资阳市2012—2013学年度高中二年级第一学期期末质量检测理科数学答案

资阳市2012—2013学年度高中二年级第一学期期末质量检测

理科数学参及评分意见

一、选择题：本大题共12个小题，每小题5分，共60分． 1－5.DBCDA；6－10. CACAB；11－12.DA

二、填空题：本大题共4个小题，每小题4分，共16分．

41 13. 9；14. ；15.；16. ①②④⑤．

33三、解答题：本大题共6个小题，共74分． 17．解: (Ⅰ)由频率分布直方图得第七组频率为： 1-(0.008×2＋0.016×2＋0.04×2＋0.06)×5＝0.06，

∴第七组的人数为0.06×50＝3. ··························································································· 3分 由各组频率可得以下数据：

组别 样本数 一 2 二 4 三 10 四 10 五 15 六 4 七 3 八 2 ····································································································································· 6分 (Ⅱ)第二组中四人可记为a、b、c、d，其中a为男生，b、c、d为女生，第七组中三人可记为1、2、3，其中1、2为男生，3为女生，所以基本事件有12个．

实验小组中恰有一男一女的事件有1b,1c,1d,2b,2c,2d,3a，共7个， ································· 10分

7因此实验小组中恰有一男一女的概率是. ······································································· 12分

1218. (Ⅰ)证明: 易知AP⊥BP，由AA1⊥平面PAB， 得AA1⊥BP， 且AP∩AA1＝A， 所以BP⊥平面PAA1， 又BP平面A1PB,

故平面A1PB平面A1AP. ··································································································· 6分 (Ⅱ)解：由题意A1AAP,A1AAB,A1ABP,APPB,PBA1P, ································ 9分 又在三棱锥A1APB的6条棱中，任取2条棱的基本事件有15种.

51故互相垂直的概率P． ·························································································· 12分

15319. 解:(Ⅰ)茎叶图 ·········································································································· 3分 统计结论：(写出以下任意两个即可) ·········································································· 5分 ①甲批树苗比乙批树苗高度整齐；

②甲批树苗的高度大多数集中在均值附近，乙批树苗的高度

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分布较为分散；

③甲批树苗的平均高度小于乙批树苗的平均高度；

④甲批树苗高度的中位数为27 cm，乙批树苗高度的中位数为28.5 cm.

1(Ⅱ)x甲＝[37＋21＋31＋20＋29＋19＋32＋23＋25＋33]＝27，

101································· 7分 x乙＝[10＋30＋47＋27＋46＋14＋26＋10＋44＋46]＝30. ·

10∴甲批树苗中高度高于平均数27的是：37,31,29,32,33，共5株，

乙批树苗中高度高于平均数30的是：47,46,44,46共4株． ······································ 9分 新的样本有9株树苗，从中任取2株的基本事件有36个， 其中“一株来自甲批，一株来自乙批”为事件A，

包含的基本事件有5×4＝20个，················································································ 11分

205∴P(A)＝＝. ········································································································ 12分

36920. （Ⅰ）证明： 以B为原点建立坐标系，得下列坐标： B(0,0,0),A(1,0,0),C(0,0,1),E(0,1,0)

2222a,0,1a),N(a,a,0)， 222222··························································································· 2分 MN(0,a,a1)， ·

2222设MNBCBE，即(0,， a,a1)=（0,0,1）+（0,1,0）

222222∴=a，=a-1，所以当=a，=a-1时，MN∥平面BCE，又MN不在

2222 F(1,1,0),M(平面BCE内，NM∥平面BCE． ···················································································· 5分

22（Ⅱ）解：由（Ⅰ）知，MN(0,a,a1)

2222∴|MN|2(0,a,a1)2a22a1，∴|MN|a22a1．

22∵|MN|a22a1＝(a∴当a221)， 222时，MN的长最小． ················································································ 8分 22111（Ⅲ）当a时，MN的中点为G(,,)，

2244111111········································································ 10分 GB(,,),GA(,,) ·

244244GAGB1=. ·所求二面角的余弦值cos·························································· 12分

3|GA||GB|资阳高二数学（理科）答案第2页（共5页）

（用其它方法做的可参照给分）

21．解: 设事件Ａ为“方程x22axb20有实根”,当a≥0, b≥0时,

方程x22axb20有实根的条件为a≥b. ····························································· 2分 (Ⅰ)基本事件共有12个(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0), (2,1),(2,2),(2,3),(3,0),(3,1)(3,2),(3,3),(4,0), (4,1), (4,2), (4,3)其中第一个数表示a的取值,第二个数表示b的取值.事件Ａ中包含14个基本事件,故事件Ａ发生的概率为

