Given that the function y = cosx ^ 2 + asinx-a ^ 2 + 2A + 5 has the maximum value of 2, try to find the value of real number a

Given that the function y = cosx ^ 2 + asinx-a ^ 2 + 2A + 5 has the maximum value of 2, try to find the value of real number a

The derivation will come out
y=kx-bx=0
Given that the function y = cos2x + asinx-a2 + 2A + 5 has the maximum value 2, try to find the value of real number a
Let SiNx = t, then y = f (T) = - t (T) = - T 2 + at-a2 + 2A + 6, t ∈ [-1, 1, 1], the symmetry axis is t = A2, the symmetry axis is t = A2, when A2 ＜ − 1, that is, a ＜ - 2, ymax = f (- 1) = -a2 + A + A + 5 = 2, a = 1 ± 132 (rounding) when − 1 ≤ A2 ≤ 1, that is - 2 ≤ a ≤ 2 ≤ 2, that is - 2 ≤ 2 ≤ a ≤ 2, ymax = f (A2) = f (A2) = f (A2) = 34a2 + 2A + 2A + 2A + 2A + 2A + 2A + 6 + 6 = 2, that is, at this time, a = 4 (rounding) or a = 43. When A2, that is a \\\\\\\\\+ 5 = 2, a = 3 + 212 or a = 3 − 212, so a = 3 + 212 or a = − 43
If the image of the function y = 4x + 3-K passes through the origin, then K=______ .
The coordinate of the origin is (0, 0), that is, when x = 0, y = 0, it is substituted into the function y = 4x + 3-K to get k = 3
If the equation LG (AX-1) - LG (x-3) = 1 of X has a solution, what is the value range of a
The correct answer is a less than or equal to 0 or a = 4, please write down the process
lg(ax-1)=lg（x-3）+lg10
lg(ax-1)=lg[10(x-3)]
lg(ax-1)-lg[10(x-3)]=0
lg{(ax-1)/[10(x-3)]}=0
(ax-1)/[10(x-3)]=1
ax-1=10x-30
(a-10)x=-29
When A-10 is not equal to 0, that is, when a is not equal to 10, the equation has a solution. The solution is x = - 29 / (A-10)
Because LG (AX-1) - LG (x-3) = 1 has a solution,
So LG (AX-1) = LG (x-3) + LG10 = LG [10 (x-3)],
AX-1 = 10 (x-3), and AX-1 > 0, x-3 > 0,
(a-10)x=-29，
Because x > 3,
So - 29 / (A-10) > 3,
When a > 10, 3 (A-10) 3,
So - 29 / (A-10) > 3,
When a > 10, 3 (A-10)
If y = 4x + B, the area of the triangle formed by its image and two coordinate axes is 16, then the value of B is（
The intersection of image and two axes (O, b) (- B / 4,0)
s=b²/8=16
b=±8√2
Can you write the process? thank you!
Given the function y = LG (AX2 + 2aX + 1): (1) if the domain of definition of the function is r, find the value range of a; (2) if the domain of definition of the function is r, find the value range of A
(1) When a ≠ 0, there should be a ＞ 0 and △ = 4a2-4a ＜ 0, and the solution is a ＜ 1. Therefore, the value range of a is [0,1). (2) if the value range of the function is r, then AX2 + 2aX + 1 can take all positive integers, a ＞ 0 and △ = 4a2-4a ≥ 0. The solution is a ≥ 1, so the value range of a is [1, + ∞)
A mathematical problem. It is known that the function f (x) is a monotone odd function defined on R, and f (1) = = u 2
(1) Verify F (x) is a monotone decreasing function. (2) solve the inequality f (2) + F (2)_ The x power of 4_ 1）>0
Errr... Forget (1) and (2) is based on one, add: (1): because it is an odd function, so f (1) = - 2, so f (- 1) = - (- 2) = 2. - 1F (1) is a monotone function, so it is a decreasing function (2): let 2 ^ x = y 4 ^ x = (2 ^ 2) ^ x = 2 ^ 2x = (2 ^ x) ^ 2 = y ^ 2 (the square of Y) be transformed into: F (y) + F (Y-Y ^ 2
If the value range of function y = LG (ax2-x + 1) is r, then the value range of a is r___ .
Let f (x) = ax2-x + 1 satisfy the condition, i.e. (0, + ∞) ⊊ {y | y = f (x)}, when a = 0, f (x) = ax2-x + 1 = - x + 1, which satisfies the condition. When a ≠ 0, if f (x) = ax2-x + 1 satisfies the condition, then when a < 0, it satisfies the condition, when a > 0, it satisfies the discriminant △ = 1-4a ≥ 0, i.e. 0 < a ≤ 14 So the answer is: 0 ≤ a ≤ 14
It is known that y = f (x) is an odd function defined on (- 2,2) and a decreasing function if f (m-2) + F (2m-1) > 0
Finding the value range of real number m
Y = f (x) is defined in (- 2,2)
∴-2＜m-2＜2
-2＜2m-1＜2
The solution is 0 ＜ m ＜ 3 / 2
Moreover, f (x) is an odd function and a decreasing function in the domain of definition
f(m-2)+f(2m-1)>0
f(m-2)＞-f(2m-1)=f(1-2m)
∴m-2＜1-2m
The solution is m < 1
In conclusion, 0 < m < 1
Who knows the logarithm function, where the base of logarithm is one eighth and the true number of logarithm is 16
Find the formula
It is stipulated that the first number after log is the base number and the second number is the true number
log 1/8 16
=1og 2^(-3) 2^4
=- 4 / 3 log 2 2 (the power of the true number can be used as the numerator directly in advance, and the power of the base number can be used as the denominator in advance)
= - 4/3
Log (1 / 8) 16 where 1 / 8 is written in the lower right corner of log
Let log (1 / 8) 16 = X
Then the x power of (1 / 8) = 16
Solving equation (1 / 2) to the power of 3x = (1 / 2) to the power of - 4
3x=-4
x=-4/3
That is log (1 / 8) 16 = x = - 4 / 3