1. It is known that a > 0, X belongs to [0,2 π faction], y = 1 / 2cos 2x - asin x + B + 1 / 2, the maximum value is 0 and the minimum value is - 4

1. It is known that a > 0, X belongs to [0,2 π faction], y = 1 / 2cos 2x - asin x + B + 1 / 2, the maximum value is 0 and the minimum value is - 4

XCF

Here is an overview of your problem. Given the function f (x) = 2cos ^ 2x + radical 3sin2x + A, find the maximum and minimum value of F (x) and find the minimum value of real number a Given the function f (x) = 2cos ^ 2x + Radix 3sin2x + a x ∈ (0, π / 2), and / F (x) / < 2 (1), find the maximum and minimum of F (x) and (2) find the minimum value of real number a

When 2x + π / 3 = π / 2 + 2K π, f (x) takes the maximum value of F (x), f (max) = 2 (max) = 2 + 2 + 1 + a = 3 + a = 2x + 2 + 2 + 2K π π, f (x) takes the maximum value, f (max) = 2 + 1 + a = 3 + a when 2x + π / 3 = π / 2 + 2 + 2K π, when 2x + π / 3 = - π / 2 + 2 + 2K π, f (x) takes the minimum value, f (min) = - 2 + 1 + a = - 1 + A, find the minimum value of a, a is not a change, a is not a change, a is not a change, a is not a change, f (min) = - 2 + 1 + a = - quantity, no most

Let x + y = 5, xy = minus 3, find the value of (2x-3y-2xy) - (x-4y + XY)

（2x-3y-2xy）-（x-4y+xy）
=2x-3y-2xy-x+4y-xy
=（2-1)x+(-3+4)y+(-2-1)xy
=x+y-3xy
Because x + y = 5, xy = - 3
So the original formula = 5 + 9 = 14

Given | x-y-1 | + (XY + 2) ° = 0, find the value of (- 2XY + 2x + 3Y) - (3xy + 2y-2x) - (x + 4Y + XY)

|X-y-1 | + (XY + 2) 2 = 0, find (- 2XY + 2x + 3Y) - (3xy + 2y-2x) - (x + 4Y + XY) | x-y-1 + (XY + 2) A2 = 0, the sum is zero only when both of them are 0, so x-y-1 = 0, XY + 2 = 0 = > X-Y = 1.xy = - 2 (- 2XY + 2x + 3Y) - (3xy + 2y-2x) - (x

If x + y = 5, xy = 3, find the value of (2x-3y-2xy) - (x-4y + XY)

Simplify the original equation
2x-3y-2xy-x+4y-xy
=x-3xy+y
=5-3*3
=-4

It is known that the mean of 2,6, - 2x, 3Y is 3: - 5, - 3x, 12,4y is 3. Find the value of XY (x + y)

∵ the average of 2,6, - 2x, 3Y is 3: - 5, - 3x, 12, 4Y is 3
∴2+6-2X+3y=12；—5—3x+12+4y=12
∴-2X+3y=4①；-3x+4y=5②
① X 3 - 2 × 2, y = 2
Substituting y = 2 into ① gives: x = 1
Put y = 2 x = 1 in
xy(x+y)=1×2×﹙1﹢2﹚=6

Given that the mean of 2,6-2x, 3Y is 3; the mean of - 5, - 3x, 12, 4Y is 3, find XY (x + y)

(2+6-2x+3y)/3=3
(-5-3x+12+4y)/4=3
By solving the equations, the following results are obtained
x=-11
y=-7
xy=-11*(-7)=77
x+y=-7-11=-18

Given 3x-y = 1 / 16, xy = 2, find 6x ^ 4Y ^ 3-2x ^ 3Y ^ 4

6x^4y^3-2x^3y^4
=2x^3y^3*(3x-y)
=2*2^3*1/16
=2*8*1/16
=1

If X-Y = 3, xy = 5, then the value of the algebraic expression (3x-4y + 5xy) - (2x-3y) + 5xy is

（3x-4y+5xy）-（2x-3y）+5xy
=3x-4y+5xy-2x+3y+5xy
=(x-y)+10xy
=3+50
=53

Given 3x-y = 1 / 2, xy = 2, find the value of 3x ^ 4Y ^ 3-x ^ 3Y ^ 4

3x^4y^3-x^3y^4
=(xy)^3*(3x-y)
=2^*1/2
=4