HSC 2013 MX2 Marathon (archive) (3 Viewers)

[bleakarcher]

Re: HSC 2013 4U Marathon

How did you get that ?

Yeah, that's why I wasn't sure about it. Pretty sure it's wrong now.

 

[Sy123]

Re: HSC 2013 4U Marathon

How did you get that ?

Yeah, that's why I wasn't sure about it. Pretty sure it's wrong now.

I'm pretty sure that step is legitimate.
We are doing the same operation to both sides, if f(x) = g(x), then f'(x) = g'(x)?

or am I missing something?


EDIT: wtf man what am I thinking

Oh and Realise, nice work for my question I had the same method.

 

[bleakarcher]

Re: HSC 2013 4U Marathon

I'm pretty sure that step is legitimate.
We are doing the same operation to both sides, if f(x) = g(x), then f'(x) = g'(x)?

or am I missing something?


EDIT: wtf man what am I thinking

Oh and Realise, nice work for my question I had the same method.

Well, at the time that I wrote the solution I rushed and thought that too. On second thought, how could f '(x+f(x))=[1+f '(x)]*f '(x+f(x) and f '(x+f(x))=f '(x)?

 

[Sy123]

Re: HSC 2013 4U Marathon

Well, at the time that I wrote the solution I rushed and thought that too. On second thought, how could f '(x+f(x))=[1+f '(x)]*f '(x+f(x) and f '(x+f(x))=f '(x)?

Because from the very beginning f(x) = 2012, and f'(x) = 0
so f(x+f(x)) = f(x+0)=f(x).

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

I'm pretty sure that step is legitimate.
We are doing the same operation to both sides, if f(x) = g(x), then f'(x) = g'(x)?

or am I missing something?


EDIT: wtf man what am I thinking

Oh and Realise, nice work for my question I had the same method.

lol then what would be the point of using the chain rule in the first place

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Because from the very beginning f(x) = 2012, and f'(x) = 0
so f(x+f(x)) = f(x+0)=f(x).

Unless you made a typo, I cant see how that works?

 

[Sy123]

Re: HSC 2013 4U Marathon

Unless you made a typo, I cant see how that works?

Yep its wrong.

Since from the beginning f(x) = c
then f(x) = f(x+f(x)) = f(x+c)

Since the inside is linear, f'(x) = f'(x+c)

 

[seanieg89]

Re: HSC 2013 4U Marathon

 

[asianese]

Re: HSC 2013 4U Marathon

I smell Lipschitz.

 

[seanieg89]

Re: HSC 2013 4U Marathon

Some random calculus true/false practice for multiple choice:

1. Every continuous function is differentiable.

2. Every differentiable function is continuous.

3. Every differentiable function has a continuous derivative.

4. If a differentiable f attains a maximum at some c in the interval a <= c <= b, then f'(a)=0.

5. Which of these is a stronger statement: f is differentiable on the interval (a,b) vs f has an definite integral over the interval (a,b).

6. If a function f defined on R has a maximum at a, then f''(a) > 0.

7. If a < b, f(a) < 0, and f(b) > 0, then f(c)=0 for some c between a and b.

8. If f is differentiable at a and f'(a) is nonzero, then f has a differentiable inverse function near a.


Some justification would be nice unless you think it is obvious.

 

[seanieg89]

Re: HSC 2013 4U Marathon

I smell Lipschitz.

You don't need to know the name of such functions or any properties thereof to answer this question.

 

[asianese]

Re: HSC 2013 4U Marathon

You don't need to know the name of such functions or any properties thereof to answer this question.

I know

 

[seanieg89]

Re: HSC 2013 4U Marathon

For the UNSW problem, one would have to prove that f is differentiable before one could differentiate it!

Also I don't quite get your reasoning for this question Sy, why must iterations of f applied to an arbitrary a converge, and why must this limit be in the range of f? I also don't really see how you used continuity?

 

[bleakarcher]

Re: HSC 2013 4U Marathon

For the UNSW problem, one would have to prove that f is differentiable before one could differentiate it!

Also I don't quite get your reasoning for this question Sy, why must iterations of f applied to an arbitrary a converge, and why must this limit be in the range of f? I also don't really see how you used continuity?

lol, there's something else wrong with my attempt.

 

[study1234]

Re: HSC 2013 4U Marathon

A harder 3u question (from Cambridge 3u):

A projectile is fired vertically upwards with speed V from the surface of the Earth. Assuming that acceleration = -kx^-2, prove that k = gR^2, where g is gravitational acceleration and R is the radius of the Earth. Thus, given that R = 6400 and g= 9.8/x^2, find the least value of V so that the projectile will never return.

 

[Sy123]

Re: HSC 2013 4U Marathon

For the UNSW question, I have been able to simplify the problem into proving:



k=2012

If this can be done then I have a nice proof in mind.

 

[Registered User]

Re: HSC 2013 4U Marathon

Hint: expand and try to get a telescoping sum.

 

[rolpsy]

Re: HSC 2013 4U Marathon

n(n+1)n!

lol

 

[asianese]

Re: HSC 2013 4U Marathon

Methinks he meant

 

[Registered User]

Re: HSC 2013 4U Marathon

n(n+1)n!

lol

Yeah, I meant

Your answer is wrong.