The value range of the function y = cos ^ 2x + 3cosx + 2, XS ∈ [0, π / 2) is

The value range of the function y = cos ^ 2x + 3cosx + 2, XS ∈ [0, π / 2) is

y=cos²x+3cosx+2
=cos²x+3cosx+9/4-9/4+2
=(cosx+3/2)²-1/4
∵x∈[0,π/2）
∴cosx∈(0,1]
That is, cosx + 3 / 2 ∈ (3 / 2,5 / 2]
(cosx+3/2)²∈(9/4,25/4]
∴y∈(2,6]
That is, the value range of the function is (2,6]
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The known function f (x) = sin2x, G (x) = cos (2x + π) 6) The line x = t (t ∈ R) intersects the graph of functions f (x) and G (x) at M and N points respectively (1) When t = π At 4, calculate the value of | Mn |; (2) Find | Mn | at t ∈ [0, π 2] The maximum value of

(1) In the function f (x), G (x), we get the following result: we get the ∵ | | | | | (f (π 4) | (f (π 4) | | (f (π 4) − g (π 4) | = | sin (2 × π 4) − cos (2 × π 4 + π 6) | | | (1 | Cos2 π 3 | = 32. (2) the | of | Mn | | | | Mn | | | | | f (π 4) | (π f (π 4) | (4) | cos (2T + π 6) | = | 32sin2t − 32cos2t | = 3 | sin (2t − π 6) | ∵ t

Let f (x) = sin2x + 2 √ 3 cos? X Let f (x) = sin2x + 2 √ 3 cos? X Find the maximum value of F (x). Please be more detailed

F (x) = sin2x + 2 √ 3cos? X = sin2x + √ 3 (2cos? X-1) + √ 3 = sin2x + √ 3cos2x + √ 3 = 2 (sin2xcos60? + cos2xsin60? + √ 3 = 2Sin (2x + 60?) + √ 3 ∵ 1 ≤ sin (2x + 60) ≤ 1. Therefore, when sin (2x + 60) = 1, the maximum value of F (x) is

If the function f (x) = x? - ax + 4 has a zero point on [1,4], then the value range of real number a is?

x^2-ax+4=0
A = (x ^ 2 + 4) / x = x + 4 / X is obtained
From the mean inequality, x + 4 / x > = 2 √ (x * 4 / x) = 4. When x / = 4 / x, that is, x = 2, the equal sign is taken, so a > = 4
The maximum value of X + 1 / X is obtained at the endpoint of [1,4]:
x=1,a=5,
x=4,a=5
Therefore, the value range of a is [4,5]

Given that a is a real number, the function f (x) = 2aX 2 + 2x-3-a, if y = f (x) has a zero point on the interval [- 1,1], find the value range of A

Look at the picture:

Given that a is a real number, the function f (x) = 2aX 2 + 2x-3-a, if the function y = f (x) has a zero point on the interval [- 1,1], find the value range of A Why does the equation f (x) = 0 have two distinct real roots on [- 1,1] when a = 5? The equation f (x) = 0 has two distinct real roots on the interval [- 1,1]. Combining with the image, we get the following conclusion A>0 f(1)>=0 f(-1)>=0 f(-1/2a)=0 f(-1/2a)

When a > 0, the opening is upward, and f (1) and f (- 1) must be > 0. The lowest point is the position of the symmetry axis X = - B / 2a, where the axis of symmetry is x = - 1 / 2a, and a < 0 is the same
A>0
f(1)>=0
f(-1)>=0
F (- 1 / 2a) < 0 and
A<0
f(1)<=0
f(-1)<=0
f(-1/2a)>0

Given the function f (x) = e ^ x + X? - x, if the function y = | f (x) - t | - 3 has four zeros, then the value range of real number T is obtained

F '(x) = e ^ x + 2x-1, when x0,
Therefore, f (x) takes the minimum value f (0) = 1 at x = 0,
Therefore, when T1, y = | f (x) - t | - 3 takes the maximum value | f (0) - t | - 3 = | 1-t- 3 at x = 0,
To make y = 0 have four roots,
Only | 1-T | 3 > 0,
The solution is t > 4 (omit t

The minimum positive period of the function f (x) = sin2x sin (2x Pai / 3) is

3.14/2

The known function f (x) = sin2x-2sin2x (I) find the minimum positive period of function f (x); The minimum value of (x) and (x) is obtained

(I) because f (x) = sin2x - (1-cos2x)=
2sin（2x+π
4）-1，
So the minimum positive period of the function f (x) is t = 2 π
2=π
(II) from (I), when 2x + π
4=2kπ−π
2，
That is, x = k π − π
When 8 (K ∈ z), the minimum value of F (x) is −
2−1；
Therefore, when the function f (x) takes the minimum value, the set of X is: {x | x = k π - π
8，k∈Z}

The function f (x) = sin2x-2 √ 3cos square x ＋√ 3 to find the minimum period and monotone interval of a function

f(x)=sin2x－2√3cos²x＋√3=sin2x+√3cos2x=2sin（2x+π/3）
So t = 2 π / 2 = π
The minimum positive period of F (x) is π
Let 2K π - π / 2