2016屆山東省萊蕪市第一高三上學(xué)期第一次摸底考試數(shù)學(xué)理試題

2016屆山東省萊蕪市第一高三上學(xué)期第一次摸底考試數(shù)學(xué)理試題2015.9本試卷分第Ｉ卷和第II卷兩部分,共4頁.滿分150分,考試用時120分鐘.考試結(jié)束后,將本試卷和答題卡一并交回.注意事項：1.答題前,考生務(wù)必用0.5毫米黑色簽字筆將自己的姓名、座號、考生號、縣區(qū)和科類填寫在答題卡和試卷規(guī)定的位置上.山東聯(lián)盟提供。2.第I卷每小題選出答案后,用2B鉛筆把答題卡上對應(yīng)題目的答案標(biāo)號涂黑；如需改動,用橡皮擦干凈后,再選涂其它答案標(biāo)號,答案寫在試卷上無效.3.第II卷必須用0.5毫米黑色簽字筆作答,答案必須寫在答題卡各題目指定區(qū)域內(nèi)相應(yīng)的位置,不能寫在試卷上；如需改動,先劃掉原來的答案,然后再寫上新的答案；不能使用涂改液、膠帶紙、修正帶,不按以上要求作答的答案無效.4.填空題請直接填寫答案,解答題應(yīng)寫出文字說明、證明過程或演算步驟.參考公式：錐體的體積公式：SKIPIF1<0,其中SKIPIF1<0是錐體的底面積,SKIPIF1<0是錐體的高.如果事件SKIPIF1<0互斥,那么SKIPIF1<0；如果事件A,B獨(dú)立,那么SKIPIF1<0.第Ｉ卷（共50分）一.選擇題：本大題共10小題,每小題5分,共50分.在每小題給出的四個選項中,只有一項是符合題目要求的.SKIPIF1<0．（原創(chuàng)）若復(fù)數(shù)SKIPIF1<0,(SKIPIF1<0是虛數(shù)單位),則SKIPIF1<0的共軛復(fù)數(shù)是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0．（原創(chuàng)）設(shè)全集為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則“SKIPIF1<0”是“SKIPIF1<0”的A．充分不必要條件B．必要不充分條件C．充要條件D．既不充分也不必要條件SKIPIF1<0．（原創(chuàng)）函數(shù)SKIPIF1<0的定義域為A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0SKIPIF1<0．（改編）若實數(shù)SKIPIF1<0,SKIPIF1<0滿足約束條件SKIPIF1<0,則SKIPIF1<0的取值范圍是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0．（原創(chuàng)）已知矩形SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0．（原創(chuàng)）SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0．（原創(chuàng)）有下列4個命題：兩個平面垂直,過一個平面內(nèi)任意一點作交線的垂線,則此直線必垂直于另一平面;平面SKIPIF1<0內(nèi)兩條不平行的直線都平行于另一平面SKIPIF1<0,則SKIPIF1<0;SKIPIF1<0兩條直線和一個平面所成的角相等,則這兩條直線平行;直線SKIPIF1<0不平行于平面SKIPIF1<0,則平面SKIPIF1<0內(nèi)不存在與直線SKIPIF1<0平行的直線.其中正確命題的個數(shù)是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0．（改編）如圖，該程序框圖的算法思路源于我國古代數(shù)學(xué)專著《九章算術(shù)》中的“更相減損術(shù)”,執(zhí)行此程序框圖,若輸入的SKIPIF1<0,SKIPIF1<0分別為SKIPIF1<0,SKIPIF1<0,則輸出的SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0開始輸入m、nm=nm>n輸出mm=m-nn=n-m結(jié)束否是否是開始輸入m、nm=nm>n輸出mm=m-nn=n-m結(jié)束否是否是第8題圖SKIPIF1<0．（原創(chuàng)）定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0,且當(dāng)SKIPIF1<0時,SKIPIF1<0,則SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0（改編）設(shè)函數(shù)SKIPIF1<0的零點為SKIPIF1<0,函數(shù)SKIPIF1<0的零點為SKIPIF1<0,則A.SKIPIF1<0,SKIPIF1<0B.SKIPIF1<0,SKIPIF1<0C.SKIPIF1<0,SKIPIF1<0D.SKIPIF1<0,SKIPIF1<0二.