學(xué)必求其心得，業(yè)必貴于專精學(xué)必求其心得，業(yè)必貴于專精學(xué)必求其心得，業(yè)必貴于專精2017—2018學(xué)年度高三年級第一學(xué)期期中抽測數(shù)學(xué)一、填空題：本大題共14個小題，每小題5分,共70分.將答案填在答題紙上.1.設(shè)集合，，則．2.已知復(fù)數(shù)滿足,其中為虛數(shù)單位，則復(fù)數(shù)的實部為．3。函數(shù)的最小正周期為．4。已知一組數(shù)據(jù)：87，，90，89,93的平均數(shù)為90，則該組數(shù)據(jù)的方差為．5.雙曲線的離心率為．6。從2個黃球,3個紅球中隨機(jī)取出兩個球,則兩球顏色不同的概率是．7。執(zhí)行如圖所示的算法流程圖，則輸出的值為．8.各棱長都為2的正四棱錐的體積為．9.已知公差不為零的等差數(shù)列的前項和為，且,若成等比數(shù)列,則的值為．10。如圖,在半徑為2的扇形中，，為弧上的一點，若，則的值為．11.已知函數(shù)（為自然對數(shù)的底數(shù)），若，則實數(shù)的取值范圍是．12。已知實數(shù)滿足，則的最小值為．13。已知是圓上的動點，點,若直線上總存在點，使點恰是線段的中點，則實數(shù)的取值范圍是．14。已知函數(shù),若存在,使,則實數(shù)的取值范圍是．二、解答題(本大題共6小題，共90分。解答應(yīng)寫出文字說明、證明過程或演算步驟.）15.已知的內(nèi)角所對應(yīng)的邊分別為，且.（1)求角的大小；（2）若，，求的面積.16。如圖，在三棱錐中，，,為的中點，為上一點，且平面,求證：（1）直線平面；(2）平面平面.17。如圖，有一塊半圓形空地，開發(fā)商計劃建一個矩形游泳池及其附屬設(shè)施，并將剩余空地進(jìn)行綠化，園林局要求綠化面積應(yīng)最大化，其中半圓的圓心為，半徑為,矩形的一邊在直線上，點在圓周上,在邊上，且，設(shè).（1）記游泳池及其附屬設(shè)施的占地面積為,求的表達(dá)式；（2）求符合園林局要求的的余弦值.18.如圖,在平面直角坐標(biāo)系中,橢圓的左頂點為，離心率為，過點的直線與橢圓交于另一點，點為軸上的一點。(1）求橢圓的標(biāo)準(zhǔn)方程；（2）若是以點為直角頂點的等腰直角三角形，求直線的方程。19。已知數(shù)列的前項和為，滿足，，數(shù)列滿足,，且。（1）求數(shù)列和的通項公式;（2）若，數(shù)列的前項和為，對任意的，都有，求實數(shù)的取值范圍。（3）是否存在正正數(shù)，使成等差數(shù)列？若存在，求出所有滿足條件的；若不存在,請說明理由。20。已知函數(shù)（，是自然對數(shù)的底數(shù)）.（1)若函數(shù)在區(qū)間上是單調(diào)減函數(shù),求實數(shù)的取值范圍;（2）求函數(shù)的極值;（3）設(shè)函數(shù)圖像上任意一點處的切線為，求在軸上的截距的取值范圍.21.【選做題】本題包括A，B，C，D四小題，請選定其中兩小題，并在相應(yīng)的答題區(qū)域內(nèi)作答,若多做,則按作答的前兩小題評分，解答時應(yīng)寫出文字說明,證明過程或演算步驟.A．【選修4-1：幾何證明選講】如圖，是圓的切線,切點為，是過圓心的割線且交圓于點,過作圓的切線交于點，。求證：.B.【選修4—2:矩陣與變換】已知矩陣，若直線在矩陣對應(yīng)的變換作用下得到的直線過點，求實數(shù)的值。C.【選修4—4：坐標(biāo)系與參數(shù)方程】在極坐標(biāo)系中，圓的方程為,以極點為坐標(biāo)原點,極軸為軸正半軸建立平面直角坐標(biāo)系，設(shè)直線的參數(shù)方程為(為參數(shù)），若直線與圓恒有公共點，求實數(shù)的取值范圍.D．【選修4-5：不等式選講】設(shè)均為正數(shù)，且,求證：。【必做題】第22題、第23題，每題10分，共計20分，請在答題卡指定區(qū)域內(nèi)作答，解答時應(yīng)寫出文字說明、證明過程或演算步驟。22.如圖，在三棱錐中，兩兩互相垂直，點分別為棱的中點，在棱上，且滿足，已知,.（1）求異面直線與所成角的余弦值；(2）求二面角的正弦值.23。某同學(xué)在上學(xué)路上要經(jīng)過三個帶有紅綠燈的路口，已知他在三個路口遇到紅燈的概率依次是，遇到紅燈時停留的時間依次是40秒、20秒、80秒，且在各個路口是否遇到紅燈是相互獨(dú)立的.