HSC 2015 MX1 Marathon (archive) (5 Viewers)

InteGrand · Apr 24, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Bae's theorem ?

You don't need to worry about it for HSC (pretty sure Bayes' Theorem and conditional probability aren't in the HSC syllabus).

InteGrand · Apr 24, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$e^{e^x}$

+C ^__^

braintic · Apr 24, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
this shows that humans as a species tend to have a relatively poor probabilistic sense, in that we can't necessarily make fairly accurate predictions about the possibilities of events that can occur, possibly due to the effect of past experiences which cloud our ability to make these correct judgements.

Having taught this a number of times, I can say that the reason people don't predict the answer well for this question is that they focus on themselves. That is, they imagine that THEY THEMSELVES (as part of the group) have to share a birthday with someone else in the group. They don't consider that the shared birthdays might not involve themselves.

InteGrand · Apr 24, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I checked with another tutor today Integrand, and he reckons you did everything right except for the value of 0.098

P(A|B) = (0.10 x 0.98 )/0.116

Sorry, you're right, the answer is ~0.845. I realised I subbed in a wrong value for Bayes' Theorem.

I said in my original working that "Pr(B|A) = Pr(positive result | person is non-carrier) = 0.098 (from the first part)".

But actually, the first part asked something else; it asked for the probability that a randomly selected result is false positive (without the "| person is non-carrier" part).

What I should have said was Pr(B|A) = Pr(positive result | person is non-carrier) = Pr(test is wrong) = 10% = 0.1.

If you sub. in this value along with the other values I gave into Bayes' Theorem, you get the right answer of 49/58.

Gabriel Moussa · Apr 28, 2015

Re: HSC 2015 3U Marathon

Prove that Sin^-1(x) + Cos^-1(x) = pi/2

Drsoccerball · Apr 28, 2015

Re: HSC 2015 3U Marathon

Gabriel Moussa said:
Prove that Sin^-1(x) + Cos^-1(x) = pi/2

Our teacher said he's going to make a video on these proofs but if you are too impatient :
Label sides of right angled triangle as :
$a,b,c$

$sin\theta = \frac{b}{c}$

$\therefore sin^{-1}\frac{b}{c}=\theta$

$$ Find the other angle of the triangle $ (90-\theta)$

$\therefore cos(90 - \theta)= \frac{b}{c}$
$Adding the two results of the resulting inverses:$

$sin^{-1}x +cos^{-1}x=\frac{\pi}{2} $ where $ x= \frac{b}{c}$

Kaido · Apr 28, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Our teacher said he's going to make a video on these proofs but if you are too impatient :
Label sides of right angled triangle as :
$a,b,c$

$sin\theta = \frac{b}{c}$

$\therefore sin^{-1}\frac{b}{c}=\theta$

$$ Find the other angle of the triangle $ (90-\theta)$

$\therefore cos(90 - \theta)= \frac{b}{c}$
$Adding the two results of the resulting inverses:$

$sin^{-1}x +cos^{-1}x=\frac{\pi}{2} $ where $ x= \frac{b}{c}$

Dayum, cool teacher, does he have videos online?

Drsoccerball · Apr 28, 2015

Re: HSC 2015 3U Marathon

Kaido said:
Dayum, cool teacher, does he have videos online?

He uploads them to moodle sometimes when we skip something like proofs and things that will take too long in class
We're abit behind we just started the projectile motion chapter
But for extension 2 were starting mechanics probs in like 2 days

