题目内容
对任意的x∈(0,1),下列不等式恒成立的是( )A.
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B.
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C.
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D.
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【答案】分析:作出y=
与y=2x-1的图象,即可判断A、B的正误;对C、D可采用特值法判断,如令x=
可排除C,x=
可排除D,从而得到答案.
解答:解:作出y=
与y=2x-1的在∈(0,1)上的图象,可知y=
的图象在y=2x-1的图象的上方,故
,即A正确,从而B错误;
令x=
,tan(
)=tan(
)=1>
,可排除C;
再令x=
,有0<
,从而排除D.
故选A.
点评:本题考查指数函数的单调性与特殊点与正切函数的单调性,难点在于对A、B的分析,着重考查数形结合法与特值法,属于中档题.
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解答:解:作出y=
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令x=
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再令x=
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故选A.
点评:本题考查指数函数的单调性与特殊点与正切函数的单调性,难点在于对A、B的分析,着重考查数形结合法与特值法,属于中档题.
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