8767 #30 #49 #56 #63

Forum for the GRE subject test in mathematics.
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8767 #30 #49 #56 #63

Post by chival » Wed Nov 04, 2009 1:42 am

Hey everyone,

i'm 4 days away from taking the test, and still having some problems with GR8767:

all i get is the Integral = f^2(x)/2|(a->b), don't know why it's "necessarily zero"

Answer is 35, but i created a group of Z28, and a subgroup {0,4,8,12,16,20,24}, i don't know where is wrong with this example.

seriously jz cant get it, i chose B

it says the inverse integral is on (0,1), but what if at some point x0 that f(x0)<0?
doesn't that make f-1(y) on (f(x0), 1)?

ANY advice will be appreciated, thanks a lot.

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Joined: Fri Oct 30, 2009 7:58 pm

Re: 8767 #30 #49 #56 #63

Post by mrb » Wed Nov 04, 2009 4:59 am

#30: I didn't actually try to compute the integral normally. Just look at or think about the graph of f. It is symmetric reflected about the peak of the semicircle. And wherever you have a given value for f'(x) on the left hand side, if y is the corresponding point on the other side, then f'(y) = -f'(x). So it's easy to see that the integral of the two sides cancel each other out. In other words, f' is positive on the left side of the graph and negative on the right side.

#49: in Z_28, 14 is its own inverse, so it doesn't meet the criterion in the problem statement.

#56: Yeah this should just be a fairly simple Taylor polynomial thing, but I kept screwing it up and I'm not sure I ever actually got the right answer...

#63: you can't have f(x0) < 0. Think about what would happen... you would have f(x0) < 0, and since f is strictly decreasing, it would stay smaller than f(x0). But then since that part of the graph would never approach 0, the integral over that part would be negative infinity, which contradicts the problem statement.

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Re: 8767 #30 #49 #56 #63

Post by chival » Wed Nov 04, 2009 7:54 am

Thanks so much!

#30: sorry i totally forgot about the hint on the top of the page "questions 28-30 are based on the following information - about f(x) being a semicircle"

#49 i thought it was referring to the SUBGROUP that contains no element of its own inverse

#56 well.. i'll do the thing again later

#63 oh yah!!! you got me on this one!!! it's so right f(x) cant be negative

i guess i should be more careful, no wonder i cant find other ppl asking questions about these problems :P

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Re: 8767 #30 #49 #56 #63

Post by mirkin » Fri Apr 09, 2010 1:22 pm

RE: #56

You simply take f(x) = root of x
and then you find the difference using the taylor series: P(1.01) = (f'''(c)*(0.01)^3)/3! - where we use 3 for the derivative, power, and factorial because the polynomial is given by a power of 2. The difference b/w f(1.01)-p(1.01) then = P(1.01) where we use c=1 (since 1< c <1.01)

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