HSC 2012 MX2 Marathon (archive) (1 Viewer)

seanieg89 · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
$P(x)=\sum_{r=1}^{2n}\frac{x^{r}}{r!}\\ P'(x)=\sum_{r=1}^{2n-1}\frac{x^{r}}{r!}\\ P''(x)=\sum_{r=1}^{2n-2}\frac{x^{r}}{r!}\\ We~will~now~show~by~induction~that~P(x)~has~no~real~roots.\\ Let~n=1~to~show,\\ P(x)=\frac{x^2}{2}+x+1\\ P'(x)=x+1=0,~x=-1\\ P(-1)=\frac{1}{2}.~\therefore ~for~n=1,~P(x)~has~no~real~roots\\ P''(x)=1.~\therefore ~the~function~is~concave~up.\\ Assume~that~P(x)~has~no~real~roots~for~n=k.\\ i.e.~\sum_{r=1}^{2k}\frac{x^{r}}{r!}> 0\\ To~prove~true~for~n=k+1\\ TBP:~P(x)=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}> 0\\ LHS=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}\\ =\sum_{r=1}^{2k}\frac{x^{r}}{r!}+\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{2k+2}}{(2k+2)!}\\ taking~the~second~derivative\\ P''(x)=\sum_{r=1}^{2k}\frac{x^{r}}{r!}>0.\\ this~isnt~correct~cos~it~isnt~showing~that~the~min~stat~point~is~>~0~but~i~dunno~where~to~go.$

Very good! Definitely on the right track using concavity. Keep working using this approach, you are actually very close to done.

sophisms · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

i am so going to fail maths ext 2 after seeing this thread. I don't even get what's going on :/

asianese · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

Dw sophisms I thought the exact same thing...but you'll get used to it! Promise!

barbernator · Jul 16, 2012

Re: 2012 HSC MX2 Marathon

still got no idea on how to finish it off lol

seanieg89 · Jul 16, 2012

Re: 2012 HSC MX2 Marathon

$We aim to prove that $p_n(x)$ is positive everywhere for each non-negative integer $n$. (The subscript simply tells us what we are taking n to be).\\So the argument can be structured as follows:\\i) $p_0(x)=1>0$ everywhere.\\ii) Observe that any real poly of odd degree with positive leading coefficient tends to $\pm\infty$ as $x \rightarrow \pm \infty$.\\iii) Assume $p_k>0$ everywhere for some non-negative integer $k$. From ii) and the fact that $p_{k+1}''=p_k>0$, we can show that $p_{k+1}$ has a unique stationary point and that this stationary point must be the minimum of p on the entire real line. (It is a little bit of work to prove this, but the fact is relatively obvious from graphing such a function).\\iv) Now if $\alpha$ is this stationary point, then $p_{k+1}'(\alpha)=0$. So $p_{k+1}(\alpha)=p_{k+1}(\alpha)-p_{k+1}'(\alpha)=\alpha^{2k}/(2k)!>0$. This completes the induction.$

sophisms · Jul 16, 2012

Re: 2012 HSC MX2 Marathon

asianese said:
Dw sophisms I thought the exact same thing...but you'll get used to it! Promise!

don't think i'll ever get use to it :L My maths must be really bad :'(

Godmode · Jul 18, 2012

Re: 2012 HSC MX2 Marathon

Alright! Time for a (possibly ridiculous) Probability question.

Jack and Michael are playing a game of blackjack, using 8 standard decks.

i. How many two card selections are there:
a) in total?
b) that will have blackjack?
ii. What is the probability of:
a) only one player being dealt blackjack?
b) both players being dealt blackjack?
iii. What is the probability of being dealt blackjack, where one card is a King?
iv. If Both Jack and Michael are dealt blackjack, what is the probability (out of all possible 2 card selections for both players, [blackjack or not] dealt to both players) that Jack will beat Michael (where 10 < Jack < Queen < King and let's say Spades = 4 points > Clubs = 3 > Diamonds = 2 > Hearts = 1, and the suit of both cards count, and most points wins)

For those who have never played Blackjack, (My teacher said he once had a Yr 12 class where barely anyone had even seen a deck of cards lol) Having Blackjack is when the sum of both card's numbers is 21 [Ace = 1 or 11, 10 = Jack = Queen = King]

Note: I have no idea how to do this question, I just made it up on the spot. Additionally, probability is not my forte, so don't rely on me to come up with the answers any time soon lol.

For part i. , I just mean a single hand, but for part iv. , I mean for both hands at the same time ( it is a bit confusing lol).

Carrotsticks · Jul 18, 2012

Re: 2012 HSC MX2 Marathon

This question is possibly more complex than you thought. My reasons:

1. In a game of blackjack, you also have to count the fact that cards are given to another player at the same time (so there are fewer cards in the deck).

2. Remember that you can 'hit' in Blackjack, and we also have to determine the probability of one of them 'busting out', thus guaranteeing a win for the other player assuming they haven't already Blackjacked.

Godmode · Jul 18, 2012

Re: 2012 HSC MX2 Marathon

Just to make it slightly easier, reduce the deck number to 1, and we are only considering it to be a win if both players are below 22. (im not sure if this makes it easier or harder)

EDIT: And yes, this is becoming worse and worse as I think about it.

What's the worst we can get in MX2 probability?

lolcakes52 · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

$$find $ \int \frac{\sin{x}-\cos{x}}{2+\sin{2x}} dx$
I made this one up myself.... I hope its alright....

