# Questions from the GRE Math Subject Test Practice Book

Tags:
1. Mar 30, 2006

### island-boy

Hello, I'm preparing for the GRE Math Subject Test, as such, I am using ETS's free practice book which can be downloaded from the gre.com website. The answers are given for the questions, but the solutions are not. I am having difficulty getting the solutions to the following questions. Help and input is appreciated:

I gathered all the questions into one thread so as not to spam the forums:

----------
24) Which of the ff sets of vectors is a basis for the subspace of Euclidean 4-space consisting of all vectors that are orthogonal to both (0,1,1,1) and (1,1,1,0)?
A) {(0,-1,1,0)}
B) {(1,0,0,0), (0,0,0,1)}
C) {(-2,1,1,-2), ((0,1,-1,0)}
D) {(1,-1,0,1), (-1,1,0,-1), (0,1,-1,0)}
E) {(0,0,0,0),(-1,1,0,-1),(0,1,-1,0)}

-------------
32)When 20 children in a classroom line up for lunch, Pat insists on being somewhere ahead of Lynn, If Pat's demand is to be satisfied, in how many ways can the children line up?

-------------
37) Summation of (k^2)/k! from k = 1 to infinity is ____

-------------
44) Let f denote the function defined for all x>0 by f(x) = x^(x/2), which of the folowing statement is false?

answer is the derivative f'(x) is positive for all x>0.
I was able to get why the other choices are true, but I can't understand why this is false. Isn't the derivative [x^(x/2) x lnx]/2? which is postive for all x>0?

----------------
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 __

----------
50)How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]

--------------
54) The inside of a certain water tank is a cube measuring 10 ft on each edge and having vertical sides and no top. Let h(t) denote the water levele in feet, above the floor of the tank at time t seconds. Starting at t=0, water pours into the tank at a constan rate of 1 cubic foot per second and simultaneously, water is removed from the tank at a rate of 0.25 h(t) cubic ft per second as t -> infinity, what is the limit of the volume of the tank?

answr is 400 cubic ft.

--------------
56)For every set S and every metric d on S, which of the ff is a metric on S?

a) 4 + d
b) e^d - 1
c)d - |d|
d)d^2
e)d^1/2

--------------
57) Let R be the field of real numbers and R(x) the ring of polynomials in x with coefficent in R. Which of ff subsets of R(x) is a subring of R(x)?

answer: I was able to get why the others are subrings, but I can't get why: All polynomials whoose degree is an even integer, together with the zero polynomial, is not a subring.

------------------
59)A cyclic group of order 15 has an element x such that the set {x^3, x^5, x^9} has exactly 2 elements. THe number of elements in the set {x^13n: n a positive integer} is:

------------
60) If S is a ring with the property that s = s^2, which of the ff must be true?
I. s + s = 0 for all s in S
II (s+t)^2 = s^2 + t^2 for all s, t in S
II S is commutative.

answer is all 3 are true.

------------
61) What is the greatest integer that divides p^4 -1 for every prime number p greater than 5?

--------------
63)At how many points in the xy plane do the graphs of y= x^12 and y =2^x intersect?

answer is 3. Shouldn't it be 2 since y=x^12 is just a parabola (am not sure of this) and y = 2^x is just similar to the graph of y = e^x?

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

answer is I and II only. I know this has something to do with the mean value theorem.

_________
that's it. sorry for the long post and thanks for the help!

2. Mar 30, 2006

### dextercioby

Here's a proof for 37)

$$S_{2}= \sum_{k=1}^{\infty } \frac{k^{2}}{k!} =...?$$

Consider $$S_{1}= \Sum_{k=1}^{\infty} \frac{k}{k!} = \Sum_{k=1}^{\infty} \frac{1}{(k-1)!}$$

making the sub $k-1=n$, we get

$$S_{1}= \Sum_{n=0}^{\infty} \frac{1}{n!} = e$$

But one can prove that

$$S_{2}-S_{1}= \Sum_{k=1}^{\infty} \frac{k^{2}-k}{k!} = \Sum_{k=1}^{\infty} \frac{k(k-1)}{k!} =\Sum_{k=1}^{\infty} \frac{k-1}{(k-1)!}$$

Make the sub $k-1=n$

and then get

$$S_{2}-S_{1}=S_{1} \Rightarrow S_{2}=2S_{1}=2e$$

Daniel.

