江蘇省宿遷市2022-2023學(xué)年高二上學(xué)期期末調(diào)研測(cè)試數(shù)學(xué)試題（含解析）

2022-2023學(xué)年江蘇省宿遷市高二年級(jí)上學(xué)期調(diào)研測(cè)試數(shù)學(xué)一、單選題(本大題共8小題，共40.0分.在每小題列出的選項(xiàng)中，選出符合題目的一項(xiàng))1.在等差數(shù)列{SKIPIF1<0}中，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的值為()A.18 B.20 C.22 D.24【答案】B【解析】【分析】根據(jù)等差數(shù)列通項(xiàng)公式相關(guān)計(jì)算求出公差，進(jìn)而求出首項(xiàng).【詳解】設(shè)公差為SKIPIF1<0，由題意得：SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0.故選：B2.若直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0垂直，則SKIPIF1<0的值為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)兩直線(xiàn)垂直與斜率之間的關(guān)系即可求解.【詳解】SKIPIF1<0直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0垂直，當(dāng)SKIPIF1<0時(shí)不滿(mǎn)足，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0．故選:D.3.若直線(xiàn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的一條切線(xiàn)，則實(shí)數(shù)SKIPIF1<0的值為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)導(dǎo)數(shù)的幾何意義分析運(yùn)算．【詳解】SKIPIF1<0，則SKIPIF1<0，設(shè)直線(xiàn)l與曲線(xiàn)C的切點(diǎn)SKIPIF1<0，則直線(xiàn)l的斜率SKIPIF1<0，由于直線(xiàn)SKIPIF1<0斜率為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，即切點(diǎn)為SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0．故選：C.4.體育館等建筑的屋頂一般采用曲面結(jié)構(gòu)．如圖所示，某建筑的屋頂采用雙曲面結(jié)構(gòu)，該建筑屋頂外形弧線(xiàn)可看作是雙曲線(xiàn)上支的部分，其漸近線(xiàn)方程為SKIPIF1<0，上焦點(diǎn)坐標(biāo)為SKIPIF1<0，那么該雙曲線(xiàn)的標(biāo)準(zhǔn)方程為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】設(shè)雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0，根據(jù)題意求出SKIPIF1<0、SKIPIF1<0的值，即可得出所求雙曲線(xiàn)的標(biāo)準(zhǔn)方程.【詳解】解：設(shè)雙曲線(xiàn)的標(biāo)準(zhǔn)方程為SKIPIF1<0，因?yàn)樵撾p曲線(xiàn)的漸近線(xiàn)方程為SKIPIF1<0，則SKIPIF1<0，又因?yàn)樵撾p曲線(xiàn)的上焦點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，SKIPIF1<0，因此，該雙曲線(xiàn)的方程為SKIPIF1<0.故選：B.5.圓SKIPIF1<0與圓SKIPIF1<0的公切線(xiàn)條數(shù)為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】先判斷兩圓的位置關(guān)系，進(jìn)而確定公切線(xiàn)的條數(shù)．【詳解】由圓SKIPIF1<0，可得圓SKIPIF1<0的圓心為SKIPIF1<0，半徑為1，由圓SKIPIF1<0

，可得圓SKIPIF1<0的圓心為SKIPIF1<0，半徑為SKIPIF1<0，∵圓SKIPIF1<0與圓SKIPIF1<0的圓心距SKIPIF1<0，∴圓SKIPIF1<0與圓SKIPIF1<0相離，故有SKIPIF1<0條公切線(xiàn)．故選：D.6.已知數(shù)列SKIPIF1<0是各項(xiàng)均為正數(shù)的等比數(shù)列，若SKIPIF1<0，SKIPIF1<0是方程SKIPIF1<0的兩個(gè)根，則SKIPIF1<0的值為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】由韋達(dá)定理，可得SKIPIF1<0，后由等比數(shù)列性質(zhì)結(jié)合對(duì)數(shù)運(yùn)算性質(zhì)可得答案.【詳解】由韋達(dá)定理，可得SKIPIF1<0，由等比數(shù)列性質(zhì)可得SKIPIF1<0，SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0.故選：B7.