HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

nightweaver066 · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
can u please reveal the solution to this question?

$(1+x)^{n^2} = \binom{n^2}{0} + {n^2}{1}x + ... +\binom{n^2}{n^2}x^{n^2} ... (1)$

$[(1 + x)^n]^n$

$= [\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + ... + \binom{n}{n}x^n]^n ... (2)$

$$Coefficient of $ x^2 $ in $ (1) = \binom{n^2}{2}$

$Coefficient of $ x^2 $ in $ (2) = \binom{n}{0}^{n-1} \binom{n}{2} \times \binom{n}{n-1} $ (Selecting the n-1 brackets, then 1 remaining for the $ x^2 $)$

$+ \binom{n}{0}^{n-2} \binom{n}{1}^2 \times \binom{n}{n-2} $ (Selecting the n-2 brackets, then 2 remaining for the x)$$

$= \binom{n}{1} \binom{n}{2} + \binom{n}{1}^2 \binom{n}{2} $ As $ \binom{n}{r} = \binom{n}{n-r}$

$= \binom{n}{1} \binom{n}{2} [1 + \binom{n}{1}]$

HeroicPandas · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
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:

$\binom{n}{2}$

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

$\binom{n}{1} \cdot \binom{n}{1}$

So we combine both of these to get:

$\binom{n}{2} \binom{n}{1}^2$

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:

$\binom{n}{1} \binom{n}{2}$

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

$\binom{n^2}{2} = \binom{n}{1} \binom{n}{2} \left (1+\binom{n}{1}\right )$

nightweaver066 said:
$(1+x)^{n^2} = \binom{n^2}{0} + {n^2}{1}x + ... +\binom{n^2}{n^2}x^{n^2} ... (1)$

$[(1 + x)^n]^n$

$= [\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + ... + \binom{n}{n}x^n]^n ... (2)$

$$Coefficient of $ x^2 $ in $ (1) = \binom{n^2}{2}$

$Coefficient of $ x^2 $ in $ (2) = \binom{n}{0}^{n-1} \binom{n}{2} \times \binom{n}{n-1} $ (Selecting the n-1 brackets, then 1 remaining for the $ x^2 $)$

$+ \binom{n}{0}^{n-2} \binom{n}{1}^2 \times \binom{n}{n-2} $ (Selecting the n-2 brackets, then 2 remaining for the x)$$

$= \binom{n}{1} \binom{n}{2} + \binom{n}{1}^2 \binom{n}{2} $ As $ \binom{n}{r} = \binom{n}{n-r}$

$= \binom{n}{1} \binom{n}{2} [1 + \binom{n}{1}]$

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

asianese · Feb 3, 2013

Re: HSC 2013 3U Marathon Thread

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

$\text{Prove that in any triangle with sides } a, b \text{ and } c, \text{ that }\\ a^2=b^2+c^2-2bc \cos A \\ \text{ where A is the angle opposite side a. }$

asianese · Feb 4, 2013

Re: HSC 2013 3U Marathon Thread

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

$\text{Consider the parabola } x^2=4y \text{ and the circle } x^2+(y-4)^2=r^2. \\ \text{ The circle's radius is increasing at } r \text{ units per time and initially, } r=1. \\ \text{i) Find the relationship between } r \text{ and } t. \fbox{2}\\ \text{ii) The circle's radius increases until it first becomes a tangent to the parabola. Find what radius this will occur at. } \fbox{2} \\ \text{iii) Find the time taken to reach this point. Leave your answer in exact form. } \fbox{1} \\ \text{iv) Find the rate of change of the circle's area when the circle becomes a tangent to the circle.} \fbox{2}$

nightweaver066 · Feb 4, 2013

Re: HSC 2013 3U Marathon Thread

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

$\text{Consider the parabola } x^2=4y \text{ and the circle } x^2+(y-4)^2=r^2. \\ \text{ The circle's radius is increasing at } r \text{ units per time and initially, } r=1. \\ \text{i) Find the relationship between } r \text{ and } t. \fbox{2}\\ \text{ii) The circle's radius increases until it first becomes a tangent to the parabola. Find what radius this will occur at. } \fbox{2} \\ \text{iii) Find the time taken to reach this point. Leave your answer in exact form. } \fbox{1} \\ \text{iv) Find the rate of change of the circle's area when the circle becomes a tangent to the circle.} \fbox{2}$

Just did it.

i. $r = e^t$

ii. $r = 2\sqrt{3}$

iii. $t = ln(2\sqrt{3})$

iv. $24 \pi u^2/s$

Are these the answers?

