HSC 2015 MX2 Marathon (archive) (12 Viewers)

FrankXie · Feb 3, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$$ Find the equation(s) of the tangent(s) drawn from the origin to the curve $ y=3-\frac{1}{x^2}.$

VBN2470 · Feb 3, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ Find the equation(s) of the tangent(s) drawn from the origin to the curve $ y=3-\frac{1}{x^2}.$

$$ y=\pm{2x}$$

Ekman · Feb 3, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ Find the equation(s) of the tangent(s) drawn from the origin to the curve $ y=3-\frac{1}{x^2}.$

$y=mx $ is the equation of a line that is tangential to the curve$$

$\therefore \frac{dy}{dx} = m $ for the equation: $ y=mx$

$\frac{dy}{dx}=\frac{2}{x^3} $ for the given equation: $ y=3-\frac{1}{x^2}$

$\\$(Equate the two equations since they must have the same gradient at a tangent): $ m=\frac{2}{x^3}$

$\therefore x^3=\frac{2}{m}$

$\therefore y=\frac{2}{x^2} $ (however, this is a function, and not a tangent) $$

$\frac{2}{x^2} = 3-\frac{1}{x^2}$

$x=\pm1 $ and $ y=\pm2$

$m=\frac{2}{(\pm1)^3}=\pm2$

$\\y=\pm2x $ is always a tangent of the given equation and passes through 0,0$$

seanieg89 · Feb 3, 2015

Re: HSC 2015 4U Marathon

Player A is serving to Player B in a tennis match.

He wins each point with probability 0 < p < 1.

Given that both players are equally likely to win a service game if the score is 0-30, find:

a) the value of p

b) the probability that A wins a service game from 0-0.

braintic · Feb 3, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Player A is serving to Player B in a tennis match.

He wins each point with probability 0 < p < 1.

Given that both players are equally likely to win a service game if the score is 0-30, find:

a) the value of p

b) the probability that A wins a service game from 0-0.

Does this question allow for the possibility of multiple deuces?

seanieg89 · Feb 3, 2015

Re: HSC 2015 4U Marathon

braintic said:
Does this question allow for the possibility of multiple deuces?

Yep, that's the fun part

.

Ekman · Feb 3, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Yep, that's the fun part .

Oh geez, now im scared to do Ext 2 probability now

braintic · Feb 3, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Yep, that's the fun part .

Hm ... I got a massive equation which I had to type into Wolfram Equation Solver.
It gave an approximate answer of p=0.660566
Do I need to check my equation?

InteGrand · Feb 3, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Player A is serving to Player B in a tennis match.

He wins each point with probability 0 < p < 1.

Given that both players are equally likely to win a service game if the score is 0-30, find:

a) the value of p

b) the probability that A wins a service game from 0-0.

Let $q\equiv 1-p$ (this is the probability of player B winning a given point).

a) Let D_A be the probability that A wins the game given that the game has just reached deuce.

Then $D_A = \Pr(\text{A wins first two points of deuce}) + \Pr(\text{A wins exactly one of the first two points of deuce})\cdot D_A )$

i.e. $D_A = p^2 + 2pq\cdot D_A \Rightarrow D_A = \frac{p^2}{1-2pq}$ .

Now, the probability that A wins the service game from 0-30, P_A, is given by

$P_A=\Pr(\text{A wins next 4 points})+\Pr(\text{A wins 3 of the next 4 points (i.e. goes to a deuce))}) \cdot D_A$

$\Rightarrow P_A=p^4+4qp^3\cdot \frac{p^2}{1-2pq}$

Since B is equally likely to win from 0-30, P_A must equal ½.

$\Rightarrow \frac{1}{2}=p^4+4qp^3\cdot \frac{p^2}{1-2pq}$

From here, I'm getting a sixth-degree polynomial in p which I'm not seeing how to solve. Maybe a different approach needed?

braintic · Feb 3, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$\Rightarrow \frac{1}{2}=p^4+4qp^3\cdot \frac{p^2}{1-2pq}$

From here, I'm getting a sixth-degree polynomial in p which I'm not seeing how to solve. Maybe a different approach needed?

