Set a = {- 2 less than or equal to x less than or equal to 2}. B = {x less than or equal to a}. If a intersects B not equal to an empty set, then the value of real number a

Set a = {- 2 less than or equal to x less than or equal to 2}. B = {x less than or equal to a}. If a intersects B not equal to an empty set, then the value of real number a

Draw the number axis, a > = - 2
Given the set P = {x 2 ≤ x ≤ 5}, q = {x k + 1 ≤ 2k-1}, find the value range of real number k when p ∩ q = & # 8709
2k-15 or q is an empty set
Get K4
First of all, we should ensure that 2k-1 > = K + 1, then we should ensure that 2k-15 is k > 4
Given the set P = {x ∣ 2 ≤ x ≤ 5}, q = {x ∣ K + 1 ≤ x ≤ 2k-1}, finding P ∩ q = empty set is the value range of real number K
If there are questions, K + 1 > 5, k > 4
Or 2k-1
Given that the real number a ≠ 0, the function f (x) = ax (X-2) ^ 2 (x belongs to R) has a maximum value of 32, find the value of a?
f(x)=ax(x-2)^2=ax^3-4ax^2+4ax
f'(x)=3ax^2-8ax+4a=3a(x-2/3)(x-2)
(1)a>0,
∵x∈(-∞,2/3）,f'(x)>0,x∈(2/3,2）
f'(x)
If f (x) is derived, f '(x) = a (X-2) ^ 2 + 2aX (X-2) = a (X-2) (3x-2), Let f' (x) = 0, x = 2 or x = 2 / 3
1. When a > 0, then x = 2 / 3 is the maximum, f (2 / 3) = 32, that is, a * (2 / 3) * (16 / 9) = 32, then a = 27
2. When a
Given the function f (x) = 2x1 − x, the monotonicity of function y = f (AX) (a < 0) is judged and proved by the definition of function monotonicity
∵ f (x) = 2x1 − x, ∵ f (AX) = 2ax1 − x, let x1 ＜ X2, then f (x1) - f (x2) = 2ax11 − x1-2ax21 − x2 = 2A (x1 − x2) (1 − x1) (1 − x2) ∵ x1-x2 ＜ 0, a ＜ 0, ∵ 2A (x1-x2) ＞ 0, when x1 ＜ x2 ∈ (- ∞, 1), (1-x1) (1-x2) ＞ 0, when x1 ＜ x2 ∈ ((∞, 1), when x1 ＜ x2 ∈
Let the image of quadratic function f (x) = ax ^ 2 + BX + C intersect with X axis at (- 1,0) and satisfy [f (x) - x] * [f (x) - (x ^ 2 + 1) / 2] ≤ 0
(1) Finding the value of F (1)
(2) Finding the analytic expression of F (x)
(3) Prove ∑ (1 / F (k)) > 2n / (n + 2)
.
(1) Because [f (x) - x] * [f (x) - (x ^ 2 + 1) / 2] ≤ 0 is constant, if x = 1 is added, we can get (a + B + C-1) ^ 2 ≤ 0, so (a + B + C-1) ^ 2 = 0, we can get f (1) = 1. (2) from F (- 1) = 0, f (1) = 1, we can get b = 1 / 2, a + C = 1 / 2. Take these two results out of that inequality, we can get [ax ^ 2 + (1 / 2-1) x + C] * [...]
It is known that the domain of function f (x) is all real numbers where x is not equal to 0. For any x1, X2 in the domain, f (x1x2) = f (x1) + F (x2)
And when x > 1, f (x) > 0, we prove that f (x) is even function (2) and f (x) is increasing function on (0, + infinity)
(1) It is proved that the domain of definition of function f (x) is any real number where x is not equal to 0. For any X1 and X2 in the domain, f (x1x2) = f (x1) + F (x2), f (- x) = f (- 1) + F (x), f (x) = f (- 1) + F (- x), f (- x) - f (x) = f (x) - f (- x) f (- x) = f (x) f (x) is an even function (2) f (x) = f (...)
