萬(wàn)載中學(xué)宜春中學(xué)高三數(shù)學(xué)上學(xué)期第一次月考試題 理-人教版高三數(shù)學(xué)試題

屆高三年級(jí)第一次聯(lián)考數(shù)學(xué)（理）試卷萍鄉(xiāng)中學(xué)萬(wàn)載中學(xué)宜春中學(xué)考試時(shí)間：120分總分：150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(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求.1．設(shè)SKIPIF1<0的共軛復(fù)數(shù)是SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<02．已知SKIPIF1<0均為單位向量，它們的夾角為SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．43．從某地高中男生中隨機(jī)抽取100名同學(xué)，將他們的體重（單位：kg）數(shù)據(jù)繪制成頻率分布直方圖（如圖）．由圖中數(shù)據(jù)可知體重的平均值為（）A．64.5B．59.5C．69.5D．504、宜春為“月亮文化節(jié)”招募了20名志愿者，他們的編號(hào)分別是1號(hào)、2號(hào)、…、19號(hào)、20號(hào)．若要從中任意選取4人再按編號(hào)大小分成兩組去做一些預(yù)備服務(wù)工作，其中兩個(gè)編號(hào)較小的人在一組，兩個(gè)編號(hào)較大的在另一組．那么確保5號(hào)與14號(hào)入選并被分配到同一組的選取種數(shù)是（）A．16 B.21 C.24 5．已知兩定點(diǎn)SKIPIF1<0和SKIPIF1<0，動(dòng)點(diǎn)SKIPIF1<0在直線SKIPIF1<0上移動(dòng)，橢圓SKIPIF1<0以SKIPIF1<0為焦點(diǎn)且經(jīng)過(guò)點(diǎn)SKIPIF1<0，記橢圓SKIPIF1<0的離心率為SKIPIF1<0，則函數(shù)SKIPIF1<0的大致圖像是（）6．如圖是一幾何體的三視圖，正視圖是一等腰直角三角形，且斜邊BD長(zhǎng)為2；側(cè)視圖一直角三角形；俯視圖為一直角梯形，且AB=BC=1，則異面直線PB與CD所成角的正切值是（）A.1B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<07．已知實(shí)數(shù)x∈，執(zhí)行如圖所示的流程圖，則輸出的x不小于55的概率為（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<08.若方程的任意一組解SKIPIF1<0都滿足不等式SKIPIF1<0，則SKIPIF1<0的取值范圍是（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<09．已知F1，F(xiàn)2分別為雙曲線的左右焦點(diǎn)，P為雙曲線上除頂點(diǎn)外的任意一點(diǎn)，且△F1PF2的內(nèi)切圓交實(shí)軸于點(diǎn)M，則|F1M|?|MF2|的值為（）A．b2B．a(chǎn)2C．c2D．10．已知函數(shù)SKIPIF1<0，若存在正實(shí)數(shù)SKIPIF1<0，使得方程SKIPIF1<0在區(qū)間（2，+SKIPIF1<0）上有兩個(gè)根SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0二、填空題：本大題共4小題，每小題5分，共20分.11．設(shè)隨機(jī)變量SKIPIF1<0，且SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的值為．13、已知數(shù)列{an}為等差數(shù)列，公差d≠0{an}的部分項(xiàng)組成的數(shù)列ak1,ak2,…,akn恰為等比數(shù)列，其中k1=1,k2=5,k3=17，則kn=.14．已知是定義在［－1，1］上的奇函數(shù)且，當(dāng)、［－1，1］，且時(shí)，有，若對(duì)所有、恒成立，則實(shí)數(shù)m的取值范圍是.三.選做題：請(qǐng)?jiān)谙铝袃深}中任選一題作答.若兩題都做，則按第一題評(píng)閱計(jì)分.本題共5分.15．