Different equation to mine, but Wolfgram gave the same answer.

FrankXie · Feb 3, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Let $q\equiv 1-p$ (this is the probability of player B winning a given point).

a) Let D_A be the probability that A wins the game given that the game has just reached deuce.

Then $D_A = \Pr(\text{A wins first two points of deuce}) + \Pr(\text{A wins exactly one of the first two points of deuce})\cdot D_A )$

i.e. $D_A = p^2 + 2pq\cdot D_A \Rightarrow D_A = \frac{p^2}{1-2pq}$ .

Now, the probability that A wins the service game from 0-30, P_A, is given by

$P_A=\Pr(\text{A wins next 4 points})+\Pr(\text{A wins 3 of the next 4 points (i.e. goes to a deuce))}) \cdot D_A$

$\Rightarrow P_A=p^4+4qp^3\cdot \frac{p^2}{1-2pq}$

Since B is equally likely to win from 0-30, P_A must equal ½.

$\Rightarrow \frac{1}{2}=p^4+4qp^3\cdot \frac{p^2}{1-2pq}$

From here, I'm getting a sixth-degree polynomial in p which I'm not seeing how to solve. Maybe a different approach needed?

I don't really know about the tennis rule, but I doubt 4qp^3 might be wrong, because if B win the first game over ?

InteGrand · Feb 3, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
I don't really know about the tennis rule, but I doubt 4qp^3 might be wrong, because if B win the first game over ?

According to the tennis scoring system, Player B needs to win two (not just one) of the next four points to win the game (without going to deuce).

seanieg89 · Feb 4, 2015

Re: HSC 2015 4U Marathon

Oh sorry guys, yeah I doubt you can find a nice surdic expression for p. Characterisation as the root of a certain polynomial is fine

.

The main point was to find a way to handle to concept of deuce, questions with that theme (a very elementary random walk with stopping rule) appear in several HSC/Trial questions from memory.

Drsoccerball · Feb 4, 2015

Re: HSC 2015 4U Marathon

I wish i knew the rules of tennis...

seanieg89 · Feb 4, 2015

Re: HSC 2015 4U Marathon

Dw, they obviously don't use any games/sports in HSC questions without explaining relevant rules lol.

FrankXie · Feb 4, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$Solve the equation $ x^4-8x^3+16x^2-8x+1=0. $ Leave your answer in exact form.$

Drsoccerball · Feb 4, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$Solve the equation $ x^4-8x^3+16x^2-8x+1=0. $ Leave your answer in exact form.$

Are we given any information about the roots?

FrankXie · Feb 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Are we given any information about the roots?

no

InteGrand · Feb 4, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$Solve the equation $ x^4-8x^3+16x^2-8x+1=0. $ Leave your answer in exact form.$

$x^4-8x^3 + 16x^2 -8x + 1 = 0,( $so $x \neq 0)$

$\fbox{\div $ $x^2 }\left ( x^2 + \frac{1}{x^2}\right )-8\left ( x+\frac{1}{x} \right ) + 16 =0$

Using $x^2 + \frac{1}{x^2}=\left ( x+\frac{1}{x} \right )^2-2$ , the equation becomes

$\left ( x + \frac{1}{x}\right )^2 -8\left ( x+\frac{1}{x} \right ) + 14 =0$ .

Discriminant is $\Delta =64-4(14)=8\Rightarrow \sqrt{\Delta}=2\sqrt{2}$ .

Therefore, by the quadratic formula, $x+\frac{1}{x}=\frac{8\pm 2\sqrt{2}}{4}=\frac{4\pm \sqrt{2}}{2}$ .

Letting $c_+ =\frac{4+\sqrt{2}}{2},c_-=\frac{4-\sqrt{2}}{2}$ , we will get two quadratic equations:

$x+\frac{1}{x}=c_{\pm}$ .

These can be solved by the quadratic formula and using square roots of complex numbers where necessary.

FrankXie · Feb 4, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$You may use $ \cos3\theta=4\cos^3\theta-3\cos\theta. $ Solve the equation $ 4x^3-9x-1=0. $ Leave your answer in exact trigonometric form.$

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