Please help you solve a math problem, a freshman in high school

cos [0,1], so cos2 -4 is negative, and the upper and lower limits of 4m-2mcos are 2m, 4m, if m is 0 or negative, it is obvious that f(cos2 -4) + f(4m-2mcos) 0 f(0), so m must be positive. Then f(cos2 -4) + f(4m-2mcos) = f(4m-2mcos)-f(4-cos2), which requires (4m-2mcos)-4-cos2) 0, that is, m (4-cos2) (4-2cos) = (2+cos) 2, because [0, 2] is constant, so take cos = 1 substitution, m 3 2.

Since f(x) is an odd function, so f(0)=0, f(cos -4)=-f(4-cos) so f(4m-2mcos)>f(4-cos) and the function f(x) is an increasing function on r, so 4m-2mcos >4-cos, and the solution is cos -2mcos +4m-4>0, again [0, 2], so (-2m) 2-4*(4m-4)>=0, so that (m-2) 2>=0, and cos +2>2m

So the range of m is >

f(0)=0, f(cos -4)>-f(4m-2mcos) odd functions. f(cos -4) > f(2mcos -4m) increment function. cos²α-4>2mcosα-4mcosα-2<0

cosα+2>2m

It should be like this, note that both sides of the inequality divide by the same negative number, and the inequality sign is reversed.

From the odd function, we know that f(0)=0

cos²α-4<0

As long as 4m-2mcos >4-cos are constant.

1, g is taken, h(t)=-1 2gt 2+v0t+h0, where v0=, h0=18. This question is entirely about physics.

2. It must be at the highest point. That is, when v=0, the formula is obtained.

At t=, there is a maximum height of about 29m

(1) This is obtained according to the law of upward throwing motion in physics, and at this stage it can be directly used to be the displacement caused by gravitational acceleration (2) to transform the practical problem into a mathematical problem, that is, to find what is the maximum value of the independent variable of the quadratic function, what is the maximum value, and write the formula as the vertex formula;

Solution: (1)h(t)=-2t2+4 3 t+19=-2(t- 3 )2+25, when t= 3, h(t) achieves the maximum value of 25 3 seconds after the fireworks burst out is the best moment for it to burst, when the height from the ground is 25 meters

Solution: (1) Let the height of the fireworks from the ground at t seconds be h m, then it can be known from the principle of object motion: h(t)=

Note: This is derived from knowledge of physics. If you look at the diagram, you will understand, but if you don't understand the diagram, I can't (2) make an image of the function h(t)= (as shown in the figure on the right).

Obviously, the vertex of the function image is the highest point of the fireworks, and the abscissa of the vertex is the best moment for the fireworks to burst, and the ordinate is the height from the ground at this time.

Thanks to the knowledge of quadratic functions, for h(t)=, we have:

(1) The displacement of the uniform acceleration linear motion formula x=1 2*a*t 2+v0*t, because the direction of acceleration is opposite to the direction of the initial velocity, a is negative and v0 is positive, there is 1 2*a= time; Actually, a=-g), v0 is the initial velocity given, so there is h(t)=; As for adding 18, do you think there is a saying that the place where it starts to rush out is 18m high?

2) Set this time to t, when the initial speed decreases to 0, the fireworks will start to slide, so reaching the maximum height at this time is the best moment to burst. Since the initial velocity is at the beginning, and the gravitational acceleration is, you can calculate how many seconds it decreases to 0; The formula is.

**If you don't understand, you can ask.

(2) The process of finding the value range of the function Solution: y=(1 2) (2x)-(1 2) x+1 Let t=(1 2) x, -3<=x<=2, so 1 4<=t<=8 then y=t 2-t+1=(t-1 2).

If x belongs to (1,3).

You can think of it from the point of view of a quadratic function image.

Since (x-1)(x-3)=y 0, the range of x should be below the x-axis.

The first two are conditional.

t=2π/w=π

w=2f(x)=sin(2x+φ)

The axis of symmetry of sinx is where the maximum value is taken.

i.e. sin(2x+) = 1

2x+φ=kπ+π/2

x=-π/6

So =k +5 6

From the range, take k=-1

So f(x) = sin(x- 6).

The center of symmetry of sinx is the intersection point with the x-axis.

i.e. sin(x-6)=0

x-π/6=kπ

x=kπ+π/6

You might as well take k=0

So a symmetry center of f(x) is (6,0).

1.If one of the axes of symmetry of the function y=sin(2x+ ) 2) is the straight line x=- 4, then =0

2.Evaluation range.

1)y=2cosx+1/2cosx-1

2)y=|sinx|+sin|x|

x+ = 2+k, substitute x=- 4, and solve the shift.

cosx+1 2cosx-1=5 2cosx-1, because cosx is greater than -1 and less than 1, so the range of y is [-7 2,3 2].

The second question, because |sinx|The range is [0,1], sin|x|The range is [-1,1], so the range of y is [0,2], you can see it by drawing a diagram.

The essence of physics is related to the knowledge of physics, allowing students to explore the functions of time and height.

Relation: f(x) = Age.

2) The process of finding the range of the function pants noise.

Solution: Mozhen y=(1 2) (2x)-(1 2) x+1 makes t=(1 2) x,3<=x<=2, so 1 blind pure coarse 4<=t<=8 then y=t 2-t+1=(t-1 2).

1, g is taken, h(t)=-1 2gt 2+v0t+h0, where v0=, h0=18. This question is entirely about physics.

2. It must be the most balanced time to shoot the high beam point. That is, when v=0, it is necessary to dismantle the envy, and the formula is obtained.

At t=, there is a maximum height of about 29m