HSC 2013 MX2 Marathon (archive) (13 Viewers)

Sy123 · Jul 8, 2013

Re: HSC 2013 4U Marathon

$Prove that$

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

Sy123 said:
$i) Prove$

$\int_1^2 f(x) \ dx = \lim_{n \to \infty} \left(\frac{1}{n} \sum_{k=1}^{n}f \left(1 + \frac{k}{n} \right ) \right )$

$ii) Hence evaluate$

$\lim_{n \to \infty} \left(\frac{n}{2n^2 + 2n + 1} + \frac{n}{2n^2 + 4n + 4} + \frac{n}{2n^2 + 6n + 9} + \frac{n}{2n^2 + 8n+ 16} + \dots + \frac{n^2}{2n^3 + 3n^2} \right )$

ii) It is obvious that one simply needs to pick the right function f(x), it is clear then that we need to find some function f so that:

$\frac{1}{n}f( \left(1+ \frac{k}{n} \right ) = \frac{n}{2n^2 + 2kn + k^2}$

This may not always give the correct answer, because it is possible the sum inside the limit may be manipulated or simplified further, but we can simply verify our result once we find a good guess as to what the function is.

$f(1+k/n) = \frac{n^2}{2n^2 + 2kn + k^2} = \frac{1}{2 + \frac{2k}{n} + \frac{k^2}{n^2}} = \frac{1}{1 + (1+ k/n)^2}$

$\therefore \ \ f(x) = \frac{1}{1+x^2} \ \ $?$$

We can verify this by simply putting 1+k/n back in.

So f(x) = 1/(1+x^2), the answer then is $\tan^{-1} (2) - \frac{\pi}{4}$

RealiseNothing · Jul 9, 2013

Re: HSC 2013 4U Marathon

Find, for some positive integer $n$ :

$\sum_{k=1}^{n} \frac{n-k+1}{\sin^2(\frac{k\pi}{2n+1})}$

Sy123 · Jul 9, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Find, for some positive integer $n$ :

$\sum_{k=1}^{n} \frac{n-k+1}{\sin^2(\frac{k\pi}{2n+1})}$

Using De'Moivers Theorem:

$\frac{\sin(2n+1)x}{\sin^{2n+1}x} = \binom{2n+1}{1} \cot^{2n} x - \binom{2n+1}{3} \cot^{2n-2}x + \dots + (-1)^n$

$\sin(2n+1)x = 0$

$x= \frac{k\pi}{2n+1}$

$z= \cot^2 x$

$\binom{2n+1}{1} z^n - \binom{2n+1}{3} z^{n-1} + \dots + (-1)^{n} = 0$

With roots:

$z_k = \cot^{2} \frac{k\pi}{2n+1}$

$\theta_k = \frac{k\pi}{2n+1}$

Sum of roots:

$\sum_{k=1}^n \cot^2 \theta_k = \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{n}{3} (2n-1)$

$\cot^2 x + 1 = \csc^2 x$

$\sum_{k=1}^n \csc^2 \theta_k - \sum_{k=1}^n 1 = \frac{n}{3} (2n-1)$

$\therefore \sum_{k=1}^{n} \csc^2 \theta_k = \frac{2n}{3} (n+1)$

$\sum_{k=1}^{n}(n+1) \csc^{2} \theta_ k - k\csc^2 \theta_k = \frac{2n}{3} (n+1)^2 - \sum_{k=1}^{n} k \csc^2 \theta_k$

....

Finishing later.

EDIT: Is this sum even possible?

RealiseNothing · Jul 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
EDIT: Is this sum even possible?

It is possible. I wouldn't recommend splitting the summation up though, it is easy keeping it together.

RealiseNothing · Jul 11, 2013

Re: HSC 2013 4U Marathon

Something easier than usual:

$$Show that this equation has no real roots$~ \geq2$

$x^4 - x^3 - 2x^2 + 1 = 0$

braintic · Jul 12, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Something easier than usual:

$$Show that this equation has no real roots$~ \geq2$

$x^4 - x^3 - 2x^2 + 1 = 0$

Simplifies to (x^2 - 1)^2 = x^3.
A graph of y = (x^2 - 1)^2 and y=x^3 on the same axes, and a substitution of x=2 into each equation, shows the result.

btx3 · Jul 12, 2013

Re: HSC 2013 4U Marathon

braintic said:
Simplifies to (x^2 - 1)^2 = x^3.
A graph of y = (x^2 - 1)^2 and y=x^3 on the same axes, and a substitution of x=2 into each equation, shows the result.

good to know brah but leave it to a student you know

asianese · Jul 12, 2013

Re: HSC 2013 4U Marathon

^ don't hate, appreciate.

luvglee4lyfe · Jul 12, 2013

Re: HSC 2013 4U Marathon

btx3 said:
good to know brah but leave it to a student you know

l3l this.

Sy123 · Jul 12, 2013

Re: HSC 2013 4U Marathon

btx3 said:
good to know brah but leave it to a student you know

A whole day and no one answered it, its up to anyone by then.