已知雙曲線(xiàn)SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0的直線(xiàn)SKIPIF1<0與圓SKIPIF1<0相切，直線(xiàn)SKIPIF1<0與雙曲線(xiàn)左右支分別交于SKIPIF1<0兩點(diǎn)，且SKIPIF1<0，若雙曲線(xiàn)SKIPIF1<0的離心率為SKIPIF1<0，則SKIPIF1<0的值為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】過(guò)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，利用雙曲線(xiàn)的定義和性質(zhì)、離心率的計(jì)算公式求解即可.【詳解】過(guò)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，由題意可得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0是SKIPIF1<0中點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，由雙曲線(xiàn)定義可得SKIPIF1<0，即SKIPIF1<0①，SKIPIF1<0②，SKIPIF1<0③，①②③聯(lián)立可得SKIPIF1<0.故選：A8.已知SKIPIF1<0則(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】注意到SKIPIF1<0，SKIPIF1<0.后構(gòu)造函數(shù)SKIPIF1<0，可判斷b與c大小.【詳解】注意到SKIPIF1<0，SKIPIF1<0.則SKIPIF1<0.令SKIPIF1<0，其中SKIPIF1<0.則SKIPIF1<0，得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.則SKIPIF1<0，又函數(shù)SKIPIF1<0R上單調(diào)遞增，則SKIPIF1<0，即SKIPIF1<0.故SKIPIF1<0．故選：D【點(diǎn)睛】方法點(diǎn)睛：比較代數(shù)式大小的常見(jiàn)方法有：(1)利用函數(shù)單調(diào)性；(2)利用中間量；(3)構(gòu)造函數(shù).二、多選題(本大題共4小題，共20.0分.在每小題有多項(xiàng)符合題目要求)9.已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0，則下列說(shuō)法正確的是()A.SKIPIF1<0 B.SKIPIF1<0為SKIPIF1<0中的最大項(xiàng)C.SKIPIF1<0 D.SKIPIF1<0【答案】AC【解析】【分析】根據(jù)題意，先由SKIPIF1<0求得SKIPIF1<0，然后根據(jù)等差數(shù)列求和，以及性質(zhì)逐一判斷，即可得到結(jié)果.【詳解】對(duì)于A：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，經(jīng)檢驗(yàn)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，A正確；對(duì)于B：令SKIPIF1<0，則SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0和SKIPIF1<0為SKIPIF1<0中的最大項(xiàng)，B錯(cuò)誤；對(duì)于C：SKIPIF1<0，C正確；對(duì)于D：SKIPIF1<0SKIPIF1<0，D錯(cuò)誤．故選:AC10.已知函數(shù)SKIPIF1<0，下列說(shuō)法正確的是()A.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在單調(diào)遞增區(qū)間B.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在兩個(gè)極值點(diǎn)C.SKIPIF1<0是SKIPIF1<0為減函數(shù)的充要條件D.SKIPIF1<0，SKIPIF1<0無(wú)極大值【答案】AC【解析】【分析】由題，SKIPIF1<0，設(shè)SKIPIF1<0.A選項(xiàng)，判斷當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上有無(wú)解即可；B選項(xiàng)，判斷當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是否有兩根即可；C選項(xiàng)，由充要條件定義驗(yàn)證即可判斷選項(xiàng)正誤；D選項(xiàng)，由A選項(xiàng)分析可判斷選項(xiàng)正誤.【詳解】由題，SKIPIF1<0，設(shè)SKIPIF1<0.A選項(xiàng)，當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，方程SKIPIF1<0的判別式SKIPIF1<0，則SKIPIF1<0兩根為SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的解為SKIPIF1<0，則此時(shí)SKIPIF1<0存在單調(diào)遞增區(qū)間SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的解為SKIPIF1<0，則此時(shí)SKIPIF1<0存在單調(diào)遞增區(qū)間SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的解為SKIPIF1<0，則此時(shí)SKIPIF1<0存在單調(diào)遞增區(qū)間SKIPIF1<0.