New Zealand exam leaves students in tears

[@maxwelllittle5291]

I guessed the cylinder one as h = 2r because it would be the closest to a sphere, which is the most efficient shape in terms of minimizing surface area for a given volume.

[@jellyfrancis]

That's pretty nifty 👍

[@davidseed2939]

the minimum SA of a capped cylinder is well known to be at 2r=h ie the vertical section is a square. V=πr²h SA = 2 πr² + 2 πrh = 2 πr² + 2 V/r d(SA)/dr = 4πr +2V(-1)/r² set to zero and substitute V 4πr=2(πr²h)/r² 2r=h

[@jellyfrancis]

@@davidseed2939 nah he's talking about the efficiency... it's a different concept that sphere is the shape that has minimized area for a corresponding volume

[@Yusso]

Close to a cube I would say. A cube has always the minimum surface area compared to all rectangular shapes (cuboids) with the same volume.

[@vincentpelletier57]

@maxwelllittle5291 It is an interesting intuitive understanding, but remember intuition is not a proof, so a derivation like the one in the video is required. Especially if this is the type of exam where we need to show our work. The strength of this intuitive understanding is as a confirmation: we calculate the answer, then think about it, and it makes sense, reinforcing our belief that we did not make a mistake. If the calculation had yielded something like h = 10r, it would look like a drinking straw, and we could intuitively say this is doubtful and then spend time looking for the mistake in the calculation.

[@everquill4275]

So, as someone who took this exam, this is why we were upset with these problems. The first problem (the numerically equivalent solution) was not an issue for me. I’d been doing some level 3 complex numbers in our curriculum previously, so I was pretty familiar with how equations could be manipulated in this kind of context. However, this was a merit question, not an excellence, and it followed a hard excellence question in the algebra booklet (which I am surprised was not also covered in this video, being that it was an interesting proof question). However, the second minimisation question was truely awful. It was the first excellence questions on a truely difficult calculus paper, and it required knowledge we had not been taught. I spent around 45minutes trying to do it, as I thought I had prepared myself for any minimisation/maximisation questions for the standard. However, it asked for knowledge outside of the standard, which is why many were “left in tears”. Also, for everyone saying that these questions are easy/ NZ math is bad, remember this was a country wide standard test- not an advanced math test. The government failed to standardise the test fairly, which is why many were upset

[@gazorper]

Life is not a race to the bottom to accommodate the lowest common denominator. It is a continuous striving to accomplish more. These questions are hard for for someone who wants a C, takes some work and thinking if you want a B, and is not really hard for someone who wants an A, as you should've been practicing (for any grade, actually) before the exam. It sounds more like the students in NZ have become lazy across the board, so at least that's standard, right?

[@stevenz933]

@@michaelz6555 I agree exactly. You learn that the minimum of a parabola is -b/2a and for this example you have a parabola that is symmetric with the y-axis so simply set that equal to zero which implies -b = 0. For this problem we have -b = 6k - 2 = 0 or k = 1/3.

[@thomasdalton1508]

@@michaelz6555 Why do people keep claiming it is a parabola? It has an r^-1 term in it. It is not a parabola.

[@thomasdalton1508]

@@everquill4275 Thank you for sharing your personal experience. What was it that you weren't taught for the second question? The syllabus only talks about differentiating polynomials, but the power rule is the same whether the exponent is positive or negative. Did your teacher just never mention that or give you any examples with negative exponents?

[@thomasdalton1508]

@@stevenz933 That's the first question. He said the first question was fine. It was the second question that he said he hadn't been taught.

[@martincohen8991]

For the question posed at the start, if f(x) is the function you want f(x)=f(-x) so (x^2-2x)/(4x-1)=(x^2+2x)/(-4x-1). You can cancel the x since x e 0, so (x-2)/(4x-1)=(x+2)/(-4x-1) for which x=+/- sqrt{2}/2. Solving z=(k-1)/(k+1) gives k=(1-z)/(1+z)=1/3.

[@salnikau]

The first question is trivial: in order for a parabola to have 2 symmetric solutions the peak should be at x=0, so f'(0)=0 which is -6k+2=0

[@craigabbott6402]

This is what I saw too. To have equal but opposite roots, it must reflect about the Y axis, which means that it must have its minimum at x=0 so f’(0) =0 and the rest of the solution follows.

[@douglaswolfen7820]

True, but don't forget that things are only trivial if you're already very familiar with them. You have to understand them well and recognise them when you see them That might make the question easy for the best students, but I wouldn't expect this to be easy for most students

[@AdamantineAxe]

❌ "You didn't show your work."

[@Speak22wastaken]

Except question one was on the algebra exam, so you have to use an algebraic method for marks, best way is to use difference of two squares, in the exam I used the aspects of the quadratic formula

[@jffrysith4365]

@@Speak22wastaken as someone from nz you will receive 0 marks if you do it 'incorrectly' or with the wrong reasoning. So even though this is a really cool solution it would fail the test... NZ has a garbage highschool education system lol.

