If x2 PX + AB = (x + a) (x + b), then p = () A. -a+bB. -a-bC. a-bD. a+b

If x2 PX + AB = (x + a) (x + b), then p = () A. -a+bB. -a-bC. a-bD. a+b

∵ (x + a) (x + b), = x2 + (a + b) x + AB, = x2 PX + AB, ∵ - P = a + B, then p = - a-b

U = {1,2}, a = {x | X & # 178; + PX + q = 0}, CUA = {1}, then p + Q=_____

Two equations are solved, one is P & # 178; - 4q = 0, the other is to bring 2 into 4 + 2p + q = 0

Given a = {1, 2, x2-5x + 9}, B = {3, X2 + ax + a}, if a = {1, 2, 3}, 2 ∈ B, find the value of real number a

∵ a = {1, 2, 3}, 2 ∈ B, ∵ x2-5x + 9 = 3, X2 + ax + a = 2, the solution is: a = - 23 or a = - 74, so the value of real number a is: a = - 23 or a = - 74

When x = 1, the value of the algebraic formula PX3 + QX + R is 9; when x = - 1, find the value of the algebraic formula PX3 + QX + R + 9

Bring x = 1 into PX ^ 3 + QX + r = 9
p+q+r=9
p+q=9-r
Then x = - 1 is brought into the required formula PX ^ 3 + QX + R + 9 = - P-Q + R + 9 = - (P + Q) + R + 9 = - (9-r) + R + 9
=-9+r+r+9=2r

The function f (x) = 4x ^ 2-2 (P-2) x-2p ^ 2-P + 1 has at least real number C in the interval [- 1,1]. What is the negation of F (c) > 0?

For a quadratic function f (x) = 4x & # 178; - 2 (P-2) x-2p & # 178; - P + 1 has an opening upward, so f (- 1) = - 2p & # 178; + P + 1 f (1) = - 2p & # 178; - 3P + 9 has an axis of symmetry of (P-2) / 4 when (P-2) / 4 ≥

It is known that for any angle θ, y = - Sin ^ 2 θ - 2msin θ - 2m - 1 is always less than 0

Let t = sin θ 0 ≤ t ≤ 1
f(t)=-t^2-2mt-2m-1
f(0)

Mathematics of senior one: if sin ^ 2 θ + MSIN θ + 2m-4

You can use the variable separation method to do, (sin θ + 2) M0 is constant, both sides divide sin θ + 2 to simplify
M

Given the function f (x) = cos square x-2cosxsinx-sin square x, find the maximum and minimum of F (x)

According to the meaning of the title:
f(x)=(cosx)^2-(sinx)^2-2sinxcosx
=cos2x-sin2x=√2cos(2x+π/4)
So the maximum value of F (x) is √ 2 and the minimum value is - √ 2

Let a ≥ 0, if the maximum value of y = cos2x asinx + B is 0 and the minimum value is - 4, try to find the values of a and B, and find the maximum and minimum values of Y and the corresponding x values

The deformation of the original function is y = - (SiNx + A2) 2 + A24 + B + 1, ∵ - 1 ≤ SiNx ≤ 1, a ≥ 0. If 0 ≤ a ≤ 2, when SiNx = - A2, ymax = 1 + B + A24 = 0. ① when SiNx = 1, ymax = - (1 + A2) 2 + 1 + B + A24 = - A + B = - 4 ② If a is greater than 2, A2 ∈ (1, + ∞) ymax = - (1 − A2) 2 + 1 + B + A24 = a + B = 0 & nbsp; & nbsp; and (3) Ymin = - (1 + A2) 2 + 1 + B + A24 = − a + B = − 4. When a = 2 is obtained from (3) 4, A2 = 1 (rounding off), there is only one set of solutions a = 2, B = - 2

It is known that the function f (x) = root sign (a square + b square / 4) cos (2x + π / 6) + A, (a > 0, b > 0), and the maximum value of F (x) is 1 + A, and the minimum value is - 1 / 2
(1) Find the value of a and B
(2) Find the monotone increasing interval of function f (x)

Given the function f (x) = √ [(A & # 178; + B & # 178;) / 4] cos (2x + π / 6) + A, (a > 0, b > 0), and the maximum value of F (x) is 1 + A, and the minimum value is - 1 / 2
(1) Find the value of a and B
(2) Find the monotone increasing interval of function f (x)
(1) . MAXF (x) = f (- π / 12) = √ [(A & # 178; + B & # 178;) / 4] + a = 1 + A, i.e., √ (A & # 178; + B & # 178;) = 2, a & # 178; + B & # 178; = 4. (1)
Minf (x) = f (5 π / 12 + π / 6) = - √ [(A & # 178; + B & # 178;) / 4] + a = - 1 / 2, that is √ (A & # 178; + B & # 178;) - 2A = 1, √ (A & # 178; + B & # 178;) = 2A + 1,
a²+b²=4a²+4a+1,3a²+4a-b²+1=0.(2)
Substituting B & # 178; = 4-A & # 178; into (2), we get 3A & # 178; + 4A - (4-A & # 178;) + 1 = 4A & # 178; + 4a-3 = (2a-1) (2a + 3) = 0, so a & # 8321; = 1 / 2;
If a = - 3 / 2, then the maximum value is 1 + a = 1-3 / 2 = - 1 / 2 = the minimum value
That is, a = 1 / 2, B = ± √ (4-1 / 4) = ± (√ 15) / 2
(2) From - π / 2 + 2K π ≤ 2x + π / 6 ≤ 2K π, we get - 2 π / 3 + 2K π ≤ 2x ≤ 2K π - π / 6,
The simple increasing interval is: - π / 3 + K π ≤ x ≤ K π - π / 12