1. Derive f(x) directly.
f'(x)=(2*x*2^x-x^2*2^x*ln2)/(2^(2x))
Because there must be an inverse function, which simply increases after 0, after finding the extreme value, there will be a y corresponding to two x, so the point where the derivative is 0 is the value of a.
So, there is x=ln2/2.
This is probably the solution. You can do the math yourself.
2. Well, this question is not authentic. This problem should be the maximum value of 1/(c+ 1)+9/(9+a). Otherwise it is obvious that at least the maximum value will be greater than 9. sorry
Judging from the threshold of quadratic function value, discriminant = 0, a >；; 0, corresponding to 16-4ac=0, so ac=4, so a>0, c>0, ac=4.
Then c=4/a, and when you bring it into the expression, you get1/(4/a+1)+9/(a+9) = a/(4+a)+9/(9+a).
F(a)=a/(4+a)+9/(9+a)
This problem is equivalent to finding the maximum value of f(a). (write here. Spit out a few words Ask for guidance again. Well, I really don't want to write about the process. I only got 40 reward points. Alas, alas, my life is hard. )
Then take the derivative of a, there are
f'(a)=4/(4+a)^2-9/(9+a)^2
Then make this derivative equal to 0, and that's the answer.
Kindly, I continued to write down this process:
f'(a)=[4*(9+a)^2-9*(4+a)^2]/[(4+a)^2*(9+a)^2]
=(-5a^2+ 180)/[(4+a)^2*(9+a)^2]
Therefore, when a is from 0 to+infinity, the derivative is greater than 0, and reaches the maximum when a=6, and the derivative is 0.
So, just bring in A = 6 and C = 4/6 = 2/3.
So the answer is .6/5. Ha ha.
3. this question. Shit. It is too long. I strongly demand an increase in reward points.
All right, let's get started:
For example, the real geometric series: let an = A 1 * Q (N- 1).
This relationship includes:
a 1*q^3*a 1*q^(2n-5)= 10^2n
So there are:
(a 1*q^(n- 1))^2=( 10^n)^2
That is, (an) 2 = (10 n) 2
An>0, so an =10 n.
Actually, it's not very rigorous here. I am too lazy to think about it. Look here for yourself when you come back. This expression actually has no requirement for n= 1, so in fact an=a 1.
(n= 1)
10^n
(n & gt 1)
I won't discuss it with you. If it is different from the answer, it may be that it is not beautiful here. sorry After all, if I tell you the general meaning, it will be over. Ha ha. Meditation 100. )
Then find the general term bn = 2 (n-1) * LG (an) = n * 2 (n-1).
Can you make this thing? All right. Keep writing. sn=∑bn。
sn=2^0+2*2^ 1+3*2^2+……+n*2^(n- 1)
2 * Serial number =
2^ 1+2*2^2+……+(n- 1)*2^(n- 1)+n*2^n
Two forms of subtraction.
sn=n*2^n-(2^0+2^ 1+2^2+……+2^(n- 1))=n*2^n-2^n+ 1
So sn = n * 2 n-2 n+ 1.
Again, the question of a 1 should be considered. In doing so, I acquiesced that a 1 conforms to the general formula. So strictly speaking, we should discuss the problem of a 1. Let's take care of the rest ourselves. Ha ha.