Official BOS Trial 2015 Thread (1 Viewer)

kawaiipotato · Oct 3, 2015

Trebla said:
FYI I wrote that question lol

Oh sorry

thx for not putting (2-b, a) in

Trebla · Oct 3, 2015

As a trap, I would've thought some people correctly read that as PQ is represented by a+ib and then just obtain the result by multiplication of i as their answer without considering the complex number represented by P.

Ekman · Oct 3, 2015

How was the graph question in question 12?

Ekman · Oct 3, 2015

Oh and also, to let everyone's expectations of me down, im not getting above 70. I counted my losses and it's not happening.

Drsoccerball · Oct 3, 2015

Trebla said:
FYI I wrote that question lol

Im calling the child abuse hotline...
Btw you look like vanoss

Paradoxica · Oct 3, 2015

Ekman said:
How was the graph question in question 12?

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)

Drsoccerball · Oct 3, 2015

For those who don't know what vanoss looks like :

porcupinetree · Oct 3, 2015

Ekman said:
Oh and also, to let everyone's expectations of me down, im not getting above 70. I counted my losses and it's not happening.

inb4 69%

DatAtarLyfe · Oct 3, 2015

Drsoccerball said:
For those who don't know what vanoss looks like :

Are both u and ekman azns? Some reason i thought u guys were arab lol

Drsoccerball · Oct 3, 2015

porcupinetree said:
inb4 69%

Me and ekman made a bet that if either of us gets above 70 were coming to next years BOS trials and streaking.

Drsoccerball · Oct 3, 2015

DatAtarLyfe said:
Are both u and ekman azns? Some reason i thought u guys were arab lol

Wow just coz i have a beard im arab ?

Im neither asian nor am i Arab.

DatAtarLyfe · Oct 3, 2015

Only reason i thought for ekman was coz the baklava as his dp

Ekman · Oct 3, 2015

DatAtarLyfe said:
Are both u and ekman azns? Some reason i thought u guys were arab lol

Drsoccerball said:
Wow just coz i have a beard im arab ? Im neither asian nor am i Arab.

I am asian, drsoccerball can confirm this.

InteGrand · Oct 3, 2015

Paradoxica said:
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)

$\textsl{Critical point} refers to a point where the derivative is zero or undefined. So $(-1,-1)$ is a critical point because $y^\prime$ is undefined here, but $(1,1)$ is also a critical point, as $y^\prime = 0$ here.$

Drsoccerball · Oct 3, 2015

Ekman said:
I am asian, drsoccerball can confirm this.

Technically this is true... for both of us

Drsoccerball · Oct 3, 2015

InteGrand said:
$\textsl{Critical point} refers to a point where the derivative is zero or undefined. So $(-1,-1)$ is a critical point because $y^\prime$ is undefined here, but $(1,1)$ is also a critical point, as $y^\prime = 0$ here.$

YESSSSS BUDDDDYYYYY I can now get 26% rather than 25

Paradoxica · Oct 3, 2015

Hm. Well I was never taught that critical points are both of those, only one of those.

Another reason to take over the maths faculty at my school, I suppose....

InteGrand · Oct 3, 2015

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

$Clearly neither $x$ nor $y$ can be $0$, since the argument of the log cannot be $0$. Since $x$ and $y$ must have the same sign for the log to be defined, and we showed that the curve is symmetric about $y = -x$ in a question, we just can focus on sketching in the first quadrant, and then reflect this part of the graph into the third quadrant later.$

$So considering $x,y>0$ from the equation for $\mathcal{C}:x^3 - y^3 = 3\ln x + 3\ln y \Longleftrightarrow x^3 - 3\ln x = y^3 + 3\ln y,$ for $x,y>0$, we see that as $x \to 0^+$, the left-hand expression goes to $+\infty$, so the right-hand expression must as well. This clearly happens if and only if we have $y \to +\infty$. Then if we reflect the curve into the third quadrant about $y=-x$, we get an asymptote of the $x$-axis as $y\to -\infty$.$

Somehow LaTeX doesn't seem to be working now if using $'s for symbolic expressions, even though it always has before. Test: $Test: $x$.$

Ekman · Oct 3, 2015

InteGrand said:
$Clearly neither $x$ nor $y$ can be $0$, since the argument of the log cannot be $0$. Since $x$ and $y$ must have the same sign for the log to be defined, and we showed that the curve is symmetric about $y = -x$ in a question, we just can focus on sketching in the first quadrant, and then reflect this part of the graph into the third quadrant later.$

$So considering $x,y>0$ from the equation for $\mathcal{C}:x^3 - y^3 = 3\ln x + 3\ln y \Longleftrightarrow x^3 - 3\ln x = y^3 + 3\ln y,$ for $x,y>0$, we see that as $x \to 0^+$, the left-hand expression goes to $+\infty$, so the right-hand expression must as well. This clearly happens if and only if we have $y \to +\infty$. Then if we reflect the curve into the third quadrant about $y=-x$, we get an asymptote of the $x$-axis as $y\to -\infty$.$

Somehow LaTeX doesn't seem to be working now if using $'s for symbolic expressions, even though it always has before. Test: $Test: $x$.$

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

porcupinetree · Oct 3, 2015

InteGrand said:
$\textsl{Critical point} refers to a point where the derivative is zero or undefined. So $(-1,-1)$ is a critical point because $y^\prime$ is undefined here, but $(1,1)$ is also a critical point, as $y^\prime = 0$ here.$

Thank goodness

Official BOS Trial 2015 Thread (1 Viewer)

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