This latex is ****ed up today

Last edited: Mar 30, 2006
3. Mar 30, 2006

### island-boy

hey thanks dex :)
I was able to understand your proof. Though, it would be hard of me to think of that solution during the actual exams :( I'm getting discouraged, I only got around 60% percentile scoring (based on the GRE scaled score) after 2 practive exams...

4. Mar 30, 2006

For question 44, I got $$0.5 x^{\frac{x}{2}} (1+lnx)$$ when I differentiated. As long as you remember that ln(x) is negative for $$0<x<1$$, you can see why the differentiated result is negative for some positive values of x!

Last edited: Mar 30, 2006
5. Mar 30, 2006

### island-boy

pizza,

you're right! ln x is negative for 0<x<1...DOH! I got confused over that. Now, I get why the derivative is negative.

Thanks for the heads up there :)

2 questions down, 12 to go.

6. Mar 30, 2006

### nocturnal

#32)
There are 20! ways to line up 20 students. For each way we must have that Pat is either ahead of lynn or behind her. It follows that the number of arrangements with Pat ahead of Lynn is equal to the number of arrangements with Pat behind Lynn. So the number of ways with Pat ahead of Lynn is 20!/2.

7. Mar 30, 2006

Here are my thoughts for Question 54... We need to introduce the variable for volume since some of the information given in the question talks about volume.

Let v represent the volume of water in the tank (in cubic feet) and let y=h(t). Since the water will definitely cover the floor of the tank, so v=100y.

Setting up a differential equation based on the information provided, $$\frac{dv}{dt} = 1 - 0.25 y$$ Since we know the relationship between v and y, we can change the differential equation until the variable present on the right-hand side is v.

Solve the differential equation, with initial conditions v=0 when t=0.

In the end, the solution should be something like $$v = 400 - 400 e^{-0.0025t}$$ If t tends to infinity, we can easily deduce what the limit of v would be!

Last edited: Mar 30, 2006
8. Mar 30, 2006

### HallsofIvy

24. If (a,b,c,d) is orthogonal to both (0,1,1,1) and (1,1,1,0) then we must have (a,b,c,d).(0,1,1,1)= b+ c+ d= 0 and (a,b,c,d).(1,1,1,0)= a+ b+ c= 0. Subtracting those two equations, a- d= 0 or a= d. Now solve a+ b+ c= 0 for c: c= -a- b. We can choose a and b to be anything and then solve for c and d so this subspace is two dimensional.
A) {(0,-1,1,0)}
No, that's only one vector

B) {(1,0,0,0), (0,0,0,1)}
a is not equal to d in either

C) {(-2,1,1,-2), ((0,1,-1,0)}
In the first, a= d= -2, c= 1= 2-1= -a+b. In the second, a= d= 0, c=-1= 0-1= a-b. Yes, both satisfy the required equations.

D) {(1,-1,0,1), (-1,1,0,-1), (0,1,-1,0)}
No, there are 3 vectors not 2.
E) {(0,0,0,0),(-1,1,0,-1),(0,1,-1,0)}
No, there are 3 vectors not 2.

9. Mar 30, 2006

To tell you the truth, I would not know how to solve Q50 if the answer was not provided...

Anyway, the 4 functions which satisfy this equation are probably $$f(x) = x, f(x) = -x, f(x) = \mid x \mid, and f(x) = -\mid x \mid$$, all of which are defined for $$-1 \leq x \leq 1$$. If you draw the graphs of the 4 functions, you can convince yourself that they really are different.

All the best!

10. Mar 30, 2006

### nocturnal

#59)
Let $G$ be a cyclic group with $|G| = 15$, and suppose $x \in G$ such that $|\{x^3, x^5, x^9\}| = 2$. Since $\langle x \rangle$ is a subgroup of $G$, $| \langle x \rangle |$ divides $|G|$, therefore $| \langle x \rangle |$ is either 1,3,5, or 15. However, since $|\{x^3, x^5, x^9\}| = 2$ we must have $|\langle x \rangle| = 3$.
Note that $\{x^{13n}: n \in \mathbb{Z}^+ \} = \langle x^{13} \rangle$. Since $| \langle x \rangle | = 3$, $x^3 = e$, where $e$ is the identity element of $G$. Therefore $\langle x^{13} \rangle = \langle x \rangle$, and we already know that $| \langle x \rangle | = 3$.