（1）（坐標(biāo)系與參數(shù)方程選做題）在極坐標(biāo)系中，定點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上運(yùn)動(dòng)，當(dāng)線段SKIPIF1<0最短時(shí)，點(diǎn)SKIPIF1<0的極坐標(biāo)為．（2）（不等式選講選做題）對(duì)于任意SKIPIF1<0SKIPIF1<0恒成立，則實(shí)數(shù)a的取值范圍______.四．解答題：本大題共6小題，共75分.解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟.16．（本小題滿分12分）已知SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0．（1）求SKIPIF1<0的最小正周期和單調(diào)增區(qū)間；（2）在SKIPIF1<0中，SKIPIF1<0分別為SKIPIF1<0的對(duì)邊，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，求邊SKIPIF1<0．17．（本小題滿分12分）某商場(chǎng)共五層，從五層下到四層有3個(gè)出口，從三層下到二層有4個(gè)出口，從二層下到一層有4個(gè)出口，從一層走出商場(chǎng)有6個(gè)出口。安全部門(mén)在每層安排了一名警員值班，負(fù)責(zé)該層的安保工作。假設(shè)每名警員到該層各出口處的時(shí)間相等，某罪犯在五樓犯案后，欲逃出商場(chǎng)，各警員同時(shí)接到指令，選擇一個(gè)出口進(jìn)行圍堵。逃犯在每層選擇出口是等可能的。已知他被三樓警員抓獲的概率為SKIPIF1<0．（Ⅰ）問(wèn)四層下到三層有幾個(gè)出口？（Ⅱ）天網(wǎng)恢恢，疏而不漏，犯罪嫌疑人最終落入法網(wǎng)。設(shè)抓到逃犯時(shí)，他已下了SKIPIF1<0層樓，寫(xiě)出SKIPIF1<0的分布列，并求SKIPIF1<0．18．（本小題滿分12分）SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0如圖正三棱柱SKIPIF1<0中，底面邊長(zhǎng)為SKIPIF1<0，側(cè)棱長(zhǎng)為SKIPIF1<0，若經(jīng)過(guò)對(duì)角線SKIPIF1<0且與對(duì)角線SKIPIF1<0平行的平面交上底面于SKIPIF1<0。SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（1）試確定SKIPIF1<0點(diǎn)的位置，并證明你的結(jié)論；（2）求二面角SKIPIF1<0的大?。?9、（本小題滿分12分）數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，．SKIPIF1<0（Ⅰ）求數(shù)列SKIPIF1<0的通項(xiàng)公式；

（Ⅱ）若SKIPIF1<0，SKIPIF1<0，求不超過(guò)SKIPIF1<0的最大的整數(shù)值20．（本小題滿分13分）已知橢圓C的中心在原點(diǎn)，焦點(diǎn)在軸上，離心率為，短軸長(zhǎng)為4。（Ⅰ）求橢圓C的標(biāo)準(zhǔn)方程；（Ⅱ），是橢圓C上兩個(gè)定點(diǎn)，A、B是橢圓C上位于直線PQ兩側(cè)的動(dòng)點(diǎn)。若直線AB的斜率為，求四邊形APBQ面積的最大值；當(dāng)A、B兩點(diǎn)在橢圓上運(yùn)動(dòng)，且滿足∠APQ=∠BPQ時(shí)，直線AB的斜率是否為定值，說(shuō)明理由。BBAPQOxy20題21．（本小題滿分14分）設(shè)SKIPIF1<0．（1）若SKIPIF1<0對(duì)一切SKIPIF1<0恒成立，求SKIPIF1<0的最大值．（2）設(shè)SKIPIF1<0，且SKIPIF1<0是曲線SKIPIF1<0上任意兩點(diǎn)，若對(duì)任意的SKIPIF1<0，直線AB的斜率恒大于常數(shù)SKIPIF1<0，求SKIPIF1<0的取值范圍；

