High school math questions with additional marks .

1.(1) by the title: f(1)=f(1)+f(1) so f(1)=02)f(1 2)=f( 2 2 * 2 2 )=2f( 2 2) =2f(x) + f(2-x) = f(2x-x 2) by the title:

Because the function f(x) is a subtraction function defined on r+.

f(2x-x^2)>f(1/2)

So 2x-x 2<1 2

x>02-x>0

So the range of x: 02. By the title: a1 + a6 = 0

i.e. 2a1+5d=0 a1=-5d2

So the general formula: an=nd-7d 2

am=md-7d/2 am+1=md-5d/2 a2m=2md-7d/2

So (md-5d 2) 2=(md-7d 2)(2md-7d 2)2m 2-11m+12=0

m=43.Let the general term of an be an=a1+(n-1)da3=a1+2d s3=(3a1+3d) s5=5a1+10d b3=1 (3a1+3d).

By the title: (a1+2d) (3a1+3d)=1 28a1+13d=21

a1==d1=1

an=nsn=(n+1)n/2

So bn=2 (n+1)n

Glad to solve the problem for you!

Please ask the students in your class with good grades, that way, you will be easier to understand, and when you encounter similar problems in the future, you will draw inferences and solve them easily!

1）∵f（xy）=f（x）+f（y）

f（1）=f(1)+f(1)

f（1）=0

2) f(x)+f(2-x)=f(2x-x) and 2=2f(2 2)=f(2 2)+f(2 2)=f((2 2))=f(1 2).

And the function f(x) is a subtraction function defined on r+.

2x-x²＜1/2

0

It is possible to find a derivative.

f(x)'=[x root number(x 2+1)]-aBecause x is at [0, positive infinity], f(x) is a monotonic function, so f(x).'Evergrande is less than 0 or constant

Let y=x root number (x 2+1).

Because x belongs to [0, positive infinity], 0<=y<1

And because a>0,0<=y<1

So when a is greater than or equal to 1, f(x).'Constant less than zero, the function monotonically decreases, if x1, x2 belongs to 0 to positive infinity and x1 is greater than x2, then f(x1)-f(x2)=......This method is similar to the answer you provided, split into two functions, f(x)=g(x)-h(x), and y1 and y2 in the answer

The first part of the red line, I feel that it should be y>0, x>=0 combined with the given answer, and a>0.

g(x) and h(x) are both increment functions.

f(x1)-f(x2)=g（x1）-g（x2）-h（x1）-h（x2）

g（x1）-g（x2）>0

h（x1）-h（x2）>0

If f(x1)-f(x2) is greater than or less than 0 (because it is monotonic), then g(x) is faster or slower than h(x).

According to the answer, g(x) is a hyperbola with y=x as the asymptote, and the combination of numbers and shapes, h(x) must increase faster than g(x) to ensure that f(x) is monotonous, so the slope of h(x) is greater than or equal to the slope of y=x, that is, a>=1 I think it is easier to find the derivative method of the first method, if you have not learned, you can use the second method.

Triangle Inner Angle Sum = 180 degrees.

a+b+c=180

c=180-a-b

tanc=tan(180-a-b)= tan(-a-b)=-tan(a+b)= -(tana+tanb)/(1-tana*tanb)= -5t/(1-4t^2)= 5t/(4t^2-1)

Because it is an acute triangle, tanb = t>0, tanc = 5t (4t 2-1) > 0, so 4t 2-1>0, t>1 2

Each angle of an acute triangle is between 0 and 90 degrees, and from the monotonically increasing nature of the tangent function in this interval, it can be known that the larger the tangent of the angle, the larger the tangent, and since the tangent of a must be greater than the tangent of b, a must be greater than b, and then compare a and c

If a is greater than or equal to c, then the tangent of a is greater than or equal to the tangent of c, then.

4t>=5t/(4t^2-1)

4>=5/(4t^2-1)

4t^2-1>=5/4

4t^2>=9/4

2t>=3/2

4t>=3

That is, when a is the maximum internal angle, the minimum value of the tangent of the maximum internal angle is 3

If c is greater than or equal to a, then.

4t<=5t/(4t^2-1)

t<=3/4

5t/(4t^2-1)=1/(4t/5-1/5t)

Under the condition of t>1 2, 4t 5 increases with the increase of t, 1 5t decreases with the increase of t, so -1 5t increases with the increase of t, so 4t 5-1 5t increases with the increase of t, and as the denominator increases, the numerator = 1 remains unchanged, and the whole fraction becomes smaller, that is, 1 (4t 5-1 5t) decreases with the increase of t, and the minimum value of 3 is taken when t = 3 4, that is, when c is the maximum angle, the minimum value of the tangent of the maximum inner angle is also 3

To sum up, the minimum tangent of the maximum inside angle of this triangle is 3

c=5t/(4t2-1)

It's all acute angles. So t>1 2

a-c=4t-5t (4t2-1)>0 t<1 2 (rounded) or t>=3 4

a>c>b amin=4*3/4=3

a-c<0 c>a 1 2< t<=3 4c=5t (4t2-1) derivative monotonically reduced.

cmin=5*3/4/(4*9/16-1)=3

1) It is easy to get p1 (1,1), p2 (2,3 2).