An alternative approach is simply letting it be f(x), finding f'(x) factorising x out and solving the quadratic.
======

$The lengths of a triangle are the first three terms in an arithmetic sequence with first term 1 and common difference d. Find the set of possible values of d$

RealiseNothing · Jul 12, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A whole day and no one answered it, its up to anyone by then.

An alternative approach is simply letting it be f(x), finding f'(x) factorising x out and solving the quadratic.
======

$The lengths of a triangle are the first three terms in an arithmetic sequence with first term 1 and common difference d. Find the set of possible values of d$

Unless the triangle is degenerate, $d>0$

Now by triangle inequality:

$1+1+d >1+2d$

$1>d$

Hence:

$1>d>0$

asianese · Jul 12, 2013

Re: HSC 2013 4U Marathon

Where's that question with P Q and f(sinx)?? It was a nice one using the classic fact that

$\int_0^a f(x) \mathop{dx} = \int_0^a f(a-x) \mathop{dx}.$

Sy123 · Jul 12, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Unless the triangle is degenerate, $d>0$

Now by triangle inequality:

$1+1+d >1+2d$

$1>d$

Hence:

$1>d>0$

The upper bound is correct, but d can be negative.

asianese said:
Where's that question with P Q and f(sinx)?? It was a nice one using the classic fact that

$\int_0^a f(x) \mathop{dx} = \int_0^a f(a-x) \mathop{dx}.$

Integration marathon

asianese · Jul 12, 2013

Re: HSC 2013 4U Marathon

lol awks.

RealiseNothing · Jul 12, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
The upper bound is correct, but d can be negative.

Oh right, I was thinking that the side with length 1 was the shortest side. Also I just realised d=0 is possible, because you know, what even is an equilateral triangle?

RealiseNothing · Jul 12, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A whole day and no one answered it, its up to anyone by then.

An alternative approach is simply letting it be f(x), finding f'(x) factorising x out and solving the quadratic.
======

$The lengths of a triangle are the first three terms in an arithmetic sequence with first term 1 and common difference d. Find the set of possible values of d$

$1>d>\frac{-1}{3}$

Using triangle inequality twice.

Sy123 · Jul 12, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$1>d>\frac{-1}{3}$

Using triangle inequality twice.

Yep.

I particularly like this question, it came from an old HSC paper:

$If we can consider the Earth to be a sphere of uniform shape and density, then the gravitational acceleration towards the centre of the Earth above the Earth's surface is proportional to$ \ \ x^{-2} \ \ $Where$ \ \ x \ \ $is the distance from the Earth's centre$

$$Below the Earth's surface, the gravitational acceleration is proportional to$ \ \ x$

$Consider a tunnel dug through through the centre of the Earth to the other side, so along a diameter$

$A projectile is fired with initial velocity$ \ \ U \ \ $through the hole$

$Prove that the motion is oscillatory if$

$U^2 < 2gR$

$Where$ \ \ R \ \ $ is the radius of the earth, and$ \ \ g \ \ $is the gravitational acceleration at the surface of the Earth$

Sy123 · Jul 13, 2013

Re: HSC 2013 4U Marathon

$P(x) \ $is a polynomial of a degree at least 2 such that$ \ \ P'(a) = 0 \ $. Show that when$ \ P(x) \ $is divided by$ \ (x-a)^2 \ $the remainder is$ \ \ P(a)$

seanieg89 · Jul 14, 2013

Re: HSC 2013 4U Marathon

$$g(t)$ is a continuous function and $A,B$ are positive constants such that:\\ \\ $g(t)\leq A+B\int_0^t g(s)\, ds $\\ \\ for all $t\geq 0.$\\ \\Prove that $g(t)\leq Ae^{Bt}$\\ \\for all $t\geq 0.$$

Sy123 · Jul 15, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$$g(t)$ is a continuous function and $A,B$ are positive constants such that:\\ \\ $g(t)\leq A+B\int_0^t g(s)\, ds $\\ \\ for all $t\geq 0.$\\ \\Prove that $g(t)\leq Ae^{Bt}$\\ \\for all $t\geq 0.$$

First find the function f(t) such that the equality holds:

$f(t) = A + B\int_0^t f(s) \ ds$

$f'(t) = B(F'(t) - F'(0))$

For F the primitive of f:

$\frac{f'(t)}{f(t)} = B$

$\ln f(t) = Bt + c$

$f(0) = A + B \int_0^0 f(s) \ ds = A$

$c = \ln A$

$\therefore f(t) = e^c e^{Bt}$

$f(t) = Ae^{Bt}$

$$Now fix for some function$ \ \ g(t) \leq f(t)$

$\therefore \ $if$ \ \ g(t) < A + B \int_0^t g(s) \ ds$

$g(t) \leq Ae^{Bt}$

I have a feeling this argument is not valid but I cant see it right now, its a bit late

HSC 2013 MX2 Marathon (archive) (13 Viewers)

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