三校聯(lián)考卷答案選擇題題號(hào)12345678910答案DCABACBBAB填空題11、4；12、-1；13、SKIPIF1<0；14、SKIPIF1<0;15、（1）SKIPIF1<0；（2）SKIPIF1<0三、解答題16、解：（1）SKIPIF1<0SKIPIF1<0所以SKIPIF1<0的最小正周期SKIPIF1<0由SKIPIF1<0得單調(diào)增區(qū)間為：SKIPIF1<0（2）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0又SKIPIF1<0SKIPIF1<0由余弦定理SKIPIF1<0得：SKIPIF1<0即SKIPIF1<0SKIPIF1<0或SKIPIF1<017．解：（1）設(shè)四層下到三層有SKIPIF1<0個(gè)出口，恰好被三樓的警員抓獲，說(shuō)明五層及四層的警員均沒(méi)有與他相遇。SKIPIF1<0，解得SKIPIF1<0（2）SKIPIF1<0可能取值為0,1,2,3,4,5SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0所以，分布列為SKIPIF1<001234pSKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<018、解：（1）SKIPIF1<0為SKIPIF1<0的中點(diǎn).連結(jié)SKIPIF1<0與SKIPIF1<0交于SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0則SKIPIF1<0為SKIPIF1<0的中點(diǎn)，SKIPIF1<0為平面SKIPIF1<0與平面SKIPIF1<0的交線，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0∵SKIPIF1<0//平面SKIPIF1<0∴SKIPIF1<0//SKIPIF1<0，∴SKIPIF1<0為SKIPIF1<0的中點(diǎn)。（2）過(guò)SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，由正三棱柱的性質(zhì)，SKIPIF1<0平面SKIPIF1<0，連結(jié)SKIPIF1<0，在正SKIPIF1<0中，SKIPIF1<0是SKIPIF1<0的中點(diǎn)，SKIPIF1<0又在直角三角形SKIPIF1<0中，SKIPIF1<0SKIPIF1<0所以可得SKIPIF1<0.則SKIPIF1<0為二面角SKIPIF1<0的大小，可求得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（2）解法（二）（空間向量法）建立如圖所示空間直角坐標(biāo)系，則SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0。設(shè)SKIPIF1<0是平面SKIPIF1<0的一個(gè)法向量，則可得SKIPIF1<0，所以SKIPIF1<0.所以可得SKIPIF1<0又平面SKIPIF1<0的一個(gè)法向量SKIPIF1<0設(shè)SKIPIF1<0則SKIPIF1<0又可知二面角SKIPIF1<0是銳角,所以二面角SKIPIF1<0的大小是SKIPIF1<019、解：（Ⅰ）因?yàn)?，所以①?dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則，②當(dāng)時(shí)，，所以，即，SKIPIF1<0所以，而，所以數(shù)列是首項(xiàng)為，公比為的等比數(shù)列，所以SKIPIF1<0．SKIPIF1<0（Ⅱ）由(1)知SKIPIF1<0SKIPIF1<0SKIPIF1<0所以SKIPIF1<0SKIPIF1<0，故不超過(guò)的最大整數(shù)為SKIPIF1<0．20、解：（Ⅰ）設(shè)C方程為由已知b=離心率得所以，橢圓C的方程為（Ⅱ）①由（Ⅰ）可求得P、Q的坐標(biāo)為，，則，設(shè)AB(),直線AB的方程為，代人得由△>0,解得，由根與系數(shù)的關(guān)系得四邊形APBQ的面積故，當(dāng)②∠APQ=∠BPQ時(shí)，PA、PB的斜率之和為0，設(shè)直線PA的斜率為，則PB的斜率為，PA的直線方程為與聯(lián)立解得，同理PB的直線方程，可得所以所以直線AB的斜率為定值21．解：（1）∵SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0的解為SKIPIF1<0，∴SKIPIF1<0，∵SKIPIF1<0對(duì)一切x∈R恒成立，∴SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．（2）設(shè)SKIPIF1<0是任意的兩實(shí)數(shù)，且SKIPIF1<0SKIPIF1<0，故SKIPIF1<0∴不妨令函數(shù)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0恒成立