HSC 2015 MX2 Marathon (archive) (1 Viewer)

Axio · Aug 27, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
$(x^3-\frac{1}{2})^2=x^6-x^3+\frac{1}{4} \geqslant 0$
$(x^2-\frac{1}{2})^2=x^4-x^2+\frac{1}{4} \geqslant 0$
$(x-\frac{1}{2})^2=x^2-x+\frac{1}{4} \geqslant 0$
Adding these three inequalities yields:
$x^6+x^4+x^2-x^3-x^2-x+\frac{1}{4}+\frac{1}{4}+\frac{1}{4} > 0$
The strict inequality is true since all three inequalities are zero at different values of x.
Adding $\frac{1}{4}$ only furthers the strictness of the inequality, which leads to:
$x^6+x^4-x^3-x+1 > 0$
Which after re-arranging gives the desired inequality.
Now, my question is, how many marks would I get (compared to whatever would be given) in the HSC for doing it this way instead of calculus? (I have nothing against calculus, it's just that I will always try to solve a problem in the most elegant way I can (quickly) think of)

I would have thought that unless they asked you to use calculus or it had 'hence' in the question (and a previous part involved using calculus) that you would get full marks...

Drsoccerball · Sep 12, 2015

Re: HSC 2015 4U Marathon

LOOOL my teacher expected us to do this :
$pqr+2(pr+qr+pq)+4(p+q+r)+8 =64$
From AM-GM
$p+q+r \geq 3(pqr)^{1/3} $ and $ pr+qr+pq \geq 3(pqr)^{2/3}$

$pqr +3 \cdot 2(pqr)^{2/3} +3 \cdot 2^2 (pqr)^{1/3} +8 \leq 64$

$((pqr)^{1/3}+2)^3 \leq 64$

$\therefore pqr \leq 8$

kawaiipotato · Sep 12, 2015

Re: HSC 2015 4U Marathon

New question: question b
http://imgur.com/XPZ7Ass
http://imgur.com/bqCRBsv

Also, would anyone know what paper this is from?

dan964 · Sep 12, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
New question: question b
http://imgur.com/XPZ7Ass
http://imgur.com/bqCRBsv

Also, would anyone know what paper this is from?

I suspect it is from a catholic/cssa paper or Independent (one of those).

Drsoccerball · Sep 12, 2015

Re: HSC 2015 4U Marathon

dan964 said:
I suspect it is from a catholic/cssa paper or Independent (one of those).

Its a cssa paper i think either 2011 or 12

Ekman · Sep 13, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
New question: question b
http://imgur.com/XPZ7Ass
http://imgur.com/bqCRBsv

Also, would anyone know what paper this is from?

$$ I) $ \frac{x^{2n-1}}{\sqrt{1-x^2}} - \frac{x^{2n+1}}{\sqrt{1-x^2}} = x^{2n-1}\left(\frac{1}{\sqrt{1-x^2}} - \frac{x^2}{\sqrt{1-x^2}}\right) = x^{2n-1} \sqrt{1-x^2}$

$$ II) $ \frac{d}{dx}\left( -x^{2n}\sqrt{1-x^2} \right) = \frac{x^{2n+1}}{\sqrt{1-x^2}} - 2nx^{2n-1}\sqrt{1-x^2}$

$\therefore $ (By integrating both sides between 0 to 1) $ 0 = I_{2n+1} - 2n\int_{0}^{1} x^{2n-1}\sqrt{1-x^2}$

$\therefore $ (Using Part I) $ 0 = I_{2n+1} - 2n I_{2n-1} + 2n I_{2n+1}$

$\therefore I_{2n+1} = \frac{2n}{2n+1} I_{2n-1}$

$$ III) $ I_{2n+1} = \frac{2n}{2n+1} I_{2n-1}, I_{2n-1} = \frac{2n-2}{2n-1} I_{2n-3} \dots I_{3} = \frac{2}{3} I_{1}, I_{1}= 1$

$\therefore I_{2n+1} = \frac{2^n \cdot n!}{1 \times3 \times \dots (2n-1)}$

$$ IV) $ \int_{0}^{1} \sum_{n=1}^{\infty} C_{n} \frac{x^{2n+1}}{\sqrt{1-x^2}} = C_{1} \cdot I_{3} + C_{2} \cdot I_{5} + \dots = \sum_{n=1}^{\infty} \frac{1}{(2n+1)^2}$

$\int_{0}^{1} \frac{x}{\sqrt{1-x^2}} = 1 $ $ \therefore \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} + \int_{0}^{1} \sum_{n=1}^{\infty} C_{n} \frac{x^{2n+1}}{\sqrt{1-x^2}} = 1 + \sum_{n=1}^{\infty} \frac{1}{(2n+1)^2} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$

$$ V) $ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} + \int_{0}^{1} \sum_{n=1}^{\infty} C_{n} \frac{x^{2n+1}}{\sqrt{1-x^2}} = \int_{0}^{1} \frac{x+ \sum_{n=1}^{\infty}C_{n}x^{2n+1}}{\sqrt{1-x^2}} = \int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} = \left[\frac{(\sin^{-1}x)^2}{2}\right]_{0}^{1} = \frac{\pi^2}{8}$

$\therefore \frac{\pi^2}{8} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$

$$ VI) $ S= 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \dots$

$\therefore S =\frac{\pi^2}{8} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots$

$$ We know that $ \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots = \frac{1}{2^2} + \frac{1}{2^2 \cdot 2^2} + \frac{1}{2^2 \cdot 3^2} + \dots = \frac{1}{2^2} \left(1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \dots \right) = \frac{1}{4} \cdot S$

$\therefore S = \frac{\pi^2}{8} + S\frac{1}{4} \Rightarrow S = \frac{\pi^2}{6}$

Edit: Please do excuse the lack of dx's in each integral.

