HSC 2013 Maths Marathon (archive) (1 Viewer)

RealiseNothing · Oct 21, 2013

Re: HSC 2013 2U Marathon

Drewk said:
I found this Question in a trial paper
Integrate 7^x

never seen it before and the process is quite long, are we meant to know some simple formula or something ??

$\int 7^x dx = \frac{1}{\ln(7)} \int 7^x \ln(7) dx = \frac{1}{\ln(7)} 7^x + C$

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
Yes blame it on the infallible calculator <> hahaha my calculator sometimes loses its mind too.

Okay my incompetence at using my calculator

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
Yes blame it on the infallible calculator <> hahaha my calculator sometimes loses its mind too.

Triage said:
Okay my incompetence at using my calculator

lol

ForOneEon · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$$By considering triangle PBA, show that$ \ \sin\alpha = \frac{x^2 - 16}{4x}$

A three line variant of the answer (despite it being just a rearrangement of the aforementioned answers is):
$\ \cos 90 - \alpha = \frac{x^2 + 2^2 - (2\sqrt{5})^2}$
$\ \sin \alpha = \frac{x^2 + 4 - 20}{4x}$
$\ \sin \alpha = \frac{x^2 - 16}{4x}$

God, can't split the lines apart!

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Here's the continuation of the question:

$Hence, given that$ \ \cos\alpha = \frac{x^2 - 32}{4x}, \ $show that the$

$$value of$ \ x \ $is a solution of$ \ x^4 - 56x^2 + 640 = 0$

Najji · Oct 21, 2013

Re: HSC 2013 2U Marathon

ForOneEon said:
A three line variant of the answer (despite it being just a rearrangement of the aforementioned answers is):
$\ \cos 90 - \alpha = \frac{x^2 + 2^2 - (2\sqrt{5})^2}$
$\ \sin \alpha = \frac{x^2 + 4 - 20}{4x}$
$\ \sin \alpha = \frac{x^2 - 16}{4x}$

God, can't split the lines apart!

Which formula is this? And why did you make it cos 90-a??

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Najji said:
Which formula is this? And why did you make it cos 90-a??

It is the cosine formula..

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

ForOneEon · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$Given that$ \ \cos\alpha = \frac{x^2 - 32}{4x}, \ $show that the$

$$value of$ \ x \ $is a solution of$ \ x^4 - 56x^2 + 640 = 0$

$\ \sin \alpha = \frac{x^2 - 16}{4x}$
$\ \cos\alpha = \frac{x^2 - 32}{4x}$
$\ (\cos\alpha)^2 + \ (\sin \alpha)^2 = 1$
$1 = \frac{(x^2 - 16)^2}{16x^2} + \frac{(x^2 - 32)^2}{16x^2}$
$16x^2 = (x^2 - 16)^2 +(x^2 - 32)^2$
$16x^2 = x^4 - 32x^2 + 256 + x^4 - 64x^2 + 1024$
$0 = 2x^4 - 112x^2 + 1280$
$0 = x^4 - 56x^2 + 640$

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

ForOneEon said:
$\ \sin \alpha = \frac{x^2 - 16}{4x}$

$\ \cos\alpha = \frac{x^2 - 32}{4x}$

$\ (\cos\alpha)^2 + \ (\sin \alpha)^2 = 1$

$1 = \frac{(x^2 - 16)^2}{16x^2} + \frac{(x^2 - 32)^2}{16x^2}$

$16x^2 = (x^2 - 16)^2 +(x^2 - 32)^2$

$16x^2 = x^4 - 32x^2 + 256 + x^4 - 64x^2 + 1024$

$0 = 2x^4 - 112x^2 + 1280$

$0 = x^4 - 56x^2 + 640$

Leave more lines empty makes the solution

more

clear

mahoo_mk · Oct 21, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
$\int 7^x dx = \frac{1}{\ln(7)} \int 7^x \ln(7) dx = \frac{1}{\ln(7)} 7^x + C$

$\frac{1}{\ln(7)} \int 7^x \ln(7) dx = \frac{1}{\ln(7)} 7^x + C$

what did you do there boss?

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:

D = (-4,1)

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
D = (-4,1)

Not quite...According to the answer anyways

(I got the same answer as you, so solutions incorrect?)

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Not quite...According to the answer anyways

Oh fark...

Shazer2 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:

Can't you just find difference of C from A, and apply that from B to D such that D(-4, 1)

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Answer says (4,3) :')

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Answer says (4,3) :')

There are two answers as there are two different points which can make ABCD a parallelogram.

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Yeah, then that should be the answer as 3 great minds have come up with the same result.

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

How can you tell which pair of points is the diagonal and which is the side of the parallelogram?

bedpotato · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Answer says (4,3) :')

This is what I got.

You use midpoints since the diaganols of parallelograms bissect each other.

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