R上的函数f(x),g(x).函数y=f(g(x))有不动点.则函数y=g(f(x))不可能是A．X^2+X-(1/5)B.X^2+X+(1/5)C.X^2-(1/5)D.X^2+(1/5)

Question

R上的函数f(x),g(x).函数y=f(g(x))有不动点.则函数y=g(f(x))不可能是
A．X^2+X-(1/5)
B.X^2+X+(1/5)
C.X^2-(1/5)
D.X^2+(1/5)

Answer 1

y=f(g(x))有不动点,即方程 y=f(g(x))=x有解
A.x^2+x-(1/5) = x => x^2 = 1/5 ,有解
B.x^2+x+(1/5) = x => x^2 = -1/5 ,无解
C.x^2-(1/5) = x => (x-1/2)^2 = 9/20 ,有解
D.x^2+(1/5) = x => (x-1/2)^2 = 1/20 ,有解
答案为B