2020年馬鞍山市數(shù)學(xué)高考一模試卷帶答案

2020/r/n年馬鞍山市數(shù)學(xué)高考一模試卷帶答案/r/n選擇題/r/n1./r/n在''一帶一路”知識(shí)測(cè)驗(yàn)后，甲、乙、丙三人對(duì)成績(jī)進(jìn)行預(yù)測(cè)./r/n甲：我的成績(jī)比乙高./r/n乙：丙的成績(jī)比我和甲的都高./r/n丙：我的成績(jī)比乙高./r/n成績(jī)公布后，三人成績(jī)互不相同且只有一個(gè)人預(yù)測(cè)正確，那么三人按成績(jī)由高到低的次序/r/n為/r/nA./r/n甲、乙、丙/r/nB./r/n乙、甲、丙/r/nC./r/n丙、乙、甲/r/nD./r/n甲、丙、乙/r/n2./r/n(-/r/n)/r/n(?)=/r/n()/r/nA./r/n3+1/r/nB/r/n?/r/n一/r/n3/r/n一/r/n1/r/nC/r/n?/r/n一/r/n3+1/r/nD.3-1/r/n3./r/n將編號(hào)為/r/n1,2,3,4,5,6/r/n的六個(gè)小球放入編號(hào)為/r/n1,2,3,4,5,6/r/n的六個(gè)盒子，每個(gè)盒子放一/r/n個(gè)小球，若有且只有三個(gè)盒子的編號(hào)與放入的小球編號(hào)相同，則不同的放法種數(shù)是（）/r/nA./r/n40/r/nB./r/n60/r/nC./r/n80/r/nD/r/n?/r/n100/r/n4./r/n一個(gè)頻率分布表（樣本容量為/r/n30）/r/n不小心被損壞了一部分，只記得樣本中數(shù)據(jù)在/r/n［20,60/r/n）/r/n上的頻率為/r/n0.8,/r/n則估計(jì)樣本在/r/n［40,50/r/n）/r/n、/r/n［50,60/r/n）/r/n內(nèi)的數(shù)據(jù)個(gè)數(shù)共有（）/r/n分組/r/n[10.20)/r/n(20,30)/r/n[30.40)/r/n頻數(shù)/r/n3/r/n4/r/n5/r/nD.17/r/nD./r/n(/r/n3,4/r/n)/r/n第/r/n1/r/n次到第/r/n14/r/n次的考試成/r/nA./r/n14/r/nD.17/r/nD./r/n(/r/n3,4/r/n)/r/n第/r/n1/r/n次到第/r/n14/r/n次的考試成/r/n9/r/n5./r/n函數(shù)/r/n/(x)=ln(x+l)--/r/n的一個(gè)零點(diǎn)所在的區(qū)間是()/r/nA/r/nA./r/n(/r/n0,1)/r/nB./r/n(1,2/r/n)/r/nC./r/n(/r/n2,3/r/n)/r/n如圖是某高三學(xué)生進(jìn)入高中三年來的數(shù)學(xué)考試成績(jī)莖葉圖,/r/n績(jī)依次記為下圖是統(tǒng)計(jì)莖葉圖中成績(jī)?cè)谝欢ǚ稇z/r/n1/r/n內(nèi)考試次數(shù)的一個(gè)算法流/r/n程圖，那么算法流程圖輸出的結(jié)呆是（）/r/n莖/r/n葉/r/n7/r/n9/r/n8/r/n6385/r/n9/r/n398841/r/n10/r/n31/r/n11/r/n4/r/nC.9/r/nB.8/r/nD.10/r/n已知向量/r/na/r/n，/r/n5/r/n滿足同=血，|引=/r/n1/r/n，/r/n且|萬+引/r/n=2,/r/n則向量°與乙的夾角的余弦值為（）/r/nD./r/nV2/r/n4/r/nV2/r/n

/r/n?/r/n2/r/nD./r/nV2/r/n4/r/n已知函數(shù)/r/n/（x）/r/n=/r/n（x-/r/n3）e/r/nx/r/n+a（2\nx-x+l）/r/n在/r/n（l/r/n、+/r/ns）/r/n上有兩個(gè)極值點(diǎn)，且/⑴在/r/n（1,2）/r/n上單調(diào)遞增，則實(shí)數(shù)/r/nQ/r/n的取值范闈是（）/r/nA./r/n（匕/r/n+s）/r/nB./r/n（e,2e/r/n2/r/n）/r/nC./r/n（2，，4^c）/r/nD./r/n（匕/r/n2€‘）U（2￡‘，+<x））/r/n22/r/n己知雙曲線/r/nC/r/n：/r/n二—匚/r/n=1（/r/n。>/r/n0/r/n上>/r/n0）/r/n的焦距為/r/n2c,/r/n焦點(diǎn)到雙曲線/r/nC/r/n的漸近線的距/r/ncr/r/nZ?"/r/n離為/r/n週/r/nC,/r/n則雙曲線的漸近線方程為（）/r/n2/r/nA/r/n?/r/n〉‘=±屈/r/nB/r/n?/r/ny=/r/nC.y=±A-/r/nD/r/n?/r/ny=/r/n±2x/r/n下表提供了某廠節(jié)能降耗技術(shù)改造后在生產(chǎn)/r/nA/r/n產(chǎn)品過程中記錄的產(chǎn)量/r/nX/r/n（噸）與相應(yīng)的生產(chǎn)能耗）’（噸）的幾組對(duì)應(yīng)數(shù)據(jù)，根據(jù)表中提供的數(shù)據(jù)，求岀關(guān)于％的線性回歸方程為/r/ny=0/r/n?/r/n7x+0/r/n?/r/n35,/r/n則下列結(jié)論錯(cuò)誤的是（）/r/nX/r/n3/r/n4/r/n5/r/n6/r/ny/r/n2.5/r/nt/r/n4/r/n4.5/r/nA./r/n產(chǎn)品的生產(chǎn)能耗與產(chǎn)量呈正相關(guān)/r/nB./r/n回歸直線一定過/r/n（4.5,3/r/n?/r/n5）/r/nC.A/r/n產(chǎn)品每多生產(chǎn)/r/n1/r/n噸，則相應(yīng)的生產(chǎn)能耗約增加/r/n0.7/r/n噸/r/nD./r/n/的值是/r/n3.15/r/n11./r/n已知銳角三角形的邊長分別為/r/n2,3,X,/r/n則/r/nX/r/n的取值范圍是（）/r/nA/r/n?/r/ny/5<X<y/13/r/nB/r/n?/r/nV13<X<5/r/n

