題目列表(包括答案和解析)
touch
rely
loss
agriculrure
B.閱讀理解:(30分)
Now we can see a man and his wife at the breakfast table. They are not speaking to each other.
They haven’t spoken to each other at the breakfast table for years. The husband is reading his newspaper. We can’t see his face. The wife looks very worried as she gets a cup of tea ready for him. Today she is using a new kind of tea for the first time. The husband picks up his cup. He isn’t interested. He tastes his tea. Suddenly he puts down his newspaper. Something is different! Can it be the tea? He takes another taste. It’s wonderful. He smiles. He looks at his wife and says in surprise, “Doris, when did you cut your hair?” Doris is pleased. She answers, “Two months ago.” Doris asks, “Herbie, when did your hair begin to become white?” He answers, “A long time ago.” Doris says, “We have been together for many years, but we never cared about each other.” Now they aren’t worried any longer. Breakfast is different. Has a new kind of tea changed their lives?
36. This story happens______________________.
A. before breakfast B. after breakfast
C. at home D. in a teahouse
37. In the passage, we can see ________________________.
A. Doris is drinking tea B. Herbie likes the new kind of tea
C. Doris is reading a newspaper D. Herbie is very young and good-looking
38. Herbie and Doris lived ______________ before this day.
A. a wonderful B. an unhappy
C. an enjoyable D. a friendly
39. Which of the following statements is true?
A. They are good friends. B. They have just got married.
C. They like to talk about their hair. D. They are no longer young.
40. From the passage, we think it may be ______________.
A. a radio programme B. a short film
C. a computer game D. a beautiful painting
1.解析:,故選A。
2.解析:抽取回族學生人數(shù)是,故選B。
3.解析:由,得,此時,所以,,故選C。
4.解析:∵∥,∴,∴,故選C。
5.解析:設公差為,由題意得,;,解得或,故選C。
6.解析:∵雙曲線的右焦點到一條漸近線的距離等于焦距的,∴,又∵,∴,∴雙曲線的漸近線方程是,故選D.
7.解析:∵、為正實數(shù),∴,∴;由均值不等式得恒成立,,故②不恒成立,又因為函數(shù)在是增函數(shù),∴,故恒成立的不等式是①③④。故選C.
8.解析:∵,∴在區(qū)間上恒成立,即在區(qū)間上恒成立,∴,故選D。
9.解析:∵
,∴此函數(shù)的最小正周期是,故選C。
10.解析:如圖,∵正三角形的邊長為,∴,∴,又∵,∴,故選D。
11.解析:∵在區(qū)間上是增函數(shù)且,∴其反函數(shù)在區(qū)間上是增函數(shù),∴,故選A
12.解析:如圖,①當或時,圓面被分成2塊,涂色方法有20種;②當或時,圓面被分成3塊,涂色方法有60種;
③當時,圓面被分成4塊,涂色方法有120種,所以m的取值范圍是,故選A。
13.解析:將代入結果為,∴時,表示直線右側區(qū)域,反之,若表示直線右側區(qū)域,則,∴是充分不必要條件。
14.解析:∵,∴時,,又時,滿足上式,因此,。
15.解析:設正四面體的棱長為,連,取的中點,連,∵為的中點,∴∥,∴或其補角為與所成角,∵,,∴,∴,又∵,∴,∴與所成角的余弦值為。
16.解析:∵,∴,∵點為的準線與軸的交點,由向量的加法法則及拋物線的對稱性可知,點為拋物線上關于軸對稱的兩點且做出圖形如右圖,其中為點到準線的距離,四邊形為菱形,∴,∴,∴,∴,∴,∴向量與的夾角為。
17.(10分)解析:(Ⅰ)由正弦定理得,,,…2分
∴,,………4分
(Ⅱ)∵,,∴,∴,………………………6分
又∵,∴,∴,………………………8分
∴。………………………10分
18.解析:(Ⅰ)∵,∴;……………………理3文4分
(Ⅱ)∵三科會考不合格的概率均為,∴學生甲不能拿到高中畢業(yè)證的概率;……………………理6文8分
(Ⅲ)∵每科得A,B的概率分別為,∴學生甲被評為三好學生的概率為!12分
19.(12分)解析:(Ⅰ)∵,∴,
,,……………3分
(Ⅱ)∵,∴,
∴,
又,∴數(shù)列自第2項起是公比為的等比數(shù)列,………………………6分
∴,………………………8分
(Ⅲ)∵,∴,………………10分
∴!12分
20.解析:(Ⅰ)∵∥,,∴,∵底面,∴,∴平面,∴,又∵平面,∴,∴平面,∴!4分
(Ⅱ)∵平面,∴,,∴為二面角的平面角,………………………6分
,,∴,又∵平面,,∴,∴二面角的正切值的大小為!8分
(Ⅲ)過點做∥,交于點,∵平面,∴為在平面內(nèi)的射影,∴為與平面所成的角,………………………10分
∵,∴,又∵∥,∴和與平面所成的角相等,∴與平面所成角的正切值為!12分
解法2:如圖建立空間直角坐標系,(Ⅰ)∵,,∴點的坐標分別是,,,∴,,設,∵平面,∴,∴,取,∴,∴!4分
(Ⅱ)設二面角的大小為,∵平面的法向量是,平面的法向量是,∴,∴,∴二面角的正切值的大小為。………………………8分
(Ⅲ)設與平面所成角的大小為,∵平面的法向量是,,∴,∴,∴與平面所成角的正切值為。………………………12分
21.解析:(Ⅰ)設拋物線方程為,將代入方程得
所以拋物線方程為。………………………2分
由題意知橢圓的焦點為、。
設橢圓的方程為,
∵過點,∴,解得,,,
∴橢圓的方程為。………………………5分
(Ⅱ)設的中點為,的方程為:,
以為直徑的圓交于兩點,中點為。
設,則
∵
………………………8分
∴
………………………10分
當時,,,
此時,直線的方程為。………………………12分
22.(12分)解析:(Ⅰ)∵是偶函數(shù),∴,
又∵∴,,………………………2分
由得,,
∵時,;時,;時,;∴時,函數(shù)
湖北省互聯(lián)網(wǎng)違法和不良信息舉報平臺 | 網(wǎng)上有害信息舉報專區(qū) | 電信詐騙舉報專區(qū) | 涉歷史虛無主義有害信息舉報專區(qū) | 涉企侵權舉報專區(qū)
違法和不良信息舉報電話:027-86699610 舉報郵箱:58377363@163.com