Question

x∈(-1,1)恒成立,2x²+(a-1)x-a(a-1)

Answer 1

y=2x^2+(a-1)x-a*(a-1)为开口向上的抛物线,由韦达定理得

x1+x2=-(a-1)/2

x1x2=-a(a-1)/2

(1)a=1,顶点在(0,0),显然是不符合要求

(2)a>1时,对称轴在x=-(a-1)/41即满足题意

x2=1时,x1=-(a-1)/2-1=-(a+1)/2

x1x2=-(a+1)/2 = -a(a-1)/2

a+1=a^2-a

a^2-2a-1=0

a=1+-V2 取正解 1+V2

a>1+V2

(2)a0处,只要左解x1