HSC 2015 Maths Marathon (archive) (4 Viewers)

kawaiipotato · Sep 28, 2015

Re: HSC 2015 2U Marathon

flavurr said:
Q. If the continuous function f(x) is an even function, prove that f'(x) is an odd function

let y = f(x) be an even function
So, y = f(x) = f(-x)

Differentiating wrt. x,
dy/dx = f'(x) = f'(-x) * d(-x)/dx
f'(x) = -f'(-x)
which is odd

Drsoccerball · Sep 28, 2015

Re: HSC 2015 2U Marathon

Next question: Prove that every odd function is divisible by x.

psyc1011 · Sep 28, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Next question: Prove that every odd function is divisible by x.

You mean polynomial.

Consider f(x) = sin (x)...

braintic · Sep 28, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Next question: Prove that every odd function is divisible by x.

You mean every odd polynomial.

(Although it is always true if you consider Taylor series)

Perhaps a better more general question is "Prove that if an odd function is defined for x=0, then it must pass through the origin".

BlueGas · Sep 28, 2015

Re: HSC 2015 2U Marathon

braintic said:
You mean every odd polynomial.

(Although it is always true if you consider Taylor series)

Perhaps a better more general question is "Prove that if an odd function is defined for x=0, then it must pass through the origin".

psyc1011 said:
You mean polynomial.

Consider f(x) = sin (x)...

How do you guys know it's polynomial, lol, is this even 2U? And Taylor series dayuuum.

braintic · Sep 28, 2015

Re: HSC 2015 2U Marathon

BlueGas said:
How do you guys know it's polynomial, lol, is this even 2U? And Taylor series dayuuum.

The fact that it should be a general polynomial means it is not 2U.

But my alternative question is 2U.

Drsoccerball · Sep 28, 2015

Re: HSC 2015 2U Marathon

braintic said:
The fact that it should be a general polynomial means it is not 2U.

But my alternative question is 2U.

Yeah i was trying to say what you said

my bad

leehuan · Sep 29, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Firstly is my answer correct?

I missed this. Anyway, logic:

Total outcomes = 50*49*48*47*46

Favourable outcomes = Let us consider the spades suit first.

There is only one royal flush in the spades suit. That is, 10-J-Q-K-A of spades. So that's 1 favourable outcome
We have 4 suits altogether. So that's 4 favourable outcomes

Hence, required probability = 4/(50*49*48*47*46)

rand_althor · Oct 1, 2015

Re: HSC 2015 2U Marathon

braintic said:
Prove that if an odd function is defined for x=0, then it must pass through the origin

For an odd function f(-x) = -f(x). If x = 0, we have f(0) = -f(0) -> 2f(0) = 0 -> f(0) = 0. Therefore odd functions defined for x = 0 pass through the origin (0,0).

leehuan · Oct 2, 2015

Re: HSC 2015 2U Marathon

Break time for 2U.

$\\ $1. Use algebra to solve ${ x }^{ 2 }=\sqrt { x } \\ $2. Sketch the curves $y={ x }^{ 2 }$ and $ y=\sqrt { x } $, clearly showing any points of intersection \\ 3. Calculate the area bound between the curves $ y={ x }^{ 2 }$ and $ y=\sqrt { x }$

rand_althor · Oct 2, 2015

Re: HSC 2015 2U Marathon

leehuan said:
Break time for 2U.

$\\ $1. Use algebra to solve ${ x }^{ 2 }=\sqrt { x } \\ $2. Sketch the curves $y={ x }^{ 2 }$ and $ y=\sqrt { x } $, clearly showing any points of intersection \\ 3. Calculate the area bound between the curves $ y={ x }^{ 2 }$ and $ y=\sqrt { x }$

$\begin{align*}\textup{1. }x^2&=\sqrt{x} \\x^2-\sqrt{x}&=0 \\\sqrt{x} \left (\sqrt{x^3}-1 \right ) &=0 \\\sqrt{x}&=0\to x=0 \\\sqrt{x^3}-1&=0 \to \sqrt{x^3}=1 \to x^3 = 1\to x=1 \\\textup{Solutions are }x&=0,1. \\\end{align*}$

$\begin{align*}\textup{3. A}&=\int^1_0 \sqrt{x}-x^2 \, \mathrm{d}x \\&=\left [\frac{2}{3}\sqrt{x^3}-\frac{x^3}{3} \right ]^1_0 \\&=\frac{1}{3}\left ( 2\sqrt{1^3}-1^3 \right ) \\&=\frac{1}{3} \left (2-1 \right ) \\&=\frac{1}{3}\textup{ units}^2\end{align*}$

kawaiipotato · Oct 13, 2015

Re: HSC 2015 2U Marathon

A point starts at A(0,0). It forms a staircase by moving to the right, cos(18) metres. It then moves up, cos^(2)(18) metres. Then it moves to the right cos^(3)(18) metres and then moves up cos^(4)(18) metres and so on infinitely. Find the angle that the line connecting A to the top of the staircase makes with the horizontal provided the angle is less than 90.

rand_althor · Oct 13, 2015

Re: HSC 2015 2U Marathon

kawaiipotato said:
A point starts at A(0,0). It forms a staircase by moving to the right, cos(18) metres. It then moves up, cos^(2)(18) metres. Then it moves to the right cos^(3)(18) metres and then moves up cos^(4)(18) metres and so on infinitely. Find the angle that the line connecting A to the top of the staircase makes with the horizontal provided the angle is less than 90.