綜上：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在單調(diào)遞增區(qū)間.故A正確；B選項(xiàng)，由A選項(xiàng)分析可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在兩個(gè)極值點(diǎn)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在唯一極值點(diǎn)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在唯一極值點(diǎn)1.故B錯(cuò)誤.C選項(xiàng)，當(dāng)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上恒成立，得SKIPIF1<0為SKIPIF1<0上的減函數(shù)；若SKIPIF1<0為SKIPIF1<0上的減函數(shù)，則SKIPIF1<0在SKIPIF1<0上恒成立，得SKIPIF1<0，則SKIPIF1<0.綜上，SKIPIF1<0是SKIPIF1<0為減函數(shù)的充要條件.故C正確.D選項(xiàng)，由A選項(xiàng)分析可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則此時(shí)SKIPIF1<0有極大值SKIPIF1<0.故D錯(cuò)誤.故選：AC11.平行于拋物線(xiàn)對(duì)稱(chēng)軸的光線(xiàn)經(jīng)拋物線(xiàn)壁的反射，光線(xiàn)匯聚于焦點(diǎn)處，這就是“焦點(diǎn)”名稱(chēng)的來(lái)源SKIPIF1<0運(yùn)用拋物線(xiàn)的這一性質(zhì)，人們?cè)O(shè)計(jì)了一種將水和食物加熱的太陽(yáng)灶SKIPIF1<0反過(guò)來(lái)，從焦點(diǎn)處發(fā)出的光線(xiàn)，經(jīng)過(guò)拋物線(xiàn)反射后將變成與拋物線(xiàn)的對(duì)稱(chēng)軸平行的光線(xiàn)射出，運(yùn)用這一性質(zhì)，人們制造了探照燈SKIPIF1<0如圖所示，已知拋物線(xiàn)SKIPIF1<0，SKIPIF1<0為坐標(biāo)原點(diǎn)，一條平行于SKIPIF1<0軸的光線(xiàn)SKIPIF1<0從點(diǎn)SKIPIF1<0射入，經(jīng)過(guò)SKIPIF1<0上的點(diǎn)SKIPIF1<0反射后，再經(jīng)過(guò)點(diǎn)SKIPIF1<0反射后，沿直線(xiàn)SKIPIF1<0射出，經(jīng)過(guò)點(diǎn)SKIPIF1<0，SKIPIF1<0為拋物線(xiàn)焦點(diǎn)，SKIPIF1<0為拋物線(xiàn)SKIPIF1<0上一點(diǎn)，則下列說(shuō)法正確的是()A.SKIPIF1<0的最小值為SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0平分SKIPIF1<0【答案】BCD【解析】【分析】過(guò)SKIPIF1<0作SKIPIF1<0垂直SKIPIF1<0的準(zhǔn)線(xiàn)，垂足為SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0垂直SKIPIF1<0的準(zhǔn)線(xiàn)，垂足為SKIPIF1<0，再根據(jù)拋物的焦半徑公式逐一分析各個(gè)選項(xiàng)即可得出答案.【詳解】解：過(guò)SKIPIF1<0作SKIPIF1<0垂直SKIPIF1<0的準(zhǔn)線(xiàn)SKIPIF1<0，垂足為SKIPIF1<0，所以SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0垂直SKIPIF1<0的準(zhǔn)線(xiàn)SKIPIF1<0，垂足為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0三點(diǎn)共線(xiàn)時(shí)，取等號(hào)，故選項(xiàng)A錯(cuò)誤；因?yàn)镾KIPIF1<0平行SKIPIF1<0軸，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以過(guò)SKIPIF1<0的直線(xiàn)為SKIPIF1<0，聯(lián)立SKIPIF1<0,得SKIPIF1<0，所以SKIPIF1<0，故選項(xiàng)B正確；因?yàn)镾KIPIF1<0可得SKIPIF1<0，或SKIPIF1<0，即SKIPIF1<0，代入SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故選項(xiàng)C正確；因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0平分SKIPIF1<0，故選項(xiàng)D正確．故選：BCD.12.若圓SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在直線(xiàn)SKIPIF1<0上，則()A.圓SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0B.圓SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0C.