/r/nC.2<X<5/5/r/nD./r/ny/s<x<5/r/n設(shè)集合/r/nM/r/n={x|log/r/n2/r/n(x-l)<0},/r/n集合/r/nN={x\x>-2],/r/n則/r/nM\jN=(/r/n)/r/nA.{x|-2<x<2}/r/nb./r/n{x|x>-2}c.{x|x<2}/r/nD.{x|l<x<2}/r/n二、填空題/r/n若雙曲線匚一二=/r/n1(°>0*>0/r/n)/r/n兩個(gè)頂點(diǎn)三等分焦距，則該雙曲線的漸近線方程/r/nCT/?-/r/n是/r/n?/r/n如圖，一輛汽車在一條水平的公路上向正西行駛，到力處時(shí)測(cè)得公路北側(cè)一山頂/r/nD/r/n在/r/n西偏北/r/n30°/r/n的方向上，行駛/r/n600m/r/n后到達(dá)/r/nB/r/n處，測(cè)得此山頂在西偏北/r/n75/r/n。的方向上，仰角為/r/n30°,/r/n則此山的高度/r/nCD/r/n二/r/nm/r/n?/r/n項(xiàng)的系數(shù)是/r/n54,/r/n則/r/nn=/r/nTOC\o"1-5"\h\z/r/n函數(shù)/r/n/(x)/r/n=/r/n^yiogZx-T/r/n的定義域?yàn)?r/n?/r/n(/r/nx/r/n3/r/n+/r/n-y/r/n的展開式中/的系數(shù)是_.(用數(shù)字填寫答案)/r/n18./r/n在等腰梯形/r/nABCD/r/n中啟知/r/n\\DC,AB=2,BC/r/n=1,/r/nZABC/r/n=60/r/n,點(diǎn)/r/nE/r/n和點(diǎn)/r/nF/r/n分別在/r/n—,2——1—/r/n，/r/n線段/r/nBC/r/n和/r/nCD/r/n匕且/r/nBE=-BC、DF=—DC,/r/n則刁/r/nE/r/n?喬的值為—?/r/n3/r/n6/r/n設(shè)復(fù)數(shù)/r/nZ=-l-/(/r/n//r/n虛數(shù)單位)，乙的共緬復(fù)數(shù)為/r/nZ/r/n，/r/n則/r/n|(1/r/n一/r/n2/r/n)/r/n?石=/r/n■/r/n若函數(shù)/r/nf(x)=x/r/n2/r/n-x+l+alnx/r/n在/r/n(0,/r/n乜)上單調(diào)遞增，則實(shí)數(shù)。的最小值是/r/n三、解答題/r/n隨著移動(dòng)互聯(lián)網(wǎng)的發(fā)展，與餐飲美食■相關(guān)的手機(jī)/r/n4PP/r/n軟件層出不窮，現(xiàn)從某市使用/r/n4/r/n和/r/n3/r/n兩款訂餐軟件的商家中分別隨機(jī)抽取/r/n100/r/n個(gè)商家，對(duì)它們的“平均送達(dá)時(shí)間”進(jìn)行統(tǒng)計(jì)，得到頻率分布直方圖如下：/r/n已知抽取的/r/n100/r/n個(gè)使用/r/n4/r/n未訂餐軟件的商家中，甲商家的“平均送達(dá)時(shí)間”為/r/n18/r/n分鐘，現(xiàn)從使用/r/n4/r/n未訂餐軟件的商家中“平均送達(dá)時(shí)間”不超過/r/n20/r/n分鐘的商家中隨機(jī)抽取/r/n3/r/n個(gè)商家進(jìn)行市場(chǎng)調(diào)研，求甲商家被抽到的概率；/r/n試估計(jì)該市使用/r/n4/r/n款訂餐軟件的商家的“平均送達(dá)時(shí)間”的眾數(shù)及平均數(shù)；/r/n如果以''平均送達(dá)時(shí)間〃的平均數(shù)作為決策依據(jù)，從/r/n4/r/n和/r/n3/r/n兩款訂餐軟件中選擇一款訂餐，你會(huì)選擇哪款？/r/n點(diǎn)/r/n點(diǎn)/r/nP/r/n滿足/r/nBP=2BA./r/n求動(dòng)點(diǎn)/r/nP/r/n的軌跡方程；/r/n設(shè)/r/nQ/r/n為直線/r/nl:x=3/r/n上一點(diǎn)，/r/n0/r/n為坐標(biāo)原點(diǎn)，且/r/nOP/r/n丄/r/n02,/r/n求/r/nZOQ/r/n面枳的最小值./r/n(遼寧省葫蘆島市/r/n2018/r/n年二模)直角坐標(biāo)系/r/nxOy/r/n中，直線/的參數(shù)方程為/r/n兀=/r/n2+/r/ntcosa/r/n./r/n?(/為參數(shù))，在極坐標(biāo)系(與直角坐標(biāo)系/r/nxOy/r/n取相同的長度單位，且以原/r/ny=l/r/n+/r/ntsina/r/n點(diǎn)為極點(diǎn)，以/r/nX/r/n軸正半軸為極軸)中，圓/r/nC/r/n的方程為/r/nQ=6cosO./r/n求圓/r/nC/r/n的直角坐標(biāo)方程：/r/n設(shè)圓/r/nC/r/n與直線/交于點(diǎn)/r/n若點(diǎn)/r/nP/r/n的坐標(biāo)為/r/n(/r/n2,1),/r/n求/r/n\PA\+\PB\/r/n的最小值./r/nAABC/r/n在內(nèi)角/r/nA/r/n、/r/nB/r/n、/r/nC/r/n的對(duì)邊分別為/r/na,b,c,/r/n已知/r/na=bcosC+csinB./r/n(/r/nI/r/n)/r/n求/r/nB/r/n；/r/n(II)/r/n若/r/nb=2,/r/n求/r/nAABC/r/n面積的最大值./r/n紅隊(duì)隊(duì)員甲、乙、丙與藍(lán)隊(duì)隊(duì)員/r/nA/r/n、/r/nB/r/n、/r/nC/r/n進(jìn)行圍棋比賽，甲對(duì)/r/nA,/r/n乙對(duì)/r/nB,/r/n丙對(duì)/r/nC/r/n各一盤，已知甲勝/r/nA,/r/n乙勝/r/nE,/r/n丙勝/r/nC/r/n的概率分別為/r/n0.6,0.5,0.5,/r/n假設(shè)各盤比賽結(jié)果相互獨(dú)立./r/n(/r/nI/r/n)/r/n求紅隊(duì)至少兩名隊(duì)員獲勝的概率：/r/n(ID/r/n用纟表示紅隊(duì)隊(duì)員獲勝的總盤數(shù)，求§的分布列和數(shù)學(xué)期/r/n型/r/nEg./r/n26/r/n-/r/n如圖,四棱錐一邊中，/r/nABHDC，/r/n，/r/nA—/r/n扣/r/n"2,/r/nPD=PB=y/6/r/n9/r/nPD1.BC/r/n?