一些难题

bluesky8UID: 2691497

Hi everyone, I have a question from GR0568and some other interesting ones from the previous real ETS’s exams. Hope we allcan benefit from discussing them. Thank you!

GR0568_#62
Let K be a nonempty subset of Rn, wheren>1. Which of the following statements must be true?
I. If K is compact, then every continuousreal-valued function defined on K is bounded.
II. If every continuous real-valuedfunction defined on K is bounded, then K is compact.
III. If K is compact, then K is connected.
(A) I only
(B)II only
(C) III only
(D) I and II only
(E) I, II, and III
[answer is D. For II, we can givecounterexample such as f(x)=sin(pi*x) where x is (-1,1), which f(x) is [-1,1].However, the number of counterexample is limited if x belongs to an open set. So,I’m not sure that if every continuous real-valued function defined on K isbounded then will K must be a closed set?]

The following questions are from theprevious real ETS’s exam.

1. A fair coin is tossed and when eitherfour consecutive heads or tails appear the process will be stopped. What’s theprobability of two consecutive heads or tails?

2. For any ε>0, exists δ>0, s.t. |f(x)-f(x0)|> ε whenever |x- x0|>δ. Which one isthis statement equivalent to: (A) f is unbounded; (B) lim|f(x)|=∞, for |x|->∞.
[Although it is obvious that A is the answer,don’t know how to prove that B is wrong.]

3. Given f’’(x)<0, f’(0)=0
T=f(0)+2f(2)+4f(4)+f(6), I=∫f(x)dx whereintegral from 0 to 6, R=2f(2)+2f(4)+2f(6)
Question: List T, I and R from small tobig.
[It is easy to know that T>R and I>R. But, don’t know how to compare T and I]

4. a, b, c and d are natural numbers. Theiraverage value is 100 and a<b<c<d.
question: what is the maximum or minimumnumber of a+d?
[don’t know how to solve this problem.which theory will be suitable to use?]

5. f is a function with two linear parts.f(0)=f(2)=0, 0<x<1, max(f)=1. choose the range of the length of f:
A. (2*2^0.5, 1+5^0.5)
B. (2*2^0.5, 1+5^0.5]
C. [2*2^0.5, 1+5^0.5)
D. [2*2^0.5, 1+5^0.5]

6. A 10x10cm square card is folded. Theupper half part of the square is blue and the lower half part of the square isred. A circle with radius 1 cm is placed so that it is entirely on the square. Whatis the probability that the circle lies entirely in the red part?
[The answer maybe is 3/8. However, don’tknow how to get it.]

nankailiqiUID: 2389849

The first question 2 is correct.   You can only prove it by using contradicting method. And it is easy to prove that every set K , not compact set in Rn, means it must be a set that is not closed or unbounded. Either of these two case will have examples that some continuous function on it is not nesscerily bounded. If not closed like interval (0,1)  then 1/x is continuous but unbounded. If not bound [0,infinite) then x is an example.

windsxfUID: 2687198

For the previous ETS questions...
1. I'm not sure if you stated the problem clearly... If according to what you said, the probability should be 1, since under any circumstance, whenever the process stops, there must already exist a consecutive-four heads or tails... then it's trivial to have two consecutive heads or tails.
2. try to give a counterexample, just some thoughts, say set f(x)=1/x or 1/x^2
4.set b=a+t1, c=b+t_2, d= c+ t_3
so t_1, t_2,t_3 are all natural numbers greater than 0.  
so we have 400=a+b+c+d=4a+3t_1+2t_2+t_3.    and a+d=2a+t_1+t_2+t_3
noice that 200=2a+1.5t_1+t_2+0.5t_3, then simply we have a+d=200+1/2(t_3-t_1)
now the problem is quite solvable.  Note that 4a+3t_1+2t_2+t_3=400,a>=0  t_1,t_2,t_3 >= 1
So t_3-t_1<=395-1=394, when t_3=395, t_1=t_2=1,a=0
t_3-t_1>=1-393/3=-130 when t_3=t_2=1, t_1=131,a=1   i.e  a+d = 200+1/2(t_3-t_1)  is in [135,397]
5. Draw a coordinate system, it's easy to find that it's actually asking the range of the sum of the other two sides of a triangle,which has one side equals 2 and height equals 1 on this side.  choose C
6.   We only consider the location of the center of the circle. Suppose the square is described by (0,0),(0,10),(10,0),(10,10), then all possible location of the center of the circle is bounded by a square (1,1),(1,9),(9,1),(9,9).  
Similarly, in order to be placed entirely in red part, the center of the circle is bounded within a rectangle (1,1)(1,4)(9,4)(9,1).   Hence the probability of the circle to be in red part is just the area of rectangle over area of square.  Namely probability = 3*8/(8*8)=3/8

3. I think there's a typo in your problem . T should be f(0)+2f(2)+2f(4)+f(6), in that case I>T>R.  Easy to get f'(x) <0  for x>0, so f is strictly decreasing for x>0. so it's trivial that T>R. For I and T,
T=f(0)+2f(2)+2f(4)+f(6)=(f(0)+f(2))+(f(2)+f(4))+(f(4)+f(6))<I_[0 to 2]+I_[2 to 4]+I_[4 to 6]=I, since f is strictly concave (f''<0).

vinnieUID: 2104933

Which area for NO.1?

It seems I didn't cover it yet...