Official BOS Trial 2015 Thread (2 Viewers)

[InteGrand]

On Wolfram, the graph looks like it intercepts the y-axis and x-axis. I did what you did, but wolfram graphed something weird.

It's mainly due to difficulty of computing values that satisfy that equation near x = 0. Of course the curve shouldn't actually cut or touch the axes anywhere, because we said x and y can't be 0.

 

[Paradoxica]

It's mainly due to difficulty of computing values that satisfy that equation near x = 0. Of course the curve shouldn't actually cut or touch the axes anywhere, because we said x and y can't be 0.

You can't even compute values algebraically.... For that, you would need the Product Log function, and that is beyond high school...

 

[kawaiipotato]

Who wrote that graph question?

 

[Trebla]

For the graphs question, you're meant to work out the behaviour of the graph as x approaches 0 in a similar way to how you should've worked out the behaviour as x approaches infinity, which is by separating out the x and y in the equation.

Once you've done that use the fact that it has a stationary point at (1,1), it is symmetric about y=-x and the fact there is an asymptote of y=x to work out the rest of the graph.

 

[Trebla]

Who wrote that graph question?

Carrot wrote that question

 

[Ekman]

Carrot wrote that question

When can we expect the results?

 

[InteGrand]

You can't even compute values algebraically.... For that, you would need the Product Log function, and that is beyond high school...

I meant it's probably hard for the computer to obtain numerical solutions (by this I mean approximate solutions of course) when x is very small, or they are too erratic around there. Lambert W function wouldn't help us too much anyway as we would then need to approximate that to obtain numerical values.

 

[Mr_Kap]

When can we expect the results?

On Wednesday, 16 December 2015

 

[Drsoccerball]

On Wednesday, 16 December 2015

Speaking about that... im going to be in Africa then... Would i be able to get my results from somewhere other than inside Australia? (For HSC obviously...)

 

[leehuan]

Speaking about that... im going to be in Africa then... Would i be able to get my results from somewhere other than inside Australia? (For HSC obviously...)

HSC results are released on studentsonline

 

[Paradoxica]

It's awfully nice of you people (Carrotstick, Trebla, and other people) to buy the food, print out all those papers, and hold this entire event.
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Emphasis on awful
(jks we all love you carrotsticks and trebla)

 

[leehuan]

It's awfully nice of you people (Carrotstick, Trebla, and other people) to buy the food, print out all those papers, and hold this entire event.
.
.
.
.
Emphasis on awful
(jks we all love you carrotsticks and trebla)

Wow. The damage is already done.

 

[Drsoccerball]

For the ex 1 paper 11e im pretty sure dummy variable are ex 2?

 

[porcupinetree]

For the ex 1 paper 11e im pretty sure dummy variable are ex 2?

Also, I think the whole idea of finding an 'equation' for I (where I is the integral) is a mainly 4U concept

 

[Drsoccerball]

Also, I think the whole idea of finding an 'equation' for I (where I is the integral) is a mainly 4U concept

Straight up it was too long so i skipped it... I was really not bothered during the ex 1 exam...I circled random answers for mc after reading the first one...Anyways... why was question 12 and 14 impossible ?

 

[Carrotsticks]

Straight up it was too long so i skipped it... I was really not bothered during the ex 1 exam...I circled random answers for mc after reading the first one...Anyways... why was question 12 and 14 impossible ?

Is this Extension 1 or 2?

 

[Carrotsticks]

For the ex 1 paper 11e im pretty sure dummy variable are ex 2?

Also, I think the whole idea of finding an 'equation' for I (where I is the integral) is a mainly 4U concept

It is toe-ing the syllabus line, but I think it is very well within the bounds of Extension 1 knowledge-wise, as no 'extra theory' is required to do the problem.

It is simply a more difficult deduction that needs to be made.

 

[Carrotsticks]

When can we expect the results?

A bit more than a week from now.

How were we supposed to figure out there are asymptotes on the axes?

As well as that, there is no critical point at (1,1). Wolfram Alpha says so. The critical point is (-1,-1)

When x approaches negative infinity, y can approach zero. Since the curve is symmetric in y=-x, we can immediately deduce that the y axis is an asymptote.

 

[Carrotsticks]

FYI I wrote that question lol

To add to this, Trebla writes most of the MC usually. I modify and replace some of them only for whatever reason I see fit.

ie: Q2 and 4 of MX2 MC was mine. The rest was pretty much Trebla's.

 

[pororo]

does anyone know how to do question 15cii and iii?