Last edited: Mar 30, 2006
11. Mar 30, 2006

### Nimz

#59)
One of these three possibilities must be true:
1) x^3 = x^5
2) x^3 = x^9
3) x^5 = x^9
Once you've convinced yourself that option 2 is the correct one, the rest follows pretty easily.

#60)
Property II is easy enough to verify. (s+t)2 = s+t, since s+t $\in$ S. And s+t = ss + tt.
Property III can be verified in a similar manner: sstt = st = (st)2 = stst. You should be able to finish from here.
For Property I, consider what happens to -s.

#63)
Think about where y=2^x and y=x^2 intersect. Since 2^x tends toward zero at -infinity and grows much faster than x^2 for large x, the number of intersections must be odd unless the functions are tangent to each other at some point. Can you see why? The reasoning is the same for the functions given, and the case against the functions being tangent is even better.

12. Mar 30, 2006

### island-boy

hey guys, thanks for the outpouring of help! I really appreciate the help here.

I do have some additional questions and comments for some of the solutions though

I was able to get your explanation, althought I think that solution would be V = 400 - Ce^(-0.0025t) instead of V = 400 - 400e^(-0.0025t) since h(0) is not given in the problem so we can't set v= 0 when t = 0. Your explanation did help me remember my ODE class though :)

I understood the solution although I'm wondering why we can conclude that the dimension of the orthogonal vectors is 2. Is it because the Euclidean 4-space is given and the there are two vectors in question (thus dimension of orthogonal vectors = 4-2) or is it because there are only two vectors in question? What I'm saying is, what if the question asks that you give the basis for the subspace of the Euclidean 4-space consisting of vectors that are orthogonal to say (1,1,1,0), (1,0,1,1), and (1,1,0,1)? Should our answer consist of 1 vector or 3?

I was confused by this. How can we conclude that the number of intersections to be odd? also, isn't the functions being compared with are y = x^12, not y =x^2? or will the 2 have the same graph?

Again a big thank you to everyone.

Last edited: Mar 30, 2006
13. Mar 30, 2006

Well, the graphs of $$y=x^{12}$$ and $$y=x^2$$ are not identical, but they are similar in shape! From my understanding, if we draw the sketches of the curves mentioned in your question on the same diagram, it becomes immediately apparent that there are at least 2 intersection points. Question is, will the graph of $$y=2^x$$ rise sharply enough when x is large to cut the other curve a third time? On observing the sketch carefully, you should discover that if there is a large value of x where $$2^x>x^{12}$$, there can be a third intersection point. At this stage, we just have to substitute a few large values of x (e.g. 100, 150) to see if the inequality can be satisfied (as it is difficult to solve the inequality by conventional methods). If yes, then there is a third intersection point! And this is the case.
We can argue that there can be no more than three intersection points because the curve of $$y=2^x$$ will rise even more sharply after the third intersection point, something Nimz mentioned.
P/S I am intrigued by Q47 and Q61. If anyone discovers the solution, care to share it with me?

Last edited: Mar 30, 2006
14. Mar 30, 2006

### island-boy

I was able to make some headway in 56)
----------
56)For every set S and every metric d on S, which of the ff is a metric on S?

a) 4 + d
b) e^d - 1
c)d - |d|
d)d^2
e)d^1/2

---------
we can conlude that a) 4-d is not a metric since if d is a metric, then the least value of 4-d is 4, which makes it not a metric.

c) d - |d| is also not a metric since for all metric d, d-|d| =0 always.

I'm thinking that d) d^2 can't be a metric since if d is a metric (i.e., equal to every positive real), then d^2 can't be equal to every positive real.

I can understand why e) d^1/2 can be a metric, although I'm still not sure why b)e^d - 1 is not a metric.

for 61)
------------
61) What is the greatest integer that divides p^4 -1 for every prime number p greater than 5?

