6. Given the function $f(x)=\left\{\begin{array}{l}x^{2}, x<0, \\ -x^{2}, x \geqslant 0,\end{array}\right.$ if for $\forall x \in\left(t^{2}-4, t^{2}\right)$, the inequality $f(x+t)<4 f(x)$ always holds, then the range of the real number $t$ is
A. $\left(\frac{1-\sqrt{17}}{2}, \frac{1+\sqrt{17}}{2}\right)$
B. $(0,1)$
C. $\left[\frac{1-\sqrt{17}}{2}, \frac{1+\sqrt{17}}{2}\right]$
D. $[0,1]$
Solve
$f(x+t)2 x \Rightarrow t>x \Rightarrow t \geqslant t^{2}$ $\Rightarrow t^{2}-t \leqslant 0 \Rightarrow t \in[0,1]$, so choose $D$.