In the first year of high school, the math function problem should be processed, and the details wil

1) From f(x) is an odd function: f (0)=0, substituting the function formula, we get a=-2;

2) f (x+1)-f (x)=-2^(x+1)+1/2^(x+1)+2^x-1/2^x=-2^x-12^(x+1)<

0, knowing that the function f(x) is a subtraction function.

3) To make f(t 2 2t) f(2t 2 k) <0, i.e. f(t 2 2t)<-f(2t 2 k)=f(k 2t 2), and f(x) is a subtraction function, then you only need to.

t 2 2t>k 2t 2, that is, k<3t 2-2t constant can be established, then k<-1 3

1) Because the function f(x) 2 x+a 2 (x+1)+2 is an odd function with the domain r defined.

Therefore, f(0)=-1+a 2+2=0,a=-22)f(x)=-2 x-1 2 x+2, take x1,x2,(02 x1,2 (x1+x2)>2 0=1,f(x1)-f(x2)>0

Therefore the function f(x) is a subtraction function on (0,+), and the resulting function f(x) is a subtraction function on r.

3)f(t^2-2t)<-f(2t^2-k)=f(k-2t^2),t^2-2t>k-2t^2

k < (3t 2-2t) min, apparently [3t 2-2t] min = -1 3k range (-1 3).

This question can do that.

f( x+t ) 2f(x) does not directly algebra the function, but treats 2f(x) as f( 2x).

Because 2f(x)=2x=f(2x).

Then f(x) is the odd function on r, and when x 0, f(x)=x is obtained.

f(x) increases monotonically on r (this is not proven).

For any x at [ t , t+2 ], the inequality f( x+t ) 2f(x) is constant.

Equivalent. For any x at [ t , t+2 ], the inequality x+t 2x is constant.

Re-separation variables: t ( 2-1)x constant true.

x at [ t , t + 2 ].

Just t ( 2-1) (t + 2) is all you need.

Solution t 2

Solution: According to the known straight line oc: y=x straight line ab: y=-x+6024When

It's easy to divide it into three sections.

Let 4-x=0, x=4, then f(4-x) over (4,1) and the inverse function image over (1,4).

The power of 3 5 power, first cubic and then 5 power, ordinary calculators can't calculate, computer calculators calculate yes.

t=2^x,t∈(-2]

1+2 x+a*4 x=at 2+t+1 is greater than 0 on (- 2) and the opening is a 0 upwards, then the axis of symmetry is 0

1, no root, 1-4a<0, 1 42There are roots, impossible.

So 1 4

(1) x1, x2 belongs to (0, + infinity), x12 or t<-2 (rounded) a=t-root(t2-4)b=t+root(t2-4).

3)g(x)=t-1/x -4x g'(x)=1/x^2 -4 =(1-4x^2)/x^2 x>=1 g'(x)<0

g(x) at [1,+infinity) is a subtractive function g(1)=t-1 -4<0 t<5

f(x)+g(x)=2 x+x Replace x with -x.

f(-x)+g(-x)=2^(-x)-x

f(x) is an odd function and g(x) is an even function.

So -f(x)+g(x)=2 (-x)-x simultaneous f(x)+g(x)=2 x+x and -f(x)+g(x)=2 (-x)-x solve the equation.

The addition of the two formulas gives g(x)=(2 x+2 (-x)) 2 subtracts the two formulas to obtain f(x)=(2 x-2 (-x)) 2+x2)f(-x)= 2-g(-x)= 2-g(x)= 2-g(x), so f(x) is an even function.

Question 1. Let A be 1, B be q, and C be q2, then 1+q+q2=125 Solving q is what is sought.

Question 2. The area of the middle one is equal to 1 5 of the sum of the areas of the other 10 rectangles, then the area of the middle one is 1 6 of the whole area

So the middle group accounts for all 1 6, so it is 180 * 1 6 = 30.

Thank you.. Happy New Year and success in your studies.

The first one asks, the products drawn out are also proportional, and then there is a formula, calculate it to know B. The second one is 6 5x=180

1. f(q)=q+,0<=f(q+<

2(1-(q+cong meng<=1

2.1<=y=2x 2-1 "Stool book=7

2<=2x^2<=8

1 "Zao Zhenghong = x< = 2, x has 4 ranges: [-2,-1], 2,1], 1,2], 1,2].