147········································································································ 6分 .·

2010 (Ⅱ)试验的全部结果所构成的区域为{(a,b)|0≤a≤4,0≤b≤3}.构成事件Ａ的区域为{(a,b)|0≤a≤4,0≤b≤3},即如变式答图P(A)的阴影区域所示. ………………………………………9分

1343252所以所求的概率为P(A).…………12分

34822. 解法1：（Ⅰ）连结BG，则BG是BE在面ABD的射影，

即∠EBG是A1B与平面ABD所成的角.设F为AB中点，连结EF、FC，∵D、E分别是CC1、A1B的中点，又DC⊥平面ABG，

∴CDEF为矩形. 连结DF，G是△ADB的重心，∴ＧFD. ···································· 2分

１ 在直角三角形EFD中，EF2＝FG·FD＝FD2，

３126∵EF＝1∴FD＝3.于是ED＝2，EG＝＝，

33∵FG＝ED＝2，∴AB＝22，A1B＝23，EB＝3， ∴sinEBG=

6127EG=＝，∴cosEBG 333EB3∴A1B与平面ABD所成的角是cosEBG7. ························································· 5分 3（Ⅱ）解法一：∵ED⊥AB，ED⊥EF，又EF∩AB＝F，∴ED⊥面A1AB， 又EDØ面AED，∴平面AED⊥平面A1AB，且面AED∩面A1AB＝AE. 作A1K⊥AE，垂足为K，

∴A1K⊥平面AED.即A1K是A1到平面AED的距离. ················································· 7分 在△A1AB1中，A1K＝

A1AA1B122226， AB1323∴A1到平面AED的距离为26. ················································································· 9分 3资阳高二数学（理科）答案第3页（共5页）

解法二：连结A1D，VA1ADEVDA1AE.∵ED⊥AB，ED⊥EF，又EF∩AB＝F，

∴ED⊥平面A1AB，设A1到平面AED的距离为h， ·················································· 7分

1则 SAEDhSA1AEED，又SA1AEA1AAB2，

4162226, ∴h. SAEDAEED2236226即A1到平面AED的距离为.················································································· 9分

31（Ⅲ）由(Ⅰ)知VDEABCVAEFCDVBEFCDABSEFCD, ··········································· 11分

3∵ABC 是等腰直角三角形, AB＝22,∴ACBC2,∴CF2,又EF1,

114∴VDEABCABSEFCD2212 ······························································· 14分

333解2: (Ⅰ)连接BG,则BG是BE在面ABD 内的射影,即A1BG是A1B与平面ABD所成的角. 如图所示,建立空间直角坐标系Cxyz,设CACB2a,

2a2a1则A(2a,0,0),B(0,2a,0),D(0,0,1),A1(2a,0,2),E(a,a,1),G(,,)

333aa2且GE(,,),BD(0,2a,1). ················································································· 2分

33322由题设GEBD,从而有GEBDa20. 解得a1.

33241∴BA1(2,2,2),BG(,,).

33314BA1BG73∴cosA1BG. 13|BA1||BG|232137故A1B与平面ABD 所成角的余弦值为cosA1BG. ··········································· 5分

3 (Ⅱ)由(Ⅰ)有A(2,0,0),A1(2,0,2),E(1,1,1),D(0,0,1).故 AEED(1,1,1)(1,1,0)0,AA1ED(0,0,2)(1,1,0)0. ∴ED平面AA1E. 又ED平面AED,∴平面AED⊥平面AA1E.又面AED面AA1EAE, 作A1KAE,垂足为K,则A1K平面AED,

即A1K是A1到平面AED的距离. ·················································································· 7分 连接EB1,在RtA1AB1中, A1K故A1到平面AED的距离为(Ⅲ) ∵VDEABCVEABDA1AA1B122226. AB132326. ·················································································· 9分 31111VCABDVEABDVDABCEGABDFCDABAC

3232资阳高二数学（理科）答案第4页（共5页）

221由(Ⅰ)知A(2,0,0),B(0,2,0),E(1,1,1),G(,,), ·························································· 11分

3332216, AB22,GE(1)2(1)2(1)23333 ∵ABC 是等腰直角三角形,∴ACBC2,∵AA12,D是CC1的中点,CD1, ∴ADBD=5,设F为AB中点，则DFAB,且DF3, ····························· 13分 161114∴VDEABC=·············································· 14分 223122.

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