共50条信息
设\(f\left(x\right)=\begin{cases}{\left(x-a\right)}^{2},x\leqslant 0 \\ x+ \dfrac{1}{x}+a+4,x > 0\end{cases} \),若\(f\left(0\right) \)是\(f\left(x\right) \)的最小值,则\(a\)的取值范围为( )
给出下列两个命题: 命题\(p:\)若在边长为\(1\)的正方形\(ABCD\)内任取一点\(M\),则\(\left| MA \right|\leqslant 1\)的概率为\(\dfrac{\pi }{4}.\)命题\(q\):若函数\(f\left( x \right)=x+\dfrac{4}{x}\),则\(f\left( x \right)\)在区间\(\left[ 1,\dfrac{3}{2} \right]\)上的最小值为\(4.\)那么,下列命题中为真命题的是\((\) \()\)
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