HSC 2013 MX2 Marathon (archive) (1 Viewer)

Trebla · Apr 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
What I did:

Construct a polynomial in the form:

$x^3-ax^2+bx-c$

Now let the roots of this polynomial be our three unknown integers, denoted by $\alpha,~\beta,~\gamma$

$a=\alpha +\beta +\gamma = 4$ as it is the sum of the roots.

$b=\alpha^2 +\beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma) = 4^2-2b = 16-2b = 10$

So $b=10$

$\alpha^3+\beta^3+\gamma^3=4(\alpha^2+\beta^2+ \gamma^2) - 3(\alpha + \beta + \gamma) + 3c$

$16 = 4(10) - 3(4) + 3c$

$-12=3c$

$c=-4$

So we have the polynomial $x^3-4x^2+3x+4$

Now we want $\alpha\beta\gamma$ which is just the product of the roots, which is just -4 by observing the polynomial.

@ Sy123, this was what I was thinking of

Sy123 · Apr 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
What I did:

Construct a polynomial in the form:

$x^3-ax^2+bx-c$

Now let the roots of this polynomial be our three unknown integers, denoted by $\alpha,~\beta,~\gamma$

$a=\alpha +\beta +\gamma = 4$ as it is the sum of the roots.

$b=\alpha^2 +\beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma) = 4^2-2b = 16-2b = 10$

So $b=10$

$\alpha^3+\beta^3+\gamma^3=4(\alpha^2+\beta^2+ \gamma^2) - 3(\alpha + \beta + \gamma) + 3c$

$16 = 4(10) - 3(4) + 3c$

$-12=3c$

$c=-4$

So we have the polynomial $x^3-4x^2+3x+4$

Now we want $\alpha\beta\gamma$ which is just the product of the roots, which is just -4 by observing the polynomial.

Ah yes your way is a bit faster than my one.

Sy123 · Apr 3, 2013

Re: HSC 2013 4U Marathon

$Find$

$I=\int \frac{dx}{1+\sqrt{x}}$

shongaponga · Apr 3, 2013

Re: HSC 2013 4U Marathon

$\\$ The points P, Q, R and S lie on an ellipse and have coordinates $ \\ (a $ cos $ p, b $ sin $ p), (a $ cos $ q, b $ sin $ q), (a $ cos $ r, b $ sin $ r) $ and $ (a $ cos $ s, b $ sin $ s) $ respectively; where $0 \leq p < q < r < s < 2\pi $; and a and b are both positive.$$

$\\ $Given that neither of the lines PQ or SR are vertical, show that these lines are parallel if and only if $ r + s - p - q = 2\pi.$

Sy123 · Apr 3, 2013

Re: HSC 2013 4U Marathon

shongaponga said:
$\\$ The points P, Q, R and S lie on an ellipse and have coordinates $ \\ (a $ cos $ p, b $ sin $ p), (a $ cos $ q, b $ sin $ q), (a $ cos $ r, b $ sin $ r) $ and $ (a $ cos $ s, b $ sin $ s) $ respectively; where $0 \leq p < q < r < s < 2\pi $; and a and b are both positive.$$

$\\ $Given that neither of the lines PQ or SR are vertical, show that these lines are parallel if and only if $ r + s - p - q = 2\pi.$

Use gradient formula, equate gradients, then turn the sum into products using those formulae that are easily derivable, we end up with

$\frac{2\sin((q-p)/2) \cos((q+p)/2)}{-2\sin((q+p)/2) \sin((q-p)/2)} = \frac{2\sin((s-r)/2) \cos((s+r)/2}{-2\sin((s+r)/2) \sin((s-r)/2)}$

Note that:

$\frac{q-p}{2} \neq 0, \pi, 2\pi$

$\frac{s-r}{2} \neq 0, \pi, 2\pi$

Since they are all less than 2pi, and the points do not coincide.

We end up with:

$\cot\frac{p+q}{2} = \cot \frac{s+r}{2}$

Cotangent is periodic by pi, and since all angles are less than 2pi we only take the first case, which is: (note p+q =/= s+r)

$\pi + \frac{p+q}{2} =\frac{s+r}{2}$

Rearrange and we are done.

