HSC 2013-14 MX1 Marathon (archive) (6 Viewers)

RealiseNothing · May 10, 2013

Re: HSC 2013 3U Marathon Thread

Find:

$\sum_{k=0}^{2r}(-1)^k \binom{n}{k}$

Where:

$n \geq 2r+1$

omgiloverice · May 15, 2013

Re: HSC 2013 3U Marathon Thread

The 3U thread needs bumping as well...

$The curve C has equation $ xy = \frac{1}{2} . $The tangents to C at the distinct points$ \ P(p,\frac{1}{2p}) \ and \ Q (q,\frac{1}{2q} ), $where p and q are positive, intersect at T$

$Show that T is the point$

$(\frac{2pq}{p+q}, \frac{1}{p+q}).$

omgiloverice · May 17, 2013

Re: HSC 2013 3U Marathon Thread

$$A piston (shown above) has a rod (AB) that is rotating at$ \ \frac{\pi}{4} $rads/s$

$\\\ a)$Prove that the distance of AC is given by$ \ x=\sqrt{144 - 4sin^2\theta}+2cos\theta$

$$ b) Hence find the velocity of C at 270^{\circ }$

anomalousdecay · May 17, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
Find:

$\sum_{k=0}^{2r}(-1)^k \binom{n}{k}$

$\sum_{k=0}^{2r}(-1)^k \binom{n}{k}$

$= \frac{a(z^y-1)}{z-1}$

$= \frac{1((-1)^k-1)}{-2}$

$= \frac{1-(-1)^k}{2}$

$\text{Since the limiting sum is 2r, the final power is even.}$

$\text{Therefore,} \sum_{k=0}^{2r}(-1)^k \binom{n}{k}$

$= 0 = \text{zero}$

I could be wrong since we only just started Series in 2-unit.

Trebla · May 19, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
Find:

$\sum_{k=0}^{2r}(-1)^k \binom{n}{k}$

Is k supposed to vary from 0 to 2r or 0 to n?

Sy123 · May 19, 2013

Re: HSC 2013 3U Marathon Thread

$$i) Show that$ \ \ (n+1)! - n! = n \cdot n!$

$$ii) Hence prove that$ \ \ \sum_{k=1}^{n} k \cdot k! = (n+1)! - 1$

$iii) Similarly, find an expression for$

$\sum_{k=1}^{n} k! (k^2 + k + 1)$

RealiseNothing · May 19, 2013

Re: HSC 2013 3U Marathon Thread

Trebla said:
Is k supposed to vary from 0 to 2r or 0 to n?

That one.

RealiseNothing · May 19, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$i) Show that$ \ \ (n+1)! - n! = n \cdot n!$

$$ii) Hence prove that$ \ \ \sum_{k=1}^{n} k \cdot k! = (n+1)! - 1$

$iii) Similarly, find an expression for$

$k! (k^2 + k + 1)$

I'm just going to do the last part and leave the first two if some one else wants to write up a solution.

If we expand the brackets and take individual summations we get:

$\sum_{k=1}^{n} k!k^2 + \sum_{k=1}^{n} k!k + \sum_{k=1}^{n} k!$

We know the 2nd summation from previous parts and so it is:

$\sum_{k=1}^{n} k \dot k! = (n+1)! - 1$

Now lets look at the 1st summation:

$\sum_{k=1}^{n} k!k^2$

$\sum_{k=1}^{n} k(k!k)$

$1(2!-1!) + 2(3!-2!) + 3(4!-3!) + ... + n[(n+1)!-n!]$

If we break this up into individual series we get:

$(2!-1!) + (3!-2!) + (4!-3!) + ... + [(n+1)!-n!]$

$(3!-2!) + (4!-3!) + ... + [(n+1)!-n!]$

$(4!-3!) + ... + [(n+1)!-n!]$

Adding all these up gives:

$(n+1)! - 1! + (n+1)! - 2! + (n+1)! - 3! + ... + (n+1)! - n!$

$n(n+1)! - \sum_{k=1}^{n} k!$

But the 3rd summation was:

$\sum_{k=1}^{n} k!$

So they somewhat cancel out when we add the 1st and 3rd summation, and after adding the 2nd summation we are left with:

$n(n+1)! + (n+1)! - 1$

$(n+1)(n+1)! - 1$

Sy123 · May 19, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
I'm just going to do the last part and leave the first two if some one else wants to write up a solution.

If we expand the brackets and take individual summations we get:

$\sum_{k=1}^{n} k!k^2 + \sum_{k=1}^{n} k!k + \sum_{k=1}^{n} k!$

We know the 2nd summation from previous parts and so it is:

$\sum_{k=1}^{n} k \dot k! = (n+1)! - 1$

Now lets look at the 1st summation:

$\sum_{k=1}^{n} k!k^2$

$\sum_{k=1}^{n} k(k!k)$

$1(2!-1!) + 2(3!-2!) + 3(4!-3!) + ... + n[(n+1)!-n!]$

If we break this up into individual series we get:

$(2!-1!) + (3!-2!) + (4!-3!) + ... + [(n+1)!-n!]$

$(3!-2!) + (4!-3!) + ... + [(n+1)!-n!]$

$(4!-3!) + ... + [(n+1)!-n!]$

Adding all these up gives:

$(n+1)! - 1! + (n+1)! - 2! + (n+1)! - 3! + ... + (n+1)! - n!$

$n(n+1)! - \sum_{k=1}^{n} k!$

But the 3rd summation was:

$\sum_{k=1}^{n} k!$

So they somewhat cancel out when we add the 1st and 3rd summation, and after adding the 2nd summation we are left with:

$n(n+1)! + (n+1)! - 1$

$(n+1)(n+1)! - 1$

Yep nice work.

