2024年全國(guó)甲卷高考數(shù)學(xué)（理數(shù)）真題試題（原卷版+含解析）

2024年高考全國(guó)甲卷數(shù)學(xué)（理）一、單選題1．設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．10 D．SKIPIF1<02．集合SKIPIF1<0，則?AA∩B=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．若實(shí)數(shù)SKIPIF1<0滿足約束條件SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．25．已知雙曲線SKIPIF1<0的上、下焦點(diǎn)分別為SKIPIF1<0，點(diǎn)SKIPIF1<0在該雙曲線上，則該雙曲線的離心率為（

）A．4 B．3 C．2 D．SKIPIF1<06．設(shè)函數(shù)SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處的切線與兩坐標(biāo)軸圍成的三角形的面積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的大致圖像為（

）A． B．C． D．8．已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．已知向量SKIPIF1<0，則（

）A．“SKIPIF1<0”是“SKIPIF1<0”的必要條件 B．“SKIPIF1<0”是“SKIPIF1<0”的必要條件C．“SKIPIF1<0”是“SKIPIF1<0”的充分條件 D．“SKIPIF1<0”是“SKIPIF1<0”的充分條件10．設(shè)SKIPIF1<0是兩個(gè)平面，SKIPIF1<0是兩條直線，且SKIPIF1<0.下列四個(gè)命題：①若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0

②若SKIPIF1<0，則SKIPIF1<0③若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0

④若SKIPIF1<0與SKIPIF1<0和SKIPIF1<0所成的角相等，則SKIPIF1<0其中所有真命題的編號(hào)是（

）A．①③ B．②④ C．①②③ D．①③④11．在SKIPIF1<0中內(nèi)角SKIPIF1<0所對(duì)邊分別為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．已知b是SKIPIF1<0的等差中項(xiàng)，直線SKIPIF1<0與圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，則SKIPIF1<0的最小值為（

