I find it absolutely hilarious that 1/3 is the most intuitive guess. If they don't require showing your work I'd have gotten this correct while having NO idea why it's correct.
I would have thought 3/8 would be the most intuitive guess, since it's halfway between 1/4 and 1/2. But yeah, funny that it worked out to be a somewhat natural, guessable answer.
This is probably the most natural answer due to the fact that actually solving this would eat up an ungodly amount of time. Spending 3 minutes on 1 problem on the sat is not a good strat
Wow, I usually like this channel because I can spend a few minutes on a novel problem and afterwards I come out feeling a bit smarter than I did before for having figured it out. Not this time. This time I just felt like I should have known better but I did not. It's one of the harder ones I've seen here. Thank you, Presh.
Wow, this is the kind of problem that we called "a happy idea" one in my college years. Thanks for sharing! Just one small detail of little importance: in 1:16 when you do the u substitution, you must also change the integration limits, and instead of 4 and 8 should be f(4) and f(8). Anyway after evaluating you write -1/f(8) - (-1)/f(4), which is correct.
We can say that the mean of the square and the mean of -f'/f^2 = (-1/f)' are respectively 1/4 and 1/2, so the variance is 0, which implies that (-1/f)' is constant, i.e. -1/f is affine. So -1/f(6) = 1/2( -1/f(4) - 1/f(8)) = 1/2(-2-4) = -3, i.e. f(6) = 1/3.
The university entrance exam in Vietnam is composed of 50 questions (for each subject). These hard problems, and other problems with the same difficulty, usually appear at questions 40-50. The first 40 questions are quite easy, but then you are just hit by the shear confusion after question 40 and so on. I am taking the entrance exam this year, can’t wait 😝 Edit: I took the entrance exam 3 months ago and got the results not too long ago. I managed to get a 26.85/30, which placed me in the top 8% of students in my country). However, I couldn’t land my position at my dream university. Fortunately, I barely squeezed myself into an equally prestigious university, and now I’m studying Computer Engineering at VNU University of Information Technology.
Exacly from q45. Q41 to 45 is still pretty much doable, it is from the 46th onwards that the difficulty kicks in real hard. Unless you have some sort of galaxy brain, your best bet is to randomly choose answers and pray. These last questions are designed so that it will eat up most of your exam time before you figure out what to do. And good luck on your exam.
I'm about to take this exam too and I'm a little bit nervous now. Although I'm quite confident about maths, solving 5 hard questions from 45 to 50 in just 30 minutes ( that's the time you have only when you can solve the first 45 questions in 20 minutes) is impossible for me, and honestly, for most of the students so the only way to get 10 mark is to choose randomly. Therefore, I think the Ministry of Education should consider increasing the time. By the way, good luck in the exam
Well do you take the HSA exam? I think if you are really confident about the entrance exam then you would have no problem with HSA. The math part have less time but they are way easier than 5 last questions of the entrance exam.
Yes, I had succeeded! I used another method, using the Cauchy-Schwarz inequality. The video implicitely assumes f' continuous, and so will I for simplicity (but it's not really essential, one would only need to be careful with phrasing). The third equality can be interpreted as ||f'/f^2||=1, where the norm the classical one in L^2([4,8]). Then, we have by C-S
@@cristianalvarez3811 Thanks! Since, I have figured that deep down, all three valid solutions I have seen (Presh's, mine and the 0-variance I have seen in comments) have the same underlying idea to establish that f'/f^2 is constant. It is that f'/f^2 doesn't lose any energy by being projected into the 1-dimensional space of constants, therefore it is constant
That kind of problem is 1 of 50 multiple choice questions in the graduation and entrance exam to university in Vietnam. It is considered as one of the most challenge questions. For those who can guess the correct answer without doing maths, congratulations. You are genious.
Nice problem and nice exposition. A point about motivation here: this solution is a direct consequence of the famous Cauchy-Schwarz Inequality. For vectors this is (v.w)^2
I applied the mean value theorem for definite integrals, made a guess on the function f(x) as f(x) = A/x+B and solved for A and B. Then verified that the function satisfies the initial conditions and finally obtained f(6)=1/3.
The most interesting thing here is that it really doesn't seem from the problem statement that there's only a single such f. It doesn't really look like there's enough information. The fact that not only is f(6) determined but so is f(x) for any x between 4 and 8 is surprising.
@Piyush, Answer: 336 You probably meant "Find the _last_ three digits". You'll need Euler's Little Theorem and Euler's totient function. For a calculation, see my reply in the thread started by TaismoFanBoy (top comment), where you've posted the same question.
Wow that used a bit more tricks than I thought. Great problem!!! I loved how simple the solution was and it made sense all the way through. Great teaching.
Dude, I could pass a fully multiple choice test, without lifting a pencil. If the correct answer is presented, I can recognize it. I just cant find it.
Well, this is indeed a multiple choice question, 4 answers. This type of question is usually asked in Vietnam's graduation test which occurs after finishing grade 12. The questions in which are multiple choices A, B, C, D.
Great! Thank you Presh. Another, may be simpler, solution is to call g=1/f and use Schwarz inequality for the functions g' and 1, which in this case becomes an equality and thus g' is proportional to 1, thus constant. g(x)=a.x+b; Using g(4)=4 and g(8)=8, we get g(x)=6-x, thus f(x)=1/(6-x), thus f(6)=1/3.
Oh yeah, i had this problem in my 2nd term math test a few months ago. This was the only problem that i couldn't finish in time even though i've seen similar problems like this one before. Only after i had exited the exam room, did i figure out the value of k, which was really frustrating. Appreciate the video though, keep up the good work!
The squares inside the integral is screaming Cauchy’s inequality and as soon as we try it it fall in the equality case, which gives you the function. f(x)=2/(12-x). Vietnam’s national math Olympiad is way harder than this. I see now the solution in the video looks complicated because it has embedded in it the proof of Cauchy’s inequality via the quadratic polynomial.
It may be a hard question for students, who are not Vietnamese. In high school, we have to learn to solve these kinds of calculus questions if we want to reach 9+ marks on the national university entrance test. But these questions are so hard for most of us. Some students don't know how to solve them in just a few minutes then they choose the answer randomly. Sometimes, the answer is right, but it does not mean you understand or solve it by yourself to get the right answer. The creator of this video has solved the question in a fantastic way and it's easy to understand for most of us. Great job.
I got it correct! Unlike in the proof you’ve shown, I used the mean value theorem to deduce that the average value of [f’(x)]^2 /[f(x)]^4 from x = 4 to x = 8 was 1/4. Without any justification, I assumed the integrand was a constant function and went on from there. I can’t believe I got it right!
I also used the mean value theorem but to do an intelligent guess of the function as f(x)=A/x+B. Then solved for A an B. You always need to justify in mathematics even if they are guesses, then prove your hypothesis. Note that you need to verify that your obtained function satisfies the initial given conditions.
