Question

不等式(a－2)x²+2(a－2)x－4＜0对任意x∈(0,1)恒成立，求实数a的取值范围

Answer 1

a-2≤0
4a²-8a+8+16a-32＜0
4a²+8a-16＜0
a²+2a-4＜0
（-1-√5）＜a＜（-1+√5）
a≤-2
-1-√5＜a≤-2
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Answer 2

case 1: a=2
(a-2)x^2+2(a-2)x-4
=-4
case 1: true

case 2: a>2
f(x) =(a－2)x^2+2(a－2)x－4
f'(x) =2(a-2)x + 2(a-2)
f'(x)=0
=> x= -1 (min)
f(1) =(a-2)+2(a-2)－4
= 3a-10
f(1) ≤0
3a-10≤0
a≤ 10/3
solution for case 2: 2<a≤ 10/3

case 3: a<2
f(x) =(a－2)x^2+2(a－2)x－4
f'(x) =2(a-2)x + 2(a-2)
f'(x)=0
=> x= -1 (max)
f(0) = -4 <0
solution for case 3: a<2

(a-2)x^2+2(a-2)x－4＜0对任意x∈(0,1)恒成立
case 1 or case 2 or case 3
ie
a≤ 10/3