HSC 2013 Maths Marathon (archive) (4 Viewers)

anthony_wheels · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$Prove that the volume of a sphere with radius$ \ \ r \ \ $is$ \ \ \frac{4\pi r^3}{3}$

asianese · Oct 21, 2013

Carrotsticks said:
Remember the time you actually tried to expand it?

I actually remember this clearly. It was gr8

ColdLipstick · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$g'(x) = f'(x) (\ln f(x) + 1)$

$\therefore \ \ g'(a) = f'(a) (\ln f(a) + 1) \ \ \fbox{1}$

$From the original equation$

$g(a) = f(a) \ln f(a)$

$g(a) + f(a) = f(a) (\ln f(a) +1)$

$$therefore, due to$ \ \ \fbox{1} \ \ \frac{g'(a)}{f'(a)} = \ln f(a) + 1 \ \ $if$ \ \ f'(a) \neq 0$

$\therefore \ \ g(a) + f(a) = f(a) \times \frac{g'(a)}{f'(a)}$

$g'(a) = 0$

$\therefore \ \ f(a) + g(a) = 0 \ \ $if$ \ \ f'(a) \neq 0$

I don't think the result holds if f'(a) = 0 though, an example is

$y= (x^2+1) \ln (x^2+1)$

Umm may i ask?

Did you justify g(a) is a minimum value, not maximum?

Correct me if I misinterpreted you.

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

Find the equation of the tangent to the curve y=pi * tan (y/x) + x - 4 at P(4,pi)

I stole this question from cambridge, so I'm not sure if that's against protocol but I enjoyed it.

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
Find the equation of the tangent to the curve y=pi * tan (y/x) + x - 4 at P(4,pi)

I stole this question from cambridge, so I'm not sure if that's against protocol but I enjoyed it.

That question looks gross...

I have a tasty one:
Marco is a mob boss who needs $50000 to purchase firearms imported by an organisation which does not need to be named. The manager of a bank has agreed to an interest rate of 36% compounded monthly, the way Marco got the manager of the bank to agree to this large rate is not important, but what is important is that Marco invests $15000. How many months is it until Marco's investment has grown enough to purchase the firearms?

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
That question looks gross...

I have a tasty one:
Marco is a mob boss who needs $50000 to purchase firearms imported by an organisation which does not need to be named. The manager of a bank has agreed to an interest rate of 36% compounded monthly, the way Marco got the manager of the bank to agree to this large rate is not important, but what is important is that Marco invests $15000. How many months is it until Marco's investment has grown enough to purchase the firearms?

I'm not sure If I'm reading this question right, it seem like a really easy compounding question?

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
I'm not sure If I'm reading this question right, it seem like a really easy compounding question?

Yes, just use a log to solve for "n"

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Triage said:
I'm not sure If I'm reading this question right, it seem like a really easy compounding question?

Or guess and check, if you're into that.

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:

$QK = 6 - x$

$KP = 6 + x$

$$Since the sheet of paper has been folded over from P (now L),$ \ KP = KL$

$\therefore \ KL = 6+x$

$$By Pythagoras,$ \ (6 + x)^2 - (6 - x)^2 = QL^2$

$QL^2 = 36 + 12x +x^2 - 36 + 12x - x^2$

$\therefore \ QL^2 = 24x$

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$QK = 6 - x$

$KL = 6+x$

$$By Pythagoras,$ \ (6 + x)^2 - (6 - x)^2 = QL^2$

$QL^2 = 36 + 12x +x^2 - 36 + 12x - x^2$

$\therefore \ QL^2 = 24x$

Part 1/2i) Show KL = 6+x

Triage · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Part 1/2i) Show KL = 6+x

Isn't it given. The page is folded along that line ?

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Part 1/2i) Show KL = 6+x

If you 'unfold' the sheet from L, P will replace L. From there it is basically given that KL (or KP) is 6 + x

andybandy · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$QK = 6 - x$

$KL = 6+x$

$$By Pythagoras,$ \ (6 + x)^2 - (6 - x)^2 = QL^2$

$QL^2 = 36 + 12x +x^2 - 36 + 12x - x^2$

$\therefore \ QL^2 = 24x$

How did you get KL as 6 + x ?

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

andybandy said:
How did you get KL as 6 + x ?

KL = KP because its just been folded over. I'll adjust my solution to make it more clear.

andybandy · Oct 21, 2013

Re: HSC 2013 2U Marathon

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

It helps when you graph it.

andybandy · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
KL = KP because its just been folded over. I'll adjust my solution to make it more clear.

Ah i see! thankyou

spanky125 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sy123 · Oct 21, 2013

Re: HSC 2013 2U Marathon

ColdLipstick said:
Umm may i ask?

Did you justify g(a) is a minimum value, not maximum?

Correct me if I misinterpreted you.

Would you need to?
The proof only uses that g'(a) = 0, which is a sufficient to prove the result

mahoo_mk · Oct 21, 2013

Re: HSC 2013 2U Marathon

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

gradient of the line is 3
find the point on the curve where the tangent has the same gradient, (3) making it parallel
once you have that point use perpendicular distance formula

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