help with GRE math subj practice problems

[ihaveanswers]

I have managed to get solutions for almost all of the problems on the GRE math subject practice test posted on the ETS site with the exception of problems 51,58,64,65. I'm also not sure about 47. I would greatly appreciate solutions to these questions (posted below) and am willing to provide solutions to the other problems in gratitude. the practice test gives answers but no explanations (thanks guys). Here we go:

47) Let x and y be uniformly distributed, independent random variables on [0,1]. The probability that the distance between x and y is less than 1/2 is:

ans: 3/4

51) Let D be the region in the xy plane in which the series sum (k=1 to infinity) ((x + 2y)^k)/k converges. Then the interior of D is:

ans: an open region between two parallel lines

I tried using the ratio test and got the half plane y<1/2(1-x) . why is this wrong?

58) Let f be a real-valued function defined and continuous on the set of real numbers R. Which of the following must be true of the set S={f(c): 0<c<1}
I. S is a connected subset of R
II. S is an open subset of R
III. S is a bounded subset of R

ans: I and III only.

This problem confuses me the most I think. since the preimage 0<c<1 is an open set and since f is cts, shouldn't S also be open? Isn't that a significant theorem, that continuous functions map open sets to open sets?

64) Suppose that f is a continuous real-valued function defined on the closed interval [0,1]. Which of the following must be true?
I. There is a constant C>0 such that |f(x) - f(y)| <=C for all x,y in [0,1]
II. There is a constant D>0 s.t. |f(x) - f(y)| <=1 for all x,y in [0,1] that satisfy |x-y| <= D
III. There is a constant E>0 s.t. |f(x) - f(y)| <= E|x-y| for all x,y in [0,1]

ans: I and II only

II follows directly from the epsilon-delta definition of continuity, doesn't it? And III is def of lipschitz, which is stronger than just continuity. I follows from the extreme value theorem I believe, although what about tan(X+pi/2)? Isn't this is a cts unbounded function on [0,1] in the extended real number system?

65) Let p(x) be the polynomial x^3 + ax^2 +bx +c, where a,b, and c are real constants. If p(-3) = p(2) = 0 and p'(-3)<0, which of the following is a possible value of c?

ans: -27

I did not know where to even start on this one. Whoever solves this one gets extra genius points.

Thank you so much for helping if you can! I'm pretty nervous about the test....

[mamal]

Ok you seem to be so nervous about the test. My frist advice is to be calm . here we go!

47)
to obtain the solution you need to draw the x-y coordinate and indentify the four limiting lines. they are x=0,x=1,y=0 and y=1. it is obvious that the answer should be located between these four constraints. that is, it should be within a square whoes vetexes are (0,0), (1,0), (0,1) and (1,1). Let us call the area of such space as S1. On the other hand, we have further restrictions. Let us denote the numbers as X and Y. since X and Y should be selected so that distance between them is smaller than 1/2, they should be located between the lines X-Y=1/2 and Y-X=1/2; call this area S2. Now the probability is easily calculated as P=S2/S1=3/4.

51)
we have S=sum (k=1 to inf) Z^k/k, where Z=x+2y. it is clear that such a series is convergent everywhere provided that |Z|<1. now substitude Z=x+2y and you get D={(x,y), |x+2y|<1}, it is clear that D is nothing but an OPEN (due to '<' and not '<=') region between TWO (due to || operator) parrallel lines.

58)
this also perplexes me! Well I also think you might be right. I'' let you know if I changed my mind!

64)
Ok this is somewhat hard! since the third statement implies that the inequality condition should hold for all x,y in [0,1], devide both sides by |x-y| and take the limit as y->x, that would naturally lead to the satement that LIMIT |[f(x)-f(y)]/[x-y]| (as y->x) = |f'(x)|<=E. this means that f' should be bounded which is not neccessairly true for a continuous function.

65)
we know that:
I) p(-3)=-27+9a-3b+c=0
II) p(2)=8+4a+2b+c=0
III) p'(-3)=27-6a+b<0
now let S=m*(I)+n*(II)+(III)=-27m+8n+a(9m+4n-6)+b(-3m+2n+1)+(m+n)*c. it is all clear that S<0. now select m and n, such that S is free of a and b, and only contains c. this can easily be achieved by solving the following set of equations:
i) 9m+4n-6=0
ii) -3m+2n+1=0
which leads to m=8/15 and n=3/10. inserting m and n into S, one obtains that c<-18. thus the only true choice would be (A)

Hope my solutions can help you dude! By the way are you going to take test in Nov. 3 ?

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[benjiemoe]

5Cool Let f be a real-valued function defined and continuous on the set of real numbers R. Which of the following must be true of the set S={f(c): 0<c<1}
I. S is a connected subset of R
II. S is an open subset of R
III. S is a bounded subset of R


]I Must be true because continuous functions preserve connectivity (intermediate value theorem)

II is absolutely false: a counter-example is take f to be a constant function, then S is just a point, which is obviously closed. Property II is called an open mapping which is quite distinct from continuity (which states that the inverse image of an open set is open). of course if the function is invertible and its inverse is continuous (homeomorphism) it is an open mapping.

III. a continuous function on a compact inverval, say [0,1] achieves an absolute maximum and minimum, so bounded on [0,1] implies bounded on (0,1)

Problem 65 can better be solved by writing p(x)= (x+3)*(x-2)*(x-a) for some (unknown) root a

(since -3 and 2 are roots). then multiply this out and get p(x)= x^3 + (1-a)x^2 - (6+a)x+6a

then p'(x) = 3x^2 +2(1-a)x -(6+a)
p'(-3) = 27-6(1-a)-(6+a) = 15+5a, set this less than zero, get a<-3
and c= -3*2*a (last term in a cubic is minus product of roots) c<-18 so -27 is the only choice that works

[seangia]

my attempt solution:
http://sfmathgre.blogspot.com/2010/05/g ... omial.html