I got it right! I started with 1/3. I say f(4) = 1/4 and f(8) = 1/2. I literally just guessed that the 1 as the numerator would stay the same. I knew 6 was directly in between 4 and 8, and then just guessed for what was directly between 2 and 4 for the denominator. Normally an unacceptable method, but it worked lol
as a vietnamese teacher has said "For the questions that u do not know the answer, do not hesitate to pick C", and if this question is from an uni-entrance test in vietnam, i'll sure go for the C.
Nice problem. I solved it with a little guesswork, as I had a suspicion that f'(x)^2 / f(x)^4 would turn out to be a constant function. Using that as an assumption, I solved for f(x) and checked that the solution satisfied all the requirements.
@@fabricekahn4752 Well, I noticed that the problem would be easier if the function were monotonic (since then we could guarantee f(6) is between f(4) and f(8)), and the nicest possible type of monotonic function would be if the function we needed to integrate was constant. A little wishful thinking sometimes helps. But also, I've just seen enough problems sort of like this to have an idea what they might be going for.
Dear foreign friends, this was considered a bit hard compared to later questions in the national university entrance test which took place in July every year. There are other ones that you wish to never encounter😉 From Vietnam with love
Don’t enhance too much our national education bro , the national examination event take place in many developed countries such as Japan , China , Korea also hard more hundred times than these one it just a drop in the ocean dude :)) From vn
way more convenient method: given the 3 main conditions that the function has to fulfill we can create a polynomial function with 3 terms that satisfies them. starting with a simple matrix resembling the first two conditions as a system of linear equations, we are already almost finished, leaving the last condition as an equation that can be solved by plugging the other values from the established function in and getting the last value. that way you also get a correct function alongside the solution for f(6)
@@jensraab2902 that is irrelevant, the task is to find a function that suits the requirements, regardless of the type of the function. Therefore, if there is a polynomial (3n) in my case, then it's a solution to the problem. And it makes the problem a lot easier.
I love it. If you know your calculus rules the first third of the problem is easy to figure out, then it evolves into somewhat more of a thinking problem. Bit hard to put on an exam paper, tho?
@@mike1024. this is one of the 50 math questions MANDATORY to Vietnamese students in order to graduate. Of course this is impossible to do in a time crunch but you don’t have to get all questions right
i find the "hard" vietnamese function questions are usually all tricky nowadays (i mean, all the questions you can find in the entrance exam). you can always finish them by some taught technique. but, the others (geometry) are quite interesting. i hope you can bring some of them to the viewers.
It's the kind of question you WON'T be able to do unless you've seen or done it before, atleast once. They are pretty linear once you've solved a couple of them though, and you'll start finding them easy. Don't know whether to call it good or bad coz there's no creativity involved , just a standardized approach. Once again, math is all about practice and experience, as this question proves.
just a note that in your intermediate step, when you substitute u = f(x), your upper and lower limits of the integral should change as part of your working. ultimately it did not matter as you substituted u = f(x) back.
I solved it in a different way. My strategy was to look for functions that satisfy the boundary conditions and has a parameter that I can calibrate using the integral. After a long search I found x/(16+a(x^2-12x+32)). This function satisfy the boundary conditions and is easy to integrate in the defined integral. After evaluating the integral and setting the answer to 1 you find a = -1/2. Substituting and reducing gives the same function as the video shows.
The Euler-Lagrange Equations are an interesting guess that turns out to give you the right function. It's like a variational calculus problem too, which is cool.
Presh, there's a slight notation error in 1:58. When you take the integral of both sides, you should have taken the right hand side inside brackets, for the operation to take effect on the whole of it. The next step stands correctly anyway.
There's a slightly more intuitive way to solve this. Start by substituting y=1/f, then dy/dx = -1/f^2 df/dx, or (dy/dx)^2 = [f'(x)]^2/[f(x)]^4. Now y is a continuously differentiable function where y(4)=4 and y(8)=2. Now dy/dx is the slope of that function, and (dy/dx)^2 is the square of the slope which is always positive. Let's start with y being the simplest continuously differentiable function over the interval, which is a straight line descending between (4,4) and (8,2). Then, dy/dx is just the slope of the line, which is -1/2, and (dy/dx)^2 is 1/4. Then, the integral of dx/4 is x/4 which, when evaluated between the interval [4,8] is equal to 1. This is the same value as the integration criterion stipulated in the problem. For this function, y(6)=3, therefore f(6)=1/3. What's left is to prove that the function is unique, that is, there either are no other functions for which the integral equal to 1 is satisfied or that all functions integrate to 1. This is not rigorous, but let's say you "bend" the line just a bit like you were pulling on a rubber band. In that case you would have a set of values for (dy/dx)^2 that were steeper than the value 1/4, and an integration for which the value would be slightly greater than 1. From there, it follows that all functions not being the straight descending line yield integrals greater than 1. It has the feel of a variational calculus problem, but I don't think it really is. If the integral were equal to anything other than 1, I would conclude there would have to be an infinity of continuously differentiable functions that would address the criteria, therefore there would be an associated infinity of solutions for f(6).
Actually, it _is_ a calculus of variations problem. I used your method, then to prove the answer was unique, I showed the integral of (dy/dx)^2 from 4 to 8 is extremal when y is a linear function of x by using the calculus of variations. Since the integral of (dy/dx)^2 from 4 to 8 takes on this extremal value, the solution for y must be unique, as any variation would increase the value of that integral.
These are the types of math problems that always bugged me. Integrations on polar functions were the worst -- you applied one of twenty trigonometric identities, and if you picked the right one, it simplified after a half page of calculations. If you picked the wrong one, you usually knew after a page of calculations.
If you let g(x) = 1/f(x) then the integrand becomes (g'(x))^2 by a bit of manipulation involving the quotient rule...then you have g(4) = 4 and g(8) = 2. While there's no guarantee that g(x) is linear, you can start by seeing if there's a linear function that satisfies the 3 conditions given, which there is (namely g(x)=6-0.5*x; g'(x) is then -0.5 which means that the integral is the integral from 4 to 8 of 0.25 dx which does in fact equal 1)...this means that f(x)=1/g(x) = 1/(6-0.5x) or 2/(12-x) if you don't like fractions in the denominator. Which then yields the answer of f(6)=1/3...this is how I did it, though I suppose the jump to try looking for a linear function for g(x) was a bit or a stretch.
@@swayamjain1322 yes we can? it just said f(x) ≠ 0 in [4,8]. The function f(x) = 2/(12 - x) if you calculate the domain, it's every real number except 12
f(8)=1/2 f(4)=1/4 it's an increasing function So i guessed f(6) =1/3(becoz it looked like linearly increasing function) but I really like the explanation
It doesn't necessarily has to be an increasing function. It could be a weirder function with increasing and decreasing parts within the domain. Nice guesswork
In VN, this is usually a multiple-choice question, and there are 2 types of people, one will calculate, one will just choose an answer base on gut-feeling.
