Urgent questions

2009-08-26 2:45 am
For Q.1, Decide whether the given statements are true or false. If a statement is true, prove it. If it is false, give a counter-example.

1a) If 0<p<q, then Integral [upper limit: p, lower limit: 0] f(x)dx < integral [upper limit: q, lower limit: 0] f(x)dx.
b) If 0<p<q, then log p< log q.
c) If n^3 - n is divisible by 4, then n must be divisible by 4.
d) If n is a positive integer, then n^3 - n is divisible by 3.

2. Given that f(x) = x/(x + 1), find f^2(x) i.e. f(f(x)). Find also f^3(x) and suggest a possible form for f^n(x). Prove that your result is correct by M.I.
更新1:

&int;[0,q] f(x) dx = ∫[0,p] f(x) dx + ∫[p,q] f(x) dx Why do we need to add ∫[p,q] f(x) dx? Also, why do ∫[0,p] f(x) dx , the upper limit is 0 and the lower limit is p/q? Could u explain 1b)' again?

更新2:

Sorry! Sould be Could u explain 1a)' again? not 1b)

回答 (1)

2009-08-26 3:39 am
✔ 最佳答案
1 (a) False
∫[0,q] f(x) dx = ∫[0,p] f(x) dx + ∫[p,q] f(x) dx
If f(x) is negative in the (p,q). Then ∫[0,p] f(x) dx > ∫[0,q] f(x) dx
(b) True. Because log is strictly monotonic increasing
(c) False
n^3 - n = n(n^2-1)=(n-1)n(n+1)
We can see that there should be two even factor no matter what the n is.
(d) True. From (c) We see that there should be one multiple of 3 in the expression.
2 f(x)=x/(x+1)
f(f(x))
= {x/(x+1)]/{[x/(x+1)]+1}
=x/(2x+1)
f^[3](x)
=[x/(2x+1)]/{[x/(2x+1)]+1}
=x/(3x+1)
Generally, f^[n](x)=x/(nx+1).
When n=1, there is nothing to prove.
When n=k, assume f^[k](x)=x/(kx+1).
f^[k+1](x)=[x/(kx+1)]/{[x/(kx+1)]+1}
=x/[(k+1)x+1]
By M.I. f^[n](x)=x/(nx+1).

2009-08-25 19:42:37 補充:
(c) should be

We can see that only what n is odd then n^3-n is divisible by 4. So n is not divisible by 4


收錄日期: 2021-04-26 13:50:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090825000051KK01802

檢視 Wayback Machine 備份