HSC 2013-14 MX1 Marathon (archive) (3 Viewers)

Status
Not open for further replies.

JJ345

Member
Joined
Apr 27, 2013
Messages
78
Gender
Female
HSC
N/A
Re: HSC 2013 3U Marathon Thread

What I did here was use sum of the following geometric series: (1+x)^m-1 +(1+x)^m+.........+(1+x)^m+n-1 =[ (1+x)^m+n -(1+x)^m-1/x]

Then I equated coefficients of x^m-1 and rewrote them in the form (m+n)!/m!n!....Does it work??
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

What I did here was use sum of the following geometric series: (1+x)^m-1 +(1+x)^m+.........+(1+x)^m+n-1 =[ (1+x)^m+n -(1+x)^m-1/x]

Then I equated coefficients of x^m-1 and rewrote them in the form (m+n)!/m!n!....Does it work??
Yep that worked for me

Well done

========



 
Last edited:

JJ345

Member
Joined
Apr 27, 2013
Messages
78
Gender
Female
HSC
N/A
Re: HSC 2013 3U Marathon Thread

This is a pretty neat question (I think it appeared in some ancient HSC paper)--> Nothing tricky, just slightly more conceptual =)

A woman travelling along a straight flat road passes three points at intervals of 200m. From these points she observes the angle of elevation of the top of the hill to the left of the road to be respectively 30°, 45°, and again 45°. Find the height of the hill.
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Re: HSC 2013 3U Marathon Thread

This was the last question of a 4U HSC paper in the 90s. (or 80s..)
 

JJ345

Member
Joined
Apr 27, 2013
Messages
78
Gender
Female
HSC
N/A
Re: HSC 2013 3U Marathon Thread

This was the last question of a 4U HSC paper in the 90s. (or 80s..)
Probably, can't remember...but looking at my method I think its alot simpler than a typical 4U Q8. from the 1980s, I found it more 3U style.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Yep that worked for me

Well done

========



Take the identity:





Then equate the co-efficients of x^k on both sides.

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



 

dunjaaa

Active Member
Joined
Oct 10, 2012
Messages
473
Gender
Male
HSC
2014
Re: HSC 2013 3U Marathon Thread

Sketch the curve and develop and inequality, with x=pi and you should obtain that result
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread













===



 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread



















 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread



 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Yep that's pretty much it





Difficulty: Q14
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

I had a go at making a parabola question, its not as brute force as it looks.











Difficulty: Q13
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread











Difficulty: Q13
 
Last edited:

obliviousninja

(╯°□°)╯━︵ ┻━┻ - - - -
Joined
Apr 7, 2012
Messages
6,624
Location
Sydney Girls
Gender
Female
HSC
2013
Uni Grad
2017
Re: HSC 2013 3U Marathon Thread

Sy, next to the questions you post, can you write its difficulty, ie question number it belongs to (section of the paper).
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread

Sy, next to the questions you post, can you write its difficulty, ie question number it belongs to (section of the paper).
I'll try

Most if not all questions I post will be at the higher end of difficulty in terms of HSC, for both the 2U and 4U marathons
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 3U Marathon Thread









Difficulty: Q14
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 3)

Top