------------

I'm thinking we can factor p^4 -1 to (p^2+1) (p+1)(p-1)
To get the answer of 240, we have to conclude that the given factors are divisible by 2^4 x 3 x 5.
We can easily see that each factor is divisible by 2 for each p...for them to be divisible by 2 and 3 and 5, I'm not so sure...

15. Mar 30, 2006

### island-boy

hey, now I understand why there are 3 intersections.

for 47), someone posted this answer in another forum where I was posting. Here's what he has to say although I wasn't quite able to get his explanation:
---------
47/
This is just a problem in the real plane. I.e double integrals and such:

Note that if x=0 then we need 0<y<1/2
If x=1/2 then y can take any value i.e. we need 0<y<1
If x=1 then we need 1/2<y<1

Note that we can interchange x and y. I.e. if y=1/2 we need 0<x<1 etc.

So, look at the plot of x versus y. You should find that y starts from (0,1/2) and goes up linearly to (1/2, 1). We want everything below this area.

Then we have a line from (1/2, 0) to (1, 1/2). We want everything above this area. Hope that makes sense.

Calculate the area covered by the above, between (0, 1).

I.e. 3/8 * 2 = 3/4.
-----------------

here's what I was able to come up with on my own. I understand that there is a need to graph the problem and use double integrals to solve them, since x and y are uniform distributions, then we let x = 1/m, y = 1/n. Thus the problem we need to solve is to graph 1/m - 1/n <= 1/2 in the m-n plane and use double integrals to solve them...although I'm not sure yet how to solve this.

16. Mar 30, 2006

Trial and error

61) What is the greatest common factor of $$p^4 -1$$ for every prime number p greater than 5?

Let me share my thoughts on Q61. I may not have gotten the answer, and my method may not be the best, but we all learn from each other don't we? BEWARE: LONG POST AHEAD

Since the question only involves prime numbers greater than 5, we can conclude that all numbers p will end with the digits 1, 3, 7 or 9 (since the only prime number ending with the digit 5 is 5)

After that, I carried out some factorization like what island-boy did $$p^4-1 = (p-1)(p+1)(p^2+1)$$. Then, I determined the last digit of each of the factors for all possible last digits of p. And the results are as follows:

p ...1 ...3 ...7 ...9

p-1 ...0 ...2 ...6 ...8

p+1 ...2 ...4 ...8 ...0

p^2+1 ...2 ...0 ...0 ...2

(This means if p were 79, then the last digit of p-1 will be 8, that of p+1 will be 0, and that of p^2+1 will be 2)

By observing the last digits of the factors, we can immediately conclude that the highest common factor of $$p^4 -1$$ is at least 40, as $$p^4-1$$ will be divisible by 10 and 4 (2 copies of 2).

Next, we look at $$p^4-1=(p^2+1)(p^2-1)$$ and we can see that $$p^4-1$$ is actually a product of 2 numbers, one of which is greater than the other by 2. As long as $$p^2$$ is not a multiple of 3, then $$(p^2+1)(p^2-1)$$ will definitely be a multiple of 3. (Try this with any combination of numbers!) And we know that $$p^2$$ is not a multiple of 3, as the only factors of $$p^2$$ are $$1, p, p^2$$, and p cannot be 3. So, the highest common factor of $$p^4 -1$$ is now at least 120.

Finally, I complete the square for $$p^2+1$$ and my expression becomes $$p^4-1=(p+1)^3(p-1)-2p(p+1)(p-1)$$. My intention was to prove that the expression on the right hand side is a multiple of 16, so the highest common factor of $$p^4 -1$$ will be 240 (the correct answer), but I was not able to do so. Whilst it can be proven that $$(p+1)^3(p-1)$$ is a multiple of 16, I cannot do the same for $$2p(p+1)(p-1)$$, being only able to prove that it is a multiple of 8. BUT I have made an observation. I have realised that prime numbers greater than 5 are either 1 unit higher or 1 unit lower than a multiple of 4 (e.g. 101 and 100, 43 and 44). If my observation is valid for all such prime numbers, then $$2p(p+1)(p-1)$$ will definitely be a multiple of 16 and the answer of 240 can be obtained.

Final thoughts: If this method is used, how do we prove that there are no higher common factors of $$p^4-1$$?

