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宣城市2022-2023學(xué)年度第一學(xué)期期末調(diào)研測(cè)試高二數(shù)學(xué)試題注意事項(xiàng):1.答卷前,考生務(wù)必將自己的姓名、考生號(hào)等填寫在答題卡和試卷指定位置上.2.回答選擇題時(shí),選出每小題答案后,用鉛筆把答題卡對(duì)應(yīng)題目的答案標(biāo)號(hào)涂黑.如需改動(dòng),用橡皮擦干凈后,再選涂其他答案標(biāo)號(hào).回答非選擇題時(shí),將答案寫在答題卡上.寫在本試卷上無效.3.考試結(jié)束后,將本試卷和答題卡一并交回.一、單選題(本題共8小題,每小題5分,共40分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.在數(shù)列SKIPIF1<0中,已知SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.5【答案】C【解析】【分析】由遞推關(guān)系依次求SKIPIF1<0即可.【詳解】因?yàn)楫?dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,故選:C.2.已知直線SKIPIF1<0的傾斜角為SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由直線方程求得SKIPIF1<0,可判斷出SKIPIF1<0為鈍角,再利用同角三角函數(shù)的基本關(guān)系可求得SKIPIF1<0的值.【詳解】由題意可知,直線SKIPIF1<0的斜率為SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0為鈍角,則SKIPIF1<0,由同角三角函數(shù)的基本關(guān)系可得SKIPIF1<0,解得SKIPIF1<0.故選:A.【點(diǎn)睛】本題考查直線的傾斜角,考查直線傾斜角與斜率的關(guān)系,考查三角函數(shù)值的求法,是基礎(chǔ)題.3.數(shù)學(xué)與建筑的結(jié)合造就建筑藝術(shù)品,如吉林大學(xué)的校門是一拋物線形水泥建筑物,如圖.若將該大學(xué)的校門輪廓(忽略水泥建筑的厚度)近似看成拋物線SKIPIF1<0的一部分,且點(diǎn)SKIPIF1<0在該拋物線上,則該拋物線的焦點(diǎn)坐標(biāo)是()A.SKIPIF1<0 B.(0,-1) C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)點(diǎn)SKIPIF1<0的坐標(biāo)求得SKIPIF1<0,由此求得拋物線的焦點(diǎn)坐標(biāo).【詳解】依題意SKIPIF1<0在拋物線SKIPIF1<0上,所以SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0,且拋物線開口向下,所以拋物線的焦點(diǎn)坐標(biāo)為SKIPIF1<0.故選:A4.如圖所示,在平行六面體SKIPIF1<0中,SKIPIF1<0為SKIPIF1<0與SKIPIF1<0交點(diǎn),若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)空間向量基本定理結(jié)合空間向量線性運(yùn)算求解.【詳解】由題意可得:SKIPIF1<0SKIPIF1<0.故選:D.5.已知等比數(shù)列SKIPIF1<0的各項(xiàng)都是正數(shù),其公比為4,且SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)等比數(shù)列的性質(zhì)求解即可.【詳解】解:根據(jù)等比數(shù)列性質(zhì),有SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,因?yàn)榈缺葦?shù)列SKIPIF1<0的公比為SKIPIF1<0,所以,SKIPIF1<0.故選:C6.古希臘數(shù)學(xué)家阿波羅尼奧斯(約公元前262~公元前190年)的著作《圓錐曲線論》是古代世界光輝的科學(xué)成果,他證明過這樣一個(gè)命題:平面內(nèi)與兩定點(diǎn)距離的比為常數(shù)k(k>0,k≠1)的點(diǎn)的軌跡是圓,后人將這個(gè)圓稱為阿波羅尼斯圓.在平面直角坐標(biāo)系中,設(shè)A(﹣3,0),B(3,0),動(dòng)點(diǎn)M滿足SKIPIF1<0=2,則動(dòng)點(diǎn)M的軌跡方程為A.(x﹣5)2+y2=16 B.x2+(y﹣5)2=9C.