[@anonymouscheesepie3768]

let u = (k-1)/(k+1) x^2 - 2x = 4ux - u x^2 - (2 + 4u)x + u = 0 For a quadratic that has two unique but opposite solutions, the coefficient on the x-term is zero. Hence 2 + 4u = 0 u = -1/2 Then we can resubstitute for k (k-1)/(k+1) = -1/2 k - 1 = (-1/2)k - 1/2 3/2k = 1/2 k = 1/3

[@migBdk]

I solved it that way in my first try. Then I realised I could do even faster by not making the substitution. From my other comment: I solved the first problem in a different and faster way (maybe also easier): Begin with cross-multiplying again, to get (x^2 -2x)(k+1) = (k-1)(4x-1) Realise that this equation is quadratic in x. Remember that in the solution formula for the quadratic equation the denominator is -b plus or minus sqrt(D). Obviously if and only if b=0, we get two solutions with opposite signs and equal value. And b=0 only if the terms of first order in x on both sides are equal. You can then write the new equation with -2x*(k+1) = (k-1)*4x, which are just the first order terms of x on both sides. Reduce out the x factor to get -2(k+1) = (k-1)*4 solve for k to get k = 1/3.

[@disgruntledtoons]

Your first question solution was too complicated out of the gate. We all remember that if the roots of a quadratic are equal in magnitude and opposite in sign, then the quadratic has no linear term. Therefore, select the value of k so that the product of (k-1)/(k+1) with 4x equals -2x to cancel out the -2x on the left side of the equation. (k-1)/(k+1) therefore equals -1/2, giving k = 1/3.

[@Guyofburning]

This was used in the second solution, except they show why there is no linear term.

[@barrymamadousaidou5164]

I have the same approach ; it is quite straightforward.

[@Illusioner_]

I think the first solution was easier, since it's based only on the fundamental theorem of algebra, which everyone knows, regardless of what you have learned about quadratics

[@paulstelian97]

@@Illusioner_ The fundamental theorem of algebra (namely that you have as many complex roots, potentially overlapping, as the degree of the equation) isn't exactly trivial and known before you learn quadratics...

[@kniels2856]

When actually sitting an exam, many people find it difficult to have such "insights" even though they know the curriculum completely. In my experience with students at a variety of levels, very close to none would solve the problem (in an exam situation) as you propose. In fact, I would argue that even the solution in the video in the video requires some realizations or "insights" that most students would simply not have under pressure. Insights are almost always obvious in hindsight and much easier to come by after years of experience (which is what it feels like is going on in this comment section). In this instance, I feel fairly confident that the majority of students that come up with the correct answer do so by simply brute-forcing. I.e. doing the the cross-multiplication, isolating, plugging into the solution for a quadratic equation, setting the derived solution for one root equal to minus the solution to the other root, and then solving for k.

[@laszloliptak611]

For the first problem I would always note at the beginning that the denominators cannot be 0 (so x can't be 1/4 and k can't be -1). Then after cross-multiplying one gets a quadratic equation in x, and the condition requires that the linear term vanishes. One then still needs to check that the resulting value of k will give real solutions and not the forbidden value of 1/4. For the second problem, it is not enough to use the second derivative test, because it will only show that you have a local minimum, not necessarily a global one. One needs to argue, e.g., that since there is only one critical point and the domain is an interval, the one local minimum must be a global one.

[@chrisglosser7318]

Prob 2: “given a rectangle with perimeter p what is the maximum area?” A = (p/4)^2 (a square)

[@arnoldn2017]

I would’ve been crying too, with joy, it took me around one second to understand to get the solution of this equation.

[@zortnet]

right? thats early high school math!

[@chintansatani6514]

Too much late.... Just scrolled and got answer in nano sec😂

[@andreare7766]

Quite!

[@emad3241]

rookie numbers, i calculated the solution before the video was released

[@bobross7473]

Stop bragging and/or lying. You’re not special.

[@MirlitronOne]

I was shown the solution of "The Baked Bean Can" problem (as we called it) as a schoolboy in the 1970s. Decades later, as a fully-qualified materials scientist, I had to solve a problem for calculating the average cell size in a cellular polymer. Guess what? It came down to the same maths. Nothing is wasted.

[@AbeIJnst]

Shouldn't you be checking back your answer, both for k and x, since you're cross multiplying? Denominators can't be zero.

[@bobross7473]

With k=1/3 there’s no zero denominator

[@AbeIJnst]

@@bobross7473 Neither with x=±√0.5. But in my days I would've had a point deducted for not mentioning x≠0.25 and k≠-1 and another one for not checking back on those conditions.

[@ArandelaGriffe]

1/ A simpler way for students could be setting F(x) = K, solve the equation and find K then solve equation K = f(k), then find k. Problem is solve without fear... 2/ Instinct : Put a sphere inside as Euclide will said; Do math : find the same result. THANKS EUCLID, ^^

[@kimba381]

The second of these is a standard type of calculus problem. Indeed, when I was teaching, this specific problem: minimise the surface area of a cylinder, was always worked in class.

[@RAFAELSILVA-by6dy]

The critical value must be a minimum, as the surface area is unbounded as h tends to zero. There cannot be a maximum. It's poor technique to plug in V = 500ml at the outset. As the questions become more advanced, the numbers can become unmanageable compared to the algebra. Keeping numbers out also allows you to check for dimensional consistency throughout. So, if there is an error, it's much easier to find in algebraic expressions.

