## 51, 54, 55, 64, 65 - GR0568

Forum for the GRE subject test in mathematics.
moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

### 51, 54, 55, 64, 65 - GR0568

51. If [x] denotes the greatest integer not exceeding x, then Int_0^Inf [x]e^(-x) dx =?

A) e/(e^2-1)
B) 1/(e-1)
C) (e-1)/e
D) 1
E) +Inf

I'm unsure how to answer this question fully since I dont know how to deal with [x]. I tried approximating it by just replacing it with x and integrating that way and came up with 1. However this is the wrong answer, the correct answer is infact 1/(e-1).

54. The four shaded circles in Figure 1 above are congruent and each is tangent to the large circle and to two of the other shaded cricles. Figure 2 is the result of replacing each of the shaded circles in Figure 1 by a figure that is geometrically similar to Figure 1. What is the ratio of the area of the shaded portion of Figure 2 to the area of the shaded portion of Figure 1?

A) 1/2Root(2)
B) 1/(1+Root(2))
C) 4/(1+Root(2))
D) ( Root(2)/(1+Root(2)) )^2
E) ( 2/(1+Root(2)) )^2

So the number we are looking for is Area of Shaded Region in Figure 2 / Area of Shaded Region in Figure 1. Just by inspection alone we can deduce that the ration must be lower then 1 so we can eliminate C as an answer.

I did some of my own calculations (hard to explain without the figures) but I determined the answer to be 4R^2 where R is the radius of the 4 inscribed circles in Figure 1 (Letting the radius of the main circle to be 1). What did everyone else get?

55. For how many positive integers k does the ordinary decimal respresentation of the integer k! end in exactly 99 zeros.

A) None
B) One
C) Four
D) Five
E) Twenty-four

I'm unsure how to start this problem on so many levels. At first I though I could count how many orders of 10 were being multiplied until there were 99 but then I realized the other factors could produce extra zeroes as well.

64. For each positive integer n, let f_n be the function defined on the interval [0,1] by f_n(x) = x^n /(1+x^n) which of the following statements is true?

I. The sequence converges pointwise.
II. The sequence converges uniformly
III. lim_n^Inf Integral_0^1 f_n(x) dx = Integral_0^1 lim_n^Inf f_n(x) dx

I know the function converges pointwise and not uniformly since convergence depends on the value of x. My problem is that I'm unsure to k now for certain when its ok to pass the limit for it. I know if it was converging uniformly it would be alright since f is continuous on the interval, is there any rule here that I should be made aware of?

65. Which of the following statements are true about the open interval (0,1) and the closed interval [0,1]?

I. There is a continuous function from (0,1) onto [0,1]
II. There is a continuous function from [0,1] onto (0,1)
III. There is a continuous bijective function from (0,1) to [0,1]

So, I know there is a bijection between the two sets however its not continuous from what I recall. If III is correct then both I and II are correct therefore D is wrong as a possible answer (I and III). How did you guys go about checking each of these individually though?

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
# 55 - solved http://www.mathematicsgre.com/viewtopic.php?f=1&t=121
64. For each positive integer n, let f_n be the function defined on the interval [0,1] by f_n(x) = x^n /(1+x^n) which of the following statements is true?

I. The sequence converges pointwise.
II. The sequence converges uniformly
III. lim_n^Inf Integral_0^1 f_n(x) dx = Integral_0^1 lim_n^Inf f_n(x) dx
let f(x)=o if 0<x<1 and,
f(x)=1/2 if x=1 then f_{n} converges pointwise to f but not converges uniformly to f. so II is incorrect

III is correct because of Lebeques integral

So I, III are correct

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
Brief Solutions.

Q. 51 is an infinite geometric series
= integral(1..2) e^-x dx + integral(2..3) 2 e^-x dx + integral(3..4) 3 e^-x dx + ...
= (-e^-2 + e^-1) + 2(-e^-3 + e^-2) + 3(-e^-4 e^-3) + ...
=e^-1 + e^-2 + ... = 1/e[1 / [1 - 1/e] ]
=which equals 1/(e - 1).

Q. 54 Consider Figure 1. Let A be the point (r, 0) (lies on the X-Axis), B be the point (0, r) (lies on the Y-Axis), C be the point (r, r), (center of the shaded circle), D be the point (R cos 45, R sin 45) (the point at which the shaded circle is tangent to the bigger circle). Use the distance formula | CD | = | AC | = | BC |, to find the ratio r / R. Then the required answer is simply (4 r^2 / R^2)^2.