(x+5)2+y2=16 D.x2+(y+5)2=9【答案】A【解析】【分析】首先設(shè)SKIPIF1<0,代入兩點(diǎn)間的距離求SKIPIF1<0和SKIPIF1<0,最后整理方程.【詳解】解析:設(shè)SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,可得:(x+3)2+y2=4(x﹣3)2+4y2,即x2﹣10x+y2+9=0整理得SKIPIF1<0,故動(dòng)點(diǎn)SKIPIF1<0的軌跡方程為SKIPIF1<0.選A.【點(diǎn)睛】本題考查了軌跡方程的求解方法,其中屬于直接法,一般軌跡方程的求解有1.直接法,2.代入法,3.定義法,4.參數(shù)法.7.已知正四面體ABCD的棱長(zhǎng)為a,點(diǎn)E,F(xiàn)分別是BC,AD的中點(diǎn),則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)向量的線性運(yùn)算得出SKIPIF1<0,SKIPIF1<0,根據(jù)正四面體的性質(zhì)得出SKIPIF1<0,且SKIPIF1<0、SKIPIF1<0、SKIPIF1<0三向量?jī)蓛蓨A角為SKIPIF1<0,即可通過向量數(shù)量積的運(yùn)算率得出答案.【詳解】SKIPIF1<0四面體ABCD是正四面體,SKIPIF1<0,且SKIPIF1<0、SKIPIF1<0、SKIPIF1<0三向量?jī)蓛蓨A角SKIPIF1<0,SKIPIF1<0點(diǎn)E,F(xiàn)分別是BC,AD的中點(diǎn),SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,故選:C.8.已知雙曲線SKIPIF1<0的左右焦點(diǎn)分別為SKIPIF1<0,直線SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0且與該雙曲線的右支交于SKIPIF1<0兩點(diǎn),若△SKIPIF1<0的周長(zhǎng)為SKIPIF1<0,則該雙曲線離心率的取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)雙曲線定義及焦點(diǎn)三角形周長(zhǎng)、焦點(diǎn)弦的性質(zhì)有SKIPIF1<0,即可求離心率范圍.【詳解】根據(jù)雙曲線定義知:△SKIPIF1<0的周長(zhǎng)為SKIPIF1<0,而SKIPIF1<0,所以SKIPIF1<0,而△SKIPIF1<0的周長(zhǎng)為SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,雙曲線離心率的取值范圍是SKIPIF1<0.故選:A二、多選題(本題共4小題,每小題5分,共20分.在每小題給出的選項(xiàng)中,有多項(xiàng)符合題目要求.全部選對(duì)的得5分,部分選對(duì)的得2分,有選錯(cuò)的得0分.)9.已知等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則()A.數(shù)列SKIPIF1<0是遞減數(shù)列 B.SKIPIF1<0C.SKIPIF1<0時(shí),n的最大值是18 D.SKIPIF1<0【答案】BC【解析】【分析】根據(jù)等差數(shù)列的性質(zhì)和前n項(xiàng)求和公式可得SKIPIF1<0、SKIPIF1<0,結(jié)合通項(xiàng)公式和前n項(xiàng)求和公式計(jì)算,依次判斷選項(xiàng)即可.【詳解】設(shè)等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0.A:由SKIPIF1<0,可得SKIPIF1<0所以等差數(shù)列SKIPIF1<0為遞增數(shù)列,故A錯(cuò)誤;B:SKIPIF1<0,故B正確;C:SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以n的最大值是18,故C正確;D:SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,故D錯(cuò)誤.故選:BC.10.圓SKIPIF1<0,直線SKIPIF1<0,點(diǎn)SKIPIF1<0在圓SKIPIF1<0上,點(diǎn)SKIPIF1<0在直線SKIPIF1<0上,則下列結(jié)論正確的是()A.圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱B.SKIPIF1<0的最大值是9C.從SKIPIF1<0點(diǎn)向圓SKIPIF1<0引切線,切線長(zhǎng)的最小值是3D.