How are you all preparing of the upcoming exams?
How are you all preparing of the upcoming exams?
Hello all,
With the mGRE fast approaching, I was wondering how you all are preparing for the exam? It would be great if some of you could share some tips or advice. Personally, I have already worked through 3 of the practice exams, as well as the Princeton Review book. I tried going through some Calculus books, but the problems seem rather trivial and not of the mGRE flavor.
Hope to hear from you all soon, and good luck studying.
With the mGRE fast approaching, I was wondering how you all are preparing for the exam? It would be great if some of you could share some tips or advice. Personally, I have already worked through 3 of the practice exams, as well as the Princeton Review book. I tried going through some Calculus books, but the problems seem rather trivial and not of the mGRE flavor.
Hope to hear from you all soon, and good luck studying.

 Posts: 13
 Joined: Fri May 11, 2018 4:28 am
Re: How are you all preparing of the upcoming exams?
Try to solve all problems in the Stewart's calculus book. If you solve them all quickly and understand them thoroughly, I believe calculus questions in mGRE would be a piece of cake (This is the only source I do calculus problems)
About Algebra and Topology, I do not know what to say. Hope someone would give a more comprehensive answer.
About Algebra and Topology, I do not know what to say. Hope someone would give a more comprehensive answer.
Re: How are you all preparing of the upcoming exams?
You can post i teresting problems and we can discuss more efficient solutions
Re: How are you all preparing of the upcoming exams?
I did a lot of problems from Stewart but they seem rather trivial so I stopped bothering. The mGRE questions seem to have a different flavor.chauchitrung1996 wrote:Try to solve all problems in the Stewart's calculus book. If you solve them all quickly and understand them thoroughly, I believe calculus questions in mGRE would be a piece of cake (This is the only source I do calculus problems)
About Algebra and Topology, I do not know what to say. Hope someone would give a more comprehensive answer.
Re: How are you all preparing of the upcoming exams?
You should pay more attention to those. There are very interesting problems here or there
Re: How are you all preparing of the upcoming exams?
I suppose I should clarify, which version of Stewart are you all using exactly?DDswife wrote:You should pay more attention to those. There are very interesting problems here or there
Re: How are you all preparing of the upcoming exams?
I don’t know. Any, I guess. I am a tutor. And students come (once in while) with very interesting questions
Re: How are you all preparing of the upcoming exams?
I solved a while ago this exercise, and I think that it’s GRE like
f(x) = x^2 + a x + 1
Find all possible a such that f(x) >= 0 for all x, x in R
a) (oo,2]
b) [2,2]
c) [2,oo)
d) [0,oo)
Enjoy
f(x) = x^2 + a x + 1
Find all possible a such that f(x) >= 0 for all x, x in R
a) (oo,2]
b) [2,2]
c) [2,oo)
d) [0,oo)
Enjoy
Re: How are you all preparing of the upcoming exams?
This seems very GRE like, I believe the answer should be C. But what is the most efficient method in achieving the solution?DDswife wrote:I solved a while ago this exercise, and I think that it’s GRE like
f(x) = x^2 + a x + 1
Find all possible a such that f(x) >= 0 for all x, x in R
a) (oo,2]
b) [2,2]
c) [2,oo)
d) [0,oo)
Enjoy
Re: How are you all preparing of the upcoming exams?
I think that the most efficient method is using the options, Try 3 and 3, 2 and 2, eventually, 0,but I would dismiss 0 just by inspection since is part of all the options. It won’t be easy to do that without a calculator, but test points and see.
Last edited by DDswife on Sun Jul 29, 2018 9:16 pm, edited 2 times in total.
Re: How are you all preparing of the upcoming exams?
Alternatively, think of the definition of abs, and where the vertex is.
I mea: you have a piecewise function whose 2 “sides” are piece of quadratics. See if the vertex are included or not
Check where the roots are 0.
