6. Given the equation of the curve $\Gamma$ is $x|x|+y|y|=1$, if the line $l$ passing through the point $A(1,0)$ intersects the curve $\Gamma$ at exactly three different points, then the range of the inclination angle of the line $l$ is $\qquad$ .
Answer
As shown in the figure, the tangent line of curve $\Gamma$ at point $A$ is $x=1$.
Consider the line $l$ being tangent to $y^{2}-x^{2}=1$, let $C\left(x_{0}, y_{0}\right), 2 y y^{\prime}-2 x=0 \Rightarrow y^{\prime}=\frac{x}{y}$, then $A C: y-y_{0}=\frac{x_{0}}{y_{0}}\left(x-x_{0}\right)$, substituting point $(1,0)$ yields $-y_{0}^{2}=x_{0}-x_{0}^{2} \Rightarrow x_{0}=-1, y_{0}=\sqrt{2}$, thus $k_{A C}=\frac{\sqrt{2}}{-1-1}=-\frac{\sqrt{2}}{2}$. Combining with the graph, the range of the slope of line $l$ is $(-\infty,-1) \cup\left(-1,-\frac{\sqrt{2}}{2}\right)$, so the range of the inclination angle of line $l$ is $\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right) \cup\left(\frac{3 \pi}{4}, \pi-\arctan \frac{\sqrt{2}}{2}\right)$.