Function f (x) = 1 / 2 (SiNx + cosx) - 1 / 2 SiNx cos x

Function f (x) = 1 / 2 (SiNx + cosx) - 1 / 2 SiNx cos x

When SiNx > = cosx, it is 2K π + π / 4=

The known function f (x) = loga1 − MX The image of X − 1 (a > 0, a ≠ 1) is symmetric about the origin (1) Find the value of M; (2) The monotonicity of function f (x) on (1, + ∞) is determined and proved according to the definition

(1) ∵ the image of the function f (x) = loga1 − mxx − 1 (a ﹥ 0, a ≠ 1) is symmetric about the origin ᙽ the function is odd and satisfies f (- x) + F (x) = 0, that is, loga1 + MX − x − 1 + loga1 − mxx − 1 = 0 holds for any X in the definition domain, that is, loga (1 + MX − x − 1 · 1 − mxx − 1) = loga1, 1 − m2x21 −

The known function f (x) = loga1 − MX The image of X − 1 (a > 0, a ≠ 1) is symmetric about the origin (1) Find the value of M; (2) The monotonicity of function f (x) on (1, + ∞) is determined and proved according to the definition

(1) ∵ the image of the function f (x) = loga1 − mxx − 1 (a ﹥ 0, a ≠ 1) is symmetric about the origin ᙽ the function is odd and satisfies f (- x) + F (x) = 0, that is, loga1 + MX − x − 1 + loga1 − mxx − 1 = 0 holds for any X in the definition domain, that is, loga (1 + MX − x − 1 · 1 − mxx − 1) = loga1, 1 − m2x21 −

Given the function f (x) = loga (x + 1) (a > 1), if the image of function y = g (x) and the image of function y = f (x) are symmetric about the origin 1. Write the analytic formula of G (x). 2. Find the solution set a of inequality 2F (x) + G (x) ≥ 0. 3. Whether there exists m belongs to positive real number, so that the solution set of inequality f (x) + 2g (x) ≥ loga (m) is exactly A. if there is, find the value of M. if not, please explain the reason Please give the detailed process of the third question

If (x0, Y0) is a point on the image of the function y = g (x), then the coordinates of its symmetrical point about the origin are (- x0, - Y0). On the image with y = f (x), there is - Y0 = loga (- x0 + 1) = loga [1 / (1-x0)] and then replace x0 with X, G (x), and Y0 has g (x) = loga [1 / (1-x)] (x1 then (x + 1) ^

The known function f (x) = loga1 − MX The image of X − 1 (a > 0, a ≠ 1) is symmetric about the origin (1) Find the value of M; (2) The monotonicity of function f (x) on (1, + ∞) is determined and proved according to the definition

(1) ∵ function f (x) = loga1 − MX
The image of X − 1 (a > 0, a ≠ 1) is symmetric about the origin
The function is an odd function, which satisfies f (- x) + F (x) = 0, that is, loga1 + MX
−x−1+loga1−mx
X − 1 = 0 holds for any X in the defined domain,
That is, loga (1 + MX)
−x−1•1−mx
x−1)=loga1，1−m2x2
1 − x2 = 1 holds for any X in the defined domain,
ν M2 = 1, M = ± 1, M = 1 does not conform to the question, so the value of M is - 1;
(2) When 0 < a < 1, f (x) is an increasing function of (1, + ∞); when a > 1, f (x) is a decreasing function of (1, + ∞)
From (1), f (x) = loga1 + X
x−1，（x＞1）
Let t = 1 + X
If x − 1, then 1 ＜ x1 ＜ X2, then T1 = 1 + x1
x1−1，t2=1+x2
x2−1，
T1-T2 = 1 + x1
x1−1-1+x2
x2−1=2(x2−x1)
(x1 − 1) (x2 − 1) > 0, T1 ＞ T2,
The function T = 1 + X
X − 1 is a decreasing function over (1, + ∞)
According to the monotonicity rule of composite functions, it is obtained that when 0 < a < 1, f (x) is an increasing function of (1, + ∞);
When a > 1, f (x) is a decreasing function of (1, + ∞)

