Hard Proofs Question (1 Viewer)

[tickboom]

Does anybody have any ideas on how to approach part v) of this question? I've managed to do parts i), ii), iii) and iv), but v) is eluding me!

 

[beetree1]

it's questions like these that make me question why I opted this subject

 

[mani88]

Surely that question is not in the syllabus

 

[tickboom]

Surely that question is not in the syllabus

I thought the same thing when I saw it! But the person who shared it with me confirmed it was on a (very hard) 4U math exam.

 

[dasfas]

Lmao this is literally harder than some Uni level maths

 

[tickboom]

Lmao this is literally harder than some Uni level maths

Yeah when I first saw it I also thought it was a Uni question. But so far I've only needed high school concepts. Part i) requires induction. Parts ii), iii) and iv) require you to compare the shaded area under the curve (i.e. the integral) with the shaded area under rectangles.

Part v) I just can't work out what relevant rules with limits will help.

 

[mani88]

I thought the same thing when I saw it! But the person who shared it with me confirmed it was on a (very hard) 4U math exam.

What 4U exam???

 

[tickboom]

What 4U exam???

It's taken from a bespoke exam written at Baulkham Hills. Apparently a concoction of difficult questions to help prepare for the hardest possible things that might pop up on the final exam. I'm thankful I didn't get a question like this when I sat the HSC!

 

[Nav123]

Definitely tricky I think it has something to do with squeeze theorem but not exactly sure.

 

[tickboom]

Definitely tricky I think it has something to do with squeeze theorem but not exactly sure.

Yes I was thinking squeeze theorem might help too. Especially given part iv) involved finding a lower and upper bound. But I can't work out how to get those bounds as a function of n, in order to then work out their limit as n goes to infinity ... But I suspect there may be some way to do it by making use of all the results from the prior sections.

It's interesting that the result for C is very close to the value of the midpoint between the upper and lower bounds calculated in part iv), but not exactly the same.

 

[Fabrizio]

I thought the same thing when I saw it! But the person who shared it with me confirmed it was on a (very hard) 4U math exam.

Since a lot of the harder topics that usually get used in Q16 got removed like conics, we will get either a proof or some weird complex number/vector proof hybrid question as the hardest questions. Mechanics, integration are just too easy and its hard to find a completely unique question

 

[Trebla]

Mechanics, integration are just too easy and its hard to find a completely unique question

It is definitely possible to have very difficult questions in either of those topics. See Q15 and Q16a) of the 2020 BoS trials.

The only topic that I would consider “hard” that is removed from the syllabus is Harder Ext1 where they had the freedom to ask any question they really want to (provided it’s within the Adv/Ext1/Ext2 syllabus). Most of the harder questions in the past HSC exams came from that. Very rarely did they come from Conics, Volumes, Graphs or Circular motion (all removed for new syllabus).

 

[username_2]

It's taken from a bespoke exam written at Baulkham Hills. Apparently a concoction of difficult questions to help prepare for the hardest possible things that might pop up on the final exam. I'm thankful I didn't get a question like this when I sat the HSC!

Do you have all of the questions from this "bespoke exam written at Baulkham Hills"? Can you share if possible? Thanks in advance.

 

[tickboom]

Do you have all of the questions from this "bespoke exam written at Baulkham Hills"? Can you share if possible? Thanks in advance.

Nah I only have this one question. A student submitted it to me for help through my website www.tickboom.study. I asked them where it came from because at first I was convinced it couldn't have been a high school question.

 

[fan96]

I think there's a possibility that there's some algebraic trick you can do (independent of the previous parts) to find once you know that it actually exists.

Part v) specifically states that the limit exists (with a DO NOT PROVE attached). That would be unnecessary and possibly misleading if the squeeze theorem was intended to be used, as it proves the existence of the limit anyway. I also suspect that the inequality in part i) might be too loose to use the squeeze theorem on.

It looks like parts i) to iv) is a proof that the limit exists using the monotone convergence theorem, with some parts omitted (as the theorem isn't 4U knowledge).

 

[tickboom]

I think there's a possibility that there's some algebraic trick you can do (independent of the previous parts) to find once you know that it actually exists.

Part v) specifically states that the limit exists (with a DO NOT PROVE attached). That would be unnecessary and possibly misleading if the squeeze theorem was intended to be used, as it proves the existence of the limit anyway. I also suspect that the inequality in part i) might be too loose to use the squeeze theorem on.

It looks like parts i) to iv) is a proof that the limit exists using the monotone convergence theorem, with some parts omitted (as the theorem isn't 4U knowledge).

I've been doing some research on the monotone convergence theorem, and as far as I can understand, it is focused on the convergence of a sequence (i.e. showing that the limit of the values in a sequence is some value). However, in this question I think we are more focused on the convergence of the series (being the sum of the items in the sequence). So I'm still really stuck on this.

 

[Trebla]

I've been doing some research on the monotone convergence theorem, and as far as I can understand, it is focused on the convergence of a sequence (i.e. showing that the limit of the values in a sequence is some value). However, in this question I think we are more focused on the convergence of the series (being the sum of the items in the sequence). So I'm still really stuck on this.

A sum of n terms can be treated like a sequence. For example define the sequence {s_n} where

 

[tickboom]

A sum of n terms can be treated like a sequence. For example define the sequence {s_n} where

Yes that's a good point. I tried taking that path, and I used the technique where you state S_n+1 in terms of S_n, substitute C for both S_n+1 and S_n, and then solve for C, but unfortunately it didn't lead anywhere.

 

[Paradoxica]

I'd like to speak to the author of this question grabs knuckle dusters

 

[Paradoxica]

So I solved (v) with two applications of the Stolz–Cesàro theorem, I have given up on attempting a more within-the-syllabus proof.