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HSC 2015 MX2 Marathon (archive) (2 Viewers)

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Sy123

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Re: HSC 2015 4U Marathon

A prime number is a positive integer greater than 1 that is divisible by only 1 and itself.

i) Show that (n! + 1) is not divisible by any number from 2 to n inclusive

ii) Hence show that the number of primes is infinite
 

Sy123

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Re: HSC 2015 4U Marathon

 
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Sy123

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Re: HSC 2015 4U Marathon

What exacly is this asking o_O do we just square the original equation to have it in terms of y_n ?
yes, but the equation must be only in terms of y_n, and the terms must be linear (so no sqrt(y_n), y_n^2 etc.)
 
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Re: HSC 2015 4U Marathon

What if its not symmetrical?
f(4-x) = f(x)
I'm not sure what another method would be if it wasn't symmetrical besides finding the roots. But if you find the roots they would be conjugates of each other (real coefficients, surd roots come in conjugates) which indicate a symmetrical property. I would use the dreaded shell method otherwise
 
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Drsoccerball

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Re: HSC 2015 4U Marathon

f(4-x) = f(x)
I'm not sure what another method would be if it wasn't symmetrical besides finding the roots. But if you find the roots they would be conjugates of each other (real coefficients, surd roots come in conjugates) which indicate a symmetrical property. I would use the dreaded shell method otherwise
But the question forces slices and there been questions like that you'd just have to do the long way
 

Sy123

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Re: HSC 2015 4U Marathon

A prime number is a positive integer greater than 1 that is divisible by only 1 and itself.

i) Show that (n! + 1) is not divisible by any number from 2 to n inclusive

ii) Hence show that the number of primes is infinite




 
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Sy123

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Re: HSC 2015 4U Marathon

My teacher didn't solve questions like this because it wasn't in the syllabus :(
The way I gave the question made it HSC worthy, the first part guides you through, just use the ordinary logic you know from Volumes.
 

InteGrand

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Re: HSC 2015 4U Marathon

i) Using the perpendicular distance to line formula, the required distance r of (t, tn) to the line is:

(we can drop the absolute value signs as ttn).

Note this is the radius of a typical disc rotated about the axis of rotation (the line y = x).

ii) For a given t, denote by (a, a) the closest point to (t, tn) on the line y = x. Denote by the distance along the line we have travelled from the origin to get to the point . Note that and (by Pythagoras' Theorem).

We find a as a function of t.

If D denotes the distance from to , then .

To find the a that will minimise D, we need to minimise D2, so we need the derivative of the RHS above with respect to a to be 0:

.

So .

Consider a typical disc at a point , it has radius r, thickness . So its volume is given by .

Total volume of the solid is:

.

Substitute (this is one-to-one for , so it is safe to use), , when , when , and .

Then .

This is now easy (albeit somewhat tedious) to evaluate.
 
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InteGrand

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Re: HSC 2015 4U Marathon

We do it for the general case , .

This is a rectangular hyperbola, so eccentricity is .

If you consider a left-right opening hyperbola , 'a' represents the distance of the point where the hyperbola cuts its horizontal axis to the origin.

So in the case of , the a we would use in "focal length = ae" is the distance of the point where the hyperbola cuts the line y = x to the origin. The point in question is , so the distance in question is .

Then our focal length d is given by .

As the focus lies on the line y = x, we need the coordinates (f, f) of the point on this line which is d units from the origin. This is given by: .

So the focus is at . In your question, we had . So the focus would be at .

(Obviously, there's also a focus at , or (-4,-4) in your example.)
 
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