直線(xiàn)SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0D.直線(xiàn)SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0【答案】ABD【解析】【分析】A選項(xiàng)根據(jù)點(diǎn)到直線(xiàn)的距離公式可求解，B選項(xiàng)當(dāng)SKIPIF1<0與圓相切時(shí)符合題意，C選項(xiàng)利用對(duì)稱(chēng)性可以判斷，D選項(xiàng)當(dāng)SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0時(shí)符合題意.【詳解】對(duì)于A，圓心到直線(xiàn)的距離為SKIPIF1<0，故SKIPIF1<0，圓SKIPIF1<0上存在點(diǎn)SKIPIF1<0使得SKIPIF1<0

，A正確；對(duì)于B，過(guò)SKIPIF1<0作圓SKIPIF1<0的切線(xiàn)，切點(diǎn)為SKIPIF1<0，

則SKIPIF1<0

，故當(dāng)SKIPIF1<0與圓相切時(shí)，

SKIPIF1<0，B正確；對(duì)于C，設(shè)點(diǎn)SKIPIF1<0關(guān)于直線(xiàn)SKIPIF1<0的對(duì)稱(chēng)點(diǎn)為點(diǎn)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，故C錯(cuò)誤；對(duì)于D，當(dāng)SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，故D正確．故選：ABD.三、填空題(本大題共4小題，共20.0分)13.在數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0__________．【答案】46【解析】【分析】利用累加法求解即可.【詳解】由SKIPIF1<0，則有SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<014.過(guò)點(diǎn)SKIPIF1<0的直線(xiàn)SKIPIF1<0，被直線(xiàn)SKIPIF1<0，SKIPIF1<0所截得的線(xiàn)段SKIPIF1<0的中點(diǎn)恰好在直線(xiàn)SKIPIF1<0上，則直線(xiàn)SKIPIF1<0的方程為_(kāi)_________．【答案】SKIPIF1<0【解析】【分析】先求出線(xiàn)段SKIPIF1<0的中點(diǎn)，在求出直線(xiàn)SKIPIF1<0的斜率，最后用點(diǎn)斜式即可求出直線(xiàn)SKIPIF1<0的方程.【詳解】設(shè)SKIPIF1<0中點(diǎn)為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0在直線(xiàn)SKIPIF1<0上，由SKIPIF1<0在直線(xiàn)SKIPIF1<0上，聯(lián)立可得SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0中點(diǎn)為SKIPIF1<0，所以直線(xiàn)SKIPIF1<0的斜率SKIPIF1<0，所以SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0．15.已知橢圓SKIPIF1<0的左、右焦點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0軸的垂線(xiàn)，交橢圓于點(diǎn)SKIPIF1<0，若直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，則橢圓SKIPIF1<0的離心率為_(kāi)_________．【答案】SKIPIF1<0##SKIPIF1<0【解析】【分析】利用橢圓的標(biāo)準(zhǔn)方程和離心率計(jì)算公式求解即可.【詳解】由題意可得SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0軸，且SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0①，又SKIPIF1<0②，①②聯(lián)立得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0(舍去)，故答案為：SKIPIF1<016.若不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是__________．【答案】SKIPIF1<0【解析】【分析】當(dāng)SKIPIF1<0時(shí)顯然成立，當(dāng)SKIPIF1<0時(shí)，構(gòu)造SKIPIF1<0，則原不等式等價(jià)于SKIPIF1<0，利用導(dǎo)函數(shù)求單調(diào)性可得SKIPIF1<0對(duì)SKIPIF1<0恒成立，再構(gòu)造SKIPIF1<0求SKIPIF1<0最大值即可.