）A．2 B．3 C．4 D．SKIPIF1<0二、填空題13．SKIPIF1<0的展開(kāi)式中，各項(xiàng)系數(shù)的最大值是．14．已知甲、乙兩個(gè)圓臺(tái)上、下底面的半徑均為SKIPIF1<0和SKIPIF1<0，母線長(zhǎng)分別為SKIPIF1<0和SKIPIF1<0，則兩個(gè)圓臺(tái)的體積之比SKIPIF1<0．15．已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．16．有6個(gè)相同的球，分別標(biāo)有數(shù)字1、2、3、4、5、6，從中不放回地隨機(jī)抽取3次，每次取1個(gè)球.記SKIPIF1<0為前兩次取出的球上數(shù)字的平均值，SKIPIF1<0為取出的三個(gè)球上數(shù)字的平均值，則SKIPIF1<0與SKIPIF1<0差的絕對(duì)值不超過(guò)SKIPIF1<0的概率是．三、解答題17．某工廠進(jìn)行生產(chǎn)線智能化升級(jí)改造，升級(jí)改造后，從該工廠甲、乙兩個(gè)車(chē)間的產(chǎn)品中隨機(jī)抽取150件進(jìn)行檢驗(yàn)，數(shù)據(jù)如下：優(yōu)級(jí)品合格品不合格品總計(jì)甲車(chē)間2624050乙車(chē)間70282100總計(jì)96522150(1)填寫(xiě)如下列聯(lián)表：優(yōu)級(jí)品非優(yōu)級(jí)品甲車(chē)間乙車(chē)間能否有SKIPIF1<0的把握認(rèn)為甲、乙兩車(chē)間產(chǎn)品的優(yōu)級(jí)品率存在差異？能否有SKIPIF1<0的把握認(rèn)為甲，乙兩車(chē)間產(chǎn)品的優(yōu)級(jí)品率存在差異？(2)已知升級(jí)改造前該工廠產(chǎn)品的優(yōu)級(jí)品率SKIPIF1<0，設(shè)SKIPIF1<0為升級(jí)改造后抽取的n件產(chǎn)品的優(yōu)級(jí)品率.如果SKIPIF1<0，則認(rèn)為該工廠產(chǎn)品的優(yōu)級(jí)品率提高了，根據(jù)抽取的150件產(chǎn)品的數(shù)據(jù)，能否認(rèn)為生產(chǎn)線智能化升級(jí)改造后，該工廠產(chǎn)品的優(yōu)級(jí)品率提高了？（SKIPIF1<0）附：SKIPIF1<0SKIPIF1<00.0500.0100.001k3.8416.63510.82818．記SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，且SKIPIF1<0．(1)求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)和為SKIPIF1<0．19．如圖，在以A，B，C，D，E，F(xiàn)為頂點(diǎn)的五面體中，四邊形ABCD與四邊形ADEF均為等腰梯形，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的中點(diǎn)．(1)證明：SKIPIF1<0平面SKIPIF1<0；(2)求二面角SKIPIF1<0的正弦值．20．設(shè)橢圓SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0軸．(1)求SKIPIF1<0的方程；(2)過(guò)點(diǎn)SKIPIF1<0的直線與SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，直線SKIPIF1<0交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，證明：SKIPIF1<0軸．21．已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的極值；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．22．在平面直角坐標(biāo)系SKIPIF1<0中，以坐標(biāo)原點(diǎn)SKIPIF1<0為極點(diǎn)，SKIPIF1<0軸的正半軸為極軸建立極坐標(biāo)系，曲線SKIPIF1<0的極坐標(biāo)方程為SKIPIF1<0.(1)寫(xiě)出SKIPIF1<0的直角坐標(biāo)方程；(2)設(shè)直線l：SKIPIF1<0（SKIPIF1<0為參數(shù)），若SKIPIF1<0與l相交于SKIPIF1<0兩點(diǎn)，若SKIPIF1<0，求SKIPIF1<0的值.23．實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0．(1)證明：SKIPIF1<0；(2)證明：SKIPIF1<0．2024年高考全國(guó)甲卷數(shù)學(xué)（理）一、單選題1．設(shè)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．10 D．SKIPIF1<0【答案】A【解析】根據(jù)SKIPIF1<0，則SKIPIF1<0.故選A2．集合SKIPIF1<0，則?AA∩B=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則A∩B=1,4,9，?故選D3．若實(shí)數(shù)SKIPIF1<0滿足約束條件SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，作出可行域如圖：根據(jù)SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0的幾何意義為SKIPIF1<0的截距的SKIPIF1<0，則該直線截距取最大值時(shí)，SKIPIF1<0有最小值，此時(shí)直線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0.故選D.4．等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】B【分析】由SKIPIF1<0結(jié)合等差中項(xiàng)的性質(zhì)可得SKIPIF1<0，即可計(jì)算出公差，即可得SKIPIF1<0的值.【解析】由SKIPIF1<0，則SKIPIF1<0，則等差數(shù)列SKIPIF1<0的公差SKIPIF1<0，故SKIPIF1<0.故選B.5．已知雙曲線SKIPIF1<0的上、下焦點(diǎn)分別為SKIPIF1<0，點(diǎn)SKIPIF1<0在該雙曲線上，則該雙曲線的離心率為（

）A．4 B．3 C．2 D．SKIPIF1<0【答案】C【分析】由焦點(diǎn)坐標(biāo)可得焦距SKIPIF1<0，結(jié)合雙曲線定義計(jì)算可得SKIPIF1<0，即可得離心率.【解析】根據(jù)題意，SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0.故選C.6．設(shè)函數(shù)SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處的切線與兩坐標(biāo)軸圍成的三角形的面積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】借助導(dǎo)數(shù)的幾何意義計(jì)算可得其在點(diǎn)SKIPIF1<0處的切線方程，即可得其與坐標(biāo)軸交點(diǎn)坐標(biāo)，即可得其面積.【解析】SKIPIF1<0，則SKIPIF1<0，即該切線方程為SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以該切線與兩坐標(biāo)軸所圍成的三角形面積SKIPIF1<0.故選A.7．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的大致圖像為（

