Question

若函数fx=(x^2+ax-2)/(x^2-x+1)的值域为[-2,2],求a的值

Answer 1

函数fx=(x^2+ax-2)/(x^2-x+1)的值域为[-2,2]
-2<=(x^2+ax-2)/(x^2-x+1)<=2
(x^2-x+1）=(x-1/2)^2+3/4>=3/4
-2x^2+2x-2<=x^2+ax-2<=2x^2-2x+2
由-2x^2+2x-2<=x^2+ax-2得
3x^2+(a-2)x>=0
3(x+(a-2)/6)^2-(a-2)^2/12>=0
由于该不等式对任意x都成立则有-(a-2)^2/12>0,又(a-2)^2/12>=0
得(a-2)^/12=0
a=2
由x^2+ax-2<=2x^2-2x+2得
x^2-(a+2)x+4>=0
(x-(a+2)/2)^2+4-(a+2)^/4>=0
由于该不等式对任意x都成立则有4-(a+2)^/4>=0得-6<=a<=2
综上所述a=2