High School Math Solving Detailed Process Thank you 20

Two +1 3; 8、-2/3； 9、an=2^(n-2)（a1=1）；

5 sub-questions: a(n+1)=a1*q n,sn=(a1-a1*q n) (1-q)=a1*(1-q n) (1-q),s2n=a1*(1-q 2n) (1-q);

then tn=(17sn-s2n) a(n+1)=[17*a1*(1-q n)-a1*(1-q 2n)] [a1*q n*(1-q)].

17-17q^n+q^2n)/[q^n(1-q)]=[(17/q^n)-17+q^n]/(1-q)=[(17/x)-17+x]/(1-q)；…x=q^n；

Because 1-q<0, it is clear that when (17 x)-17+x takes the minimum value, tn is maximum;

When (17 x)=x, i.e., x= (17), the value of the above equation is the smallest, i.e., q 2n=17, and substituting q= 2 yields: 2 n=17, n 4;

i.e., the fourth term of tn is the largest;

15 sub-questions: (1) Let the first term of the series be a1 and the common ratio is q, then sn=a1*(1-q n) (1-q)=[a1 (1-q)]-a1 (1-q)]*q n;

Comparing the expressions sn and y, we can see that -a1 (q-1)=-1= r; b=q；

2)b=2=q, then y=2 x-1, sn=2 n-1=a1*(1-2 n) (1-2)=a1*(2 n-1);

a1=1；an=q^(n-1)=2^(n-1)；bn=(n+1)/[4*2^(n-1)]=(n+1)/2^(n+1)；

tn=σbn=σ[(n+1)/2^(n+1)]；2*tn=σ[(n+1)/2^n]；

2tn-tn=[2/2^1-(n+1)/2^(n+1)]+1/2^n]……

tn=1-[(n+1)/2^(n-1)]+1/2)-(1/2^n)]；

tn=(3/2)-(2n+3)/2^n；

This is a transcendent equation that cannot be solved exactly. You can only find an approximate solution.

x+1>0, and x+1 is not equal to 1

Then x>-1 and x is not equal to 0

1) If -10

In this case, log(x+1)2 is a decreasing function and greater than 0, then x+1 log(x+1)2-1 is an increasing function, then the solution is unique.

It is not difficult to see that x=3 meets the requirements.

So the solution is x=3

x, and log2 (x+1) are both increments.

Therefore, f(x)=x+log2 (x+1)-1 is also an increasing function, and at most there is only one f(0)=-1<0

f(1)=1+1-1=1>0

Therefore there is a unique solution in the (0,1) interval.

It can be solved numerically: x=

20. (1) a(n+1)-an=2/[a(n+1)+an-1] =a(n+1)-an]*[a(n+1)+an-1]=2

bn=(an-1/2)^2=an^2-an+1/4;

b(n+1)=(a(n+1)-1/2)^2=a(n+1)^2-a(n+1)+1/4

b(n+1)-bn=[a(n+1)^2-an^2]-[a(n+1)-an]

a(n+1)-an]*[a(n+1)+an-1]

A series is a series of equal differences with a tolerance of 2.

2) b1=(a1-1/2)^2=(1-1/2)^2=1/4

bn=b1+(n-1)d=1/4+(n-1)*2=2n-2+1/4=(8n-7)/4

an≥1，∴an-1/2≥0

an-1/2=√bn =>an=1/2+√bn=[1+√(8n-7)]/2

3) am=k=[1+√(8m-7)]/2 =>2k-1=√(8m-7)

2k-1)^2=8m-7 =>m=[(2k-1)^2+7]/8

That is, as long as the value of m is m=[(2k-1) 2+7] 8, am=k

21.(1) f(x)=alnx+x 2-12x+11 => bridge circle f'(x)=a/x+2x-12

x=4 is the extreme point, then f'(4)=0=a/4+8-12=a/4-4 =>a=16

2) f(x)=16lnx+x^2-12x+11 f'(x)=16 x+2x-12=2(x 2-6x+8) Yubump x=2(x-2)(x-4) x

Define the field as x>0 and let f'(x) 0, which can be solved to be x4 or x2

That is, the monotonic increase interval of the function f(x) is (0,2][4,+).

