HSC 2012 MX2 Marathon (archive) (1 Viewer)

nightweaver066 · Jan 6, 2012

Same thing as MX1 thread, just MX2 this time.

Question:
$A fifth degree polynomial, P(x), has a triple root at x = 0 and P(x) - 1 has a triple root at x = 1. Find P(x)$

largarithmic · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
Same thing as MX1 thread, just MX2 this time.

Question:
$A fifth degree polynomial, P(x), has a triple root at x = 0 and P(x) - 1 has a triple root at x = 1. Find P(x)$

Didn't one of these exist ages ago?

Anyway let P(x) = x^3(ax^2 + bx + c).

Now by the remainder theorem, P(1) = 1. Differentiate P(x) - 1, and you find that P'(x) is has a double root at x = 1, so P'(1) = 0 and P''(1) = 0. Put that into the above to get:
a+b+c = 1 [1]
5a + 4b + 3c = 0 [2]
20a + 12b + 6c = 0 [3]

Now combine [1] and [2] to get 2a + b = -3, and [2] and [3] to get 10a + 4b = 0. Combine these two to get b = -15 and then a = 6, from these c = 10. So P(x) = 6x^5 - 15x^4 + 10x^3.

I'll post one in a sec.

largarithmic · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

Here's one I just found (havent done it yet) but it looks decently cool:

$Let $T_n = (\sqrt{a+1} + \sqrt{a})^n$ where n is a positive integer and a is a given positive integer.$

$$(i) When n is even, prove by induction that $T_n$ can be written in the form $A_n + B_n\sqrt{a(a+1)}$ where An and Bn are integers depending on a and n which satisfy $A_n^2 = a(a+1)B_n^2 + 1.$

$$(ii) In the case when n is odd, shower by considering $(\sqrt{a+1} + \sqrt{a})T_m$ where m is even, or otherwise, that Tn can be written in the form $C_n\sqrt{a+1} + D_n \sqrt{a}$ where Cn and Dn are integers depending on a and n and $(a+1)C_n^2 = aD_n^2 + 1$

$(iii) Deduce that for each n, Tn can be written as the sum of the square roots of two consecutive integers.$

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

You guys will love this one! This one is very nice to do.

$\textup{Let} \;0<\theta< \pi$

$\textup{a) Solve the equation} \; z^2=\cos \theta + i\sin \theta$

$\textup{b) Let} \; u_{1} \; \textup{and} \; u_{2} \; \textup{be the roots of the equation} \; (u+1)^2=\cos \theta + i\sin \theta \; \\ \;\textup{where}\; \textup{Im}(u_{1})<0}$

$\; \textup{i) Find} \; u_{1} \textup{and} \; u_{2}}$

$\; \textup{ii) Prove that} \; \frac{u_{2}}{u_{1}}=-i\tan\frac{\theta}{4}$

$\textup{iii) Hence find all integers n for which} \; (\frac{u_{2}}{u_{1}})^n \; \textup{is a real number.}$

$\textup{iv) Prove that} \; (u_{1})^{12}=2^{12}\cos^{12}\frac{\theta}{4}(\cos 3\theta+i\sin 3\theta)$

$\textup{v) Hence, find all values of} \; \theta \; \textup{for which} \; (u_{1})^{12}-(u_{2})^{12} \; \textup{is a real number.}$

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
Here's one I just found (havent done it yet) but it looks decently cool:

$Let $T_n = (\sqrt{a+1} + \sqrt{a})^n$ where n is a positive integer and a is a given positive integer.$

$$(i) When n is even, prove by induction that $T_n$ can be written in the form $A_n + B_n\sqrt{a(a+1)}$ where An and Bn are integers depending on a and n which satisfy $A_n^2 = a(a+1)B_n^2 + 1.$

$$(ii) In the case when n is odd, shower by considering $(\sqrt{a+1} + \sqrt{a})T_m$ where m is even, or otherwise, that Tn can be written in the form $C_n\sqrt{a+1} + D_n \sqrt{a}$ where Cn and Dn are integers depending on a and n and $(a+1)C_n^2 = aD_n^2 + 1$

$(iii) Deduce that for each n, Tn can be written as the sum of the square roots of two consecutive integers.$

lol do the random upside down question marks and exclamation marks mean anything?

largarithmic · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
lol do the random upside down question marks and exclamation marks mean anything?

I don't really know how tex works on this forum... I think Ive fixed it now though!