/r/n（/r/n1/r/n）/r/n求證：平面丄平面/r/nPBC/r/n；/r/n（/r/n2）/r/n在線段/r/nPC/r/n上是否存在點(diǎn)/r/nM/r/n，/r/n使得平面與平面所成銳二面角為扌？若存/r/nCM/r/n在，求一的值；若不存在，說明理由./r/nCP/r/n【參考答案】/r/n***/r/n試卷處理標(biāo)記，請(qǐng)不要?jiǎng)h除/r/n一、選擇題/r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n利用逐一驗(yàn)證的方法進(jìn)行求解./r/n【詳解】/r/n若甲預(yù)測(cè)正確，則乙、丙預(yù)測(cè)錯(cuò)誤，則甲比乙成績(jī)高，丙比乙成績(jī)低，故/r/n3/r/n人成績(jī)由高到低依次為甲，乙，丙；若乙預(yù)測(cè)正確，則丙預(yù)測(cè)也正確，不符合題意；若丙預(yù)測(cè)正確，則甲必預(yù)測(cè)錯(cuò)誤，丙比乙的成績(jī)高，乙比甲成績(jī)高，即丙比甲，乙成績(jī)都高，即乙預(yù)測(cè)正確，不符合題意，故選/r/nA./r/n【點(diǎn)睛】/r/n本題將數(shù)學(xué)知識(shí)與時(shí)政結(jié)合，主要考查推理判斷能力.題目有一定難度，注重了基礎(chǔ)知識(shí)、邏輯推理能力的考查./r/nB/r/n解析：/r/nB/r/n【解析】/r/n【分析】/r/n先分別對(duì)分子和分母用乘法公式化簡(jiǎn)，再分子分母同時(shí)乘以分母的共軌復(fù)數(shù)，化簡(jiǎn)即得最后結(jié)果./r/n【詳解】/r/n由題意得，復(fù)數(shù)/r/n（「“（嚴(yán)+”/r/n二/r/nzlijl=/r/n（T+3i）i/r/n=_3_i/r/n故應(yīng)選/r/ne/r/ni/r/n-i-ii/r/n【點(diǎn)睛】/r/n本小題主要考查復(fù)數(shù)的乘法和除法的運(yùn)算，乘法的運(yùn)算和實(shí)數(shù)的運(yùn)算類似，只需要記住/r/ni/r/n2/r/n=-l-/r/n除法的運(yùn)算記住的是分子分母同時(shí)乘以分母的共軌復(fù)數(shù)，這一個(gè)步驟稱為分母實(shí)數(shù)化，分母實(shí)數(shù)化的主要目的是將分母變?yōu)閷?shí)數(shù)，然后將復(fù)數(shù)的實(shí)部和虛部求出來?屬于基礎(chǔ)題./r/nA/r/n解析：/r/nA/r/n【解析】解：三個(gè)小球放入盒子是不對(duì)號(hào)入座的方法有/r/n2/r/n種，由排列組合的知識(shí)可得，不同的放法總數(shù)是：/r/n2C/r/n；/r/n=40/r/n種./r/n本題選擇/r/nA/r/n選項(xiàng)./r/nB/r/n解析：/r/nB/r/n【解析】/r/n【分析】/r/n計(jì)算出樣本在/r/n［20,60/r/n)/r/n的數(shù)據(jù)個(gè)數(shù)，再減去樣本在/r/n［20,40/r/n)/r/n的數(shù)據(jù)個(gè)數(shù)即可得出結(jié)呆./r/n【詳解】/r/n由題意可知，樣本在/r/n［/r/n20,60/r/n)/r/n的數(shù)據(jù)個(gè)數(shù)為/r/n30x0.8=24,/r/n樣本在/r/n［20,40/r/n)/r/n的數(shù)據(jù)個(gè)數(shù)為/r/n4+5=9,/r/n因此，樣本在/r/n［40,50)/r/n、/r/n［50,60)/r/n內(nèi)的數(shù)據(jù)個(gè)數(shù)為/r/n24-9=15./r/n故選：/r/nB./r/n【點(diǎn)睛】/r/n本題考查利用頻數(shù)分布表計(jì)算頻數(shù)，要理解頻數(shù)、樣本容量與頻率三者之間的關(guān)系，考查計(jì)算能力，屬于基礎(chǔ)題./r/n./r/nB/r/n解析：/r/nB/r/n【解析】/r/n【分析】/r/n先求出根據(jù)零點(diǎn)存在性定理得解./r/n【詳解】/r/n/、?/r/n由題得/r/n/(l)=ln2-y=ln2-2<0,/r/n2/r/n/(2)=ln3--=ln3-l>0,/r/n所以/(!)/⑵/r/n<0,/r/n2/r/n所以函數(shù)/r/n/(x)=ln(x+l)-/r/n—的一個(gè)零點(diǎn)所在的區(qū)間是/r/n(1,2/r/n)/r/n./r/nX/r/n故選/r/nE/r/n【點(diǎn)睛】/r/n本題主要考查零點(diǎn)存在性定理，意在考查學(xué)生對(duì)該知識(shí)的理解掌握水平，屬于基礎(chǔ)題./r/n./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/n根據(jù)流程圖可知該算法表示統(tǒng)計(jì)/r/n14/r/n次考試成績(jī)中人于等于/r/n90/r/n的人數(shù)，結(jié)合莖葉圖可得答案./r/n【詳解】/r/n根據(jù)流程圖所示的順序，可知該程序的作用是累計(jì)/r/n14/r/n次考試成績(jī)超過/r/n90/r/n分的次數(shù).