Give a try, and tell me what you finally did. You might think of other ways to solve it.
This is what I do when I play with these little problems. I try to find different ways, and see what is the best way to go. I have never taking the exam. I don’t know if I ever will. But I have learnt a lot out of doing that. Whatever you find out today will be useful if you have a similar problem during the exam
I mea: you have a piecewise function whose 2 “sides” are piece of quadratics. See if the vertex are included or not
Check where the roots are 0.
Give a try, and tell me what you finally did. You might think of other ways to solve it.
This is what I do when I play with these little problems. I try to find different ways, and see what is the best way to go. I have never taking the exam. I don’t know if I ever will. But I have learnt a lot out of doing that. Whatever you find out today will be useful if you have a similar problem during the exam
Re: How are you all preparing of the upcoming exams?
If you womder where I got this question from, it was in a Chinese website. It is in Chinese, of course. But I was able to follow the Math to understand what it meant. This is school level, according to what a Chinese frind of mine, who was the one that sent me the problems, told me. I believe that this is why they do good in the test. They are used to solve these weird questions since they are young
Last edited by DDswife on Mon Jul 30, 2018 12:10 am, edited 1 time in total.
Re: How are you all preparing of the upcoming exams?
Thanks for all the posts. Do you have any more examples of such problems? They seem like great practice.DDswife wrote:If you womder where I got this question from, it was in a Chinese website. It is in Chinese, of course. But I was able to follow the Math to understand what it meant. This is school level, according to what a Chinese frind of mine, who was the one that sent me the problems, told me. I believe that this is why they do good in the test. They are used to solve these weird questions since they are young,
Re: How are you all preparing of the upcoming exams?
http://m.ishare.iask.sina.com.cn/f/63995575.html
I don’t understand what most of them mean, but you can guess and try to solve some of them
I don’t understand what most of them mean, but you can guess and try to solve some of them
Re: How are you all preparing of the upcoming exams?
I think the best way to solve this question is to do the relatively simple computations. There's a small catch in the question.
Assume that f(x) is greater than or equal to zero.
Case I: x is >= 0. In this case, f(x) = x^2 + ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. So, D <= 0 iff a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) < 0 (I have omitted the denominator) if and only if sqrt(a^2  4) < a. We must have a is greater than 0 over here, and by squaring, it is clear that the inequality is true (the other root is clearly negative). So, the last case is true if a is positive. So, a must be between 2 and infinity (inclusive). We must include all possibilities since f(x) >= 0 in 3 cases!
Case I: x is <= 0. In this case, f(x) = x^2  ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. Again, a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if a < sqrt(a^2  4). If a is positive, squaring, we see that this isn't true. If a is negative, this is always true, as sqrt(a^2  4) > 0. Considering the second root, a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if  a > sqrt(a^2  4) > 0. Again, a can't be positive. If a is negative, can square both sides, and see the inequality is true. So, a must be between  infinity and 2 (inclusive).
Both cases will be simultaneously true iff a is in [2,2].
Please correct me if I am wrong.
It's easy to ignore the subcase that double roots exists, but are in half in which we're not interested. This is what makes it to be an annoying mGRE like problem. On than that, we simply have to play with the cases, and use the basics of logic, which makes it a good problem to give in a precalculus course.
I haven't done it by playing with the answers by first trying to plug in 2, 2 (and 0). I guess the difficult thing in this approach is to figure out the behavior if a > 2 or a < 2, without doing any computations. I find it very hard, and that also takes a minute or two. It's better to write down the problem, but it's easy to screw it up if one wants to solve it in 2ish minutes. I screwed it up before by not considering the subcase in each case that I mentioned in the previous paragraph.
Assume that f(x) is greater than or equal to zero.
