0  442947  442955  442961  442965  442971  442973  442977  442983  442985  442991  442997  443001  443003  443007  443013  443015  443021  443025  443027  443031  443033  443037  443039  443041  443042  443043  443045  443046  443047  443049  443051  443055  443057  443061  443063  443067  443073  443075  443081  443085  443087  443091  443097  443103  443105  443111  443115  443117  443123  443127  443133  443141  447090 

25. She was over the age limit and ________, her application for the job was rejected.

  A. as a result  B. in conclusion    C. worse still   D. what’s more

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24. After five years of preparation, Shanghai is presenting the world _______ many say is the greatest Special Olympics.

   A. when     B. what        C. where      D. which

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23.  ---He suggests the number of cars should be limited to stop air pollution.

--- __________, the idea is not very practical.

A.  Sounds good as it         B. As it sounds good

C. As good it sounds         D. Good as it sounds

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22.  Scientists are trying to develop a special material, _______ they will make use of in space.

A. it     B. that         C. what       D. one

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第一节:单项填空:(共15小题;每小题1分,满分15分)

21.  The education of _______young is always _______ hot and serious topic in modern society.

A. the; /    B. a; the       C. /; the       D. the; a

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4.已知非零实数a、b满足,求的值 

解法一:所给等式左边的分子、分母同除以a,则已知等式化为关于的方程,可求出 

解:由题设得

解这个关于的方程得 

 

解法二:已知等式的左边的分子、分母都具有asinα+bcosα的结构,可考虑引入辅助角求解 

解:∵

 

其中,即 

∴由题设得

,即 (k∈Z) 

因此, 

解法三:在已知等式的左边,分子与分母同时除以acos得: 

=tanθ,则

评注:解法一利用了集中变量的思想,是一种基本方法解法二通过模式联想,引入辅助角,解法三通过联想两角和的正切公式,利用了换元法,实质上是综合了解法一和解法二的优点

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2.已知αβγ组成公差为的等差数列,求tanα·tanβ+tanβtanγ+tanγtanα的值 

分析:条件的使用形式较多,可以把αβγ通过条件置换成β=α=,γ=α+π;也可以用等差中项公式β=,但两种形式置换后的结果比较复杂化切为弦虽是一种常用方法,但在这里效果不明显如果换个角度使用条件,即把条件变为β-α=,γ-β=,γ-α=π,两边取正切后可分别出现所求式中的tanαtanβ、tanβtanγ、tanγtanα,然后将它们整体代入,便可使问题解决 

解:由条件得β-α=,两边取正切得tan (β-α)=tan,即,化简可得tanαtanβ=(tanβ-tanα)-1      ① 

同理,由γ-β=得tanγtanβ=(tanγ-tanβ)-1     ② 

γ-α=π得tanγtanα= (tanα-tanγ)-1       ③ 

以上三式相加得tanαtanβ+tanβtanγ+tanγtanα=-3

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