HSC 2013-14 MX1 Marathon (archive) (5 Viewers)

asianese · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
:O i never knew Cambridge 3U uses LaTeX..

LOL. Mate. All of Bill Pender's books are beautifully typeset! You only appreciate it later in life (lol).

Makematics · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

OOOO i remember that question

HeroicPandas · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
LOL. Mate. All of Bill Pender's books are beautifully typeset! You only appreciate it later in life (lol).

damnn i never knew until now...lol

asianese · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

I only really had a look at the CHALLENGE questions he had for both 2U and 3U recently (post HSC) and I realised there are some very beautiful questions there, that many teachers/school inevitably don't get to. Or you can use MIF and just stop learning, but that's a different story.

QUESTION:

Simplify
$\sin^{-1}(x-y)+\sin^{-1}(y-x).$

HeroicPandas · Jun 24, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
I only really had a look at the CHALLENGE questions he had for both 2U and 3U recently (post HSC) and I realised there are some very beautiful questions there, that many teachers/school inevitably don't get to. Or you can use MIF and just stop learning, but that's a different story.

QUESTION:

Simplify
$\sin^{-1}(x-y)+\sin^{-1}(y-x).$

$$Let $ z = \sin^{-1}(x-y)+\sin^{-1}(y-x) \\ $Sine both sides $ \\ \\ \therefore \sin(z) = \sin\left ( \sin^{-1}(x-y)+\sin^{-1}(y-x) \right ) \\ $Let \alpha = sin^{-1}(x-y) \rightarrow sin(\alpha) = $\frac{x-y}{1}, ($Then draw a triangle) \\$ $Simiarly, let \beta = sin^{-1}(y-x)\rightarrow sin(\beta)=$\frac{y-x}{1}$ $($Then draw a triangle$) \\ \\ \therefore sin(z) = sin(\alpha+\beta) \\ ~~~~~~~~~~~~= sin(\alpha)cos(\beta) + sin(\beta)cos(\alpha) \\ ~~~~~~~~~~~~= (x-y)\sqrt{1-(y-x)^2} + (y-x)\sqrt{1-(x-y)^2} ($Using triangles and sin(sin^{-1}\psi )) = \psi) \\ ~~~~~~~~~~~= 0 \\$
wasted my time because u had tricked me...

$OR\dots \dots \dots \\ \\ $Since, \\ $\underline{sin^{-1}(-\varphi ) = -sin^{-1}(\varphi)}\\ \\ \therefore sin^{-1}(x-y) + sin^{-1}(y-x) = sin^{-1}(x-y) - sin^{-1 }(x-y) = 0\dots \dots \\ \\$

I've touched MIF at school, and i will never return to it...

asianese · Jun 24, 2013

Re: HSC 2013 3U Marathon Thread

Lol. I successfully tricked you. $\sin^{-1}(-x)=-\sin^{-1}(x)$ .

And also in your last line of latex, don't bold the 0, this implies it is the 0 vector - irrelevant at the moment but still.

Makematics · Jun 24, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
I only really had a look at the CHALLENGE questions he had for both 2U and 3U recently (post HSC)

as in the extension questions?? yeh they're awesome

HeroicPandas · Jun 24, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$A^{n+1} - A^nB + B^{n+1} - B^nA = (A-B)(A^n - B^n)\geq 0 \\($because \dots if A $>B,\therefore A-B>0, \therefore A^n-B^n >0, \therefore (A-B)(A^n - B^n)>),$ same thing happens when A$<B$) \\ $ $ And equality occurs when A = B$ \\ \\ $If ($A-B)(A^n-B^n) \geq 0 \\\\ A^{n+1}- A^nB + B^{n+1} - B^nA \geq 0 \\ \\ \therefore A^{n+1}+ B^{n+1} \geq A^nB+B^nA$

Is this ok?

Did i do it correctly?

omgiloverice · Jun 25, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
Did i do it correctly?

I guess~ I can't really find any logical faults. But another way you can do it is by dividing $A^{n}-B^{n}$ by $A-B$

And you'll get $(A-B)(A-B)(A^{n-1}+A^{n-2}B+...+b^{n-1})$ which then makes the whole process so much more clearer

study1234 · Jul 1, 2013

Re: HSC 2013 3U Marathon Thread

In the expansion of (1+3x+ax^2)^n, where n is a positive integer, the coefficient of x^2 is 0. Find, in terms of n, the value of the coefficient of x^3.

Sy123 · Jul 5, 2013

Re: HSC 2013 3U Marathon Thread

$A projectile with a velocity$ \ \ V \ \ $at an angle$ \ \ \theta \ \ $to the horizontal$

$It lands on an inclined plane at an angle$ \ \ \alpha \ \ $to the horizontal as shown in the diagram$

$Prove that maximum range achieved can only be done so if$

$\theta = \frac{\alpha}{2} + \frac{\pi}{4}$

HeroicPandas · Jul 6, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$A projectile with a velocity$ \ \ V \ \ $at an angle$ \ \ \theta \ \ $to the horizontal$

$It lands on an inclined plane at an angle$ \ \ \alpha \ \ $to the horizontal as shown in the diagram$

$Prove that maximum range achieved can only be done so if$

$\theta = \frac{\alpha}{2} + \frac{\pi}{4}$

omgiloverice said:
I guess~ I can't really find any logical faults. But another way you can do it is by dividing $A^{n}-B^{n}$ by $A-B$

And you'll get $(A-B)(A-B)(A^{n-1}+A^{n-2}B+...+b^{n-1})$ which then makes the whole process so much more clearer

ooo thats a nice approach!

Sy123 · Jul 6, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
View attachment 28319

ooo thats a nice approach!