【詳解】當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0恒成立；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0恒成立；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0對(duì)SKIPIF1<0恒成立，構(gòu)造SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由單調(diào)性可知SKIPIF1<0，整理得SKIPIF1<0對(duì)SKIPIF1<0恒成立，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，綜上，實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】導(dǎo)函數(shù)中常用的兩種轉(zhuǎn)化方法：一是利用導(dǎo)數(shù)研究含參函數(shù)的單調(diào)性，?；癁椴坏仁胶愠闪?wèn)題；二是函數(shù)的零點(diǎn)、不等式證明常轉(zhuǎn)化為函數(shù)的單調(diào)性、極(最)值問(wèn)題處理，本題的關(guān)鍵是利用同構(gòu)的思路，將不等式變形為SKIPIF1<0，再構(gòu)造函數(shù)，問(wèn)題就會(huì)迎刃而解.四、解答題(本大題共6小題，共70.0分.解答應(yīng)寫(xiě)出文字說(shuō)明，證明過(guò)程或演算步驟)17.已知正項(xiàng)等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，SKIPIF1<0，且_________．請(qǐng)?jiān)冖賁KIPIF1<0；②SKIPIF1<0是SKIPIF1<0與SKIPIF1<0等差中項(xiàng)；③SKIPIF1<0，三個(gè)條件中任選一個(gè)補(bǔ)充在上述橫線(xiàn)上，并求解下面的問(wèn)題：(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)若SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)等比數(shù)列基本量的計(jì)算即可逐一求解，(2)根據(jù)裂項(xiàng)求和即可求解.【小問(wèn)1詳解】選SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，不符合題,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0則SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0則SKIPIF1<0負(fù)值舍去SKIPIF1<0,則SKIPIF1<0選SKIPIF1<0由題知SKIPIF1<0即SKIPIF1<0，有SKIPIF1<0,即SKIPIF1<0負(fù)值舍去,那么SKIPIF1<0選SKIPIF1<0即SKIPIF1<0同SKIPIF1<0有SKIPIF1<0【小問(wèn)2詳解】SKIPIF1<0，則SKIPIF1<0SKIPIF1<018.已知函數(shù)SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0處有極值．(1)求函數(shù)SKIPIF1<0的解析式；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的最值．【答案】(1)SKIPIF1<0；(2)最小值為SKIPIF1<0，最大值為SKIPIF1<0．【解析】【分析】(1)因SKIPIF1<0在SKIPIF1<0處有極值，則SKIPIF1<0，得SKIPIF1<0，后檢驗(yàn)SKIPIF1<0滿(mǎn)足題意即可；(2)由(1)，利用導(dǎo)數(shù)可求得SKIPIF1<0在SKIPIF1<0上的最值．【小問(wèn)1詳解】由題，SKIPIF1<0.因SKIPIF1<0在SKIPIF1<0處有極值，則SKIPIF1<0.又SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，因SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0.得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則函數(shù)SKIPIF1<0在SKIPIF1<0處有極大值，滿(mǎn)足題意，故SKIPIF1<0.【小問(wèn)2詳解】當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0.故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.則SKIPIF1<0，SKIPIF1<0.故函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，最小值SKIPIF1<0.19.已知圓SKIPIF1<0：SKIPIF1<0，直線(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0．(1)若直線(xiàn)SKIPIF1<0被圓SKIPIF1<0所截得的弦長(zhǎng)為SKIPIF1<0，求直線(xiàn)SKIPIF1<0的方程；(2)若直線(xiàn)SKIPIF1<0與圓SKIPIF1<0交于另一點(diǎn)SKIPIF1<0，與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，且SKIPIF1<0為SKIPIF1<0的中點(diǎn)，求直線(xiàn)SKIPIF1<0的方程．【答案】(1)SKIPIF1<0或SKIPIF1<0(2)SKIPIF1<0．