[@davidhowe6905]

Very good points; yet many of the students I used to teach (in the UK) had done Physics Advanced level (typically at age 18), in which dimensional analysis was explicitly omitted from the syllabus!

[@RAFAELSILVA-by6dy]

@@davidhowe6905 I've noticed that the A-level syllabus allows things like x = sin(t), which gets students off on the wrong foot altogether. I spot so many of my errors when I feel that the dimensions have gone awry. Also, the relationship between physical quantities and units is fundamental. I came at physical from a pure maths background and these things were really important insights for me in getting to grips with physics.

[@davidhowe6905]

​@@RAFAELSILVA-by6dy Yes, they try to simplify things even by using incorrect ideas; another example is saying velocity = delta x/delta t so as to avoid derivatives. Even though many of my students did calculus in Maths A-level, they often found it conceptually difficult to apply it to physical problems. Also, more than once, when I met students who had progressed from my course, they thanked me for teaching dimensional analysis; it saves SO MUCH time if you have made an algebra error!

[@hjs6102]

Second problem, same for squares, cubes etc. If you want minimal surface, you must have equal side length. Third bionic formula...

[@paulstelian97]

Problem is how do you do exact equal side length when the sides have different shapes.

[@mihailghinea]

​​@@paulstelian97a cylinder is either an infinite number of circles stacked upon eachother, or a rotated rectangle(in our case a square rotated around one of its axis that is parallel to one of its sides).

[@hjs6102]

@@paulstelian97 I suppose, in a paralelogram, for example, the base side and the height must be of identical length to maximise the area, but I haven't a proof for it. Would be worth a second thought ;)

[@paulstelian97]

@ The closest to sphere/circle tends to be the best, but always needs to be proven. So the best parallelogram for a certain perimeter will be the square, as that will be the one with the highest area. Although I wouldn’t be surprised if any rhombus also hits that optimal area.

[@MercuriusCh]

The first question doesn't have a complete answer, because you are always obligated to check for existence of solution) Good happens, and k=1/3 leads to x^2 = 1/2 => x = +-1/sqrt(2). And the denominator of original question does not turn into 0.

[@StyrbjörnMårtensson]

Er, the first problem is much easier that that: The equation needs to be possible to write on the form x^2-a^2 = 0 (with roots +-a) which means you only need to rewrite the equation as a second-order polynomial with the x-coefficient being zero, which happens when 6k-2=0, i.e. k = 1/3. (One also needs to verify that the coefficient representing a^2 > 0.)

[@grawuka6900]

For the first question, depending on the curriculum content, you may need to also check that the solution for k gives real solutions for x to have full points. For example: changing (4x-1) to (4x+1) gives the same solution for k but no real solution for x.

[@Moleculor]

8:19 What is the 'simplification' process that takes you from 500/(π((250/π)^(2/3))) to 2*(250/π)^(1/3)?

[@go_gazelle]

r = (250/π)^(1/3) r = 5 * (2/π)^(1/3) Solve for h. h = 500 / (πr²) h = (500/π) / [ 25 * (2/π)^(2/3) ] h = (20/π) * (2/π)^(-2/3) h = [ 5 * 2² * π^(-1) ] * 2^(-2/3) * π^(2/3) h = 5 * 2^(4/3) * π^(-1/3) h = 5 * 2 * 2^(1/3) * π^(-1/3) h = 10 * (2/π)^(1/3)

[@ChimezieFredAnaekwe]

The lid aspect of the question is actually confusing unless there were given the formula. Because the surface area of the Lid can is πR²-πr² which must be incorporated into the surface area function.

[@taflo1981]

7:47 I always tell my students that the second derivative test should not automatically be the method of choice, especially if you're looking for *global* extrema. Checking the sign of the first derivative outside the stationary point or just comparing the value at the point with the limits of the function towards the bounds of the range is usually easier and works even if the second derivative test fails. In this case, the argument for having a *global* minimum at the value r=R that we determined should be either of the following: (a) The range for r is all strictly positive real numbers, and the surface area clearly tends to infinity if r tends to either 0 or infinity. Since the function for the surface area is furthermore continuous, it has to have a global minimum. Since the function is also everywhere differentiable (even continuously so), r=R is the only candidate, and we don't even need to check the second derivative. (b) The derivative is clearly negative for rR, so the surface area strictly decreases all the way until r=R and strictly increases afterwards. As the range is one single interval, we have a global minimum at r=R.

[@Bens963]

Thanks for the upload, I haven't done math in quite some time but these were fun and interesting to solve

[@lollol-tt3fx]

at 4:37 you could also see it in another way(transforming parabolas)

[@cm5754]

The first also has solutions +-0 with k = 1. The problem did not say the values of the solutions were distinct, only that they had opposite signs.

[@thomasdalton1508]

Apparently the problem with the second question is that they are only supposed to cover differentiation of polynomials in this year and differentiation of other functions isn't covered until the next year and technically that isn't a polynomial since it includes r^-1 but the rule for differentiating negative exponents is exactly the same as for positive exponents, so I don't understand why they are separated out. It seems like poor drafting of the national curriculum to me - it says polynomials when what it means is powers.