根據(jù)莖葉圖可得超過/r/n90/r/n分的次數(shù)為/r/n9./r/n故選：/r/nC./r/n【點(diǎn)睛】/r/n本題主要考查了循壞結(jié)構(gòu)，以及莖葉圖的認(rèn)識(shí)，解題的關(guān)鍵是弄清算法流程圖的含義，屬于基礎(chǔ)題./r/n7./r/nD/r/n解析：/r/nD/r/n【解析】/r/n【分析】/r/n_/r/n1/r/n一廠/r/nd-b/r/n根據(jù)平方運(yùn)算可求得/r/n0/r/n小=-,利用/r/ncos/r/n>=麗■求得結(jié)果./r/n【詳解】/r/n由題意可知：/r/n”+/r/n=5’/r/n+275+^'/r/n=3+2&/r/n仍=/r/n4,/r/n解得：/r/na-b/r/n=/r/n_/r/nr/r/nab/r/n1/r/nJI/r/ncosv/r/n〉=/r/n—/r/n7TT7=/r/n—/r/n==/r/n—/r/n\a\\b\/r/n2近/r/n4/r/n本題正確選項(xiàng)：/r/nD/r/n【點(diǎn)睛】/r/n本題考查向屋夾角的求解問題，關(guān)鍵是能夠通過平方運(yùn)算求得向量的數(shù)量枳./r/n8/r/n./r/nC/r/n解析：/r/nC/r/n【解析】/r/n【分析】/r/nX/r/n求得函數(shù)的導(dǎo)數(shù)廣/r/n(0/r/n=(兀-/r/n2)/r/n?(芒亠)，根據(jù)函數(shù)/“)在/r/n(1,?)/r/n上有兩個(gè)極值點(diǎn)，轉(zhuǎn)化為/r/nxe/r/nx/r/n-a=/r/n0/r/n在/r/n(he)/r/n上有不等于/r/n2/r/n的解，令/r/ng(x)=xe\/r/n利用奧數(shù)求得函數(shù)的單調(diào)性，得到/r/na>g(l)=e/r/n且/r/ndHg/r/n(/r/n2/r/n)/r/n=/r/n方，又由//r/nXx)/r/n在/r/n(1,2)/r/n上單調(diào)遞增，得到/r/nf\x)>0/r/n在/r/n(1,2)/r/n上恒成立，進(jìn)而得到/r/n>/r/nxe/r/nv/r/n在/r/n(1,2)/r/n上恒成立，借助函數(shù)/r/ng(x)=xe/r/nx/r/n在/r/n(l,g)/r/n為單調(diào)遞增函數(shù)，求得/r/no>g(2)=2K,/r/n即可得到答案./r/n【詳解】/r/n由題意，函數(shù)/r/nf(x)=(x-/r/n3)e'/r/n+6/(2/r/nInx-x/r/n+1),/r/n7/r/na/r/nxp/r/nx/r/n—/r/na/r/n可得廣/r/n(x)=/r/n+(x-3)e/r/nx/r/n+d(——1)=(x—2)(/——)=(x—2)/r/n?(/r/n)，/r/nX/r/nX/r/nX/r/n又由函數(shù)/r/n/(x)/r/n在/r/n(1,*Q/r/n)/r/n上有兩個(gè)極值點(diǎn)，/r/nX/r/n則/r/n/'/r/n(/r/nX/r/n)/r/n=0,/r/n即/r/n(/r/nX—/r/n2)/r/n?(蘭=￡)=/r/n0/r/n在/r/n(1,+s)/r/n上有兩解，/r/nX/r/n即/r/nxe/r/nx/r/n-a=/r/n0/r/n在在/r/n(1,2/r/n)/r/n上有不等于/r/n2/r/n的解，/r/n令/r/ng(x)=xe/r/nx/r/nf/r/n則/r/ng'(x)=(x+l)/r/n『/r/n>0,(x>l),/r/n所以函數(shù)/r/ng(x)=xe/r/nx/r/n在/r/n(l,*o)/r/n為單調(diào)遞增函數(shù)，/r/n所以/r/na>g(l)=e/r/n且/r/noHg/r/n(/r/n2/r/n)/r/n=2//r/n,/r/n又由/r/n/W/r/n在/r/n(1,2)/r/n上單調(diào)遞增，則/r/nr(x)>0/r/n在/r/n(1,2)/r/n上恒成立,/r/n即/r/n(/r/nx/r/n-2)/r/n?/r/n二上)/r/nX0/r/n在/r/n(1,2)/r/n上恒成立，即/r/nXe/r/nx/r/n-a</r/n0/r/n在/r/n(1,2)/r/n上恒成立，/r/nx/r/n即/r/na/r/nX/r/nxe/r/nx/r/n在/r/n(1,2)/r/n上恒成立，/r/n又由函數(shù)/r/ng(x)=x//r/n在/r/n(/r/nl,*o)/r/n為單調(diào)遞增函數(shù)，所以/r/na/r/n>/r/ng/r/n⑵/r/n=2,/r/n，綜上所述，可得實(shí)數(shù)/r/na/r/n的取值范圍是/r/na>2e\/r/n即/r/nae(2e\+^),/r/n故選/r/nC./r/n【點(diǎn)睛】/r/n本題主要考查導(dǎo)數(shù)在函數(shù)中的綜合應(yīng)用，著重考查了轉(zhuǎn)化與化歸思想、邏輯推理能力與計(jì)算能力，對(duì)導(dǎo)數(shù)的應(yīng)用的考查主要從以下幾個(gè)角度進(jìn)行：/r/n(/r/n1)/r/n考查導(dǎo)數(shù)的幾何意義，求解曲線在某點(diǎn)處的切線方程；/r/n(/r/n2)/r/n利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間，判斷單調(diào)性：已知單調(diào)性，求參數(shù)；/r/n(/r/n3)/r/n利用導(dǎo)數(shù)求函數(shù)的最值(極值)，解決函數(shù)的恒成立與有解問題，同時(shí)注意數(shù)形結(jié)合思想的應(yīng)用./r/n9/r/n./r/nA/r/n解析：/r/nA/r/n【解析】/r/n【分析】/r/n利用雙曲線/r/nC/r/n：/r/n亠_舌=/r/n1(/r/n。>/r/n0">0)/r/n的焦點(diǎn)到漸近線的距離為孚，求出/r/na,b/r/n的關(guān)系式，然后求解雙曲線的漸近線方程./r/n【詳解】/r/n雙曲線/r/nC/r/n：/r/n二—買/r/n=1(°>0/r/n上>/r/n0)/r/n的焦點(diǎn)/r/n(/r/nc,0)/r/n到漸近線/r/nbx+ciy=/r/n0/r/n的距離為/r/n空/r/nc/r/n,/r/ncrb-/r/n2/r/n可得：/r/n//r/n1/r/n"/r/n=雖,/r/n可得/r/n匕=匹,/r/n-=y/3,/r/n則/r/nC/r/n的漸近線方程為/r/ny=±y/3x./r/n2/r/nc/r/n2/r/na/r/n故選/r/nA./r/n【點(diǎn)睛】/r/n