I like the premise of the question, although i found it slightly easy.

asianese · Feb 4, 2013

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 · Feb 4, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I'm not a fan of it

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

$Prove using the Binomial Theorem/Combinatorics that$

$\sum_{j=m}^{n} \binom{j}{m} \equiv \binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1}$

Solution I was looking for:

$Consider the expression$

$(1+x)^m + (1+x)^{m+1}+ (1+x)^{m+2}+ \dots + (1+x)^{n}$

$=(1+x)^m (1+(1+x)+(1+x)^2+ \dots + (1+x)^{n-m})$

$=(1+x)^m \frac{1-(1+x)^{n-m+1}}{1-(1+x)}$

$=(1+x)^m \frac{(1+x)^{n-m+1} -1}{x}$

$=\frac{1}{x} ((1+x)^{n+1} - (1+x)^m)$

So now we have the equality:

$(1+x)^m + (1+x)^{m+1}+ (1+x)^{m+2}+ \dots + (1+x)^{n} = \frac{1}{x} ((1+x)^{n+1} - (1+x)^m)$

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:

$\sum_{j=m}^{n} \binom{j}{m} \equiv \binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m}+ \dots + \binom{n}{m} = \binom{n+1}{m+1}$

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

twinklegal19 · Feb 5, 2013

Re: HSC 2013 3U Marathon Thread

$\\\text{Use Mathematical Induction to show that }\\\\ \sin\theta+ \sin2\theta+\sin3\theta+...+\sin{n\theta} =\frac{\sin{\frac{1}{2}(n+1)\theta}\sin{\frac{1}{2}n\theta}}{\sin{\frac{1}{2}\theta}}$

Sy123 · Feb 7, 2013

Re: HSC 2013 3U Marathon Thread

$\int \frac{\cos \left(\frac{1}{x} \right)}{x^2} \ dx$

asianese · Feb 7, 2013

Re: HSC 2013 3U Marathon Thread

Did you mean from 1 to k?

Carrotsticks · Feb 8, 2013

Re: HSC 2013 3U Marathon Thread

twinklegal19 said:
$\\\text{Use Mathematical Induction to show that }\\\\ \sin\theta+ \sin2\theta+\sin3\theta+...+\sin{n\theta} =\frac{\sin{\frac{1}{2}(n+1)\theta}\sin{\frac{1}{2}n\theta}}{\sin{\frac{1}{2}\theta}}$

$$By inspection, the result is true for all $ n \in \mathbb{N} \qquad \square.$

=)

asianese · Feb 8, 2013

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.

twinklegal19 · Feb 8, 2013

Carrotsticks said:
$$By inspection, the result is true for all $ n \in \mathbb{N} \qquad \square.$

=)

My favourite method!

asianese said:
^ 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.

Nah you don't need to use it, 3U trig is all you'll ever need =)

Sy123 · Feb 8, 2013

Re: HSC 2013 3U Marathon Thread

$Considering the identity$

$(2+x)^n = (1+(1+x))^n$

$Prove that$

$\binom{n}{k} \cdot \frac{\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}}{\binom{k}{0}+\binom{k}{1}+\dots+\binom{k}{k}} = \binom{n}{0} \binom{n}{k} + \binom{n}{1} \binom{n-1}{k} + \dots + \binom{n}{k} \binom{k}{k}$

HeroicPandas · Feb 8, 2013

Re: HSC 2013 3U Marathon Thread

lfcserg said:
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

Sy123 · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

$Prove using the binomial theorem$

$\binom{n^2}{n} > n^n$

RealiseNothing · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Prove using the binomial theorem$

$\binom{n^2}{n} > n^n$

Note: True for $n \geq 2$

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:

$\binom{n^2}{n} > n$

We can see this as $n^2 > n$ for $n\geq 2$ and since $\binom{k^2}{k^2-1} = k^2$ , it follows the result is true. Now:

$\binom{n^2}{n} (n-1)^n > n(n-1)^n = n^n$

$\binom{n^2}{n} > \frac{n^n}{(n-1)^n} = (\frac{n}{n-1})^n$

$\binom{n^2}{n} > (1+\frac{1}{n-1})^n$

Substitute our minimum value in, $n=2$

$\binom{4}{2} = 6 > (1+1)^2 = 4$

So it is true for our minimum value, and since as $n$ gets larger $\binom{n^2}{n}$ gets larger and $(1+\frac{1}{n-1})^n$ gets smaller, the inequality will hold for all $n\geq2$ .

#

hayabusaboston · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
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 · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

hayabusaboston said:
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

Drongoski · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\int \frac{\cos \left(\frac{1}{x} \right)}{x^2} \ dx$

$= - \int cos (\frac {1}{x}) d(\frac {1}{x}) = - sin (\frac {1}{x}) + C$

HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

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