[@itsphoenixingtime]

It's irresponsible. You cannot just casually throw in a question that doesn't test what is stated for the year and then expect students to solve it. Good on those who somehow knew, but this feels like an irresponsible division of the curriculum. The power rule is always taught for all exponents, polynomial or not. Negative exponents pop up a lot in these problems where there is optimisation. Regardless, I don't blame these students for crying. It's easy to me but that's because I was taught all the differentiation already. Imagine just walking in and suddenly you see a negative exponent and you weren't even taught how to do it???

[@thomasdalton1508]

@itsphoenixingtime Yes, if you've written the curriculum wrong you have to stick with it just in case some teachers have bizarrely not taught the power rule for all exponents at the same time, but the harm should be minimal. They'll have completed the first half of the problem before getting to the differentiation so will get half the marks for the question anyway. It is normal for people to cry after exams. It's a natural response to the emotional release that comes after a high pressure situation ends. The fact that the exam contained a difficult question is not the reason people were crying. All exams contain difficult questions.

[@itsphoenixingtime]

@@thomasdalton1508 Could be that or they are just taken aback by the fact they are just thrown something they are not expected. Which I don't blame them for. Exams, as much as we hate to say it, are treated as a "do this and only this thing", because of how rigorous they are. So when they're only taught + exponents and they see a -, they have no clue. And it's not like they would inherently know the rule applies for all exponents... I also hope whatever you said is true and this is only a few marks, so the damage is minimal. If this applied to most of the questions it would be even worse.

[@Speak22wastaken]

Im from nz, I took this exact exam, we learn the power rule in year 12, which includes negative powers, also since these are excellence questions getting one of them correct instantly makes you pass the paper, there are 6 excellence questions per paper, and you still don’t need them to pass, these shouldn’t be easy

[@orionspur]

Calculus helps a lot, but is not strictly necessary for either problem. Parabolas have an obvious minimum.

[@mikoaj1349]

Problem 1: construct the quadratic equation with a parameter k as you did. And then the solution follows directly from Vieta's formulas: for an equation ax2 + bx + c = 0 we know that the sum of its roots - which in this case is 0, as they are the same number with opposite signs - equals -b/a. Although do remember that x can't equal 1/4 (otherwise the denominator in the original equation would be 0), so formally we should check whether the solution to the equation with k=1/3 isn't 1/4. In such case the answer would be: there's no such k.

[@SuperPassek]

We can solve the second problem with Cauchy-Schwarz inequality. When r^2 h = c, a constant, r^2 + rh = r^2 + rh/2 + rh/2 >= 3 (r^2 * rh/2 * rh/2)^(1/3) = 3 c^(2/3), so it is the minimum value, and the minimum value is achieved when r^2 = rh/2, which means h = 2r.

[@saurishjayam5679]

Wow, nice solution. I would never have thought of using Cauchy-Schwarz inequality in this type of question.

[@qipengliu2627]

or just lagrange-multipliers it....

[@ashish_45playz42]

isn't that am-gm?

[@SuperPassek]

@@ashish_45playz42 Oops, you are right. Sorry for the confusion.

[@HedwigBleicher]

It might be helpful to keep in mind that there can't be a maximum solution. By reducing the diameter and extending the height it is possible to increase the surface indefinitely while maintaining the volume. Same applies to an increased diameter with an ever smaller height.

[@schifoso]

Parents outraged. Students in tears. When will it ever end?

[@AlejandroCastilloRamirez]

when students, study properly and we stop the culture of "you are excellent just by existing"

[@sandikay3323]

They weren't whining like the ones you're commenting on. Those questions were extremely difficult even for advanced math students.

[@kevinmadden1645]

In other words no trophies for just going through the motions .

[@matrboyd6076]

You are back. These are very nice problems, not very difficult, but I have to say bravo to you for posting them, and bravo to New Zeeland for testing kids with 2 great problems.

[@m1k1_15]

thats lietrally beginer level quadratic formula question

[@setredid]

For the first question it is also important to account for +0 and -0, since it is numerically equal with opposite signs. Since k=1/3 and k=1(solved by subbing x=0)

[@noakidoesmaths]

thanks for covering my channel lmao ♥

[@hevado01]

Quadratic formula is clear: opposite roots when b=0. Cross multiplication and combining equal powers of x gives b=0 for k=1/3

[@ritumammathsclasses8452]

That's such an easy question for even an avg Jee Student in India...just apply componendo and dividendo....We get a simple quadratic in X...Now put its Sum of root equals 0..we get the ans 😊

[@person5551]

Yeah New Zealand has pretty terrible education. I mean I'm sure it's far from the worst but yeah.

[@bobross7473]

You’re not special

[@BytebroUK]

Not sure - does that cover the clause in the question which said about "including the lid"? I may need to revisit the vid!

[@afesfrytuykjhyj]

As someone who took this exam, I can say the news is exaggerating the exam a lot. I found these questions quite easy, "top students" were not crying and teachers at my school at first thought they were referring to a very different question in the paper. For the second question, students couldn't recognize a/x as ax^-1 and use the product rule and some teachers hadn't "taught" their students. The teachers at my school feel as though the paper with the second question was far easier than the paper with the first question as the calculus paper was only basic applications and understanding of integration and differentiation with formulae given to you that you could instantly use to get the answer whereas the algebra paper you still had to put in some effort to get the answer even with the formulae sheet such as rearranging, understanding of how quadratics are modeled as shown and general understanding of what the question is asking (NCEA maths is an odd mix of some English and maths).

