HSC Mathematics Marathon (1 Viewer)

[largarithmic]

This is a fantastic question the first one obviously doesn't (choose example f[x] = x), but for the second you basically integrate what should be its derivative (well you dont actaully integrate it because you dont know its differentiable but thats what it amounts to):

For some x,y real and n a positive integer, let

Then "integrating" the function's derivative:


Now the choice of x and y is irrelevant of n: so sending n to infinity, the right hand side of that inequality goes to zero, so by the squeezetheorem/sandwich principle, for all x and y, ; and if we swap the positions of x and y, : combining these together we must obtain for all x and y, proving the function is constant.

 

[artosis]

lolololololol ^ state ranker right there. gg bro
whats ur name so i can look out for u on the state rankers list for mx1 and mx2

dude are u smarter than your teacher???

 

[apollo1]

This is a fantastic question the first one obviously doesn't (choose example f[x] = x), but for the second you basically integrate what should be its derivative (well you dont actaully integrate it because you dont know its differentiable but thats what it amounts to):

For some x,y real and n a positive integer, let

Then "integrating" the function's derivative:


Now the choice of x and y is irrelevant of n: so sending n to infinity, the right hand side of that inequality goes to zero, so by the squeezetheorem/sandwich principle, for all x and y, ; and if we swap the positions of x and y, : combining these together we must obtain for all x and y, proving the function is constant.

jesus f***ing christ.

 

[seanieg89]

Tiny signs mistake in your first line of working but otherwise perfect

 

[math man]

lol what a nice question from a uni book...part 2 is done very easily knowing the mean value theorem

 

[largarithmic]

its a pretty cool question coz its like, calculus but not lol. and if my name is on the list, it could easily be inferred from my surname

anyway I came across this awesome geometry problem the other day, its neat but pretty hard. hint is that its probably doable by year 8 or 9 students try it in spare time or something, its different style to HSC questions / I couldnt be bothered getting a question to put up thats hscish

ABCDEF is a convex hexagon with equal opposite sides: i.e. AB = DE, BC = EF, CD = FA. Let A1 be the midpoint of FB, B1 the midpoint of AC, C1 the midpoint of BD, D1 the midpoint of CE, E1 the midpoint of DF and F1 the midpoint of EA. Prove that A1D1, B1E1 and C1F1 are concurrent.

 

[largarithmic]

Actually for seanieg89 and mathman, try this question (its of very similar flavour and pretty good as far as I remember it):

Let f be a function that takes real numbers to non-negative reals, and define .

Prove that f satisfies the following inequality for all reals x and all nonnegative y:
if and only if g(x) is nondecreasing.

 

[MATHSCAPE]

jesus f***ing christ.

+1 lol... ridiculous.

 

[seanieg89]

edit: by the way mathman, how does MVT give a quicker solution than largarithmics to that earlier question?

 

[math man]

Actually for seanieg89 and mathman, try this question (its of very similar flavour and pretty good as far as I remember it):

Let f be a function that takes real numbers to non-negative reals, and define .

Prove that f satisfies the following inequality for all reals x and all nonnegative y:
if and only if g(x) is nondecreasing.

yep i like it...but is that from a uni book or q8 lol?

 

[math man]

divide through by x-y



now you will notice the LHS is the MVT:



Therefore we get:



now f'(c) is a constant, say K, hence this becomes:



Edit: we know x>y, hence x-y>0, which means K>0, hence f is not constant, so yeah you are right..MVT contradicts it lol

 

[seanieg89]

And how does that imply that f is constant? (Also you have made the assumption that f is differentiable everywhere, this follows from the given inequality though.)

 

[deterministic]

Can't part 2 be done by simply applying differentiation through first principles. Eg. let x=a+h, y=a for h>0, then for all a:

Then choosing for x=a, y=a+h, h>0 so for all a

Thus f'(a)=0 for all a and hence f is constant

 

[seanieg89]

Yeah that works deterministic, some sort of squeeze theorem is needed to make the argument rigorous though. (We cannot write things like f'(a) before we know that the defining limit exists.)

edit: mathman, you haven't really found a contradiction...in your working "K" is not necessarily positive. MVT is most useful for passing between local properties and global properties of a given function, not so much in these sorts of question.

 

[math man]

Yeah that works deterministic, some sort of squeeze theorem is needed to make the argument rigorous though. (We cannot write things like f'(a) before we know that the defining limit exists.)

edit: mathman, you haven't really found a contradiction...in your working "K" is not necessarily positive. MVT is most useful for passing between local properties and global properties of a given function, not so much in these sorts of question.

but my contradiction is that for this to work K has to be 0, but i proved it cant be 0, which contradicts f being constant

 

[largarithmic]

I don't really pretend to understand mean value theorem, but I really don't get how your proof is supposed to work math man. Like how do you know that f'(c) or K is a constant? You've chosen a particular c for a particular pair y<x, but wouldnt the desired value of f'(c) vary for the pairs of (x,y) you choose?

I don't really understand what's going on here though, I know barely any analysis and probably because of the fact you can pretty easily do very nonrigorous things, more discrete maths is a bit more comfortable for me

 

[math man]

I don't really pretend to understand mean value theorem, but I really don't get how your proof is supposed to work math man. Like how do you know that f'(c) or K is a constant? You've chosen a particular c for a particular pair y<x, but wouldnt the desired value of f'(c) vary for the pairs of (x,y) you choose?

I don't really understand what's going on here though, I know barely any analysis and probably because of the fact you can pretty easily do very nonrigorous things, more discrete maths is a bit more comfortable for me

lol...well the MVT states for some function that is both differentiable and continuous on the interval a to b there lies some c in between such that:



and this is pretty easy to picture...you draw a diagram of a continuous function on a to b and work out the gradient between those two points which is the LHS above, and you see that somewhere in that interval the gradient at some point c is the same

In first yr maths MVT is a wonderful thing applied to lots of areas in calculus...eg we use it to prove the arc length and surface area of curves

 

[largarithmic]

lol...well the MVT states for some function that is both differentiable and continuous on the interval a to b there lies some c in between such that:



and this is pretty easy to picture...you draw a diagram of a continuous function on a to b and work out the gradient between those two points which is the LHS above, and you see that somewhere in that interval the gradient at some point c is the same

In first yr maths MVT is a wonderful thing applied to lots of areas in calculus...eg we use it to prove the arc length and surface area of curves

Yeah I know what the MVT is, but I don't really get how your proof of seanieg89's question actually works (like how do you assume f'(c) is constant , presumably as you range x and y)

 

[seanieg89]

you haven't really shown that K>0 mathman.

 

[math man]

you haven't really shown that K>0 mathman.

i know it doesn't ....as i said...i proved K can't be zero so my proof contradicts the question lol....so MVT fails here

Edit: well K can be zero but i can't prove it here using the MVT..so to stop any further questions MVT fails on this question