设\(f(x)\)是定义在\(\mathbb{R}\)上的奇函数,且当\(x\geqslant 0\)时,\(f(x)=x^2\),若对任意的\(x\in\left[t,t+2\right]\),不等式\(f(x+t)\geqslant 2f(x)\)恒成立,则实数\(t\)的取值范围是$(\qquad) \(
\)\mathrm{A}.\left[\sqrt{2},+\infty\right) $ \(\qquad \mathrm{B}.\left[2,+\infty \right)\) \(\qquad \mathrm{C}.\left(0,2\right]\) \(\qquad \mathrm{D}.\left[-\sqrt{2},-1\right]\cup\left[\sqrt{2},\sqrt{3}\right]\)
解析:
由题易知$$
f(x)=
\begin{cases}
x^2,&x\geqslant 0,\
-x^2,&x<0.
\end{cases}$$
因此\(f(x)\)为单调递增函数,而$$2f(x)=f\left(\sqrt{2}x\right).$$因此题中不等式等价于$$
f(x+t)\geqslant f(\sqrt{2}x).$$再结合\(f(x)\)的单调性可知$$
\forall x\in\left[t,t+2\right],x+t\geqslant \sqrt{2} x.$$即有$$t\geqslant \left(\sqrt{2}-1\right)\left(t+2\right).$$
解得\(t\)的取值范围为\(\left[\sqrt{2},+\infty \right)\),因此\(\rm A\)选项正确.
