Questions from 2007 and 2008 Paper (1 Viewer)

Official · Feb 22, 2009

http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/pdf_doc/maths-07.pdf
2007 paper

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2008exams/pdf_doc/2008HSC-mathematics.pdf
2008 paper

Here's the link to the 2 papers; can anyone please be kind enough to do these questions for me (all worked):

2007: Question - 10b
2008: Question - 9c and 10b

Thanks millions

Timothy.Siu · Feb 22, 2009

lol i might do them later, its late now...

but for 10b (2008) just use the fact they're similar triangles and u shud be fine =)

cutemouse · Feb 23, 2009

lol, can't be bothered downloading that PDF. Could you post the question?

vds700 · Feb 23, 2009

not sure if its right, but anyways

Juliaan · Feb 23, 2009

and your doing your HSC in 2011; dayumm.

vds700 · Feb 23, 2009

Juliaan said:
and your doing your HSC in 2011; dayumm.

maybe hes accelerated

Official · Feb 23, 2009

Juliaan said:
and your doing your HSC in 2011; dayumm.

Ah, only for 2U maths tho lol

Juliaan · Feb 23, 2009

When i was in year 10, i didn't even know what the HSC was ~

Official · Feb 23, 2009

fair nuff =p, HSC doesn't really come into most pplz minds anyway until yr 11 +

Official · Feb 23, 2009

vds700 said:
not sure if its right, but anyways

thank you kind sir

edit: Oops the answer seems to be wrong grr; actual answer: 11kb^4 / 12

vds700 · Feb 23, 2009

Official said:
thank you kind sir

edit: Oops the answer seems to be wrong grr; actual answer: 11kb^4 / 12

ah damn, dunno what i did wrong. Someone else might be able to solve it

Timothy.Siu · Feb 23, 2009

i got what vds got...i'm not sure what i put in exam though..

Trebla · Feb 23, 2009

At vds700's file, when you subbed in x = b, you forgot about the k.

$f'(x) = k\left(b^2x-\dfrac{x^3}{3}\right)\\f(x) = \dfrac{kb^2x^2}{2}-\dfrac{kx^4}{12}+c\\f(b)=0\\\Rightarrow \dfrac{kb^4}{2}-\dfrac{kb^4}{12}+c=0\\c = - \dfrac{5kb^4}{12}$ which is f(0) $\\$Therefore the distance below the beam from the x-axis at x = 0 is $\dfrac{5kb^4}{12}$

Though it still doesn't agree with the supposed "actual" answer...lol

Timothy.Siu · Feb 23, 2009

Trebla said:
When you subbed in x = b, you forgot about the k.

$f'(x) = k\left(b^2x-\dfrac{x^3}{3}\right)\\f(x) = \dfrac{kb^2x^2}{2}-\dfrac{kx^4}{12}+c\\f(b)=0\\\Rightarrow \dfrac{kb^4}{2}-\dfrac{kb^4}{12}+c=0\\c = - \dfrac{5kb^4}{12}$ which is f(0) $\\$Therefore the distance below the beam from the x-axis is $\dfrac{5kb^4}{12}$

Though it still doesn't agree with the "actual" answer...lol

lol i actually got what you got =P

jpmeijer · Feb 23, 2009

for 2007 Q10b

part (i)

at P,
$N_{1}= \frac{L_{1}}{x^{2}}, N_{2}= \frac{L_{2}}{(m-x)^{2}}$
$N = N_{1} + N_{2}$
$N = \frac{L_{1}}{x^{2}} + \frac{L_{2}}{(m-x)^{2}}$

part (ii)

now we want to minimise the noise level N, so we differentiate (with respect to x)
remember when you are doing this that L₁, L₂ and m are all constants, so you treat them just like you would any number like 2, 5 or 6

$\frac{dN}{dx} = \frac{-2L_{1}}{x^{3}} + \frac{2L_{2}}{(m-x)^{3}}$
$\frac{dN}{dx} = \frac{-2L_{1}(m-x)^{3} + 2L_{2}x^{3}}{x^{3} + (m-x)^{3}}$

now we find any stationary points:

let dN/dx = 0

$\Rightarrow 2L_{2}x^{3} = 2L_{1}(m-x)^{3}$
$L_{2}x^{3} = L_{1}(m-x)^{3}$

now we take the cube root of both sides:
$\sqrt[3]{L_{2}}x = \sqrt[3]{L_{1}}(m-x)$

and now expand the RHS and make x the subject, giving

$x = \frac{\sqrt[3]{L_{1}}m}{\sqrt[3]{L_{2}}+\sqrt[3]{L_{1}}}$

Don't know if this is enough...
You might need to keep going and show that this is in fact a minimum... ie find the 2nd derivative, sub this answer in and show that it is > 0
Anyone else know?

PS: Many apologies if there are any mistakes in there... this is my 1st time using LaTeX

lyounamu · Feb 23, 2009

Official said:
http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/pdf_doc/maths-07.pdf
2007 paper

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2008exams/pdf_doc/2008HSC-mathematics.pdf
2008 paper

Here's the link to the 2 papers; can anyone please be kind enough to do these questions for me (all worked):

2007: Question - 10b
2008: Question - 9c and 10b

Thanks millions

Lol I remember these questions.

Good old times.....

Well done to those who answered the questions.

lyounamu · Feb 23, 2009

Uh...are you sure, Official?

I think I got what Trebla got...

(I don't remember exactly though)

vds700 · Feb 23, 2009

Trebla said:
At vds700's file, when you subbed in x = b, you forgot about the k.

$f'(x) = k\left(b^2x-\dfrac{x^3}{3}\right)\\f(x) = \dfrac{kb^2x^2}{2}-\dfrac{kx^4}{12}+c\\f(b)=0\\\Rightarrow \dfrac{kb^4}{2}-\dfrac{kb^4}{12}+c=0\\c = - \dfrac{5kb^4}{12}$ which is f(0) $\\$Therefore the distance below the beam from the x-axis at x = 0 is $\dfrac{5kb^4}{12}$

Though it still doesn't agree with the supposed "actual" answer...lol

oops thats what i did wrong, sorry

Official · Feb 23, 2009

Umm, I'm not sure but I got the answers from this site:

http://www.pasthsc.com.au/Downloads/Mat2U/Maths2U08QuickAnswers.pdf

lyounamu · Feb 23, 2009

Official said:
Umm, I'm not sure but I got the answers from this site:

http://www.pasthsc.com.au/Downloads/Mat2U/Maths2U08QuickAnswers.pdf

Hey, I am sure the answer is the one posted up by Trebla. I personally did that question during my HSC and I got that question right.

Questions from 2007 and 2008 Paper (1 Viewer)

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