Q. 55: Simple: Number of integers with one zero in their factorial

1 ! = 1 (No zeros)
2 ! = 2
3 ! = 6
4 ! = 24
5 ! = 120 (One zero)
6 ! = 720
7 ! = 5040
8 ! = 40320
9 ! = 362880
10 ! = 3628800 (Two zeros)

Obviously as soon as your factorial gets multiplied by a 5 or a 10 (or their multiple) one more zero is added to the end. So for example 5! - 9! end in 1 zero, 10! - 14! end in two zeros...

The answer is 5.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
65. Which of the following statements are true about the open interval (0,1) and the closed interval [0,1]?

I. There is a continuous function from (0,1) onto [0,1]
II. There is a continuous function from [0,1] onto (0,1)
III. There is a continuous bijective function from (0,1) to [0,1]
We do have the following theorem: there is a real-valued continuous function f :[0,1]------->[0,1] such that
i) f is onto
ii) f is increasing ( and hence one-to one)

(You guy can search this theorem by using key word " Cantor set, and generalized Cantor set"

Since (0,1) is subset of [0,1] so let g :[0,1]--------(0,1) which is defined as the following :
g(x)=x if x in (0,1)
g(0)=g(1)=1/2
then g is continuous and onto

let h= g(f(x)) then h is continuous and onto since both f and g are continuous and onto
so II is correct

Correct me if I am wrong : D

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
A continuous function takes a compact set to a compact set.

[0,1] is compact and (0,1) is not, so there is no continuous function that is onto.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
CoCoA,
Yes, you are right, based on your explanation, II is incorrect.
for III if f : (0,1)--------> [0,1] is continuous and bijection then the inverse of f : f^(-1) : [0,1]------->(0,1) is continuous and bijection so f^(-1)[0,1] =(0,1) , therefore must be compact . This is a contradiction since (0,1) is not compact.

Hence , I is correct

But if you guys know how to construct a function that satisfies I, please post here so that we can discuss. Thanks

CoCoA
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
A couple of examples that satisfy I:

First: f(x) = { 2|x-1/4| if x in (0,1/2)
{ 1 - 2|x-3/4| if x in [1/2,1)

Second: f(x) = [sin(2pi*x)+1]/2

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm
Nameless - could you explain in further detail what you meant for 55 that III is correct because of Lebegue's integral? From what I have looked up online its another version of summing over intervals to approximate integrals. How does this factor into the statement of passing the limit through the integral?

Amateur - When you say (r,r) is in the middle of the shaded circle which circle are you referring to?

EDIT: After trying to work through your explination I cannot seem to follow your steps. Assuming (r,r) is the middle of the bottom left shaded circle, how do you know |BC|=|AC|=|CD| is the condition that gives us the correct ratio?

I guess I'm just a little unsure how you got your final answer to be in terms of r and R since I'm not completley sure I understand where you are putting them on the figure.

PS: Thanks to all of you for your help on these problems. I should have saw the geometric series on that integral (kicking myself).

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
Thanks for your compliment.

All points in the figure are in the first quadrant.

r is the radius of the shaded circle in the first quadrant,
R is the radius of the unshaded bigger circle in the first quadrant,

Here are the details. For the first quadrant,

[ (r - 0)^2 + (r - r)^2 ]^(0.5) = [ (R/2^0.5 - r)^2 + (R/2^0.5 - r)^2 ]^(0.5)

Therefore r / R = 1 / (2^0.5 + 1)

The ratio of the area of the shaded portion in Figure 1 to its unshaded portion is 4 r^2 / R^2 = (2 / 2^0.5 + 1)^2

Let the area of the smallest circle in Figure 2 be r1. Therefore the ratio of r1 to r is the same as that of r to R. So the ratio of the area of the shaded portion of Figure 2 to that in Figure 1 will be 16 r1^2 / 4 r^2 = 4 r1^2 / r^2 = 4 r^2 / R^2 = (2 / 2^0.5 + 1)^2 which is choice E.

P.S> It took me one complete day to come up with this solution.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm
Ok, I think I understand the proof now, I was getting confused because I was unsure how you new that the distance between Rcos45,Rsin45 = r. But it obviously ends at the tip of the shaded circle with radius r.

