Ask for a review paper for the first year of high school, and ask for help for a math problem for th

1. Solution: 2 a

1 (1-2) a, i.e., -1 a

Similarly -1 a

1 [1-(-1)] a, answer: -1, 2, proof: when a is a set of single elements.

a=1/(1-a)

a(1-a)=1

a-a²=1

a²-a+1=0

No solution. a cannot be a single element set.

(1) Let the proportionality coefficient be k, then.

y 2ln2 k ln(x m) ln(2m) Substituting x m(e 1), y 2 into the above equation yields.

2 2ln2 k ln(me) ln(2m) , solution k 2, substitution , get.

y 2ln2 2 ln(x m) ln(2m), the relationship between the maximum speed y of the rocket and the fuel mass x is .

y＝2ln［1＋（x／m）］．

2) It is proposed to substitute v=12km s=12000m s into the function relation:

12000=2000ln(1+m m), that is, ln(1+m m)=6 to the exponent of e: 1+m m=e 6 =>m m=e 6-1 (I calculated about 402).

That is, when the mass of fuel is about 402 times the mass of the rocket, the maximum speed of the rocket can reach 12km s

Solution: Let x1=x2=0, then there is f(0+0)+f(0-0)=2f(0)f(0), that is, 2f(0)=2[f(0)] 2, so f(0)=0 or f(0)=1;

If f(0)=0, then there is f(x1)+f(x1)=2f(x1)f(0)=0, i.e., f(x1)=0, which is contrary to the function y=f(x) is not constant zero, so f(0)=1;

Let x1=0, then there is f(x2)+f(-x2)=2f(0)f(x2), that is, f(x2)+f(-x2)=2f(x2), that is, f(-x2)=f(x2), so the function is even.

Let x1 and x2 be equal to 0, then we get f(0)+f(0)=2f(0)f(0), and we get f(0)=0 (rounded) or 1

Let x1=0 and x2=x, then we get f(x)+f(-x)=2f(0)f(x).

By f(0)=1, so f(x)+f(-x)=2f(0)f(x)=2f(x).

i.e. f(x) = f(-x).

So the function is even.

f(x)=sinx+√3/2cosx+1/2sinx=√3(√3/2sinx+1/2cosx)=√3(sinxcosπ/6+cosxsinπ/6)=√3sin(x+π/6)

The minimum positive period is 2 w=2 1=2, the center of symmetry is (-6+2k, 0), and the increasing interval is (-2, 3, +2k, 1 3 +2k).

f(x)=2 (x-1) is a subtraction function in the defined domain, so the maximum and minimum values on [-6,-2] are :

Maximum: f(x)max=f(-6).

Minimum: f(x)min=f(-2).