问题 选择题

已知偶函数y=f(x)在区间[-1,0]上是增函数,且满足f(1-x)+f(1+x)=0,下列判断中错误的是(  )

A.f(5)=0

B.函数f(x)在[1,2]上单调递减

C.函数f(x)的图象关于直线 x=1对称

D.函数f(x)的周期是T=4

答案

对于A,令x=0代入题中等式,得f(1-0)+f(1+0)=0

∴f(1)=0,结合函数为偶函数得f(-1)=f(1)=0

再令x=2代入题中等式,,得f(1-2)+f(1+2)=0,得f(3)=-f(-1)=0

结合函数为偶函数得f(-3)=f(3)=0

最后令x=4,f(1-4)+f(1+4)=0,得f(5)=-f(-3)=0,故A项正确;

对于B,因为偶函数y=f(x)图象关于y轴对称,在区间[-1,0]上是增函数,

所以y=f(x)在区间[0,1]上是减函数,

设F(x)=f(1+x),得F(-x)=f(1-x)

因为f(1-x)+f(1+x)=0,得f(1+x)=-f(1-x),

所以F(x)=f(1+x)是奇函数,图象关于原点对称.由此可得y=f(x)图象关于点(1,0)对称.

∵区间[1,2]和区间[0,1]是关于点(1,0)对称的区间,且在对称的区间上函数的单调性一致

∴函数f(x)在[1,2]上单调递减,故B项正确;

对于C,由B项的证明可知,y=f(x)图象关于点(1,0)对称,

若f(x)的图象同时关于直线 x=1对称,则f(x)=0恒成立,

这样与“在区间[-1,0]上f(x)是增函数”矛盾,故C不正确;

对于D,因为f(x)=f(1-(1-x))=-f(1+(1+x))=-f(x+2)

所以f(x+2)=-f(x+4),可得f(x+4)=f(x),函数f(x)的周期是T=4,D项正确

故选:C

写作题

书面表达

       假定你是李华,有一位美国朋友Smith托你在北京找工作.你看到21世纪英文报("21st

Century")上登载了一则招聘启事,认为对他很合适.请用英文写一封短信,用Email发给Smith,告知此事.广告原文如下:

                                      Foreign Teachers Wanted! 

     The Education Department of the Ladder Company is running an English course for children and

adults. 

     Requirements :

     1. Native English speaker

     2. University degrees in education or related field preferred

     3. Foreign expert certification(证明)

     If you are interested,  please call us :

     Tel : 861068019433

     Email : liecbj @ hotmail.com

     Beijing Ladder Information Company Limited

     注意:

    1. 开头语已为你写好;

    2. 介绍须包括所有内容,但不要逐条译成英语;

    3. 文章的词数需100词左右.

Dear Smith,

     I know you want to find a job in Beijing._______________________________________________________________________________________________________________________________

______________________________________________________________

                                                                                                                                      Good luck,

                                                                                                                                         Li Hua

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