設(shè)是函數(shù)的反函數(shù).則與的大小關(guān)系為 查看更多

 

題目列表(包括答案和解析)

設(shè)隨機變量,記,則

A.        B.     C.       D. 的大小關(guān)系不定

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閱讀下面的短文,然后按照要求寫一篇150詞左右的英文短文。

The Use of Examinations

A good education should make people think for themselves. But the exam can never do anything like that.

First, examinations are second to none because it brings anxiety to students. As soon as a child begins school, he enters a world of terrible competition.

Second, exams do not interest students to read or to learn widely. They do not help students to look for more and more knowledge, but try to put as much as possible into the students’ head.

Third, they lower the standards of teaching, for they do not give any freedom to teachers to decide what to teach. Teachers themselves have no choice but to train their students in exam techniques. The most successful students are not always the best educated but the best trained in the technique of exams.

Fourth, they can tell you nothing about a person’s true ability. The exams are often nothing more than a subjective assessment(主觀評價) by some examiners.

However, if there are no exams, few students or even no one will do any revision or study!

【寫作內(nèi)容】

1.       概括短文的內(nèi)容要點,該部分的詞數(shù)大約30;

2.       就“中學(xué)生學(xué)習(xí)成績與能力的相互關(guān)系”為題發(fā)表你的看法,至少包括以下的內(nèi)容要點,該部分的詞數(shù)大約120。

(1)    你每次考試之前的心情如何?

(2)    你是如何備考的?

(3)    你認(rèn)為考試成績能反映你的學(xué)習(xí)能力嗎?

【寫作要求】

你可以使用實例或其他論述方法支持你的觀點,也可以參照閱讀材料的內(nèi)容,但不得直接引用原文中的句子。

   【評分標(biāo)準(zhǔn)】

    概括準(zhǔn)確、語言規(guī)范、內(nèi)容合適、篇章連貫。

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此題要求改正所給短文中的錯誤。對標(biāo)有題號的每一行做出判斷:
如無錯誤,該行右邊橫線上畫一個勾(√);
如有錯誤(每行只有一個錯誤),則按下列情況改正:
此行多一個詞:把多余的詞用斜線(\)劃掉,在該行右邊橫線上寫出該詞,并也用斜線劃掉。
此行缺一個詞:在缺詞處加一個漏字符號(∧),在該行右邊橫線上寫出該加的詞。
此行錯一個詞:在錯的詞下劃一橫線,在該行右邊橫線上寫出改正后的詞。
注意:原行沒有錯的不要改。
(請在答題卡的指定位置作答)
第三節(jié)書法展示(滿分5分)
請抄寫下面的文章,注意本題是按照書法的好壞評分的,格式要求與原文一致。
建議:先在草稿紙上練習(xí),找找手感,再答卷。
My Hometown
Welcome to my hometown which is located on the Bohai Bay. It is very beautiful! Living here makes you feel quite good. Miles of beaches stretch along the coast. You can lie on the beach, bathing in the sun and enjoying the fresh air. The blue sky and beautiful scenery here will make you relaxed and excited.
People here are very friendly and considerate. They are ready to offer help to people from the country and abroad. There are plenty of tidy hotels and inns for visitors to stay in. Most importantly, you can enjoy all kinds of tasty special foods. I am sure you will have a good time here. 
(請在答題卡的指定位置作答)

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請閱讀下列應(yīng)用文及其相關(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。下面是一篇應(yīng)用文及其應(yīng)用場合的信息, 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, 請閱讀下列六則卡通人物的性格介紹:
Why do you remember Hello Kitty, Snoopy and all the other cartoon characters? What makes you love them? Well, maybe because they're like the people around you. Think about it! You may find a friend or classmate who is like them.
A. As sweet as Hello Kitty: She likes to eat cake. She loves to make new friends. She likes to ask friends to her parties. Her smile is so lovely.
B. As clever as Snoopy: He went to school when he was nine. He learned to use a typewriter in two years! He thinks a lot. He is so clever that you like to be with him.
C. As sarcastic (諷刺的) as Garfield: He sits happily in the seat and says sharp words to you. Sometimes he is not nice. He doesn't really like you? He thinks you're a fool? No, in his heart, he loves you. He is a friend with hard words but a warm heart.
D. As naughty as MashiMaro: He doesn't look like a good boy. He has sleepy eyes and looks naughty. He always plays tricks. So you may get angry with him and don't like him very much. His mind is active and full of ideas. He tries to be big and catch your eyes. But, he' s still a child.
E. As confident as Prince of Tennis: He has faith in himself and always wants to win.
F. As friendly as Mickey: He is clever and kind. He has a good heart and is glad to help others. Everyone likes to turn to him for help whenever they are in trouble.
請閱讀Susan, Tom, John, Bob, George的個性描述,然后匹配與他們個性相當(dāng)?shù)目ㄍㄈ宋铩?br />【小題1】Tom is a warm-hearted boy, and he cares for others very much. If you meet with any difficulty, you can tell him, and he will surely help you.
【小題2】Susan has many friends and she likes to stay with them in many kinds of parties. On her face there is always a smile.
【小題3】John likes thinking and can always find ways to solve any difficult problems.
【小題4】Bob seems to be very rude and impolite and his words are unpleasant to hear, but his inner heart is full of concerns for others.
【小題5】George believes in himself, and he always wants to win the first place in almost everything.

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一個四棱錐和一個三棱錐恰好可以拼接成一個三棱柱.這個四棱錐的底面為正方形,且底面邊長與各側(cè)棱長相等,這個三棱錐的底面邊長與各側(cè)棱長也都相等.設(shè)四棱錐、三棱錐、三棱柱的高分別為,,則(  )

A.           B.                  C.              D.

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1.解析:,故選A。

2.解析:抽取回族學(xué)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(shè)公差為,由題意得,,解得,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)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(dāng)時,圓面被分成2塊,涂色方法有20種;②當(dāng)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:將代入結(jié)果為,∴時,表示直線右側(cè)區(qū)域,反之,若表示直線右側(cè)區(qū)域,則,∴是充分不必要條件。

學(xué)科網(wǎng)(Zxxk.Com)14.解析:∵,∴時,,又時,滿足上式,因此,。

學(xué)科網(wǎng)(Zxxk.Com)15.解析:設(shè)正四面體的棱長為,連,取的中點,連,∵的中點,∴,∴或其補角為所成角,∵,,∴,∴,又∵,∴,∴所成角的余弦值為。

學(xué)科網(wǎng)(Zxxk.Com)16.解析:∵,∴,∵點的準(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

19.(12分)解析:(Ⅰ)∵,∴

 ,,……………3分

(Ⅱ)∵,∴

,

,∴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,交于點,∵平面,∴在平面內(nèi)的射影,∴與平面所成的角,………………………10分

學(xué)科網(wǎng)(Zxxk.Com),∴,又∵,∴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.解析:(Ⅰ)設(shè)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(shè)橢圓的方程為,

∵過點,∴,解得,,,

∴橢圓的方程為!5分

(Ⅱ)設(shè)的中點為,的方程為:,

為直徑的圓交兩點,中點為

設(shè),則

  

………………………8分

………………………10分

當(dāng)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

得,,

時,;時,;時,;∴時,函數(shù)


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