.∴SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，也滿(mǎn)足上式．所以SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，符合上式當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，也符合上式，綜上，SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0.（2）由（1）可得：SKIPIF1<0∴SKIPIF1<0SKIPIF1<0兩式相減：SKIPIF1<0SKIPIF1<0∴SKIPIF1<0當(dāng)n為奇數(shù)時(shí)，不妨設(shè)SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0單調(diào)遞減，SKIPIF1<0當(dāng)n為偶數(shù)時(shí)，不妨設(shè)SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0單調(diào)遞增，SKIPIF1<0∴SKIPIF1<0的最小值為SKIPIF1<0，最大值為1.18．（2023春·浙江·高三校聯(lián)考開(kāi)學(xué)考試）設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0滿(mǎn)足：SKIPIF1<0，記SKIPIF1<0．(1)證明：SKIPIF1<0是等比數(shù)列，并求SKIPIF1<0的通項(xiàng)公式；(2)求SKIPIF1<0的最大值．【答案】(1)證明見(jiàn)解析，SKIPIF1<0；(2)SKIPIF1<0.【分析】（1）根據(jù)SKIPIF1<0與SKIPIF1<0之間的關(guān)系可得，SKIPIF1<0，進(jìn)而可推得SKIPIF1<0，即SKIPIF1<0，求出SKIPIF1<0，可得出SKIPIF1<0，即可得出SKIPIF1<0，進(jìn)而得出SKIPIF1<0；（2）作差可得SKIPIF1<0，通過(guò)研究函數(shù)SKIPIF1<0的性質(zhì)，即可得出SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，進(jìn)而求出SKIPIF1<0的值，即可得出答案.【詳解】（1）由已知可得，SKIPIF1<0①，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0②，①-②整理可得，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是首項(xiàng)為SKIPIF1<0，公比為SKIPIF1<0的等比數(shù)列，所以SKIPIF1<0，則SKIPIF1<0；（2）由（1）可知，SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，所以要求SKIPIF1<0的最大值，先比較SKIPIF1<0與SKIPIF1<0的大小，令SKIPIF1<0，則SKIPIF1<0，根據(jù)函數(shù)的單調(diào)性，可知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增.且SKIPIF1<0時(shí)，有SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，所以SKIPIF1<0單調(diào)遞增.又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0時(shí)，有SKIPIF1<0，即SKIPIF1<0單調(diào)遞減，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0最大，此時(shí)SKIPIF1<0.19．（2023春·江西宜春·高三江西省豐城中學(xué)?？奸_(kāi)學(xué)考試）如圖，已知橢圓SKIPIF1<0的離心率為SKIPIF1<0，其左、右頂點(diǎn)分別為SKIPIF1<0．過(guò)點(diǎn)SKIPIF1<0的直線(xiàn)SKIPIF1<0與該橢圓相交于SKIPIF1<0兩點(diǎn)．(1)求橢圓SKIPIF1<0的方程；(2)設(shè)直線(xiàn)SKIPIF1<0與SKIPIF1<0的斜率分別為SKIPIF1<0．試問(wèn)：是否存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0？若存在，求SKIPIF1<0的值；若不存在，請(qǐng)說(shuō)明理由．【答案】(1)SKIPIF1<0；(2)存在，SKIPIF1<0.【分析】（1）根據(jù)題意求出SKIPIF1<0，即可得解；（2）方法一：設(shè)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立方程，利用韋達(dá)定理可求得SKIPIF1<0兩點(diǎn)的坐標(biāo)，再根據(jù)SKIPIF1<0三點(diǎn)共線(xiàn)，即可得出結(jié)論.方法二：根據(jù)當(dāng)直線(xiàn)SKIPIF1<0垂直于SKIPIF1<0軸時(shí)，得出SKIPIF1<0的值，在證明直線(xiàn)斜率不存在時(shí)，SKIPIF1<0也為這個(gè)值即可.