If there are exactly two integers in the solution set of inequality (2x-1) ^ 2 ≤ ax ^ 2 about X, then the value range of real number a,

If there are exactly two integers in the solution set of inequality (2x-1) ^ 2 ≤ ax ^ 2 about X, then the value range of real number a,

(4-A) x ^ 2-4x + 1 ≤ x ^ 2-4x / (4-A) + 1 / (4-A) ≤ 0 [X-2 / (4-A)] ^ 2-4 / (4-A) ^ 2 + 1 / (4-A) ≤ 0 {X-2 / (4-A) + √ [A / (4-A) ^ 2]} {X-2 / (4-A) - √ [A / (4-A) ^ 2]} ≤ 0. To make the solution set of X have exactly two integers, we must make 2 ≤ 2 / (4-A) + √ A / i4-ai-2 / (4-A) - √ A / i4-ai
For real numbers x, y, a new operation "*" is defined. For rational numbers x, y, a new operation "*" is defined. For rational numbers x, y, a new operation "*": X * y = ax + by + XY, where a and B are constants. On the right side of the equation are the usual addition and multiplication operations. Given 2 * 1 = 7, (- 3) * 3 = 3, then the value of 1 / 3 * 6. (bivariate linear equation ~)
From the meaning of the question: 2A + B + 2 * 1 = 7-3a + 3B + (- 3) * 3 = 3, that is, 2A + B = 5 (1) - 3A + 3B = 12 (2) (2) / (- 3) get: A-B = 4 (3) (1) + (3) get: 3A = 9, a = 3 bring a = 3 into (3) get: B = - 1, so 1 / 3 * 6 = 1 / 3A + 6B + 1 / 3 * 6 = 1 / 3 * 3 + 6 * (- 1) + 2 = 1-6 + 2 = - 3, hope to adopt
Given that real numbers a, B, X and y satisfy a + B = x + y = 2, ax + by = 5, find ay + BX
ruti
∵a+b=2,x+y=2.
∴(a+b)*(x+y)=2*2
ax+ay+bx+by=4
∵ax+by=5
∴ay+bx=4-5=1
For multiple choice and fill in the blanks, special value method is recommended; otherwise, conventional method is recommended
It is known that the real number ABXY satisfies ax + by = 3, ay BX = 5
What is the value of (a ^ 2 + B ^ 2) (x ^ 2 + y ^ 2) if a, B, X and y satisfy ax + by = 3 and ay BX = 5
ax+by=3
ay-bx=5
In theory, it's just the expansion of the following thing
So 9 + 25 = 34
The function f (x) defined on R satisfies f (x + y) + F (X-Y) = 2F (x) f (y) and f (1 / 2) = 0, f (0) ≠ 0
(1) To prove that f (x) is an even function
(2) It is proved that f (x) is a periodic function
(3) If f (x) is a monotone function in [0,1], find the values of F (1 / 3) and f (1 / 6)
High school mathematics will say thank you!
(1) If we substitute x = y = 0 into f (x + y) + F (X-Y) = 2F (x) f (y), then we substitute x = 0 into f (0 + y) + F (0-y) = 2F (0) f (0), that is, 2f (0) = 2F (0) f (0), f (0) = 1, then substitute x = 0 into f (0 + y) + F (0-y) = 2F (0) f (y), f (y) + F (- y) = 2F (y), f (- y) = f (y), so f (x) is even function. (2) substitute y = 1 / 2 into f (x +
(1) Let y = 0, the equation becomes: F (x) + F (x) = 2F (x) f (0), so f (0) = 1;
It is proved that even function (- F, y) = f (y) + F (y) = 0;
(2) Let y = 1 / 2, equating to: F (x + 1 / 2) + F (x-1 / 2) = 0;
If X-1 / 2 = Z, then f (z) = - f (Z + 1) = f (Z + 2), periodic function, minimum positive period T = 2;
(3) Let x = 1 / 3... Expand
(1) Let y = 0, the equation becomes: F (x) + F (x) = 2F (x) f (0), so f (0) = 1;
Then x = 0, the equation is: F (y) + F (- y) = 2F (0) f (y), so f (y) = f (- y), even function is proved;
(2) Let y = 1 / 2, equating to: F (x + 1 / 2) + F (x-1 / 2) = 0;
If X-1 / 2 = Z, then f (z) = - f (Z + 1) = f (Z + 2), periodic function, minimum positive period T = 2;
(3) Let x = 1 / 3; y = 1 / 6, the equation is: F (1 / 3 + 1 / 6) + F (1 / 3-1 / 6) = 2F (1 / 3) f (1 / 6),
That is, f (1 / 6) [2F (1 / 3) - 1] = 0. Because of monotonicity, and f (0) = 1 ＞ f (1 / 2) = 0, i.e. single decreasing function, f (1 / 6) ＞ f (1 / 3) ＞ f (1 / 2) = 0;
f(1/3)=1/2;
Let x = y = 1 / 3, the equation is: F (1 / 6 + 1 / 6) + F (1 / 6-1 / 6) = 2F (1 / 6) f (1 / 6),
F (1 / 6) = √ 3 / 2
I really forgot that. I have been working for several years since I graduated from university
(1) By substituting x = y = 0 into f (x + y) + F (X-Y) = 2F (x) f (y), f (0) + F (0) = 2F (0) f (0), i.e
2F (0) = 2F (0) f (0). From F (0) ≠ 0, f (0) = 1,
Then we substitute x = 0 into f (0 + y) + F (0-y) = 2F (0) f (y),
F (y) + F (- y) = 2F (y), f (- y) = f (y), so f (x) is an even function.