So r/R is 1/(1+root(2))

thus 4(r/R)^2 = (2/(1+root(2)))^2 = K, which is the ratio of shaded area to the area of the big circle in figure 1.

Thus the area in figure2 compared to figure1 is: 4pi*r^2*k / 4pi*r^2 = k = (2/(1+root(2)))^2

Thanks for the help. Now I just need to do it in 2-3 mins for the GRE test heh.

gaucho85
Posts: 11
Joined: Thu Sep 18, 2008 5:58 pm
I'm not sure if this is the same solution as described here for 54. If is is, sorry to repeat.

I'll call the biggest circle, "the big circle", the middle circle of which there are 4 in either figure "the medium circle" and the circle for which there are 16 in figure 2, the "small circle"

Consider 1 medium circle and the 4 small circles inside it.

If you let the radius of one of the small circles be 1, then the distance from the center of the upper left small circle to the center of the lower left small circle is 2.
Similarly the distance from the lower left circle's middle the the lower right circle's middle is 2.
Thus the distance from the upper left circle's center to the lower right circle's center is root 8 (right triangle).

thus, the radius of a medium circle is half of the sum of root 8 plus two (the two being the distance from the center of the upper left and lower right's circles to the edge of the medium circle).

So the small circle's radius is 1, and the middle circle's radius is 1 + root 2.

thus the ratio of area is 4*1/(1 + root 2)^2, or (2/(1+root 2))^2

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm
Not the same way, but seems slightly less time consuming and much easier to understand .

ngoc
Posts: 6
Joined: Mon Nov 03, 2008 5:54 am
But doesn't continuous function maps open set to open sets? Then answer (I) can't be right? (This is the part I'm confused over).

As for the examples given by CoCoA, I'm not sure why it works, since the end points 0 and 1 (of the image of f(0,1) are not included. i.e.: there is no x s.t. f(x) = 0 or 1.

(Clearly something is wrong with my argument, since the GRE answer is B (i.e.: (I) only)). If you can please point out where I went wrong, it's much appreciated, thank you.

ngoc
Posts: 6
Joined: Mon Nov 03, 2008 5:54 am
Ah no, take that back. U is open when f.inverse(U) is open -- that's why. Ok. No more worries. Thanks guys.

sachem
Posts: 9
Joined: Wed Oct 29, 2008 8:01 pm
I'm a little bit confused about question 65 still and would much appreciate it if someone could help me out.

I'm a little confused about I.

I understand II - since any continuous function maps a compact set to a compact set, there can be no continuous function that maps [0,1] to (0,1) since the latter is not compact.

I also understand III for the same reason - if we have a continuous one to one, onto (bijective) fucntion (0,1) to [0,1], there is a bijection given by f^(-1) from [0,1] to (0,1). But then (0,1) is not compact....contradiction.

Now I go to part I. I am thinking about the fact that if f:X1->X2 is continuous, then the set f^(-1)(C) is closed in X1 for every closed subset C of X2.

So, since [0,1] is closed, I look at f^(-1)([0,1]) and say that this set must be closed. But it seems to me that f^(-1)([0,1])=(0,1) which is open...so I say that I is true.

Where did I go wrong here?

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
f^(-1)([0,1])=(0,1)
why you have this if f is not one-to-one?

Since f is onto so we ONLY have f([0,1])=(0,1).

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
Q. 55: Simple: Number of integers with one zero in their factorial

1 ! = 1 (No zeros)
2 ! = 2
3 ! = 6
4 ! = 24
5 ! = 120 (One zero)
6 ! = 720
7 ! = 5040
8 ! = 40320
9 ! = 362880
10 ! = 3628800 (Two zeros)

Obviously as soon as your factorial gets multiplied by a 5 or a 10 (or their multiple) one more zero is added to the end. So for example 5! - 9! end in 1 zero, 10! - 14! end in two zeros...

The answer is 5.
The solution above was solved. today, I looked at it again and the solution was wrong
suppose that x is positive integer then the number of zeros of x! is
[x/5]+[x/25]+....+[x/5^t]+...
where [y] is the integer part of y
by the assumption we have
[x/5]+[x/25]+....+[x/5^t]+...=99

solve this equation we have 400<=x<425
then there are 24 numbers
Last edited by Nameless on Thu Nov 06, 2008 7:20 pm, edited 2 times in total.

