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        1. (Ⅱ)記.規(guī)定.求數(shù)列的前項和. 和平區(qū)2008-2009學(xué)年度第二學(xué)期高三年級 查看更多

           

          題目列表(包括答案和解析)

          (14分)已知數(shù)列中,當(dāng)有:

          (Ⅰ)設(shè)數(shù)列滿足,證明散列為等比數(shù)列,并求數(shù)列的通項公式;

          (Ⅱ)記,規(guī)定,求數(shù)列的前項和

           

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          0    9.2    9.5    8.8    9.6    9.7
          現(xiàn)從上面6個分值中隨機(jī)的一個一個地不放回抽取,規(guī)定抽到數(shù)9.6或9.7,抽取工作即停止.記在抽取到數(shù)9.6或9.7所進(jìn)行抽取的次數(shù)為ξ,求ξ的分布列及數(shù)學(xué)期望.

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          在進(jìn)行一項擲骰子放球游戲中,規(guī)定:若擲出1點,甲盒中放一球;若擲出2點或3點,乙盒中放一球,若擲出4點或5點或6點,丙盒中放一球,前后共擲3次,設(shè)x,y z 分別表示甲,乙,丙3個盒中的球數(shù).
          (1)求x,y,z依次成公差大于0的等差數(shù)列的概率;
          (2)記ξ=x+y,求隨機(jī)變量ξ的概率分布列和數(shù)學(xué)期望.

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          數(shù)列{2n-1}的前n項組成集合An={1,3,7,…,2n-1}(n∈N*),從集合An中任取k(k=1,2,3,…,n)個數(shù),其所有可能的k個數(shù)的乘積的和為Tk(若只取一個數(shù),規(guī)定乘積為此數(shù)本身),記Sn=T1+T2+…+Tn.例如:當(dāng)n=1時,A1={1},T1=1,S1=1;當(dāng)n=2時,A2={1,3},T1=1+3,T2=1×3,S2=1+3+1×3=7.
          (Ⅰ)求S3
          (Ⅱ)猜想Sn,并用數(shù)學(xué)歸納法證明.

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          奇瑞公司生產(chǎn)的“奇瑞”轎車是我國民族汽車品牌.該公司2009年生產(chǎn)的“旗云”、“風(fēng)云”、“QQ”三類經(jīng)濟(jì)型轎車中,每類轎車均有舒適型和標(biāo)準(zhǔn)型兩種型號.某周產(chǎn)量如下表:
          車型 旗云 風(fēng)云 QQ
          舒適 100 150 x
          標(biāo)準(zhǔn) 300 y 600
          若按分層抽樣的方法在這一周生產(chǎn)的轎車中抽取50輛進(jìn)行檢測,則必須抽取“旗云”轎車10輛,“風(fēng)云”轎車15輛.
          (Ⅰ)求x,y的值;
          (Ⅱ)在年終促銷活動中,獎給了某優(yōu)秀銷售公司2輛舒適型和3輛標(biāo)準(zhǔn)型“QQ”轎車,該銷售公司又從中隨機(jī)抽取了2輛作為獎品回饋消費者.求至少有一輛是舒適型轎車的概率;
          (Ⅲ)今從“風(fēng)云”類轎車中抽取6輛,進(jìn)行能耗等各項指標(biāo)綜合評價,并打分如下:9.0、9.2、9.5、8.8、9.6、9.7,現(xiàn)從上面6個分值中隨機(jī)的一個一個地不放回抽取,規(guī)定抽到數(shù)9.6或9.7,抽取工作即停止.記在抽取到數(shù)9.6或9.7所進(jìn)行抽取的次數(shù)為ξ,求ξ的分布列及數(shù)學(xué)期望.