1.f(x1x2)=f(x1)+f(x2)
Substituting X1 = - 1, X2 = 0, we get f (- 1) = 0
If X1 = - 1, X2 = X2 is substituted, f (- x2) = f (x2) + 0 is proved
Two
Let 0 < X1 < x2
f(x1)-f(x2)
=f(x1)-f(x1×x2/x1)
=f(x1)-[f(x1)+f(x2/x1)]=-f(x2/x1)
Because 0 < X1 < X2, so... Expand
1.f(x1x2)=f(x1)+f(x2)
Substituting X1 = - 1, X2 = 0, we get f (- 1) = 0
If X1 = - 1, X2 = X2 is substituted, f (- x2) = f (x2) + 0 is proved
Two
Let x2 < X10
f(x1)-f(x2)
=f(x1)-f(x1×x2/x1)
=f(x1)-[f(x1)+f(x2/x1)]=-f(x2/x1)
Because 0 ＜ x1 ＜ X2, so x2 / x1 ＞ 1, f (x2 / x1) ＞ 0
∴f(x1)-f(x2)＜0
So f (x) is closed in (0, + ∞) monotonically increasing order
Given that the maximum value of the function f (x) = loga (x + 1) (a > 0, a = / 1) in the interval [1,7] is 1 / 2 larger than the minimum value, find a
emergency
Because f (x) = loga (x + 1) (a > 0, a = / 1), there may be the following two cases: (1) if a > 1, then f (x) increases monotonically in the interval [1,7], then when x = 1, f (x) has a minimum value of loga2; when x = 7, f (x) has a maximum value of loga8
When a & gt; 1, f (x) = loga (x + 1) increases monotonically in the interval [2,8] = f (8) - f (2) = loga9-loga3 = loga3 = 1 / 2, and when a = 9 A & lt; 1, f (x) = loga (x + 1)
If 0
It is known that the domain of function f (x) is any real number where x is not equal to 0. For any x 1 x 2 in the domain, f (x 1. X 2) = f (x 1) + F (x 2), and if x > 1
F (x) > 0, f (2) = 1 prove that f (x) is even function
2. F (x) is an increasing function on (0 infinity)
3. Solving the inequality f (2x squared-1)
By substituting X1 = x2 = 1, we get the following results
If f (1) = f (1) + F (1), then f (1) = 0
Substituting X1 = - 1 and X2 = - 1, we get the following results
f(1)=f(－1)＋f(－1)
f(－1)＋f(－1)=0
Then:
f(－1)=0
Then:
f(－x)=f(－1)＋f(x)
f(－x)=f(x)
So the function f (x) is even
2. Let X1 > x2 > 0, then X1 / x2 > 1, that is, f (x1 / x2) > 0
And f (x1) = f (x1 / x2 * x2) = f (x1 / x2) + F (x2)
That is, f (x1) - f (x2) = f (x1 / x2) > 0
That is, f (x1) > F (x2)
So the function is an increasing function on (0, + OO)
3.f(4)=f(2)+f(2)=2
f(2x^2-1)
Given that the maximum value of the function f (x) = loga (x ^ 2 + x-1) in the interval [1,2] is greater than the minimum value by 2, find a
First of all, we should discuss the value range of A. ① a ＞ 1 symmetry axis X = - 1 / 2x increases in the interval [1,2]. F (x) is an increasing function. ② 0 ＜ a ＜ 1 x decreases in the interval [1,2]. ② 0 ＜ a ＜ 1 x decreases in the interval [1,2]
(x ^ 2 + x-1) = (x + 1 / 2) ^ 2-5 / 4 increases in the interval [1,2]
When x = 1 (x ^ 2 + x-1) = 1
When x = 2 (x ^ 2 + x-1) = 5
So a ^ 2 = 5-1 = 4
A=2