GR 0568 Q's: 8, 27, 55, 61

Forum for the GRE subject test in mathematics.
PieceOfPi
Posts: 25
Joined: Mon Aug 02, 2010 3:16 pm

GR 0568 Q's: 8, 27, 55, 61

Postby PieceOfPi » Thu Sep 16, 2010 7:40 pm

Hi,

I have a few more questions from GR 0568 that I would greatly appreciate if anyone could answer:

(I highlighted the correct answer in green.)

#8. What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?

(A) 1/2 (B) 1 (C) sqrt(2), (D) pi, (E) (1 + sqrt(2))/4

My guess: Since I would get an isosceles triangle, maybe there is some kind of identity that I can use?

#27 Consider the two planes x + 3y - 2z = 7 and 2x + y - 3z = 0 in \mathbb{R}^3. Which of the following sets is the intersection of these planes?

(A) Empty set
(B) \{(0, 3, 1)\}
(C) \{(x, y, z): x = t, y = 3t, z = 7 - 2t, t \in \mathbb{R}^3\}
(D) \{(x, y, z): x = 7t, y = 3 + t, z = 1 + 5t, t \in \mathbb{R}^3\}
(E) \{(x, y, z): x - 2y - z = -7\}

My guess: Since there should be an infinitely many solutions and those planes are not parallel, the options (A), (B), and (E) are eliminated. Somehow, though, I'm stuck on finding a line that would work for both equations.

Also, why do I not get a right answer by setting those two equations equal? (It seems like I'm forgetting something basic from vector calculus, but I haven't really found a satisfactory answer to this question yet...)

EDIT:Oh wait, this one was actually pretty easy. All I need to do was find one point (which was given in the answer for (B)), and then take cross product of those two normal vectors for those planes (i.e. (1, 3, -2) x (2, 1, -3)), and that should give me the equations of the line given in D!

#55 For how many positive integers k does the ordinary decimal representation of the integer k! end in exactly 99 zeros?

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

My guess: Not much clue on this one.

#61 Which of the following sets has the greatest cardinality?

(A) \mathbb{R}
(B) The set of all functions from \mathbb{Z} to \mathbb{Z}.
(C) The set of all functions from \mathbb{R} to \{0, 1\}.
(D) The set of all finite subsets of \mathbb{R}.
(E) The set of all polynomials with coefficients in \mathbb{R}.

My guess: Not so sure about this one either. Except that I know I can eliminate (B) because I think (B) is countable. How do I even find the cardinality of functions, anyway?

-----

That's it for now. If anyone can even give me a hint on just one of these questions above, I would be happy to hear it from you.

Thanks!

PP

Chapel
Posts: 59
Joined: Mon Sep 13, 2010 11:25 pm

Re: GR 0568 Q's: 8, 27, 55, 61

Postby Chapel » Thu Sep 16, 2010 11:15 pm

For 61, remember that you are not looking at the cardinality of the function (because that doesn't make any sense), rather you are looking for the cardinality of the set of functions. In set theory it often helps to think of functions as ordered pairs (x, f(x)).

For a further hint, read the Wikipedia article on "Function Space".

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: GR 0568 Q's: 8, 27, 55, 61

Postby owlpride » Thu Sep 16, 2010 11:20 pm

A couple of hints:

#8: The area of a triangle = 1/2 * (length of side A)*(length of side B)*(sine of angle between A and B). If you need to guess, you can immediately eliminate B, C and D because the values are too large.

#55: Count the number of 2s and 5s in the prime factorization.

#61: The power set of the real numbers has cardinality strictly greater than the cardinality of the reals. (The remaining answer choices have the same cardinality as the reals.)

PieceOfPi
Posts: 25
Joined: Mon Aug 02, 2010 3:16 pm

Re: GR 0568 Q's: 8, 27, 55, 61

Postby PieceOfPi » Thu Sep 16, 2010 11:39 pm

Chapel wrote:For 61, remember that you are not looking at the cardinality of the function (because that doesn't make any sense), rather you are looking for the cardinality of the set of functions. In set theory it often helps to think of functions as ordered pairs (x, f(x)).

For a further hint, read the Wikipedia article on "Function Space".


Thanks! I see, so the cardinality of (B) is \mathbb{Z}^{\mathbb{Z}}, (C) is 2^{\mathbb{R}}, and since (D) is a proper subset of the power set of \mathbb{R}. So that eliminates everything but (C) and (E). (E) is a slightly interesting one because I guess we can think of this as n-tuples of real coefficients (a_0, a_1, ... , a_n), but not sure how to find the cardinality of this set...

Maybe I need to do a bit of review of set theory and cardinality?

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: GR 0568 Q's: 8, 27, 55, 61

Postby owlpride » Thu Sep 16, 2010 11:51 pm

Chapel wrote:Thanks! I see, so the cardinality of (B) is \mathbb{Z}^{\mathbb{Z}}, (C) is 2^{\mathbb{R}}, and since (D) is a proper subset of the power set of \mathbb{R}. So that eliminates everything but (C) and (E)

Be careful. A proper subset of a power set can still have the same cardinality as the power set.