[@jannemec2483]

Square has higher area/border than and rectangle, so for cube it will be even more powerfull. And so because cilinder has just cut corners, it still aplies, that for max volume/surface, it should be made from a square and therefore h=d=2r

[@MegaSuperEnrique]

2nd equation is useful not just for heat transfer, but for minimum metal used per can. I would think that would be a great real-world example, but there seem to be few cans, cups, bottles where height = diameter.

[@afesfrytuykjhyj]

This question has been paraphrased many times in textbooks, worksheets, practice exams, etc for NCEA L2 (the qualification gained from this exam). Many of them use heat transfer and minimum metal to reduce costs as context.

[@lixiaonong1646]

Wait is that NaokiDoesMath at 7:20?? i remember watching his ncea past paper speedruns and they were so good and informative

[@Whiplash_MUSIC-hj7vp]

I'm a first year uni student, and I solved the first problem pretty quickly by expanding and then taking the derivative and subbing in x=0, since the vertex of the quadratic would have to be on the y axis to get the desired result. Understanding the middle term must be 0 is definitely a lot more elegant tho lol.

[@satyareddy6382]

Though I cannot justify how well their exam went but if that was my question paper I would be so happy, like bro the question that made students cry yeah I understood it immediately and if they think its bad i have to write every step( I could do half the math in my head but our teachers don’t allow that) :)

[@Yusso]

I think the cylindrical cup problem can be solved easier. In order to have the minimal surface the cup must be close to a cube. That would mean 2r = h or r = h/2. V = 500 = πr²h, substitute r with h/2 500 = π(h/2)²h = π(h²/4)h = π(h³/4) 500/π = h³/4 2000/π = h³ h = ∛(2000/π) ≈ 8.6 r = h/2 ≈ 4.3 After typing this I realized it's easier to substitute h with 2r 500 = πr²2r 500/2π = r³ r = ∛(500/2π) ≈ 4.3 h = 2r ≈ 8.6

[@kimba381]

You assumed that h = 2r, you have to demonstrate it. Once you have established that as a fact, the rest is trivial

[@paulstelian97]

That is a guess-and-check solution though, even if you can prove it. But those solutions are good because you can do it with fewer tools and thus you can do that in e.g. high school.

[@Yusso]

@@kimba381 The minimum surface area for a given volume of a cylinder occurs when its height equals its diameter. Same with a cuboid, minimum surface occurs when you have a cube. So those are facts for sure. The question is do you need to prove that every time?

[@Yusso]

@@paulstelian97 It's not a guess, it's a fact.

[@kimba381]

@@Yusso In a Calculus exam, yes you do,

[@9adam4]

With the left side being f(x), I solved f(x) = f(-x) for x, then subbed in to find k.

[@hugh081]

Don't even need the quadratic formula, we know from the FTA that the resultant polynomial must be of the form x^2-a^2 so the coefficient of x is 0.

[@migBdk]

I solved the first problem in a different and faster way (maybe also easier): Begin with cross-multiplying again, to get (x^2 -2x)(k+1) = (k-1)(4x-1) Realise that this equation is quadratic in x. Remember that in the solution formula for the quadratic equation the denominator is -b plus or minus sqrt(D). Obviously if and only if b=0, we get two solutions with opposite signs and equal value. And b=0 only if the terms of first order in x on both sides are equal. You can then write the new equation with -2x*(k+1) = (k-1)*4x, which are just the first order terms of x on both sides. Reduce out the x factor to get -2(k+1) = (k-1)*4 solve for k to get k = 1/3.

[@rayxxkaiser3586]

I can't figure out the first one because I am not still familiar yet but I can fix the second one alone because I'm studying Pearson but running through the Cambridge IAL for the first volume. Set Once acquire Ar = 2pi()r^3+2pi()r^2l (Substitute 2pi()r^2l=V=500), we will obtain the equation with only Area A and Radius r. Then, we examine (or the memory of similar questions) r->0 A->infinity, r=1,A=2pi()+1000, r->infinity, A->infinity, so that it is a curve equation with stationary point(s, but of course, there is only 1 stationary point demonstrated afterward in the video). The curve must concave downward so that dA/dr = 0 is the stationary pt, and as it is only the stationary pt, we don't need to check d^2A/dr^2 > 0, but directly find out the value of r, and backward substitution let l and A can be found. Now studying chemistry and then physics and finally back to maths for some past paper training soon.

[@KyriZee]

what level have these questions being asked at? They seem pretty standard IGCSE math questions or typical for any y10 - y11 high school pupil.

[@franciscook5819]

I solved this then skimmed the video. I suspect what you were meant to do is equate the "+x" and "-x" equations (so the -2x and +4x terms swap signs). That gives x=±1/√2. Equating that to the "k" expression gives k=1/3. For clarity ... (x²-2x)/(4x-1)=(x²+2x)/(-4x-1) and you can immediately cancel an "x" from both sides, cross multiply to get -4x²+7x+2=4x²+7x-2 or 8x²=4 so x=±1/√2. Sub into the original equation gives -1/2=(k-1)/(k+1) so k=1/3

[@handanyldzhan9232]

(x^2 - 2x)/(4x-1) = (k-1)/(k+1) = (x^2 + 2x)/(-4x-1) Let's subtract both the top and bottom sides = -4x/8x = -1/2 = (k-1)/(k+1). Then k = 1/3.