/r/n本題考查雙曲線的簡(jiǎn)單性質(zhì)的應(yīng)用，構(gòu)建出/r/nd#/r/n的關(guān)系是解題的關(guān)鍵，考查計(jì)算能力，屬于中檔題./r/n10./r/nD/r/n解析：/r/nD/r/n【解析】/r/n由題意,/r/n3+4+5+6/r/n由題意,/r/n=4.5,/r/n4/r/nTy=0.7x40.35,/r/n:./r/n7=0.7x4.5+0.35=3.5,/r/nAt=4x3.5/r/n?/r/n2.5/r/n?/r/n4-4/r/n?/r/n5=3,/r/n故選/r/nD./r/n11/r/n?/r/nA/r/n解析：/r/nA/r/n【解析】/r/n試題分析：因?yàn)槿切问卿J角三角形，所以三角形的三個(gè)內(nèi)角都是銳角，則設(shè)邊/r/n3/r/n對(duì)的銳/r/n角為角根據(jù)余弦定理得/r/ncos6Z=/r/n2+V/r/n'~/r/n3/r/n~/r/n>0,/r/n解得/r/nx>>/5/r/n:設(shè)/r/nx/r/n邊對(duì)的銳角為/r/n4x/r/n+/r/nV/r/n-X/r/n2/r/n一/r/n0,/r/n根據(jù)余弦定理得/r/nCOS0=/r/n—-/r/n一/r/n>0,/r/n解得/r/n0/r/n<x</r/n伍,/r/n所以實(shí)數(shù)/r/nX/r/n的取值范/r/n12/r/n圍是/r/n序/r/n<X</r/n屈,/r/n故選/r/nA./r/n考點(diǎn)：余弦定理./r/n12./r/nB/r/n解析：/r/nB/r/n【解析】/r/n【分析】/r/n求解出集合/r/nM/r/n，/r/n根據(jù)并集的定義求得結(jié)呆./r/n【詳解】/r/nM/r/n=/r/n|x|log/r/n2/r/n(x-/r/n1)/r/nv0}/r/n=/r/n{科/r/n0/r/nvx-/r/n1/r/nv1}/r/n=/r/n{屮/r/nvxv2}.?.MuN={x|xn-2}/r/n本題正確選項(xiàng)：/r/nB/r/n【點(diǎn)睛】/r/n本題考查集合運(yùn)算中的并集運(yùn)算，屬于基礎(chǔ)題./r/n二、填空題/r/n【解析】【分析】由題意知漸近線方程是再據(jù)得出與的關(guān)系代入漸近線方程即可【詳解】/r/n??/r/n?雙曲線的兩個(gè)頂點(diǎn)三等分焦距.??乂/r/n??????/r/n漸近線方程是故答案為【點(diǎn)睛】本題考查雙曲線的兒何性質(zhì)即雙曲線的漸近線方程為屬于基礎(chǔ)題/r/n解析：/r/ny/r/n=/r/n±2>/2x/r/n【解析】/r/n【分析】/r/n由題意知，漸近線方程是/r/ny=±-x/r/nf/r/n2d=/r/n：/r/nx2c,/r/n再據(jù)/r/nc/r/n2/r/n=a/r/n2/r/n+b/r/n2/r/n,/r/n得出/r/nb/r/n與/r/nd/r/n的關(guān)/r/na/r/n3/r/n系，代入漸近線方程即可./r/n【詳解】/r/n??/r/n?雙曲線二—匚=/r/n1/r/n(a/r/n>/r/n0,b/r/n>0)/r/n的兩個(gè)頂點(diǎn)三等分焦距，/r/nCTb?/r/nc=3a./r/n又/r/nc/r/n2/r/n=a/r/n2/r/n+b/r/n2/r/n,:/r/n?b=2/r/n忑/r/nci/r/n???/r/n漸近線方程是/r/ny=±-x=/r/n±2/r/n屈/r/n，/r/n故答案為/r/ny/r/n=±2/r/nJIx./r/na/r/n【點(diǎn)睛】/r/n本題考查雙曲線的幾何性質(zhì)即雙曲線/r/n2/r/n-厶/r/n=1/r/n(a>0,b>0)/r/n的漸近線方程為/r/ny=±—x/r/na~b~/r/na/r/n屬于基礎(chǔ)題./r/n1006/r/n【解析】試題分析:由題設(shè)可知在中由此可得由正弦定理可得解之得乂/r/n因?yàn)樗詰?yīng)填/r/n1006/r/n考點(diǎn)：正弦定理及運(yùn)用/r/n解析：/r/n100\/r/n心/r/n【解析】/r/n試題分析油題設(shè)可知在中/r/n嚴(yán)血/r/n=30/r/n°/r/n=/r/n,由此可得/r/nGCB/r/n=/r/n斗/r/n5°,/r/n由/r/nCB/r/n_/r/n600/r/n正弦定理可得血/r/n30/r/n：/r/n血/r/n45/r/n：/r/n，解之得/r/nCB/r/n=/r/n30072/r/n，又因?yàn)?r/nZCBD/r/n=/r/n30/r/n［所以/r/nCD/r/n=/r/nCB/r/ntan30'/r/n=100^/6/r/n，應(yīng)填/r/n100/r/nv/r/n6/r/n考點(diǎn)：正弦定理及運(yùn)用./r/n【解析】【分析】利用通項(xiàng)公式即可得出【詳解】解：/r/n(/r/nl+3x)n/r/n的展開/r/n式中通項(xiàng)公式:/r/nTr+1(3x)r=3rxr//r/n含有/r/nx2/r/n的系數(shù)是/r/n54/.r=2/./r/n54/r/n可得/r/n6.*.6nEN*/r/n解得/r/nn=4/r/n故答案為/r/n4/r/n【點(diǎn)睛】本題考/r/n解析：/r/n4/r/n【解析】/r/n【分析】/r/n利用通項(xiàng)公式即可得出./r/n【詳解】/r/n解：/r/n(l+3x)/r/n"的展開式中通項(xiàng)公式：/r/n=/r/nQ/r/n；/r/n(3x)/r/n『=/r/n30h./r/n??/r/n?含有/r/n*/r/n的系數(shù)是/r/n54,/r/n?/r/n"=2./r/n

