HSC 2013 MX2 Marathon (archive) (7 Viewers)

Helvetica · Feb 2, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
Yeh man!

$$ Since P(x) has roots \alpha_1, \alpha_2, ..., \alpha_n \therefore P(x) = (x - \alpha_1)(x - \alpha_2)\dots(x - \alpha_n) \\ \\$ $ Log both sides and applying the useful rule: log_e ab = log_e a + log_e b\\ \\ ln(P(x)) = ln(x-\alpha_1) + \dots +ln(x-\alpha_n) \\ \\ $ $Differentiating both sides with respect to x \\ \\ \therefore \frac{P'(x)}{P(x)} = \frac{1}{x-\alpha_1} + \frac{1}{x-\alpha_2} +\dots +\frac{1}{x-\alpha_n} \\ \\$ $ Times both sides by P(x) and zu end up with ze answer!$

Which latex generator did you use? The one you used looks much better than http://www.codecogs.com/latex/eqneditor.php

iBibah · Feb 2, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
Let m and n be the roots of the equation $t^{2}+bt+c=0$ . Show that b and c are the roots of the equation $t^{2}+(m+n-mn)t-mn(m+n)=0$ .

Is this actually a 4U question?

asianese · Feb 3, 2013

Re: HSC 2013 4U Marathon

No it's more like harder 2u...or just 2u.

iBibah · Feb 3, 2013

Re: HSC 2013 4U Marathon

asianese said:
No it's more like harder 2u...or just 2u.

Thought so.

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

An easier one

$Find all solutions to$

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

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
I think the first part should be: $$i) Prove that$ \ \ \ \frac{I_{n}}{I_{n-n}} = \frac{n-1}{n} \ \ \ \ \fbox{2}$
(i) Let $u = sin^{n-1}x \therefore du = (n-1)(sin^{n-2}x)(cosx)$ and $dv = sinx \therefore v = -cosx$
$\Rightarrow \left [ -cosxsin^{n-1}x \right ]_{0}^{\pi } + \int_{0}^{\pi }(1-sin^{2}x)sin^{n-2}xdx \Rightarrow (n-1)\left ( \int_{0}^{\pi }sin^{n-2}x - \int_{0}^{\pi }sin^{n}x \right ) \therefore \frac{I_{n}}{I_{n-2}}=\frac{n-1}{n}$
(ii) I got this: $I_{n}=\frac{n-1}{n}.\frac{n-2}{n-1}.\frac{n-3}{n-2}...\frac{4}{3}.\frac{\pi }{2}$ don't know if it helps. Any way, how do you do (ii)?

For the second one, consider what you did for the RHS of your expression, but do that for I instead

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
For the second one, consider what you did for the RHS of your expression, but do that for I instead

oh shit, accidentally deleted the post

Can you please tell me how you do your latex? The website I use doesn't allow spacing.
Edit: I get it, you put $

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
oh shit, accidentally deleted the post
Can you please tell me how you do your latex? The website I use doesn't allow spacing...

See how you can bold text using the [.B] amd [./B] (without the dots), like this: bold (to see the tags quote this post)
So for latex, use [.tex] and [./tex] (without the dots), like this: $$THIS IS LATEX$ \ \ \frac{\pi^2}{6}$

There is a short guide on how to use it

http://community.boredofstudies.org/showthread.php?t=234259

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
See how you can bold text using the [.B] amd [./B] (without the dots), like this: bold (to see the tags quote this post)
So for latex, use [.tex] and [./tex] (without the dots), like this: $$THIS IS LATEX$ \ \ \frac{\pi^2}{6}$

There is a short guide on how to use it

http://community.boredofstudies.org/showthread.php?t=234259

Ok, thank you

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

Can I post conics questions?

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
Can I post conics questions?

Go right ahead

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

$$Prove that$ \ \ \frac{a+b}{2} \geq \sqrt{ab}$

$Using the above result, hence prove that$

$\frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove that$ \ \ \frac{a+b}{2} \geq \sqrt{ab}$