I did it in 3 seconds First it gives you f(4)=1/4 and f(8)=1/2 Since it is asking for f(6) which is inbetween 4 and 8 all u need to do is find what fraction is in between 1/2 and 1/4
At 3:08 you assumed that the function is continuously differentiable as we can only make the statement that if the integral of a non-negative function is 0 the function is 0 for a continuous function. In General a function can still satisfy this property and be non-zero on a measure zero set but since its continuous the measure zero set will be open and hence be the null set. I believe we can go around this assumption by using cauchy-schwarz inequality and use the fact that equality is satisfied iff the 2 vectors are linearly dependent which finally gives rise to a simple differential equation.
The k trick is remarkable. It implies in general that for any g(x) if [Int_a^b g(x) dx]^2 >= (b-a) Int_a^b [g(x)]^2 dx, then g(x) must be a constant. I had noted in the problem that if 1/f(x) is linear then a solution is found, but it had escaped me that it is the unique solution because 1/f(x) MUST be linear.
1:20 This is technically not correct: You would have to tag the limits along with the u-sub. In terms of "u" certainly the boundaries cannot be 4 and 8. Then at 1:28 it continues correctly. I know, it is more of a "notational" issue but I felt I needed to bring it up
There's a simpler way to do this problem. Based on the given data, we know that f(x) is always positive in [4,8]. Now, we can write the RHS of the integral, i.e, 1 as (1/4)*(integration of dx from 4 to 8). Move it to LHS and now RHS is zero. Equate the integrand to zero. And we end up with 3:20 but squared terms on both sides. We know f(x) is positive so we solve for that.
At 3:10, why should the integrand be identically equal to zero? It may be zero but there's no condition that implies it should necessarily be zero. Can't it be possible that the integral of non-zero integrand may have equal values at 4 and 8?
Which is what tests are testing, so all good, right ? They don't test your intelligence, they just want to see if you put your time in school to good use.
Or you can use intuition. If i took this test i'd be able to give the correct answer, although i couldn't explain it at all. f(4)=1/4 f(8)=1/2 so f(6)=1/3.
Nope, that f'^2 /f^4 gives intuition that maybe f is 1/linear as then f' is 1/(linear)^2, and then you can check by the conditions on f(4) and f(8) by taking 1/f linear and see that the solution satisfied the integral condition too, if this is a mcq, then this is enough to tell you that if there is enough info to determine f(6), it follows that f(6) must be 1/3
Before I watched the video I looked at the thumbnail and thought "If f(4) is 1/4 and f(8) is 1/2 then f(6) is probably 1/3. Glad to see that my guess was right. :)
I have a question: at 2:50, you use k=-1/2 so that f(x) can be easily determined for this condition. But why can't k take any other value? Of course it will be very difficult, maybe impossible, to find another function that would satisfy this new condition... but this does not mean that another function cannot exist...and, if it does exist, it could well be that an alternative function would have a different value for f(6), and the answer to the question would not be unique. So my question: is the given function the ONLY one that would satisfy the initial given conditions? If yes, how to prove it?
Try substituting k with another value and you will see this can only lead to the initial expansion at 2:03, no conclusion can be drawn. The value k has to be something that makes the integrand equal to a certain value (in this case 0) so that we can separate it from the integral and calculate the function.
Nice. I never would have thought of doing that. Letting g(x) = 1/f(x) simplifies the problem somewhat to: g(4) = 4, g(8) = 2 and ∫[4,8] g'(x)² dx = 1. More generally, if g(a) = A, g(b) = B and ∫[a,b] g'(x)² dx = C, the same procedure shows that g(x) is linear = ((B-A)x+(Ab-aB))/(b-a), the slope is -k and C = (B-A)²/(b-a).
Swear to god I've seen question like this so many times that I lost count. This question and it's friends are quite the final boss in the university entrance exam, and I didn't beat them when I did the exam. What an experience
With two complex numbers satisfy |z|=1 and |w|=2. When |z+ⅈw ̅+6+8ⅈ| gets the minimum value, what is the value of |z-w|?. A question from 2021 entrance exam in Viet Nam.
The solution is really brilliant. After some thought I can motivate the substitution u(x) = 2 * f'(x) / f(x)^2. Then the given information can be written as: int_4^8 u(x) dx = 4 and int_4^8 u(x)^2 dx = 4. However, I still find it hard to conclude from here that u(x) = 1 is the only solution, without using some smart trick.
I got it using the cauchy-schwarz inequality. In this case, int(g^2) >= (1/4)int(g)^2 with equality if and only if g is constant. Substitute in g = (1/f)' and we get an equality.
Cool problem! Here's my solution: Let g=1/f, then g'=-f'/f^2. So $g'^2=f'^2/f^4$, which is what we want! Now we have: g(4)=4,g(8)=2 We have to deal with integral g'^2 Since (g'+a)^2 >= 0, we have g'^2 + a^2 >= -2ag' Integrating all, we have 1+4a^2 >= 4a If a=1/2 we have strict equality, so we would need g'+1/2=0 which means g'=-1/2 So, g is linear with slope -1/2 which means g(6)=3 which means f(6)=1/3 I think it's the same as yours but I found thinking in terms of g=1/f more intuitive
*@Presh Talwalkar* Your headline is wrong for the content of video. Someone not being able to solve this difficult problem *does not mean* that they could pass the math test in Vietnam. The question simply should have been "Can you solve this problem on a math test given in Vietnam?"
Does anyone agree that Presh's methid seems to come out of nowhere and Indont see why anyone would think of it..why integrate the square root of the integrand?
As with all the best math, I stumbled on a nearly identical solution to this accidentally. I was actually going in the reverse, trying to find a find an easy function g(x) so that the integral from 4 to 8 of (ag(x) + b)^2 would be 1. Then if you took g(x) = f'(x)/f(x)^2 (which I was now calling F'(x), where F(x) = -1/f(x)), this simplified down to (a + 2b)^2 exactly as you had but knowing the integral of F'(x) instead of f'(x)/f(x)^2 certainly felt more naturally motivated. Took me a bit to then realize that I could choose a and b so the RHS were zero.
when you said "pause the video if you'd like to give this problem a try" At that moment I didn't have a paper or a pencil. So i just said if one of them was 1, one was 1/2 and the other was 1/4‚ i suggested and said of course it would be 1/3 now I think of myself as a mathematician ;)
I understand you’re probably joking with this comment, but since a couple others have commented similar ideas, I’m echoing my response to theirs: That only works because f(x) happened to be a linear function. If f(x) ended up being more complex (with an x^2 term, or something else), then you wouldn’t have been able to interpolate the results in that manner. No way to know without going through the steps in the video.
