It is known that f (x) = (x ^ 2-5) / 2x, f (3 + 2Sin θ)

It is known that f (x) = (x ^ 2-5) / 2x, f (3 + 2Sin θ)

Derivation: F (x) = (2x * 2x - (x ^ 2-5) * 2) / (4x ^ 2) = 0.5 + 5 / (2x ^ 2) > 0
So f (x) increases monotonically because of the range of 3 + 2Sin θ [1,5]
The maximum value of F (3 + 2Sin θ) is f (5) = 2
Therefore, m ^ 2 + 3m-2 > 2 means that M is (- infinite, - 4), (1, + infinite)
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F (x) = 1 + 2Sin (2x - π / 3), X belongs to [π / 4, π / 2], find the maximum value of F (x) and find the corresponding X

Analysis:
Y = sin a is an increasing function on a ∈ [π / 4, π / 2] and a decreasing function on a ∈ [π / 2, π]
So let 2x - π / 3 = a
Then when a = π / 2, there is a maximum value, where 2x - π / 3 = π / 2
The solution is: x = 5 π / 12 ＜ π / 2. Therefore, when x = 5 π / 12, the function has the maximum value, f (x) = 3
When a is symmetric in the symmetric interval, the function value remains unchanged (i.e. x = π / N and x = π - π / N, a ≠ 0)
Therefore, X ∈ [π / 4, π / 2] is introduced into sin (2x - π / 3)
When x = π / 4, a = π / 6
When x = π / 2, a = 2 π / 3
π / 6 is closer to the x-axis than 2 π / 3, so sin (π / 6) < sin (2 π / 3)
So sin (π / 6) is smaller
So when x = π / 4, the function has a minimum value, f (x) = 2

Given the function f (x) = 2Sin ^ 2x-3, find the minimum positive period of function f (x) The known function f (x) = 2Sin ^ 2x-3 1. Find the minimum positive period of function f (x) 2. Find the maximum value of function f (x) and the value of X when the maximum value is obtained

1,f(x)=1-2cos2x-3=-2cos2x-2
So the minimum positive period T = 2 π / 2 = π
When cos2x = - 1, f (x) reaches the maximum
From 2x = π + 2K π, x = π / 2 + K π K ∈ Z is obtained
So the maximum value of F (x) is - 2 × (- 1) - 2 = 0, where x = π / 2 + K π K ∈ Z
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Monotone increasing decreasing interval of y = √ 2Sin (2x + pi / 4)

y=√2sin(2x+π/4)
Obviously, when 2x + π / 4 is in the range 2K π - π / 2 to 2K π + π / 2, y increases monotonically,
However, y decreases monotonically from 2K π + π / 2 to 2K π + 3 π / 2,
So y's
The monotone increasing interval is (K π - 3 π / 8, K π + π / 8)
The monotone decreasing interval is (K π + π / 8, K π + 5 π / 8), and K is an integer

The monotone increasing interval of the function y = log (1 / 2) [2Sin (2x + π / 3) - 1] is to find the detailed solution,

t=2sin(2x+π/3)-1
Y = log (1 / 2) t is a decreasing function when t > 0
The increasing interval of the original function is required
That is, the decreasing interval of T = 2Sin (2x + π / 3) - 1 when t > 0
∴ sin(2x+π/3)>1/2
∴ 2kπ+π/2≤2x+π/3

F (x) = (SiNx + cosx) ^ 2 / [2 + 2Sin (2x) - (COS (2x)) ^ 2]

2 + 2Sin (2x) - (COS (2x)) ^ 2 does not = 0,
1 + 2Sin (2x) + (sin ((2x)) ^ 2 does not = 0,
[sin (2x) + 1] ^ 2 does not = 0,
Sin (2x) not = - 1,
2X not = 2K π + π, (k is an integer)
X does not = k π + π / 2. (k is an integer)
So the definition domain of F (x) is: {x | X does not = k π + π / 2, K is an integer}

Given the function f (x) = √ 3cos2x + 2Sin (x + π / 2) cos (x + π / 2) (1), the function is expressed as a sinusoidal function (2) and the function is solved in R Monotone increasing interval

f(x)=√3cos2x+2sin(x+π/2)cos(x+π/2)
=√3cos2x+sin(2x+π)
=√3cos2x-sin2x
=2(√3/2cos2x-1/2sin2x)
=2(sinπ/3cos2x-cosπ/6sin2x)
=2sin(π/3-2x)
Monotonically increasing interval: 2K π - π / 2

Given x ∈ R, find the function f (x) = (SiNx + 3) (COS) Given x ∈ R, find the value range of function f (x) = (SiNx + 3) (cosx-3)

F (x) = sinxcosx-3 (SiNx cosx) - 9, let a = SiNx cosxa? = sin? 2A + cos? A-2sinacosa = 1-2sinxcosx = (1-A?) / 2Y = f (x) = (1-A?) / 2-3a-9a = SiNx cosx = √ 2 (SiNx * √ 2 / 2 + cosx * √ 2 / 2) = √ 2 (sinxcos π / 4 + cosxsi

Given that the opposite sides of the inner angles a, B and a in the acute triangle ABC are a, B, C, the definition vector M = (SINB, - radical 3), n = (cos2b, 4cos ^ 2 (B / 2-2)) And M / / N, Q: if B = 1, find the maximum area of the triangle ABC

M = (SINB, - √ 3), n = (cos2b, 4cos ^ 2 (B / 2) - 2) = (cos2b, 2cosb), from m ∥ n, SINB / cos2b = - √ 3 / (2cosb), - tan2b = - √ 3,2b = 2 π / 3, or 5 π / 3, B = π / 3 or 5 π / 6. B = 1, according to the sine theorem, a = Sina / SINB, C = sinc / SINB, ﹤ s △ ABC = (1 / 2) acsinb = sinasinc /

(1 / 2) in the triangle ABC, the known angles ABC are ABC, vector M = (2sinb, - radical 3), n = (cos2b, B-1 of 2cos Square), and m

In the triangle ABC, the known angles ABC are ABC, vector M = (2sinb, - radical 3), n = (cos2b, B-1 of 2cos Square), and m
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