）A． B．C． D．【答案】B【分析】利用函數(shù)的奇偶性可排除A、C，代入SKIPIF1<0可得SKIPIF1<0，可排除D.【解析】SKIPIF1<0，又函數(shù)定義域?yàn)镾KIPIF1<0，故該函數(shù)為偶函數(shù)，故A、C錯(cuò)誤，又SKIPIF1<0，故D錯(cuò)誤.故選B.8．已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】先將SKIPIF1<0弦化切求得SKIPIF1<0，再根據(jù)兩角和的正切公式即可求解.【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故選B.9．已知向量SKIPIF1<0，則（

）A．“SKIPIF1<0”是“SKIPIF1<0”的必要條件 B．“SKIPIF1<0”是“SKIPIF1<0”的必要條件C．“SKIPIF1<0”是“SKIPIF1<0”的充分條件 D．“SKIPIF1<0”是“SKIPIF1<0”的充分條件【答案】C【分析】根據(jù)向量垂直和平行的坐標(biāo)表示即可得到方程，解出即可.【解析】A，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即必要性不成立，A錯(cuò)誤；B，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，解得SKIPIF1<0，即必要性不成立，B錯(cuò)誤；C，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，即充分性成立，C正確；D，當(dāng)SKIPIF1<0時(shí)，不滿足SKIPIF1<0，所以SKIPIF1<0不成立，即充分性不立，D錯(cuò)誤.故選C.10．設(shè)SKIPIF1<0是兩個(gè)平面，SKIPIF1<0是兩條直線，且SKIPIF1<0.下列四個(gè)命題：①若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0

②若SKIPIF1<0，則SKIPIF1<0③若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0

④若SKIPIF1<0與SKIPIF1<0和SKIPIF1<0所成的角相等，則SKIPIF1<0其中所有真命題的編號(hào)是（

）A．①③ B．②④ C．①②③ D．①③④【答案】A【分析】根據(jù)線面平行的判定定理即可判斷①；舉反例即可判斷②④；根據(jù)線面平行的性質(zhì)即可判斷③.【解析】①，當(dāng)SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0既不在SKIPIF1<0也不在SKIPIF1<0內(nèi)，因?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0，①正確；②，若SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0不一定垂直，②錯(cuò)誤；③，過(guò)直線SKIPIF1<0分別作兩平面與SKIPIF1<0分別相交于直線SKIPIF1<0和直線SKIPIF1<0，因?yàn)镾KIPIF1<0，過(guò)直線SKIPIF1<0的平面與平面SKIPIF1<0的交線為直線SKIPIF1<0，則根據(jù)線面平行的性質(zhì)定理知SKIPIF1<0，同理可得SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0平面SKIPIF1<0，SKIPIF1<0平面SKIPIF1<0，則SKIPIF1<0平面SKIPIF1<0，因?yàn)镾KIPIF1<0平面SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，又因?yàn)镾KIPIF1<0，則SKIPIF1<0，③正確；④，若SKIPIF1<0與SKIPIF1<0和SKIPIF1<0所成的角相等，如果SKIPIF1<0，則SKIPIF1<0，④錯(cuò)誤；①③正確，故選A.11．在SKIPIF1<0中內(nèi)角SKIPIF1<0所對(duì)邊分別為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】利用正弦定理得SKIPIF1<0，再利用余弦定理有SKIPIF1<0，再利用正弦定理得到SKIPIF1<0的值，最后代入計(jì)算即可.【解析】因?yàn)镾KIPIF1<0，由正弦定理得SKIPIF1<0.由余弦定理可得:SKIPIF1<0，即:SKIPIF1<0，根據(jù)正弦定理得SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0為三角形內(nèi)角，則SKIPIF1<0，則SKIPIF1<0.故選C.12．已知b是SKIPIF1<0的等差中項(xiàng)，直線SKIPIF1<0與圓SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，則SKIPIF1<0的最小值為（

