HSC 2013 MX2 Marathon (archive) (1 Viewer)

nightowl · Aug 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Solve for all$ \ \ x \ $and$ \ y$

$\frac{1}{x} + \frac{1}{y} = - 1$

$x^3+y^3 = 4$

$\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{1}{x^3}+\frac{1}{y^3}+3\left(\frac{1}{x^y}+\frac{1}{xy^2}\right)$

$-1=\frac{x^3+y^3}{x^3y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)$

$-1=\frac{4}{x^3y^3}-\frac{3}{xy}$

$-x^3y^3=4-3x^2y^2$

Let $z=xy$

Factorise $z^3-3z^2+4=0$

$(z+1)(z-2)^2=0$

$y=-\frac{1}{x} \text{ or }y=\frac{2}{x}$

The rest is just solving quadratics.

AnimeX · Aug 21, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots < 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 3$

I i'm wondering, shouldn't this be <3? because that never reaches 3?

Are you allowed to say =3?

Carrotsticks · Aug 21, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
I i'm wondering, shouldn't this be <3? because that never reaches 3?

Are you allowed to say =3?

The expression on the right is an infinite series, which is EQUAL to 3.

If I took a partial sum that added a finite number (say N) of terms, then you could say it's less than 3.

Kinda the same reason why 0.99999..... = 1, contrary to popular belief that 0.999999...... < 1.

Menomaths · Aug 21, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
I i'm wondering, shouldn't this be <3? because that never reaches 3?

Are you allowed to say =3?

Let x=0.9999999.....
10x = 9.99999999....
10x-x = 9 since the 9's cancel each other
9x = 9
x = 1
Therefore 1 = 0.999999.....
Yay for first contribution to 4U marathon

AnimeX · Aug 21, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
The expression on the right is an infinite series, which is EQUAL to 3.

If I took a partial sum that added a finite number (say N) of terms, then you could say it's less than 3.

Kinda the same reason why 0.99999..... = 1, contrary to popular belief that 0.999999...... < 1.

Not quite getting this.

So does that mean this is wrong? (changing the places of < sign and = sign)

$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots < 3$

Carrotsticks · Aug 21, 2013

Re: HSC 2013 4U Marathon

I'll short-name it as L, M and R.

It should be L < M = R.

So L is less than M, but M is equal to R, so therefore L is less than R.

RealiseNothing · Aug 21, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
Not quite getting this.

So does that mean this is wrong? (changing the places of < sign and = sign)

$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots < 3$

$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} \dots$

I took each term, and made it correspond to a slightly larger term so I could establish an upper bound for the series:

$\frac{1}{4} > \frac{1}{6}$

$\frac{1}{8} > \frac{1}{24}$

etc

So you end up with my result.

chris_power_96 · Aug 21, 2013

Re: HSC 2013 4U Marathon

Find 1 +xcosx+x^2cos2x+...+x^ncosnx

JohnMaximus · Aug 21, 2013

Re: HSC 2013 4U Marathon

Not sure if 4U marathon or harder 3U marathon.

asianese · Aug 21, 2013

Re: HSC 2013 4U Marathon

JohnMaximus said:
Not sure if 4U marathon or harder 3U marathon.

$Harder 3U $ \subset $ 4U.$

obliviousninja · Aug 21, 2013

Re: HSC 2013 4U Marathon

Any got any 2013 trial papers?