（1）求這名同學(xué)在上學(xué)路上在第三個路口時首次遇到紅燈的概率；（2)記這名同學(xué)在上學(xué)路上因遇到紅燈停留的總時間為隨機(jī)事件,求的概率分布與期望.試卷答案一、填空題1．2．3．64．5．6．7．48．9．8810．11．12．13．14．二、解答題15．（1）因為,由正弦定理,得．因為，所以．即，所以．因為，所以又因為,所以．（2)由余弦定理及得，，即．又因為，所以，所以．16。（1）因為平面，平面，平面平面，所以．因為平面,平面，所以平面．(2）因為為的中點，，所以為的中點．又因為，所以，又，，所以．又平面，，所以平面．因為平面,所以平面平面．17．（1）由題意,，，且為等邊三角形,所以，，,．（2）要符合園林局的要求，只要最小,由（1）知，,令，即,解得或（舍去)，令,，當(dāng)時，是單調(diào)減函數(shù);當(dāng)時，是單調(diào)增函數(shù)，所以當(dāng)時，取得最小值.答：符合園林局要求的的余弦值為。18．（1）由題意可得：即從而有，所以橢圓的標(biāo)準(zhǔn)方程為：．（2）設(shè)直線的方程為,代入，得，因為為該方程的一個根,解得,設(shè)，由，得：，即：由，即,得，即，即，所以或，當(dāng)時，直線的方程為，當(dāng)時,代入得，解得，此時直線的方程為.綜上，直線的方程為，.19．（1）當(dāng)時，,所以．當(dāng)時，，,兩式相減得,又，所以,從而數(shù)列為首項，公比的等比數(shù)列，從而數(shù)列的通項公式為．由兩邊同除以,得，從而數(shù)列為首項,公差的等差數(shù)列，所以，從而數(shù)列的通項公式為．（2）由（1）得，于是，所以，兩式相減得，所以，由（1）得，因為對，都有，即恒成立，所以恒成立,記，所以,因為，從而數(shù)列為遞增數(shù)列，所以當(dāng)時,取最小值，于是．（3）假設(shè)存在正整數(shù)，使（)成等差數(shù)列，則，即，若為偶數(shù)，則為奇數(shù)，而為偶數(shù)，上式不成立.若為奇數(shù)，設(shè)，則，于是，即，當(dāng)時，,此時與矛盾；當(dāng)時，上式左邊為奇數(shù),右邊為偶數(shù),顯然不成立.綜上所述，滿足條件的不存在．20．（1）函數(shù)的導(dǎo)函數(shù)，則在區(qū)間上恒成立，且等號不恒成立，又，所以在區(qū)間上恒成立，記,只需，即解得．經(jīng)檢驗，時,是上的單調(diào)減函數(shù)，又,所以實數(shù)的取值范圍是．（2）由,得，①當(dāng)時，有；,所以函數(shù)在上單調(diào)遞增，在上單調(diào)遞減,所以函數(shù)在取得極大值,沒有極小值．②當(dāng)時，有;,所以函數(shù)在上單調(diào)遞減，在上單調(diào)遞增,所以函數(shù)在取得極小值，沒有極大值．綜上可知:當(dāng)時，函數(shù)在取得極大值，沒有極小值；當(dāng)時，函數(shù)在取得極小值，沒有極大值．（3)設(shè)切點為，則曲線在點處的切線方程為，當(dāng)時，切線的方程為,其在軸上的截距不存在．當(dāng)時,令，得切線在軸上的截距為，令，,考慮函數(shù)，則,列表如下:↗極大值↘↘極小值↗所以．故切線在軸上的截距的取值范圍是．?dāng)?shù)學(xué)Ⅱ(附加題)參考答案21A．∵是圓的切線,∴，連結(jié)，則，∵是圓的切線，∴，又，∴,∴,則，而，∴，∴，由得，代入得，故．21B．矩陣，得，所以,將點代入直線得。21C．由（為參數(shù)），可得直線的普通方程為，由得，所以，圓的標(biāo)準(zhǔn)方程為，因為直線與圓恒有公共點，所以，又因為，所以，解之得,所以，實數(shù)的取值范圍為．21D．證明：因為，所以，因為,當(dāng)且僅當(dāng)時等號成立,所以．22．(1)如圖,以為原點,分別以方向為軸、軸、軸正方向建立空間直角坐標(biāo)系。依題意可得：，，，，，，所以,，所以.因此異面直線與所成角的余弦值為。（2）平面的一個法向量為.設(shè)為平面的一個法向量，又，則即不妨取，則，所以為平面的一個法向量，從而,設(shè)二面角的大小為，則.因為，所以.因此二面角的正弦值為。23．(1)設(shè)這名同學(xué)在上學(xué)路上在第三個路口時首次遇到紅燈為事件,因為事件等于事件“這名同學(xué)在第一和第二個路口沒有遇到紅燈，在第三個路口遇到紅燈”,所以.答:這名同學(xué)在上學(xué)路上在第三個路口時首次遇到紅燈的概率為。（2）的所有可能取值為0，40，20，80，60，100，120，140（單位:秒).的分布列是：；；；；;；；。所以.