填空題：本大題共5小題,每小題5分,共25分,答案須填在答題卡題中橫線上.SKIPIF1<0（改編）在等差數(shù)列SKIPIF1<0中,已知SKIPIF1<0,則SKIPIF1<0.SKIPIF1<0（原創(chuàng)）由曲線SKIPIF1<0與SKIPIF1<0圍成的封閉圖形的面積是________.SKIPIF1<0（原創(chuàng)）在SKIPIF1<0的展開式中,SKIPIF1<0的系數(shù)為________（用數(shù)字作答）.SKIPIF1<0（原創(chuàng)）SKIPIF1<0,SKIPIF1<0,SKIPIF1<0分別是SKIPIF1<0的三邊,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的面積是________.SKIPIF1<0（改編）觀察下列等式SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0照此規(guī)律,SKIPIF1<0SKIPIF1<0.三．解答題：本大題共6小題,共75分,解答應(yīng)寫出文字說明、證明過程或演算步驟.SKIPIF1<0（原創(chuàng)）（本小題滿分12分）已知函數(shù)SKIPIF1<0（Ⅰ）求SKIPIF1<0的單調(diào)遞減區(qū)間;（Ⅱ）將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個單位,再將得到的圖象上各點的橫坐標(biāo)伸長到原來的SKIPIF1<0倍,縱坐標(biāo)不變,得到函數(shù)SKIPIF1<0的圖象,求函數(shù)SKIPIF1<0在SKIPIF1<0上的值域.SKIPIF1<0（改編）（本小題滿分12分）如圖,三棱柱SKIPIF1<0中,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0、SKIPIF1<0分別是線段SKIPIF1<0、SKIPIF1<0的中點.（Ⅰ）求證：SKIPIF1<0∥平面SKIPIF1<0;（Ⅱ）求平面SKIPIF1<0與平面SKIPIF1<0所成銳二面角的余弦值.SKIPIF1<0（原創(chuàng)）（本小題滿分12分）已知數(shù)列SKIPIF1<0是遞增的等比數(shù)列,SKIPIF1<0,SKIPIF1<0.（Ⅰ）求數(shù)列SKIPIF1<0的通項公式;（Ⅱ）若SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0.SKIPIF1<0（原創(chuàng)）（本小題滿分12分）中秋節(jié)吃月餅是我國的傳統(tǒng)習(xí)俗.設(shè)有兩種月餅禮盒,甲禮盒中裝有2個五仁月餅,2個豆沙月餅,2個蓮蓉月餅；乙禮盒中裝有3個五仁月餅,3個豆沙月餅.這12個月餅外觀完全相同,從中隨機(jī)選取4個.（Ⅰ）設(shè)事件SKIPIF1<0為“選取的4個月餅中恰有2個五仁月餅,且這2個五仁月餅選自同一個禮盒”,求事件SKIPIF1<0發(fā)生的概率;（Ⅱ）設(shè)SKIPIF1<0為選取的4個月餅中豆沙月餅的個數(shù),求隨機(jī)變量SKIPIF1<0的分布列和數(shù)學(xué)期望.SKIPIF1<0（改編）（本小題滿分13分）已知一工廠生產(chǎn)某種產(chǎn)品的年固定成本為100萬元,每生產(chǎn)1千件需另投入27萬元.設(shè)該工廠一年內(nèi)生產(chǎn)這種產(chǎn)品SKIPIF1<0千件并全部銷售完,每千件的銷售收入為SKIPIF1<0萬元,且SKIPIF1<0（Ⅰ）寫出年利潤SKIPIF1<0（萬元）關(guān)于年產(chǎn)量SKIPIF1<0（千件）的函數(shù)關(guān)系式;（Ⅱ）年產(chǎn)量為多少千件時,該工廠在這種產(chǎn)品的生產(chǎn)中所獲得的年利潤最大?(注：年利潤SKIPIF1<0年銷售收入SKIPIF1<0年總成本)SKIPIF1<0（原創(chuàng)）（本小題滿分14分）設(shè)函數(shù)SKIPIF1<0,其中SKIPIF1<0.（Ⅰ）SKIPIF1<0時,求曲線SKIPIF1<0在點SKIPIF1<0處的切線方程;（Ⅱ）討論函數(shù)SKIPIF1<0的單調(diào)性;（Ⅲ）當(dāng)SKIPIF1<0時，證明對SKIPIF1<0,都有SKIPIF1<0.數(shù)學(xué)試題（理科）評分標(biāo)準(zhǔn)選擇題：本大題共10小題,每小題5分,共50分.1.B2.C3.A4.C5.A6.B7.D8.C9.C10.A二.填空題：本大題共5小題,每小題5分,共25分.11.1412.SKIPIF1<013.12014.SKIPIF1<015.SKIPIF1<0三．解答題：本大題共6小題,共75分.SKIPIF1<0（本小題滿分12分）【解析】解法一：（Ⅰ）SKIPIF1<0SKIPIF1<0SKIPIF1<0………………….4分由SKIPIF1<0,SKIPIF1<0,………………….5分得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0,SKIPIF1<0.…………………6分（=2\*ROMANⅡ）將SKIPIF1<0的圖象向左平移SKIPIF1<0個單位,得到SKIPIF1<0SKIPIF1<0,………………….7分再將SKIPIF1<0圖象上各點的橫坐標(biāo)伸長到原來的SKIPIF1<0倍,縱坐標(biāo)不變,得到SKIPIF1<0,………………….8分SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.………………….9分SKIPIF1<0SKIPIF1<0.