HSC 2016 MX1 Marathon (archive) (1 Viewer)

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davidgoes4wce

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Re: HSC 2016 3U Marathon

When asked to determine the x^3 term in the Binomial expansion. This was the answer from Fitzpatrick answers in his main text:



Would it be ok to state the other terms mC0 and nC0 in the answer?

 

leehuan

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Re: HSC 2016 3U Marathon

Why is there a nC0 on the LHS?

But on the RHS I'd argue you actually should and Fitzpatrick is inaccurate (although not incorrect, obviously).
 

davidgoes4wce

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Re: HSC 2016 3U Marathon



I didn't quite understand the last time

I also assume in this question we have to rote memorise the binomial expansion to do any of these questions more efficiently?

 

leehuan

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Re: HSC 2016 3U Marathon

If we are trying to extract the coefficient of x^r in:
1+(1+x)+(1+x)^2+...+(1+x)^n

We firstly note that the term in (1+x)^n with x^r is nCr x^r
We then note that the term before (1+x)^n is (1+x)^(n-1), so the term with x^r is (n-1)Cr
Then we can extract the next term, (n-2)Cr from (1+x)^(n-2)

Now, this question will naturally assume the conditions 1 <= r <= n

We note that if we spread out the thing further, it would look something like
1 + (1+x) + (1+x)^2 + ... + (1+x)^(r-1) + (1+x)^r + (1+x)^(r+1) + ... + (1+x)^n

We note that the 1 + (1+x) + (1+x)^2 + ... + (1+x)^(r-1) component will not even have a term including x^r, so we disregard it when trying to extract the coefficient

Hence, the last term that is included in the coefficient of x^r comes from (1+x)^r, which naturally must just be rCr x^r
 

leehuan

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Re: HSC 2016 3U Marathon

NEXT QUESTION



Note: If you haven't been taught the resulting integral after the substitution, consult your formula sheet.
 
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leehuan

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Re: HSC 2016 3U Marathon

NEXT QUESTION

 

leehuan

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Re: HSC 2016 3U Marathon

Part (a) should be undefined.Left hand limits is not equal to the Right Hand Limit. (Try x=4.01 and Try x=3.99 as proof)

Part (b) limit as x approaches infinity, approaches 1/2 (by Dominating terms)
Your margin of error is too large. Try 4.0001 and 3.9999 instead. Tbh, the limit can actually be evaluated without testing both sides.

In terms of extracurricular maths (or maybe even 4u??), yep happy with the answer for b). Within the 3u course they want you to use lim x->infinity (1/x) = 0 though.

So you divide top and bottom by 1/x^3 to apply the limit. Because naturally it follows that lim x->infinity (1/x)^n = 0 for |n|>=1
 
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leehuan

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Re: HSC 2016 3U Marathon

There's hundreds of errors in MIF. Hate to sound rude but like I said, nobody really cares just to look here to pick one out. You're not informing anyone anything even though your intentions are clearly pure.
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NEXT QUESTION:
 
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Carrotsticks

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Re: HSC 2016 3U Marathon

Your margin of error is too large. Try 4.0001 and 3.9999 instead. Tbh, the limit can actually be evaluated without testing both sides.
What is 'too large'? What makes your margin of error 'better' than another one?

And as you mentioned, these questions are ideally done without any sort of numerical analysis.
 

davidgoes4wce

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Re: HSC 2016 3U Marathon

There's hundreds of errors in MIF. Hate to sound rude but like I said, nobody really cares just to look here to pick one out. You're not informing anyone anything even though your intentions are clearly pure.
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NEXT QUESTION:
 

leehuan

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Re: HSC 2016 3U Marathon

What is 'too large'? What makes your margin of error 'better' than another one?

And as you mentioned, these questions are ideally done without any sort of numerical analysis.
Nothing apart from the fact the magnitude of the values I chose are more closer to that of the given value (and equal in sign, since this isn't the case of a limit to 0).
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I'll give some benefit of the doubt; it's not intuitive to realise that direct substitution of x=4 results in 0/0 because the calculator will say "math error", which can be interpreted as a/0 (a is any complex number except 0+0i). But when giving it to a 3U student arbitrarily I would have hoped that they'd then realise (x-4) is a factor of both the numerator and denominator, and thus perform polynomial long division (or any other correct approach) to find a limit that can be evaluated on the spot.

But that's what I would've done, which is like you said, avoidance of numerical analysis.
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NEXT QUESTION:


(This is actually very simple 2U. But I'm leaving it here because apparently it was hard in the VCE and 2U has an unanswered question I think?)
 
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