HSC 2015 MX1 Marathon (archive) (1 Viewer)

braintic · Oct 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
The answer is n=3

From Kinney-Lewis videos:

https://www.youtube.com/watch?v=hQlEigQ5ieI

That video is WRONG. The second expression should be (2n-1)!/(n-1)!

davidgoes4wce · Oct 5, 2015

Re: HSC 2015 3U Marathon

OK this was in Kinney Lewis's solutions:

The R.H.S denominator is wrong. Very unlike Kinney-Lewis to make a mistake but we are all humans aren't we?

kawaiipotato · Oct 5, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Let $A,B,C$ be real numbers such that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}=\frac{An+B}{2^n}+C$ holds for every positive integer $n$. Find the value of $A+B+C$.$

$$ let $$ S $ = \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}$

$$ 2S $ = 1 +\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{n}{2^{n-1}}$

$$ 2S - S $ = $ S $ = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ... + \frac{1}{2^{n-1}} - \frac{n}{2^{n}}$

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

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

$= 2 - \frac{2 - n}{2^n} \equiv \frac{An + B}{2^{n}} + C$

$\therefore $ A = 1 , B = -2, C = 2$

braintic · Oct 5, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
$$ let $$ S $ = \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^n}$

$$ 2S $ = 1 +\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{n}{2^{n-1}}$

$$ 2S - S $ = $ S $ = 1 + \frac{1}{2} + \frac{1}{2^{2}} + ... + \frac{1}{2^{n-1}} - \frac{n}{2^{n}}$

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

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

$= 2 - \frac{2 - n}{2^n} \equiv \frac{An + B}{2^{n}} + C$

$\therefore $ A = 1 , B = -2, C = 2$

Alternatively - sub in n=1, 2, 3 and solve the resulting 3 equations simultaneously.

InteGrand · Oct 5, 2015

Re: HSC 2015 3U Marathon

Also, the question asked to find A + B + C, so make sure to write the value of that at the end.

rand_althor · Oct 6, 2015

Re: HSC 2015 3U Marathon

$$Solve for $x$: $2^{3x}-\frac{8}{2^{3x}}-6\left (2^x-\frac{1}{2^{x-1}} \right )=1$

Paradoxica · Oct 6, 2015

Re: HSC 2015 3U Marathon

substitute 2^x = u, and factorise out u^-3 on the left hand side. the numerator can be factorised as (u^2 - 2)^3
collect the entire expression as a perfect cube. this is now a root of unity equation, and the only real solution is (u^2 - 2)/u = 1. Solving this quadratic yields u=-1 and u=2. An exponential with a positive base cannot be negative and hence our only valid solution is 2^x = 2 and hence x = 1.

Paradoxica · Oct 6, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Also, the question asked to find A + B + C, so make sure to write the value of that at the end.

the solution it asks makes it sound like you took the question from brilliant.org....

InteGrand · Oct 6, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
the solution it asks makes it sound like you took the question from brilliant.org....

Lol what do you mean? I didn't even post that question.

kawaiipotato · Oct 6, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Lol what do you mean? I didn't even post that question.

Just downloaded the app. He's referencing the style of questions they ask. They always ask to find the pronumerals and then give the result when we add them

InteGrand · Oct 6, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
Just downloaded the app. He's referencing the style of questions they ask. They always ask to find the pronumerals and then give the result when we add them

Did he just reply to the wrong person then? He replied to me when I didn't even post the question.

kawaiipotato · Oct 6, 2015

Re: HSC 2015 3U Marathon

Most likely he did and thought you asked the question.

davidgoes4wce · Oct 6, 2015

Re: HSC 2015 3U Marathon

I had another go at another 'inequality induction' question. Keen to know what Bored of Studies forumites think this would get in a scale of 0 to 100%?

leehuan · Oct 6, 2015

Re: HSC 2015 3U Marathon

Hm. I would say it is ok.

However, for the inductive step, this is what I would do personally.

RTP: 3(1+2k)>2k+3
...
k>0
This is true by definition.

Therefore 3^1 . 3^k > 3(1+2k) > 2k+3 by inductive assumption
Therefore 3^(k+1) > 2k+3
__________________________
But remember, induction can be done in a variety of ways.

davidgoes4wce · Oct 6, 2015

Re: HSC 2015 3U Marathon

This question is SHM from Fitzpatrick:

The displacement x metres at time t seconds of a point moving in a straight line is given by x=a cos (nt+E). Find the form that this expression takes it initially:

b) x=0 and the velocity is negative

This is the solution:

I didn't quite understand why acceleration was positive. I tried to do the derivative of velocity and get something along the lines of:
acceleration= -a*n^2 cos (nt+Pi/2)

leehuan · Oct 6, 2015

Re: HSC 2015 3U Marathon

I just ate and my brain isn't fully functioning but I'm lost. Which question? Q8? Cause I can't find acceleration in Q8

davidgoes4wce · Oct 6, 2015

Re: HSC 2015 3U Marathon

leehuan said:
I just ate and my brain isn't fully functioning but I'm lost. Which question? Q8? Cause I can't find acceleration in Q8

The displacement x metres at time t seconds of a point moving in a straight line is given by x=a cos (nt+E). Find the form that this expression takes it initially:

Q8b)
x=0 and the velocity is negative

I understand everything but not why a>0

leehuan · Oct 6, 2015

Re: HSC 2015 3U Marathon

a is the amplitude not the acceleration

davidgoes4wce · Oct 6, 2015

Re: HSC 2015 3U Marathon

leehuan said:
a is the amplitude not the acceleration

Of course it is! It must be early Monday morning for me.

rand_althor · Oct 6, 2015

Re: HSC 2015 3U Marathon

$$Let $\alpha=\frac{m\pi}{n}$ where $m$ and $n$ are positive relatively prime integer numbers. Find $m+n$ if $0 \leq \alpha \leq \frac{\pi}{2}$ and $\sin{\alpha}=\frac{1-\sqrt{2}}{\sqrt{6-3\sqrt{2}}-\sqrt{2+\sqrt{2}}}$.$

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