• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Status
Not open for further replies.
Joined
Apr 1, 2011
Messages
1,012
Location
District 12
Gender
Male
HSC
2013
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
 
Last edited:
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
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.
 
Last edited:

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
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

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
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.
 
Last edited:
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top