）A．2 B．3 C．4 D．SKIPIF1<0【答案】C【分析】結(jié)合等差數(shù)列性質(zhì)將SKIPIF1<0代換，求出直線恒過(guò)的定點(diǎn)，采用數(shù)形結(jié)合法即可求解.【詳解】因?yàn)镾KIPIF1<0成等差數(shù)列，所以SKIPIF1<0，SKIPIF1<0，代入直線方程SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，故直線恒過(guò)SKIPIF1<0，設(shè)SKIPIF1<0，圓化為標(biāo)準(zhǔn)方程得：SKIPIF1<0，設(shè)圓心為SKIPIF1<0，畫(huà)出直線與圓的圖形，由圖可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最小，SKIPIF1<0，此時(shí)SKIPIF1<0.

故選C二、填空題13．SKIPIF1<0的展開(kāi)式中，各項(xiàng)系數(shù)的最大值是．【答案】5【分析】先設(shè)展開(kāi)式中第SKIPIF1<0項(xiàng)系數(shù)最大，則根據(jù)通項(xiàng)公式有SKIPIF1<0，進(jìn)而求出SKIPIF1<0即可求解.【解析】根據(jù)題展開(kāi)式通項(xiàng)公式為SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，設(shè)展開(kāi)式中第SKIPIF1<0項(xiàng)系數(shù)最大，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，故SKIPIF1<0，所以展開(kāi)式中系數(shù)最大的項(xiàng)是第9項(xiàng)，且該項(xiàng)系數(shù)為SKIPIF1<0.答案為：5.14．已知甲、乙兩個(gè)圓臺(tái)上、下底面的半徑均為SKIPIF1<0和SKIPIF1<0，母線長(zhǎng)分別為SKIPIF1<0和SKIPIF1<0，則兩個(gè)圓臺(tái)的體積之比SKIPIF1<0．【答案】SKIPIF1<0【分析】先根據(jù)已知條件和圓臺(tái)結(jié)構(gòu)特征分別求出兩圓臺(tái)的高，再根據(jù)圓臺(tái)的體積公式直接代入計(jì)算即可得解.【解析】根據(jù)題可得兩個(gè)圓臺(tái)的高分別為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.答案為：SKIPIF1<0.15．已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．【答案】64【分析】將SKIPIF1<0利用換底公式轉(zhuǎn)化成SKIPIF1<0來(lái)表示即可求解.【解析】由題SKIPIF1<0，整理得SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0答案為：64.16．有6個(gè)相同的球，分別標(biāo)有數(shù)字1、2、3、4、5、6，從中不放回地隨機(jī)抽取3次，每次取1個(gè)球.記SKIPIF1<0為前兩次取出的球上數(shù)字的平均值，SKIPIF1<0為取出的三個(gè)球上數(shù)字的平均值，則SKIPIF1<0與SKIPIF1<0差的絕對(duì)值不超過(guò)SKIPIF1<0的概率是．【答案】SKIPIF1<0【分析】根據(jù)排列可求基本事件的總數(shù)，設(shè)前兩個(gè)球的號(hào)碼為SKIPIF1<0，第三個(gè)球的號(hào)碼為SKIPIF1<0，則SKIPIF1<0，就SKIPIF1<0的不同取值分類(lèi)討論后可求隨機(jī)事件的概率.【解析】從6個(gè)不同的球中不放回地抽取3次，共有SKIPIF1<0種，設(shè)前兩個(gè)球的號(hào)碼為SKIPIF1<0，第三個(gè)球的號(hào)碼為SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0為：SKIPIF1<0，因此有2種，若SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0為：SKIPIF1<0，SKIPIF1<0，因此有10種，當(dāng)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0為：SKIPIF1<0，SKIPIF1<0，因此有16種，當(dāng)SKIPIF1<0，則SKIPIF1<0，同理有16種，當(dāng)SKIPIF1<0，則SKIPIF1<0，同理有10種，當(dāng)SKIPIF1<0，則SKIPIF1<0，同理有2種，共SKIPIF1<0與SKIPIF1<0的差的絕對(duì)值不超過(guò)SKIPIF1<0時(shí)不同的抽取方法總數(shù)為SKIPIF1<0，因此所求概率為SKIPIF1<0.答案為：SKIPIF1<0三、解答題17．