………………….11分SKIPIF1<0SKIPIF1<0.SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0.………………….12分解法二：（Ⅰ）SKIPIF1<0SKIPIF1<0SKIPIF1<0………………….4分下同解法一.SKIPIF1<0（本小題滿分12分）【解析】解法一：（Ⅰ）連接SKIPIF1<0,交SKIPIF1<0于SKIPIF1<0,連接SKIPIF1<0、SKIPIF1<0,SKIPIF1<0四邊形SKIPIF1<0為平行四邊形,SKIPIF1<0SKIPIF1<0為線段SKIPIF1<0的中點.SKIPIF1<0SKIPIF1<0為SKIPIF1<0的中點,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.………………….1分SKIPIF1<0為SKIPIF1<0的中點,SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0…………………2分SKIPIF1<0四邊形SKIPIF1<0為平行四邊形.SKIPIF1<0∥SKIPIF1<0.……………….3分SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0.SKIPIF1<0∥平面SKIPIF1<0.………………….5分（Ⅱ）SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0∥SKIPIF1<0,SKIPIF1<0SKIPIF1<0平面SKIPIF1<0.又SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0.又SKIPIF1<0SKIPIF1<0,以SKIPIF1<0為坐標(biāo)原點,SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分別為SKIPIF1<0軸、SKIPIF1<0軸、SKIPIF1<0軸建立空間直角坐標(biāo)系.………………….6分則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0.………………….7分設(shè)平面SKIPIF1<0的一個法向量為SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0SKIPIF1<0SKIPIF1<0令SKIPIF1<0,得SKIPIF1<0SKIPIF1<0SKIPIF1<0.………………….9分又平面SKIPIF1<0的一個法向量SKIPIF1<0………………….10分SKIPIF1<0SKIPIF1<0………………….11分SKIPIF1<0平面SKIPIF1<0與平面SKIPIF1<0所成銳二面角的余弦值為SKIPIF1<0.………………….12分解法二：（Ⅰ）取SKIPIF1<0中點SKIPIF1<0,連接SKIPIF1<0,SKIPIF1<0.SKIPIF1<0SKIPIF1<0為線段SKIPIF1<0的中點,又四邊形SKIPIF1<0為平行四邊形.SKIPIF1<0SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.………………….2分SKIPIF1<0SKIPIF1<0,SKIPIF1<0分別是SKIPIF1<0,SKIPIF1<0的中點,SKIPIF1<0SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.………………….4分SKIPIF1<0SKIPIF1<0,SKIPIF1<0、SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0∥平面SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0∥平面SKIPIF1<0.………………….5分（Ⅱ）同解法一.SKIPIF1<0（本小題滿分12分）【解析】解法一：（Ⅰ）由SKIPIF1<0即SKIPIF1<0…………………2分消SKIPIF1<0得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,SKIPIF1<0SKIPIF1<0或SKIPIF1<0………………….4分SKIPIF1<0SKIPIF1<0是遞增數(shù)列,SKIPIF1<0SKIPIF1<0…….5分SKIPIF1<0SKIPIF1<0.…….6分（Ⅱ）SKIPIF1<0…….7分SKIPIF1<0SKIPIF1<0………………….8分SKIPIF1<0SKIPIF1<0………………….9分SKIPIF1<0………………….10分SKIPIF1<0………………….11分SKIPIF1<0SKIPIF1<0………………….12分解法二：（Ⅰ）因為SKIPIF1<0是等比數(shù)列,SKIPIF1<0,所以SKIPIF1<0…….1分又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0是方程SKIPIF1<0的兩根，SKIPIF1<0SKIPIF1<0或SKIPIF1<0……….3分SKIPIF1<0SKIPIF1<0是遞增數(shù)列,SKIPIF1<0SKIPIF1<0……….4分SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.