I'll label each corner of the staircase with B, C, D, E, ... So we have:
$$B$ \left (\cos{18}, ~ 0 \right )$
$$C$ \left (\cos{18}, ~ \cos^2{18} \right )$
$$D$ \left (\cos{18} + \cos^3{18}, ~ \cos^2{18} \right )$
$$E$ \left (\cos{18} + \cos^3{18}, ~ \cos^2{18} + \cos^4{18} \right )$
Clearly there is a geometric series here, and a limiting sum exists since -1 < cos(18) < 1 and -1 < cos²(18) < 1. The staircase will eventually end in one point, which we'll call n. The co-ordinates of n are:
$n \left (\cos{18} + \cos^3{18} + \cos^5{18} + ..., ~ \cos^2{18} + \cos^4{18} + \cos^6{18} + ... \right )$
Which we simplify to:
$n\left (\frac{\cos{18}}{1-\cos^2{18}},~\frac{\cos^2{18}}{1-\cos^2{18}} \right ) \to ~ n \left (\frac{\cos{18}}{\sin^2{18}},~\frac{\cos^2{18}}{\sin^2{18}} \right )$
Since A is the origin, the line passing through the top of the staircase and A has the form y=mx. Solving for m:
$y=mx \to \frac{\cos^2{18}}{\sin^2{18}} = m \cdot \frac{\cos{18}}{\sin^2{18}} \to m=\cos{18}$
Now, for the angle the line makes with the x-axis, we use m=tan(θ):
$m=\tan{\theta} \to \theta = \tan^{-1}{m}=\tan^{-1}{(\cos{18})} \approx 0.58^{c}=33.44^{\circ}$

rand_althor · Oct 13, 2015

Re: HSC 2015 2U Marathon

There is a circular table with 6 chairs. Alice, Bob, and Eve each pick a random chair to sit down in. Compute the probability that each person has an empty chair on either side of them.

flavurr · Oct 14, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
There is a circular table with 6 chairs. Alice, Bob, and Eve each pick a random chair to sit down in. Compute the probability that each person has an empty chair on either side of them.

feel like i did this horribly wrong but is it 12/84?

InteGrand · Oct 14, 2015

Re: HSC 2015 2U Marathon

flavurr said:
feel like i did this horribly wrong but is it 12/84?

$The answer is $\frac{2}{^5 P_2}=\frac{1}{10}$.$

$Since the table is symmetrical at the start, we can fix Alice to be in some seat. Now, of the five remaining seats, Bob and Eve must be sitting in the two seats that are two seats away from Alice (for otherwise, there would be at least two people next to each other, which we do not want). Since there are $^5 P_2$ ways to place Bob and Eve into these five seats, and only 2 favourable seating arrangements (either Bob is in the favourable seat that is to the left of Alice, or Eve is), the answer is $\frac{2}{^5 P_2}=\frac{1}{10}$.$

Mr_Kap · Oct 14, 2015

Re: HSC 2015 2U Marathon

Can someone teach me how to do this question without permutations...it is 2u after all. I can't remember Perms and Combinations from 3u ..as I learnt it just before the trials and when i realised i was gonna drop 3u i didn't really pay attention. Although, i think perms and combs is a good 3unit topic, it seemed hard but enjoyable.

InteGrand · Oct 14, 2015

Re: HSC 2015 2U Marathon

Mr_Kap said:
Can someone teach me how to do this question without permutations...it is 2u after all. I can't remember Perms and Combinations from 3u ..as I learnt it just before the trials and when i realised i was gonna drop 3u i didn't really pay attention. Although, i think perms and combs is a good 3unit topic, it seemed hard but enjoyable.

Oh yeah sorry, forgot this was the 2U thread (Q looked like a perms and combs Q so subconsciously thought about it in a perms and combs way I guess).

InteGrand · Oct 14, 2015

Re: HSC 2015 2U Marathon

Basically, fix Alice somewhere at the start due to symmetry. Now, there's 5 seats left and we need Bob and Eve to sit in two specific seats (the ones two seats away from Alice). Call these Seats 1 and 2. As there's 5 seats left, there's a 1/5 chance of a given seat being selected. We can have Bob sit in Seat 1 and Eve sit in Seat 2, or vice versa. The probability of Bob sitting in Seat 1 then Eve in Seat 2 is (1/5)•(1/4) = 1/20 (the 1/4 is the probability of Eve picking the other seat, Seat 2, given that Bob has picked Seat 1, since there's 4 seats left now for Eve to choose from). The other option is just reversing Bob and Eve and so is the same probability of 1/20. Hence the answer is 1/20 + 1/20 = 1/10.

Mr_Kap · Oct 14, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
Basically, fix Alice somewhere at the start due to symmetry. Now, there's 5 seats left and we need Bob and Eve to sit in two specific seats (the ones two seats away from Alice). Call these Seats 1 and 2. As there's 5 seats left, there's a 1/5 chance of a given seat being selected. We can have Bob sit in Seat 1 and Eve sit in Seat 2, or vice versa. The probability of Bob sitting in Seat 1 then Eve in Seat 2 is (1/5)•(1/4) = 1/20 (the 1/4 is the probability of Eve picking the other seat, Seat 2, given that Bob has picked Seat 1, since there's 4 seats left now for Eve to choose from). The other option is just reversing Bob and Eve and so is the same probability of 1/20. Hence the answer is 1/20 + 1/20 = 1/10.

so does seating Alice not matter to the probability in this case because it is a circle, and circles can start from anywhere?

HSC 2015 Maths Marathon (archive) (4 Viewers)

Well-Known Member

Well-Known Member

#truth

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Active Member

New Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)