## GR1268 Complete Solutions!

Forum for the GRE subject test in mathematics.
DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### GR1268 Complete Solutions!

Hello mathematicians!

I have just finished writing up full solutions to the GR1268 questions. I hope they are of some help to you! Ivanjam
Posts: 60
Joined: Tue Mar 17, 2015 2:29 am

### Re: GR1268 Complete Solutions!

DMAshura wrote:Hello mathematicians!
I have just finished writing up full solutions to the GR1268 questions. I hope they are of some help to you! Great job! Thank you for the time, patience and hard work that you put into it!

Posts: 96
Joined: Fri Mar 27, 2015 6:42 pm

### Re: GR1268 Complete Solutions!

good job. I'd like to point out though that your solution to #6 is way too complicated. Just raise everything to the 6th power.

#13 seems pretty handwavy. This is a standard MVT problem: (f(3) - f(0))/(3-0) = (5 - f(0))/3 = f'(c) >= -1, so f(0) <= 8. (your example -x+8 shows this bound is tight.)

DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### Re: GR1268 Complete Solutions!

Nice solution for #6; that's significantly easier. I just remember that when I took the test, I had the graph of x^(1/x) fresh on my mind, having just tutored another calculus student who was analyzing it, so I knew its maximum was e^(1/e) offhand and the rest took me about a minute. Your solution is significantly more elegant, though. I'll also update #13 -- what you see there is the logic I followed, but MVT does make it more precise.

I've gone ahead and updated the pdf. The "thank you" at the beginning for improvements to elegance is for you. Please let me know if there are any other solutions that can be improved.

Thanks so much!

Posts: 96
Joined: Fri Mar 27, 2015 6:42 pm

### Re: GR1268 Complete Solutions!

Using calculus isn't a bad idea for #6 if it were something a bit more complex like order e^pi and pi^e. Then you could look at x/logx. But for integer stuff, it's overkill.

I didn't read all the solutions in too much detail, but an alternative for #1 is (once you have -3/2*sin(3x)/x) -9/2* sin(3x)/(3x) -> -9/2*1 since sinx/x -> 1. In calculus classes, we teach students not to "cheat" and use L'H since the limit sinx/x is what comes up in the derivatove of sinx and it's circular reasoning.

On #19, we both had the idea to approach along different paths through the origin, but converting to polar is somewhat overkill IMO. It's better to let x = 0, so z = y, and the limit is y^2/y^2 = 1, and then let x = y, so z = x + ix, and the limit is (2x^2)/(2ix^2) = 1/i.

Similarly, on #27, I just did it through repeated squaring (i.e., writing 10 in binary). (1+i)^2 = 2i, (1+i)^4 = (2i)^2 = -4, (1+i)^8 = (-4)^2 = 16, (1+i)^10 = 16*2i = 32i.

On #56: Off the top of my head, I know that 0 < logx/x <= 2/sqrt(x) for x > 1 (this is a good way to show logx/x -> 0 without using L'H). (The idea is 0 < logx/x < 1 since e^x >= x + 1 > x (which you can prove using MVT or Taylor's theorem), so replace x by sqrt(x) to see 0 < log(sqrtx)/sqrtx < 1, or 0 < 1/2log(x)/sqrt(x) < 1, or 0 < logx/x < 2/sqrt(x)). So logx <= 2sqrt(x).

(II) is false since \sum_1^n k^r = O(n^(r+1)) (this bound is tight) in general.

|sinx|/|x| <= 1, so |sinx| <= |x|
|sinx - x| <= |sinx| + |x| <= 2|x| <= 2|x|^3 for |x| > 1. Make the constant bigger if necessary for [-1, 1] since |sinx-x| attains a max on the compact set [-1, 1].

DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### Re: GR1268 Complete Solutions!

Good alternative for #1, although for the purposes of doing the GRE quickly, I prefer L'Hôpital's Rule as being a bit more algorithmic and "mindless" in this case. An interesting point, though, is that you can show that d/dx sin x = cos x without making reference to the derivative of sin x / x if you look at it geometrically: http://www.animations.physics.unsw.edu. ... gonometric

On #19, if you happen to think of approaching along the line x=y that's good; I figured the advantage of polar notation is that you sidestep the guess-and-check in figuring out which way to approach the axis from. For many people, I think, their first instinct is to try along the x-axis, and then along the y-axis; once you've found that both equal 1, you're either led to believe that the limit might be 1 if you're not careful, or you try another direction but have wasted time on the first two.