Drsoccerball · Sep 19, 2015

Re: HSC 2015 4U Marathon

$$ If $ 1,w,w^2,w^3,w^4 $ are roots of $ w^5-1=0$

$$ Given that $ (1-w)(1-w^2)(1-w^3)(1-w^4)=5 $ and $ (1-w)(1-w^4) =2-2cos\frac{2 \pi}{5}$

$$ Hence show that $ sin\frac{\pi}{5} sin\frac{2 \pi}{5} =\frac{\sqrt{5}}{4}$

Drsoccerball · Sep 20, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ If $ 1,w,w^2,w^3,w^4 $ are roots of $ w^5-1=0$

$$ Given that $ (1-w)(1-w^2)(1-w^3)(1-w^4)=5 $ and $ (1-w)(1-w^4) =2-2cos\frac{2 \pi}{5}$

$$ Hence show that $ sin\frac{\pi}{5} sin\frac{2 \pi}{5} =\frac{\sqrt{5}}{4}$

Got it dw

asdfghjklqwerty · Sep 20, 2015

Re: HSC 2015 4U Marathon

https://gyazo.com/5b037e94325321a00dbc213793ccd457

hey guys for this question can someone explain to me why the restrictions on the locus of N are x<0 and y>0? thanks

Drsoccerball · Sep 21, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
https://gyazo.com/5b037e94325321a00dbc213793ccd457

hey guys for this question can someone explain to me why the restrictions on the locus of N are x<0 and y>0? thanks

It because you have y^2 =-3x/8 therefore since theres y^2 that means x has to be negative. However the range should be all real y except y=0.

braintic · Sep 21, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
https://gyazo.com/5b037e94325321a00dbc213793ccd457

hey guys for this question can someone explain to me why the restrictions on the locus of N are x<0 and y>0? thanks

Could you supply the answer to that part.

asdfghjklqwerty · Sep 21, 2015

Re: HSC 2015 4U Marathon

https://gyazo.com/02364a9ea1544a4b485cf7f921b7f7d0
here are the answers. just dont get why y>0 unless answers wrong

Drsoccerball · Sep 21, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
https://gyazo.com/02364a9ea1544a4b485cf7f921b7f7d0
here are the answers. just dont get why y>0 unless answers wrong

Its because pq is less than 0 as p and q are in opposite quadrants

asdfghjklqwerty · Sep 23, 2015

Re: HSC 2015 4U Marathon

i know this is a rlly easy q but i cant seem to get it.
how do u integrate cosec 2x using t formula?
thanks

Drsoccerball · Sep 23, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
i know this is a rlly easy q but i cant seem to get it.
how do u integrate cosec 2x using t formula?
thanks

let t= tanx not tanx/2 and work from there

VBN2470 · Sep 25, 2015

Re: HSC 2015 4U Marathon

Prove that if $n$ is a positive integer such that $n>3$ , then at least one of $n$ , $n+2$ or $n+4$ is composite.

dan964 · Sep 25, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Prove that if $n$ is a positive integer such that $n>3$ , then at least one of $n$ , $n+2$ or $n+4$ is composite.

It surfices to show first, the each prime number bigger than 3 is either 6K + 1 or 6K+5 where K is an integer.
Firstly all numbers can be written in the form 6K+n where K=>0, 0<=n<6
All numbers of the form 6K+2 and 6K+4 are composite, except 2 (divisible by 2) and 6K+3 is also composite, except 3 (divisible by 3).

Thus the question only concerns prime numbers which are of the form 6K+1 or 6K+5 (and not 2 or 3 obviously), as composite numbers already satisfy the result:
If n=6K+1, n+2 = 6K+3 which is composite (divisble by 3)
If n=6K+5, n+4 = 6K+9 which is composite (divisble by 3)

VBN2470 · Sep 25, 2015

Re: HSC 2015 4U Marathon

Yep, you're method is fine. Here's mine (similiar):

Since we have $n>3$ , we have the following:

Case 1: Let $n=3k$ for some positive integer $k>1$ . Then $n$ is divisible by 3 and we are done.

Case 2: Let $n=3k+1$ for some positive integer $k>1$ . Then $n+2=3k+3$ is divisible by 3 and we are done.

Case 3: Let $n=3k+2$ for some positive integer $k>1$ . Then $n+4=3k+6$ is divisible by 3 and we are done.

Hence, at least one of $n$ , $n+2$ or $n+4$ is composite.

VBN2470 · Sep 25, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION:

Prove that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Drsoccerball · Sep 25, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION:

Prove that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

I remember remembering this in year 8

dont know how to prove it tho

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