问题 选择题
已知f(x)=
x+1,x≤-1
x2,-1<x<2,若f(x)=3,则x
2x,x≥2
的值是(  )
A.2B.2或
3
2
C.±
3
D.
3
答案

因为在每段定义域对应的解析式上都有可能使得f(x)=3成立,所以将原不等式转化为:

x+1=3
x≤-1
x2=3
-1<x<2
2x=3
x≥2
,从而得x=
3
.故选D.

完形填空
完形填空。
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