====

$Find without integrating$

$\int_b^a \sqrt{a^2-x^2} \ dx$

$a>b>0$

Buwwy · May 21, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Find without integrating$

$\int_b^a \sqrt{a^2-x^2} \ dx$

$a>b>0$

You could consider the graph of $y=\sqrt{a^2-x^2}$ which is just a semi-circle.

$y_{1}=\sqrt{a^2-b^2}$

$\int_b^a \sqrt{a^2-x^2} \ dx=Area(sector)-Area(triangle)$

$=\frac{1}{2}a^{2}\theta - \frac{1}{2}b\sqrt{a^{2}-b^{2}}$

$=\frac{1}{2}(a^{2}\cos^{-1}\frac{b}{a} - b\sqrt{a^{2}-b^{2}})$

Is that right?

Sy123 · May 21, 2013

Re: HSC 2013 3U Marathon Thread

Buwwy said:
You could consider the graph of $y=\sqrt{a^2-x^2}$ which is just a semi-circle.

View attachment 28135

$y_{1}=\sqrt{a^2-b^2}$

$\int_b^a \sqrt{a^2-x^2} \ dx=Area(sector)-Area(triangle)$

$=\frac{1}{2}a^{2}\theta - \frac{1}{2}b\sqrt{a^{2}-b^{2}}$

$=\frac{1}{2}(a^{2}\cos^{-1}\frac{b}{a} - b\sqrt{a^{2}-b^{2}})$

Is that right?

Something like that, nice work.

hayabusaboston · May 21, 2013

Re: HSC 2013 3U Marathon Thread

there are 20 men on a hockey team. There are 2 goalies, 6 defensemen, and 12 forwards. How many lineups can be created if there must be 1 goalie, two defensemen and three forwards, and if two players insist on being in the same line together (like, defense or forward) or they dont want to play at all, while three others refuse to be in the same line (defense line or forward line)?

study1234 · May 21, 2013

Re: HSC 2013 3U Marathon Thread

(From Yr 12 Cambridge 3u)

A large vertical wall stand on horizontal ground. The nozzle of a water hose is positioned at a point C on the ground at a distance d from the wall and the water jet can be pointed in any direction from P. Also the water issues from the nozzle with speed V .
If V = 2√(gd) prove that the portion of the wall that can be reached by the jet is a parabolic segment of height (15d)/8 and area (5√15(d^2))/2.

nightweaver066 · May 21, 2013

Re: HSC 2013 3U Marathon Thread

$\int \sqrt{1-sin2x}dx$

hayabusaboston · May 22, 2013

Re: HSC 2013 3U Marathon Thread

hayabusaboston said:
there are 20 men on a hockey team. There are 2 goalies, 6 defensemen, and 12 forwards. How many lineups can be created if there must be 1 goalie, two defensemen and three forwards, and if two players insist on being in the same line together (like, defense or forward) or they dont want to play at all, while three others refuse to be in the same line (defense line or forward line)?

Fuuuck my question must be tough if even realisenothing and sy123 havent answered it yet LOLOL

Makematics · May 22, 2013

Re: HSC 2013 3U Marathon Thread

hayabusaboston said:
Fuuuck my question must be tough if even realisenothing and sy123 havent answered it yet LOLOL

what if they havent seen it?

hayabusaboston · May 22, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
what if they havent seen it?

They've been active on the site in the last 48hrs, they would have seen it.

RealiseNothing · May 22, 2013

Re: HSC 2013 3U Marathon Thread

hayabusaboston said:
They've been active on the site in the last 48hrs, they would have seen it.

Doesn't mean I've tried it.

Sy123 · May 22, 2013

Re: HSC 2013 3U Marathon Thread

hayabusaboston said:
there are 20 men on a hockey team. There are 2 goalies, 6 defensemen, and 12 forwards. How many lineups can be created if there must be 1 goalie, two defensemen and three forwards, and if two players insist on being in the same line together (like, defense or forward) or they dont want to play at all (1), while three others refuse to be in the same line (defense line or forward line)? (2)

There is a fault in your question.

And I only just got to look at this question now.

edit: I wasn't thinking properly and made a mistake, I was confused with the conditions and I thought that the team couldn't play in some conditions but I see you resolved that in your question, so yes there is no mistake

Ok I'll try with this attempt, first name the conditions in the question, (1) and (2).

There then initiates four cases that we only observe for Defense (D) and Forwards (F)

It's possible to do this because your conditions don't make sense if they apply for people among different lines.

$$Case 1$ \ \ D : (1) \ \ F: (1)$

$$Case 2$ \ \ D: (1) \ \ F: (2)$

$$Case 3$ \ \ D: (2) \ \ F: (1)$

$$Case 4$ \ \ F: (1) \ \ F: (2)$

We will simply ignore all arrangements for goalies, I'll multiply my answer by arranging defence and forward by 2 at the end.

Case 1:

$(^2 C_2 + ^3C_1) \cdot (^{12}C_3) = 880$

Case 2:

$(^2C_2 + ^4C_2) \cdot (^{12}C_3 - (^3C_3 + ^3C_2 \cdot ^9C_1)) = 1344$

Case 3:

$(^6C_2 - ^3C_2) \cdot (^{10}C_1 + ^{10}C_3) = 1560$

Case 4:

$(^6C_2) \cdot (^{10}C_1 \cdot ^2C_2 + ^{10}C_3 - (^3C_3 + ^3C_2 \cdot ^7C_1)) = 1620$

So my final answer is:

10808

Is it? It might be wrong because I'm quite weak with permutations and combinations.

hayabusaboston · May 22, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
There is a fault in your question.

And I only just got to look at this question now.

Well if there's a fault, naturally you would have mentioned what the fault was. Since u havent, we can all assume u are making shit up.

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