【詳解】（1）依題意可知SKIPIF1<0，SKIPIF1<0，所以橢圓的方程為：SKIPIF1<0；（2）(方法一)設(shè)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立方程組SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，所以點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，同理，可解得點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，由SKIPIF1<0三點(diǎn)共線(xiàn)，得SKIPIF1<0，化簡(jiǎn)有SKIPIF1<0，由題知SKIPIF1<0同號(hào)，所以SKIPIF1<0，故存在SKIPIF1<0，使得成立．(方法二)當(dāng)直線(xiàn)SKIPIF1<0垂直于SKIPIF1<0軸時(shí)，SKIPIF1<0點(diǎn)的坐標(biāo)分別為SKIPIF1<0，所以此時(shí)直線(xiàn)與的斜率分別為SKIPIF1<0，有SKIPIF1<0，由此猜想：存在SKIPIF1<0滿(mǎn)足條件，下面證明猜想正確．當(dāng)直線(xiàn)SKIPIF1<0不垂直于SKIPIF1<0軸時(shí)，設(shè)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，SKIPIF1<0聯(lián)立方程組SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，由此可得猜想正確，故存在SKIPIF1<0，使得成立．【點(diǎn)睛】本題考查了橢圓的離心率及橢圓的方程，考查了直線(xiàn)與橢圓的位置關(guān)系的應(yīng)用，計(jì)算量較大，有一定的難度.20．（2022·北京·統(tǒng)考模擬預(yù)測(cè)）如圖所示，過(guò)原點(diǎn)O作兩條互相垂直的線(xiàn)OA，OB分別交拋物線(xiàn)SKIPIF1<0于A，B兩點(diǎn)，連接AB，交y軸于點(diǎn)P．(1)求點(diǎn)P的坐標(biāo)；(2)證明：存在相異于點(diǎn)P的定點(diǎn)T，使得SKIPIF1<0恒成立，請(qǐng)求出點(diǎn)T的坐標(biāo)，并求出SKIPIF1<0面積的最小值．【答案】(1)SKIPIF1<0；(2)證明見(jiàn)解析，SKIPIF1<0，8.【分析】（1）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，聯(lián)立直線(xiàn)和拋物線(xiàn)方程得到韋達(dá)定理，化簡(jiǎn)SKIPIF1<0即得解；（2）當(dāng)SKIPIF1<0與x軸平行時(shí)，SKIPIF1<0，設(shè)SKIPIF1<0，由題得SKIPIF1<0，化簡(jiǎn)即得SKIPIF1<0.求出SKIPIF1<0,即得解.【詳解】（1）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的斜率必存在，設(shè)SKIPIF1<0與拋物線(xiàn)聯(lián)立可得SKIPIF1<0，∴SKIPIF1<0，可知：SKIPIF1<0.∵SKIPIF1<0，∴SKIPIF1<0∵SKIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0，即SKIPIF1<0.（2）由SKIPIF1<0，可知：SKIPIF1<0，當(dāng)SKIPIF1<0與x軸平行時(shí)，SKIPIF1<0，∴存在點(diǎn)T在y軸上，設(shè)SKIPIF1<0，SKIPIF1<0，∴TP為SKIPIF1<0的角平分線(xiàn)，有SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴存在SKIPIF1<0，使得：SKIPIF1<0恒成立，∴SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0軸時(shí)，SKIPIF1<0面積的最小值為8.21.（2023春·廣西柳州·高三統(tǒng)考階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0恒成立，求實(shí)數(shù)k的取值范圍．【答案】(1)答案見(jiàn)解析(2)SKIPIF1<0【分析】（1）求SKIPIF1<0的導(dǎo)函數(shù)，對(duì)導(dǎo)函數(shù)中的參數(shù)SKIPIF1<0分類(lèi)討論，在每種情況下通過(guò)導(dǎo)函數(shù)的正負(fù)得出函數(shù)的單調(diào)性；（2）將函數(shù)恒成立問(wèn)題轉(zhuǎn)化為關(guān)于函數(shù)最值的不等式，解不等式得出參數(shù)取值范圍.【詳解】（1）SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；②當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減；③當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增．綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增．（2）設(shè)SKIPIF1<0，則題意等價(jià)于SKIPIF1<0時(shí)，SKIPIF1<0恒成立，所以SKIPIF1<0，故SKIPIF1<0.SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0為增函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0為減函數(shù)；當(dāng)SKIPIF1<0時(shí)