I tried solving this way: Notice that f(8) - f(4) = 1/2 - 1/4 = 1/4 = 1/4 * 1 = 1/4 * (integral from 4 to 8). Therefore, 1/4 times the Integral from 4 to 8 is the same as f(8) - f(4). I then claimed that f(x) is the antiderivative of 1/4 times the integrand (according to the Fundamental Theorem of Calculus part 2) Differentiate both sides, I get 1/4 * integrand = f'(x), which is a differential equation, solved simply by separation of variables. Then I isolated y and expressed it as a function of x, solved for c using f(8) and f(4) and went to evaluate f(6) but didn't get the answer. 2 different c values resulted from the given initial values. I honestly don't know what my mistake was. If anyone can enlighten me that would be great.
I think you lost an opportunity here: calling u=-1/f and rewriting the conditions and the question would make the rest of the solution a lot less mysterious. Also, finding a value at a certain point inside the interval is only possible when you realize the function as an extreme case of an inequality (where equality is achieved).
@@leif1075 Note that f(x) has to go from 1/4 to 1/2, so it needs positive slope somewhere between x=4 to 8. At the same time 1/f(x)^4 starts off as 256, so if the slope is large near x=4, then the integral of f'(x)^2/f(x)^4 is going to be WAY larger than 1. So a natural question to ask is if I limit f'(x)^2/f(x)^4 to be small, say equal to 1/4 throughout so that its integral is 1, then will f(x) be able to rise fast enough? What will the value of f(x) be at x=8? f'(x)^2/f(x)^4=1/4 means f'(x)/f(x)^2 = 1/2. With some thought you'll see that f has to be of the form 2/(c-x). For f(4)=1/4, c should be equal to 12. Then f(x) = 2/(12-x), so f(8)=1/2, exactly what we wanted. We got lucky and we can find f(6)=1/3. Now the question is how to prove that this is the only possible function. This I couldn't do by myself but looking at the video it kind of made sense although it was still tricky. If I can prove that f'(x)/f(x)^2 - 1/2 has to be equal to 0, I would be done. So integrate (f'(x)/f(x)^2 - 1/2)^2. Turns out that has to be zero because of the given conditions, and so our lucky answer is the only answer.
1:43 Did you choose to do the square of the original integrand? If that is the case, then why use k? Sorry It has been a while since I have taken Calculus.
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I find it absolutely hilarious that 1/3 is the most intuitive guess. If they don't require showing your work I'd have gotten this correct while having NO idea why it's correct.
Ha my thoughts exactly. Immediately I guessed 1/3, as 6 is the midpoint between the givens (4 and 8)
Same here! My answer would have stated that all working was classified and on a need to know basis only.
I would have thought 3/8 would be the most intuitive guess, since it's halfway between 1/4 and 1/2. But yeah, funny that it worked out to be a somewhat natural, guessable answer.
and f(2) =? ( ignoring the interval limit)
This is probably the most natural answer due to the fact that actually solving this would eat up an ungodly amount of time. Spending 3 minutes on 1 problem on the sat is not a good strat
Wow, I usually like this channel because I can spend a few minutes on a novel problem and afterwards I come out feeling a bit smarter than I did before for having figured it out. Not this time. This time I just felt like I should have known better but I did not. It's one of the harder ones I've seen here. Thank you, Presh.
Yes, my eyes glazed over about f(8) way through.
Don't youbgius agree there must be another more intuitive way to solve than what Presh did??
Wow, this is the kind of problem that we called "a happy idea" one in my college years. Thanks for sharing!
Just one small detail of little importance: in 1:16 when you do the u substitution, you must also change the integration limits, and instead of 4 and 8 should be f(4) and f(8). Anyway after evaluating you write -1/f(8) - (-1)/f(4), which is correct.
What you just said is all greek to me, and I'm part greek.
@@toyin.dindoinlove9236 that's because calculus is a British and German discovery...
What is a happy idea? When was the test Presh mentioned given..I college or high school?
@@leif1075 It's a trick that solves the problem, but hard to come with.
@@NestorAbad Oh is that an idiom or idiomatic expression in Vietnamese you mean? Thanks for enlightening me.
We can say that the mean of the square and the mean of -f'/f^2 = (-1/f)' are respectively 1/4 and 1/2, so the variance is 0, which implies that (-1/f)' is constant, i.e. -1/f is affine. So -1/f(6) = 1/2( -1/f(4) - 1/f(8)) = 1/2(-2-4) = -3, i.e. f(6) = 1/3.
No
These types of questions reminds me of my mentor's advice that is "Every question is not meant to be answered" 🙂🙂
Is it Mohit tyagi sir?
Watching this video makes me feel like being a 12th grader again (I'd graduated in 2006). Many heartfelt thanks!
Can you plz help me to solve this question
Find the the three digits of
2^22^222^2222^22222^222222^22222222222
@Piyush,
Answer: 336.
For the calculation/proof, see my reply in the thread started by TaismoFanBoy (top comment), where you posted the same question.
The university entrance exam in Vietnam is composed of 50 questions (for each subject). These hard problems, and other problems with the same difficulty, usually appear at questions 40-50. The first 40 questions are quite easy, but then you are just hit by the shear confusion after question 40 and so on. I am taking the entrance exam this year, can’t wait 😝
Edit: I took the entrance exam 3 months ago and got the results not too long ago. I managed to get a 26.85/30, which placed me in the top 8% of students in my country). However, I couldn’t land my position at my dream university. Fortunately, I barely squeezed myself into an equally prestigious university, and now I’m studying Computer Engineering at VNU University of Information Technology.
All the best bruda
Exacly from q45. Q41 to 45 is still pretty much doable, it is from the 46th onwards that the difficulty kicks in real hard. Unless you have some sort of galaxy brain, your best bet is to randomly choose answers and pray. These last questions are designed so that it will eat up most of your exam time before you figure out what to do.
And good luck on your exam.
I'm about to take this exam too and I'm a little bit nervous now. Although I'm quite confident about maths, solving 5 hard questions from 45 to 50 in just 30 minutes ( that's the time you have only when you can solve the first 45 questions in 20 minutes) is impossible for me, and honestly, for most of the students so the only way to get 10 mark is to choose randomly. Therefore, I think the Ministry of Education should consider increasing the time. By the way, good luck in the exam
^gl man
Well do you take the HSA exam? I think if you are really confident about the entrance exam then you would have no problem with HSA. The math part have less time but they are way easier than 5 last questions of the entrance exam.
Yes, I had succeeded! I used another method, using the Cauchy-Schwarz inequality. The video implicitely assumes f' continuous, and so will I for simplicity (but it's not really essential, one would only need to be careful with phrasing). The third equality can be interpreted as ||f'/f^2||=1, where the norm the classical one in L^2([4,8]). Then, we have by C-S
Nice one! Very good argument, I also tried using CS but eventually just found the solution presented in the video.
@@cristianalvarez3811 Thanks! Since, I have figured that deep down, all three valid solutions I have seen (Presh's, mine and the 0-variance I have seen in comments) have the same underlying idea to establish that f'/f^2 is constant. It is that f'/f^2 doesn't lose any energy by being projected into the 1-dimensional space of constants, therefore it is constant
@@fabricenonez9098 of you don't remember caucny Schwartz surely there is another way of solving it?? WHY would anyone think to sub 1/f for u..