徐州市2017—2018學(xué)年度高三年級摸底考試數(shù)學(xué)I參考答案一、填空題1．2．3．64．5．6．7．48．9．8810．11．12．13．14．二、解答題15．（1）因為,由正弦定理，得．················2分因為，所以．即，所以．····························································································4分因為，所以．················································································6分又因為，所以．···············································································7分（2）由余弦定理及得，，即．··································································································10分又因為,所以，····················································································12分所以．·································································14分16。（1）因為平面，平面,平面平面，所以．·························································3分因為平面，平面,所以平面．···························································································6分（2）因為為的中點，，所以為的中點．又因為，所以，············································································8分又，，所以．···························································10分又平面，,所以平面．···························································································12分因為平面，所以平面平面．············································14分17．(1）由題意，，，且為等邊三角形,所以，,·····································································2分,．····················································6分（2）要符合園林局的要求，只要最小，由(1）知，，令，即，解得或（舍去),·························································10分令，,當(dāng)時，是單調(diào)減函數(shù)；當(dāng)時，是單調(diào)增函數(shù)，所以當(dāng)時，取得最小值.答：符合園林局要求的的余弦值為。··························································14分18．(1）由題意可得：即從而有,所以橢圓的標(biāo)準(zhǔn)方程為：．····································································4分（2）設(shè)直線的方程為,代入,得，因為為該方程的一個根，解得,·······································6分設(shè),由，得：,即：····························································10分由，即，得，即，即，所以或，··························································································14分當(dāng)時，直線的方程為，當(dāng)時，代入得，解得，此時直線的方程為。綜上,直線的方程為，。··························································16分19．（1）當(dāng)時，，所以．當(dāng)時，，，兩式相減得，又，所以，從而數(shù)列為首項，公比的等比數(shù)列，從而數(shù)列的通項公式為．由兩邊同除以,得，從而數(shù)列為首項，公差的等差數(shù)列，所以，從而數(shù)列的通項公式為．·····························································4分(2）由（1）得，于是，所以,兩式相減得，所以，由（1）得，·················································································8分因為對，都有，即恒成立，所以恒成立，記，所以，············································································································10分因為,從而數(shù)列為遞增數(shù)列，所以當(dāng)時，取最小值，于是．······················································12分（3）假設(shè)存在正整數(shù),使(）成等差數(shù)列,則,即,若為偶數(shù),則為奇數(shù)，而為偶數(shù)，上式不成立。若為奇數(shù)，設(shè)，則,于是，即，當(dāng)時，,此時與矛盾；當(dāng)時,上式左邊為奇數(shù),右邊為偶數(shù)，顯然不成立。綜上所述,滿足條件的不存在．····································································16分20．（1)函數(shù)的導(dǎo)函數(shù)，則在區(qū)間上恒成立，且等號不恒成立,（表述不對吧，可以去掉，后面再檢驗?）又,所以在區(qū)間上恒成立,·········································2分記,只需即解得．經(jīng)檢驗,時，是上的單調(diào)減函數(shù)，又，所以實數(shù)的取值范圍是．··············································4分我的機(jī)子這題答案后三行亂碼，答案我算的是a小于三分之一，原答案忽略了0（2）由，得，①當(dāng)時，有；，所以函數(shù)在上單調(diào)遞增，在上單調(diào)遞減，所以函數(shù)在取得極大值,沒有極小值．②當(dāng)時，有；，所以函數(shù)在上單調(diào)遞減，在上單調(diào)遞增，所以函數(shù)在取得極小值,沒有極大值．綜上可知:當(dāng)時，函數(shù)在取得極大值，沒有極小值；當(dāng)時，函數(shù)在取得極小值，沒有極大值．·········································································································································10分（3）設(shè)切點為，則曲線在點處的切線方程為，當(dāng)時，切線的方程為，其在軸上的截距不存在．當(dāng)時，令，得切線在軸上的截距為，············································································12分令,，考慮函數(shù),則，列表如下：↗極大值↘↘極小值↗所以．故切線在軸上的截距的取值范圍是．······························16分

徐州市2017—2018學(xué)年度高三年級摸底考試（第21(A)（第21(A)題）21A．∵是圓的切線，∴,連結(jié)，則，∵是圓的切線,∴，又,∴，∴，則，而，∴，∴，································5分由得，代入得，故．···················································································10分21B．矩陣，得,·························································5分所以，將點代入直線得。··································································10分21C．由（為參數(shù))，可得直線的普通方程為，由得,所以,圓的標(biāo)準(zhǔn)方程為,·························································5分因為直線與圓恒有公共點，所以，又因為,所以，解之得，所以,實數(shù)的取值范圍為．·······