Given that f (x) is a quadratic function defined on R, then the solution set of the equation f [f (x)] = a cannot be a quadratic function i 'm sorry, A{1,2,3,4} B{1,2,4,8} C{0,2,4,} D{1,2}

Answer: C
The main points of this question are as follows:
The root of the equation f (x) = a is the solution of the compound variable f (x) of the equation f [f (x)] = a (note that this solution is not the solution of the equation f [f (x)] = a)
If the equation f (x) = a has no solution, then the equation f [f (x)] = a has no solution
If the equation f (x) = a has one solution, then the equation f [f (x)] = a has two solutions
If the equation f (x) = a has two solutions, then the equation f [f (x)] = a has four solutions

If the function f (x) defined on R is odd and f (x + 2) = 0 has five real roots, then the sum of these real roots is
Why and is the center of symmetry.

The sum of five roots of odd function with respect to the origin symmetry f (x) = 0 f (x + 2) is the center of symmetry x = - 2

F (x) is an odd function on the domain of definition, f (3) = 2, f (x + 4) = f (x) for all real numbers x, then f (2009) =? (please write the procedure)

Because f (x) = f (x + 4), f (2009) = f (2005) = f (2001) =... = f (2009-4 * 502) = f (1)
Because f (3) = 2 and f (x + 4) = f (x), f (- 1) = f (3) = 2
Because f (x) is an odd function, f (1) = - f (- 1) = - 2
So f (2009) = - 2

Given that f (x) is a quadratic function defined on R, then the solution set of the equation f [f (x)] = a cannot be a quadratic function i 'm sorry, A{1,2,3,4} B{1,2,4,8} C{0,2,4,} D{1,2}