Senior 1 Mathematics. How to do it Can you explain it Thank you

f(1)=f(1+0)=f(1)+f(0), so f(0)=0f(-1+1)=f(1)+f(-1)=0, so f(1)=-f(-1), so it is an odd function.

f(1)=-2 , f(x+1)=f(x)+f(1)=f(x)-2, so f(x+1) is always less than f(x) and is a decreasing function.

Because it is a decreasing function, bring in the endpoint f(-2)=f(-1-1)=f(-1)+f(-1)=4

f(4)=f(2)+f(2)=-f(-2)-f(-2)=-8 value range [-8,2].

1) Because f(x)=f(x+0)=f(x)+f(0), so f(0)=0, and because f(0)=f(x+(-x))=f(x)+f(-x)=0

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

So f(x) is an odd function.

2) If x1>x2

Then f(x1)-f(x2)=f(x1)+f(-x2)=f(x1-x2) Since x1-x2>0, f(x1-x2) < 0, i.e., f(x1)-f(x2)<0.

So it's a subtraction function.

3) Since it is a subtraction function, the maximum value is f(-2) = f(-1) + f(-1) = 4

The minimum value is f(4)=-f(-4)=-(f(-2)+f(-2))=-8

tanatanc=2+√3

tana+tanc=3+√3

Equations can be solved.

But the easier way is to:

tana+tanc=3+ 3=1+2+ 3=1+tanatanc, so tanatanc-tana-tanc+1=(tana-1)(tanc-1)=0, tana=1 or tanc=1, when tana=1, then a=45°, and b=60°, so c=75°;

When tanc = 1, then c = 45°, and b = 60°, so a = 75°.

Now I understand.

Let f(x)=ax 2+bx+c

So the solution set for ax 2+bx+c>-2x is (1,3) so a<0

and c a=3,-(b+2) a=3

c=3a，b=-3a-2，a<0

f（x）+6a=ax^2+bx+c+6a=ax^2-(3a+2)x+9a=0

Discriminant δ = (3a+2) 2-36a 2=0, so a=2 3 or -2 9

Because a<0, a=-2 9

f（x）=-2/9x^2-4/3x-2/3

The above question refers to the combination of two functional images. Analysis reveals that the newly combined images are in the first interval before below the x-axis and then above the x-axis.

So choose from it. In the analysis of the negative half axis engraving selection A

A should be chosen

Because when x is slightly greater than 0, fx is less than 0 and gx is greater than 0, multiplied by less than zero. When x is very much greater than 0, both fx and gx are greater than zero, then the multiplication is also greater than zero.

The part where x is less than 0 can also be discussed as described above

Therefore, I choose A

d Obviously, f(x)=0 when f(x)=0, so f(x)=f(x)*g(x) has three intersections with the x-axis.

Odd function multiplied by even function = odd function.

So choose D

I choose AOdd functions x odd functions are even functions.

In the second figure, x cannot be equal to 0, so the next big formula x cannot be 0, so choose b

d is an odd function. Odd multiplication is an odd function.

Substituting x=4 and y=3 into it, we get m=1

f（-x）=-x+4/x=-f（x）

So f(x) is an odd function.

Let x1, x2 (0, 2), and x2 >x1, then x=x2-x1 0

y=f（x2）-f（x1）

x2-4/x2-x1+4/x1

x2-x1-(4x1-4x2/x1x2)=△x+4△x/x1x2

x1 and x2 belong to (0, 2).

x1x2＞0

So y 0

So f(x) is an increasing function on (0, 2).

, m = odd function.

is the increment function, g=x is the increment function, f(x)=h+g is the increment function.