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Given that the function F X = x2-2ax + 3A + 1 decreases monotonically on (- ∞, 1), then the value range of real number a is
F (x) = x ^ 2-2ax + 3A + 1, let X1 > X2, and x1, X2 belong to (- ∞, 1) f (x1) - f (x2) = X1 ^ 2-2ax1 + 3A + 1 - (x2 ^ 2-2ax2 + 3A + 1) = [(x1-a) ^ 2 + 3A + A-A ^ 2] - [(x2-a) ^ 2 + 3A + A-A ^ 2] = (x1-a) ^ 2 - (x2-a) ^ 2 = (x1 + x2-2a) (x1-x2) because of monotonic decreasing, so the X1 + x2-2ax of F (x1) 0
Let an odd function f (x) be an increasing function over [- 2,2]; (1) find the domain of the function y = f (2x + 1); (2) find the solution set of the inequality f (2x + 1) + F (x) > 0
1>,-2
It is known that the proposition p: function y = loga (1-2x) increases monotonically in the domain of definition; proposition q: inequality (A-2) x2 + 2 (A-2) x-4 ＜ 0 holds for any real number X. if P ∨ q is a true proposition, the value range of real number a is obtained
∵ proposition p function y = loga (1-2x) increases monotonically in the domain of definition; ∵ 0 ＜ a ＜ 1 (3 points) and ∵ proposition q inequality (A-2) x2 + 2 (A-2) x-4 ＜ 0 holds for any real number x; ∵ a = 2 (2 points) or a − 2 ＜ 0 △ = 4 (a − 2) 2 + 16 (a − 2) ＜ 0, (3 points) that is - 2 ＜ a ≤ 2 (1 point) ∵ P ∨ q is true proposition, and the value range of ∵ A is - 2 ＜ a ≤ 2 (5 points)
Given the function f (x) = loga (2x + B / 2x-b), the domain of definition and value of the function are obtained
t=(2x+b)/(2x-b)
f(x)=loga(t)
The domain of definition is t > 0, that is, x > | B | / 2, or X
RELATED INFORMATIONS
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