问题 解答题

设函数f(x)=|x-1|+|x-a|(a<0)

(Ⅰ)若a=-1,解不等式f(x)≥6;

(Ⅱ)如果∃x0∈R,f(x0)<2,求a的取值范围.

答案

(Ⅰ) 当a=-1时,f(x)=|x-1|+|x+1|(a<0),

不等式f(x)≥6等价于

x<-1
1-x-(1+x)≥6
,或 
-1≤x<1
1-x+x-1≥6
,或 
x≥1
x-1+x+1≥6

解得 x≤-3 或 x≥3,

故原不等式的解集为{ x|x≤-3,或 x≥3}.

(Ⅱ)如果∃x0∈R,f(x0)<2,则f(x)的最小值小于2,

函数f(x)=

-2x+a-1    (x≤a)
1-a   (a<x<1)
2x-(a+1)  (x≥1)

故函数f(x)的最小值为  1-a,由

a <0
1-a<2

解得-1<a<0,

故a的取值范围为(-1,0).

写作题

书面表达。

    A. 国有国法,家有家规,学校有学校的规章制度。请你就你们学校的学生在校期间必须穿校服的规定发表看法。 

    注意:1. 描述准确,观点阐述清晰、明确,语言简练。   2. 字数在40-50词左右 

                                                                                                                                                               

                                                                                                                                                               

                                                                                                                                                               

    B. 请根据下图提示,用英语写一篇短文,先简要描写下图,然后再发表你自己的看法和意见。

                            

    注意:1. 字数在80词左右    2. 可适当增加细节,以使行文连贯 

                                                                                                                                                               

                                                                                                                                                               

                                                                                                                                                               

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