2.已知f(x)是奇函数,定义域为{x|x属于R,x不等于0},又f(x)在区间(0,+无穷)上是增函数,且f(-1)=0,且满足f(x)>0的x的取值范围是A.(1,+无穷)B.(0,1)C.(-1,0)∪(1,+无穷)D.(-无穷,-1
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![2.已知f(x)是奇函数,定义域为{x|x属于R,x不等于0},又f(x)在区间(0,+无穷)上是增函数,且f(-1)=0,且满足f(x)>0的x的取值范围是A.(1,+无穷)B.(0,1)C.(-1,0)∪(1,+无穷)D.(-无穷,-1](/uploads/image/z/1614192-24-2.jpg?t=2.%E5%B7%B2%E7%9F%A5f%28x%29%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0%2C%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BA%EF%BD%9Bx%EF%BD%9Cx%E5%B1%9E%E4%BA%8ER%2Cx%E4%B8%8D%E7%AD%89%E4%BA%8E0%EF%BD%9D%2C%E5%8F%88f%EF%BC%88x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%880%2C%2B%E6%97%A0%E7%A9%B7%EF%BC%89%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E4%B8%94f%EF%BC%88-1%EF%BC%89%3D0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3f%EF%BC%88x%EF%BC%89%3E0%E7%9A%84x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AFA.%EF%BC%881%2C%2B%E6%97%A0%E7%A9%B7%EF%BC%89B.%EF%BC%880%2C1%EF%BC%89C.%EF%BC%88-1%2C0%EF%BC%89%E2%88%AA%EF%BC%881%2C%2B%E6%97%A0%E7%A9%B7%EF%BC%89D.%EF%BC%88-%E6%97%A0%E7%A9%B7%2C-1)
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2.已知f(x)是奇函数,定义域为{x|x属于R,x不等于0},又f(x)在区间(0,+无穷)上是增函数,且f(-1)=0,且满足f(x)>0的x的取值范围是A.(1,+无穷)B.(0,1)C.(-1,0)∪(1,+无穷)D.(-无穷,-1
2.已知f(x)是奇函数,定义域为{x|x属于R,x不等于0},
又f(x)在区间(0,+无穷)上是增函数,且f(-1)=0,且满足f(x)>0的x的取值范围是
A.(1,+无穷)
B.(0,1)
C.(-1,0)∪(1,+无穷)
D.(-无穷,-1)∪(1,+无穷)
2.已知f(x)是奇函数,定义域为{x|x属于R,x不等于0},又f(x)在区间(0,+无穷)上是增函数,且f(-1)=0,且满足f(x)>0的x的取值范围是A.(1,+无穷)B.(0,1)C.(-1,0)∪(1,+无穷)D.(-无穷,-1
答案是 C f(-1)=0,f(x)为奇函数,故 f(1)=0 .f(x)在区间(0,+无穷)上是增函数 ,且f(1)=0 ,因此在(1,+无穷)上f(x)恒大于0 ,在(0,1)上f(x)恒小于0 ,由于f(x)为奇函数,故在区间(-1,0)上f(x)恒大于 0
简单C
C