題目列表(包括答案和解析)
(本題13分)已知函數(shù)
(1)已知一直線經(jīng)過原點且與曲線相切,求的直線方程;
(2)若關(guān)于的方程有兩個不等的實根,求實數(shù)的取值范圍。
第二節(jié) 書面表達(滿分25分)
假如你是張林,剛進入高三學(xué)習(xí),需要購買一本有關(guān)英語書面表達的書,請根據(jù)下列提示,寫信給你在外文書店工作的朋友Jack,請他幫你挑選。
要 求 | |
書的內(nèi)容要求 | 1.含寫作技巧:如文章結(jié)構(gòu),過渡詞使用; 2.有多種寫作題材及其范文; 3.最好有近三年高考的書面表達題。 |
其它 | 1.大小同英語課本,厚度不限。 2.最新版本,供2007級高三學(xué)生使用 3.把書郵寄到學(xué)校。 |
要求:1.詞數(shù)100個左右。
2.不可逐字翻譯,但可適當(dāng)增加內(nèi)容,使行文連貫。
3.參考詞匯:過渡詞transitional words 版本edition
注:信的開頭和結(jié)尾已為你寫好,不計入總詞數(shù)。
Dear Jack,
How are you getting along these days?
I’m in Senior Grade 3 now, and I’m writing to you ask for a favor.
Thank you very much, my dear friend. I wish I hadn’t put you to so much trouble.
Yours,
Zhang Lin
閱讀下列應(yīng)用文及相關(guān)信息,并按照要求匹配信息。請在答題卡上將對應(yīng)題號的相應(yīng)選項字母涂黑。
下面是一篇有關(guān)書籍介紹的應(yīng)用文,請閱讀下列應(yīng)用文和相關(guān)信息,并按照要求匹配信息。
首先請閱讀下列應(yīng)用文:
A
I Am a Pencil
Sam Swope's job was teaching writing to third-graders in New York City. His students were from 21 countries, speaking 11 languages, with different backgrounds. But there were a few things they had in common. Family troubles, for one. Money struggles. And poetry. Every single student, with the help of this creative teacher, came forth with awesome writing. Swope leaves the reader with the inspiring conviction (堅信) that deep within each of us lives a poet.
B
Between a Rock and a Hard Place
Aron Ralston, 28, went hiking in a remote Utah canyon without telling anyone. An unexpected catastrophe struck. With enough supplies only for a day, Ralston knew his situation was full of danger. Sure enough, after five days he was in a fight against death. That was when he carried out a courageous plan - using a pocket knife to cut off his trapped arm. His amazing survival story rests at a place among the classics of the genre (體裁).
C
Our Brother's Keeper
Author Jedwin Smith spent 30 years trying to repress (克制) all memories of his brother, Jeff, who was killed in Vietnam. But in Our Brother's Keeper he tells what happens when the Internet brings him into contact with several of his brother's old Marine friends, including the guy who held Jeff in his arms as he died. First via e-mail, and then in person, Smith gets to know these men.
D
The All Americans
With his graduation from West Point, Henry Romanek sailed toward Omaha Beach on the eve of Dday. It was June 1944, and he was about to face the bloodiest battle of his life. Just yesterday, it seemed, he was a standout soldier on the Army team. Now, he was a leader of youngsters in battle, fighting, quite literally, for his country and the future of the free world. In this book, Lars Anderson retraces Romanek's life and that of three other soldiers.
E
Copies in Seconds
With the push of a button, anyone can make copies of almost anything - unlike the old days, when papers had to be rewritten long-hand, carbon-copied out of fussy mimeograph machines (蠟紙油印機). In Copies in Seconds, David Owen showed how a shy engineer named Chester Carlson perfected his xerography machine (靜電復(fù)印機) and shopped it around until finally hooking up with the Haloid Corporation. That partnership led to the Xerox copier and changed the face of work forever.
