It is known that f (x) is an odd function defined on (- 1,1), which decreases monotonically on the interval [0,1], and f (1-A) + F (1-A & sup2;) < 0. The value range of real number a is obtained

It is known that f (x) is an odd function defined on (- 1,1), which decreases monotonically on the interval [0,1], and f (1-A) + F (1-A & sup2;) < 0. The value range of real number a is obtained

Odd function, f (- x) = - f (x)
Monotonically decreasing in the interval [0,1], then monotonically decreasing in (- 1,0);
Let x > 0, - x0 > F (x)
So the function is decreasing on (- 1,1)
f（1-a）+f（1-a²）＜0
f(1-a)＜-f（1-a²）=f（a²-1）
That is to find the solution set of F (1-A) < f (A & sup2; - 1)
So: 1-A > A & sup2; - 1
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