Mathematics in the first year of high school requires a process. Thank you

Solution: (1) Let x=-1, y=1, then f(0)-f(1)=-1(-1+2+1) is known

f（0）=-2

2) Let y=0, then f(x)-f(0)=x(x+1) and f(0)=-2

f（x）=x2+x-2

3) The inequality f(x)+3 2x+a, that is, x2+x-2+3 2x+a, that is, x2-x+1 a, because when 0 x 1 2, 3 4 x2-x+1 1, and x2-x+1=(x-1 2 )2+3 4 a is constant, so a=,g(x)=x2+x-2-ax=x2+(1-a)x-2 axis of symmetry x=a-1 2, and g(x) is a monotonic function on [-2,2], so there is a-1 2 -2, or a-1 2 2,b=,crb=

a∩crb=．

1) According to f(1+0)-f(0)=1*(1+0+1)=2f(1)-f(0)=2

f(1)=0, then f(0)=-2

According to f(x+0)-f(0)=x(x+0+1), then f(x)=f(0)+x+x=x +x-20, then a=[1, positive infinity).

g(x)=f(x)-ax=x +x-2-ax=x -(a-1)x-2 is monotonic on [-2,2].

i.e. (a-1) 2 less than or equal to -2 or (a-1) 2 greater than or equal to 2, i.e. a, "-3 or a"5

then b = (negative infinity, -3] u[5, + positive infinity).

crb=（-3,5）

then AI(CRB) = [1,5).

Question 10: Answer D

Use the exponentiation formula:

cos²(x/2)

1+cosx)/2

Then 1+cosx=3 2

The solution is cosx=1 2

[1-log6(3)]=log6(2)

1-log6(3)] 2=log6(2)*log6(2)numerator=log6(2)[log6(2)+log6(18)]=log6(2)*log6(36)=2log6(2)=log6(2) = denominator.

Numerator denominator = 1

i.e. original = 12)2x=log2(12).

x=(1/2)log2(12)=log2(√12)=log2(2√3)