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
Here's one I just found (havent done it yet) but it looks decently cool:

$Let $T_n = (\sqrt{a+1} + \sqrt{a})^n$ where n is a positive integer and a is a given positive integer.$

$$(i) When n is even, prove by induction that $T_n$ can be written in the form $A_n + B_n\sqrt{a(a+1)}$ where An and Bn are integers depending on a and n which satisfy $A_n^2 = a(a+1)B_n^2 + 1.$

$$(ii) In the case when n is odd, shower by considering $(\sqrt{a+1} + \sqrt{a})T_m$ where m is even, or otherwise, that Tn can be written in the form $C_n\sqrt{a+1} + D_n \sqrt{a}$ where Cn and Dn are integers depending on a and n and $(a+1)C_n^2 = aD_n^2 + 1$

$(iii) Deduce that for each n, Tn can be written as the sum of the square roots of two consecutive integers.$

Haven't I seen this question before...?

largarithmic · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Haven't I seen this question before...?

You might've, but I kinda doubt it? Ive never seen it posted on these forums before, and it was written for a UK test so not in the "HSC circulation" per se.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
You might've, but I kinda doubt it? Ive never seen it posted on these forums before, and it was written for a UK test so not in the "HSC circulation" per se.

Was it from STEP?

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
You might've, but I kinda doubt it? Ive never seen it posted on these forums before, and it was written for a UK test so not in the "HSC circulation" per se.

This is such a Sydney Grammar type question lol.

I'll give it a try once I get back home from work.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
This is such a Sydney Grammar type question lol.

I'll give it a try once I get back home from work.

You go on BOS from work...? I assume you are using a smartphone.

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
You go on BOS from work...? I assume you are using a smartphone.

haha nah, using computers is part of our work, so I occasionally hop on Facebook or check my email etc.

Office job.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
haha nah, using computers is part of our work, so I occasionally hop on Facebook or check my email etc.

Office job.

Risky...I would never do that. Get back to work! :/

meilz92 · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

im serious, got like 95 in MX2 back in 09

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

tywebb said:
$\ \newline(a)\text{ If }y=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-n^2}\text{ for }0<x<1\text{ use the fact that }\pi^2=\sum_{n=1}^\infty{6\over n^2}\text{ to show }\newline\ \newline{\hskip0.25in} \pi^2+y'+y^2=0\newline\ \newline (b)\text{ Hence show }y=\pi\cot\pi x\text{ for }0<x<1\newline\ \newline{\hskip0.25in}\text{ and }\cot x=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-\pi^2n^2}\text{ for }0<x<\pi\newline\ \newline (c)\text{ Find }\int_0^x(\cot t-{1\over t})\ \!dt\text{ and hence show }\newline\ \newline{\hskip0.25in}\sin x=\textstyle x\prod_{n=1}^\infty(1-{x^2\over\pi^2n^2})\text{ for }0<x<\pi\newline\ \newline (d)\text{ Let }x=\textstyle{\pi\over k}\text{ to show }\prod_{n=1}^\infty{(kn)^2\over(kn)^2-1}={\pi/k\over\sin(\pi/k)}\text{ for }k>1\newline\ \newline (e)\text{ Let }k=2\text{ to prove Wallis' Product, }\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi\over2}$

This is an excellent question.

Combine it with the 2010 HSC Question 8 (Proof for Basel Problem) to make a super question.

largarithmic · Jan 6, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Was it from STEP?

yup, STEP III 2011. STEP is a pretty good source of questions, I know grammar occasionally takes questions from it

nightweaver066 · Jan 8, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
You guys will love this one! This one is very nice to do.

$\textup{Let} \;0<\theta< \pi$

$\textup{a) Solve the equation} \; z^2=\cos \theta + i\sin \theta$

$\textup{b) Let} \; u_{1} \; \textup{and} \; u_{2} \; \textup{be the roots of the equation} \; (u+1)^2=\cos \theta + i\sin \theta \; \\ \;\textup{where}\; \textup{Im}(u_{1})<0}$

$\; \textup{i) Find} \; u_{1} \textup{and} \; u_{2}}$

$\; \textup{ii) Prove that} \; \frac{u_{2}}{u_{1}}=-i\tan\frac{\theta}{4}$

$\textup{iii) Hence find all integers n for which} \; (\frac{u_{2}}{u_{1}})^n \; \textup{is a real number.}$

$\textup{iv) Prove that} \; (u_{1})^{12}=2^{12}\cos^{12}\frac{\theta}{4}(\cos 3\theta+i\sin 3\theta)$

$\textup{v) Hence, find all values of} \; \theta \; \textup{for which} \; (u_{1})^{12}-(u_{2})^{12} \; \textup{is a real number.}$

This took an unbelievable amount of time to type.. Better be correct..