Here's a "proper" sample argument that's a bit more involved.

(B) The set of all functions from Z to Z. This set is uncountable because it is possible to embed the real numbers into it. For example, you can index the positions in the decimal expansion of a real number and encode their values as function values. The number

99.123555

might correspond to the function

f(2)=9, f(1)=9, f(-1)=1, f(-2)=2, f(-3)=3, f(-4)=5, f(-5)=5, f(-6)=5, f(x)=0 everywhere else.

On the other hand, the following argument shows that the cardinality of the set of functions is not bigger than the cardinality of the reals by encoding each function as a real number. First, put the integers into bijective correspondence with the natural numbers and consider maps from N to N instead. (I don't want to encode signs.) Then write all natural numbers in base 9 without using the digit 0. (I want to use it as a delimiter). Then encode a function as a real number as follows:

0.f(1) 0 f(2) 0 f(3) 0 f(4) 0 f(5) 0 ...


Obviously, coming up with 5 arguments like this not an option on the real GRE. Here are a few rules of thumb:

- Cardinalities usually only increase one step at a time. It would be highly unlikely that the set of functions Z -> Z has cardinality greater than the reals.

- Finite lists and operations don't usually change infinite cardinalities. The set of finite subsets of an infinite set has the same cardinality as the set, but the set of all subsets (= the power set) does not. Similarly, the set of finite tuples of real numbers has the same cardinality as the set of real numbers (but allowing infinite tuples might increase the cardinality).

- Power sets are often encoded as indicator functions. (As in (C).)
Last edited by owlpride on Thu Sep 16, 2010 11:52 pm, edited 1 time in total.

PieceOfPi
Posts: 25
Joined: Mon Aug 02, 2010 3:16 pm

Re: GR 0568 Q's: 8, 27, 55, 61

Postby PieceOfPi » Thu Sep 16, 2010 11:51 pm

owlpride wrote:A couple of hints:

#8: The area of a triangle = 1/2 * (length of side A)*(length of side B)*(sine of angle between A and B). If you need to guess, you can immediately eliminate B, C and D because the values are too large.


Oh that's right! I totally forgot about the fact about cross product |A \times B| = |A||B|\sin\theta = area of a parallelogram (provided that 0 \leq \theta \leq 180).

#55: Count the number of 2s and 5s in the prime factorization.


Hm. So in order to have 99 consecutive 0's, I must have 99 2's and 5's, right? So if k! is such number and k is a multiple of 5, then (k+1)!, (k+2)!, (k+3)!, and (k+4)! also have 99 consecutive 0's because they will still have only 99 5's. And that's why there are five integers that satisfy this condition?

Thanks!

madrespek
Posts: 4
Joined: Tue Oct 05, 2010 1:52 am

Re: GR 0568 Q's: 8, 27, 55, 61

Postby madrespek » Tue Oct 05, 2010 1:54 am

hints
27) find normal vectors of the two planes and take the cross product. This vector must be parallel to the line that is the intersection of two planes

or you could just find two points on that line

soshdog
Posts: 4
Joined: Sat Oct 09, 2010 11:31 am

Re: GR 0568 Q's: 8, 27, 55, 61

Postby soshdog » Sat Oct 09, 2010 11:50 am

#8
You can set up a point on the circle in the first quadrant and label it (x1,y1), and your other point would be (-x1,y1) So the area would be 1/2*(2x)*y = xy. So actually you are trying to maximize f(x,y)=xy. You can notice that that areas of rectangles (since now our function can be seen as maximizing a rectangle of length x and width y) generally are maximized when they are squares (although not always, it depends on the problem). When x and y are equal is when they are 1/root2. Thus xy=1/2

You can also use lagrange multipliers to verify these results. Again you are trying to maximize f(x,y)=xy subject to the contraint x squared + y squared = 1. The gradient of f is <y,x>. Lambda times the gradient of the constraint is lambda<2x,2y>. Thus x=y. Thus x squared + x squared = 1. So 2*x squared = 1. Thus x squared = 1/2. Thus x=y=1/root2. Thus xy=1/2 (same as above)

Mathemagician
Posts: 10
Joined: Thu Aug 05, 2010 2:29 pm

Re: GR 0568 Q's: 8, 27, 55, 61

Postby Mathemagician » Fri Oct 22, 2010 12:47 am

#8 should be very quick. Your circle is radius 1. Say it's centered at (0, 0) and say one of its sides goes to (1, 0); the question is just where the third point (x, y) should be. This establishes that the base of your triangle is 1 and its height is y; surely y is maximized at (0, 1). 1/2*1*1 = 1/2.

dearcatanddog
Posts: 1
Joined: Wed Sep 03, 2014 10:56 pm

Re: GR 0568 Q's: 8, 27, 55, 61

Postby dearcatanddog » Thu Oct 02, 2014 7:52 am





Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: No registered users and 4 guests