I make videos where I solve Math problems. Every problem I solve here I am seeing for the first time so you can hear what my first thoughts are, and the general thought process that goes into solving (or not solving) a fresh problem.
I didn't remember the proper methodology for taking a second derivative of a parametric equation, so I did it the "hard" way, and solved for t in terms of x and substituted that into y to get a y as a function of x, then just found the second derivative the old fashioned way. It was messy, but I got the right answer in the end. Once you explained the actual method (which was correct), I solved it again the "easy" way, and spent the rest of the video screaming at the screen about your algebra! Oh my god! (6 t^3 + 6 t^2)/(t + 1) is just 6 t^2 !!!! dx/dt = 2t + 2 dy/dt = 12 t^3 + 12 t^2 y' = dy/dx = 6 t^2 y'/dt = 12 t y'' = y'/dt / dx/dt = 12t / (2t + 2) y''(2) = 4
I got confused because none of the answers matched the answer I got. Thats because answer E is written wrong. There is no 4x present in the expression. Thank you for your videos.
GRE questions are normally way above my head. But in this one I think using what I know it was OK. Using parametric differentation dy/dx is what you had, but factor out 12t² in numerator you end up with 6t². All I know, not really knowing how it so, what you proved, is the second derivative is found by differentating dy/dx and dividing by dx/dt, that leads to 6t/t+1. If the point is at x=8, setting to t² +2t =8 leads to t=2 (other solution is t= -4, not valid as t>o). Put that in to second derivative gives 4. I love watching these videos where you literally solve everything from your mathematical knowledge and prowess, things most probably, including me, just say "well that is what I have been taught to do", not knowing why and how it was derived.
The question was Q9 STEP 2 2016 if anybody interested and looking up the mark scheme. I loved your absolute from first principles approach to the question using calculus. Purity of mathematical knowledge. Most students sitting the exam would have had the benefit of having done mechanics for the previous 2 years and not have to work out the formulas from first priciples, but just know them, such as SUVAT, work done, conservation of momentum and kinetic energy. Your answer for first part is correct, but could have used loss of kinetic energy of bullet (1/2mu²) equals the work done against resistance by block (Ra). Leads to same answer you got for a. But your answers to the second part are not completely correct. In the second part the mark scheme suggested using conservation of momentum as you mentioned in video but did not use, giving velocity v of bullet/block system as v=mu/(m+M). Then using initial KE of bullet and final KE energy bullet/block system and setting difference equal to work done by reistance force inside the block, leads to the answers. Which are b=aM/(m+M) and c= amM/(m+M)². STEP question love using letters like a for distance to discombobulate you when used to having a as acceleration, s is distance etc. Thanks again for doing question, I thought it was really interesting.
Huh… I’ll have to look at this again some time to see where I went wrong then. It was a little while ago that I recorded this, so off the top of my head I’m not even sure if it would be a conceptual mistake or just some kind of algebra error.
So from what I can tell, and from the other solutions to this problem that I've seen, I did actually get the correct answer to the second part. I just didn't simplify it and write it in terms of the quantity found in the first part, the distance a. I guess this might not count as fully correct because I didn't express the answer the way they wanted, but I think it's easy enough to factor out the quantity a in the final expressions that I have!
Hi, the intended method is the mark scheme is even simpler. As you said this is a hidden quadratic, let y=cos²x, you end up with 9y² - 12y +7, the complete square, 9(y-2/3)²+3. Minimum is 3, when y=2/3 like you had.
Yes I mentioned that this can be done more directly, but I find completing the square to be more work than taking a derivative. Which means more room for mistakes!
Your approach is so elegant! I went through the process of finding all the inflection points of f(x) by setting f'(x) = 0, which came up with cos2(x) = 0, cos2(x) = 1, or cos2(x) = 2/3, and then I had to plug them all back in to find which one was the minimum.
The function I’m taking a derivative of is a parabola with a positive coefficient, so it’s definitely a minimum and not a maximum. Maybe I could have mentioned this while going through the work.
I would love to do a format like that too! It’s getting really hard to find the time to record even these short problems though. I’ve done some longer format things in the past (contest papers rather than entrance exams) but that was back when I wasn’t so busy with real life stuff.
I solved it without using the cross product. First I set up the equation of the straight line L that results from the intersection of the two given planes. This means solving the system of equations resulting from the two plane equations.This results in the following equation for the straight line L: (0,-1,4) + t*(1,0,-1). So that means, a plane that is perpendicular to this straight line L must have the coefficients a=1, b=0, c=-1 and this is option A).