Tmalak · Aug 23, 2013

Re: HSC 2013 4U Marathon

2013 Trials PLEASEtrollface.jpg

Tmalak · Aug 23, 2013

Re: HSC 2013 4U Marathon

Trials

Tmalak · Aug 23, 2013

Re: HSC 2013 4U Marathon

Sy123 · Aug 24, 2013

Re: HSC 2013 4U Marathon

$$Prove for some positive$ \ x \ $and$ \ a \geq 1$

$x^a + \frac{1}{x^a} \geq x^{a-1} + \frac{1}{x^{a-1}}$

$For what values of$ \ x \ $and$ \ a \ $does equality hold?$

Sy123 · Aug 25, 2013

Re: HSC 2013 4U Marathon

$$For positive real values$ \ a, b, c \ $it is given$ \ a+b+c \geq 3\sqrt[3]{abc} \ \ \ $and$ \ a+b+c = 1$

$$i) Prove$ \ \frac{1}{a} +\frac{1}{b} + \frac{1}{c} \geq 9$

$$ii) Hence find the minimum value of$ \ \ \left( \frac{1}{a} - 1 \right ) \left(\frac{1}{b}- 1 \right ) \left( \frac{1}{c} - 1 \right )$

HeroicPandas · Aug 25, 2013

Re: HSC 2013 4U Marathon

$\rightarrow \\ (\sqrt{x}-\frac{1}{\sqrt{x}}) \geq 0 \\ ~~~~~\therefore x+\frac{1}{x} \geq 2 \dots [1]\\ \\ (i) LHS = \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \\ \\ ~~~~~~~~~~~=(a+b + c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \\ \\ ~~~~~~~~~~~= 3 + a + \frac{1}{a} + b + \frac{1}{b} + c + \frac{1}{c} \\ \\~~~~~~~~~~~\geq 3 + 2 + 2 + 2 (using [1]) \\ \\ ~~~~~~~~~~~=9$

$\left ( \frac{1}{a}-1 \right )\left ( \frac{1}{b}-1 \right )\left ( \frac{1}{c}-1 \right ) = \frac{1}{abc} - \left ( \frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc} \right ) + \frac{1}{a}+ \frac{1}{b} + \frac{1}{c} -1 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\geq \frac{1}{abc} - \left ( \frac{a+b+c}{abc} \right ) + 9 -1 \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~($as a+b +c =1 and using part i)\\ \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= $\frac{1}{abc} - \frac{1}{abc} + 8 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=8 \\ \\ \therefore \left ( \frac{1}{a}-1 \right )\left ( \frac{1}{b}-1 \right )\left ( \frac{1}{c}-1 \right ) \geq 8 \\ \\ \therefore The minimum value of \left ( \frac{1}{a}-1 \right )\left ( \frac{1}{b}-1 \right )\left ( \frac{1}{c}-1 \right )$ is 8$

i stuffed up the latex typing lol

i thought of another method: (does it work?)

$3\sqrt[3]{abc} \leq a+b+c \\ a\rightarrow \left ( \frac{1}{a}-1 \right ) \\b\rightarrow \left ( \frac{1}{b} -1\right )\\c\rightarrow \left ( \frac{1}{c}-1 \right ) \\ \\ \therefore \left ( \frac{1}{a}-a \right )\left ( \frac{1}{b} -1\right )\left ( \frac{1}{c}-1 \right ) \leq \left ( \frac{\frac{1}{a}+\frac{1}{b}+ \frac{1}{c}-3}{3} \right )^3 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\geq \left ( \frac{9-3}{3} \right )^3 ($using part i) \\ \\ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 8 \\ \\ \therefore LHS_{min} = 8$

asianese · Aug 25, 2013

Re: HSC 2013 4U Marathon

As I said...just use the align or eqnarray function...~~~~~~ is a manual function that is the antithesis of LaTeX.

Sy123 · Aug 26, 2013

Re: HSC 2013 4U Marathon

Yep HeroicPandas, that's pretty much it, that is 7 marks from a school trial (ascham 2005)

===

$$Suppose$ \ \ a,b,c > 0 \ $and$ \ abc= 1$

$$It is known that$ \ a+b+c > \frac{1}{a}+ \frac{1}{b} + \frac{1}{c}$

$Prove that exactly one of the numbers$ \ a,b,c \ \ $is greater than 1$

HeroicPandas · Aug 26, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep HeroicPandas, that's pretty much it, that is 7 marks from a school trial (ascham 2005)

damnn 7 marks

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