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          一、選擇題(每小題5分,共50分)

            1.C  2.B  3.D  4.A  5.C  6.B  7.A  8.C  9.B  10.D

          二、填空題(每小題4分.共24分)

            11.5  12.4   13.3825   14.6ec8aac122bd4f6e 15.6ec8aac122bd4f6e   16.3

          三.解答題(本大題共6小題,共76分)

          17.(本題12分)

          解:(Ⅰ)∵6ec8aac122bd4f6e

          6ec8aac122bd4f6e          ………………………(2分)

          6ec8aac122bd4f6e    ………………………(3分)

          6ec8aac122bd4f6e              ……………………(4分)

          ∵在6ec8aac122bd4f6e中,6ec8aac122bd4f6e

          6ec8aac122bd4f6e                             ………………………(5分)

          (Ⅱ)設(shè)6ec8aac122bd4f6e分別是6ec8aac122bd4f6e6ec8aac122bd4f6e的對邊,

          ∵?6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e    ①                                          ……………………(6分)

          由正弦定理:6ec8aac122bd4f6e,得 6ec8aac122bd4f6e             ……………………(7分)

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e    ②                                          ……………………(8分)

          由①②解得6ec8aac122bd4f6e                                    ……………………(9分)

          由余弦定理,得6ec8aac122bd4f6e                       ………………(10分)

                           6ec8aac122bd4f6e

                           6ec8aac122bd4f6e                                       ………………(11分)

          6ec8aac122bd4f6e,即邊6ec8aac122bd4f6e的長為6ec8aac122bd4f6e。                              ……………………(12分)

           

          18.(本題12分]

          解:(Ⅰ)∵6ec8aac122bd4f6e是偶函數(shù),

          6ec8aac122bd4f6e,即6ec8aac122bd4f6e            ……(2分)

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e                                  ………………………………(4分)

          6ec8aac122bd4f6e對一切6ec8aac122bd4f6e恒成立。

          6ec8aac122bd4f6e                                    ……………………………………(6分)

          (Ⅱ)由6ec8aac122bd4f6e                            ………………(7分)

          6ec8aac122bd4f6e

            6ec8aac122bd4f6e                                         …………………(8分)

            6ec8aac122bd4f6e

          錯誤!不能通過編輯域代碼創(chuàng)建對象。6ec8aac122bd4f6e                                           ……………………(10分)

          6ec8aac122bd4f6e6ec8aac122bd4f6e                                    …………………(11分)

          6ec8aac122bd4f6e6ec8aac122bd4f6e

          ∴若使方程6ec8aac122bd4f6e有解,則6ec8aac122bd4f6e的取值范圍是6ec8aac122bd4f6e6ec8aac122bd4f6e          ………………(12分)

           

          19.(本題12分)

          解:(Ⅰ) ∵6ec8aac122bd4f6e分別是6ec8aac122bd4f6e的中點,

          6ec8aac122bd4f6e6ec8aac122bd4f6e                                  ……………………(1分)

          6ec8aac122bd4f6e平面6ec8aac122bd4f6e,6ec8aac122bd4f6e平面6ec8aac122bd4f6e

          6ec8aac122bd4f6e平面6ec8aac122bd4f6e                                   …………………………(2分)

          6ec8aac122bd4f6e平面6ec8aac122bd4f6e6ec8aac122bd4f6e平面6ec8aac122bd4f6e,6ec8aac122bd4f6e平面6ec8aac122bd4f6e

          6ec8aac122bd4f6e                                          …………………………(4分)

          6ec8aac122bd4f6e6ec8aac122bd4f6e的中點,

          6ec8aac122bd4f6e6ec8aac122bd4f6e的中點.

          6ec8aac122bd4f6e                               ………………………(5分)

          6ec8aac122bd4f6e

          ∴四邊形6ec8aac122bd4f6e是平行四邊形                         …………………………(6 分)

          (Ⅱ)當(dāng)6ec8aac122bd4f6e時,平面6ec8aac122bd4f6e平面6ec8aac122bd4f6e                   …………………(8分)

          6ec8aac122bd4f6e上取一點,6ec8aac122bd4f6e連接6ec8aac122bd4f6e

          6ec8aac122bd4f6e當(dāng)6ec8aac122bd4f6e時,

          6ec8aac122bd4f6e,6ec8aac122bd4f6e

          6ec8aac122bd4f6e

                 6ec8aac122bd4f6e

                 6ec8aac122bd4f6e

          即當(dāng)6ec8aac122bd4f6e時,6ec8aac122bd4f6e,6ec8aac122bd4f6e                    ……………………(9分)

