题目内容
7.已知关于x的方程ax2-(a+1)x+2=0在区间[0,1]上有实数根,求实数a的取值范围.分析 讨论a的取值,结合一元二次方程根的分布,建立不等式关系即可得到结论.
解答 解:若a=0,则方程等价为x=2,在区间[0,1]上没有实数根,不满足条件.故a≠0.
若方程ax2-(a+1)x+2=0在区间[0,1]上只有一个实数根,
设f(x)=ax2-(a+1)x+2,
则满足①$\left\{\begin{array}{l}{a≠0}\\{△=0}\\{0≤-\frac{-(a+1)}{2a}≤1}\end{array}\right.$,或②$\left\{\begin{array}{l}{a≠0}\\{△>0}\\{f(0)f(1)≤0}\end{array}\right.$,
由①得$\left\{\begin{array}{l}{a≠0}\\{{a}^{2}-6a+1=0}\\{a≥1或a≤-1}\end{array}\right.$,即$\left\{\begin{array}{l}{a≠0}\\{a=3+2\sqrt{2}或a=3-2\sqrt{2}}\\{a≥1或a≤-1}\end{array}\right.$,解得a=3+2$\sqrt{2}$.
由②得$\left\{\begin{array}{l}{a≠0}\\{a>3+2\sqrt{2}或a<3-2\sqrt{2}}\\{2×1≤0}\end{array}\right.$,此时不等式无解.
若方程ax2-(a+1)x+2=0在区间[0,1]上有两个实数根,
则$\left\{\begin{array}{l}{a>0}\\{△>0}\\{f(0)≥0,f(1)≥0}\\{0≤-\frac{-(a+1)}{2a}≤1}\end{array}\right.$,即$\left\{\begin{array}{l}{a>0}\\{a>3+2\sqrt{2}或a<3-2\sqrt{2}}\\{1≥0,2≥0}\\{a≥1或a≤-1}\end{array}\right.$,解得a>3+2$\sqrt{2}$.
若a<0,∵f(0)=2>0,
∴若x的方程ax2-(a+1)x+2=0在区间[0,1]上有实数根,
则f(1)≤0,
即a-a-1+2≤0,
即1≤0,此时不等式不成立,
综上a≥3+2$\sqrt{2}$.
点评 本题主要考查一元二次方程根的分布问题,利用函数和方程之间的关系是解决本题的关键.注意要进行分类讨论.
A. | y=x-x2 | B. | y=|x+1| | C. | y=-$\frac{1}{x}$ | D. | y=x2-2x |