直線SKIPIF1<0被圓SKIPIF1<0截得的弦長(zhǎng)取值范圍為SKIPIF1<0【答案】CD【解析】【分析】根據(jù)SKIPIF1<0不在直線SKIPIF1<0上判斷A;根據(jù)SKIPIF1<0判斷B;根據(jù)SKIPIF1<0時(shí),切線長(zhǎng)最小求解判斷C;根據(jù)直線SKIPIF1<0過定點(diǎn)SKIPIF1<0,再結(jié)合弦長(zhǎng)公式判斷D.【詳解】解:對(duì)于A選項(xiàng),SKIPIF1<0圓SKIPIF1<0,∴圓心SKIPIF1<0,半徑SKIPIF1<0,∵SKIPIF1<0,∴圓SKIPIF1<0不關(guān)于直線SKIPIF1<0對(duì)稱,故A選項(xiàng)錯(cuò)誤;對(duì)于B選項(xiàng),由圓心SKIPIF1<0到直線SKIPIF1<0的距離為:SKIPIF1<0,SKIPIF1<0SKIPIF1<0的最小值是SKIPIF1<0,故SKIPIF1<0,故B選項(xiàng)錯(cuò)誤;對(duì)于C選項(xiàng),從SKIPIF1<0點(diǎn)向圓SKIPIF1<0引切線,當(dāng)SKIPIF1<0時(shí),切線長(zhǎng)最小,最小值是SKIPIF1<0,故C正確;對(duì)于D選項(xiàng),直線SKIPIF1<0過定點(diǎn)SKIPIF1<0,該定點(diǎn)在圓C內(nèi),所以直線SKIPIF1<0被圓SKIPIF1<0截得的弦長(zhǎng)最長(zhǎng)時(shí),所截弦長(zhǎng)為過點(diǎn)SKIPIF1<0和圓心的圓SKIPIF1<0的直徑,即弦長(zhǎng)的最大值為8,最短的弦長(zhǎng)為垂直與該直徑的弦長(zhǎng),SKIPIF1<0和圓心SKIPIF1<0的距離為SKIPIF1<0,最短弦長(zhǎng)為SKIPIF1<0,故直線SKIPIF1<0被圓SKIPIF1<0截得的弦長(zhǎng)取值范圍為SKIPIF1<0,D正確.故選:CD.11.如圖,在長(zhǎng)方體SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,E為棱SKIPIF1<0的中點(diǎn),則()A.SKIPIF1<0面SKIPIF1<0 B.SKIPIF1<0C.平面SKIPIF1<0截該長(zhǎng)方體所得截面面積為SKIPIF1<0 D.三棱錐SKIPIF1<0的體積為SKIPIF1<0【答案】ABD【解析】【分析】對(duì)于A:根據(jù)長(zhǎng)方體的性質(zhì)得出SKIPIF1<0,即可證明;對(duì)于B:根據(jù)底面SKIPIF1<0是正方體,得出SKIPIF1<0,根據(jù)三垂線定理結(jié)合長(zhǎng)方體性質(zhì)即可證明;對(duì)于C:根據(jù)長(zhǎng)方體對(duì)稱性易知平面SKIPIF1<0截該長(zhǎng)方體所得截面面積為SKIPIF1<0,根據(jù)已知得出SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,即可根據(jù)余弦定理得出SKIPIF1<0,即可根據(jù)同角三角函數(shù)公式得出SKIPIF1<0,即可根據(jù)三角形面積公式得出答案驗(yàn)證;對(duì)于D:根據(jù)已知直接利用三棱錐的體積公式得出答案;【詳解】對(duì)于選項(xiàng)A:連接SKIPIF1<0,SKIPIF1<0為長(zhǎng)方體,SKIPIF1<0,SKIPIF1<0,∴四邊形SKIPIF1<0是平行四邊形,SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0面SKIPIF1<0,故選項(xiàng)A正確;對(duì)于選項(xiàng)B:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0在平面SKIPIF1<0上的投影為SKIPIF1<0,SKIPIF1<0,故選項(xiàng)B正確;對(duì)于選項(xiàng)C:根據(jù)長(zhǎng)方體對(duì)稱性易知平面SKIPIF1<0截該長(zhǎng)方體所得截面面積為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,由SKIPIF1<0,SKIPIF1<0可得SKIPIF1<0,則SKIPIF1<0,故C錯(cuò)誤;對(duì)于選項(xiàng)D:三棱錐SKIPIF1<0的底面積SKIPIF1<0,高為SKIPIF1<0,則三棱錐SKIPIF1<0的體積為SKIPIF1<0,故D正確;故選:ABD.12.已知SKIPIF1<0為坐標(biāo)原點(diǎn),SKIPIF1<0,SKIPIF1<0分別是漸近線方程為SKIPIF1<0的雙曲線SKIPIF1<0的左、右焦點(diǎn),SKIPIF1<0為雙曲線SKIPIF1<0上任意一點(diǎn),SKIPIF1<0平分SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,則()A.雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0B.雙曲線SKIPIF1<0的離心率為SKIPIF1<0C.點(diǎn)SKIPIF1<0到兩條漸近線的距離之積為SKIPIF1<0D.若直線SKIPIF1<0與雙曲線SKIPIF1<0的另一支交于點(diǎn)SKIPIF1<0為SKIPIF1<0的中點(diǎn),則SKIPIF1<0【答案】BCD【解析】【分析】不妨設(shè)SKIPIF1<0為雙曲線SKIPIF1<0的右支上一點(diǎn),延長(zhǎng)SKIPIF1<0,SKIPIF1<0交于點(diǎn)SKIPIF1<0,進(jìn)而得SKIPIF1<0,SKIPIF1<0,再結(jié)合雙曲線的定義,中位線定理得SKIPIF1<0,SKIPIF1<0,進(jìn)而判斷AB;設(shè)SKIPIF1<0,則SKIPIF1<0,再直接計(jì)算點(diǎn)SKIPIF1<0到兩條漸近線的距離之積判斷C;設(shè)SKIPIF1<0,SKIPIF1<0,根據(jù)點(diǎn)差法求解判斷D.【詳解】解:不妨設(shè)SKIPIF1<0為雙曲線SKIPIF1<0的右支上一點(diǎn),延長(zhǎng)SKIPIF1<0,SKIPIF1<0交于點(diǎn)SKIPIF1<0,如圖,因?yàn)镾KIPIF1<0平分SKIPIF1<0,且SKIPIF1<0,即SKIPIF1<0,所以,在SKIPIF1<0與SKIPIF1<0中,SKIPIF1<0,所以,SKIPIF1<0≌SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0根據(jù)雙曲線的定義得,SKIPIF1<0,在SKIPIF1<0中,SKIPIF1<0為其中位線,SKIPIF1<0所以,SKIPIF1<0,因?yàn)殡p曲線SKIPIF1<0的漸近線方程為SKIPIF1<0,所以SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0所以雙曲線SKIPIF1<0的標(biāo)準(zhǔn)方程為SKIPIF1<0,離心率為SKIPIF1<0,所以A不正確,B正確;設(shè)SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0所以,點(diǎn)SKIPIF1<0到兩條漸近線的距離之積為SKIPIF1<0,所以C正確;設(shè)SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0在雙曲線SKIPIF1<0上,所以SKIPIF1<0①,SKIPIF1<0②,①-②并整理得,SKIPIF1<0,即SKIPIF1<0,因?yàn)镾KIPIF1<0所以SKIPIF1<0,所以D正確.故選:BCD.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛:本題解題的關(guān)鍵在于延長(zhǎng)SKIPIF1<0,SKIPIF1<0交于點(diǎn)SKIPIF1<0,進(jìn)而結(jié)合幾何關(guān)系得到SKIPIF1<0為SKIPIF1<0的中點(diǎn),進(jìn)而求得雙曲線的解析式.三、填空題(本題共4小題,每小題5分,共20分.)13.若直線SKIPIF1<0與直線SKIPIF1<0平行,則SKIPIF1<0______.【答案】SKIPIF1<0【解析】【分析】根據(jù)兩條直線平行列方程,由此求得SKIPIF1<0的值.【詳解】解:將直線SKIPIF1<0變形為SKIPIF1<0,因?yàn)橹本€SKIPIF1<0與直線SKIPIF1<0平行,所以SKIPIF1<0,解得SKIPIF1<0.故答案為:SKIPIF1<014.數(shù)列SKIPIF1<0是等差數(shù)列,且SKIPIF1<0,SKIPIF1<0,那么SKIPIF1<0______.【答案】SKIPIF1<0【解析】【分析】根據(jù)等差數(shù)列的通項(xiàng)公式,進(jìn)而寫出數(shù)列SKIPIF1<0的通項(xiàng)公式,可得答案.【詳解】解:令SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的公差為SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0,所以SKIPIF1<0.故答案為:SKIPIF1<0.15.若圓SKIPIF1<0與圓SKIPIF1<0恰有兩條公切線,則實(shí)數(shù)a的取值范圍為______.