That kind of problem is 1 of 50 multiple choice questions in the graduation and entrance exam to university in Vietnam. It is considered as one of the most challenge questions. For those who can guess the correct answer without doing maths, congratulations. You are genious.
hey there fellow Vietnamese
Its not even worth doing if you are on a time crunch then. If its multiple choice then guess and come back to it at the end.
proud to say i guessed 1/3 by using the initial conditions
It took me 30 seconds to get the right answer 😂
wait so no negative marking?
Nice problem and nice exposition. A point about motivation here: this solution is a direct consequence of the famous Cauchy-Schwarz Inequality. For vectors this is (v.w)^2
Yes that’s what I observed too in my comment.
I applied the mean value theorem for definite integrals, made a guess on the function f(x) as f(x) = A/x+B and solved for A and B. Then verified that the function satisfies the initial conditions and finally obtained f(6)=1/3.
The most interesting thing here is that it really doesn't seem from the problem statement that there's only a single such f. It doesn't really look like there's enough information. The fact that not only is f(6) determined but so is f(x) for any x between 4 and 8 is surprising.
You literally find hundreds of questions like this in high school.
Can you plz help me to solve this question
Find the the three digits of
2^22^222^2222^22222^222222^22222222222
@Piyush,
Answer: 336
You probably meant "Find the _last_ three digits". You'll need Euler's Little Theorem and Euler's totient function. For a calculation, see my reply in the thread started by TaismoFanBoy (top comment), where you've posted the same question.
Wow that used a bit more tricks than I thought. Great problem!!! I loved how simple the solution was and it made sense all the way through. Great teaching.
I literally gave this question more than an hour and I still couldn't solve it. This was a really good question.
If it was multiple choice, 25% of the time I would get it right everytime.
Dude, I could pass a fully multiple choice test, without lifting a pencil. If the correct answer is presented, I can recognize it. I just cant find it.
If the multiple choice was 1 correct answer over 100 options?
Well, this is indeed a multiple choice question, 4 answers.
This type of question is usually asked in Vietnam's graduation test which occurs after finishing grade 12. The questions in which are multiple choices A, B, C, D.
@west explains best
That means you'll get wrong 75% of the times.
@@ysbshankar, I mean, if you have prior knowledge of the topic, there is at least a 50% chance that you will guess it right.
Great! Thank you Presh. Another, may be simpler, solution is to call g=1/f and use Schwarz inequality for the functions g' and 1, which in this case becomes an equality and thus g' is proportional to 1, thus constant. g(x)=a.x+b; Using g(4)=4 and g(8)=8, we get g(x)=6-x, thus f(x)=1/(6-x),
thus f(6)=1/3.
If you are a high schooler in Vietnam that tool is not accessible. That's a neat insight however, thanks!
Great answer
Oh yeah, i had this problem in my 2nd term math test a few months ago. This was the only problem that i couldn't finish in time even though i've seen similar problems like this one before. Only after i had exited the exam room, did i figure out the value of k, which was really frustrating. Appreciate the video though, keep up the good work!
The squares inside the integral is screaming Cauchy’s inequality and as soon as we try it it fall in the equality case, which gives you the function. f(x)=2/(12-x). Vietnam’s national math Olympiad is way harder than this. I see now the solution in the video looks complicated because it has embedded in it the proof of Cauchy’s inequality via the quadratic polynomial.
I like it. Basically, any time a function can be determined by finite values. It's good to expect (sth)^2=0 comes up.
It may be a hard question for students, who are not Vietnamese. In high school, we have to learn to solve these kinds of calculus questions if we want to reach 9+ marks on the national university entrance test. But these questions are so hard for most of us. Some students don't know how to solve them in just a few minutes then they choose the answer randomly. Sometimes, the answer is right, but it does not mean you understand or solve it by yourself to get the right answer. The creator of this video has solved the question in a fantastic way and it's easy to understand for most of us. Great job.
Impressed with myself, before even solving it I guessed that it would end up being 1/3 and I was right.
Yeah you can guess it from the other 2 functions and just use the complicated one to corroborate
I got it correct! Unlike in the proof you’ve shown, I used the mean value theorem to deduce that the average value of [f’(x)]^2 /[f(x)]^4 from x = 4 to x = 8 was 1/4. Without any justification, I assumed the integrand was a constant function and went on from there. I can’t believe I got it right!
I also used the mean value theorem but to do an intelligent guess of the function as f(x)=A/x+B. Then solved for A an B. You always need to justify in mathematics even if they are guesses, then prove your hypothesis. Note that you need to verify that your obtained function satisfies the initial given conditions.
I was sure you were going to split the integral at 6 and do some clever canceling. This was intense!
I have an MSEE. I took ALL of the math (or so it seemed at the time). Calculus is definitely a "use it or lose it" skill.
This question reminds me of the time when I was revising for graduation exams, and it lies at 1 of the last 5 out of 50 questions in the exam
I got it right! I started with 1/3. I say f(4) = 1/4 and f(8) = 1/2. I literally just guessed that the 1 as the numerator would stay the same. I knew 6 was directly in between 4 and 8, and then just guessed for what was directly between 2 and 4 for the denominator. Normally an unacceptable method, but it worked lol
Same lol
Yeah I did the same
I did the same, and I thought "no way this cannot be the answer". BUT IT WAS
sometimes sketchy math gets you to the right answer
If engineers thought like that we would have daily plane crash
as a vietnamese teacher has said "For the questions that u do not know the answer, do not hesitate to pick C", and if this question is from an uni-entrance test in vietnam, i'll sure go for the C.
"Câu nào mình không làm được phải chọn toàn C"
Nice problem. I solved it with a little guesswork, as I had a suspicion that f'(x)^2 / f(x)^4 would turn out to be a constant function. Using that as an assumption, I solved for f(x) and checked that the solution satisfied all the requirements.
Why / How did you get this suspicion ?
@@fabricekahn4752 Well, I noticed that the problem would be easier if the function were monotonic (since then we could guarantee f(6) is between f(4) and f(8)), and the nicest possible type of monotonic function would be if the function we needed to integrate was constant. A little wishful thinking sometimes helps. But also, I've just seen enough problems sort of like this to have an idea what they might be going for.
I guessed the right answer before it began, but if my professor were to say “show your work” I’d just start crying
Dear foreign friends, this was considered a bit hard compared to later questions in the national university entrance test which took place in July every year. There are other ones that you wish to never encounter😉
From Vietnam with love
Oh you mean the "vận dụng cao" (advanced applying) shits?
