HSC 2015 Maths Marathon (archive) (1 Viewer)

braintic · Mar 28, 2015

Re: HSC 2015 2U Marathon

Evaluating a definite integral by instead evaluating the integral of the inverse function is really an Ext 1 concept, unless perhaps the diagram is given.

braintic · Mar 28, 2015

Re: HSC 2015 2U Marathon

Ekman said:
Also just to add, the question is a bit faulty as the displacement, velocity and acceleration is undefined at t=0. So when this particle starts moving, its initially location, velocity and acceleration is undefined... I understand that the question is asking about the integration between specific intervals, but in a mathematical and physical sense, the question seems a bit faulty.

And ..... just past t=0 it will be going faster than the speed of light !

braintic · Mar 28, 2015

Re: HSC 2015 2U Marathon

Ekman said:
Also drawing them out separately still wont help...

Actually it does. This can be written as a definite integral, and by drawing a diagram each log integral can be turned into the difference of a simple area and an area under an exponential function.

Drsoccerball · Mar 29, 2015

Re: HSC 2015 2U Marathon

braintic said:
Actually it does. This can be written as a definite integral, and by drawing a diagram each log integral can be turned into the difference of a simple area and an area under an exponential function.

This is exactly what i was looking for in this question

but the question was faulty with the speeds ect..

leehuan · Apr 17, 2015

Re: HSC 2015 2U Marathon

$Andrew and Jack are playing a game of Ludo. Andrew rolls the die first.\\To be able to take a token out of the yard and onto the board, a 6 must be rolled on an ordinary die. If a 6 is not rolled, then the next player has a turn to roll a 6. The person who rolls a 6 also gets to go again, i.e. they go first in moving their token.\\Find the probability that Andrew will get the first move.$

Zlatman · Apr 17, 2015

Re: HSC 2015 2U Marathon

leehuan said:
$Andrew and Jack are playing a game of Ludo. Andrew rolls the die first.\\To be able to take a token out of the yard and onto the board, a 6 must be rolled on an ordinary die. If a 6 is not rolled, then the next player has a turn to roll a 6. The person who rolls a 6 also gets to go again, i.e. they go first in moving their token.\\Find the probability that Andrew will get the first move.$

Still haven't learnt how to use LATEX, haha
http://i.imgur.com/cLHkVpU.jpg?1

(Haven't learnt Probability or Series in depth, so hopefully this is right)

Ambility · Apr 17, 2015

Re: HSC 2015 2U Marathon

leehuan said:
$Andrew and Jack are playing a game of Ludo. Andrew rolls the die first.\\To be able to take a token out of the yard and onto the board, a 6 must be rolled on an ordinary die. If a 6 is not rolled, then the next player has a turn to roll a 6. The person who rolls a 6 also gets to go again, i.e. they go first in moving their token.\\Find the probability that Andrew will get the first move.$

I may not be right, I just gave it my best shot. I'm actually an 11th grader who just started, and I haven't even touch on series yet. All this is done from stuff I learnt a little while ago on the web.

$\text{P(win first turn)}=\frac{1}{6}\\\text{P(win second turn)}=\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{36}\times\frac{1}{6}\\\text{P(win third turn)}=\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{36}\times\frac{25}{36}\times\frac{1}{6}\\\text{P(win fourth turn)}\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}\times\frac{25}{36}\times\frac{25}{36}\times\frac{1}{6}\\\text{This can be represented as the converging infinite geometric series: }\sum_{k=0.}^{\infty}\frac{5^{2(k-1)}}{6^{2k+1}}\\\text{The sum of this series is equal to }\frac{\text{the first term}}{1-\text{the common ratio}}\\=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}=P(\text{winning})$

Ambility · Apr 17, 2015

Re: HSC 2015 2U Marathon

Dangit, gazumped me!

Chris_S · Apr 17, 2015

Re: HSC 2015 2U Marathon

The gradient of the curve is given by dy/dx = 3x^2+2x. The curve passes through the point (2,13). What is the equation of the curve?