@RajSandhu-gm8iz And nothing wrong with that I don’t think! A pretty simple way to combine intuition and precise statements about two topics that are not necessarily directly related in this case.
@@mathoutloud Nope. Having done lots of these MAT questions there is virtually always a question using discriminant, question with a "hidden" quadratic, a question that looks like algebra but easier to solve using geometry, a question on graphs, random questions using tuition and logic. But really clever the way they ask them normally combining a few areas of maths as you said. Really nice to see these videos, can't believe you are able to do some many differents aspects of maths, often from first principles in areas you have done in many years.
I guess that’s the benefit of having gone through my education with a mindset that I really want to understand everything, rather than just memorizing formulas and results. By understanding what everything means I can just rederive it if I don’t remember it directly.
Hi, I saw this video last night, I did it the same way you did. Thinking about it again today, I solved it another way, but it is not any quicker at all, and maybe uses more unfamiliar concepts than the cosine rule. Just more coordinate geometry. If you set the ▲ABC up with A at the origin, then B has coordinates (6√2,0) and C is (0,6√2). If you draw horizontal lines from points D and E to the line AB, say intersecting at E and G respectively, you end up with a series of similar ▲'s. You can get work out the base lengths and heights of these triangles using ratios. That allows you to work out the coordinates of points D and E and hence the gradients of the lines AD and AE, which end up as 3 and 1/2. Finally work out angle between the lines using the formula. As I said no quicker at all but just a bit more interesting!! Why do something in a quick and straightforward way in maths when a long convuluted method also works😃😃
That’s a good way to do it if you’re like me and don’t necessarily like pure geometry/trigonometry problems. It can really help a lot sometimes to set up a coordinate system and just brute force it.
The answer is correct. You can proof this by using an equivalent characterization of projections: Let P be a projection (P°P = P) then it holds that V is the direct sum of the image of P (imP) and the kernel of P (kerP) with P restricted to imP being the identity on imP. KerP is the eigenspace to the eigenvalue 0 (follows immediately from the definition) while imP is the eigenspace to the eigenvalue 1 (since P restricted on imP is the identity). Since V is the direct sum of imP and kerP it's a direct sum of eigenspaces of P and thus P is diagonalizable.
Good solution! Just two little things i noticed: In your answer to b), n=3^2 is already sufficient, no need to go for n=3^6. And in your answer to e), your example for x(n)=7, 57, of course works, if you take the correct factorization 57=3*19, instead of 6*19. Anyways, in both cases you came up with other valid examples, so no big deal.
If you watch enough of my channel you’ll start to see that I’m definitely not immune to silly mistakes or doing things the hard way! Thanks for pointing these out!
Hi, loved your solution using vectors. I saw this question a few years ago and eventually worked out it was essentially the line ax+by-c=0 has to be a tangent to the circle with radius 1 centred on the origin. I could not proceed thou. Most elegant solution I saw online was finding the distance from the point (0,0) to the line ax+by-c=0 using the distance from a point to a line formula, d=|a(0)+b(0)-c|/√(a²+b²), and setting this equal to 1.
I often freeze as well when given even a fairly straightforward geometry problem. To me they just always seem so tedious, but I’m happy I stuck this one out.
This is asking about the dot product of the vector u = (a,b) with a unit vector v = (x,y), which is maximized when u and v are colinear, in which case you get the norm of u.
well done. Maybe for those who couldn't solve it geometrically, like in this video: You can also get the result by differentiating the function ax+by with the constraint that x² + y² = 1. Then you have to calculate a little, but in the end you also get the result.
This was my first instinct! But I thought it would end up being a bit more complicated than the approach I took. What I did is essentially the idea behind the constrained maximisation anyway.
@@mathoutloud Yes, I also think your approach is very smart. My ability to solve mathematical problems geometrically or visually is sometimes not soooo good, so I just stubbornly start calculating and try to find the solution that way ;)
Hi, my very basic way to try and solve was using the fact that the sum of exterior angles is 360°. If you had 4 acute angles in a polygon, that means 4 exterior angles > 90°, but that adds up to more than 360°, hence maximum number of acute angles in any convex polygon is 3. Is this not correct?
I think I might have been thrown off by the fact that we’re specifically asked about a 10-gon here. But the result should be the same for any convex polygon.
