共50条信息
函数\(f(x)= \dfrac{\ln (x+1)}{x-2}\)的定义域是\((\) \()\)
已知函数\(f\left( x \right){=}\begin{cases} 2^{x}{-}1{,}x{\geqslant }0 \\ 2^{{-}x}{-}1{,}x{ < }0 \end{cases}\),设\(g\left( x \right){=}kf\left( x \right){+}x^{2}{+}x(k\)为常数\()\),若\(g\left( 10 \right){=}2018\),则\(g\left( {-}10 \right)\)等于( )
给出定义:若\(m- \dfrac{1}{2} < x\leqslant m+ \dfrac{1}{2}(\)其中\(m\)为整数\()\),则\(m\)叫做离实数\(x\)最近的整数,记作\(\{x\}\),即\(\{x\}=m.\)现给出下列关于函数\(f(x)=|x-\{x\}|\)的四个命题:\(①f\)\(\left( \left. - \dfrac{1}{2} \right. \right)\)\(=\)\( \dfrac{1}{2}\);\(②f(3.4)=-0.4\);\(③f\)\(\left( \left. - \dfrac{1}{4} \right. \right)\)\(=f\)\(\left( \left. \dfrac{1}{4} \right. \right)\);\(④y=f(x)\)的定义域为\(R\),值域是\(\left[ \left. - \dfrac{1}{2}, \dfrac{1}{2} \right. \right]\).其中真命题的序号是\((\) \()\)
若函数 \(y\)\(=\) \(f\)\(( \)\(x\)\()\)的定义域是\([0,2]\),则函数 \(g\)\(( \)\(x\)\()=\)的定义域是( ).
进入组卷