Given that the points (a, b) are on the circle x2 + y2 = 1, the minimum positive period and minimum value of the function f (x) = acos2x + bsinxcosx-a2-1 are () A. 2π，-32B. π，-32C. π，-52D. 2π，-52

Given that the points (a, b) are on the circle x2 + y2 = 1, the minimum positive period and minimum value of the function f (x) = acos2x + bsinxcosx-a2-1 are () A. 2π，-32B. π，-32C. π，-52D. 2π，-52

F (x) = acos2x + bsinxcosx-a2-1 = a · 1 + cos2x2 + b2sin2x-a2-1 = 12 (bsin2x + acos2x) - 1 = B2 + A22 (BB2 + a2sin2x + AB2 + a2cos2x) - 1 = 12sin (2x + θ) - 1, (Tan θ = AB).. the minimum positive period of the function is 2 π 2 = π. When sin (2x + θ) = 1, the minimum value of the function is - 32

Given the function f (x) = sin (2x + α) + ACOS (2x + α), where a > 0 and 0 < a < π, if the image of F (x) is symmetric with respect to the line x = π / 6, and the maximum value of F (x) is 2
(1) Finding the values of a and α
(2) How to get the image of y = 2Sin (2x + π / 3) from the image of y = f (x)

f(x)=sin(2x+α)+acos(2x+α)=√（1+a²）sin(2x+α+β),tanβ=a
The maximum value of F (x) is 2, a = √ 3, β = π / 3
From the symmetry of the image of F (x) with respect to the straight line x = π / 6, 2 × π / 6 + π / 3 + α = π / 2 or 3 π / 2, and α = 5 π / 6 with respect to the range of α
So f (x) = 2Sin (2x + 7 π / 6)
2 y = 2Sin (2x + π / 3) is obtained by F (x) = 2Sin (2x + 7 π / 6) translating π / 4 units along the positive direction of X

Given that f (x) = ACOS (2x + φ) satisfies f (x) ≤ f (5 Π / 12) = 5, then the set of X when f (x) takes the minimum value is?

The maximum value of F (x) is f (5 Π / 12) = 5,
The period of F (x) is Π,
The difference between the maximum value and the minimum value is half a period,
Therefore, when 5 Π / 12 + Π / 2 = 11 / 12 Π, the minimum value is taken,
So the minimum set {x = 11 / 12 Π + K Π, K is an integer}

High school mathematics exponential function! Urgent ah
Make the image of the absolute value of the function y = (x-1 of 3), and point out that when k is the value, the absolute value of equation (x-1 of 3) = k has no solution? Has solution? How many solutions?

As shown in the figure
When K & lt; 0, the absolute value of equation (x-1 of 3) = k has no solution;
When k is 0, the absolute value of the equation (x-1 of 3) = k has one solution x = 0;
When k is (0,1), the absolute value of equation (3 to x-1) = k has two solutions;
When k is [1, + ∞), the absolute value of the equation (x-1 of 3) = k has a solution

Given the odd function f (x) defined on R, if x > 0, f (x) = (1 / 2) ^ x, find the analytic expression of F (x)
The topic should be exponential function. If you want to have a process, please make it clear,

Because of the odd function f (x) on R, then f (0) = 0, f (- x) = - f (x)
If x > 0, f (x) = (1 / 2) ^ x,
So when x

The solution set of inequality 4x-3 × 2x + 1-16 > 0 is ()
A. （-1，3）B. （-3，1）C. （3，+∞）D. （-∞，-1）∪（3，+∞）

Let t = 2x, (t ＞ 0), then inequality 4x-3 × 2x + 1-16 ＞ 0 can be transformed into t2-6t-16 ＞ 0 solution to get t ＞ 8, or t ＜ - 2 (rounding off), that is, 2x ＞ 8 solution to get x ＞ 3, so the solution set (3, + ∞) of inequality 4x-3 × 2x + 1-16 ＞ 0 should be c

If the value of F (x) = log a (2-x) on (1,2) is always negative, then the value range of a is?

Is this a base or a true number? If a is a base, then there is a lower solution!
If x belongs to (1,2), i.e. 1 ＜ x ＜ 2, then 0 ＜ 2-x ＜ 1, because the Y power of a = (real number) 2-x, and because f (x) is always negative, that is, y is negative, and the real number is greater than 0 and less than 1, only when a is greater than 1, the value of function f (x) = log a (2-x) on (1,2) is always negative, and it holds when a is greater than 1!
Of course, if you don't understand, you can use the special value substitution method! Let the exponential power y be - 1, if 0 < a < 1, then the true number of - 1 power of a is greater than 1, which is not in line with the meaning of the problem, and when a is greater than 1, then the true number of - 1 power of a is always greater than 0 and less than 1!

For any real number x, the value of the function f (x) = x ^ 2-4ax + 2A + 30 is non negative. The value range of the equation x / (a + 2) = | A-1 | + 1 about X is obtained

For any real number x, the value of function f (x) = x ^ 2-4ax + 2A + 30 is non negative, so △ = 16A ^ 2-4 (2a + 30) ≤ 0, the solution is - 5 / 2 ≤ a ≤ 3, when ① 1 ≤ a ≤ 3, from X / (a + 2) = | A-1 | + 1, the solution is x = a (a + 2), the solution is (a + 1) ^ 2 = x + 1, a = - 1 ± √ (x + 1), a = - 1 - √ (x + 1), which is not in line with the meaning of the problem

Find the minimum and maximum values of the function y = log ^ 2 base 2 index X - log base 2 index x ^ 2 - 3 in the interval [1,4]

Y = log ^ 2 base 2 index X - log base 2 index x ^ 2 - 3
=Log ^ 2 base 2 index X - 2 log base 2 index X - 3
=(log base 2 index x-1) ^ 2-4
When x = 2, take the minimum value, then y = - 4; when x = 1 or 4, take the maximum value
At this point, y = - 3

(+ 30 points) for X ∈ R, the function f (x) = x ^ 2-4ax + 2A + 30 (a ∈ R) is a non negative number, and the value range of the root of the equation x / (a + 2) = | A-1 | + 1 about X is obtained
We need the answer today
Pay attention to Baidu know the inequality and my requirements are not the same oh

The first condition is that no matter what the value of X is, f (x) 0, the lowest point of the function equation image is also 0, that is, when x = 2A, f (x) = - 4A ^ 2 + 2A + 30, 0 inequality has a solution, that is - (2a-6) (2a + 5) 0 - 2 / 5 "the absolute value of a" 3 A-1 "represents the distance from a point to 1 point on the number axis, which is 1 (| A-1 | + 1) (| A-1 | + 1)