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37 lines
1.2 KiB
37 lines
1.2 KiB
2 years ago
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#### 第一步：先求与直线相交交点的范围
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$$
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\left\{\begin{matrix}
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y=\frac{1}{2}x+\frac{1}{2} & \\
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y=ax^2-x+1 &
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\end{matrix}\right.
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$$
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即$\frac{1}{2}x+\frac{1}{2}=ax^2-x+1$
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$ax^2-\frac{3}{2}x+\frac{1}{2}=0$
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$\because 有两个交点$
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$\therefore \triangle>0$
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即$\frac{9}{4}-2a>0$
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$\therefore a<\frac{9}{8}$
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#### 第二步：与线段$AB$交点范围
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- $a>0$
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开口向上，模拟画出一个与线段$AB$有两个交点的抛物线，发现此抛物线当$x_1=-1,x_2=1$时，对应的$y_1,y_2$肯定是大于直线方程上的$0,1$
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即 $$
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\large \left\{\begin{matrix}
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a+1+1>=0& \\
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a-1+1>=1&
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\end{matrix}\right.
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$$
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解得$\frac{9}{8}>a>=1$
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- $a<0$
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开口向下，模拟画出一个与线段$AB$有两个交点的抛物线，发现此抛物线当$x_1=-1,x_2=1$时，对应的$y_1,y_2$肯定是小于直线方程上的$0,1$
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即 $$
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\large \left\{\begin{matrix}
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a+1+1<=0& \\
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a-1+1<=1&
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\end{matrix}\right.
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$$
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解得$a<=-2$
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所以答案是两部分$\frac{9}{8}>a>=1$或$a<=-2$
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