2) Let the coordinates of pn be (x(n),y(n)), then the equation for the straight line ln is y(n)-y(n-1)=1 (2 (n-1))*x(n)-x(n-1)) because x(n)-x(n-1)=1, so.

y(n)-y(n-1)=1/(2^(n-1))

So y(n)=y(n)-y(n-1)+y(n-1)-y(n-2)+y(2)-y(1)+y(1)

1/(2^(n-1))+1/(2^(n-2))+1/2+1

1-1/(2^n))/1/2)

2^n-1)/2^(n-1)

So the coordinates of pn are (n, (2 n-1) 2 (n-1)).

3) From (2), we can know that s opnxn=1 2*xn*yn=n*(2 n-1) 2 n

So sn=1+(n-1) 2 (n+1).

So sn-s(n-1)=(n+3) 2 (n+1), when n<3 is the value greater than 0, when n>3 the value is less than 0, when n=3 the value is 0, so sn has a maximum value, which is s3=9 8

I've coded the words for so long, and I hope to add points.

1) It is easy to find p1(1,1),p2(2,3 2).

2) From the inscription, the abscissa of pn is n. Let the ordinate of pn be an, which is known to be [an-a(n-1)] n-(n-1)]=1 2 (n-1), i.e., an-a(n-1)=1 2 (n-1), and thus obtained by accumulation.

an=an-a(n-1)+a(n-1)-a(n-2)+.a2-a1+a1

1/2^(n-1)+(1/2)^(n-2)..1/2+1

1-(1 2) n] (1-1 2)=2-(1 2) (n-1), i.e. the coordinates of pn are (n,2-(1 2) (n-1)).

3) Because s opnxn=1 2*|oxn|*an=n 2*[2-(1 2) (n-1)] so sn=(n+1) 2*[2-(1 2) n]-n 2*[2-(1 2) (n-1)].

1+n/2^n-(n+1)/2^(n+1)

1+(n-1)/2^(n+1) 。

Since s(n+1)-sn=n2 (n+2)-(n-1)2 (n+1).

2-n) 2 (n+2), so when n<2, sn increases, when n>2, sn decreases, and when n=2, s2=s3 takes the maximum value of 9 8.

1) Since p1 is on the straight line y=x and also on x=1, the coordinates of p1 are x=1, y=,1).

Let the analytical formula of the L2 line be: y=kx+b, L2 passes through P1(1,1), the slope of the L2 line K is 1 2, and the analytical formula is substituted to obtain B=1 2, Y=1 2X+1 2, P2 is on the X=2 line, and P2(2,3 2).

2) Let ln linear analytic formula: y=1 (2 (n-1))x+bn, bn=(n-1) 2, get, pn(n,n(2 (n-1))+n-1) 2), simplify: pn(n,(2 n-1) 2 (n-1)).

3) sn=s opn+1xn+1-s opnxn(n n*), opn is a right triangle, s opn=1 2*n*2n 2 n=n 2 2 n, sn=n 2 2 n+n+1 ,sn-s(n-1)=(n+3) 2 (n+1), when n<3 is the value greater than 0, when n>3 the value is less than 0, when n=3 the value is 0, so sn has a maximum value, s3=9 8

1) From the graph, a=1, so w=2, substituting (6, 0) into it, the initial phase is 6So f(x)=sin(2x+ 6)So g(x)=sin(2x- 6) 2

30 points is estimated that no one will do it, because the code words are too difficult.

f(x)=ax^2+bx+1，(a,b∈r；a>0) f(x)=x has , two.

(b-1)^2-4a>0

a) <2< <4, axis of symmetry x=x[0], verify x[0]>-1f'(x)=2ax+b

axis of symmetry x=-b 2a

From the title, the opening is upward, there is one greater than 2 and one less than 2, so 2 places are less than 0, that is: f(2)-2<0 (1).

Both are smaller than 4, so 4 is greater than 0

i.e.: f(4)-4>0 (2).

1)=>4a^2+2b-1<0

2)=>16a^2+4b-3>0

12a^2+6b-3<16a^2+4b-32b<4a^2

and a ≠0 and divide by a 2

b/(2a)<1

b/(2a)>-1

x[0]>-1

II)|α2, |2. Find the range of b.

(b-1)^2-4a>0

-a|=√b-1)^2-4a)/a=2(b-1)^2=4a^2+4a

2a+1)^2=(b-1)^2-1

2a+1)=√b-1)^2-1)

0 hours 2,0).

f(-2)-(2))f(0)<0

4a-2(b-1)+1<0

4a-2b+3<0

2√((b-1)^2-1)-2b+1<0

2√((b-1)^2-1)<2b-1

1<4((b-1)^2-1)<(2b-1)^2(b-1)^2>5/4 and 0<4b+1(b<-√5/2+1 or b>√5/2+1) and (-1/4√5/2+1

0:0,2).

f(2)-2)f(0)<0

4a+2(b-1)+1<0

4a+2b-1<0

2√((b-1)^2-1)<2b+3

1<4((b-1)^2-1)<(2b+3)^25/4<(b-1)^2 and 0<20b+9(b<-√5/2+1 or b>√5/2+1) and (-9/20>b)

b>√5/2+1

It seems that there is still something wrong......