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
Hi Nameless,

In my solution, I meant that, suppose k! ends in 99 zeros, where k is divisible by 5 (k mod 5 == 0). Then (k + 1)!, (k + 2)!, (k + 3)! and (k + 4)! will also end in 99 zeros since no 5 is being added to the multiple. As soon as you reach (k + 5)!, more zeros will be added to your multiple.

Regarding your solution, the equation is more precise than my inductive solution but the solution to it is 400 <= x <= 404.

Of course one flaw in my solution is that it doesn't take into account jumps in the number of zeros that occur at every power of 5. So for example number of integers "k" such that k! has exactly five zeros = 0.

But if there is at least one integer such that its factorial ends in k zeros, then the number of integers whose factorials end in "k" zeros is exactly 5.

sachem
Posts: 9
Joined: Wed Oct 29, 2008 8:01 pm
Thanks for your response -

When I said f(-1)([0,1]) i meant the preimage of [0,1], not that there is necessarily an inverse function.

So, I am still confused by it - is it not true that the preimages of closed sets are closed?

So how is it possible to have a continuous function that takes (0,1) onto [0,1] because then wouldn't the preimage of [0,1] be (0,1), an open set?

I know there is something wrong with my thinking here....because I can explicitly write a function that takes an open set to a closed set...ie f{(0,1)}=0.

EDIT: Figured out where I went wrong, I think. It is true that if a continuous function f takes X1->X2, that if C is closed in X2 then the preimage f^(-1)(C) is closed in X1. However, in the example given, our function is NOT f:R->R, but f: (0,1)->R....so, when trying to figure out if it is possible for f to take (0,1) onto [0,1], and I consider the preimage of [0,1], i need to consider the preimage not as a subset of R, but as a subset of the domain.

Topology is very confusing to me

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
Hi Nameless,

In my solution, I meant that, suppose k! ends in 99 zeros, where k is divisible by 5 (k mod 5 == 0). Then (k + 1)!, (k + 2)!, (k + 3)! and (k + 4)! will also end in 99 zeros since no 5 is being added to the multiple. As soon as you reach (k + 5)!, more zeros will be added to your multiple.

Regarding your solution, the equation is more precise than my inductive solution but the solution to it is 400 <= x <= 404.

Of course one flaw in my solution is that it doesn't take into account jumps in the number of zeros that occur at every power of 5. So for example number of integers "k" such that k! has exactly five zeros = 0.

But if there is at least one integer such that its factorial ends in k zeros, then the number of integers whose factorials end in "k" zeros is exactly 5.

Dear Amateur,

I'm so sorry for stupid mistake, YOUR SOLUTION IS ABSOLUTELY CORRECT.

I have to refresh my mind

xyzzy240
Posts: 1
Joined: Tue Mar 31, 2009 3:20 am
I'm not seeing the justification for part III of 65. It's not true in general that if f: X -> Y is a continuous bijection, then f^-1 is continuous. It's true in some cases, such as when X is compact and Y is Hausdorff, but this criterion doesn't apply here since (0, 1) is not compact.

I'm not sure how else to show the continuity of f^-1. Am I missing something obvious?

eof
Posts: 9
Joined: Sun Oct 12, 2008 3:39 pm
It's a false statement that the inverse is generally continuous. However, the easiest way to show that III is impossible is choosing an x in (0,1) which maps to 1 (or zero and using a similar argument). Then letting y<x<z we see that f(y)<1 and f(z)<1 by bijectivity, so assuming that f(y)<f(z)<1 we have by continuity a point in the interval (y,x) which maps to f(z) contradicting the fact that f was bijective. The case f(z)<f(y)<1 follows analogously.

Charles.Rambo
Posts: 9
Joined: Wed May 22, 2013 7:33 pm

### Re: 51, 54, 55, 64, 65 - GR0568

Hello. I've solved all of the GRE form 68 problems. You can view the solutions at http://rambotutoring.com/GRE-math-subje ... utions.pdf.

-Charles

redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am

### Re: 51, 54, 55, 64, 65 - GR0568

Yet another set of solutions can be found here: http://www.gigauploader.com/file/0770203183169331

Here's the post which describes the solutions:
http://www.mathematicsgre.com/viewtopic.php?f=1&t=1378