某工廠進(jìn)行生產(chǎn)線智能化升級(jí)改造，升級(jí)改造后，從該工廠甲、乙兩個(gè)車(chē)間的產(chǎn)品中隨機(jī)抽取150件進(jìn)行檢驗(yàn)，數(shù)據(jù)如下：優(yōu)級(jí)品合格品不合格品總計(jì)甲車(chē)間2624050乙車(chē)間70282100總計(jì)96522150(1)填寫(xiě)如下列聯(lián)表：優(yōu)級(jí)品非優(yōu)級(jí)品甲車(chē)間乙車(chē)間能否有SKIPIF1<0的把握認(rèn)為甲、乙兩車(chē)間產(chǎn)品的優(yōu)級(jí)品率存在差異？能否有SKIPIF1<0的把握認(rèn)為甲，乙兩車(chē)間產(chǎn)品的優(yōu)級(jí)品率存在差異？(2)已知升級(jí)改造前該工廠產(chǎn)品的優(yōu)級(jí)品率SKIPIF1<0，設(shè)SKIPIF1<0為升級(jí)改造后抽取的n件產(chǎn)品的優(yōu)級(jí)品率.如果SKIPIF1<0，則認(rèn)為該工廠產(chǎn)品的優(yōu)級(jí)品率提高了，根據(jù)抽取的150件產(chǎn)品的數(shù)據(jù)，能否認(rèn)為生產(chǎn)線智能化升級(jí)改造后，該工廠產(chǎn)品的優(yōu)級(jí)品率提高了？（SKIPIF1<0）附：SKIPIF1<0SKIPIF1<00.0500.0100.001k3.8416.63510.828【答案】(1)答案見(jiàn)詳解(2)答案見(jiàn)詳解【分析】（1）根據(jù)題中數(shù)據(jù)完善列聯(lián)表，計(jì)算SKIPIF1<0，并與臨界值對(duì)比分析；（2）用頻率估計(jì)概率可得SKIPIF1<0，根據(jù)題意計(jì)算SKIPIF1<0，結(jié)合題意分析判斷.【解析】（1）根據(jù)題意可得列聯(lián)表：優(yōu)級(jí)品非優(yōu)級(jí)品甲車(chē)間2624乙車(chē)間7030可得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以有SKIPIF1<0的把握認(rèn)為甲、乙兩車(chē)間產(chǎn)品的優(yōu)級(jí)品率存在差異，沒(méi)有SKIPIF1<0的把握認(rèn)為甲，乙兩車(chē)間產(chǎn)品的優(yōu)級(jí)品率存在差異.（2）根據(jù)題意可知：生產(chǎn)線智能化升級(jí)改造后，該工廠產(chǎn)品的優(yōu)級(jí)品的頻率為SKIPIF1<0，用頻率估計(jì)概率可得SKIPIF1<0，又因?yàn)樯?jí)改造前該工廠產(chǎn)品的優(yōu)級(jí)品率SKIPIF1<0，則SKIPIF1<0，可知SKIPIF1<0，所以可以認(rèn)為生產(chǎn)線智能化升級(jí)改造后，該工廠產(chǎn)品的優(yōu)級(jí)品率提高了.18．記SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，且SKIPIF1<0．(1)求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)和為SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）利用退位法可求SKIPIF1<0的通項(xiàng)公式．（2）利用錯(cuò)位相減法可求SKIPIF1<0.【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，而SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，∴數(shù)列SKIPIF1<0是以4為首項(xiàng)，SKIPIF1<0為公比的等比數(shù)列，所以SKIPIF1<0.（2）SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0故SKIPIF1<0所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0.19．如圖，在以A，B，C，D，E，F(xiàn)為頂點(diǎn)的五面體中，四邊形ABCD與四邊形ADEF均為等腰梯形，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的中點(diǎn)．(1)證明：SKIPIF1<0平面SKIPIF1<0；(2)求二面角SKIPIF1<0的正弦值．【答案】(1)見(jiàn)詳解；(2)SKIPIF1<0【分析】（1）結(jié)合已知易證四邊形SKIPIF1<0為平行四邊形，可證SKIPIF1<0，進(jìn)而得證；（2）作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，連接SKIPIF1<0，易證SKIPIF1<0三垂直，采用建系法結(jié)合二面角夾角余弦公式即可求解.