9367 14, 24, 31, 43

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

9367 14, 24, 31, 43

Postby moo5003 » Fri Oct 17, 2008 8:48 pm

So I just took the 9367 today and had questions on 4 problems.

14. At a 15 percent annual inflation rate, the value of the dollar would decrease by approximately one-half every 5 years. At this inflation rate, in approximately how many years would the dollar be worth 1/1,000,000 of its present value?

A) 25
B) 50
C) 75
D) 100
E) 125
Answer (D)

So, I put D down when taking this but the way I got the answer was a little less then desired. I approximated log_2(100) ~ 20 which took me awhile. I was wondering how you guys did this. I remember there being a function to tell you this answer but I forget the exact expression.

24. If A and B are events in proabability space such that 0 < P(A) = P(B) = P(A Intersect B) < 1, which of the following CANNOT be true?

A) A and B are independent.
B) A is a proper subset of B.
C) A != B
D) A intersect B = A union B
E) P(A)P(B) < P(A Intersect B)
Answer (A)

I got this wrong, I have never taken a class dealing with proabability spaces so I may just not have enough tools at my disposale. But if anyone could explain this to me I would appreciate it.

31. If
f(x) =
Root( 1 - x^2) for 0</= x </=1
x-1 for 1 < x </= 2


then the Integral from 0 to 2 of f(x) dx is?

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

So, I started by spliting the integral from 0 to 1 and then 1 to 2. My problem is that I was unsure how to integrate Root(1-x^2) I used integration by parts with little to no help.

43. Let n be an integer greater than 1. Which of the following conditions guarantee that the equation x^n = Sum from i=0 to n-1 of a_i x^i has at least one root int he interval (0,1)?

I. a_0 > 0 and Sum i=0 to n-1 of a_i < 1
II. a_0 > 0 and Sum i=0 to n-1 of a_i > 1
III. a_0 < 0 and Sum i=0 to n-1 of a_i > 1

A) None
B) I Only
C) II Only
D) III Only
E) I and III
Answer (E)

So, I was sure that this problem had to do with the two expressions

Sum of the roots = (-1)^n * -a_(n-1)
Product of the roots = a_0

But I'm unsure how their conditions imply the root is between 0 and 1.

EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Postby EmLasker » Fri Oct 17, 2008 9:07 pm

Here's an answer for 24:

To say that the two events A and B are independent means that the occurrence of event A does not affect the probability of event B - or that the probability of (A intersect B) = P(A)P(B). As an example consider flipping a coin - regardless of what you got on the first flip the probability of getting heads is 1/2 on the second flip. Letting A be heads an B be tails we see that for two coin tosses, the probability of getting heads and then tails (i.e. the probability of A intersect B) is (1/2)(1/2) = (1/4) as I'm sure you intuitively know. In this problem if we assume A and B are independent then:
P(A intersect B) = P(A)P(B), but P(A intersect B) = P(A) and P(B) = P(A), so that P(A) = P(A)P(A) which can only occur if P(A) = 1 or 0, which contradicts the assumptions stated in the problem.

EmLasker
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

Postby EmLasker » Fri Oct 17, 2008 9:12 pm

For 31, to integrate sqrt(1-x^2) we use trig substitution: let x=sin(theta) so that dx=cos(theta); change your bounds from (0 to 1) to (0 to pi/2). Then using 1-sin^2 = cos^2 you just end up integrating cos^2(theta).

JcraigMSU
Posts: 13
Joined: Wed Aug 06, 2008 9:25 am

Postby JcraigMSU » Fri Oct 17, 2008 10:18 pm

I think integrating cos^2(theta) is not much easier, however
2cos^2(theta)-1=cos(2*theta) --> cos^2(theta) = [cos(2*theta)-1]/2 which we can integrate.

prce
Posts: 9
Joined: Mon Mar 24, 2008 3:02 pm

Postby prce » Sat Oct 18, 2008 12:40 am

For 14:

2^10=1024~1000 so 2^20 ~ 1000000

Kastro
Posts: 10
Joined: Mon Oct 13, 2008 10:36 pm

Postby Kastro » Sat Oct 18, 2008 1:59 am

Edit: Reading mistake. See solution below. :)

42.

x^n = Sum from i=0 to n-1 of a_i x^i

x^n = a_0 + a_1*x + a_2*x^2 + ... + a_(n-1)*x^(n-1)

Since x^n is continuous, we know by the intermediate value theorem that if x^n is negative at x=0 (or x=1) and positive at x=1 (or x=0), there must exist a 0 in (0,1).

Consider x = 0

Then we have x^n = a_0

Consider x = 1

Then we have x^n = a_0 + a_1 + ... + a_(n-1)

The result is clear.
Last edited by Kastro on Wed Oct 29, 2008 8:54 pm, edited 1 time in total.

Wonderacle
Posts: 5
Joined: Wed Oct 01, 2008 4:26 pm

Postby Wonderacle » Sun Oct 19, 2008 9:53 pm

31

integrating can definitely bring you to the answer, however, it will be easier if you think in this way,

let y=(1-x^2)^(1/2), the integrand of the first potion, we get
x^2+y^2=1,
this is the unit circle,
therefore, the first potion of the integral is actually the potion of unit circle at the 1st quadrant,
...

i think you can handle the rest.

octave145
Posts: 4
Joined: Thu Oct 23, 2008 11:27 pm

Postby octave145 » Thu Oct 23, 2008 11:42 pm

problem 43.

Maybe I am missing something here.
The answer is supposed to be I and III.
For I:
let n=2 , a_0 = 1/3 , a_1 = 1/3
Then a_0 > 0 and the sum < 1 , satisfying the conditions.
The resulting equation is : 1/3 + (1/3)*x
It has a root a x=-1 which is not in the interval (0,1)
So this doesn't seem to guarantee a root in the interval (0,1)
but this is one of the correct answers.

For III: Yeah, using the intermediate value theorem makes sense here.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Tue Oct 28, 2008 6:30 pm

I'm still a little confused about problem 43.

if x = 0 then x^n-.... - a_0 = 0
if x = 1 then x^n.... - a_0 = 1 + Sum of the a_i's

Thus we can show that this has a 0 if: III

I dont understand how they get I since if a_0 > 0 and the sum of a_i's is between 0 and 1 then it will not always take a 0 in (0,1) yet it is an answer...

octave145
Posts: 4
Joined: Thu Oct 23, 2008 11:27 pm

Postby octave145 » Tue Oct 28, 2008 8:18 pm

I see what I was missing now with problem 43 : the ability to read the question carefully! :oops: I hate it when that happens.

If we look at the equation y(x)=x^n - ( the sum )

At x=0, y(0) = -a_0
At x=1, y(1) = 1 - ( the sum of a_i )

So for I: a_0 > 0 implies y(0) < 0 and the sum < 1 implies
y(1) > 0 . We can then apply the intermediate value theorem.

For II: a_0 > implies y(0) < 0 and the sum >1 implies y(1) < 0.
So there is no guarantee here.

For III: a_0 < 0 implies y(0) > 0 and the sum > 1 implies y(1) < 0.
Apply the intermediate value theorem.

So choice (E) is correct.

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Tue Oct 28, 2008 9:43 pm

Thanks for the clarification I understand the line of reasoning now, I was simplifying incorrectly for x=0 and x=1.




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