BoS Maths Trials 2019 (1 Viewer)

Paradoxica

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The intended solution:

Inverse Function Theorem:

Let b = f(a), f differentiable around a

(f⁻¹)'(b) = 1/f'(a)

Regarding the problem:

By the Inverse Function Theorem, (f⁻¹)'(p) = 1/f'(p)

But the slopes are equal, by the problem assumption.

So let m = f'(p) = (f⁻¹)'(p), and we have m = 1/m ⇒ m² = 1
 

Paradoxica

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Either way here's how I would do it









While this solution is correct, it is much more cumbersome to work with, especially since you are pulling it down to a specific case. If it was put in a more general form, then it would be slightly less cumbersome to write out.

But then at that point it would just be easier to use the Inverse Function Theorem to kill it in one hit.
 

integral95

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While this solution is correct, it is much more cumbersome to work with, especially since you are pulling it down to a specific case. If it was put in a more general form, then it would be slightly less cumbersome to write out.

But then at that point it would just be easier to use the Inverse Function Theorem to kill it in one hit.

I honestly don't recall learning the inverse function theorem until first year uni to be fair (and basically never used it from 2nd year and later) , so I doubt the majority of the high-school students would even consider it.

Please share some more rants pls this is hilarious.
 

Paradoxica

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I honestly don't recall learning the inverse function theorem until first year uni to be fair (and basically never used it from 2nd year and later) , so I doubt the majority of the high-school students would even consider it.

Please share some more rants pls this is hilarious.
you don't remember learning slope of inverse function at inverse point = 1/(slope of function at function point) ?

because i certainly do remember that class in 3U
 

TheOnePheeph

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you don't remember learning slope of inverse function at inverse point = 1/(slope of function at function point) ?

because i certainly do remember that class in 3U
I'm pretty sure its not in the syllabus. There have been hsc questions on it before, but I definitely haven't learned it in class either
 

Trebla

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I'm pretty sure its not in the syllabus. There have been hsc questions on it before, but I definitely haven't learned it in class either
It is in the syllabus, but in a slightly different format. Students are expected to be taught that
(dy/dx)(dx/dy) = 1
so they can then do reciprocals of derivatives (e.g to solve differential equations). This is basically how you find the derivative of the inverse function.
 

TheOnePheeph

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It is in the syllabus, but in a slightly different format. Students are expected to know that
(dy/dx)(dx/dy) = 1
so they can do reciprocals of derivatives. This is basically how you find the derivative of the inverse function.
Well damn, we skipped completely over that in class haha. Still not too difficult to derive luckily, and is a fairly basic concept of inverse functions.
 

Trebla

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Well damn, we skipped completely over that in class haha. Still not too difficult to derive luckily, and is a fairly basic concept of inverse functions.
You were probably taught it implicitly without knowing. It basically formalises the ability to take reciprocals of derivatives - for example when finding the solution to dN/dt = kN
 

TheOnePheeph

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You were probably taught it implicitly without knowing. It basically formalises the ability to take reciprocals of derivatives - for example when finding the solution to dN/dt = kN
Yeah well it is pretty much an application of that, which is how they are able to ask questions on it in the hsc. Its more we weren't specifically told:



Oh well, its not hard to derive anyway lol.
 
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Paradoxica

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Yeah well it is pretty much an application of that, which is how they are able to ask questions on it in the hsc. Its more we weren't specifically told:



Oh well, its not hard to derive anyway lol.
except based on what Trebla said, that wasn't even need anyway, I'm just an idiot who hasn't bothered reviewing the syllabus outlines

by the statement of the question, one can see that dy/dx (p,p) = dx/dy (p,p) = 1

and the conclusion follows
 

TheOnePheeph

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except based on what Trebla said, that wasn't even need anyway, I'm just an idiot who hasn't bothered reviewing the syllabus outlines

by the statement of the question, one can see that dy/dx (p,p) = dx/dy (p,p) = 1

and the conclusion follows
Ah right, I didn't see that he meant that. Yeah thats much easier than the other method haha.
 

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