练习3:方程 x^2-2x+2k-1=0的两个根分别在区间(0,1)和(1,2)内,则K的取值范围是?请给出过程,
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![练习3:方程 x^2-2x+2k-1=0的两个根分别在区间(0,1)和(1,2)内,则K的取值范围是?请给出过程,](/uploads/image/z/5123339-35-9.jpg?t=%E7%BB%83%E4%B9%A03%EF%BC%9A%E6%96%B9%E7%A8%8B+x%5E2-2x%2B2k-1%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%E5%88%86%E5%88%AB%E5%9C%A8%E5%8C%BA%E9%97%B4%280%2C1%29%E5%92%8C%281%2C2%29%E5%86%85%2C%E5%88%99K%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%3F%E8%AF%B7%E7%BB%99%E5%87%BA%E8%BF%87%E7%A8%8B%2C)
练习3:方程 x^2-2x+2k-1=0的两个根分别在区间(0,1)和(1,2)内,则K的取值范围是?请给出过程,
练习3:方程 x^2-2x+2k-1=0的两个根分别在区间(0,1)和(1,2)内,则K的取值范围是?请给出过程,
练习3:方程 x^2-2x+2k-1=0的两个根分别在区间(0,1)和(1,2)内,则K的取值范围是?请给出过程,
x^2-2x+2k-1=0
(x-1)²+2k-2=0
(x-1)²=2-2k
x-1=±√(2-2k) (k
f(x)=x^2-2x+2k-1
f(0)>0;f(1)<0;f(2)>0;
解得2k-1>0;2k-2<0;2k+1>0;
1>k>1/2
f(x) =x^2-2x+2k-1
f'(x) = 2x-2 =0
x=1 (min)
f(0) = 2k-1
f(1) = 2k-3
f(2) = 2k-1
f(0) =2k-1 >0
k > 1/2
and
f(1) = 2k-3 <0
k < 3/2
ie
1/2
△=(-2)^2-4*1*(2k-1)>0
1-(2k-1)>0
2k<2
k<1
令f(x)=x^2-2x+2k-1
则f(0)>0,f(2)>0
2k-1>0, 2k-1>0
k>1/2, k>1/2
综上得1/2
x^2的系数是1,所以抛物线y=x^2-2x+2k-1开口向上,因为y=0的根在(0,1)(1,2)内,所以
y(x=0)=2k-1>0 k>1/2
y(x=1)=1-2+2k-1=2k-2<0 k<1
y(x=2)=4-4+2k-1=2k-1>0 k>1/2
所以1/2