【答案】SKIPIF1<0【解析】【分析】由題知圓SKIPIF1<0與圓SKIPIF1<0相交,進(jìn)而根據(jù)位置關(guān)系求解即可.【詳解】解:由題知圓SKIPIF1<0的圓心為SKIPIF1<0,半徑為SKIPIF1<0,圓SKIPIF1<0的圓心為SKIPIF1<0,半徑為SKIPIF1<0,因?yàn)閳ASKIPIF1<0與圓SKIPIF1<0恰有兩條公切線,所以圓SKIPIF1<0與圓SKIPIF1<0相交,所以SKIPIF1<0,所以,SKIPIF1<0,解得SKIPIF1<0.所以,實(shí)數(shù)a的取值范圍為SKIPIF1<0故答案為:SKIPIF1<016.在四棱錐SKIPIF1<0中,SKIPIF1<0平面BCDE,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,則該四棱錐的外接球的表面積為______.【答案】SKIPIF1<0【解析】【分析】連接SKIPIF1<0,由題意可得SKIPIF1<0在直徑為SKIPIF1<0的圓上,在SKIPIF1<0中,由余弦定理可得到SKIPIF1<0,即可得到底面外接圓的半徑,再利用SKIPIF1<0平面BCDE可得球心到底面的距離,即可求解【詳解】連接SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0在直徑為SKIPIF1<0的圓上,取SKIPIF1<0的中點(diǎn)SKIPIF1<0,即四邊形SKIPIF1<0外接圓的圓心,在SKIPIF1<0中,SKIPIF1<0即SKIPIF1<0,解得SKIPIF1<0,所以四邊形SKIPIF1<0外接圓的直徑即SKIPIF1<0外接圓的直徑為SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0平面BCDE,所以四棱錐的外接球的球心SKIPIF1<0與底面SKIPIF1<0的距離為SKIPIF1<0,所以四棱錐的外接球的半徑為SKIPIF1<0,對(duì)應(yīng)的表面積為SKIPIF1<0故答案為:SKIPIF1<0四、解答題(本題共6小題,共70分,其中第17題10分,其它每題12分,解答應(yīng)寫出文字說明證明過程或演算步驟)17.已知等差數(shù)列SKIPIF1<0滿足SKIPIF1<0,且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)首先根據(jù)已知條件列方程求出SKIPIF1<0,再根據(jù)等差數(shù)列通項(xiàng)公式求SKIPIF1<0即得;(2)由題可得SKIPIF1<0,再利用裂項(xiàng)相消法求和即得.【小問1詳解】設(shè)等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0,∵SKIPIF1<0,則由SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0;【小問2詳解】由題可得SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0.18.已知在四棱錐SKIPIF1<0中,底面SKIPIF1<0為正方形,側(cè)棱SKIPIF1<0平面SKIPIF1<0,點(diǎn)SKIPIF1<0為SKIPIF1<0中點(diǎn),SKIPIF1<0.(1)求證:直線SKIPIF1<0平面SKIPIF1<0;(2)求點(diǎn)SKIPIF1<0到平面SKIPIF1<0的距離.【答案】(1)證明見解析(2)SKIPIF1<0【解析】【分析】(1)連接SKIPIF1<0交SKIPIF1<0于點(diǎn)SKIPIF1<0,連接SKIPIF1<0,進(jìn)而根據(jù)SKIPIF1<0即可證明;(2)根據(jù)題意,以SKIPIF1<0為坐標(biāo)原點(diǎn),建立如圖所示的空間直角坐標(biāo)系,利用坐標(biāo)法求解即可.【小問1詳解】證明:連接SKIPIF1<0交SKIPIF1<0于點(diǎn)SKIPIF1<0,連接SKIPIF1<0,因?yàn)榈酌鍿KIPIF1<0為正方形,所以SKIPIF1<0為SKIPIF1<0的中點(diǎn),所以,在SKIPIF1<0中,SKIPIF1<0為SKIPIF1<0的中點(diǎn),SKIPIF1<0為SKIPIF1<0的中點(diǎn),所以SKIPIF1<0;又因?yàn)镾KIPIF1<0面SKIPIF1<0,SKIPIF1<0面SKIPIF1<0,所以SKIPIF1<0平面SKIPIF1<0.