Aesytic · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

integral[(sinx-cosx)/(2+2sin2x)] = integral[(sinx-cosx)/(1+2sinxcosx+sin^2x+cos^2x)]
= integral[(sinx-cosx)/(1+(sinx+cosx)^2)]
let u=sinx + cosx
du = cosx - sinx dx
.'. integral[(sinx-cosx)/(1+(sinx+cosx)^2)] = -integral[1/(1+u^2)]
= -arctan(u)
= -arctan(sinx + cosx) + C

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

lolcakes52 said:
$$find $ \int \frac{\sin{x}-\cos{x}}{2+\sin{2x}} dx$
I made this one up myself.... I hope its alright....

$\int \frac{sinx - cosx}{2 + 2sinxcosx}dx$
$\frac{1}{2}\int \frac{sinx - cosx}{1 + sinxcosx}dx$

Notice that this is the expansion of $tan(\alpha - \beta)$

$\int tan[tan^{-1}(sinx) - tan^{-1}(cosx)]dx$

Curious whether this could take you somewhere?

Carrotsticks · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

RealiseNothing said:
$\int \frac{sinx - cosx}{2 + 2sinxcosx}dx$
$\frac{1}{2}\int \frac{sinx - cosx}{1 + sinxcosx}dx$

Notice that this is the expansion of $tan(\alpha - \beta)$

$\int tan[tan^{-1}(sinx) - tan^{-1}(cosx)]dx$

Curious whether this could take you somewhere?

I guess you could try using triangles to simplify the expression.

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Show that for x > 0:

$\cos x < 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$

Carrotsticks · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Delicious Taylor Polynomials.

bleakarcher · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
Show that for x > 0:

$\cos x < 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$

Let f(x)=cos(x)-1+(x^2/2)-(x^4/24)
When x=0, f(x)=f(0)=0
f '(x)=-sin(x)+x-(x^3/6)
f ''(x)=-cos(x)+1-(x^2/2)
f '''(x)=sin(x)-x
Note that y=x is the tangent to y=sin(x) at x=0 meaning that x>sin(x) <=> sin(x)-x<0 for all x>0.
In order to prove the statement for x>0, which is equivalent to the statement "Prove f(x)<0 for x>0", we must consider the fact that if f(x)<0 for x>0 then f(0)<=0 and f '(x)<0 for all x>0. Since f ''(0)=0 and f '''(x)<0 for all x>0 it follows by the theorem stated above that f ''(x)<0 for all x>0. Since f '(0)=0 and f ''(x)<0 for all x>0 it follows again by the theorem that f '(x)<0 for all x>0. Applying the theorem once more we deduce that f(x)<0 for all x>0 i.e. cos(x)<1-(x^2/2)+(x^4/24) for x>0

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Can you do it this way?

$cosx < 1 - \frac{x^2}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{4!}$

Since substituting in x=0 leaves both sides equal, we can say that $f'(x) < g'(x)$ for x>0.

So we end up with by differentiating:

$-sinx < -x + \frac{x^3}{3!}$

Therefore by multiplying both sides by -1:

$sinx > x - \frac{x^3}{3!}$

Sub in x=0 and both sides are equal, so we can do the same again:

$cosx > 1 - \frac{x^2}{2!}$

Sub in x=0 and both sides are equal so we differentiate again:

$-sinx > -x$

Therefore by once again multiplying both sides by -1:

$sinx < x$

Sub x=0 and both sides are equal, so differentiating again gives:

$cosx < 1$

Which is true for all x>0, and hence we conclude that:

$cosx < 1 - \frac{x^2}{2} + \frac{x^4}{24}$

Alternatively by using taylor series though, you would only have to prove that:

$-\frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \frac{x^{12}}{12!} - ..... < 0$

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

You cannot differentiate both sides of an inequality and expect know for sure what the inequality sign is
e.g. x² + 1 > x²
If you differentiate both sides you get: 2x > 2x which is clearly false

Though you are on the right track...just not in the right direction

bleakarcher · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Do you have another good question Trebla?

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher's solution is good but wasn't what I had in mind which is probably a simpler approach conceptually but requires a little messier working.

We know that for x > 0 and for some a > 0

$\cos x \leq 1\\\\\Rightarrow \displaystyle\int_0^a \cos x\,dx \leq \displaystyle\int_0^a dx\\\\\Rightarrow \sin a \leq a\\\\\Rightarrow \sin x < x\quad ($since $a$ can be considered a dummy variable and equality doesn't hold for $x > 0$)\\\\\Rightarrow \displaystyle\int_0^a \sin x\,dx < \displaystyle\int_0^a x\,dx\\\\\Rightarrow -\cos a + 1 < \dfrac{a^2}{2}\\\\\Rightarrow -\cos x + 1 < \dfrac{x^2}{2}\\\\\Rightarrow \displaystyle\int_0^a (-\cos x + 1)\,dx < \displaystyle\int_0^a \dfrac{x^2}{2}\,dx\\\\\Rightarrow -\sin a + a < \dfrac{a^3}{6}\\\\\Rightarrow -\sin x + x < \dfrac{x^3}{6}\\\\\Rightarrow \displaystyle\int_0^a (-\sin x + x)\,dx < \displaystyle\int_0^a \dfrac{x^3}{6}\,dx\\\\\Rightarrow \cos a - 1 + \dfrac{a^2}{2} < \dfrac{a^4}{24}\\\\\Rightarrow \cos x - 1 + \dfrac{x^2}{2} < \dfrac{x^4}{24}\\\\\therefore \cos x < 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$

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