Don’t enhance too much our national education bro , the national examination event take place in many developed countries such as Japan , China , Korea also hard more hundred times than these one it just a drop in the ocean dude :))
From vn
Can you plz help me to solve this question
Find the the three digits of
2^22^222^2222^22222^222222^22222222222
way more convenient method: given the 3 main conditions that the function has to fulfill we can create a polynomial function with 3 terms that satisfies them. starting with a simple matrix resembling the first two conditions as a system of linear equations, we are already almost finished, leaving the last condition as an equation that can be solved by plugging the other values from the established function in and getting the last value. that way you also get a correct function alongside the solution for f(6)
How do you know that the function is a polynomial?
@@jensraab2902 that is irrelevant, the task is to find a function that suits the requirements, regardless of the type of the function. Therefore, if there is a polynomial (3n) in my case, then it's a solution to the problem. And it makes the problem a lot easier.
I love it. If you know your calculus rules the first third of the problem is easy to figure out, then it evolves into somewhat more of a thinking problem. Bit hard to put on an exam paper, tho?
I would think this is more of one of those challenge tests that people volunteer to do than anything you'd find in a classroom.
@@mike1024. this is one of the 50 math questions MANDATORY to Vietnamese students in order to graduate. Of course this is impossible to do in a time crunch but you don’t have to get all questions right
This is one of the toughest problems I’ve seen on this channel… I think I’d have to study the solution for an hour before I truly understand it
i find the "hard" vietnamese function questions are usually all tricky nowadays (i mean, all the questions you can find in the entrance exam). you can always finish them by some taught technique. but, the others (geometry) are quite interesting. i hope you can bring some of them to the viewers.
Hi Vietnamese :))
khó vl
@@jaylenlai ko khó đâu bn
@@prodinferno năm nay tôi thi. 40 câu đầu chắc chắn dễ, 5 câu sau hên xui, 5 câu cuối ngồi cầu nguyện :V
@@jaylenlai vậy thì chúc bn đúng 5 câu cầu nguyện :)))
Nice, this method was equivalent to using the cauchy-schwarz inequality for integrals
Can you please explain the solution
I just wanna know more about how you thought about adding a const K and then squaring to follow up the procedure..
It's the kind of question you WON'T be able to do unless you've seen or done it before, atleast once. They are pretty linear once you've solved a couple of them though, and you'll start finding them easy.
Don't know whether to call it good or bad coz there's no creativity involved , just a standardized approach. Once again, math is all about practice and experience, as this question proves.
I have done my fare share of such problems but this adding arbitrary constant k trick is completely new to me! Amazing!
just a note that in your intermediate step, when you substitute u = f(x), your upper and lower limits of the integral should change as part of your working.
ultimately it did not matter as you substituted u = f(x) back.
Well, the function f(x)= 2/(12-x) satisfies all these conditions so if there is enough information to deduce what f(6) is, it's 1/3
في اللحظة 03:42 ،هناك مشكلة في اختيار قيمة x=4 ، لتعيين قيمة c, و كان يجب التحقق من قيمة c لما x=8.
I solved it in a different way. My strategy was to look for functions that satisfy the boundary conditions and has a parameter that I can calibrate using the integral. After a long search I found x/(16+a(x^2-12x+32)). This function satisfy the boundary conditions and is easy to integrate in the defined integral. After evaluating the integral and setting the answer to 1 you find a = -1/2. Substituting and reducing gives the same function as the video shows.
Cool, I solved it by seeing its patterns, simple and straight and within a minute 😎
The Euler-Lagrange Equations are an interesting guess that turns out to give you the right function. It's like a variational calculus problem too, which is cool.
Good problem. I thought for a while that there had to be an infinite number of solutions, but watching the video a second time helped.
Presh, there's a slight notation error in 1:58. When you take the integral of both sides, you should have taken the right hand side inside brackets, for the operation to take effect on the whole of it. The next step stands correctly anyway.
Would you not agree that ∫f(x) + g(x)dx is completely unambiguous in real analysis?
@@matthew-m
I don't know, I am not an expert on notation. It's just not the way I learned it.
There's a slightly more intuitive way to solve this.
Start by substituting y=1/f, then dy/dx = -1/f^2 df/dx, or (dy/dx)^2 = [f'(x)]^2/[f(x)]^4.
Now y is a continuously differentiable function where y(4)=4 and y(8)=2. Now dy/dx is the slope of that function, and (dy/dx)^2 is the square of the slope which is always positive.
Let's start with y being the simplest continuously differentiable function over the interval, which is a straight line descending between (4,4) and (8,2). Then, dy/dx is just the slope of the line, which is -1/2, and (dy/dx)^2 is 1/4. Then, the integral of dx/4 is x/4 which, when evaluated between the interval [4,8] is equal to 1. This is the same value as the integration criterion stipulated in the problem. For this function, y(6)=3, therefore f(6)=1/3.
What's left is to prove that the function is unique, that is, there either are no other functions for which the integral equal to 1 is satisfied or that all functions integrate to 1. This is not rigorous, but let's say you "bend" the line just a bit like you were pulling on a rubber band. In that case you would have a set of values for (dy/dx)^2 that were steeper than the value 1/4, and an integration for which the value would be slightly greater than 1. From there, it follows that all functions not being the straight descending line yield integrals greater than 1.
It has the feel of a variational calculus problem, but I don't think it really is. If the integral were equal to anything other than 1, I would conclude there would have to be an infinity of continuously differentiable functions that would address the criteria, therefore there would be an associated infinity of solutions for f(6).
Actually, it _is_ a calculus of variations problem. I used your method, then to prove the answer was unique, I showed the integral of (dy/dx)^2 from 4 to 8 is extremal when y is a linear function of x by using the calculus of variations. Since the integral of (dy/dx)^2 from 4 to 8 takes on this extremal value, the solution for y must be unique, as any variation would increase the value of that integral.
Just use Schwarz inequality, which in this extreme case is an equality. Then f'/f^2 is constant and f=1/(6-x). End_of_story
why would you ever think or get the idea to calculate (f'/f^2 +k)^2 ? what's the logical reason or steps to get there?
These are the types of math problems that always bugged me. Integrations on polar functions were the worst -- you applied one of twenty trigonometric identities, and if you picked the right one, it simplified after a half page of calculations. If you picked the wrong one, you usually knew after a page of calculations.
Cauchy-Schwarz
If you let g(x) = 1/f(x) then the integrand becomes (g'(x))^2 by a bit of manipulation involving the quotient rule...then you have g(4) = 4 and g(8) = 2. While there's no guarantee that g(x) is linear, you can start by seeing if there's a linear function that satisfies the 3 conditions given, which there is (namely g(x)=6-0.5*x; g'(x) is then -0.5 which means that the integral is the integral from 4 to 8 of 0.25 dx which does in fact equal 1)...this means that f(x)=1/g(x) = 1/(6-0.5x) or 2/(12-x) if you don't like fractions in the denominator. Which then yields the answer of f(6)=1/3...this is how I did it, though I suppose the jump to try looking for a linear function for g(x) was a bit or a stretch.
