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HSC 2013 MX2 Marathon (archive) (5 Viewers)

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Re: HSC 2013 4U Marathon

I don't think so, but I think its a bit unreasonable of me to ask it like that with no hint whatsoever.

Redone:



ah, I did it without substitution by completing the square and converting to partial fractions. It should be able to be integated into inverse tans ?
dunno if it works lol



I probably made an error somewhere
 
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Re: HSC 2013 4U Marathon

The integral is possible through completing the square and then doing a partial fractions.

Did you mean x=tan(t/2) by any off chance?
 

Sy123

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Re: HSC 2013 4U Marathon

The integral is possible through completing the square and then doing a partial fractions.

Did you mean x=tan(t/2) by any off chance?
Nah definitely t=tan(x/2)
I intended the person to go backwards with it, i.e. notice that the denominator is actually



With some manipulation thereon, it returns that, the integrand:



We notice that we can sub in various trigonometric equations into there, including a dx, cos x, and sin^2 x

Then integrate normally as it is in tan^-1 form.

===================

This is a really good question:

 

seanieg89

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Re: HSC 2013 4U Marathon

A lattice point is a point in the cartesian plane with integer coordinates.

Prove by induction or otherwise that the area of a polygon in the cartesian plane with vertices on lattice points is given by:

{Number of lattice points strictly inside P} - {Number of lattice points lying on the boundary of P}/2 - 1.

Difficulty: 2/5.
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

Here's a very nice question (imo) that I just came up with after an idea popped into my head last night.

i) Explain why:



Is well defined for

ii) Now integrate:

 

seanieg89

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Re: HSC 2013 4U Marathon

Here's a very nice question (imo) that I just came up with after an idea popped into my head last night.

i) Explain why:



Is well defined for

ii) Now integrate:

1. Because sin(x) is asymptotically equivalent to x as x - > 0, as is assumed without proof in the mx2 syllabus. Similarly sin(x)/(pi-x) tends to 1 as x - > pi.

2. 0. Simply apply the rule:



and the result is immediate, as the integrand simply changes sign.
 

seanieg89

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Re: HSC 2013 4U Marathon

In the following question you may assume without proof that if a continuous function f is positive at some c, then there is a small interval centred at c in which f > f(c)/2.

0. Prove that a non-negative continuous function with integral 0 over some interval of positive length must be 0 everywhere on the interval.

1. Fix real numbers a and b with a < b.

Prove that the only function g which satisfies:

g has a continuous second derivative on the interval [a,b],

g(a)=g(b)=0,

g''g >= 0 on the interval [a,b],

is zero identically on the interval [a,b].

2. Hence deduce that for any function f with f'' > 0 on the interval [a,b], and for any x1,x2,...,xn in [a,b] we have:



where the 's are arbitrary real numbers such that

.

Interpret this result geometrically.

3. By considering the function log(x) and using the result from 2, prove the inequality:



for positive and as in 2. What familiar result does this become if we take every to be 1/n?

4. There are n guards each to be stationed 1 km away from a government building. Each one can see everything within a radius of 1km from himself. Using the result proven in 2., find the maximum possible area that n guards can cover.

Difficulty: 4/5.
 
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