$Using the above result, hence prove that$

$\frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

I have done this question before but a, b, c and d must all be positive real numbers so shouldn't you write that in the question?.
$If a and b are positive$ $\therefore (\sqrt{a}-\sqrt{b})^{2}\geq 0 \Rightarrow a+b-2\sqrt{ab} \geq 0 \therefore \frac{a+b}{2}\geq \sqrt{ab}$ $QED$
$Similarly, if c and d are positive$ $\therefore \frac{c+d}{2}\geq \sqrt{cd}$ $Now add them:$ $a+b+c+d \geq 2\sqrt{ab}+2\sqrt{cd}$ $Now we have to prove that:$ $2\sqrt{ab}+2\sqrt{cd} \geq 4\sqrt[4]{abcd}$ $Using:$
$\frac{a+b}{2}\geq \sqrt{ab}$ $sub$ $a=\sqrt{ab}$ $and$ $b=\sqrt{cd}$
$\Rightarrow \sqrt{ab} + \sqrt{cd} \geq 2\sqrt{\sqrt{ab}\sqrt{cd}} = 2\sqrt[4]{abcd}$
$\therefore a+b+c+d \geq 4\sqrt[4]{abcd}$ $QED$

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Go right ahead

$Question:$
$(a)$ $P (acos\theta , bsin\theta )$ $$lies on the ellipse$ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ . $$The tangent to the ellipse at P passes through a focus of the hyperbola$ $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$(i) Find the tangent at P (Using parametric questions).$
$(ii) Show that P lies on the directrix of the hyperbola.$
$(iii) Show that the tangent to the ellipse at P has gradient$ $\pm 1$ .
$(b) (i) Sketch the graph of the rectangular hyperbola xy=1 showing clearly the coordinates of the foci and the equation of the directrices.$
$(ii)$ $P(p,\frac{1}{p})$ $and$ $Q (q,\frac{1}{q})$ $$are two points on the rectangular hyperbola xy=1.$
$Show that the chord PQ has equation$ $x+pqy-(p+q)=0$ .
$(iii) If O is the origin, show that$ $\bigtriangleup OPQ$ $has area$ $\left | \frac{p^{2}-q^{2}}{2pq} \right |$ $square units.$

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
I have done this question before but a, b, c and d must all be positive real numbers so shouldn't you write that in the question?.
$If a and b are positive$ $\therefore (\sqrt{a}-\sqrt{b})^{2}\geq 0 \Rightarrow a+b-2\sqrt{ab} \geq 0 \therefore \frac{a+b}{2}\geq \sqrt{ab}$ $QED$
$Similarly, if c and d are positive$ $\therefore \frac{c+d}{2}\geq \sqrt{cd}$ $Now add them:$ $a+b+c+d \geq 2\sqrt{ab}+2\sqrt{cd}$ $Now we have to prove that:$ $2\sqrt{ab}+2\sqrt{cd} \geq 4\sqrt[4]{abcd}$ $Using:$
$\frac{a+b}{2}\geq \sqrt{ab}$ $sub$ $a=\sqrt{ab}$ $and$ $b=\sqrt{cd}$
$\Rightarrow \sqrt{ab} + \sqrt{cd} \geq 2\sqrt{\sqrt{ab}\sqrt{cd}} = 2\sqrt[4]{abcd}$
$\therefore a+b+c+d \geq 4\sqrt[4]{abcd}$ $QED$

YesI should of put that in the question, good work anyway. I wil leave the Conics question to someone else who has learnt all of it

darkfenrir · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
YesI should of put that in the question, good work anyway. I wil leave the Conics question to someone else who has learnt all of it

Question: Have you and RealiseNothing done the whole 4u syllabus already? lol

I would do the conics quesiton but cbf typing it. Someone else can do that

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

darkfenrir said:
Question: Have you and RealiseNothing done the whole 4u syllabus already? lol

I would do the conics quesiton but cbf typing it. Someone else can do that

I have kind of done all 4U topics but completing the course doesn't mean shit so now I will just keep doing past papers and harder questions (and pay more attention to English, UMAT and science).

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

darkfenrir said:
Question: Have you and RealiseNothing done the whole 4u syllabus already? lol

I would do the conics quesiton but cbf typing it. Someone else can do that

I have already completed the 3U course which is why I am able to do most of the Harder 3U questions on here, but in terms of the 4U syllabus I only know what I have been taught in class + Integration

twinklegal19 · Feb 3, 2013

Re: HSC 2013 4U Marathon

darkfenrir said:
Question: Have you and RealiseNothing done the whole 4u syllabus already? lol

I would do the conics quesiton but cbf typing it. Someone else can do that

yeah he has (should have done the HSC last year) lol

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

$Induction Question$
$Prove:$ $(2^{b}-1).\sum_{k=0}^{a-1}2^{kb}=2^{ab}-1$

HSC 2013 MX2 Marathon (archive) (7 Viewers)

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