@@RDon40 I think so, but given the fact the GRE questions on this channel are way above my head, as someone with, I would say, a good level of knowledge for someone interested in maths who did it until aged 18 but did not study it at university, I thought I had missed something in the question.
Hi, sorry question caused you pain. I have sent you an email with intended solution. Respectfully I think you went down the wrong route using vectors rather than parametric differentiation, then finding the equation of tangents at P and at Q, if let t=p at point P and t=q at point Q, find dy/dx, find equations of tangents at these points, use given fact that pq=-1, find coordinates of intersection, prove this point lies on the given curve. Sorry again for question, intended solution was not the algebra tedium you had to endure.
Now that you type this all out, it seems obvious that this is a better way to do it! I'm surprised I didn't scrap what I was doing and just do it the way you suggested.
The question itself was OK once worked out the limits of integration. Working with the exponents was easier when you know both functions are even, so simply going from 0 to upper limit and doubling. I always kick myself for those simple errors that mess you up.
That matrix is diagonalizable though. Any standard method to diagonalize should work, and you can start by easily seeing that it has eigenvalues of 0 and 1 with eigenvectors (1,0) and (1,1).
In option D), you can take a single-element set for the set S, for example, say S = {3}. Then the set as described in option D) corresponds to the set R\{3}. Since {3} is a closed set, its complement is open and this complement is exactly R\{3}.
There are a few ways to define closure that are equivalent, and this isn’t usually the first one I think of or the one that’s most natural to me. I think that’s why l took a bit of time to recognize that option D had something a bit “off.”
Genuinely, hats off to you for ploughing on with the question. You are at a disadvantage with these questions being advanced and specialised in your field sometimes, as oppposed to, as you mention at the start of video knowing some basic high school equations, that students sitting this exam will have known. The standard projectile SUVAT formulas and equation for the trajectory of a projectile rather than deriving them from first principles like you did. Then later on they will have known the various formulas for cos2x etc. This is classic STEP mechanics question, fairly basic mechanics, then proceeds to include calculus, trig and some messy algebra. Hope it does not put you off doing other mechanics questions!! Thanks again.
I think you and I might have very different ideas of what the difficult or annoying parts are! I thought that deriving some of these equations wasn’t an issue at all, and some of the initial parts of this problem weren’t too bad. But towards the end it essentially just turns into tedious algebra which makes anyone prone to errors and doesn’t really demonstrate anything interesting mathematically. I wouldn’t have known that it would turn into that without going through the whole problem though. Keep up the submissions! Even if they are somewhat more mechanics or physics related than mathematics.
@@mathoutloud Hi, it less about difficulty and more the fact if you went straight to trajectory formula, you get to first part in a few lines. Then after the differentiating spotting the trig formula leads you to x=htanα quickly, just save times. As you said after that there is a lot of grinding algebra. The purpose for this I have read given by exam board is "testing the ability of students to work accurately and quickly", make of that what you will and it's usefullness. In A Level maths most students as well as doing pure maths will do mechanics or statistics as well. Keep up the good work.
I submitted it, from exam papers called STEP maths (the entrance exam taken for Cambridge Uiniversity Maths degree), this is taken from STEP Maths 1 2016 Q11
loved that question, took me ages to get the diagram correct thou as you had it @4.55. After that I found the height of lines from centre to the horizontal line of the spheres, as sqrt(r^2-1) and sqrt (r^2-4), then created another right angled triangle which has height as the difference of the above, base 7 and hypotenuse 2r. Then solved
Hey ! I think you have the right answer, not sure though as i don't have a correction... You probably already know it but the Gk = {exp((2*i*n*pi)/k) | n in [0, k-1]} are pretty common groups called the k-th roots of the unity as they satisfy the equation z^k = 1 and are usually denoted Uk, k being a non-zero natural number, here in France. I have the confirmation that these groups are a part of the solution. For the infinite groups, i agree that U (the unity circle or trigonometric circle) is a compact subgroup though i don't have a justification for it being the only one. Your argument with the rational / irrationnal numbers might be enough as if you have a rationnal number in a subgroup, you will end up generating one of the Uk and if you have an irrationnal in the subgroup, you will end up with the whole circle, therefore showing that you cannot generate something else than the Uk or U (as the irrationnal and rationnal numbers form a partition of the real numbers) I also want to apologize for taking so long to answer i just really wanted to find an answer to the infinite group problem but ended up finding none, so yeah, i'm really sorry D:
Hi, and thanks again for submitting this problem! I had a good time thinking about it in the time leading up to this video. I know about roots of unity of course, I thought maybe I called it that at some point, but maybe I did actually forget about the terminology at the time. I still think about this in the back of my head from time to time, nothing too detailed, and I’m sure if I spent some dedicated time on it then I could flesh out the details. Let me know or if you have any others you want me to look at! It’s a bit easier if you send them to my email though, that way I don’t spoil the problem for myself before recording a video by reading it here.
