It is known that the image of the function f (x) = loga (1-mx / x-1), (a > 0 and a ≠ 1) is symmetrical about the origin. (1 find the value of M (2) Judge the monotonicity of F (x) on (1, + infinity) and prove it according to the definition. ⑶ when a is equal to 3, the inequality f (x) < 3 ^ x-t is constant for any x ∈ [2,3], and find the value range of T; Mainly the third question,

It is known that the image of the function f (x) = loga (1-mx / x-1), (a > 0 and a ≠ 1) is symmetrical about the origin. (1 find the value of M (2) Judge the monotonicity of F (x) on (1, + infinity) and prove it according to the definition. ⑶ when a is equal to 3, the inequality f (x) < 3 ^ x-t is constant for any x ∈ [2,3], and find the value range of T; Mainly the third question,

Substitute a = 3 into the formula calculated above and put it between the inequalities of the third Subquestion. According to the meaning of the question, it is always true between 9 and 27, that is, the inequalities on the left and right sides are always true, and then the value range of t can be obtained
The value range is the answer with M

Given the vector a = (SiNx, 3 / 2), B = (cosx, - 1). (1) when a is parallel to B, find the square of 2cosx - sin2x (2) Find the value range of F (x) = (a + b) B on [- π / 2,0]

(1)(3/2)cosx=-sinx、2sinx=-3cosx、4(sinx)^2=4-4(cosx)^2=9(cosx)^2、(cosx)^2=4/13
2(cosx)^2-sin2x
=2(cosx)^2-2sinxcosx
=2(cosx)^2+3(cosx)^2
=5(cosx)^2
=20/13
(2)a+b=(sinx+cosx,1/2)
f(x)=sinxcosx+(cosx)^2-1/2
=(1/2)sin2x+(1/2)cos2x
=(√2/2)sin(2x+π/4)
-π/2

Given vector a = (SiNx; 3 / 2), vector b = (cosx, _1), ① when a / / B, 2cos ^ 2x - the value of sin2x ② find f (x) = (a + b) B at [- π / 2,0] emergency Want process

1 vector a = (SiNx; 3 / 2) = - 3 / 2 (- 2 / 3sinx, - 1), vector b = (cosx, - 1)
a//b
-2/3sinx=cosx
tanx=-3/2
2cos^2x—sin2x=2(cosx)^2-sin2x=cos2x+1+sin2x
=[1+(tanX)^2]/[1-(tanx)^2]+1+2tanx/[1-tanX)^2]
=[1+(-3/2)^2]/[1-(-3/2)^2]+1+2(-3/2)/[1-(-3/2)^2]=4/5
2 F (x) = (a + b) B at [- π / 2,0
a+b=([sinx-3/2cosx],1/2)
|a+b|=[(sinx-1.5cosx)^2+1/4]^0.5
k1=(1/2)/(sinx-3/2cosx)=1/(2sinx-3cosx)
|b|=[(cosx)^2+1]^0.5
k2=-1/cosX
tanN=|(k2-k1)/(1+k1k2)|=2/(sinx+3cosx)
cosN=
F(x)=(a+b)b=|a+b||b|cosN=

Given vector a = (SiNx, 3 / 2), B = (cosx, - 1) (1) if vector A / / vector B, find the value of 2cos square x-sin2x (2) (2) If (a + b) is multiplied by B = (root 2) / 4 and X ∈ (0, π / 2), find the value of X

1) Because a / / B, we can get - sinx-3 / 2 * cosx = 0 from the condition of vector collinearity, which is simplified to TaNx = - 3 / 2, so 2 (cosx) ^ 2-sin2x = [2 (cosx) ^ 2-2sinxcosx] / [(SiNx) ^ 2 + (cosx) ^ 2] (add denominator 1) = (2-2tanx) / [(TaNx) ^ 2 + 1] (the molecular denominator is the same as dividing by (co

The trigonometric function knows that the vector a is [SiNx, 3 / 2] and B is [cosx. Negative 1] [1]. When they are collinear, find 2cos ² X minus sin2x [2] Find the value range of [a vector plus B vector] multiplied by B vector on [negative half π. 0] Two questions altogether

sinx*(-1)=3/2*cosxsinx/cosx=-3/2tanx=-3/22(cosx)^2-sin2x=cos2x+1-sin2x=[1-(tanx)^2]/[1+(tanx)^2]-2(tanx)/[1+(tanx)^2]+1=[1-(tanx)^2-2(tanx)]/[1+(tanx)^2]+1=[1-9/4+3]/[1+9/4]+1=(3/4)/(13/4)+1=3/13+1=16...

Known vector a = (1 + sin2x, SiNx cosx), B = (1, SiNx + cosx) Let the function f (x) = a · B (1) If x ∈ [π / 4, π / 2], find the maximum and minimum values of F (x) (2) If f（ θ)= 8 / 5, find SiN4 θ Value of

F (x) = vector A. vector B. = (1 + sin2x) * 1 + (SiNx cosx) * (SiNx + cosx). = 1 + sin2x - (COS ^ 2x sin ^ 2x). = 1 + sin2x cos2x. = 1 + √ 2Sin (2x - π / 4). ‡ f (x) = √ 2Sin (2x - π / 4) + 1. (1) x ∈ [π / 4, π / 2], ‡ (2x - π / 4) ∈ [π / 4,3 π / 4] make 2x - π / 4 = π / 2, that is