[@tvrashid]

500 ml = volume Surface Area = 2πr²+2πrh or 2πr(r+h) Volume = πr²h = 500ml or 500cm³ For minimal surface area, Diameter and hieght of the cylinder must equal. That means 2r=h πr²h=πr²2r= 500 cm³ 2πr³= 500 cm² r³=(250/π)^(1/3) r = 4.30 cm Height = 8.60 cm Volume of cylinder = 500 cm³ Surface Area = 348.52 cm²

[@hassanawdi3793]

In the first problem why k+1=1 gives different answer? The whole equation is equal to x^2-r^2 so k+1=1 and -6k+2=0

[@moolahman9043]

as someone who took this exam this is the one question i couldnt solve. thanks for explaining how to do it!

[@amoghopprasad8286]

Bruh the 1st one was an average 10th standard question in india💀

[@amadeus6834]

The equation is solveable for a bit easier; let f(x) is the left side of the equation, and we know f(x) = f(-x) for some x (if x=/= 1/4 or - 1/4), so we have this equation: (x^2 - 2x) / (2x - 1) = ((-x)^2 - 2*(-x)) / (2*(-x) - 1) This equation has 3 solutions, x=-1, 0 and 1. So we know what x values can be, so we can calculate k for them.

[@williamhuang8309]

Daily Dose of NZQA moments

[@SrinivasR-k7m]

As someone who wrote few entrance exams in India, I cannot believe that these questions are actually hard🙄

[@Player-fg4ub]

thats enough there "srinivasR". as a person who "wrote few entrance exams in india", you should be pretty smart yourself, shouldn't you? comparing to other countries standards is pretty immature in my opinion im from singapore btw

[@SrinivasR-k7m]

@Player-fg4ub Well I didn't mean in that way I never knew about other countries

[@TheJaguar1983]

For the quadratic (let u = (k-1)/(k+1) for brevity), I divided both sides by 4x-1, then subtracted u(4x-1) from both sides giving, after simplification x^2-(2+4u)x+u=0. After plugging it into the quadratic formula, I saw that for the roots to fulfil the requirements, the b term must be zero, so I just solved 2+4((k-1)/(k+1))=0 for k.

[@KekusMagnus]

The power rule is the same regardless of thd exponent, how are negative exponents a problem

[@mihailghinea]

For the 2nd question you really do not need dif calculus... you just need to proove that the smallest perimeter to area ratio, for a rectangle is obtained for the square. Then if you spin that square around one of it's simetry axis that is parallel to one of the sides, you will get the cylinder with the smallest SA to V ratio.

[@chromadl7275]

If you think the level 2 exams were unusually difficult, the NCEA Level 3 and scholarship also was very strange

[@edenli6421]

Question 2 was considered too difficult, out of the standard. Because it expected students to be able to differentiate a negative power, but many weren't taught that as technically it's for the next year level of students.

[@avisoni3948]

why do you wish to take the derivative in first place instead u can use the function that the minimum value of a quadratic is when x=-b/2a here x will be the radius as the quadratic is in r and we then have a relation b/w radius and height which can then be used in the volume formula to find the dimensions

[@kevinmartin7760]

Cans of corn don't have those proportions just for giggles... Although it is probably tweaked to account for the extra metal used in forming the seams.

[@cuzx1]

Amazing as always. Are you interested in looking at the mathematics game rougelike “calculate it”

[@RAFAELSILVA-by6dy]

Multiplying through by r gets rid of the 1/r term - but that then requires implicit differentiation - which I guess won't have been taught either.

[@brianwade4179]

For the first problem, that diversion through the quadratic formula was superfluous.

[@mdioxd9200]

For problem 1 the way you solve it is: - turn that whole thing into a polynomial - LET'S SUPPOSE we have 2 solutions - then sum of roots is blablabla and solve Thing is, you never check that your result actually gives you 2 solutions. In french that kind of reasoning is called "analysis synthesis", here you've done the analysis, but not the synthesis part, you've checked that "if my solutions exist then they'd have this form" which is the analysis part (suppose the existence of the solution and then deduce the form of the solution) but then you didn't do the synthesis part (prove rhe existence of the solution by showing that your supposed solution actually works) The reasoning is actually incomplete

[@happygamer787]

The way I thought about it is if we take a certain parabola ax^2 + bx + c = 0 and look at what equal roots with different signs really means, we can see that if the roots have this property, the parabola is also symmetrical through the y-axis. Now if we go back to the basic formula we can see that changing a doesn't break this formula's symmetry through the y-axis, as this first x has an even power. The same concept applies for c, as x^0 = 1 and 0 is also even. But only for b does changing it's value mess up the symmetry of the parabola, this is because it's respective x has an odd power> So to eliminate the asymmetry of the parabola, and by ways of doing this also make the roots have the given property, b must be 0. Now just solve for b = 0 and you're done.

[@ejd53]

The first one is just simplification, it's not that difficult, and the second requires very basic calculus. When I attended high school (a very long time ago), these questions would be rather common and nothing to get upset about. So, what is the problem? Is it students not studying properly, or a disconnect between the examination authority's expectations and what is actually being taught to the students?