I do not have idea about this. Old curriculum and graduated earlier perhaps. I feel old. Huhuhu. It was so simple back then.
I thought I was smart. I was wrong. But then I graduated in 1970, so maybe just old. Even at my age I love these challenging videos.
Pretty interesting pattern of results for the even values of x
f(0) = 1/6
f(2) = 1/5
f(4) = 1/4
f(6) = 1/3
f(8) = 1/2
f(10) = 1/1
You cant find for x=0 1 2 3 9 10 because x lies in 4 to 8
Straightforward from Cauchy Schwartz inequality.
@@swayamjain1322 yes we can? it just said f(x) ≠ 0 in [4,8]. The function f(x) = 2/(12 - x) if you calculate the domain, it's every real number except 12
That’s what I thought!
Nothing surprising because 1/f literally turns out to be linear
f(8)=1/2 f(4)=1/4 it's an increasing function
So i guessed f(6) =1/3(becoz it looked like linearly increasing function) but I really like the explanation
Yahi log bade hokar desh dubayenge
It doesn't necessarily has to be an increasing function. It could be a weirder function with increasing and decreasing parts within the domain.
Nice guesswork
In VN, this is usually a multiple-choice question, and there are 2 types of people, one will calculate, one will just choose an answer base on gut-feeling.
I mean, there wouldn’t be a third type would it? :))
to think that i used to solve stuff like this in high school test is just unreal
I did it in 3 seconds
First it gives you f(4)=1/4 and f(8)=1/2
Since it is asking for f(6) which is inbetween 4 and 8 all u need to do is find what fraction is in between 1/2 and 1/4
At 3:08 you assumed that the function is continuously differentiable as we can only make the statement that if the integral of a non-negative function is 0 the function is 0 for a continuous function. In General a function can still satisfy this property and be non-zero on a measure zero set but since its continuous the measure zero set will be open and hence be the null set.
I believe we can go around this assumption by using cauchy-schwarz inequality and use the fact that equality is satisfied iff the 2 vectors are linearly dependent which finally gives rise to a simple differential equation.
The k trick is remarkable. It implies in general that for any g(x) if [Int_a^b g(x) dx]^2 >= (b-a) Int_a^b [g(x)]^2 dx, then g(x) must be a constant. I had noted in the problem that if 1/f(x) is linear then a solution is found, but it had escaped me that it is the unique solution because 1/f(x) MUST be linear.
Glad I dropped out of EE and went into construction....simple trig and some Calc for volumes made for fun times
1:20 This is technically not correct: You would have to tag the limits along with the u-sub. In terms of "u" certainly the boundaries cannot be 4 and 8. Then at 1:28 it continues correctly. I know, it is more of a "notational" issue but I felt I needed to bring it up
I just had flashbacks. dont use this stuff anymore but I was having flashbacks as you went through it.
There's a simpler way to do this problem. Based on the given data, we know that f(x) is always positive in [4,8]. Now, we can write the RHS of the integral, i.e, 1 as (1/4)*(integration of dx from 4 to 8). Move it to LHS and now RHS is zero. Equate the integrand to zero. And we end up with 3:20 but squared terms on both sides. We know f(x) is positive so we solve for that.
At 3:10, why should the integrand be identically equal to zero? It may be zero but there's no condition that implies it should necessarily be zero. Can't it be possible that the integral of non-zero integrand may have equal values at 4 and 8?
At 3:10 the integrand function is a _square_ function and the value of integral is 0. So the _square_ function is the nul function.
@@pageegapIt is not necessary for the function to be a null function in this case.
The only way you'd be able to solve this kind of problem in a test is when you already know how to solve it
Which is what tests are testing, so all good, right ?
They don't test your intelligence, they just want to see if you put your time in school to good use.
Facts
Or you can use intuition. If i took this test i'd be able to give the correct answer, although i couldn't explain it at all. f(4)=1/4 f(8)=1/2 so f(6)=1/3.
@@Pikachu-vj5jr hence the word “solve”, not guess.
Nope, that f'^2 /f^4 gives intuition that maybe f is 1/linear as then f' is 1/(linear)^2, and then you can check by the conditions on f(4) and f(8) by taking 1/f linear and see that the solution satisfied the integral condition too, if this is a mcq, then this is enough to tell you that if there is enough info to determine f(6), it follows that f(6) must be 1/3
Before he started the solution I was thinking I knew the answer but it seemed too easy but for once it was actually the answer.
It was an amazingly beautiful solution...
I very much enjoyed the innovation of the solution....
Thank you very much for sharing.
I did think of the f'/f^2 = -d(1/f), but did not know how to go further. Hard problem and an elegant solution!
Before I watched the video I looked at the thumbnail and thought "If f(4) is 1/4 and f(8) is 1/2 then f(6) is probably 1/3. Glad to see that my guess was right. :)
I have a question: at 2:50, you use k=-1/2 so that f(x) can be easily determined for this condition. But why can't k take any other value? Of course it will be very difficult, maybe impossible, to find another function that would satisfy this new condition... but this does not mean that another function cannot exist...and, if it does exist, it could well be that an alternative function would have a different value for f(6), and the answer to the question would not be unique.
So my question: is the given function the ONLY one that would satisfy the initial given conditions? If yes, how to prove it?
Try substituting k with another value and you will see this can only lead to the initial expansion at 2:03, no conclusion can be drawn. The value k has to be something that makes the integrand equal to a certain value (in this case 0) so that we can separate it from the integral and calculate the function.
Nice. I never would have thought of doing that. Letting g(x) = 1/f(x) simplifies the problem somewhat to: g(4) = 4, g(8) = 2 and ∫[4,8] g'(x)² dx = 1.
More generally, if g(a) = A, g(b) = B and ∫[a,b] g'(x)² dx = C, the same procedure shows that g(x) is linear = ((B-A)x+(Ab-aB))/(b-a), the slope is -k and C = (B-A)²/(b-a).
1:25: the last two 8s there should be 1/2s, and the last two 4s, 1/4s.
hint is derivative of 1/f(x) is -f'(x)/(f(x))^2. cauchy Schwartz may be helpful. most likely we may reach solution cauchy Schwartz inequality.
Love the way you solved it
Now this one really could have turned into a hard flashback for some student..
Swear to god I've seen question like this so many times that I lost count. This question and it's friends are quite the final boss in the university entrance exam, and I didn't beat them when I did the exam. What an experience
Where can i find more questions like this
With two complex numbers satisfy |z|=1 and |w|=2. When |z+ⅈw ̅+6+8ⅈ| gets the minimum value, what is the value of |z-w|?. A question from 2021 entrance exam in Viet Nam.
How on Earth did you know to do all that?! Vietnam university entrants are actually this brilliant?!
It is for the people who want to have high score. Most Vietnamese student can't solve it.
If they aim for top unis in VN they would try to solve this, for everyone else questions like these would be left blank.
This is amazing!!!