It is known that the function f (x) = loga (1-mx) / (x-1) is an odd function (a > 0 and a ≠ 1) (1) to find the value of M (2) is judged to be in the interval (1, + ∞) Monotonicity and proof I can do the first question, mainly the second one

(1) By odd function
Then f (- x) = - f (x)
Then f (- x)
=loga[(1+mx)/-x-1]
=-f(x)
=loga[(x-1)/(1-mx)]
1-m^2x^2=1-x^2
(1-m^2)x^2=0
m=±1.
When m = 1, the real number = - 10 and is a minus function
Then loga t is on R +
When 0

Given the point a (cosx, 1 + cos2x), B (- λ + √ 3 * SiNx, cosx), X ∈ (0, π), vector a = (1,0). (1) if vector Ba is collinear with a, find the value of real number; if vector BA ⊥ a, find the range of real number λ

BA=(cosx-(√3)sinx+λ,1+cos2x-cosx)
(1) If Ba and a are collinear, then 1 + cos 2x - cosx = 0, and the binding x ∈ (0, π) is the same
The solution is: x = π / 3, or x = π / 2
(2) If BA ⊥ a, then cosx - (√ 3) SiNx + λ = 0
==> λ=(√3)sinx-cosx=2sin(x-π/6)
x∈(0,π) ==> x-π/6∈(-π/6,5π/6) ==> sin(x-π/6)∈(-1/2,1)
==> λ∈(-1,2).

Known vector a = (cosx-3, SiNx), B = (cosx, sinx-3), f (x) = a * B Given the vector a = (cosx-3, SiNx), B = (cosx, sinx-3), f (x) = a * B (1) if x ∈ [2 π, 3 π], find the monotone increasing interval of function f (x) (2) If x ∈ (π / 4, π / 2) and f (x) = - 1, find the value of tan2x

(1) (cosx-3, SiNx), B = (cosx, sinx-3), B = (cosx, sinx-3), B = (cosx, sinx-3), f (x) = a * b = C * b = cos * x-3 cosx + SiNx + sinx-3-3 SiNx = 1-3 √ 2 / 2cosx = 1-3 √ 2Sin (x + π / 4) = 1-3 √ 2Sin (x + π / 4)

Given the vector M = (- 1, SiNx), n = (- 2, cosx), the function f (x) = 2Mn If the angles a and B of △ ABC are a, B, f (A / 2) = 24 / 5, f (B / 2 + n / 4) = 64 / 13, a + B = 11, then find the value of A

F (x) = 2Mn = 4 + 2sinxcosx = sin2x + 4, f (A / 2) = 24 / 5, Sina = 4 / 5,
If f (B / 2 + n / 4) = 64 / 13, CoSb = 12 / 13, SINB = 5 / 13
From the sine theorem a / Sina = B / SINB = C / sinc, 5A / 4 = 13b / 5 and a + B = 11 are obtained
A = 52 / 7

In two hours, we know vector a = (m, 1), vector b = (SiNx, cosx), function f (x) = vector a times vector B, and satisfy the following conditions [urgent help] answer in two hours: given vector a = (m, 1), vector b = (SiNx, cosx), function f (x) = vector a times vector B, and satisfy f (Wu / 2) = 2, problem: (1) find the analytic formula of function y = f (x) and find its minimum positive period. (2) in the triangle ABC, if f (Wu / 12) = root sign 2sina, and AC = root 2, BC = root 3, find the size of angle B

f(x)=(m,1)*(sinx,cosx)=msinx+cosx
f(π/2)=m=2
So f (x) = 2sinx + cosx
Minimum positive period T = 2 π
(2)
f(π/12)=2sin15°+cos15°=√2sinA
AC/BC=sinB/sinA=√6/3
sinB=√6/3*(2sin15°+cos15°)/√2