GR0568 #9 #12

Forum for the GRE subject test in mathematics.
Post Reply
Posts: 10
Joined: Sat Sep 19, 2009 5:55 pm

GR0568 #9 #12

Post by sjin » Sun Sep 20, 2009 9:01 pm

J= Integration of sqrt(1/(1-X^4)) from 0 to 1
K= Integration of sqrt(1/(1+X^4)) from 0 to 1
L= Integration of sqrt(1/(1-X^8)) from 0 to 1

Which of the following is true for the definite integrals shown above?

Ans: J<L<1<K

I used the histogram to guess the relationship but is there any way to calculate the exact value?

Let A be a 2*2 matrix for which there is a constant k such that the sum of the entries in each row and each column is k. Which of the following must be an eigenvector of A?

Ans: (1 1)

I thought (1 0) and (0 1) are also appropriate, can any one help me clarify these questions?

Posts: 5
Joined: Sun Sep 13, 2009 2:00 pm

Re: GR0568 #9 #12

Post by Jefferythewind » Mon Sep 21, 2009 4:35 am


What histogram are you talking about?

as for #9 of course there is a way to calculate the exact values. You can use this:

int (a2 - x2)1/2 dx = (1/2) [a2 arcsin (x/a) + x (a2 - x2)1/2] + C

for J and L, and i'm not really sure if you can use it for K, but i am still reviewing these rules too, haha.

But i don't think it is necessary to calculate it, for me i just visualize the integral as the area under the curve for each, on the x/y axis one between 0 and 1. I think that is good enough for this problem, but of course you could do the integration out and get exact values, you just have to know the equations pretty well.

as for #12, i am still a little rusty on eigenvectors and eigen values.

Posts: 7
Joined: Mon Sep 21, 2009 7:31 am

Re: GR0568 #9 #12

Post by contrapositive » Mon Sep 21, 2009 9:07 am

For #12, let
$$A=\bigl( \begin{smallmatrix}
a&b\\ c&d
\end{smallmatrix} \bigr)$$
be the 2*2 matrix. Now a+b=k implies that b=k-a. Also, a+c=k implies that c=k-a. Finally, c+d=k implies that d=a, so our matrix A looks like
$$\bigl( \begin{smallmatrix}
a&k-a\\ k-a&a
\end{smallmatrix} \bigr)$$
Quickly multiplying it out with the given vectors, we can see that $$\underline{x}=\bigl( \begin{smallmatrix}
1\\ 1
\end{smallmatrix} \bigr)$$ is the only vector that satisfies the eigenvector equation $$A\underline{x}=\lambda\underline{x}$$,
in this case with eigenvalue

Posts: 7
Joined: Mon Sep 21, 2009 7:31 am

Re: GR0568 #9 #12

Post by contrapositive » Mon Sep 21, 2009 1:21 pm

For #9, it should be clear that over the open interval $$(0,1)$$, the following holds :

$$\[1-x^4 < 1-x^8 <1 <1+x^4\]$$.

This is simply because for any $$x\in (0,1)$$, we have that $$x^8 <x^4$$.

Since (over $$(0,1)$$) these functions are non-negative, we can take square roots of the inequality to get:

$$\[\sqrt{1-x^4} < \sqrt{1-x^8} < 1 < \sqrt{1+x^4}\]$$,

and then integrating we conclude that

$$\[\int_{0}^{1}\sqrt{1-x^4}\,dx < \int_{0}^{1}\sqrt{1-x^8}\,dx < 1 < \int_{0}^{1}\sqrt{1+x^4}\,dx\]$$.

Post Reply