【小問2詳解】解:因?yàn)镾KIPIF1<0平面SKIPIF1<0,SKIPIF1<0為正方形,SKIPIF1<0平面SKIPIF1<0,所以,SKIPIF1<0兩兩垂直,以SKIPIF1<0為坐標(biāo)原點(diǎn),建立如圖所示的空間直角坐標(biāo)系,所以,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,設(shè)平面SKIPIF1<0的法向量為SKIPIF1<0,所以,SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,設(shè)點(diǎn)P到平面MAC的距離為d,所以SKIPIF1<0,所以,點(diǎn)SKIPIF1<0到平面SKIPIF1<0的距離為SKIPIF1<0.19.已知拋物線C:SKIPIF1<0的焦點(diǎn)為F,直線l過點(diǎn)SKIPIF1<0,交拋物線于A、B兩點(diǎn).(1)若P為SKIPIF1<0中點(diǎn),求l的方程;(2)求SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)方法一:利用點(diǎn)差法求中點(diǎn)弦所在直線斜率,再根據(jù)點(diǎn)斜式得結(jié)果;注意驗(yàn)證所求直線與拋物線有兩個(gè)交點(diǎn);方法二:設(shè)中點(diǎn)弦所在直線方程,與拋物線方程聯(lián)立,利用韋達(dá)定理以及中點(diǎn)坐標(biāo)公式求中點(diǎn)弦所在直線斜率,再根據(jù)點(diǎn)斜式得結(jié)果;注意考慮中點(diǎn)弦直線斜率不存在的情況是否滿足題意;(2)由拋物線的定義轉(zhuǎn)化SKIPIF1<0,方法一:設(shè)直線l:SKIPIF1<0,與拋物線方程聯(lián)立,利用韋達(dá)定理以及二次函數(shù)性質(zhì)求最值,注意比較直線斜率不存在的情況SKIPIF1<0的值;方法二:設(shè)直線l:SKIPIF1<0,與拋物線方程聯(lián)立,利用韋達(dá)定理以及二次函數(shù)性質(zhì)求最值,此種設(shè)法已包含直線斜率不存在的情況.【詳解】解:(1)方法一:設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0,因?yàn)镾KIPIF1<0的中點(diǎn)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,∴l(xiāng)的方程為SKIPIF1<0,即SKIPIF1<0.經(jīng)檢驗(yàn),符合題意.方法二:設(shè)SKIPIF1<0,SKIPIF1<0,當(dāng)斜率不存時(shí),顯然不成立.當(dāng)斜率存在時(shí),設(shè)直線l:SKIPIF1<0,顯然SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0易知SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0的中點(diǎn)為SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,∴l(xiāng)的方程為SKIPIF1<0(2)方法一:由拋物線的定義可知SKIPIF1<0當(dāng)斜率不存在時(shí),直線l:SKIPIF1<0,SKIPIF1<0當(dāng)斜率存在時(shí),設(shè)直線l:SKIPIF1<0,顯然SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,易知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0的最小值為SKIPIF1<0綜上,SKIPIF1<0的最小值為SKIPIF1<0方法二:由拋物線的定義可知SKIPIF1<0顯然直線l不平行于x軸,設(shè)直線l:SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,易知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0時(shí),SKIPIF1<0的最小值為SKIPIF1<0【點(diǎn)睛】本題考查拋物線的定義、直線與拋物線的位置關(guān)系;考查數(shù)形結(jié)合、分類討論以及函數(shù)方程等數(shù)學(xué)思想;考查邏輯推理、直觀想象以及數(shù)學(xué)運(yùn)算等核心素養(yǎng).20.已知數(shù)列SKIPIF1<0是公差不為零的等差數(shù)列,SKIPIF1<0且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,在①SKIPIF1<0,SKIPIF1<0;②SKIPIF1<0,SKIPIF1<0;③SKIPIF1<0,SKIPIF1<0這三個(gè)條件中任選一個(gè),將序號(hào)補(bǔ)充在下面橫線處,并根據(jù)題意解決問題.