[@gabrielbarrantes6946]

You missed to verify that the solutions are real. Also, the vertex of the parabola is in -b/a 2a, so the condition happens when b is 0, no need to add the roots (small optimization)

[@itsphoenixingtime]

My logic was that any sort of quadratic equation with "roots that are numerically equal but of opposite sign" are just differences of two squares, so the equation needs to be of the form x^2 - a^2 = 0. Only thing I cared was the lack of an x term, so i expanded, collected, and found all the terms with x, and set the coefficient to 0. Then I solved to get k = 1/3, checking to see if it caused any invalid solutions [didnt other than k not being able to equal negative one and the root cannot be x = 1/4 bc of div/0 errors] The second one is a pretty standard differentiation problem for me, but I understand that they apparently were not taught differentiation of negative exponents? Isn't that a bit strange? Usually differentiation is taught with the power rule in mind which applies for all n's, so it feels weird that they would just leave it out. Either way it is crucial to know for the problem even if you did set it up. Interesting fact: To maximise the efficiency of a double-capped cylinder with a specific volume or surface area aka Case 1: Maximising the volume for a given surface area Case 2: Minimising the surface area for a given volume the one true fact that unites it is that the cylinder must have a height twice its radius, or in another way, its diameter is as tall as its height. In a more colloquial way the cylinder must be as wide as it is tall.

[@davidseed2939]

its actually quite easy. if the roots of the equation are +r and -r then b=0 so if (k-1)/(k+1) = a then x²-2x = 4ax -a x²-2x -4ax +a=0 b=0, a=-1/2… x²- (1/√2)²=0, x =±1/√2 (k-1)/(k+1)=-1/2… 2k-2=-k-1, 3k=1 k=1/3😊

[@maxhagenauer24]

Where is x^2 - 2x - 4ax + a = 0 coming from? And why can you set the initial right side of the equation to - 1/2?

[@brads2869]

I am missing something with the first question and the answer. If k=1/3 is the answer, why don't the two sides of the equation match when I substitute that value of k and random values of x. e.g. if k=1/3 and x = 1, the left side = -1/3 and the right = -1/2, which are not equal.

[@emanueller3554]

You can't choose a random value for x, you have to solve the equation. I found x=1/√2 or x=-1/√2

[@bernardliddington2633]

The issue was that the question was not set at the correct curriculum level. That’s the issue, not that those with higher maths learning can solve it.

[@geoninja8971]

I set up the first one and formed the required quadratic, but then couldn't take that last step.... of course looking at it now it's obvious. I think today's teachers are just pissed they couldn't solve it either.... the second one is a nice fun calculus prob, thanks for the brain stretch....

[@JoJoModding]

The first question is ill-posed. It asks to find the roots of an equation, but that makes no sense--equations don't have roots. Functions have roots. The question wants to ask for the values of k such that the solution set of L is {x, -x} (i.e. nonempty and invariant under negation)

[@keysistoogood]

I solved problem 1 by thinking about the shape of the parabola. If the quadratic has roots that are equal numerically but opposite sign then the point where the parabola crosses the y-axis must be the point where the slope of the graph = 0. So if you differentiate the equation and set both y and x equal to zero you can solve for k.

[@marcinbednara3825]

What about domain of the fundamental quation? Why you doesn't just use the Viete's theorem?

[@mcarston]

In the first solution to the first problem, why do we not also require the coefficients on x^2 and the constant term to match? For instance, why do we not require k+1 = 1? I don’t understand why we have only matched the linear term.

[@tvrashid]

In the given question, it was given that both roots are the same numeral with different signs. That means Root 1 * -1 = Root 2 Root 1 = -b+√b²-4ac / 2a Root 2 = -b-√b²-4ac / 2a, which equals Root 1 * -1 Hence b+√b²-4ac / 2a =( -b-√b²-4ac / 2a)*-1 -b+√b²-4ac / 2a =b+√b²-4ac / 2a Cancelling /2a in both sides gives us -b+√b²-4ac =b+√b²-4ac Cancelling +√b²-4ac both sides gives us -b=b So the equation to find k reduced to b=-b where b is the co-efficient of linear term. Therefore we need not consider a and c for solving the question.

[@mcarston]

I appreciate the response, but that is focused on the second method to the problem. I was inquiring about the first method, where he only matches coefficients on the linear term, but not the others. He completely waived his hands at that.

[@qrasy]

Ok, so he didn't label the first question with "method 1 and method 2" but the timestamp of the "first way" is around 3:26

[@qrasy]

A problem here is that if you multiply/divide the whole equation by a nonzero constant, it's still the same curve. Dividing by k+1 gives only one equation for k and leads to the intended result. If you could match both the linear and quadratic coefficients then it would lead to 2 equations for one unknown k and a contradiction. Maybe it's exactly because he doesn't want to explain the issue that he focused on another way instead.

[@jasonarmstrong4640]

As you mentioned, solving for k would require matching all three terms, but this is not what is being done here. Instead, the roots condition provides a specific relationship between the coefficients, which allows us to determine k based on the behavior of the quadratic equation's roots.