F
State of Grace
Back in the late 1950s and early 1960s, the Lynvets was just a football team in a sandy New York City neighborhood. But to most of its members - the author, Robert Timberg, included - the team was their only experience of a happy family, their only chance to rise above terrible everyday circumstances, their only shot at being heroes. The friendships these men formed sustained (維持) them throughout their lives.
請閱讀以下求書者的信息,然后匹配他們所要尋找的書籍:
1. Tom is looking for a book about the hiking stories to help him in his following outdoor activities.
2. Kate wants to find a book about the stories of the soldiers during World War II. .
3. Mark wants to research into the history of technological development.
4. Jack is researching into education in a college. He is especially interested in the teaching methods. He wants to find a book which can tell him something about how to teach students from different backgrounds.
5. John wants to find some materials about the soldiers in Vietnam War to carry on his new research.
求書者 書籍
1. Tom A. I Am a Pencil
2. Kate B. Between a Rock and a Hard Place
3. Mark C. Our Brother's Keeper
4.Jack D. The All Americans
5. John E. Copies in Seconds
F. State of Grace
2.如圖17所示,將質(zhì)量M=1.20 kg的小沙箱,用輕質(zhì)軟繩(不可伸長)懸掛起來,開始處于靜止?fàn)顟B(tài),繩的長度l=0.80 m。用槍向沙箱發(fā)射質(zhì)量m=0.05 kg的子彈,子彈以v0=100 m/s 的速度向右水平擊中小沙箱,并留在小砂箱中,小沙箱沿豎直面向上擺動。假設(shè)沙箱每次向左運動到最低點時就恰好有一顆同樣速度的子彈射入沙箱,子彈射入沙箱的過程時間極短,可忽略不計,重力加速度g取10 m/s2,不計空氣阻力。
(1)求第一顆子彈射入沙箱的過程中子彈對砂箱的沖量大;
(2)求第一顆子彈射入沙箱并相對沙箱靜止的瞬間,求砂箱對繩的拉力的大;
(3)求第一顆子彈射入沙箱后,砂箱擺動的最大高度;
(4)要使沙箱擺動的最大角度小于60°,射入沙箱的子彈數(shù)目至少為多少
1.解析:,故選A。
2.解析:∵
,
故選B。
3.解析:由,得,此時,所以,,故選C。
4.解析:顯然,若與共線,則與共線;若與共線,則,即,得,∴與共線,∴與共線是與共線的充要條件,故選C。
5.解析:設(shè)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∵,∴,∴,∴雙曲線的離心率是。故選B.
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)的最小值為,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(dāng)或時,圓面被分成2塊,涂色方法有20種;②當(dāng)或時,圓面被分成3塊,涂色方法有60種;
③當(dāng)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:做出表示的平面區(qū)域如圖,當(dāng)直線經(jīng)過點時,取得最大值5。
14.解析:∵,∴時,,又時,滿足上式,因此,,
∴。
15.解析:設(shè)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(zhǔn)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(guān)于軸對稱的兩點且做出圖形如右圖,其中為點到準(zhǔn)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(xué)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(xué)生甲被評為三好學(xué)生的概率為!12分
(理)∵,,,!9分
∴的分布列如下表:
0
1
2
3
∴的數(shù)學(xué)期望!12分
19.(12分)解析:(Ⅰ)時,
,,
由得, 或 ………3分
+
0
-
0
+
遞增
極大值
遞減
極小值
遞增
, ………………………6分
(Ⅱ)在定義域上是增函數(shù),
對恒成立,即
………………………9分
又(當(dāng)且僅當(dāng)時,)
………………………4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(biāo)系,(Ⅰ)∵,,∴點的坐標(biāo)分別是,,,∴,,設(shè),∵平面,∴,∴,取,∴,∴!4分
(Ⅱ)設(shè)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(shè)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.(Ⅰ) 解析:如圖,設(shè)右準(zhǔn)線與軸的交點為,過點
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