(2) Substituting y = 1 / 2 into f (x + y)... Expansion
(1) By substituting x = y = 0 into f (x + y) + F (X-Y) = 2F (x) f (y), f (0) + F (0) = 2F (0) f (0), i.e
2F (0) = 2F (0) f (0). From F (0) ≠ 0, f (0) = 1,
Then we substitute x = 0 into f (0 + y) + F (0-y) = 2F (0) f (y),
F (y) + F (- y) = 2F (y), f (- y) = f (y), so f (x) is an even function.
(2) Substituting y = 1 / 2 into f (x + y) + F (X-Y) = 2F (x) f (y), we get
F (x + 1 / 2) + F (x-1 / 2) = 2F (x) f (1 / 2) = 0. From F (1 / 2) = 0, f (x + 1 / 2) + F (x-1 / 2) = 0
F (x + 1 / 2) = - f (x-1 / 2), similarly, f (x-1 / 2) = f (x-1 + 1 / 2) = - f (X-1-1 / 2) = - f (x-3 / 2), so f (x + 1 / 2) = - f (X-1 / 2) = f (x-3 / 2),
Let y = x-3 / 2, then x = y + 3 / 2, x + 1 / 2 = y + 2, substituting into the above formula, f (y + 2) = f (y), so f (x) is a periodic function with period 2.
(3) Let x = y be substituted by F (x + y) + F (X-Y) = 2F (x) f (y) to get f (2x) + F (0) = 2F (x) f (x),
F (2x) = 2F (x) f (x) - 1, substituting x = 1 / 2 into f (1) = 2F (1 / 2) f (1 / 2) - 1 = - 1, i.e
f(1) =-1，
Let y = 2x be substituted by F (x + y) + F (X-Y) = 2F (x) f (y), then f (3x) + F (- x) = 2F (x) f (2x)
f(3x)=2f(x)f(2x)-f(x)=f(x)( 2f(2x)-1)
=f(x)( 4f(x)f(x)-3)=4f(x)f(x)f(x)-3f(x)
Substituting x = 1 / 3 into the above formula
F (1) = 4 f (1 / 3) f (1 / 3) f (1 / 3) - 3 F (1 / 3), let a = f (1 / 3) get
4A ^ 3-3a + 1 = 0, the solution is a = 1 / 2, that is, f (1 / 3) = 1 / 2,
Let f (2 / 3) = 1 / 6 (1 / 3) be monotone.
So f (1 / 3) = 1 / 2, f (1 / 6) = 3 / 2
The axis of symmetry equation of the function y = cos (2x + π / 3) is
I'm sure there's a mistake upstairs. Don't listen to him
Let 2x + π / 3 = k π and get x = [(3K-1) π] / 6
2x+π/3=2kπ
2x=2kπ-π/3
x=kπ-π/6 (k=Z)
T = (2π)/2 = π
One axis of symmetry is
2x + π/3 = 0 => x = -π/6
One axis of symmetry per half cycle
So the equation of axis of symmetry is
x = kT/2 - π/6 = k/2 * π - π/6
It is known that f (x) is a function defined on R. for any x, y ∈ R, f (x + y) + F (X-Y) = 2F (x) f (y),
Let f (y) = f (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (Y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) + F (y) +
（1）：f（0）=1
(2) : judge the parity of function
1) F (x) is a function defined on R. for any x, y ∈ R, f (x + y) + F (X-Y) = 2F (x) f (y),
When x = 0, y = 0
F (0) + F (0) = 2F (0) * f (0) = > F (0) = 0 or F (0) = 1
Because when f (0) = 0
When y = 0, f (x) + F (x) = 0 is unreasonable,
So f (0) = 1
2) When x = 0
The original equation becomes: F (y) + F (- y) = 2F (0) * f (y) = 2F (y) = > F (- y) = f (y)
So the function f (x) is even
A symmetry axis equation of the graph of the function y = cos (2x + π 2) is ()
A. x=-π2B. x=-π4C. x=π8D. x=π
The equation of symmetry axis of this function is 2x + π 2 & nbsp; = k π (K ∈ z). When k = 0, x = & nbsp; − π 4
The function f (x) defined on R satisfies f (x + 1) = 2F (x). If. F (x) = x (1-x) when 0 ≤ x ≤ 1, then if - 1 ≤ x ≤ 0, f (x)=______ .
When - 1 ≤ x ≤ 0, 0 ≤ x + 1 ≤ 1, f (x) = 12F (x + 1) = 12 (x + 1) [1 - (x + 1)] = - 12x (x + 1), so the answer is: - 12x (x + 1)
A symmetric axis equation of the function f (x) = cos (2x + Π / 4) is
A 、x=-∏/2
B、x=-∏/4
C、x=-∏/8
D、∏
Axis of symmetry: 2x + Π / 4 = k π (K ∈ z)
x=kπ/2-∏/8 (k∈Z)
When k = 0, x = - Π / 8
So choose C