设b>a>0为常数,函数f(x)=2x-x^2,已知当x属于[a,b]时,f(x)的值域为[1/b,1/a]时,求a,b

(2) f(x)=2x-x^2x>1单减X∈[a,b]时,g(x)=f(x)且g(x)的值域为[1/b 1/a]g(a)=1/a g(b)=1/b2a-a^2=1/aa^3-a^2-a^2+a-a+1=0(a-1)(a^2-a-1)=0a1=1 a2=(1+根号5)/2 a3直接舍去g(a)=1/b g(b)=1/a (a