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Given a = {xi3 ≤ x ≤ 7}, B = {XIX < 6}, the complete set is a real number set R, find a ∪ B, (CRA) ∪ B
A∪B={x|x≤7}
(CRA) ∪ B = {x | x < 6 or x > 7}
If a = {x | x ≤ 1}, then CRA
A={x|x≤1}
CrA = { x | x>1 }
Let the complete set be a real number set R, a = {x | ≤ x ≤ 3} B = {× | × + a < 0} if (CRA) ∩ B = B, find the real number
Let complete set be real number set R, a = {x | ≤ x ≤ 3}
If (CRA) ∩ B = B, find the value range of real number a
∵（CRA）∩B=B
Ψ B is a subset of CRA
Then a ∩ B = Φ
B=｛X|X²+a
∵（CRA）∩B=B
Ψ B is a subset of CRA
Then a ∩ B = Φ
B=｛X|X²+a
If a and C = R, find the value range of real number a and (CRA) and B
The domain of F (x) = root (x-3) - 1 / root (7-x) is set a
B = {x ∈ Z | 2 ＜ x ＜ 10}, C = {x ∈ R | x ＜ a or X ＞ a + 1}
x-3>=0
A = {x | x > = 3 and not equal to 7}
So C contains X7 or a > = 3, a + 17 or 3
Known complete set u = [x │ 0
Because X in U
How to find the monotone interval of function f (x) = alnx-ax-3
X > 0
When a = 0, the original function is a constant function and is not considered
When a ≠ 0
The first derivative of F (x) is f '(x) = A / x-a-3
If f '(x) > 0, f (x) increases monotonically
If f '(x)
If f (lgx) > F (1), then the value range of X is ()
A. (110，10)B. (0，110)∪(1，+∞)C. (110，1)D. （0，1）∪（10，+∞）
∵ f (x) is an even function, f (1) ＜ f (lgx), ∵ f (1) ＜ f (| lgx |), and ∵ f (x) is a decreasing function in the interval [0, + ∞), ∵ 1 ＞ | lgx |, ∵ 1 ＜ lgx ＜ 1, ∵ 110 ＜ x ＜ 10 ∵ X has the value range of (110, 10), so a is selected
If the function f (x) = (AX + b) / (X-B) is known, and its image is symmetric with respect to the point (- 3,2), then the value of F (2) is
f(x)=(ax+b)/(x-b)
=(ax-ab+ab+b)/(x-b),
=a+(ab+b)/(x-b)
The center is (B, a)
That is, (B, a) = (- 3,2)
therefore
a=2,b=-3
f(x)=(2x-3)/(x+3)
thus
f(2)=1/5.
The even function f (x) defined on R is a monotone increasing function in the interval [0, + ∞). If f (1) ＜ f (LNX), then the value range of X______ .
① When LNX ＞ 0, f (1) ＜ f (LNX) is equivalent to 1 ＜ LNX because f (x) is a monotone increasing function in the interval [0, + ∞), and the solution is x ＞ E
Given that the image of the function f (x) = log2 (x2 ax + A2) is symmetric with respect to x = 2, then the value of a is______ .
When a ≠ 0, △ = - 3a2 ＜ 0, the domain of definition is r, and the symmetry axis of the inner function is x = A2 ∵ the function f (x) = log2 (x2 ax + A2) is symmetric about x = 2 ∵ x = A2 = 2 ∵ a = 4, so the answer is 4
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