Well done. That is the method I had in mind.

$$The sequence$ \ \ u_1, u_2, u_3, \dots, u_n \ \ \dots$

$Is an arithmetic sequence that consists of only positive numbers$

$Show that$

$\sum_{k=2}^{n} \frac{1}{\sqrt{u_{n}} + \sqrt{u_{n-1}}} = \frac{n-1}{\sqrt{u_{n}} + \sqrt{u_1}}$

nightweaver066 · Jul 6, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Well done. That is the method I had in mind.

$$The sequence$ \ \ u_1, u_2, u_3, \dots, u_n \ \ \dots$

$Is an arithmetic sequence that consists of only positive numbers$

$Show that$

$\sum_{k=2}^{n} \frac{1}{\sqrt{u_{n}} + \sqrt{u_{n-1}}} = \frac{n-1}{\sqrt{u_{n}} + \sqrt{u_1}}$

$\sum_{k=2}^{n} \frac{1}{\sqrt{u_n}+\sqrt{u_{n-1}}}$

$=\sum_{k=2}^{n} \frac{\sqrt{u_n}-\sqrt{u_{n-1}}}{u_n - u_{n-1}}$

$=\sum_{k=2}^{n} \frac{\sqrt{u_n}-\sqrt{u_{n-1}}}{d}$

$=\frac{1}{d}(\sqrt{u_2} - \sqrt{u_1} + \sqrt{u_3} - \sqrt{u_2} + ... + \sqrt{u_n} - \sqrt{u_{n-1}})$

$=\frac{1}{d}(\sqrt{u_n} - \sqrt{u_1})$

$=\frac{u_n - u_1}{d(\sqrt{u_n}+\sqrt{u_1})}$

$=\frac{u_1 + (n-1)d - u_1}{d(\sqrt{u_n} + \sqrt{u_1})}$

$= \frac{n-1}{\sqrt{u_n} + \sqrt{u_1}}$

where are you getting these questions from? or are you making them?

asianese · Jul 6, 2013

Re: HSC 2013 3U Marathon Thread

nightweaver066 said:
where are you getting these questions from? or are you making them?

Pretty sure that came from 2011 2U HSC modified. (lol I remember it).

I think other questions come from overseas papers. (Step, HK etc)

luvglee4lyfe · Jul 6, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Lol. I successfully tricked you. $\sin^{-1}(-x)=-\sin^{-1}(x)$ .

And also in your last line of latex, don't bold the 0, this implies it is the 0 vector - irrelevant at the moment but still.

Hey asianese, I've been seeing you on the p00n! But anyways, we used vectors in Year 11 Physics and I don't really get what you're saying?

nightweaver066 · Jul 6, 2013

Re: HSC 2013 3U Marathon Thread

luvglee4lyfe said:
Hey asianese, I've been seeing you on the p00n! But anyways, we used vectors in Year 11 Physics and I don't really get what you're saying?

When studying vectors in uni, you're taught this.

asianese said:
Pretty sure that came from 2011 2U HSC modified. (lol I remember it).

I think other questions come from overseas papers. (Step, HK etc)

ahh ok cheers. Was thinking of stealing some questions in here for my student to do haha

Sy123 · Jul 7, 2013

Re: HSC 2013 3U Marathon Thread

nightweaver066 said:
$\sum_{k=2}^{n} \frac{1}{\sqrt{u_n}+\sqrt{u_{n-1}}}$

$=\sum_{k=2}^{n} \frac{\sqrt{u_n}-\sqrt{u_{n-1}}}{u_n - u_{n-1}}$

$=\sum_{k=2}^{n} \frac{\sqrt{u_n}-\sqrt{u_{n-1}}}{d}$

$=\frac{1}{d}(\sqrt{u_2} - \sqrt{u_1} + \sqrt{u_3} - \sqrt{u_2} + ... + \sqrt{u_n} - \sqrt{u_{n-1}})$

$=\frac{1}{d}(\sqrt{u_n} - \sqrt{u_1})$

$=\frac{u_n - u_1}{d(\sqrt{u_n}+\sqrt{u_1})}$

$=\frac{u_1 + (n-1)d - u_1}{d(\sqrt{u_n} + \sqrt{u_1})}$

$= \frac{n-1}{\sqrt{u_n} + \sqrt{u_1}}$

where are you getting these questions from? or are you making them?

Some questions I do genuinely make (not this one). Otherwise I will sometimes steal from STEP/Internet/modified HSC/trials.

asianese said:
Pretty sure that came from 2011 2U HSC modified. (lol I remember it).

I think other questions come from overseas papers. (Step, HK etc)

I got that from a 4U trial Q8 and removed the first part of the question.
Which just goes to show how weak of a question it was lol.

Sy123 · Jul 8, 2013

Re: HSC 2013 3U Marathon Thread

(Note I'm giving numbers here because the expression you get if it were all pro numerals would be very big and ugly)

====

Some engineers are in a laboratory in space, they are testing a new gravity simulation machine on a projectile, and assessing the effects of an instant change in gravity to a projectile in flight.
They fire a projectile with a speed of 10m/s at an angle 45 degrees to the horizontal. After 0.5 seconds they change the gravity so that the new gravity is perpendicular to the old gravity force, i.e. the inital gravity force perpendicular to the x-axis is now perpendicular to the y-axis. After another 0.5 seconds they swap it back to the initial gravity force and they let the projectile continue its flight.

All gravitational acceleration is 10m/s^2.

Calculate the final range of the projectile.

Sy123 · Jul 9, 2013

Re: HSC 2013 3U Marathon Thread

$Prove that for any triangle, the line joining the midpoints of 2 sides of the triangle, is parallel to the third side. And is half its length$

HSC 2013-14 MX1 Marathon (archive) (5 Viewers)

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