/r/n?/r/n°/r/n?/r/n3'C/r/n；/r/n=54,/r/n可得年=/r/n6,/r/n?/r/n:/r/n1/r/n）/r/n=6,”GN*./r/n解得/r/n"=4./r/n故答案為/r/n4./r/n【點(diǎn)睛】/r/n本題考查了二項(xiàng)式定理的通項(xiàng)公式，考查了推理能力與計(jì)算能力，屬于基礎(chǔ)題./r/n2+8/r/n）/r/n【解析】/r/n分析：根據(jù)偶次根式下被開方數(shù)非負(fù)列不等式解對(duì)數(shù)不等式得函數(shù)定義域詳解：要使函數(shù)有意義則解得即函數(shù)的定義域?yàn)辄c(diǎn)睛：求給定函數(shù)的定義域往往需轉(zhuǎn)化為解不等式（組）的問題/r/n解析：/r/n[2,/r/n+00/r/n）/r/n【解析】/r/n分析：根據(jù)偶次根式卜?被開方數(shù)非負(fù)列不等式，解對(duì)數(shù)不等式得函數(shù)定義域./r/n詳解：要使函數(shù)/r/n//r/n（/r/nX/r/n）/r/n有意義，則/r/nlog/r/n2/r/nx-l>0,/r/n解得即函數(shù)/r/n//r/n（/r/nX/r/n）/r/n的定義域?yàn)?r/n[2,+8/r/n）/r/n./r/n點(diǎn)睛：求給定函數(shù)的定義域往往需轉(zhuǎn)化為解不等式（組）的問題./r/n17/r/n?/r/n【解析】由題意二項(xiàng)式展開的通項(xiàng)令得則的系數(shù)是考點(diǎn)：/r/n1/r/n二項(xiàng)式定理的展開式應(yīng)用/r/n解析：/r/n35/r/n【解析】/r/n由題意，二項(xiàng)式（丘+丄/r/n）7/r/n展開的通項(xiàng)/r/n7/r/n；/r/n+]=（7/r/n；/r/n（/r/n門/r/n1（/r/n丄）「=（/r/n7/r/n；/r/n嚴(yán)/r/n4//r/n\令/r/n21—/r/n4//r/n?/r/n=5,/r/n得/r/nr=4,/r/n則疋的系數(shù)是/r/nc/r/n；/r/n=35./r/n考點(diǎn)：/r/n1./r/n二項(xiàng)式定理的展開式應(yīng)用./r/n18./r/n【解析】/r/n在等腰梯形/r/nABCD/r/n中由得所以考點(diǎn)：平面向量的數(shù)量積/r/n29/r/n解析：/r/n-/r/n【解析】/r/n在等腰梯形/r/nABCD/r/n中，由/r/nABHDC/r/n//r/nAB=/r/n2,/r/nBC/r/n=/r/nL/r/nZABC/r/n=60°,/r/n得/r/n?/r/n?/r/nI/r/n?/r/nI/r/n?/r/n?/r/nI/r/nI/r/n■/r/n■/r/n?/r/n*/r/n/I/r/n?/r/n■/r/n\/r/n//r/nI./r/n■/r/n\/r/n首而卜麗砸/r/n+1/r/n阮冠+吉殛叫疏而“+出￡嗚/r/nADBC=-/r/n//r/nab/r/n?麗/r/n=/r/n1/r/n，/r/nDC=—/r/n人氏所以/r/n首而卜麗砸/r/n+1/r/n阮冠+吉殛叫疏而“+出￡嗚/r/n=（而+|說、（而+上/r/n?考點(diǎn)：平面向量的數(shù)屋積./r/n【解析】/r/n分析：由可得代入利用復(fù)數(shù)乘法運(yùn)算法則整理后直接利用求模公/r/n式求解即可詳解：因?yàn)樗怨蚀鸢笧辄c(diǎn)睛：本題主要考查的是共輒復(fù)數(shù)的概念與運(yùn)算以及復(fù)數(shù)的乘法的運(yùn)算屬于中檔題解題時(shí)一定要注意和/r/n解析：/r/n原/r/n【解析】/r/n分析：由/r/nZ=-1-1,/r/n可得召=_/r/n1+1/r/n，/r/n代入/r/n(/r/n1—z)/r/n?/r/nZ,/r/n利用復(fù)數(shù)乘法運(yùn)算法則整理后，直接利用求模公式求解即町/r/n?/r/n詳解：因?yàn)?r/nz=-l-i,/r/n所以/r/n｛=_1+1/r/n，/r/n.?.|(l-Z/r/n)/r/n-z|=|(l+l+0-(-l+/)|=|(2+z).(-l+/)|/r/n=|-3+z|=V9+T=5/10,/r/n故答案為佰./r/n點(diǎn)睛：本題主要考查的是共軌復(fù)數(shù)的概念與運(yùn)算以及復(fù)數(shù)的乘法的運(yùn)算，屬于中檔題.解題時(shí)一定要注意尸=一/r/n1/r/n和(/r/nci/r/n+/r/nZ?z)(c/r/n+di)=(ac/r/n-Z?r/)/r/n+/r/n(ad/r/n+/r/nbe)i/r/n【解析】【分析】/r/n由函數(shù)單調(diào)遞增可得導(dǎo)函數(shù)在區(qū)間內(nèi)大于等于零恒成立根據(jù)分離變量的方式得到在上恒成立利用二次函數(shù)的性質(zhì)求得的最大值進(jìn)而得到結(jié)果/r/n【詳解】/r/n函數(shù)在上單調(diào)遞增在上恒成立在上恒成立令根據(jù)二次函數(shù)的/r/n解析:/r/nI/r/n【解析】/r/n【分析】/r/n由函數(shù)單調(diào)遞增町得導(dǎo)函數(shù)在區(qū)間內(nèi)人于等于零恒成立，根據(jù)分離變量的方式得到/r/na>x-2x/r/n2/r/n在/r/n(/r/n0,+s)/r/n上恒成立，利用二次函數(shù)的性質(zhì)求得/r/nx/r/n-2x/r/n2/r/n的最人值，進(jìn)而得到結(jié)果./r/n【詳解】/r/n函數(shù)/r/n/(x)=x/r/n2/r/n-x+l+?liix/r/n在/r/n(/r/n0,+8/r/n)/r/n上單調(diào)遞增/r/n.?.f\x)=2x-l+->0/r/n在/r/n(/r/n0,+s)/r/n上恒成立/r/na/r/n2F/r/n在/r/n(O,+a)/r/n上恒成立/r/n-X/r/n令/r/ng(x)=x-2x‘/r/n，/r/nx>0/r/n根據(jù)二次函數(shù)的性質(zhì)可知：當(dāng)/r/nx=-/r/n時(shí)，/r/ng(x)/r/n=；/r/n4/r/n'/r/n/max/r/n8/r/n???