I never claim I’m immune to the silliest of mistakes ;)
24 дня назад
Another way to see this is that Ax=x is the same thing as (A-I)x=0. But A-I is going to have a zero on the first diagonal element (and indeed the first column is all zeros) so it's not invertible (you can also see this as the determinant, here the product of the diagonals, will obviously be zero). Thus the statement that "(A-I)x=0 implies x=0" must be false, since that's the same as saying A-I is invertible.
Option B) can also be falsified directly: In a triangular matrix, the eigenvalues are directly on the main diagonal. So, among other things, 1 is an eigenvalue. This means that for A*x = 1*x, x is an eigenvector for the eigenvalue 1. And since an eigenvector is not the zero vector, it cannot follow from A*x = 1*x that x = 0. More precisely: A*x = 1*x COULD theoretically mean that x = 0, but since 1 is an eigenvalue, there must exist an vector, that is not the zero vector, and fulfills A*x = 1*x, so someone cannot conclude from A*x = x, that x must necessary be the zero vector.
Definitely! I was 99% sure that the eigenvalues of an upper-triangular are the diagonals, but for some reason I had a bit of a brain fart and confused the logic behind Ax=x with Ax=0.
The Oxford exam assumes no knowledge of complex numbers, even thou most people sitting it know them. Your solution at the end was the intended answer. In the most most basic sense of factor theorem if (x²+1) is a factor it means x^2= -1 leads to the function being zero, put that into the function, you end up with 4ⁿ - (-4)ⁿ needing to be zero, hence n needs to be even.
ooh this is a great question, it entirely threw me at first, but it really kind of tests you for your understanding of why we do what we do when we find roots of a polynomial. If something factors something else then that means when that thing that factors out is 0, the whole thing is 0. So if x^2+1=0, and it factors the equation, then the equation should equal 0 when x^2=-1. So just sub in x^2=-1 and equate it to 0. (3+1)^n-(-1+3)^n(-1-1)^n=0 4^n-(2)^n(-2)^n=0 4^n-(-1)^n4^n=0 So for this to hold n must be even.
The answer B ( even n) is right, and I just found a proof that I really like! In fact tpwards the end you kinda got the idea, as you can think about as working modulo (x^2+1), which is what you did by setting x^2 = -1. So, for the polynomial on the left, x^4 + 3, euclidean division will show you that this is equal to (x^2+1)(x^2-1) + 4, which is 4 modulo (x^2+1), and with the power of n this becomes 4^n. for the polynomial on the right, (x^2+3)(x^2-1), add and subtract 2 to each term (or again just do euclidean disision), and you will find that it is equal to (x^2+1)-4, which is just -4 modulo x^2+1, or again (-4)^n when we include the n. Thus, the whole thing is 4^n - (-4)^n modulo x^2+1, which is 0 if and only if n is even!
@@OliverGoodman-todd I did not, but it was my first instinct, and I abandonned it quickly! At first I was going to factor everything into (potentially) complex factors, like x^2+1 = (x-i)(x+i) and then continue from there, but with a quick glance I didn't see quick factors and cut the line of atack. Glad to see it would have worked, tho!
We all have to start somewhere! Basically what I’ve done is set up and equation which corresponds to the description of this number: “it is equal to one more than its cube.” So if I take the number and cube it, then add 1, I get back the original number. Once we have this equation we can ask whether there is in fact any number which satisfies it. Not every equation we write down will have a solution (or at least might not have a real solution). So if we ended up with an equation that doesn’t have a solution, then the answer to the original question is “no.” I’m just using the word “root” to mean a solution to a polynomial equation.