Sy123 · Apr 3, 2013

Re: HSC 2013 4U Marathon

$An n-sided regular polygon has its sides such that none are vertical$

$$Let$ \ \ m_1, \ m_2, \ m_3, \dots , \ m_n$

$Be the gradients of all the sides of the polygon$

$Prove that$

$m_1 m_2 + m_2 m_3 + m_3 m_4 + \dots + m_{n-1} m_n + m_n m_1 = -n$

shongaponga · Apr 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Use gradient formula, equate gradients, then turn the sum into products using those formulae that are easily derivable, we end up with

$\frac{2\sin((q-p)/2) \cos((q+p)/2)}{-2\sin((q+p)/2) \sin((q-p)/2)} = \frac{2\sin((s-r)/2) \cos((s+r)/2}{-2\sin((s+r)/2) \sin((s-r)/2)}$

Note that:

$\frac{q-p}{2} \neq 0, \pi, 2\pi$

$\frac{s-r}{2} \neq 0, \pi, 2\pi$

Since they are all less than 2pi, and the points do not coincide.

We end up with:

$\cot\frac{p+q}{2} = \cot \frac{s+r}{2}$

Cotangent is periodic by pi, and since all angles are less than 2pi we only take the first case, which is: (note p+q =/= s+r)

$\pi + \frac{p+q}{2} =\frac{s+r}{2}$

Rearrange and we are done.

Nice job Sy.

Alternatively, from the line you posted instead of simplifying down to cotangents and using the fact cot is periodic by pi, you can cross-multiply the sin's on the denominators and then bring everything to the one side to make the equation = 0. This would yield the equation to be in the form sinAcosB - cosAsinB = 0 = sin (A-B). Which then leads to the same result. Not a quicker method, just an alternative

.

twinklegal19 · Apr 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Find$

$I=\int \frac{dx}{1+\sqrt{x}}$

$\begin{align*}x&=\tan^4\theta\\dx&=4\tan^3\theta\sec^2\theta d\theta\\ \frac{1}{1+\sqrt{x}}&=\cos^2\theta\\\int {\frac{dx}{1+\sqrt{x}}}&=\int\cos^2\theta\cdot 4\tan^3\theta\sec^2\theta d\theta\\&=4\int \tan^3\theta d\theta\\&=4\int\left ( \sec^2\theta\tan\theta-\tan\theta \right )d\theta\\&=4\left ( \frac{1}{2}\tan^2\theta+\ln|\cos\theta| \right )\\&=2\sqrt{x}-2\ln\left | 1+\sqrt{x} \right |+C \end{align*}$

Sy123 · Apr 3, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\begin{align*}x&=\tan^4\theta\\dx&=4\tan^3\theta\sec^2\theta d\theta\\ \frac{1}{1+\sqrt{x}}&=\cos^2\theta\\\int {\frac{dx}{1+\sqrt{x}}}&=\int\cos^2\theta\cdot 4\tan^3\theta\sec^2\theta d\theta\\&=4\int \tan^3\theta d\theta\\&=4\int\left ( \sec^2\theta\tan\theta-\tan\theta \right )d\theta\\&=4\left ( \frac{1}{2}\tan^2\theta+\ln|\cos\theta| \right )\\&=2\sqrt{x}-2\ln\left | 1+\sqrt{x} \right |+C \end{align*}$

Yep, nice work.

RealiseNothing · Apr 3, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\begin{align*}x&=\tan^4\theta\\dx&=4\tan^3\theta\sec^2\theta d\theta\\ \frac{1}{1+\sqrt{x}}&=\cos^2\theta\\\int {\frac{dx}{1+\sqrt{x}}}&=\int\cos^2\theta\cdot 4\tan^3\theta\sec^2\theta d\theta\\&=4\int \tan^3\theta d\theta\\&=4\int\left ( \sec^2\theta\tan\theta-\tan\theta \right )d\theta\\&=4\left ( \frac{1}{2}\tan^2\theta+\ln|\cos\theta| \right )\\&=2\sqrt{x}-2\ln\left | 1+\sqrt{x} \right |+C \end{align*}$

This makes me proud :')

asianese · Apr 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
This makes me proud :')

Awh you two <3

Sy123 · Apr 3, 2013

Re: HSC 2013 4U Marathon

I just made this:

$Let$

$I_n = \int_{0}^{\pi/4}\tan^{2n}x \ dx$

$$i) Prove that$ \ \ \ I_{n} + I_{n-1} = \frac{1}{2n-1} \ \ \ n\geq 1$

$ii) Hence prove that$

$\frac{\pi}{4} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \dots$

Carrotsticks · Apr 4, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\begin{align*}x&=\tan^4\theta\\dx&=4\tan^3\theta\sec^2\theta d\theta\\ \frac{1}{1+\sqrt{x}}&=\cos^2\theta\\\int {\frac{dx}{1+\sqrt{x}}}&=\int\cos^2\theta\cdot 4\tan^3\theta\sec^2\theta d\theta\\&=4\int \tan^3\theta d\theta\\&=4\int\left ( \sec^2\theta\tan\theta-\tan\theta \right )d\theta\\&=4\left ( \frac{1}{2}\tan^2\theta+\ln|\cos\theta| \right )\\&=2\sqrt{x}-2\ln\left | 1+\sqrt{x} \right |+C \end{align*}$

