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HSC 2013 MX2 Marathon (archive) (4 Viewers)

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Sy123

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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.
 
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bleakarcher

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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

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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

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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 :p
 

Sy123

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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

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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

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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

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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

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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

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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.
 
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