a)(bx

2a)=bx

(2a

ab)x

2a

This is a uniform function.

First order ∴ coefficient 2a

ab=0，①

∴f(x)=bx

2a

∫ its range is (-∞, 4), ∴ B < 0, 2a=4.

②

∴b=-2,a=2

∴f(x)=-2x

four

2、f(x)=a

When a=0, f(x)=0, which is both odd function and even function.

When a≠0 and f(x)=f(-x)=a, then f(x) is an even function.

3.f(x)=kx-4x-8, right?

f(x)=kx-4x-8

(1) When k=0, f(x)=-4x-8, which obviously satisfies the condition.

(2) When k≠0, f(x) is a quadratic function and its symmetry axis is x = 2/k.

In order to make it a monotone function in [5,20], the symmetry axis is on the left or right side of [5,20].

Quadratic function f(x)=ax

Bronx (Bronx)

c

In order to make f(x) an even function, the coefficient b of the linear term x is 0.

f(-x)=ax-bx

c

F(x)=f(-x) is.

cut down on

Bronx (Bronx)

c=ax-bx

c

That is bx=-bx.

So b=0.

Question 3: I'll cook with you after dinner. The idea is this.

Symmetry axis x=2/k

When k < 0, the symmetry axis x = 2/k < 0, which satisfies the condition.

When k > 0, there are 2/k≤5 or 2/k≥20.

At this time, k≥2/5 or 0 < k ≤110.

To sum up, the range of K that meets the conditions is K ≤110 or k≥2/5.