I originally thought of referencing Big-O notation for parts of #56, but I was trying to steer clear since f(x) in O(g(x)) only really says that g(x) EVENTUALLY grows the same as or faster than f(x), not that it's ALWAYS above f(x). But I suppose it's fine to reference it for (II), since we're disproving it. Also, do you know a good way to show that the sum from 1 to n of k^r is a polynomial of degree r+1 without referencing discrete calculus? (Although discrete calculus *IS* pretty cool... )

Ivanjam
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Joined: Tue Mar 17, 2015 2:29 am

### Re: GR1268 Complete Solutions!

#14: Simply let x=c. Then, the integral on the RHS becomes 0, and we obtain the desired expression 3c^5 + 96 = 0.
Last edited by Ivanjam on Thu Aug 06, 2015 4:34 am, edited 1 time in total.

DMAshura
Posts: 58
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### Re: GR1268 Complete Solutions!

Ivanjam: BEAUTIFUL!! I always felt like my method was roundabout, and I *knew* there had to be a simpler way to do #14, but I couldn't put my finger on it. This does the trick!

PDF has been updated, and I've thanked you and p-adic on the last page. Thank you so much!

Posts: 96
Joined: Fri Mar 27, 2015 6:42 pm

### Re: GR1268 Complete Solutions!

The polar stuff is good when you need to prove the limit exists, and yeah, you usually try y = 0, x = 0, then start trying stuff like y = x, y = x^2, etc. My issue with polar is that there's still something to show beyond just saying "well it depends on what path you take."

I don't know of a proof of the big-oh thing off hand, but it's a well-known fact, and writing out these 3 small proofs on the exam takes quite a bit of time, so my idea is to just use some intuition to figure it out more quickly. Of course, I had to have the fact that lnx/x < 2/sqrt(x) in the back of my mind, but certain real analysis teachers telling me a million times not to use L'H taught me to figure that out. I don't know if you can also show the constant will work starting at n=1 as a byproduct of the proof of the big-oh claim.

DMAshura
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### Re: GR1268 Complete Solutions!

For the polar form, what more is there to show? If the limit depends on the path you take, it's not unique, and doesn't exist.

You're absolutely right that writing out these proofs would take way too long on the test ... That's why I said at the very beginning that such a level of detail isn't really necessary. Just perform a few of the basic calculations, couple that with good intuition and well-known facts, and you'll be able to answer most questions decently quickly. I'm just also trying to justify these answers as much as possible for the purposes of this particular guide, if that makes sense.

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### Re: GR1268 Complete Solutions!

If I were writing that claim on an undergraduate homework assignment, I'd expect to lose some partial credit for not showing specific examples. That is all.

DMAshura
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### Re: GR1268 Complete Solutions!

Fair enough. Added that into the solution for more clarity. Ivanjam
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### Re: GR1268 Complete Solutions!

#30: As usual, we start with log(x) = cx^4. Solving for c yields c = log(x)/(x^4). Similarly, equating the derivatives leads to 1/x = 4cx^3, or c = 1/(4x^4). Comparing the two expressions for c, we conclude that log(x) = 1/4, or x = e^(1/4). Substituting the latter into either expression for c, we obtain c = 1/(4e).

DMAshura
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### Re: GR1268 Complete Solutions!

Nice, Ivanjam. You've got a knack for shortcuts. Ivanjam
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### Re: GR1268 Complete Solutions!

DMAshura wrote:Nice, Ivanjam. You've got a knack for shortcuts. ...Especially after someone has already done all the work and posted the solutions, but never during the actual exam. Last edited by Ivanjam on Thu Sep 10, 2015 6:32 am, edited 1 time in total.

lp
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### Re: GR1268 Complete Solutions!

Super well done on the solutions, mate! Went through them with a fine tooth comb. Very helpful and thorough. Thanks again!
BTW, in #19, the limit along pi/4 (or x=y), should be -1, not -i.

Great job!

DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### Re: GR1268 Complete Solutions!

Thanks for noticing that typo! It has been corrected.

I've corrected a few others little by little since my last post, so let me know if you find any others. Glad you found the solutions helpful, though!

Ivanjam
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Joined: Tue Mar 17, 2015 2:29 am

### Re: GR1268 Complete Solutions!

DMAshura, did you get a chance to check out Akina's solutions on Blogspot.com? To date, she has posted solutions to the first 24 problems: http://akinaikudo.blogspot.com/2015/

lp
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Joined: Sun Aug 23, 2015 7:28 am

### Re: GR1268 Complete Solutions!

Yes, your solutions are a life saver. I would have no idea how to correctly answer a solid 20 of the questions if it weren't for your help (bachelor's in engineering, so playing catch up (and mustard!) now.

One more small thing: on my version of the solutions from a couple weeks ago, the last bullet in 63 states: "If has A only positive elements and B has only positive elements, then A•B will have only negative elements."
I think it should say "...B has only negative elements...."

Any chance you can assist with GRE 9367 #57? See my post with that heading. Thanks again!

DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### Re: GR1268 Complete Solutions!

Ivanjam: No I haven't had a chance to check those out. I'll have to take a look sometime.

lp: I actually corrected that typo recently! I'll take a look at your other post as well and see what I can do.

fluentmundo
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### Re: GR1268 Complete Solutions!

I'm new to this forum and just discovered these solutions. Thanks for them. Here are some comments on other approaches, some of them significantly faster than yours or others mentioned in this thread.

For #24, no need to eliminate D before A and C. Every homogeneous system is consistent (eliminate A), and the set of solutions is a vector space, hence closed under addition (eliminate C).

For #31, you could just write a quick expression for det(A-lambda I) -- no need to simplify it fully -- and then test the given values to see whether they're roots. This might be slightly longer than your solution, but it's still quite fast.

For #44, no need to solve the ODE -- that's a time sink. Simply analyze the slope field when x<0. It helps first to look for the equilibria, where y'=0. This gives x-xy=0, so x=0 or y=1. Since y'=x(1-y) and y is initially less than 1, y' will be negative for x<0. So the solution will flow up to the equilibrium y=1 as x tends to negative infinity.

For #53, you don't need the Leibniz rule. We are told g(x) = int_0^x f(y)(y-x)dy. This is just

int_0^x yf(y)dy - x int_0^x f(y)dy

Now differentiate, using the fundamental theorem of calculus and the product rule. (Actually, now that I look at the question again, I see it does not state that f is continuous, so strictly speaking, the fundamental theorem doesn't apply, nor does Leibniz's rule. I'm not sure yet whether the test-writers simply made a mistake, or whether the problem is vastly more interesting than it seems.)

For #56, statement III asks whether there's a constant C so that |sin(x)-x|<=C|x^3| for all x. Your solution is to think geometrically. p-adic's, earlier in this thread, uses a bound for x>1 and then modifies it for x<1. But the easiest and most straightforward approach is simply to note that |sin(x)-x| is the absolute value of the remainder of the first-order Taylor approximation of sine. Since the Taylor series for sine is alternating, the absolute value of the remainder is bounded by the absolute value of the next term, |x^3/6|. This immediately gives a C, namely 1/6, that works for all values of x. The argument requires no case analysis.

For #57, statement II, you appeal to sin(1/x) as a counterexample. But the oscillatory behavior is overkill here. The function 1/x works just as well; it is continuous on (0,1) but the f(x_n) diverge to infinity.
Last edited by fluentmundo on Wed Nov 11, 2015 5:43 am, edited 1 time in total.

DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### Re: GR1268 Complete Solutions!

fluentmundo wrote:I'm new to this forum and just discovered these solutions. Thanks for them. Here are some comments on other approaches, some of them significantly faster than yours or others mentioned in this thread.

For #24, no need to eliminate D before A and C. Every homogeneous system is consistent (eliminate A), and the set of solutions is a vector space, hence closed under addition (eliminate C).

For #31, you could just write a quick expression for det(A-lambda I) -- no need to simplify it fully -- and then test the given values to see whether they're roots. This might be slightly longer than your solution, but it's still quite fast.