心；,/r/n故實(shí)數(shù)°的最小值是;/r/n88/r/n本題正確結(jié)果：/r/n【點(diǎn)睛】/r/n本題考查根據(jù)函數(shù)在區(qū)間內(nèi)的單調(diào)性求解參數(shù)范闈的問題，關(guān)鍵是能將問題轉(zhuǎn)化為導(dǎo)函數(shù)的符號(hào)的問題，通過分離變量的方式將問題轉(zhuǎn)變?yōu)閰?shù)與函數(shù)最值之間的關(guān)系問題./r/n三、解答題/r/n(1)/r/n；；/r/n(2)40/r/n；/r/n(/r/n3)/r/n選/r/nB/r/n款訂餐軟件./r/n2/r/n【解析】/r/n【分析】/r/n⑴運(yùn)用列舉法給出所有情況，求出結(jié)果/r/n⑵由眾數(shù)結(jié)合題意求出平均數(shù)/r/n⑶分別計(jì)算出使用/r/nA/r/n款訂餐、使用/r/n3/r/n款訂餐的平均數(shù)進(jìn)行比較，從而判定/r/n【詳解】/r/n（/r/n1/r/n）/r/n使用人款訂餐軟件的商家中“平均送達(dá)時(shí)間”不超過/r/n20/r/n分鐘的商家共有/r/n100x0.006x10=6/r/n個(gè)，分別記為甲，/r/na,b,c,d,e,/r/n從中隨機(jī)抽取/r/n3/r/n個(gè)商家的情況如下：共/r/n20/r/n種./r/n{/r/n甲,/r/na,b},/r/n{/r/n甲,/r/na,c},{/r/n甲,/r/na/r/n9/r/nd},/r/n{/r/n甲,/r/na,e},/r/n{/r/n甲,/r/nb,c},/r/n{/r/n甲,/r/nb,d},/r/n{/r/n甲，/r/nb,e},/r/n{/r/n甲,/r/nc,d}{/r/n甲,/r/nc,e},/r/n{/r/n甲，/r/nd/r/n9/r/ne},{a,b,c}?{a,b/r/n9/r/nd},{a,b/r/n9/r/ne},{a/r/n9/r/nc,d},{a/r/n9/r/nc,e},{a,d,e}/r/nt/r/n{b/r/n9/r/nc,d},{b,c,e}/r/nt/r/n{b,d/r/n9/r/ne}/r/nt/r/n{c,d,e}./r/n甲商家被抽到的情況如卞：共/r/nio/r/n種./r/n{/r/n甲,/r/na,b},/r/n{/r/n甲,/r/na,c},{/r/n甲,/r/na/r/n9/r/nd},/r/n{/r/n甲,/r/na,e},/r/n{/r/n甲,/r/nb,c},/r/n{/r/n甲,/r/nb,d},/r/n{/r/n甲，/r/nb,e},/r/n{/r/n甲,/r/nc,d},{/r/n甲，/r/nc,e},/r/n{/r/n甲,/r/nd,e}/r/n記事件/r/nA/r/n為甲商家被抽到，則/r/nP（A）=-=-./r/nv/r/n202/r/n（/r/n2）/r/n依題意可得，使用/r/n4/r/n款訂餐軟件的商家中“平均送達(dá)時(shí)間”的眾數(shù)為/r/n55,/r/n平均數(shù)為/r/n15x0.06+25x0.34+350.12+45x0.04+55x0.4+65x0.04=40./r/n（/r/n3）/r/n使用/r/n3/r/n款訂餐軟件的商家中“平均送達(dá)時(shí)間”的平均數(shù)為/r/n15x0.04+25x0.2+35x0.56+45x0.14+55x0.04+65x0.02=35v40/r/n所以選/r/nB/r/n款訂餐軟件./r/n【點(diǎn)睛】/r/n本題主要考查了頻率分布直方圖，平均數(shù)和眾數(shù)，占典概率等基礎(chǔ)知識(shí)，考查了數(shù)據(jù)處理能力以及運(yùn)算求解能力和應(yīng)用意識(shí)，屬于基礎(chǔ)題./r/nr-/r/n3/r/n⑴/r/n=1（2）/r/n—/r/n?/r/n4/r/n-/r/n2/r/n【解析】/r/n【分析】/r/n（1）/r/n設(shè)出/r/nA/r/n、/r/nP/r/n點(diǎn)坐標(biāo)，用/r/nP/r/n點(diǎn)坐標(biāo)表示/r/nA/r/n點(diǎn)坐標(biāo)，然后代入圓方程，從而求出/r/nP/r/n點(diǎn)的軌跡；/r/n（/r/n2）/r/n設(shè)出/r/nP/r/n點(diǎn)坐標(biāo)，根據(jù)斜率存在與否進(jìn)行分類討論，當(dāng)斜率不存在時(shí)，求出/r/nZOQ/r/n面積的值，當(dāng)斜率存在時(shí)，利用點(diǎn)/r/nP/r/n坐標(biāo)表示/r/nAPOg/r/n的面積，減元后再利用函數(shù)單調(diào)性求出最值，最后總結(jié)出最值./r/n【詳解】/r/n解：/r/n（1）/r/n設(shè)/r/nP（x,y）,/r/n由題意得：/r/nA（x/r/np/r/ny）,B/r/n（/r/nO,y）,/r/n由/r/nBP=2BA，/r/n可得點(diǎn)/r/n4/r/n是/r/nBP/r/n的中點(diǎn),故/r/nx+0=/r/n2/r/n兀,/r/nY/r/n所以/r/n1/r/n2/r/n又因?yàn)辄c(diǎn)人在圓上，/r/n所以得_+/=/r/n1,/r/n4/r/n?/r/n故動(dòng)點(diǎn)/r/nP/r/n的軌跡方程為/r/n—+y/r/n2/r/n=l./r/n4/r/n⑵設(shè)/r/nPg,yJ,/r/n則%工/r/n0,/r/n且手+才/r/n=i,/r/n3/r/n當(dāng)不/r/n=0/r/n時(shí)，兒=±/r/n1,/r/n此時(shí)/r/n0/r/n（/r/n3,O/r/n）/r/n,S“O0=7/r/n；/r/n當(dāng)兀工/r/n0/r/n時(shí)，/r/nk/r/nop/r/n=/r/n—/r/n,/r/nA/r/n1/r/n因?yàn)?r/nOP/r/n丄/r/nOQ,/r/n???|OP|/r/n=/r/n，/r/n孫/r/nfl/r/n仔，/r/nS“/r/n。。/r/n專的四嶺普/r/n?/r/n(°<昨/r/n1)/r/n,、/r/n4/r/nz/r/n、/r/n設(shè)/r/n/(x)=/r/n——/r/n3x(0<x<l)/r/nX/r/n4/r/n因?yàn)閺V/r/nCr)/r/n=/r/n-/r/n3</r/n0/r/n恒成立,/r/n.J(x)/r/n在/r/n(/r/n0,1］/r/n上是減函數(shù)，/r/n

