HSC 2013 Maths Marathon (archive) (2 Viewers)

girlworld_club · Oct 20, 2013

Re: HSC 2013 2U Marathon

Badsiii said:
Arc length= 5pi/3 = circumference of cone
Hence radius of cone= 5/6
Radius of sector= 5 = slant height of cone
Therefore height of cone, H^2= 5^2 - 25/36
h= root(875/36)
Volume cone= 1/3pi R^2 h
= 1/3 pi 25/36 x root(875/36) ?

The answer is

125 root(35) pi / 648 cm^3

Badsiii · Oct 20, 2013

Re: HSC 2013 2U Marathon

girlworld_club said:
The answer is

125 root(35) pi / 648 cm^3

Yeh that's what I got... Mines just not simplified

girlworld_club · Oct 20, 2013

Re: HSC 2013 2U Marathon

how did you get the perpendicular height for the cone?

dw i got it.

mahmoudali · Oct 20, 2013

Re: HSC 2013 2U Marathon

$$i) Consider$ \ \ f(x) = \ln x + \ln \sqrt{x}$

$$ii) Write f(x) in the form of$ \ \ \ln (g(x))$

$iii) Find the area bounded by g(x) the y axis, y=0 and y=5 in exact form$

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
sin(2x) = 2sinx.cosx
sinx/2sinx.cosx
1/2cosx
1/2

This is a 3U double angle formula, so no marks awarded

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
This is a 3U double angle formula, so no marks awarded

Can it be done with nooby 2U methods?

Youi_ · Oct 20, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$$i) Consider$ \ \ f(x) = \ln x + \ln \sqrt{x}$

$$ii) Write f(x) in the form of$ \ \ \ln (g(x))$

$iii) Find the area bounded by g(x) the y axis, y=0 and y=5 in exact form$

How many marks is part i) lol

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Can it be done with nooby 2U methods?

It can be done by 2U methods

Badsiii · Oct 20, 2013

Re: HSC 2013 2U Marathon

Differentiate:

2yln(y)
__y

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

lebomatic said:
yeh i was wondering bout that, not sure how to do 2u method. hint please

wait..... do u know that $\lim_{x \rightarrow 0} \frac{sinx}{x}=1$ ? <--have u learnt this?

Youi_ · Oct 20, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$M = 2\alpha^2$

$L^2 = (-\alpha - 2\alpha)^2 = 9\alpha^2$

$\frac{L^2}{9} = \alpha^2$

$\frac{2L^2}{9} = 2\alpha^2 = M$

idk whats wrong with me today but where did alpha come from? please explain your solution to a tired guy

RealiseNothing · Oct 20, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
idk whats wrong with me today but where did alpha come from? please explain your solution to a tired guy

It's the root of the quadratic.

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

Prove that:

$\frac{sin\theta\cos^2\theta-sin^3\theta}{2cos^3\theta-cos\theta} = tan\theta$

andybandy · Oct 20, 2013

Re: HSC 2013 2U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" title="\frac{sin\theta cos^{2}\theta -sin^{3}\theta }{2cos^{3}\theta -cos\theta } = \frac{sin\theta (cos^{2}\theta -sin^{2}\theta) }{cos\theta (2cos^{2}\theta -1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" title="= \frac{sin\theta }{cos\theta }.\frac{cos2\theta }{cos2\theta }" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;tan\theta&space;.1&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;tan\theta&space;.1&space;=&space;tan\theta" title="= tan\theta .1 = tan\theta" /></a>

Edit: Realised this is the 3 unit method, for the 2 unit method, you would change

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(cos^{2}\theta - (1 - cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(2cos^{2}\theta - 1)}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" title="= \frac{sin\theta}{cos\theta} = tan\theta" /></a>

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

^U used 3U double angles

A hint is to never write a '1' as a '1'

andybandy · Oct 20, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
^U used 3U double angles

A hint is to never write a '1' as a '1'

I know, i corrected it using the 2 unit method, and what do you mean?

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

andybandy said:
I know, i corrected it using the 2 unit method, and what do you mean?

ignore everything i said, im n00b

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

andybandy said:
I know, i corrected it using the 2 unit method, and what do you mean?

I think he means write '1' as '2-1'

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

andybandy said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" title="\frac{sin\theta cos^{2}\theta -sin^{3}\theta }{2cos^{3}\theta -cos\theta } = \frac{sin\theta (cos^{2}\theta -sin^{2}\theta) }{cos\theta (2cos^{2}\theta -1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" title="= \frac{sin\theta }{cos\theta }.\frac{cos2\theta }{cos2\theta }" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;tan\theta&space;.1&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;tan\theta&space;.1&space;=&space;tan\theta" title="= tan\theta .1 = tan\theta" /></a>

Edit: Realised this is the 3 unit method, for the 2 unit method, you would change

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(cos^{2}\theta - (1 - cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(2cos^{2}\theta - 1)}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" title="= \frac{sin\theta}{cos\theta} = tan\theta" /></a>

It's so much easier with the 3U method, wish we could use it

For this question why can't we use $y-y_1=m(x-x_1)?$

spanky125 · Oct 20, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
It's so much easier with the 3U method, wish we could use it

For this question why can't we use $y-y_1=m(x-x_1)?$

Why bother? Just integrate it then sub in the points to find the value of C

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