Order f'(x) 0, which can be solved as 2 x 4, i.e., the monotonic reduction interval of the function f(x) is [2,4].

3) From the monotonic interval, it can be seen that the function f(x) increases monotonically on (0,2), and obtains a maximum value at x=2 to suppress the talk f(2)=16ln2-9;

Monotonically decreasing on [2,4], the minimum f(4)=32ln2-21 is obtained at x=4

Monotonically increasing on [4,+.

In order for the line y=b to have 3 intersections with y=f(x), the line needs to fall between 16ln2-9 and 32ln2-21.

That is, the value range of b is (16LN2-9,32LN2-21).

1.(1)bn+1-bn=(an+1-1/2)^2-(an-1/2)^2=(an+1+an-1)(an+1-an)=2

So bn is a series of equal differences.

2)b1=1 4 So bn=2n-7 4=(an-1 2) 2 so an=(2n-7 4) 1 2+1 2 (sorry, the third question is not very good, and the writing is a bit messy, I hope it helps you) 2(1)f'(x)=a/x+2x-12

Because x=4 is the extreme point of the function, f'(4)=0 The solution is a=16, so f(x)=16lnx+x 2-12x+11(2)f'(x)=16 x+2x-12 (x>0)let f'(x)=0 gives x=2 or 4

When x belongs to (0,2) and (4,positive infinity), f'(x) >0, f(x) single increment when x belongs to (2,4), f'(x) <0, f(x) single minus (3) f(2) = 16ln2-9f(4)=16ln4-21, so in order to satisfy the condition b, it belongs to (16ln4-21, 16ln2-9).

1(3) In addition, am=k has been solved to m=k*(k-1) 2+1 k(k-1) must be even, so the solved m is a positive integer, and there is an over reply.

（1-lgx）lgx=lg（1/100）=-2

lgx = 2 or -1 (rounded).

x=100 is not calculated with a pen, so there should be nothing wrong.

Solution: Derivative of f(x)=1 x*lnx, f'(x)=-lnx+1)/(xlnx)^2

Order f'(x)=0 gives x=1 e

At (0,1 e) f(x) increases monotonically and decreases monotonically at (1 e,1), so that at 1 e the extreme (maximum) value is obtained. f(1/e)=e

Look at the condition again is 2 1 x > x a

Take the logarithm ln on both sides to obtain: ln2 1 x>lnx a, that is: ln2*1 x>a*lnx is less than zero on (0,1).

Divide both sides by the lnx variant at the same time to get 1 x*lnxeln2

When the extreme point is the minimum:

f'(x)=1/x+a/x^2， f''(x)=-1/x^2-2a/x^3

f'(x)=0, 1 x+a x 2=0, x=-a

f(-a)=ln(-a)-a/(-a)=ln(-a)+1

If ln(-a)+1=2, then a=-e, where x=e is in the interval [1,e], f''(e) = 1 e 2>0, i.e., there is a minima.

When the boundary value x=1 is the minimum value of the function:

f(1)=ln1-a=2, then a=-2

In this case, the extreme point f(-a)=f(2)=ln2+2 2=ln2+1<2, that is, it is smaller than the boundary value, so f(1) is not the minimum value of the function.

Hence a=-e

f(x)=(a x -1) (a x +1) because f(-x)=(a -x -1) (a -x +1)=(1 a x -1) (1 a x +1)=(1- a x) (1+a x).

a x -1) (a x+1) = -f(x) so f(x) is an odd function.

f'(x)=[lna * a^x (a^x+1) -a^x-1) lna * a^x]/(a^x+1)^2

a x lna (a x+1-a x+1) (a x+1) 2=2*a x lna (a x+1) 2 It can be seen that when a > 1, f'(x)>0, f(x) is monotonically increasing;

a<1, f'(x)<0, f(x) is monotonically decreasing;

a<1, f'(x)=0, f(x) is a constant function.