For #44, no need to solve the ODE -- that's a time sink. Simply analyze the slope field when x<0. It helps first to look for the equilibria, where y'=0. This gives x-xy=0, so x=0 or y=1. Since y'=x(1-y) and y is initially less than 1, y' will be negative for x<0. So the solution will flow up to the equilibrium y=1 as x tends to negative infinity.

For #53, you don't need the Leibniz rule. We are told g(x) = int_0^x f(y)(y-x)dy. This is just

int_0^x yf(y)dy - x int_0^x f(y)dy

Now differentiate, using the fundamental theorem of calculus and the product rule. (Actually, now that I look at the question again, I see it does not state that f is continuous, so strictly speaking, the fundamental theorem doesn't apply, nor does Leibniz's rule. I'm not sure yet whether the test-writers simply made a mistake, or whether the problem is vastly more interesting than it seems.)

For #56, statement III asks whether there's a constant C so that |sin(x)-x|<=C|x^3| for all x. Your solution is to think geometrically. p-adic's, earlier in this thread, uses a bound for x>1 and then modifies it for x<1. But the easiest and most straightforward approach is simply to note that |sin(x)-x| is the absolute value of the remainder of the first-order Taylor approximation of sine. Since the Taylor series for sine is alternating, the absolute value of the remainder is bounded by the absolute value of the next term, |x^3/6|. This immediately gives a C, namely 1/6, that works for all values of x. The argument requires no case analysis.

For #57, statement II, you appeal to sin(1/x) as a counterexample. But the oscillatory behavior is overkill here. The function 1/x works just as well; it is continuous on (0,1) but the f(x_n) diverge to infinity.
Beautiful. I love some of these simplifications, especially on #44 and #53. I'll have to incorporate some of them (and credit you for your contributions of course)! Thank you! tortoise47
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### Re: GR1268 Complete Solutions!

Hey DMAshura,

#22 - how do you eliminate the dependence on z? I had a hard time visualizing a solid with the 4 equations together, much less integrating over it.

DMAshura
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### Re: GR1268 Complete Solutions!

tortoise47 wrote:Hey DMAshura,

#22 - how do you eliminate the dependence on z? I had a hard time visualizing a solid with the 4 equations together, much less integrating over it.
There really isn't a "dependence on z"; rather, z is being visualized such that it depends on x and y. Think of it kind of like finding the area of the curve bounded by x=-1, x=1, y=0, and y=x+3. When you graph these functions, you see that this is just the area under the curve y=x+3, on the interval [-1,1]. Similarly, the solid being visualized for #22 is the volume under the surface z = y+3, over the region bound by y=x² and y=2-x². Does that help?

rlightec
Posts: 25
Joined: Sun Aug 23, 2015 10:39 pm

### Re: GR1268 Complete Solutions!

Hello DMAshura,

I am ashamed to ask this as it requires a lot of work, but... is there any chance you can write solutions for the 2005 test? I heard that 2005 and 2012 test are quite similar to the current version of the exam whereas pre-2005 practice tests are easier and not very reflective of current administered tests.

Ivanjam
Posts: 60
Joined: Tue Mar 17, 2015 2:29 am

### Re: GR1268 Complete Solutions!

rlightec wrote:Hello DMAshura,
I am ashamed to ask this as it requires a lot of work, but... is there any chance you can write solutions for the 2005 test? I heard that 2005 and 2012 test are quite similar to the current version of the exam whereas pre-2005 practice tests are easier and not very reflective of current administered tests.
Charles Rambo has the solutions to the 2005 exam on his website for free: http://rambotutoring.com/GR0568-solutions.pdf

DMAshura
Posts: 58
Joined: Sat Jul 05, 2014 7:53 am

### Re: GR1268 Complete Solutions!

rlightec wrote:Hello DMAshura,

I am ashamed to ask this as it requires a lot of work, but... is there any chance you can write solutions for the 2005 test? I heard that 2005 and 2012 test are quite similar to the current version of the exam whereas pre-2005 practice tests are easier and not very reflective of current administered tests.
I'll be publishing my own solutions as well soon. rlightec
Posts: 25
Joined: Sun Aug 23, 2015 10:39 pm

### Re: GR1268 Complete Solutions!

Thank you both for your replies!