          6ec8aac122bd4f6e,6ec8aac122bd4f6e,6ec8aac122bd4f6e

          6ec8aac122bd4f6e平面6ec8aac122bd4f6e                                       ……………………(10分)

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e平面6ec8aac122bd4f6e                                 ……………………………(11分)

          6ec8aac122bd4f6e平面6ec8aac122bd4f6e

          ∴平面6ec8aac122bd4f6e6ec8aac122bd4f6e平面6ec8aac122bd4f6e                     …………………………………(12分)

           

           

           

          20.(本題12分)

          解:(Ⅰ) ∵6ec8aac122bd4f6e

          6ec8aac122bd4f6e                           ………………………………(2分)

          6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e上為減函數(shù)

          6ec8aac122bd4f6e≤O在區(qū)間6ec8aac122bd4f6e上恒成立                      …………………………(3分)

          6ec8aac122bd4f6e是開口向上的拋物線

          6ec8aac122bd4f6e6ec8aac122bd4f6e        6ec8aac122bd4f6e6ec8aac122bd4f6e      6ec8aac122bd4f6e6ec8aac122bd4f6e

          ∴只需              即                               …………………………(5分)

                  6ec8aac122bd4f6e6ec8aac122bd4f6e       6ec8aac122bd4f6e6ec8aac122bd4f6e

          6ec8aac122bd4f6e6ec8aac122bd4f6e6ec8aac122bd4f6e                             ………………………………………(6分)

          6ec8aac122bd4f6e6ec8aac122bd4f6e                                      

          6ec8aac122bd4f6e(Ⅱ)當(dāng)6ec8aac122bd4f6e時,                     

                                                

          ∴存在6ec8aac122bd4f6e,使得6ec8aac122bd4f6e

          6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e內(nèi)有且只有一個極小值點                      ……………(8分)

          6ec8aac122bd4f6e6ec8aac122bd4f6e                   

          6ec8aac122bd4f6e當(dāng)6ec8aac122bd4f6e時                      

           

          ∴存在6ec8aac122bd4f6e,使得6ec8aac122bd4f6e

          6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e內(nèi)有且只有一個極大值點                     ……………(10分)

          當(dāng)6ec8aac122bd4f6e6ec8aac122bd4f6e6ec8aac122bd4f6e時,由(Ⅰ)可知6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e上為減函數(shù)

          6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e內(nèi)沒有極值點.

          綜上可知,當(dāng)6ec8aac122bd4f6e6ec8aac122bd4f6e時,6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e內(nèi)的極值點個數(shù)為6ec8aac122bd4f6e

          當(dāng)6ec8aac122bd4f6e6ec8aac122bd4f6e6ec8aac122bd4f6e時,6ec8aac122bd4f6e在區(qū)間6ec8aac122bd4f6e內(nèi)的極值點個數(shù)為6ec8aac122bd4f6e          ………(12分)

           

          6ec8aac122bd4f6e21.(本題14分)

          解:(Ⅰ)設(shè)橢圓的長半軸長為6ec8aac122bd4f6e,短半軸長6ec8aac122bd4f6e,半焦距為6ec8aac122bd4f6e

          由離心率6ec8aac122bd4f6e,得6ec8aac122bd4f6e

          6ec8aac122bd4f6e

          6ec8aac122bd4f6e            ①                                     …………………(2分)

          ∵直線6ec8aac122bd4f6e的方程為6ec8aac122bd4f6e,原點6ec8aac122bd4f6e到直線6ec8aac122bd4f6e的距離為6ec8aac122bd4f6e

          6ec8aac122bd4f6e  ②                                     …………………(4分)

          ①代人②,解得6ec8aac122bd4f6e                            ………………………(6分)

          ∴橢圓的標(biāo)準(zhǔn)方程為6ec8aac122bd4f6e                        …………………………(7分)

          (Ⅱ) ∵6ec8aac122bd4f6e

          ∴?=6ec8aac122bd4f6e

          ∴?=?(-)=2                                    …………………(9分)

          設(shè)6ec8aac122bd4f6e,則6ec8aac122bd4f6e,即6ec8aac122bd4f6e                     ………………(10分)

          6ec8aac122bd4f6e

          ∴?=2

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