3 math problems in the first year of high school, the master helps

Only do the first. Questions 1 and 3. The second question is to determine the values of a and b by the derivative method, and then substitute it into f(x)= ax +8x+b, and then use the derivative method to find the value range.

Question 1: 1) Let x=y=0 then f(0)=2, and we can know that this function is a one-time function f(x)=x+2, so f'(x)>0 so is an increment function.

2) f(a -2a-2) = a -2a<3 solution -1 The third question: 1) Because the function f(x)=ax +bx+a satisfies the condition f(7 4+x)=f(7 4-x), the function is symmetrical with respect to 7 4, (a theorem is quoted here, if f(x) satisfies f(a+mx)=f(b-mx), then f(x) is symmetrical with respect to x=(a+b) 2) and the equation f(x)=7x+a has two equal real roots, so =0 pushes b=7 , Then by the axis of symmetry x=-7 2a=7 4 pushes out a=-2

So the analytic formula of the function is f(x)=-2x +7x-2

2) Ask if there is m,n, then you only need to find a m,n to prove the problem. Suppose 0< m < 7 4

Question 3: Substitute x=7 4+x x=7 4-x into f(x) to find an analytic formula.

Question 3. a

The area of the triangle is.

Root number pat shirt god 3

First question. Because mn=0

Collapsed by Li Yi. bsinc

2csinbcosa=0

According to. The positive attack loses string theorem.

Get. sinbsinc+2sincsinbcosa=0. 1+2cosa=0

cosa=-1/2

So a=120

Question 2. Sine theorem.

sina/a=sinc/c

So c=30 then b

S1 2Sinbac Root No. 3

x≥-2/3

Define the domain: x (-2 3, positive infinity).

x≠0, and 1+1 x≠0 and 1+1 [(1+1 x)≠0, i.e., x≠0 and x≠-1 and x≠-1 2

Define the domain: x (negative infinity, -1) (1, -1 2) (1 2, positive infinity).

And|x-1|-1≠0

i.e. x≠-2 and x≠0 and x≠2

Define the domain: x (negative infinity, -2) (2,0) (0,2) (2, positive infinity).

1、x>=0

2. First x≠0, then 1+1 x≠0, you can get, x≠-1 and x≠0, and 1+1 1+1 x≠0, you can get: x≠-1 2

Therefore, the domain is defined as x r and x≠-1 2, x≠-1, x≠03, x+2≠0, |x-1|-1≠0, so its domain is x r and x≠-2, x≠2, x≠0

Solution: 1, (a, 9) substitute y=3 x, i.e., 9=3 a, get, a=2tan2 6=tan 3=root number 3

2. B, the necessity is insufficient. Explanation: Necessity, "y=f(x) is an odd function" with respect to origin symmetry, plus absolute value, with respect to y-axis symmetry.

However, y=|f(x)|The image of the y-axis may not be the origin, so "y=f(x)' is not an odd function. Not sufficient.

3. A quarter of the cycle is 6, so the cycle is 2 3, so w=3 hopes to be helpful to the landlord

1) 3^a=9

a = 2 tan 6 a pie = tan 3 = root number 32)."y=|f(x)|Image of the y-axis"It is possible to deduce y=f(x) as an even or odd function.

y=f(x) is an odd function" can be introduced"y=|f(x)|Image of the y-axis"

So choose b3) 3 is the period 1 4, so t = 4 3t = 2 w = 4 3

w=3/2

1. (a,9) substituting y=3 x to get 9 = 3 a a = 2 tan(a 6) = tan ( 3) = root number 3

2, b3, according to the title 3 is the function f(x) period of 1 4 t = 2 w =4 3 w = 3 2

4. x [ 0 , 2 ] (exactly one period) f(x) = x 3 - x = x ( x - 1 ) x + 1 ) = 0 to get x = 0 or 1 or - 1 (rounding) - two solutions.

x [ 0,6 ] three cycles) has 6 solutions, and only d does not meet the requirements.

7. According to the title, sin a = 2 5

cos 2a = 1 - 2 sin ² a = - 3/5

1.By y=3x, substituting y gives a=2, so tan2 6 faction = root number 3

2."y=|f(x)|The image is symmetrical in relation to the y-axis"It is possible to deduce y=f(x) as an even or odd function, so it is not necessary to choose b.

3.The quarter period is 3, so the period is 4 3, so w=2 t=3 2

4. x [ 0 , 2 ] (exactly one period).

f(x) = x 3 - x = x ( x - 1 ) x + 1 ) = 0 The solution gives x = 0 or 1 or - 1 (rounding) - two solutions.

x [ 0,6 ] three periods).

Apparently compliant, d is monotonically decreasing from 0 to positive infinity.

Then the absolute values of sina and cosa are two-fifths of the root number 5 and one-fifth root number 5, respectively, so cos2a = cos square a-sin square a = negative three-fifths.

10.Apparently the union is 1, and you can hi me if you don't understand it yet.

Just solved 10If the complement is empty, then the complement is also empty, so it is i12（a+b-c）^2=a^2+b^2+c^2+2ab-2bc-2ac=3-2ab-2bc=3-2c(a+b)……1

Because (a-c)(b-c) is less than or equal to 0, and after opening it, it is ab-bc+c 2-ac=1-c(a+b) less than or equal to 0, so c(a+b) greater than or equal to 1 is less than or equal to 1, that is, the maximum is 1

14 questions did not understand asinasinbbcos square A? 15.

A is greater than or equal to 4

First find that p:x belongs to (negative infinity, 1-a 5) u(1+a 5, positive infinity) and then q:x belongs to (negative infinity, 1 2) u(1, positive infinity) from the title, it must be p contained in q, i.e., p is smaller than q, and thus it is solved.

There may be something wrong with proposition p, you take a look at the original question.

I didn't finish writing the thing behind that p......25x 2-10x+1-a 2(a>0) and then?