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HSC 2013-14 MX1 Marathon (archive) (8 Viewers)

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HeroicPandas

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Re: HSC 2013 3U Marathon Thread

For that one:

Using that identity, equate the co-efficient of x^2 on both sides:

Now to select 2 x's from the RHS, we need to either pick x from 1 (1+x)^n and 1 x from (1+x)^n, this makes x^2
And the ways we can do this is:

First pick 2 brackets from the n number of brackets:



In each of these brackets, we need to pick the x term, so that is



So we combine both of these to get:



Now we can also get x^2 via picking 1 x^2 from 1 bracket, to do this pick 1 bracket out of the n, and in that bracket pick the x^2 co-efficient:



Add both of these together, and equate co-efficent on LHS

Thanks Sy and Nightweaver

i did equate coeffs of x^2 on both sides when i first did it but i failed in finding the x^2 i nthe RHS of that identityt
 
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Re: HSC 2013 3U Marathon Thread

This is more a 2U question but the proof could be considered 3U.

 
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Re: HSC 2013 3U Marathon Thread

First time making up a question so bear with me...pls correct if wrong.

 
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nightweaver066

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Re: HSC 2013 3U Marathon Thread

First time making up a question so bear with me...pls correct if wrong.

Just did it. :)

i.

ii.

iii.

iv.

Are these the answers?

I like the premise of the question, although i found it slightly easy.
 
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Re: HSC 2013 3U Marathon Thread

hehe I'm not as volatile as Sy is. But yes those are the answers. I tried mucking about with variables (like x^2+(y-a)^2=r^2 and the parabola x^2=4ky) but it just wouldn't work out. Oh well hehe. Just some question making practice :)
 

Sy123

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Re: HSC 2013 3U Marathon Thread

I'm not a fan of it

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




Solution I was looking for:













So now we have the equality:





Equate the co-efficient of x^m on both sides
For the RHS, since we are dividing by x to both of the inside terms, we need to take the co-efficient of x^m+1 instead for both of them, so when they divide by x it becomes x^m
However (1+x)^m doesn't have x^m+1 term, therefore we do not consider it:

Therefore it follows automatically:



This is actually a famous identity and have seen it proven in many more elegant ways, but this my own way :L
 

Sy123

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Re: HSC 2013 3U Marathon Thread

 
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Re: HSC 2013 3U Marathon Thread

Did you mean from 1 to k?
 
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Re: HSC 2013 3U Marathon Thread

^ Full marks.

Btw twinkie I think you should give the formula for sincos formula (that's 'not' in the syllabus) if you're gonna give that question.
 

Sy123

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Re: HSC 2013 3U Marathon Thread







 
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HeroicPandas

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Re: HSC 2013 3U Marathon Thread

Hey guys having a bit of trouble with this question for integrating indefinite integrals by substitution:

Find the integral of x root 2x+7 dx. By using the substitution u^2 = 2x+7
u^2 = 2x + 7

Differentiating implicitly with respect to x

therefore, 2u.(du/dx) = 2

u.du = dx

Follow normal integration steps of substitution. to get rid of x, re-arrange u^2 = 2x + 7 and sub into the question
 
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Sy123

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Re: HSC 2013 3U Marathon Thread



 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread

Note: True for

This probably isn't the way you want some one to do it, but it was the first way I thought of so I'm just going to post it lol:



We can see this as for and since , it follows the result is true. Now:







Substitute our minimum value in,



So it is true for our minimum value, and since as gets larger gets larger and gets smaller, the inequality will hold for all .

#
 

hayabusaboston

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Re: HSC 2013 3U Marathon Thread

u^2 = 2x + 7

Differentiating implicitly with respect to x

therefore, 2u.(du/dx) = 2

u.du = dx

Follow normal integration steps of substitution. to get rid of x, re-arrange u^2 = 2x + 7 and sub into the question
Isnt this a 4u process? 3u kids probs wont know about it...
 

HeroicPandas

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Re: HSC 2013 3U Marathon Thread

Isnt this a 4u process? 3u kids probs wont know about it...
THe guy said to let u^2 = 2x+7 which is nice to eliminate the presence of square roots

If given this sub, we make u the subject, there will be a dirty plus-minus. How else can we do it? xD
 
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