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If the odd function f (x) is decreasing in the domain of definition, what is the range of a satisfying the inequality f (1-A) + F (1-A & # 178;) < 0?
The solution of F (x) is an odd function
That is, f (- x) = - f (x)
So from F (1-A) + F (1-A & # 178;)
If the odd function y = f (x) (x ≠ 0), when x ∈ (0, + ∞), f (x) = X-1, then the value range of X with F (x-1) < 0 is ()
A. X ＜ 0b. 1 ＜ x ＜ 2C. X ＜ 0 or 1 ＜ x ＜ 2D. X ＜ 2 and X ≠ 0
∵ when x ∈ (0, + ∞), f (x) = X-1, ∵ x ＜ 0, - x ＞ 0, f (- x) = - X-1, and ∵ y = f (x) (x ≠ 0) is an odd function ∵ f (x) = - f (- x) = x + 1; ∵ f (x) = x − 1 (x ＞ 0) x + 1 (x ＜ 0)
If the odd function y = f (x) (x ≠ 0), when x ∈ (0, + ∞), f (x) = X-1, then the value range of X with F (x-1) < 0 is ()
A. X ＜ 0b. 1 ＜ x ＜ 2C. X ＜ 0 or 1 ＜ x ＜ 2D. X ＜ 2 and X ≠ 0
∵ when x ∈ (0, + ∞), f (x) = X-1, ∵ x ＜ 0, - x ＞ 0, f (- x) = - X-1, and ∵ y = f (x) (x ≠ 0) is an odd function ∵ f (x) = - f (- x) = x + 1; ∵ f (x) = x − 1 (x ＞ 0) x + 1 (x ＜ 0)
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