Q16 style problems thread (1 Viewer)

vuhung

Member
Joined
Dec 14, 2025
Messages
53
Gender
Male
HSC
2027
I've composed a problem so student can practise with problem 16s.
Let me know what do you think.

bounding product.png
 

Attachments

Last edited:

WeiWeiMan

Well-Known Member
Joined
Aug 7, 2023
Messages
1,337
Location
behind you
Gender
Male
HSC
2026
I've composed a problem so student can practise with problem 16s.
Let me know what do you think.

View attachment 51038
(i)
Let f(x) = ln(1+x) - x + x^2/2
f(0) =ln(1) - 0 + 0 =0
f’(x) = 1/(x+1) -1 + x
= x+1 + 1/(x+1) -2
> 2 -2 {by AM/GM for x>0, no equality since x =/= 0 => x+1 =/= 1/(x+1)
= 0
f’(x) > 0 for x > 0 and f(0) = 0
Thus, f(x) > 0 for x > 0
=> 0 < ln(1+x) -x + x^2/2
=> x - x^2/2 < ln(1+x) QED

(ii)
Let x = k/n^2 for k = 1,2,3,…, n

Via x-x^2/2 < ln(1+x) < x: [given]
1/n^2 - 1/2n^4 < ln(1+1/n^2) < 1/n^2
2/n^2 - 2^2/2n^4 < ln(1+2/n^2) < 2/n^2

n/n^2 - n^2/2n^2 < ln(1+n/n^2) < n/n^2

Adding these:
(1/n^2 + 2/n^2 + … + n/n^2) - (1/2n^4 + 2^2/2n^4 + … + n^2/2n^2) < ln[(1+1/n^2)(1+2/n^2)…(1+n/n^2)] < 1/n^2 + 2/n^2 + … + n/n^2

Factorising and simplifying a bit:

1/n^2 (1 + 2 + … + n) -1/2n^4 (1 + 2^2 + … + n^2) < ln(Pn) < 1/n^2 (1+2+… + n)

I am far too lazy to prove 1+2^2 + … + n^2 = n(n+1)(2n+1)/6 so I’ll just accept it as is

Using this alongside 1+2+…+n = n(n+1)/2 (Via AP):

n(n+1)/2n^2 - n(n+1)(2n+1)/12n^2 < ln(Pn) < n(n+1)/2n^2

Exponentiating all 3:
e^[n(n+1)/2n^2 - n(n+1)(2n+1)/12n^2] < Pn < e^[n(n+1)/2n^2]

I’ll do (iii) and (iv) later since I’m doing these on phone and it’s a genuine pain to type it out

My rough ideas are:

Sandwich/Squeeze, looks like sqrt(e)
Question literally tells you what to do
 
Last edited:

WeiWeiMan

Well-Known Member
Joined
Aug 7, 2023
Messages
1,337
Location
behind you
Gender
Male
HSC
2026
Nice work Wei Wei.

You'll find some interesting findings working with part (iv), linking to part (i), Taylor expansion and the squeeze theorem.
Yeah I’ll get it done later when I’m bothered
 

WeiWeiMan

Well-Known Member
Joined
Aug 7, 2023
Messages
1,337
Location
behind you
Gender
Male
HSC
2026
is this even in the syllbus 💀
Almost certainly
(i), (ii) and (iii) can be very reasonably asked in 4U, however, (ii) is a bit annoying to do without the realistic ability to quote without proof that (1+2^2+3^2+...+n^2) = n(n+1)(2n+1)/6.

(iv) is nice to think about, however, not super HSC style I'm pretty sure. I think it's still a reasonable question since it tells you how you should arrive at the proof though.
 

ivanradoszyce

Member
Joined
Oct 18, 2023
Messages
63
Gender
Undisclosed
HSC
2018
Is this problem from a text book? If so, can you supply the title and author(s).

Thanks
 

vuhung

Member
Joined
Dec 14, 2025
Messages
53
Gender
Male
HSC
2027
I am the author of the booklet “The Last Resorts” which the problem is taken from.
You can download the booklet from my profile page.
 

ivanradoszyce

Member
Joined
Oct 18, 2023
Messages
63
Gender
Undisclosed
HSC
2018
Excellent material.

Thank-you so much for the time and effort in creating this resource.
 

vuhung

Member
Joined
Dec 14, 2025
Messages
53
Gender
Male
HSC
2027
With the trials/HSC season approaching (in one year), I've composed a problem that I think is a perfect "final boss" challenge for those aiming for the top of the state. It’s a beautiful walkthrough of the proof that ζ(3) is irrational.

Why you should try this:
  • Problem 16 Vibes: This is classic "Question 16" material - it's non-routine, deeply conceptual, and honestly a bit harder than the Q16s we’ve seen in the last few years.
  • The "Top 0.001%" Test: If you can navigate the reasoning behind these four parts without getting lost in the algebra, you’re in a very strong position for a U4.
  • Focus on Logic: The "heavy lifting" (the massive integral evaluations and LCM bounds) is already provided as hints. Your job is to connect the dots and provide the rigorous reasoning that bridges these complex results.

The Challenge:
The problem guides you through an irrationality criterion, a 3-variable optimization, and an integral construction that eventually forces a contradiction.

Good luck to everyone grinding through Extension 2! Post your solutions or where you’re getting stuck below. Let’s see who can crack the logic for the final contradiction first.

The Irrationality of ζ(3).png
 

Attachments

Last edited:

Burnt_Out

Well-Known Member
Joined
Sep 8, 2025
Messages
203
Location
6 feet under
Gender
Male
HSC
2026
well, i already did b but thanks. (also i have no idea if my answer for a would be sufficient)
 

vuhung

Member
Joined
Dec 14, 2025
Messages
53
Gender
Male
HSC
2027
Yes. Part a) is important.

but I mentioned, focus on the core reasoning.

I designed the problem so that it can be solved 20 30 mins or so, including everything, not to give you pain.

Q16s style, as I said.
 

WeiWeiMan

Well-Known Member
Joined
Aug 7, 2023
Messages
1,337
Location
behind you
Gender
Male
HSC
2026
With the trials/HSC season approaching (in one year), I've composed a problem that I think is a perfect "final boss" challenge for those aiming for the top of the state. It’s a beautiful walkthrough of the proof that ζ(3) is irrational.

Why you should try this:
  • Problem 16 Vibes: This is classic "Question 16" material - it's non-routine, deeply conceptual, and honestly a bit harder than the Q16s we’ve seen in the last few years.
  • The "Top 0.001%" Test: If you can navigate the reasoning behind these four parts without getting lost in the algebra, you’re in a very strong position for a U4.
  • Focus on Logic: The "heavy lifting" (the massive integral evaluations and LCM bounds) is already provided as hints. Your job is to connect the dots and provide the rigorous reasoning that bridges these complex results.

The Challenge:
The problem guides you through an irrationality criterion, a 3-variable optimization, and an integral construction that eventually forces a contradiction.

Good luck to everyone grinding through Extension 2! Post your solutions or where you’re getting stuck below. Let’s see who can crack the logic for the final contradiction first.


View attachment 51228
I shall do this when I am bothered 👍
 

Burnt_Out

Well-Known Member
Joined
Sep 8, 2025
Messages
203
Location
6 feet under
Gender
Male
HSC
2026
I solved it (at least I think i did)! My solutions are probably longer than it what would be needed but oh well (plus i'm doing this on word).

How did you come up with this problem anyways and should i send it (and i'm going to bed)?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top