If all solutions of the equation LG (AX) * LG (AX ^ 2) = 4 are greater than 1, the value range of a is obtained

If all solutions of the equation LG (AX) * LG (AX ^ 2) = 4 are greater than 1, the value range of a is obtained

LG (AX) * LG (AX ^ 2) = 4 (LGA + lgx) (LGA + lgx ^ 2) = 4 (LGA + lgx) (LGA + 2lgx) = 42 (lgx) ^ 2 + 3lga * lgx + (LGA) ^ 2-4 = 0, let m = lgx2m ^ 2 + 3lga * m + (LGA) ^ 2-4 = 0x1 > 1, X2 > 1, so M1 > 0, M2 > 0, so M1 + M2 > 0, M1 * M2 > 0 by Weida theorem M1 + M2 = - 3lga / 2 > 0lga0 (LGA) ^ 2 > 4LG
Lg(ax)*Lg(ax^2)=(lga+lgx)(lga+2lgx)
=2(lgx)^2+3lga*lgx+(lga)^2
All roots are greater than 1
f(x)=2(lgx)^2+3lga*lgx+(lga)^2-4=0
The root of is a positive number
Axis of symmetry: - 3lga / 4 > 0,0
Let f (x) = cos3x2cosx2-sin3x2sinx2. (I) find the minimum positive period of function f (x); (II) find the zero point of function f (x) when x ∈ [π 2, π]
(I) f (x) = cos3x2cosx2-sin3x2sinx2 = cos (3x2 + x2) = cos2x, (4 points) ∵ ω = 2, ∵ t = 2 π 2 = π, then the minimum positive period of function f (x) is π; (5 points) (II) Let f (x) = 0, that is, cos2x = 0, and ∵ x ∈ [π 2, π], (7 points) ∵ 2x ∈ [π, 2 π], (9 points) ∵
The range of function f (x) = log5 (5x + 1) is______ .
Since 5x + 1 ＞ 1, f (x) = log5 (5x + 1) ＞ log51 = 0, the range of function is (0, + ∞), so the answer is: (0, + ∞)
Let y = x2-4x + 3, X ∈ [1,4], then the range of F (x) is___ .
∵ y = x2-4x + 3 = (X-2) 2-1, x = 2 ∈ [1,4], the minimum value of this function on [1,4] is f (2) = - 1, the maximum value is f (4) = 3, and the range of function f (x) is [- 1,3]. So the answer is: [- 1,3]
Given the function f (x) = (1 / 2) ^ x, find the range of the function f (x) = f (2x) - f (x) x which belongs to (0, + infinity)
F (x) = f (2x) - f (x) = (1 / 2) ^ (2x) - (1 / 2) ^ x = (1 / 2) ^ x multiplied by [(1 / 2) ^ X-1], because (1 / 2) ^ x is greater than 0 and less than 1, then take the derivative of F (x) = 0 to get x = 1, and finally get x = 1 which is greater than - 1 / 4 and less than 0
Given the function f (x) = - 4cos ^ 2 x + 4 √ (3) SiN x cos x + 5, X belongs to R
(1) Find the maximum value of function and the set of X when taking the maximum value
(2) Find monotone increasing interval of function
When the double angle formula 2cos & # 178; X = cos2x + 12sinxcosx = sin2xf (x) = - 4cos & # 178; X + 4 √ (3) SiNx cosx + 5 = - 2 (cos2x + 1) + 2 √ 3sin2x + 5 = 2 √ 3sin2x-2cos2x + 3 = 4 (√ 3 / 2sin2x-1 / 2cos2x) + 3 = 4 (COS π / 6sin2x sin π / 6cos2x) + 3 = 4sin (2x - π / 6) + 3
Given that 0 is less than X and less than or equal to 1 / 4, find the range of function y = x ^ 2-2x + 2 / X
f(x)=y=(x²-2x+2)/x=x+2/x-2
Let x1, X2, and 0 on the domain
Let f (x) = sin (Wx + φ), where w > 0, | φ|
There is a formula cos (a + b) = cosacosb sinasinb
The range of the following functions is (1) y = x power - 2x (- 1 less than or equal to x less than or equal to 2) (2) y = x quartic power + 1
(3) Y = (x + 1), X is less than or equal to 0, - x, X is greater than 0
I can't understand what you're playing
Two
Because the square of X is equal to or greater than 0, the range y is equal to or greater than 1
I don't understand
Your question is not clear. The second question is the same as above, y is greater than or equal to 1
If SiNx, cos2x, 1 SiNx are in equal proportion sequence, then the value of SiNx is obtained
|sinx|=1
1/sinx>=1,1/sinx=1,cos2x
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