[@TJMoir]

Did it in my head more or less but I suppose when I was their age I would have found it more difficult. I found it because there is no 'damping' term like in simple harmonic motion so the coefficient of x must be zero.

[@benjaminnice]

Wouldn’t you take one of pi x r^2 off because is surface area of a cup and not a cylinder? With closed ends?

[@nothingisreal6345]

I think both questions are very simple indeed. Solve a quadratic equation and compute the derivative a polynomic function. and I don't know but the power rule for the derivate works for positive and negative values.

[@matyourin]

The h=2r is not really surprising: the cylinder tries to be as close to a sphere as possible.

[@Speak22wastaken]

I took that exam, I didn’t find any of the questions particularly difficult, ofc I was also taking the scholarship course, but the first one was a question people didn’t get because they didn’t understand the quadratic formula, and the second one is just standard optimisation, I’d also like to note that these questions were likely both excellence questions, which means if you get one of them right you will immediately pass the paper it’s on, even if you get everything else wrong, imagine the questions were easy

[@Magst3r1]

Well, I mean, can you blame people for not understanding the quadratic formula? It's never really explained how or why it works, we're just told to use it.

[@Speak22wastaken]

@@Magst3r1 Thats a blame the teachers scenario then, it should be taught

[@cmuller1441]

First x=¼ and k=1 are forbidden. Then multiply by the denominators and get a quadratic equation. X=(-b±√∆)/2a Of course a0 or it's not quadratic and no symmetrical solutions. So we need ∆=b²-4ac>0 and b=0 also -4ac>0 ...

[@BryanLu0]

For exam questions like this. It's not really necessary unless you have multiple solutions coming out

[@bobross7473]

I’m pretty sure they learn tricks to make it faster than having to use the quadratic formula. Like the vieta’s rule for sum or roots is pretty neat but I didn’t think of using that on the spot since it’s been a while that I’ve practiced these kinds of problems.

[@robertveith6383]

You wrote that wrong. 2a in the denominator must be inside grouping symbols.

[@Omar_FarukQuran]

this is too easy bro

[@qwertyuiop7007]

these are tough?

[@IIT_JEE-2027v]

U can use compenendo and divinendo for getting the answer faster in the first problem

[@mr.d8747]

*How is that hard? Just convert it into the form **_ax² +bx +c_** and use the Viète formula for the sum of the roots: -b/a = 0*

[@jayvardhankhatri4084]

Here is another solution for problem 2 without calculus we know πr²h=500 h=500/πr² putting h in SA=2πr(h+r) we get: SA=1000/r + 2πr² which can be re written as :500/r +500/r +2πr² since we know r and h are positive real numbers we can apply the AM GM inequality AM>=GM therefore: (500/r+500/r+2πr²)/3>=(500/r x 500/r x 2πr²)^1/3 splitting the 1000/r term into 500/r+500/r makes sure that the rhs of the inequality is independent of r The rhs of the inequality gives us the minimum value of the surface area but we need the dimensoins and not the actual surface area though we know that AM=GM when all the terms in the AM are equal therefore 500/r=500/r=2πr² thus r =cube root(250/π) and h =2x cube root(250/π) 🙂

[@nohaxjustxmod-sfs3984]

that thumbnail equation wasn’t too bad

[@yvesdelombaerde5909]

Same solution as minimizing the quantity of metal needed to build the can.

[@IIt_Bombay_I_am_Coming]

I don't know what level of students was tis test for , but as a JEE(everyone knows about the exam) aspirant in std.11, I found the the problems very easy, just like we do in our easiest courses, I am thinking if jee is really the hardest exam as the most basic questions for us are mind boggling ones to probably our seniors(which I assume since its on this channel) of other countries.

[@danmimis4576]

h = 2r = d because the a square (which is what we get if we intersect through its center the cylinder with a vertical plane) has the minimum area compared to other rectangles with the same perimeter.

[@jamyers1971]

Q from an old non-math person - why is cylinder radius in Centimeters?

[@TheEulerID]

I found it easier to let a = (k-1)/(k+1) then that gives us (x^2-2x)/(4x-1)=a and multiplying bot sides by 4x-1 we get x^2-2x=4ax-a -> x^2-(2+4a)x+a=0. For their to be just two equal value roots of opposite signs, 2+4a=0 and that gives a=-1/2. Substitute that into the first equation, and -1/2=(k-1)/(k+1) and if we solve for k we find k=1/3. Same method, but I prefer to have two steps with simpler terms. nb. on that quadratic equation giving just two opposite signed, but equal valued roots it's obvious that can only happen if b = 0, so the explanation using the quadratic equation seems overly complex. On the second problem I solved for the general solution volume V first. I don't like having specific dimensions used unless there is some advantages, and having equations with 500 in rather than just V just increases the opportunity for mistakes.

[@emilie375]

Students get so easily and so often in tears 😅

[@koenth2359]

Very easy, if you - understand they are talking about the two solutions for x (and not x=-k or something) - manage not to get intimidated. Done in two min. first I set the rhs to u and solved for b=0 in the resultung ax^2+bx+c expression. Got u=-1/2 so k=1/3.

[@chinmay1958]

The question in the thumbnail is too easy. The coefficient of x in the quadratic equation needs to be zero. So, -2(k+1)-4(k-1) = 0, or k = 1/3