……….5分SKIPIF1<0SKIPIF1<0.……….6分（Ⅱ）同解法一.SKIPIF1<0（本小題滿分12分）【解析】（Ⅰ）由已知有SKIPIF1<0,所以事件SKIPIF1<0發(fā)生的概率為SKIPIF1<0.……………4分（Ⅱ）SKIPIF1<0的所有可能取值為SKIPIF1<0,………………5分SKIPIF1<0………………6分SKIPIF1<0………………7分SKIPIF1<0…………………8分SKIPIF1<0…………………9分SKIPIF1<0…………………10分所以隨機(jī)變量SKIPIF1<0的分布列為SKIPIF1<001234SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0…………………11分隨機(jī)變量SKIPIF1<0的數(shù)學(xué)期望為SKIPIF1<0.…………………12分SKIPIF1<0（本小題滿分13分）【解析】（Ⅰ）SKIPIF1<0……………..3分SKIPIF1<0………………..5分（Ⅱ）當(dāng)SKIPIF1<0時,SKIPIF1<0.…………………6分令SKIPIF1<0得SKIPIF1<0SKIPIF1<0（SKIPIF1<0舍去）.…………………7分且當(dāng)SKIPIF1<0時,SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0.…………………8分所以當(dāng)SKIPIF1<0時,SKIPIF1<0SKIPIF1<0.…………………9分當(dāng)SKIPIF1<0時,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.……………11分當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0SKIPIF1<0時取等號.……………12分當(dāng)SKIPIF1<0時,SKIPIF1<0SKIPIF1<0.因為SKIPIF1<0,所以當(dāng)SKIPIF1<0時,SKIPIF1<0SKIPIF1<0.答：年產(chǎn)量為9千件時,該工廠在這種產(chǎn)品的生產(chǎn)中所獲得的年利潤最大.…………………13分SKIPIF1<0（本小題滿分14分）【解析】（Ⅰ）SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0,…………………1分 SKIPIF1<0,又SKIPIF1<0,…………………2分 SKIPIF1<0曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0.…………………3分 （Ⅱ）SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0 令SKIPIF1<0得SKIPIF1<0或SKIPIF1<0.…………………4分 當(dāng)SKIPIF1<0即SKIPIF1<0時,當(dāng)SKIPIF1<0時,SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0.…………………5分 當(dāng)SKIPIF1<0即SKIPIF1<0時,當(dāng)SKIPIF1<0時,SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0.…………………6分當(dāng)SKIPIF1<0即SKIPIF1<0時,SKIPIF1<0.…………………7分當(dāng)SKIPIF1<0即SKIPIF1<0時,當(dāng)SKIPIF1<0時SKIPIF1<0；當(dāng)SKIPIF1<0時SKIPIF1<0,當(dāng)SKIPIF1<0時SKIPIF1<0.…………………8分綜上所述：當(dāng)SKIPIF1<0時,SKIPIF1<0的增區(qū)間為SKIPIF1<0,減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0的增區(qū)間為SKIPIF1<0和SKIPIF1<0；減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0的增區(qū)間為SKIPIF1<0,無減區(qū)間；當(dāng)SKIPIF1<0時,SKIPIF1<0的增區(qū)間為SKIPIF1<0和SKIPIF1<0,減區(qū)間為SKIPIF1<0.…………………9分（Ⅲ）證法一：：①當(dāng)SKIPIF1<0時,由（Ⅱ）知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0記SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.SKIPIF1<0當(dāng)SKIPIF1<0時,SKIPIF1<0即SKIPIF1<0成立.又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.所以SKIPIF1<0.當(dāng)SKIPIF1<0時,SKIPIF1<0時SKIPIF1<0…………………11分②當(dāng)SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0S