【解析】（1）因?yàn)镾KIPIF1<0為SKIPIF1<0的中點(diǎn)，所以SKIPIF1<0，四邊形SKIPIF1<0為平行四邊形，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0平面SKIPIF1<0，SKIPIF1<0平面SKIPIF1<0，所以SKIPIF1<0平面SKIPIF1<0；（2）如圖所示，作SKIPIF1<0交SKIPIF1<0于SKIPIF1<0，連接SKIPIF1<0，因?yàn)樗倪呅蜸KIPIF1<0為等腰梯形，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，結(jié)合（1）SKIPIF1<0為平行四邊形，可得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0為等邊三角形，SKIPIF1<0為SKIPIF1<0中點(diǎn)，所以SKIPIF1<0，又因?yàn)樗倪呅蜸KIPIF1<0為等腰梯形，SKIPIF1<0為SKIPIF1<0中點(diǎn)，所以SKIPIF1<0，四邊形SKIPIF1<0為平行四邊形，SKIPIF1<0，所以SKIPIF1<0為等腰三角形，SKIPIF1<0與SKIPIF1<0底邊上中點(diǎn)SKIPIF1<0重合，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0互相垂直，以SKIPIF1<0方向?yàn)镾KIPIF1<0軸，SKIPIF1<0方向?yàn)镾KIPIF1<0軸，SKIPIF1<0方向?yàn)镾KIPIF1<0軸，建立SKIPIF1<0空間直角坐標(biāo)系，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，設(shè)平面SKIPIF1<0的法向量為SKIPIF1<0，平面SKIPIF1<0的法向量為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故二面角SKIPIF1<0的正弦值為SKIPIF1<0.20．設(shè)橢圓SKIPIF1<0的右焦點(diǎn)為SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0上，且SKIPIF1<0軸．(1)求SKIPIF1<0的方程；(2)過(guò)點(diǎn)SKIPIF1<0的直線與SKIPIF1<0交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0為線段SKIPIF1<0的中點(diǎn)，直線SKIPIF1<0交直線SKIPIF1<0于點(diǎn)SKIPIF1<0，證明：SKIPIF1<0軸．【答案】(1)SKIPIF1<0(2)證明見(jiàn)解析【分析】（1）設(shè)SKIPIF1<0，根據(jù)SKIPIF1<0的坐標(biāo)及SKIPIF1<0SKIPIF1<0軸可求基本量，故可求橢圓方程.（2）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，聯(lián)立直線方程和橢圓方程，用SKIPIF1<0的坐標(biāo)表示SKIPIF1<0，結(jié)合韋達(dá)定理化簡(jiǎn)前者可得SKIPIF1<0，故可證SKIPIF1<0軸.【解析】（1）設(shè)SKIPIF1<0，由題設(shè)有SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故橢圓方程為SKIPIF1<0.（2）直線SKIPIF1<0的斜率必定存在，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，又SKIPIF1<0，而SKIPIF1<0，故直線SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0軸.21．已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的極值；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．【答案】(1)極小值為SKIPIF1<0，無(wú)極大值.(2)SKIPIF1<0【分析】（1）求出函數(shù)的導(dǎo)數(shù)，根據(jù)導(dǎo)數(shù)的單調(diào)性和零點(diǎn)可求函數(shù)的極值.（2）求出函數(shù)的二階導(dǎo)數(shù)，就SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分類(lèi)討論后可得參數(shù)的取值范圍.【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，而SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0處取極小值且極小值為SKIPIF1<0，無(wú)極大值.（2）SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)