Solution that is a little less casework. There is symmetry between the mono increasing and mono decreases cases, so just consider the mono increasing case Clearly, the sequence 123456 is somewhat special, so we can just get that out of the way. Consider removing 1 from a sequence. The sequence after must be 23456. If the 1 were first, we would be back to our special sequence. If 1 were second, we would be double counting since we could also remove the 2. If 1 were anywhere else (4 spots left), it would work and be unique (no double counting). The same logic follows when removing 6 from a sequence. So, in total, for sequences that we remove a 1 or a 6 from there are 8 unique sequences and 2 double-counted sequences. Consider removing 2 from a sequence. The sequence after must be 13456. If the 2 were second, we would be back to our special sequence. If 2 were first or third, we would be double counting since we could also remove the 1 or 3, respectively. If 2 were anywhere else (3 spots left), it would work and be unique (no double counting). The same logic follows when removing either 3, 4, or 5 from a sequence. So, in total, for sequences that we remove a 2, 3, 4, or 5 from there are 12 unique sequences and 8 double-counted sequences. That is all the cases. Summing up, we get 1 + 8 + 2/2 + 12 + 8/2 = 26 By symmetry, total answer for mono increasing and decreasing would be 52.
Great problem It's not just that the longest cycle for a^n mod 10 is 4, but that every cycle length is a factor of 4. You can repeat a 2 cycle twice to get a 4 cycle. If there was a 3 cycle you would have to separately show that b^c^d ≡ b^c mod 3 (which it also happens to be).
Hi, all your answers were correct.If anyone is interested the question was taken from a UK maths entrance exam for The University of Cambridge. The papers are called STEP papers, STEP 1, STEP 2 and STEP 3, each paper getting a bit harder i.e STEP 3 is the most challenging. This was Q2 from STEP 1 2016. The format of the exam is you are given 3 hours per paper to solve about 5-6 questions from each paper. I did actually mange to solve this one, but took me about 45 minutes. After the differentation the simplification you did is really important so you can easily spot the values of the constants needed. Thanks for doing, thought people would find it interesting finding integrals from a differentiated function, not seen that before.
@@amritlohia8240 Obviously aware of the link between integration and differentation, but not seen that link set out with a question in this format, matching coefficients of what seems like a random differentiated function, to enable various other functions to be integrated, thought it was nice.
I had no idea how to approach this at first, but as soon as you wrote down N = 2,020 + 10,000 k, it became clear and I was able to find the answer. Nice one!
That was such a really clever solution. I'm sure most people would have found centre of circle (-a,-b) and the radius, r^2=(a^2+b^2+c). Distance from origin to centre, d^2=a^2+b^2, so r^2>d^2, leads to c>0.
5:10 ah, my favourite prime test "sounds prime" haha! I have to admit that's exactly what I was thinking as you said it. I love that little simplification at the end where you recognised the 3 sets of the same digits! It makes sense given its multiplied by 101, but I am pretty sure I'd have entirely missed that.
Everybody knows it’s a well tested tried and true method for prime testing. It definitely went through my head that I was happy I spotted the 101 multiples as well! For some reason a number like 20200 doesn’t look like a multiple of 101, but obviously it is.
I tripped myself up on this one for a second when I wrote 2(1-x)^2 = 2x^2 on my paper for some reason instead of 2(1-x)^2 = x^2 before I caught it. Still not sure where that extra coefficient in my head came from the first time.
I make mistakes like this all the time! I think it’s a useful skill to be able to spot mistakes like this through various means. Nobody is going to be perfect, so might as well get better at correcting your mistakes!
Nice I have been hooked up with the channel From what you did just now I was able to understand a bit But could you please elaborate a bit more on the determinant formula? Thank you
The formula for the determinant of a matrix is one of those things that I have no intuition for at all, I just happen to have it memorized. If you look at the Wikipedia page en.wikipedia.org/wiki/Determinant and read the section under Laplace expansion, that's the formula I always use.
hey ! i got a question that absolutely stumped me during an oral exam and i'd like to see you give it a shot. here's the question : Find all compact subgroups of (C*, x)
Could you elaborate on the notation here? What is C*? Is that all non-zero complex numbers? And what is (C*,x)? Non-zero complex numbers under multiplication?
@@mathoutloud C* is all the invertible elements of C (for the multiplication) so yes, it is C without 0 and (C*, x) is indeed non-zero complex numbers under multiplication
I've just finished recording my solution to this! You won't exactly get my totally fresh take on it because I read the question a couple days ago, but I try to give my thought process as it happened originally. Also, I'm not 100% sure I fully got the right answer, but I'm still pretty confident, although not totally convincing in communicating it. The video should be published in the next week or so, so stay tuned!