swagswagyoloswag · Apr 28, 2015

Re: HSC 2015 3U Marathon

Gabriel Moussa said:
Prove that Sin^-1(x) + Cos^-1(x) = pi/2

Method 2:
$Consider, sin^{-1}(x) + cos^{-1}(x)$

$\frac{d}{dx} [sin^{-1}(x) + cos^{-1}(x)]$

$= \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-x^{2}}}$

$= 0$

$Integrating,$

$\int \frac{d}{dx} [sin^{-1}(x) + cos^{-1}(x)] \equiv \int 0dx$

$\therefore sin^{-1}(x) + cos^{-1}(x) \equiv C$

$since equivalency $, $ let x = 0.482$

$sin^{-1}(x) + cos^{-1}(x) \equiv \frac{\pi}{2}$

InteGrand · Apr 28, 2015

Re: HSC 2015 3U Marathon

swagswagyoloswag said:
Method 2:
$Consider, sin^{-1}(x) + cos^{-1}(x)$

$\frac{d}{dx} [sin^{-1}(x) + cos^{-1}(x)]$

$= \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-x^{2}}}$

$= 0$

$Integrating,$

$\int \frac{d}{dx} [sin^{-1}(x) + cos^{-1}(x)] \int 0dx$

$= C ($a constant$)$

$\therefore sin^{-1}(x) + cos^{-1}(x) \equiv C$

$since equivalency $, $ let x = 69$

$sin^{-1}(x) + cos^{-1}(x) \equiv \frac{\pi}{2}$

$x$ must be between -1 and 1 inclusive.

swagswagyoloswag · Apr 28, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$x$ must be between -1 and 1 inclusive.

yas oopsie

InteGrand · Apr 28, 2015

Re: HSC 2015 3U Marathon

swagswagyoloswag said:
Method 2:
$Consider, sin^{-1}(x) + cos^{-1}(x)$

$\frac{d}{dx} [sin^{-1}(x) + cos^{-1}(x)]$

$= \frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-x^{2}}}$

$= 0$

$Integrating,$

$\int \frac{d}{dx} [sin^{-1}(x) + cos^{-1}(x)] \equiv \int 0dx$

$\therefore sin^{-1}(x) + cos^{-1}(x) \equiv C$

$since equivalency $, $ let x = 0.482$

$sin^{-1}(x) + cos^{-1}(x) \equiv \frac{\pi}{2}$

It's hard to show that the constant is $\frac{\pi}{2}$ if you sub. in an arbitrary value for x, so it's best to sub. in something like $x=0$ , since we know the values of $\sin ^{-1} x$ and $\cos ^{-1} x$ for x = 0.

Drsoccerball · Apr 29, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
It's hard to show that the constant is $\frac{\pi}{2}$ if you sub. in an arbitrary value for x, so it's best to sub. in something like $x=0$ , since we know the values of $\sin ^{-1} x$ and $\cos ^{-1} x$ for x = 0.

The method i did was less of that random subbing so i think the way mine is is legit

braintic · Apr 29, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
The method i did was less of that random subbing so i think the way mine is is legit

Sorry, your method is unfinished. You have proved the result only for numbers between 0 and pi/2. You now need to prove it for the remainder of the domain.
And there have been a handful of HSC questions that have required specifically the calculus method.

davidgoes4wce · Apr 30, 2015

Re: HSC 2015 3U Marathon

The velocity of a particle is moving in SHM in a straight line v^2 = 4x-x^2, where x is displacement in metres.

c) Find the maximum speed of the particle.

The solution has 4 m/s

My thinking was to let v=0
0= 4x -x^2
0=x(4-x)
x=0,4

The centre is halfway between the points x=0 and x=4, which is x=2.

letting x=2
v^2=2*(4-2)
v^2=4
v=2m/s

This question was from the Grove HSC book

photastic · Apr 30, 2015

Re: HSC 2015 3U Marathon

Hmm I also got 2m/s but I found the value of x when a = 0.

Ambility · Apr 30, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
The velocity of a particle is moving in SHM in a straight line v^2 = 4x-x^2, where x is displacement in metres.

c) Find the maximum speed of the particle.

The solution has 4 m/s

My thinking was to let v=0
0= 4x -x^2
0=x(4-x)
x=0,4

The centre is halfway between the points x=0 and x=4, which is x=2.

letting x=2
v^2=2*(4-2)
v^2=4
v=2m/s

This question was from the Grove HSC book

I don't see any problem with that. Looking at the graph of $\sqrt{4x-x^2}$ , it does make sense that the velocity is at the maximum of $2$ at $x=2$ .

davidgoes4wce · Apr 30, 2015

Re: HSC 2015 3U Marathon

Is it a case of 'another ' Margaret Grove error?

enigma_1 · Apr 30, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Is it a case of 'another ' Margaret Grove error?

That woman has problems

enigma_1 · Apr 30, 2015

Re: HSC 2015 3U Marathon

pls don't use mif

:'(

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