F (x) = 2sinx + 1, (1) let the constant ω > 0, if y = f (ω x) is an increasing function in the interval [- π / 2,2 π / 3], find the value range of ω (2) Let a = {π / 6

F (x) = 2sinx + 1, (1) let the constant ω > 0, if y = f (ω x) is an increasing function in the interval [- π / 2,2 π / 3], find the value range of ω (2) Let a = {π / 6

(1) Analysis: ∵ f (x) = 2sinx + 1F (ω x) = 2sinω x + 1 ∵ in the interval [- π / 2,2 π / 3] is an increasing function ∵ the initial phase of function f (x) is 0 ∵ the minimum point is on the left side of y-axis, the maximum point is on the right side of y-axis, the distance between them and y-axis is equal. The minimum point of function f (x) is: ω x = 2K π - π / 2 = = > x = 2K π / ω - π / (2 ω) ∵ π / (2)

When - 1 ≤ x ≤ 2, y = ax + 6 satisfies y < 10, then the value range of constant a is______ ．

When a = 0, y = ax + 6 = 6, so satisfy y < 10; when a ≠ 0, function y = ax + 6 is a linear function, it is increasing or decreasing, when - 1 ≤ x ≤ 2, y < 10. Then when x = - 1, y = ax + 6 = - A + 6 < 10, the solution is a ＞ - 4; when x = 2, y = ax + 6 = 2A + 6 < 10, the solution is a < 2; so - 4 < a < 2, and a ≠ 0

If the function f (x) = x ^ 2-4x + A / X is an increasing function in the interval (1, + ∞), then the value range of constant a is
A（-∞,-2] B（-∞,-27/4]
C(-∞,-64/27] D(-∞,-32
D（-∞，-32/27）

f(x)=x^2-4x+a/x
f'(x)=2x-4-a/x^2
=(2x ^ 3-4x ^ 2-A) / x ^ 2 > 0 is an increasing function
Because the denominator x ^ 2 > 0
Then the molecule 2x ^ 3-4x ^ 2-A > 0
Let g (x) = 2x ^ 3-4x ^ 2-a
g'(x)=6x^2-8x=2x(3x-4)=0
Because x > 1
Then x = 4 / 3 is the extremum of G (x)
Then we substitute x = 4 / 3 into f '(x)
f'(x)=2*(4/3)^3-4*(4/3)^2-a>0
a

It is known that the function f (x) = x + A / X. (a is greater than 0. It is a constant) is an increasing function in the interval (2. + 00). Then the value range of a is··

f(X)=x+a/x
F '(x) = 1-A / X2 (the square of x)
It is known that f '(x) = 1-A / x2 ≥ 0 holds for X ≥ 2
1 ≥ A / X2 is equivalent to x2 ≥ a
Let g (x) = X2 (x ≥ 2)
It is easy to know that G (x) = x2 increases monotonically in ＂ 2. + 00 ＂
∴g(X)min=g(2)=4
∴a≤g(X)min=g(2)=4
Then the value range of a is (- ∝, 4)

Given the function f (x) = | x-a | (a is a constant), if f (x) is an increasing function in the interval [1, + ∞), then the value range of a is______ ．

F (x) = | x − a | = x − ax ≥ a − x + ax ＜ a; the function is an increasing function on [a, + ∞); and f (x) is an increasing function on [1, + ∞); the value range of a ≤ 1 is (- ∞, 1]. So the answer is: (- ∞, 1)

Let f (x) = x + K / X and constant k > 0. If f (x) increases monotonically in the interval [1,4], the value range of K is obtained

The derivative of F (x) is obtained
f'(x)=1-k/x^2
For any x ∈ [1,4]
There are: F '(x) = 1-k / x ^ 2 ≥ 0
That is: K / x ^ 2 ≤ 1
k≤x^2
k≤1

SiNx = 2-3a, meaningful, find the value range of A. find the value range of function y = cos & # 178; X + 3sinx + 1

Solution 1 is determined by SiNx = 2-3a
Know - 1 ≤ SiNx ≤ 1
That is - 1 ≤ 2-3a ≤ 1
That is 3a ≤ 3 and 3a ≥ 1
That is, 1 / 3 ≤ a ≤ 1
2 by y = cos & # 178; X + 3sinx + 1
=1-sin²x+3sinx+1
=-sin²x+3sinx+2
=-（sinx-3/2）²+17/4
From - 1 ≤ SiNx ≤ 1
When SiNx = 1, y has the maximum value y = - (1-3 / 2) & # 178; + 17 / 4 = 4
When SiNx = - 1, y has the maximum value y = - (- 1-3 / 2) &# + 17 / 4 = - 2
That is, the range of function y = cos & # 178; X + 3sinx + 1 [- 2,4]

Y = sin ^ 2x + 3sinx-12. Y = - cos ^ 2x + SiNx + 3, find the range of the following functions

(1) Y = sin ^ 2x + 3sinx + 9 / 4-9 / 4-1 = (SiNx + 3 / 2) ^ 2-13 / 4, because SiNx is between [- 1,1], we get [- 3,3] (2) y = - (1-sin ^ 2x) + SiNx + 3 = sin ^ 2x + SiNx + 2 = sin ^ 2x + SiNx + 1 / 4-1 / 4 + 2 = (SiNx + 1 / 2) ^ 2 + 7 / 4, because SiNx is between [- 1,1]

Given that f (x) = sin (x + φ) + cos (x + φ) is an odd function, then one of the values of φ is ()
A. −π4B. π2C. 0D. π

Because the function f (x) = sin (x + φ) + cos (x + φ) = 2Sin (x + φ + π 4) is an odd function. Therefore, choose a

Given the function y = cos (2x - π △ 3), when x is in the back interval 0 ° to π, find the range of the function?

The range of 2x - π / 3 is [- π / 3, 5 π / 3]
Let t = 2x - π / 3, then the value range of T is [- π / 3, 5 π / 3]
Y = cost, the value range of Y is [- 1,1]