P/S Thanks island-boy for helping me to understand Q47 better!

Last edited: Mar 30, 2006
17. Mar 31, 2006

### Nimz

p = 4n+1 or 4n-1 is merely a statement that p is odd. All primes above 5 will be odd, so that observation will always hold.

When considering (p+1)(p-1)(p^2+1), either p+1 or p-1 will be a multiple of 4, while the other will only be a multiple of 2, and (p^2+1) will certainly be a multiple of 2, but not 4. This is because (4n+1)^2 = 16n^2 + 8n + 1 = 8k + 1, and (4n-1)^2 = 16n^2 - 8n + 1 = 8k + 1. Add one to 8k+1 and you get a multiple of 2, but not a multiple of 4. Thus you will always have a multiple of 16, but not always any higher power of 2.

7^4 = 2401, so the greatest common factor of p^4 - 1 can only have 2 or 5 as a factor in addition to 240. From the last paragraph, you can't have another factor of 2. 11^4 = 14641, which demonstrates you can't have another factor of 5.

18. Mar 31, 2006

### Nimz

The two vectors given are linearly independent, so the set of vectors in the Euclidean 4-space perpendicular to those two vectors has a basis of 4-2 = 2 vectors. If the two given vectors were linearly dependent, the set of perpendicular vectors would have a basis of three vectors.

19. Nov 2, 2006

### beeman1266

The reason why 64.II is true is because f continous on a closed interval implies f is uniformly continous. the formulation they give is just a particular choice for epsilon equals 1.
and the reason why e^d is not a metric is because it doesn't obey the triangle inequality
i hope that helps.
does anyone know how to do you get number 65? there's a degree 3 polynomial p s.t. p(-3)=p(2) =0 and p'(-3)<0. what are possible values of p(0): -27, -18, -6, -3, or -13. the right answer is -27.

20. Nov 3, 2006

### AKG

You know that you need a 2-dim subspace, so it's either B or C. It's easy to check that the vectors in B are not orthogonal to both (0,1,1,1) and (1,1,1,0), so its C.
How many ways are there to arrange the kids in total? 20!. By symmetry, half the arrangments have Pat somewhere before Lynn, and the other half have Lynn somewhere before Pat. So the answer is half of 20!, i.e. 20!/2.
$$\sum _1 ^{\infty} k^2/k! = \sum _1 ^{\infty}k/(k-1)! = \sum _1 ^{\infty}(1/(k-1)! + (k-1)/(k-1)!)$$

$$= \sum _0 ^{\infty} (1/k! + k/k!) = e^1 + \sum _0 ^{\infty}k/k! = e + \sum _1 ^{\infty} k/k!$$

$$= e + \sum _1 ^{\infty} 1/(k-1)! = e + \sum _0 ^{\infty}1/k! = e + e^1 = 2e$$
log(f(x)) = (x/2)log(x)
f'(x)/f(x) = (1/2)log(x) + 1/2
f'(x) = f(x)/2[log(x) + 1]

Now f(x)/2 is always positive, but log(x) + 1 can be negative. Recall

$$\lim _{x\to 0^+}\log (x) = -\infty$$
Find the region of the unit square [0,1] x [0,1] where the distance between x and y is less than 1/2. It's everything except the triangles with vertices {(0,1/2), (0,1), (1/2,1)} and {(1/2,0), (1,0), (1,1/2)}. These triangles each have area (1/2)(1/2)(1/2) = 1/8, so together their area is 1/4. Their complement, thus, has area 3/4.
This is easy. The functions are f(x) = |x|, f(x) = -|x|, f(x) = x, and f(x) = -x. You know that since f(x)2 = x2, you get |f(x)| = |x|, and by continuity you should be able to see that these are the four options.
The volume at time t, V(t), in cubic feet, is 100h(t). You're essentially given V'(t) = 1 - 0.25h(t). But since V(t) = 100h(t), V'(t) = 100h'(t), so:

100h'(t) = 1 - 0.25h(t)