$\\ $Alternatively:$ \\\\ u^2=x \Rightarrow 2udu=dx \\\\ I = 2 \int \frac{u}{1+u} du = 2 \int 1 - \frac{1}{1+u}du = 2 (u - \ln(1+u)) = 2 (\sqrt{x} - \ln|1+\sqrt{x}|)$

Carrotsticks · Apr 4, 2013

Re: HSC 2013 4U Marathon

Plus C.

seanieg89 · Apr 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I just made this:

$Let$

$I_n = \int_{0}^{\pi/4}\tan^{2n}x \ dx$

$$i) Prove that$ \ \ \ I_{n} + I_{n-1} = \frac{1}{2n-1} \ \ \ n\geq 1$

$ii) Hence prove that$

$\frac{\pi}{4} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \dots$

$I_n+I_{n-1}=\int_0^{\pi/4} \tan^{2n}(x)+\tan^{2n-2}(x)\, dx \\=\int_0^{\pi/4} \tan^{2n-2}(x)\sec^2(x)\, dx=\frac{1}{2n-1}\\ \\ $so$\\ \\ \left|\frac{\pi}{4}-\sum_{j=1}^n \frac{(-1)^j}{2j-1}\right|=\left|\frac{\pi}{4}-\sum_{j=1}^n (-1)^j(I_j+I_{j-1})\right|=I_n < \frac{1}{2n-1}\rightarrow 0.$

seanieg89 · Apr 4, 2013

Re: HSC 2013 4U Marathon

$In this question you may assume:\\ Let $(a_j)$ be a sequence of complex numbers. If there exists an $M>0$ such that:\\ \\ $\sum_{j=0}^n |a_j|\leq M$\\ \\ for all $n$, then the series\\ $\sum_{j=0}^\infty a_j$\\ converges. \\ \\ \\ Let $(a_n)$ be a positive and decreasing sequence with $a_n\rightarrow 0$. Prove that $\sum_{n=0}^\infty a_n$ does not necessarily have to converge, but $\sum_{n=0}^\infty a_n\cos(n\theta)$ converges for every $\theta$ not divisible by $2\pi$. Are all of our conditions on the sequence $a$ necessary for this result to hold?$

Difficulty rating: 4/5.

Sy123 · Apr 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$In this question you may assume:\\ Let $(a_j)$ be a sequence of complex numbers. If there exists an $M>0$ such that:\\ \\ $\sum_{j=0}^n |a_j|\leq M$\\ \\ for all $n$, then the series\\ $\sum_{j=0}^\infty a_j$\\ converges. \\ \\ \\ Let $(a_n)$ be a positive and decreasing sequence with $a_n\rightarrow 0$. Prove that $\sum_{n=0}^\infty a_n$ does not necessarily have to converge, but $\sum_{n=0}^\infty a_n\cos(n\theta)$ converges for every $\theta$ not divisible by $2\pi$. Are all of our conditions on the sequence $a$ necessary for this result to hold?$

Difficulty rating: 4/5.

A couple of questions:

When you mean decreasing, you mean |a| is decreasing right?
And in difficulty, what difficulty would you rate Q8 2010 HSC? (Just so I can compare how YOU view difficulty)

seanieg89 · Apr 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A couple of questions:

When you mean decreasing, you mean |a| is decreasing right?
And in difficulty, what difficulty would you rate Q8 2010 HSC? (Just so I can compare how YOU view difficulty)

|a|=a, remember this sequence is positive.

2 because they walk you through everything, although I will admit that there are many places for the average MX2 student to go wrong. The scale is more the calibrated to the questions I post here than the HSC questions themselves. If you can answer half of the things I post then the only thing that can hold you back in the HSC is silly mistakes/speed.

Sy123 · Apr 4, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
|a|=a, remember this sequence is positive.

2 because they walk you through everything, although I will admit that there are many places for the average MX2 student to go wrong. The scale is more the calibrated to the questions I post here than the HSC questions themselves. If you can answer half of the things I post then the only thing that can hold you back in the HSC is silly mistakes/speed.

Ok so the sequence that we have to work with is only for real numbers? Because the assumption you gave us before said a is a complex number, so I just got a little confused.

Thank you for clearing it up

seanieg89 · Apr 4, 2013

Re: HSC 2013 4U Marathon

Ah yep, a more general phrasing of the problem would use cis instead of cos, in which case we would need the assumption for all complex sequences.

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