【判断题】【消耗次数：1】
函数f(x)=x+lnx的图形在(0,+∞)内,单调递减且向上凸
①
正确
②
错误
参考答案：
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相关题目
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如果函数f(x)在区间(a,b)内恒有f(x)0,f(x)0,则函数的曲线（ ）
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单调递增且向上凸
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单调递减且向上凸
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单调递增且向下凹
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. 不是函数f(x)的驻点
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. 一定是函数f(x)的极值点
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设X的密度函数为f(x),分布函数为F(x)，且f(x)=f(-x)。那么对任意给定的a都有
①
<img class=jc-formula data-tex=f(-a)=1-\int _{ 0 }^{ a }{ f(x)dx } src=https://huaweicloudobs.ahjxjy.cn/1053A4F0A2AA674B748DEDF2A0D483FC.png style=vertical-align: middle;/>
②
<img class=jc-formula data-tex=F(-a)=\frac { 1 }{ 2 } -\int _{ 0 }^{ a }{ f(x)dx } src=https://huaweicloudobs.ahjxjy.cn/61F92082226911122757503D71FAE009.png style=vertical-align: middle;/>
③
F(a)=F(-a)
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F(-a)=2F(a)-1
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