Zlatman · Apr 17, 2015

Re: HSC 2015 2U Marathon

dy/dx = 3x^2 + 2x
y = (3x^3)/3 + (2x^2)/2 + C
y = x^3 + x^2 + C

At the point (2, 13):
13 = 8 + 4 + C
C = 1

Therefore,
y = x^3 + x^2 + 1

-----------------------------------------------------------------------------
NEW QUESTION:

The relation defining the inside shape of the side of a wine barrel is y = 35 - (x^2)/100, where:
y represents the radius of the barrel (in cm)
x represents the distance from the mid-height to the top and bottom of the barrel (in cm)

If the height of the barrel is 80cm, find the volume of the barrel in cubic centimetres.
Hence, find the capacity of the barrel in litres.

astroman · Apr 29, 2015

Re: HSC 2015 2U Marathon

Anyone know how to approach this

Differentiate:

5(log_ex)²

InteGrand · Apr 29, 2015

Re: HSC 2015 2U Marathon

astroman said:
Anyone know how to approach this

Differentiate:

5(log_ex)²

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 5(\ln x)^2 \right )=5\times 2 \ln x \times \frac{\mathrm{d} }{\mathrm{d} x}(\ln x)$ (chain rule)

$=10\ln x \times \frac{1}{x}$

$=\frac{10\ln x}{x}.$

braintic · Apr 29, 2015

Re: HSC 2015 2U Marathon

Ambility said:
I may not be right, I just gave it my best shot. I'm actually an 11th grader who just started, and I haven't even touch on series yet. All this is done from stuff I learnt a little while ago on the web.

$\text{P(win first turn)}=\frac{1}{6}\\\text{P(win second turn)}=\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{36}\times\frac{1}{6}\\\text{P(win third turn)}=\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{36}\times\frac{25}{36}\times\frac{1}{6}\\\text{P(win fourth turn)}\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}\times\frac{25}{36}\times\frac{25}{36}\times\frac{1}{6}\\\text{This can be represented as the converging infinite geometric series: }\sum_{k=0.}^{\infty}\frac{5^{2(k-1)}}{6^{2k+1}}\\\text{The sum of this series is equal to }\frac{\text{the first term}}{1-\text{the common ratio}}\\=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}=P(\text{winning})$

You don't actually need series. Just look at a pair of rolls, one for Andrew and one for Jack. Leaving answers unsimplified:

P(Andrew wins) = 6/36
P(Andrew loses) = 5/6 times 1/6 = 5/36

No matter how many rolls are required, you will eventually come to a pair of rolls where the result is decided. Ignore all the inconsequential preceding rolls.

P(Andrew wins) = 6/(6+5) = 6/11

Ambility · Apr 30, 2015

Re: HSC 2015 2U Marathon

braintic said:
You don't actually need series. Just look at a pair of rolls, one for Andrew and one for Jack. Leaving answers unsimplified:

P(Andrew wins) = 6/36
P(Andrew loses) = 5/6 times 1/6 = 5/36

No matter how many rolls are required, you will eventually come to a pair of rolls where the result is decided. Ignore all the inconsequential preceding rolls.

P(Andrew wins) = 6/(6+5) = 6/11

Huh. Interesting.

Chris_S · May 2, 2015

Re: HSC 2015 2U Marathon

Find the exact area enclosed between the curve y=e^2x and the line y=1 and x=2

Kaido · May 2, 2015

Re: HSC 2015 2U Marathon

lol, that mif Q

Chris_S · May 2, 2015

Re: HSC 2015 2U Marathon

yeah lol unfortunately my school uses it! I prefer fitzpatrick

Drsoccerball · May 2, 2015

Re: HSC 2015 2U Marathon

Chris_S said:
Find the exact area enclosed between the curve y=e^2x and the line y=1 and x=2

First sketch the graph
Since its that little triangle thingy:
$\int_0^2{e^{2x}}dx - 2 $ (area of rectangle)$$

braintic · May 5, 2015

Re: HSC 2015 2U Marathon

A trapezium is bounded by a focal chord (don't assume it is the latus rectum) of the parabola x²=4ay, vertical lines through the endpoints of the chord, and the directrix of the parabola.
Show that the area of the trapezium is given by LH/2, where L is the length of the chord, and H is the length of the side opposite the chord.

leehuan · May 10, 2015

Re: HSC 2015 2U Marathon

$PS=PV\quad (definition)\Rightarrow PS=PT+TV\\ QS=QU\\ Adding:\quad PS+QS=PQ\quad =\quad (PT+TV)+QU\\ i.e.\quad L=PV+QU\\ UV=H\quad is\quad given\\ So\quad the\quad area\quad of\quad trapezium\quad PQUV\quad is\\ A=\frac { h(a+b) }{ 2 } =\frac { UV(PV+QU) }{ 2 } =\frac { LH }{ 2 }$

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