@__Junioor__ Hey! You may have seen recently that I put up the video with your question. I’m not sure I’m fully correct, maybe you could take a look and let me know.
Man, for each k=1,2,..., you have log(k) < int_{k}^{k+1} log(x) dx < log(k+1) For some n, add all the inequalities for k=1,2,...,n-1, sum log(k) < int_{k=1}^{n} log(x) < sum log(k+1) log((n-1)!) < (log(x^x)-x)|_{1}^{n} < log(n!) log((n-1)!) < log(n^n)-n+1 < log(n!) Now you take the power of e, (n-1)! < n^ne^{1-n} < n! In other words, instead of opening each integral, taking the e power of each inequality, multiplying everything and using telescoping reasoning for the product, it is better to add all inequalities, so the sum of integrals is the total integral, take the e power and it is done. That's clean. And intuitive. And if I actually did this test, I would leave the following message for the person that made this question: "Really? "n^ne^{1-n}"? (n/e)^ne is way better!"
I really wanted to use modular arithmetic here, and I got close with considering both representations mod 4, this collapses the base 8 representation to just a, and it works nicely for the base 11 given that 11=-1 mod 4, and so you can eliminate a by setting the two equations equal. But there isn't an easy way to eliminate another variable modulo another number as far as I can see. You get some nice relationships between a, b, and c that can give you some good hints to start, but I always had to result to just start guessing numbers to see what worked. I was kind of hoping to get the numbers algebraically directly.
Would have been very cool to solve this directly! I find that some of these AIME problems boil down to having to make two or three guesses after simplifying things a lot, which isn’t too bad. But maybe there are some techniques that I don’t know or don’t think about that would allow you to skip those guesses.
Probably a bit on the difficult side, but whether it’s an appropriate level for an entrance exam is open for debate. But what I find interesting is how the style of question that’s asked is essentially the same now as it was about 70 years ago. You could have told me this was on the exam last year and I wouldn’t have questioned you!
This is definitely NOT brutal. It is an easy question and formulated in a great way to test the student's proof reasoning and style. The only thing I would change is writing (n/e)^ne instead of n^ne^{1-n}
@@mathoutloud humm background? I don't think so. The only different thing the student needs to know for this question is Int log(x) = (-1+log(x))x That's pretty much the only different thing, isn't it? Strictly increasing function, log, fatorial, exponentiation, basic integral theory, there is nothing hard about those.
Thank you so much for doing that question, I thought it was really challenging. I got to the last inequality you got to, but tried multiplying by the expression by (n-1)^(n-1) but did not help. Thank you again.
@@mathoutloud it is all relative. But this doesn't mean we can't classify things as "brutal" or "easy". I don't understand why after someone says "brutal/challenging" you kinda agree with it, and if someone else says "easy" you reply with "it's all relative". This clearly shows you don't think "it's all relative". You actually think "it is probably a bit on the difficult side" ... I never understood such ... I don't know, reaction? It is like the "issue" some people have with "It's easy to show that ..." For some reason, a lot of people don't like when the "easy" label is used. Why? Actually I see such people trying to prevent or accuse arrogance coming from "easy" label users. Something that makes NO sense. Anyway ... just some random social philosophy narrative, I guess ... I don't know why I write those, no one cares ...
@@mathoutloud what I can say is: the question and its suggestion were GREAT, the thumbnails with black background and white font is AMAZING, it is CLASSIC, it is MODEST, it is CLEAR, I freaking love the thumbnails ... I guess the titles are great too. AND something GREAR about your videos is that they are GENUINE. It is you genuinely solving an exercise. Not the result of a bunch of takes for you to look good. Just some positive opinions so I don't look that much of an a..h... Oh, I forgot the conclusion: and all this is NOT relative. It is objectively true. Haha.
@samueldeandrade8535 because context matters? This question is obviously intended for high-school students seeing as how it’s on a university entrance examination. This question is without a doubt very difficult for a typical high-school student, but if you have studied calculus and analysis at a university level for a couple years then it’s essentially trivial. Hence, relative. Perhaps my response to you takes its form due to your dismissal of someone else’s struggle with the problem. There are a lot of things that the world doesn’t need, but I’d like to include in that list condescending and elitist attitudes regarding mathematics education. If Raj thought the problem was brutal, then the problem was brutal for him. Saying “it’s not brutal” in the face of their comment is hardly a way of encouraging someone that’s excited to learn and attempt these problems.
