Given the function f (x) = LG [(a ^ 2-1) x ^ 2 + (a + 1) + 1], if the definition field of F (x) is r, then the value range of real number a is? Given the function f (x) = LG [(a ^ 2-1) x ^ 2 + (a + 1) + 1], if the definition field of F (x) is r, then the value range of real number a is? E, that one has an X

(a^2-1)x^2+(a+1)+1>0
If you don't Miss X after (a + 1)
If a ^ 2-1 > 0, 0-4 (a ^ 2-1) (a + 2) > 0
(a ^ 2-1) (a + 2) 0, so a + 21 or A0
-5 / 31 or a

Let f (x) = cos (2x + Pai / 3) + sin ^ 2x 1. Find the maximum value and minimum positive period of the function 2. Let a, B, C be the three inner angles of the triangle ABC. If CoSb = 1 / 3, f (C / 2) = - 1 / 4, and C is an acute angle, find Sina

After expansion: F (x) = - (√ 3 / 2) sin2x + (1 / 2) f (x) max = √ 3 / 2-1 / 2-1 / 2T = π 2. ∵ f (C / 2) = - 1 / 4

Let f (x) = 2cos x sin (x + Pai / 6) - Sin ^ 2x + cos ^ 2x Let f (x) = 2cos x sin (x + Pai / 6) - Sin ^ 2x + cos ^ 2x (1) Find the monotone increasing interval of function f (x); (2) When x belongs to [- Pai / 12, Pai / 6], the maximum and minimum values of function f (x) and the corresponding values of X are obtained Want to know the second step "Because x belongs to [- Pai / 12, Pai / 6], so (2x + Pai / 3) belongs to [Pai / 6,2pai / 3]," how to get there "So when 2x + Pai / 3 = Pai / 2, that is, x = Pai / 12," why is de equals Pai / 2? Then how to get to "function f (x) gets the maximum value of 1 / 2+ Radical 3 In addition, when 2x + Pai / 3 = Pai / 6, that is, x = - Pai / 12, the function f (x) obtains the minimum value of 1 / 2 + the root sign of 3 / 2

F (x) can be reduced to = 1 / 2 + radical 3 * sin (2x + Pai / 2)
There is a maximum when (2x + Pai / 2) = π / 2
1 / 2 + radical 3
=1 / 2 + radical 3 * sin (2x + Pai / 3)
Then (2x + Pai / 3) = Pai / 2
Then sin Pai / 2 = 1
So = 1 / 2 + radical 3 * 1
minimum value
When 2x + Pai / 3 = Pai / 6, that is, x = - Pai / 12, the function f (x) gets the minimum value of 1 / 2 + the root sign of 3 / 2
1 / 2 + radical 3 * 1 / 2

Find the period of the function f (x) = cos ^ 2 (x-pai / 12) + sin ^ 2 (x + Pai / 12) - 1 Teach me the process, high school one basic forget, the formula had better be listed out! Thank you, prawns!

In this paper, f (x) = [1 + cos (2x - π / 6)] / 2 + [1 + cos (2x - π / 6)] / 2 + [1-cos (2x + π / 6)] / 2 + [1-cos (2x + π / 6)] / 2-1 = 1 / 1 = 1 / 2 [cos (2x - π / 6) - cos (2x + π / 6)] = 1 / 2 (1 / 2 [(cos2xcos π / 6 + sin2xsin π / 6) - (cos2xcos π π / 6-sin2xsin 2xsin π / 6)] = sin2xsin (π π / 6)] = sin2xsin (π π π π / 6)] = sin2xsin π π π π π / 6)] = sin2xsin π/ 6 = (1

The known function FX = cos ^ 2 (x-pai / 6) - Sin ^ 2 (x) (1) Find f (12 / PAI) (2) Find the maximum value of function FX on [0, Pai / 2]

(1)f(x)= [cos(x-π/6)]^2 - (sinx)^2f(π/12) = (cos(π/12))^2 - (sin(π/12))^2=cos(π/6)=√3/2(2)f(x) =[cos(x-π/6)]^2 - (sinx)^2= [(√3/2)cosx + (1/2)sinx]^2 - (sinx)^2= (1/4)( 3(cosx)^2 + 2√3sinxcos...

The known function f (x) = x ^ 2 g (x) = X-1 Let f (x) = x ^ 2 g (x) = X-1. If x is a real number, Let f (x)

One
X? 4 or B

Let f (x) = x 3 + X (x ∈ R) (1) The parity and monotonicity of F (x) are pointed out and the reasons are given; (2) If a, B, C ∈ R, and a + b > 0, B + C > 0, C + a > 0, try to judge the sign of F (a) + F (b) + F (c)

(1) ∵ the definition domain of F (x) = X3 + X is r, which is symmetric about the origin,
And ∵ f (- x) = (- x) 3 + (- x) = - (X3 + x) = - f (x)
ν f (x) is an odd function,
∵ f ′ (x) = 3x2 + 1 ᦝ 0, ᙽ f (x) is an increasing function on R,
(2) From (1),
If a + b > 0 leads to a > - B, then f (a) > F (- b) = - f (b), that is, f (a) + F (b) > 0
Similarly, f (b) + F (c) > 0, f (c) + F (a) > 0
Therefore, f (a) + F (b) + F (b) + F (c) + F (c) + F (a) > 0,
That is, f (a) + F (b) + F (c) > 0

The known function f (x) = x(x+4)，x≥0 X (x − 4), X ＜ 0, find the values of F (1), f (- 3), f (a + 1)

F (1) = 5, (3 points)
F (- 3) = 21, (6 points)
f(a+1)＝
a2+6a+5，a≥−1
A2 − 2A − 3, a ＜ − 1. (12 points)

Given that the function f (x) = sin (x + W) + 3 ^ (1 / 2) cos (X-W) is an even function, find the value of W Given that the function f (x) = sin (x + W) + root 3cos (X-W) is an even function, find the value of W

F (x) = f (- x) sin (x + W) + sqrt (3) * cos (X-W) = sin (w-x) + sqrt (3) * cos (x + W) 2 (cos30 degree cos (x + W) - sin30 degree sin (x + W)) = 2 (cos30 degree cos (X-W) + sin30degree sin (X-W)) cos (x + W + 30 degree sin (X-W)) cos (x + W + 30 degree) = cos (x-w-30 degree) = cos (x-w-30 degree), then x + W + 30 degree = x-w-30 degree + 2K * pi, K is an integer, w = - 30 degree + 30 degree + 2K * PI, K is an integer, w = - 30 degree + 30 degree + 30 degree + 30 degree + 30 degree + 30 degree + 30 degree + 2K * PI, KP

Let f (x) = radical 3sin (Wx + a) - cos (Wx + a) (0

1. F (x) = 3sin (Wx + a) - cos (Wx + a)
When a + π / 3 = k π, f (x) is even function, while 0 < a < π, then a + π / 3 = π
If f (x) = 2coswx, and the function y = f (x), the distance between two adjacent symmetry axes of the image is π / 2
Then w = π / (π / 2) = 2
Therefore, f (x) = 2cos2x
F (π / 8) = 2cos π / 4 = radical 2
2. After the image image with y = f (x) is shifted tt / 6 units to the right, f (x) = f (x - π / 6) = 2cos (2x - π / 3)
Then, the abscissa of each point in the image is extended to 4 times of the original, and the ordinate remains unchanged
g(x)=F(x/4)=2cos(x/2-π/3)
Monotone decreasing interval of G (x)
(2k-1)π