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mathnoob · Joined: 06 Apr 2008 Posts: 5

Hi ALL, I am so glad to find such a forum and meet all you guys here! Please just help for the below problems i met。 Thank you in advance!

1）cos(97x)=x got how many solusions?

2）V,W are both 4-dim subspaces of a 7-dim space then the dimension of V intersect W can NOT be:
0,1,2,3,4

3）the point set of all the point within unit square where at least one of the coordinate of the point is irrational. This set is compact?connected?

4）find the integral of cos(t)+sqt(1+t^2)*[sin(t)]^3*[cos(t)]^t from -pi/4 to pi/4

5) how many 2*2 matrices on a q-element field are invertible（Answer：q^4-q^3-q^2+q but why??)

lime · Joined: 04 Dec 2007 Posts: 42

1. Apparently, there are no solutions for x>1 and x<-1.
Since 97~=30,8Pi,
the value of cos(97)<0
and graphically it also decreases in this point (that is also possible to check using first derivative test). Using the fact that cos(x) is the even function we could sketch out graphs for both y=cos(97x) and y=x and evaluate the number of intersetion points, which would be equal to 61.

2. If you try to put 4 balls of one color and 4 balls of another color in 7 bins, that each bin does not contain two balls of the same color, in that case at least one bin would contain two balls of different colors. Therefore, 0 is the answer.

3. Let points of this set S be the points of the square with vertices (0,0), (1,0), (1,1), (0,1).
Consider subset A of this set that
A={(x,0), where x is irrational}.
Apparently, the set A is both open and closed.
Therefore, S is disconnected. (Warning! Mistake! See further posts for more information)
Consider set family B(n) of such points in squares with side length 1-1/n (n>1). If we "insert" such "squares" in our square and add the set C of border points:
C={(x,0),(x,1),(0,y),(1,y), x,y are irrational}
we would have the open covering of our set S that has no finite subcovering.
Therefore, S is not compact.

4. I couldn't solve this integral so far. Wink

Please confirm whether it looks like this one.

5. To get the answer we should actually calculate the number of all possible 2x2 non-invertible matrixes in field q.

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mathnoob · Joined: 06 Apr 2008 Posts: 5

Hey Lime, your answers are amazing!! Thank you!

For the integral one, I am so sorry that I have a typo in my initial post, the correct one should be:

cos(t)+sqt(1+t^2)*[sin(t)]^3*[cos(t)]^3

Looking forward your reply!!

lime · Joined: 04 Dec 2007 Posts: 42

Thank you for questions.

4. Apparently, the second integral was included in order to obfuscate us. Indeed,
sqrt(1+x^2) is even function
sin(x)^3 is odd function
cos(x)^3 is even function.
Hence, integral function would be:
even*odd*even = odd function.
According to the integration interval the integral value would be equal to zero. Therefore, it would be enough to calculate only the first integral which is equal to sqrt(2).

lime · Joined: 04 Dec 2007 Posts: 42

I got very interesting reply concerning problem 3, proposing that:
"set S is connected still there is a path between any two points of it. Hence, it is path-connected. Hence, it is connected".
Nevertheless, I think that cannot be true. For instance, between two points
X(sqrt(2)/3, 0) and Y(sqrt(2)/2, 0) there is no way to create a path all the points of which will be in S. There always be a point on the way with rational coordinates.

Also, if you feel not confident with proving disconnectedness with showing that the set is both open and closed, you can figure out that it can be covered by two disjoint squares with vertices:
(0,0) (0,1) (1/2,1) (1/2,0)
(1/2,0) (1/2,1) (1,1) (1,0)
where the border x=1/2 does not belong to these squares. Therefore, S is disconnected.

You guys, correct me please if I'm wrong.

mathsubboy · Joined: 20 Jul 2008 Posts: 10

er, let me see, I think the two points X(sqrt(2)/3, 0) and Y(sqrt(2)/2, 0) can be connected by the following three segments:
seg I: X(sqrt(2)/3, 0)->A(sqrt(2)/3, sqrt(2)/3)
seg II: A(sqrt(2)/3, sqrt(2)/3)->B(sqrt(2)/2, sqrt(2)/3)
seg III: B(sqrt(2)/2, sqrt(2)/3)->Y(sqrt(2)/2, 0)

As for the proof, maybe, I think it is not fit into the definition.

Any correction is welcome!

lime · Joined: 04 Dec 2007 Posts: 42

I see I made mistake. Covering with these two squares

mathsubboy · Joined: 20 Jul 2008 Posts: 10

In fact, I can just prove the set A is closed relative to S.