Case I: x is >= 0. In this case, f(x) = x^2 + ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. So, D <= 0 iff a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) < 0 (I have omitted the denominator) if and only if sqrt(a^2  4) < a. We must have a is greater than 0 over here, and by squaring, it is clear that the inequality is true (the other root is clearly negative). So, the last case is true if a is positive. So, a must be between 2 and infinity (inclusive). We must include all possibilities since f(x) >= 0 in 3 cases!
Case I: x is <= 0. In this case, f(x) = x^2  ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. Again, a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if a < sqrt(a^2  4). If a is positive, squaring, we see that this isn't true. If a is negative, this is always true, as sqrt(a^2  4) > 0. Considering the second root, a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if  a > sqrt(a^2  4) > 0. Again, a can't be positive. If a is negative, can square both sides, and see the inequality is true. So, a must be between  infinity and 2 (inclusive).
Both cases will be simultaneously true iff a is in [2,2].
Please correct me if I am wrong.
It's easy to ignore the subcase that double roots exists, but are in half in which we're not interested. This is what makes it to be an annoying mGRE like problem. On than that, we simply have to play with the cases, and use the basics of logic, which makes it a good problem to give in a precalculus course.
I haven't done it by playing with the answers by first trying to plug in 2, 2 (and 0). I guess the difficult thing in this approach is to figure out the behavior if a > 2 or a < 2, without doing any computations. I find it very hard, and that also takes a minute or two. It's better to write down the problem, but it's easy to screw it up if one wants to solve it in 2ish minutes. I screwed it up before by not considering the subcase in each case that I mentioned in the previous paragraph.
Re: How are you all preparing of the upcoming exams?
If a > 0, f(x) will always be positive, here's a rough very informal proof:Junaid456 wrote:I think the best way to solve this question is to do the relatively simple computations. There's a small catch in the question.
Assume that f(x) is greater than or equal to zero.
Case I: x is >= 0. In this case, f(x) = x^2 + ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. So, D <= 0 iff a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) < 0 (I have omitted the denominator) if and only if sqrt(a^2  4) < a. We must have a is greater than 0 over here, and by squaring, it is clear that the inequality is true (the other root is clearly negative). So, the last case is true if a is positive. So, a must be between 2 and infinity (inclusive). We must include all possibilities since f(x) >= 0 in 3 cases!
Case I: x is <= 0. In this case, f(x) = x^2  ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. Again, a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if a < sqrt(a^2  4). If a is positive, squaring, we see that this isn't true. If a is negative, this is always true, as sqrt(a^2  4) > 0. Considering the second root, a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if  a > sqrt(a^2  4) > 0. Again, a can't be positive. If a is negative, can square both sides, and see the inequality is true. So, a must be between  infinity and 2 (inclusive).
Both cases will be simultaneously true iff a is in [2,2].
Please correct me if I am wrong.
It's easy to ignore the subcase that double roots exists, but are in half in which we're not interested. This is what makes it to be an annoying mGRE like problem. On than that, we simply have to play with the cases, and use the basics of logic, which makes it a good problem to give in a precalculus course.
I haven't done it by playing with the answers by first trying to plug in 2, 2 (and 0). I guess the difficult thing in this approach is to figure out the behavior if a > 2 or a < 2, without doing any computations. I find it very hard, and that also takes a minute or two. It's better to write down the problem, but it's easy to screw it up if one wants to solve it in 2ish minutes. I screwed it up before by not considering the subcase in each case that I mentioned in the previous paragraph.
x^2 > 0 for all x
a[x] > 0 for all x
Thus f(x) > 0 for all a >0, so there should be no upper bound on the interval. You can play around with the lower bound to find that it should be (I believe) [2, oo).
Last edited by MMDE on Mon Jul 30, 2018 10:11 am, edited 1 time in total.
Re: How are you all preparing of the upcoming exams?
That was a good one. I missed that, though it should have been obvious to me.