Solve this ODE. The related homogeneous equation is:

h'(t) = -0.0025h(t)
h(t) = Ce-t/400

A particular solution is h(t) = 4, so the total solution is:

h(t) = Ce-t/400 + 4

You can find C, but it doesn't matter, because as t goes to infinity, h(t) goes to 4, so the volume goes to 400. This is the full, "proper" way of doing it, but if you understand what's going on above, you can probably look for a simple heuristic to save you some time.
4+d is not a metric since it doesn't map (x,x) to 0. d - |d| isn't a metric because it maps every pair (x,y) to 0, and so it would only be a metric if |S| = 1, but it needs to work for any S. d2 isn't a metric because the triangle inequality fails. Observe 1/2 + 1/2 > 1, but (1/2)2 + (1/2)2 < 12. e) easily satisfies the first two properties of the metric definition, so it remains to check that it satisfies the triangle inequality. Suppose x + y > z, you want to prove:

$$x^{1/2} + y^{1/2} \geq z^{1/2}$$

Well:

$$x^{1/2} + y^{1/2} = \sqrt{(x^{1/2} + y^{1/2})^2} = \sqrt{x + 2\sqrt{xy} + y} \geq \sqrt{x + y} \geq \sqrt{z} = z^{1/2}$$

as desired. I'm doing all this off the top of my head, so I can't think of why b) fails the triangle inequality right now, so do that as an exercise for yourself.

EDIT: log(2) + log(2) > 2log(2) but (exp(log(2)) - 1) + (exp(log(2)) - 1) = 2 < 3 = exp(2log(2)) - 1.
x2 and -x2 + x have even degree, but their sum is -x, which does not have even degree, so the set of of polynomials of even degree, together with 0, is not closed under addition, hence not a ring.
If x5 = x3, then x2 = 1, so every element of that set is x, meaning that set has 1 element, and this is no good. If x5 = x9, then x4 = 1. The only element of the cyclic group of order 15 satisfying x4 = 1 is 1 itself, meaning that every element in that set is 1, again no good. But since at least one pair of the things in that set must be equal (otherwise we'd have 3 different things in that set, but we're supposed to only have 2), it must be that x3 = x9. It must also be that x isn't 1, otherwise we'd only have 1 element in that set, but we need 2. So the order of x divides 6, but isn't 1. The only choice, given that the group is cyclic of order 15, is that the order of x is 3. So there are at most 3 distinct powers of x, namely x, x2, and 1. x13 = x, x26 = x24+2 = x2 and x36 = 1, so the set in question does get all the powers of x, i.e. all 3 of them, and that's your answer.
Condition I says s = -s. Well:

-s = (-s)2 = s2 = s

II is easy:

(s+t)2 = s+t = s2 + t2

For III, apply I and II:

(s+t)2 = s2 + t2 (by II)
s2 + st + ts + t2 = s2 + t2 (distributivity)
st + ts = 0 (cancellation)
st = -ts (cancellation)
st = ts (I)

So S is commutative.
p4-1 = (p-1)(p+1)(p2+1). p is odd, so one of {p-1, p+1} is an odd multiple of 2 and the other is a multiple of 4. p2 + 1 is even, so the product is divisible by 16. p is congruent to either 1, 2, 3, or 4 (mod 5). If it's 1 or 4 (mod 5), then (p-1) or (p+1) is a multiple of 5, respectively. Otherwise, it's easily checked that p2 + 1 is a multiple of 5. So 5 also divides the product. So 5x16 = 80 divides the product. Also, one of {p-1, p+1} must be divisible by 3, otherwise p is divisible by 3, but no prime greater than 5 is so. So the 80x3 = 240 divides the product. What were the other options given? It might be easier to rule them out rather than proving that no greater factor exists.
They intersect once on the left of the y axis, and twice on the right. It should be easy to see that they intersect once, and only once, on the left. You also know that eventually, 2x >> x12. But for x = 2, 2x < x2, so at some point on the right, 2x must go below, giving you two more points of intersection. Basic understanding of these functions should assure you that there are no more than these 3.
The first follows from the extreme value theorem. The second follows because [0,1] is compact, so since f is continuous it's uniformly continuous. The last condition is that it's Lipschitz, but an arbitrary continuous real-valued function need not be Lipschitz. I guess the "classic" example is x1/2. For y = 0, |f(x)|/|x| = f(x)/x = x-1/2 is unbounded, so there is no E to bound it.

Last edited: Nov 3, 2006