Quadratic function f (x) = x 2+x, if the solution set of inequality f(-x)+f(x)≤2|x| is C.

Solve the set c first,

F(-x)+f(x)≤2|x|, bringing in simplification:

2x？ ≤2|x|

(1) when x≤0, that is, x? +x≤0, and the solution is:-1≤x≤0.

(2) when x > 0, namely x? -x≤0, the solution is: 0≤x≤ 1.

So C=[- 1, 1]

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Let m = a x (m > 0), then f (a x)-a (x+ 1) = 5 is arranged as follows:

m？ +( 1-a)m-5=0, let g(m)=m? +( 1-a)m-5

① when 0 < a < 1,

On the interval [- 1, 1], a ≤ a x ≤1/a.

Equation m? +( 1-a)m-5=0 interval [a, 1/a] has a root and must meet the following requirements:

G(a)≤0 and g( 1/a)≥0,

Solution: 0 < a ≤ 1/2.

② when a > 1,

On the interval [- 1, 1],1/a ≤ a x ≤ a.

Equation m? +( 1-a)m-5=0 interval [a, 1/a] has a root and must meet the following requirements:

G( 1/a)≤0 and g(a)≥0,

Solution: a≥5

So the range of real number A is (0, 1/2]∩[5, +∞).

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