Your reasoning only leaves 2 to oo or 0 to oo. That’s it. Try 2 and see what happens. This is a perfect square. So, it has a root, and no negative values for f(x). That elliminates the second option
Your reasoning only leaves 2 to oo or 0 to oo. That’s it. Try 2 and see what happens. This is a perfect square. So, it has a root, and no negative values for f(x). That elliminates the second option
Re: How are you all preparing of the upcoming exams?
Number 5 means this (my riend translated that one for me)DDswife wrote:http://m.ishare.iask.sina.com.cn/f/63995575.html
I don’t understand what most of them mean, but you can guess and try to solve some of them
f(x) is an odd function. For x>0, f(x)=logx (base 2)
if x<0,then f(x)=(?)
Re: How are you all preparing of the upcoming exams?
Hmm. What seems to be wrong with my argument?MMDE wrote:If a > 0, f(x) will always be positive, here's a rough very informal proof:Junaid456 wrote:I think the best way to solve this question is to do the relatively simple computations. There's a small catch in the question.
Assume that f(x) is greater than or equal to zero.
Case I: x is >= 0. In this case, f(x) = x^2 + ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. So, D <= 0 iff a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) < 0 (I have omitted the denominator) if and only if sqrt(a^2  4) < a. We must have a is greater than 0 over here, and by squaring, it is clear that the inequality is true (the other root is clearly negative). So, the last case is true if a is positive. So, a must be between 2 and infinity (inclusive). We must include all possibilities since f(x) >= 0 in 3 cases!
Case I: x is <= 0. In this case, f(x) = x^2  ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2  4. Again, a is between 2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if a < sqrt(a^2  4). If a is positive, squaring, we see that this isn't true. If a is negative, this is always true, as sqrt(a^2  4) > 0. Considering the second root, a + sqrt(a^2  4) > 0 (I have omitted the denominator) if and only if  a > sqrt(a^2  4) > 0. Again, a can't be positive. If a is negative, can square both sides, and see the inequality is true. So, a must be between  infinity and 2 (inclusive).
Both cases will be simultaneously true iff a is in [2,2].
Please correct me if I am wrong.
It's easy to ignore the subcase that double roots exists, but are in half in which we're not interested. This is what makes it to be an annoying mGRE like problem. On than that, we simply have to play with the cases, and use the basics of logic, which makes it a good problem to give in a precalculus course.
I haven't done it by playing with the answers by first trying to plug in 2, 2 (and 0). I guess the difficult thing in this approach is to figure out the behavior if a > 2 or a < 2, without doing any computations. I find it very hard, and that also takes a minute or two. It's better to write down the problem, but it's easy to screw it up if one wants to solve it in 2ish minutes. I screwed it up before by not considering the subcase in each case that I mentioned in the previous paragraph.
x^2 > 0 for all x
a[x] > 0 for all x
Thus f(x) > 0 for all a >0, so there should be no upper bound on the interval. You can play around with the lower bound to find that it should be (I believe) [2, oo).
Re: How are you all preparing of the upcoming exams?
I did the same mistake the first time I tried this exercise. For values of a greater than 2, x^2+ax+1 has only negtive roots
All the possible functions pass through (0,1). Now, the vertex is a/2. For a > 2 the vertex is less than 1. So, its roots will be on the negative side of the real line. But we don’t want this part of the parabola
For the negative side, then f(x)= x^2ax+1, whose vertex and its roots are on the positive side.
Sorry, my explanation is bad. Hope it makes sense
Check the graph, and you will see. It looks like 2 pieces of parabola whose minimum is 1, and it happens at 0
All the possible functions pass through (0,1). Now, the vertex is a/2. For a > 2 the vertex is less than 1. So, its roots will be on the negative side of the real line. But we don’t want this part of the parabola
For the negative side, then f(x)= x^2ax+1, whose vertex and its roots are on the positive side.
Sorry, my explanation is bad. Hope it makes sense
Check the graph, and you will see. It looks like 2 pieces of parabola whose minimum is 1, and it happens at 0