【解析】【分析】(1)根據(jù)點(diǎn)到直線(xiàn)的距離公式以及圓的弦長(zhǎng)公式即可求解，(2)根據(jù)中點(diǎn)坐標(biāo)公式即可根據(jù)點(diǎn)SKIPIF1<0在圓上求解SKIPIF1<0,進(jìn)而可求直線(xiàn)方程.小問(wèn)1詳解】當(dāng)直線(xiàn)斜率不存在時(shí)，SKIPIF1<0與圓相切不符合題意，舍去．當(dāng)直線(xiàn)斜率存在時(shí)，設(shè)直線(xiàn)SKIPIF1<0,即SKIPIF1<0，圓心坐標(biāo)為SKIPIF1<0，由弦長(zhǎng)為SKIPIF1<0可知，圓心到直線(xiàn)的距離為SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0則直線(xiàn)SKIPIF1<0方程為SKIPIF1<0或SKIPIF1<0【小問(wèn)2詳解】設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0為SKIPIF1<0中點(diǎn)，則SKIPIF1<0，由SKIPIF1<0在圓SKIPIF1<0上得SKIPIF1<0即SKIPIF1<0，則SKIPIF1<0．所以直線(xiàn)SKIPIF1<0即直線(xiàn)SKIPIF1<0．20.已知數(shù)列SKIPIF1<0的各項(xiàng)均為正數(shù)，前SKIPIF1<0項(xiàng)和為SKIPIF1<0，SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)設(shè)SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)SKIPIF1<0的關(guān)系可得SKIPIF1<0，進(jìn)而根據(jù)等差數(shù)列的性質(zhì)即可求解，(2)根據(jù)并項(xiàng)求和以及分組求和即可求解.【小問(wèn)1詳解】由SKIPIF1<0得SKIPIF1<0時(shí)，SKIPIF1<0兩式相減得SKIPIF1<0,整理得SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0,所以數(shù)列SKIPIF1<0是以SKIPIF1<0為公差的等差數(shù)列在SKIPIF1<0中令SKIPIF1<0解得SKIPIF1<0所以SKIPIF1<0【小問(wèn)2詳解】SKIPIF1<0令數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0為偶數(shù)，SKIPIF1<0即SKIPIF1<0所以SKIPIF1<021.設(shè)拋物線(xiàn)SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0，過(guò)SKIPIF1<0的直線(xiàn)交拋物線(xiàn)SKIPIF1<0于SKIPIF1<0兩點(diǎn)，當(dāng)直線(xiàn)SKIPIF1<0軸時(shí)，SKIPIF1<0．(1)求拋物線(xiàn)SKIPIF1<0的方程；(2)設(shè)直線(xiàn)SKIPIF1<0，SKIPIF1<0與拋物線(xiàn)SKIPIF1<0的另一個(gè)交點(diǎn)分別為點(diǎn)SKIPIF1<0，SKIPIF1<0，記直線(xiàn)SKIPIF1<0，SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)2【解析】【分析】(1)首先求出SKIPIF1<0點(diǎn)坐標(biāo)，再根據(jù)拋物線(xiàn)的定義得到方程，求出SKIPIF1<0的值，即可得解；(2)設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立直線(xiàn)與拋物線(xiàn)方程，消元、列出韋達(dá)定理，即可求出SKIPIF1<0，從而得解.【小問(wèn)1詳解】解：當(dāng)直線(xiàn)SKIPIF1<0軸時(shí)，令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，不妨取SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的方程為SKIPIF1<0；【小問(wèn)2詳解】解：設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由題可知直線(xiàn)SKIPIF1<0斜率存在且不為SKIPIF1<0，故設(shè)SKIPIF1<0的方程為SKIPIF1<0聯(lián)立SKIPIF1<0得SKIPIF1<0，則有SKIPIF1<0，SKIPIF1<0，直線(xiàn)SKIPIF1<0方程為SKIPIF1<0，聯(lián)立SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，同理可得SKIPIF1<0，因?yàn)镾KIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0．22.已知函數(shù)SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0且SKIPIF1<0；(i)求SKIPIF1<0的取值范圍；(ii)證明：SKIPIF1<0．【答案】(1)答案見(jiàn)解析；(2)(i)SKIPIF1<0；(ii)證明見(jiàn)解析【解析】【分析】(1)對(duì)SKIPIF1<0求導(dǎo)，利用導(dǎo)函數(shù)的正負(fù)討論單調(diào)性即可；(2)(i)利用SKIPIF1<0