This reminds me of when I used to do calculus. Two pages of calculations to find an answer.
It's pretty easy to see that it will be 1/3 because 6 is halfway between 4 and 8 and 1/3 is halfway between 1/2 and 1/4.
The solution is really brilliant. After some thought I can motivate the substitution u(x) = 2 * f'(x) / f(x)^2. Then the given information can be written as:
int_4^8 u(x) dx = 4 and
int_4^8 u(x)^2 dx = 4.
However, I still find it hard to conclude from here that u(x) = 1 is the only solution, without using some smart trick.
I got it using the cauchy-schwarz inequality. In this case, int(g^2) >= (1/4)int(g)^2 with equality if and only if g is constant. Substitute in g = (1/f)' and we get an equality.
I've never heard of a cauchy Schwartz inequality involving calculus? Is that we'll known..I only know the vector dot product one..
How did you go from substituting g = (1/f)’ = -1/f^2 to f(6)=1/3?
Cool problem! Here's my solution:
Let g=1/f, then g'=-f'/f^2. So $g'^2=f'^2/f^4$, which is what we want!
Now we have: g(4)=4,g(8)=2
We have to deal with integral g'^2
Since (g'+a)^2 >= 0, we have g'^2 + a^2 >= -2ag'
Integrating all, we have 1+4a^2 >= 4a
If a=1/2 we have strict equality, so we would need g'+1/2=0 which means g'=-1/2
So, g is linear with slope -1/2 which means g(6)=3 which means f(6)=1/3
I think it's the same as yours but I found thinking in terms of g=1/f more intuitive
What kind of test in Vietnam?..high school or college or advanced senior in college or prisoner on death row?
4=(int_4^8 f'/f^2 dx)^2
*@Presh Talwalkar* Your headline is wrong for the content of video. Someone not being able to solve this difficult problem *does not mean* that they could pass the math test in Vietnam. The question simply should have been "Can you solve this problem on a math test given in Vietnam?"
Wonderful! I really like this video!
Does anyone agree that Presh's methid seems to come out of nowhere and Indont see why anyone would think of it..why integrate the square root of the integrand?
As with all the best math, I stumbled on a nearly identical solution to this accidentally. I was actually going in the reverse, trying to find a find an easy function g(x) so that the integral from 4 to 8 of (ag(x) + b)^2 would be 1. Then if you took g(x) = f'(x)/f(x)^2 (which I was now calling F'(x), where F(x) = -1/f(x)), this simplified down to (a + 2b)^2 exactly as you had but knowing the integral of F'(x) instead of f'(x)/f(x)^2 certainly felt more naturally motivated. Took me a bit to then realize that I could choose a and b so the RHS were zero.
when you said "pause the video if you'd like to give this problem a try"
At that moment I didn't have a paper or a pencil. So i just said if one of them was 1, one was 1/2 and the other was 1/4‚ i suggested and said of course it would be 1/3
now I think of myself as a mathematician ;)
Writing y = 1/f(x) one gets
dy/dx = -f ' (x)/( f(x) )^2
so given integration becomes
integration (4, 8).[ (dy/dx)^2 *dx]
But how would one solve that
Nope, couldn’t pass a test in Vietnam. Very humbling question!
Hey press I am very happy to see your amazing videos.
But I want to know what software do you use to making these videos.
Please reply
That was amazing. Multiple places where one might ask how did you know to do that?
This problem used to be a part of our national test, the funny part is that i can solve it easily when i was in high school.
If
4 = 1/4 and 8 = 1/2
6 is halfway between 4 and 8.
3 is halfway between 2 and 4.
So my guess is 1/3
I understand you’re probably joking with this comment, but since a couple others have commented similar ideas, I’m echoing my response to theirs:
That only works because f(x) happened to be a linear function. If f(x) ended up being more complex (with an x^2 term, or something else), then you wouldn’t have been able to interpolate the results in that manner.
No way to know without going through the steps in the video.
@@mflboys this was my legit guess because I haven't done calculus in 10 years.
I tried solving this way:
Notice that f(8) - f(4) = 1/2 - 1/4 = 1/4 = 1/4 * 1 = 1/4 * (integral from 4 to 8).
Therefore, 1/4 times the Integral from 4 to 8 is the same as f(8) - f(4).
I then claimed that f(x) is the antiderivative of 1/4 times the integrand (according to the Fundamental Theorem of Calculus part 2)
Differentiate both sides, I get 1/4 * integrand = f'(x), which is a differential equation, solved simply by separation of variables.
Then I isolated y and expressed it as a function of x, solved for c using f(8) and f(4) and went to evaluate f(6) but didn't get the answer. 2 different c values resulted from the given initial values.
I honestly don't know what my mistake was. If anyone can enlighten me that would be great.
From Vietnamese w love
Tks you
I think you lost an opportunity here: calling u=-1/f and rewriting the conditions and the question would make the rest of the solution a lot less mysterious. Also, finding a value at a certain point inside the interval is only possible when you realize the function as an extreme case of an inequality (where equality is achieved).
HOW would that help? And why think of that at all or why do what he did and solve for the square rootnof this integrand..it comes out of nowhere??
You mean substituting u=-1/f in the original integral? You'd get integral of [ f'(x) * u^2 du ] and I don't see what you could do from there
@@leif1075 Note that f(x) has to go from 1/4 to 1/2, so it needs positive slope somewhere between x=4 to 8. At the same time 1/f(x)^4 starts off as 256, so if the slope is large near x=4, then the integral of f'(x)^2/f(x)^4 is going to be WAY larger than 1.
So a natural question to ask is if I limit f'(x)^2/f(x)^4 to be small, say equal to 1/4 throughout so that its integral is 1, then will f(x) be able to rise fast enough? What will the value of f(x) be at x=8? f'(x)^2/f(x)^4=1/4 means f'(x)/f(x)^2 = 1/2. With some thought you'll see that f has to be of the form 2/(c-x). For f(4)=1/4, c should be equal to 12. Then f(x) = 2/(12-x), so f(8)=1/2, exactly what we wanted. We got lucky and we can find f(6)=1/3.
Now the question is how to prove that this is the only possible function. This I couldn't do by myself but looking at the video it kind of made sense although it was still tricky. If I can prove that f'(x)/f(x)^2 - 1/2 has to be equal to 0, I would be done. So integrate (f'(x)/f(x)^2 - 1/2)^2. Turns out that has to be zero because of the given conditions, and so our lucky answer is the only answer.
@@FineDesignVideos what do you mean it starts at 256? How you don't know what x equals you don't know that x equals 2^9 or whatever 256 is..
@@redtoxic8701 right and why would anyone think of substituting that. Didn't anyone try the substitution u equals y squared like I did..?
i guess when k!=1/2? it can be any number
1:43 Did you choose to do the square of the original integrand? If that is the case, then why use k? Sorry It has been a while since I have taken Calculus.
I think it can be done by rolles theorum