HSC 2013 Maths Marathon (archive) (4 Viewers)

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

bree-ga said:
how do you do this? i can't seem to get it :x

$The shortest distance between 2 lines is when the gradients are parallel.$

$y = x^2 +3x +5 \ \fbox{1}$

$y = 3x - 1 \ \fbox{2}$

$m_2 = 3$

$Find the tangent of$ \ \fbox{1} \ $at the point where the gradient is 3:$

$y'_1 = 2x + 3 = 3$

$\therefore \ x = 0, \ $and after substituting back into$ \ \fbox{1} \ y = 5$

$$This point,$ \ (0, \ 5) \ $is closest to the the line$ \ y = 3x - 1.$

$$Use the perpendicular distance formula for$ \ (0, \ 5) \ $and$ \ \fbox{2}$

$\therefore \ $The shortest distance between$ \ \fbox{1} \ $and$ \ \fbox{2} \ $is$ \ \frac{6}{\sqrt{10}}$

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:

$x^3 + 3x^2$

andybandy · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

bree-ga said:
hi ty, but I got my point (0,5) and then use the perp distance formula using y=3x -1 (2) like you said (therefore, a=0, b=3, c=-1) but I don't get 6/sqrt(10)?

You have to adjust y = 3x - 1 into the form 3x - y - 1 = 0, and then use the formula. This way, a = 3, b = -1 and c = -1.

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

andybandy said:
<a href="http://www.codecogs.com/eqnedit.php?latex=y&space;=&space;x^{3}+3x^{2}+3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y&space;=&space;x^{3}+3x^{2}+3" title="y = x^{3}+3x^{2}+3" /></a>

This is the correct answer.

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

andybandy said:
<a href="http://www.codecogs.com/eqnedit.php?latex=y&space;=&space;x^{3}+3x^{2}+3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y&space;=&space;x^{3}+3x^{2}+3" title="y = x^{3}+3x^{2}+3" /></a>

Oh yeah I forgot +c

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

bree-ga said:
ahhh, thank you! so for the values of a,b,c i have to get it from equation with form ax+by+c=0 not ax^2+bx+c ?

Yep that's correct.

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$Prove the following$

$$i)$ \ \ \sin^2 x + \cos^2 x = 1$

$$ii)$ \ \ \log_a b = \frac{\log_c b}{\log_c a}$

$iii) A triangle constructed in a circle with one side$

$as a diameter is a right angled triangle$

$$ii)$ \ \ Let x=log_a b$

$a^x=b$

$Take log c of both sides$

$xlog_c a = log_c b$

$x= \frac{\log_c b}{\log_c a}$

$$Remember that$ x=log_a b$

coyney · Oct 21, 2013

Re: HSC 2013 2U Marathon

Is it just me or is the 2009 hsc paper a lot harder than the rest?

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Can someone give me their answer for iv)b), I think the answer provided is wrong
(My answers 18)

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Can someone give me their answer for iv)b), I think the answer provided is wrong
(My answers 18)

I got 72

Randox · Oct 21, 2013

Re: HSC 2013 2U Marathon

coyney said:
Is it just me or is the 2009 hsc paper a lot harder than the rest?

2003 and 09 are hard and 2010

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
I got 72

Don't think that can be correct

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Don't think that can be correct

Neither do I

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Don't think that can be correct

I got 18 as well.

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sweet, thanks.

Drewk · Oct 21, 2013

Re: HSC 2013 2U Marathon

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 ??

Triage · 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 ??

7^x = e^(xln7)

mahoo_mk · 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 ??

not a long process at all,

$\int 7^x \ dx = \int e^{x \ln 7} \ dx = \frac{e^{x \ln 7}}{ln 7} +c = \frac{7^x}{ln7} + c$

Drewk · Oct 21, 2013

Re: HSC 2013 2U Marathon

why do you use "7^x = e^(xln7)" ??

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