問題:若SKIPIF1<0,且______,求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0.注:如果選擇多個(gè)條件分別解答,按第一個(gè)解答給分.【答案】(1)SKIPIF1<0(2)答案見解析【解析】【分析】(1)根據(jù)等比中項(xiàng)性質(zhì),結(jié)合等差數(shù)列通項(xiàng)公式得SKIPIF1<0,再求通項(xiàng)公式即可;(2)根據(jù)題意求得SKIPIF1<0,再根據(jù)錯(cuò)位相減法求解即可.【小問1詳解】解:設(shè)等差數(shù)列的公差為d,因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,所以SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0(舍去).所以,SKIPIF1<0.【小問2詳解】解:選①,由SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)等式也成立,所以SKIPIF1<0,則SKIPIF1<0,所以,SKIPIF1<0,SKIPIF1<0,兩式相減得SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0.選②,由SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,所以數(shù)列SKIPIF1<0為以1為首項(xiàng)2為公比的等比數(shù)列,所以SKIPIF1<0,則SKIPIF1<0,所以,SKIPIF1<0,SKIPIF1<0,兩式相減得SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0選③,由SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0是以2為首項(xiàng),公比為2的等比數(shù)列,所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)等式也成立,所以SKIPIF1<0,則SKIPIF1<0,所以,SKIPIF1<0,SKIPIF1<0,兩式相減得SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0.21.如圖,在正三棱柱SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0是棱SKIPIF1<0的中點(diǎn).(1)證明:平面SKIPIF1<0平面SKIPIF1<0;(2)若SKIPIF1<0,求平面SKIPIF1<0與平面SKIPIF1<0的夾角余弦值的取值范圍.【答案】(1)證明見解析(2)SKIPIF1<0【解析】【分析】(1)證明SKIPIF1<0平面SKIPIF1<0即可證明結(jié)論;(2)分別取SKIPIF1<0,SKIPIF1<0的中點(diǎn)SKIPIF1<0,連接SKIPIF1<0,進(jìn)而SKIPIF1<0兩兩垂直,如圖建立空間直角坐標(biāo)系,利用坐標(biāo)法求解即可.【小問1詳解】證明:在正三棱柱中,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,所以SKIPIF1<0.因?yàn)镾KIPIF1<0,且SKIPIF1<0是棱SKIPIF1<0的中點(diǎn),所以SKIPIF1<0.因?yàn)锳B,SKIPIF1<0平面SKIPIF1<0,且SKIPIF1<0,所以SKIPIF1<0平面SKIPIF1<0.又因?yàn)镾KIPIF1<0平面SKIPIF1<0,所以平面SKIPIF1<0平面SKIPIF1<0.【小問2詳解】解:分別取SKIPIF1<0,SKIPIF1<0的中點(diǎn)SKIPIF1<0,連接SKIPIF1<0,由正三棱柱性質(zhì)得SKIPIF1<0,所以四邊形SKIPIF1<0為平行四邊形,所以SKIPIF1<0,因?yàn)镾KIPIF1<0平面SKIPIF1<0,所以SKIPIF1<0平面SKIPIF1<0因?yàn)镾KIPIF1<0平面SKIPIF1<0,所以SKIPIF1<0因?yàn)樵诘冗吶切蜸KIPIF1<0中,SKIPIF1<0,所以SKIPIF1<0兩兩垂直,如圖建立空間直角坐標(biāo)系,設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,設(shè)平面SKIPIF1<0的法向量SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,得SKIPIF1<
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