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HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

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hayabusaboston

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Re: HSC 2013 3U Marathon Thread

Considering the identity:



Then expanding both sides, equate co-efficient of on both sides knowing that , the result comes immediately.

=======


Consider (1+x)^n=nC0+nC1x+...nCnx^n, integrate twice, find ur C's, and let x=-1?
From first glance dats what it looks like.
 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread

Considering the identity:



Then expanding both sides, equate co-efficient of on both sides knowing that , the result comes immediately.
Damn I forgot you could do it that way. I had a combinations method in mind.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

Consider (1+x)^n=nC0+nC1x+...nCnx^n, integrate twice, find ur C's, and let x=-1?
From first glance dats what it looks like.
You will get on the denominators 2*3, 3*4, 4*5, .... , (n+1)(n+2) while for this one the denominators are only 2, 3, 4, .... , n+2
 

JJ345

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Re: HSC 2013 3U Marathon Thread

Hmm...this question looks familiar.
Is it something like using the binomial expansion for (1+x)^n, then multiplying both sides by x, integrating both sides (the LHS should have an expression that's slightly harder to integrate than usual), sub in x=0 to find the value of the constant of integration, then finally sub. x=-1. Not too sure, haven't done this in a while.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

Hmm...this question looks familiar.
Is it something like using the binomial expansion for (1+x)^n, then multiplying both sides by x, integrating both sides (the LHS should have an expression that's slightly harder to integrate than usual), sub in x=0 to find the value of the constant of integration, then finally sub. x=-1. Not too sure, haven't done this in a while.
A little different to what I had in mind, integrating x(1-x)^n from 0 to 1
But essentially the same thing

=====

Here is question I made:











 

Sy123

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Re: HSC 2013 3U Marathon Thread









 

dunjaaa

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Re: HSC 2013 3U Marathon Thread

Is there meant to be a (+ g) in the inside? I got the expression without the +g
 
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Sy123

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Re: HSC 2013 3U Marathon Thread

Is there meant to be a (+ g) in the inside?
I don't think so, I could be wrong but I'm getting what the question sasys.
 

seanle

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Re: HSC 2013 3U Marathon Thread

hi 2x^2 + ax + b + 3 =0 HAs real roots. Find the miniumum value of a^2 + b^2
Tried this but could not get the right answer. A cambridge extension question from year 11 book for 3 unit
 
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Re: HSC 2013 3U Marathon Thread

hi 2x^2 + ax + b + 3 =0 HAs real roots. Find the miniumum value of a^2 + b^2
Tried this but could not get the right answer. A cambridge extension question from year 11 book for 3 unit
I dont think this is right but ill have a shot anyway.
So you find a and b through sum and product of roots then add the square of both them which should get you
a^2+b^2=4c^2+4d^2+8cd+4c^2d^2-12cd+9 (c and d are the roots of the equation)
Now for a and b to me minimum, c and d approach 0, meaning the min is 9
 

seanle

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Re: HSC 2013 3U Marathon Thread

hi i tried using the discriminant let it equal zero for a minimum but i got 8 for an answer.
thank for your answer.
 

seanle

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Re: HSC 2013 3U Marathon Thread

a^2-4.2(b+3) is greater than or equal to zero

for a minimum a^2 = 8b + 24

let a^2 + b^2 = 8b + 24 + b^2 for b+3 = 0 this does give
the answer as 9 but why should i let b+ 3 =0 ?
is it a critical point?
 

seanle

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Re: HSC 2013 3U Marathon Thread

a = the square root of 8b + 24
for real roots b+3 is more than or equal to zero. So the minimum value of b is -3.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

 
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JJ345

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Re: HSC 2013 3U Marathon Thread

I'm thinking of expanding (1+x)^n, integrating, sub x=0 to find c, then let x=1 call this (1)
Repeat the process but in the final step sub x=-1 call this (2)
Then (1)-(2) will yield you a similar result---> But with 2^n on the numerator on the LHS???
 

Sy123

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Re: HSC 2013 3U Marathon Thread

I'm thinking of expanding (1+x)^n, integrating, sub x=0 to find c, then let x=1 call this (1)
Repeat the process but in the final step sub x=-1 call this (2)
Then (1)-(2) will yield you a similar result---> But with 2^n on the numerator on the LHS???
Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1)

====



 

JJ345

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Re: HSC 2013 3U Marathon Thread

Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1)

====



I think i'm missing something here, but how do we get this?
When we integrated didn't the LHS become [(1+x)^n+1/n+1] +c ?
 

seanieg89

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Re: HSC 2013 3U Marathon Thread

Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1)

====



I am quite sure the numerator should be 2^n.

Edit: In the previous question that is.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

I think i'm missing something here, but how do we get this?
When we integrated didn't the LHS become [(1+x)^n+1/n+1] +c ?
When integrating (1+x)^n, we get

Then x=0 to find c= - 1/(n+1)

Then,


EDIT: Yep I'm wrong I made a mistake, apologies


Yep well done
 
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HeroicPandas

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Re: HSC 2013 3U Marathon Thread

I found a mistake in my working out for the RHS

If i equate coefficients of x^(2n) for [3], i get
 
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