/r/n當(dāng)/r/nx/r/n=1/r/n時(shí)有最小值，即/r/nSwo/r/n3/r/n綜上：/r/nS'POQ/r/n的最小值為亍/r/n【點(diǎn)睛】/r/n本題考查了點(diǎn)的軌跡方程、橢圓的性質(zhì)等知識(shí)，求解幾何圖形的長度、面枳等的最值時(shí)，常見解法是設(shè)出變量，用變量表示出幾何圖形的長度、面積等，減元后借助函數(shù)來研究其最值./r/n23./r/n（1）/r/n（x_3）'+）F=9/r/n（2）2^7/r/n./r/n【解析】/r/n分析：/r/n（/r/n1/r/n）/r/n將。=/r/n6cos&/r/n兩邊同乘根據(jù)直角坐標(biāo)與極坐標(biāo)的對(duì)應(yīng)關(guān)系得出直角坐標(biāo)方程；/r/n（/r/n2）/r/n將直線的參數(shù)方程代入圓的普通方程，根據(jù)參數(shù)的幾何意義與根與系數(shù)的關(guān)系得出/r/n|PA|+|PB./r/n詳解：/r/n（/r/n1）/r/n由/r/nQ=/r/n6cos&,/r/n得/r/nX?'/r/n=6pcosQ/r/n,/r/n化為直角坐標(biāo)方程為/r/nx/r/n2/r/n+/r/ny/r/n2/r/n=/r/n6x/r/n,/r/n即(/r/nx-3)/r/n2/r/n+y/r/n2/r/n=9/r/n（/r/n2）/r/n將/r/n1/r/n的參數(shù)方程帶入圓/r/nC/r/n的直角坐標(biāo)方程，得尸/r/n+/r/n2/r/n（/r/ncosa—sina”/r/n一/r/n7=0/r/n因?yàn)檑獭?。，可設(shè)人心是上述方程的兩根，所以/r/n心+『/r/n因?yàn)檑獭?。，可設(shè)人心是上述方程的兩根，所以/r/n心+『/r/n2/r/n=-2(coscr-sma)/r/n2=—7/r/n又因?yàn)?r/n（/r/n2,1）/r/n為直線所過定點(diǎn),/r/n.?.|則+/r/n巴旦吐旦/r/n廠『/r/n2|/r/n=/r/n+fJ_4U/r/n=5/32-4sm2<z>V32-4/r/n=/r/n2打/r/n所以+|刖|的最小值為/r/n20/r/n點(diǎn)睛：本題考查了極坐標(biāo)方程與直角坐標(biāo)方程的轉(zhuǎn)化，參數(shù)方程的幾何意義與應(yīng)用，屬于基礎(chǔ)題./r/n24./r/n(/r/nI/r/n)/r/nB/r/n冷(/r/nII)77+1/r/n【解析】/r/n【分析】/r/n【詳解】/r/n(1)Va=bcosC+csniB/r/n:.由正弦定理知/r/nsiiiA=siiiBcosC+sinCsuiB/r/n①在三角形/r/nABC/r/n中，/r/nA=zT-(B+C)/r/nAsuiA=sin(B+C)=suiBcosC+cosBsiuC/r/n②由①和②得/r/nsuiBsmC=cosBsinC/r/n而/r/nCG(O,/r/n丸)，/r/nsinCiO,sniB=cosB/r/n又/r/nB(0,/r/n丸)，/r/nZ.B=-/r/nTOC\o"1-5"\h\z/r/n(2)/r/nSmbc=/r/n—/r/nacsinB=/r/nac,/r/n/r/n2/r/n4/r/n由斷及余弦定理得：/r/n4=/r/n宀-/r/n2.COS/r/n介/r/n2/r/n心/r/n2/r/n心孚/r/n//r/n4/r/n整理得：/r/n?c<-/r/n__/r/n,當(dāng)且僅當(dāng)/r/nQ=C/r/n時(shí)，等號(hào)成立，/r/n2—/r/n則/r/nAABC/r/n面積的最人值為/r/n-x^x^-==lxV2x(2+?)/r/n=/r/nJ2+1-/r/n2/r/n2/r/n2-V2/r/n2/r/n(I)/r/n0.55/r/n：/r/n(II)/r/n詳見解析/r/n【解析】/r/n【分析】/r/n【詳解】/r/n解：/r/n(/r/nI)/r/n設(shè)甲勝/r/nA/r/n的事件為/r/nD,/r/n乙勝/r/nB/r/n的事件為￡,丙勝/r/nC/r/n的事件為/r/nF,/r/n則/r/nD,E,F/r/n分別表示甲不勝/r/nA/r/n、/r/n乙不勝/r/nB,/r/n丙不勝/r/nC/r/n的事件./r/n因?yàn)?r/nP(D)/r/n=/r/n0.6,/r/nP(E)/r/n=/r/n0.5,P(F)/r/n=/r/n0.5/r/n,/r/nP(D)=/r/n0.4,/r/nP(E)=/r/n0.5,/r/nP(F)=/r/n0.5./r/n紅隊(duì)至少兩人獲勝的事件有：/r/ndeF/r/n、/r/ndEf/r/n、/r/nDef/r/n、/r/ndef/r/n,/r/n由于以上四個(gè)事件兩兩互斥且各盤比賽的結(jié)果相互獨(dú)立，因此紅隊(duì)至少兩人獲勝的概率/r/nP/r/n=/r/nP(DEF)/r/n+/r/nP(DEF)/r/n+/r/nP(DEF)/r/n+/r/nP(DEF)/r/n=0.6x0.5x0.5+0.6x0.5x0.5+0.4x0.5x0.5+0.6x0.5x0.5=0.55/r/n(II)/r/n由題意知；可能的取值為/r/n0,/r/n1,2,/r/n3./r/n又由/r/n(/r/nI/r/n)/r/n知/r/ndef,def/r/n、/r/ndef/r/n是兩兩互斥事件，且各盤比賽的結(jié)呆相互獨(dú)立，