$a) \; z^2 = cos\theta + isin\theta \newline z = \pm(cos\frac{\theta}{2} + isin\frac{\theta}{2})$

$b)i. \; (u + 1)^2 = cos\theta + isin\theta \newline u = \pm(cos\frac{\theta}{2} + isin\frac{\theta}{2}) - 1 \newline $As $ Im(u_1) < 0, \; u_1 = -cos\frac{\theta}{2} - isin\frac{\theta}{2} - 1$ and $ u_2 = cos\frac{\theta}{2} + isin\frac{\theta}{2} -1$

$ii)\; \frac{u_2}{u_1} = -\frac{cos\frac{\theta}{2} -1 + isin\frac{\theta}{2}}{cos\frac{\theta}{2} + 1 + isin\frac{\theta}{2}} \times \frac{cos\frac{\theta}{2} + 1 - isin\frac{\theta}{2}}{cos\frac{\theta}{2} + 1 - isin\frac{\theta}{2}} \newline = -\frac{cos^2\frac{\theta}{2} - 1 -isin\frac{\theta}{2}cos\frac{\theta}{2}+ isin\frac{\theta}{2} + isin\frac{\theta}{2}cos\frac{\theta}{2} + isin\frac{\theta}{2} + sin^2\frac{\theta}{2}}}}{cos^2\frac{\theta}{2} + 2cos\frac{\theta}{2} + 1 + sin^2\frac{\theta}{2}}$

$= -\frac{2isin\frac{\theta}{2}}{2 + 2cos\frac{\theta}{2}}$

$= -\frac{4isin\frac{\theta}{4}cos\frac{\theta}{4}}{4sin^2\frac{\theta}{4}}$

$= -itan\frac{\theta}{4}$

$iii) \; (\frac{u_2}{u_1})^n = (-itan\frac{\theta}{4})^n$

$= tan^n\frac{\theta}{4}(cos(\frac{\pi}{2}) - sin(\frac{\pi}{2})^n$

$= tan^n\frac{\theta}{4}(cos\frac{n\pi}{2} - sin\frac{n\pi}{2})$

$$For it to be a real number, $\; Im(\frac{u_2}{u_1})^n = 0$

$sin\frac{n\pi}{2} = 0$

$\frac{n\pi}{2} = k\pi$ , where k = 0, 1, 2, 3...$

$n = 2k$

$As the range is $ 0 < \theta < \pi, \; $no values of theta satisfy it$

$iv) u_1 = -cos\frac{\theta}{2} - isin\frac{\theta}{2} - 1$

$= - (2cos^2\frac{\theta}{4} + 2isin\frac{\theta}{4}cos\frac{\theta}{4})$

$= -2cos\frac{\theta}{4}(cos\frac{\theta}{4} + isin\frac{\theta}{4})$

$(u_1)^{12} = (-2)^{12}cos^{12}\frac{\theta}{4}(cos\frac{\theta}{4} + isin\frac{\theta}{4})^{12}$

$= 2^{12}cos^{12}\frac{\theta}{4}(cos3\theta + isin3\theta)$

$$As $ \frac{u_2}{u_1} = -itan\frac{\theta}{4}$

$u_2 = -itan\frac{\theta}{4} \times u_1$

$\therefore u_2 = -2isin\frac{\theta}{4}(cos\frac{\theta}{4} + isin\frac{\theta}{4})$

$= -2sin\frac{\theta}{4}(cos(\frac{\pi}{2} + \frac{\theta}{4}) + isin(\frac{\pi}{2} + \frac{\theta}{4})$

$(u_2)^{12} = 2^{12}sin^{12}\frac{\theta}{4}(cos(6\pi + 3\theta) + isin(6\pi + 3\theta))$

$= 2^{12}sin^{12}\frac{\theta}{4}(cos3\theta + isin3\theta$

$(u_1)^12 - (u_2)^12 = 2^{12}(cos3\theta + isin3\theta)(cos^{12}\frac{\theta}{4} - sin^{12}\frac{\theta}{4})$

$$For $ (u_1)^{12} - (u_2)^{12}$ to be a real number, $ Im((u_1)^12 - (u_2)^12) = 0$

$\therefore sin3\theta = 0$

$3\theta = n\pi$

$\theta = \frac{n\pi}{3}$ where n = 0, 1, 2, 3...$

$\theta = \frac{\pi}{3}, \; \frac{2\pi}{3}$

Carrotsticks · Jan 8, 2012

Re: 2012 HSC MX2 Marathon

Nightweaver, a bit of a tip.

Instead of typing or writing "Where k = 0,1,2,3,...", simply write down:

$k\in \mathbb{N}$

nightweaver066 · Jan 8, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Nightweaver, a bit of a tip.

Instead of typing or writing "Where k = 0,1,2,3,...", simply write down:

$k\in \mathbb{N}$

Thanks, i don't know how to use those symbols in latex as i type from what i know without an editor

SpiralFlex · Jan 8, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
This took an unbelievable amount of time to type.. Better be correct..

$a) \; z^2 = cos\theta + isin\theta \newline z = \pm(cos\frac{\theta}{2} + isin\frac{\theta}{2})$

$b)i. \; (u + 1)^2 = cos\theta + isin\theta \newline u = \pm(cos\frac{\theta}{2} + isin\frac{\theta}{2}) - 1 \newline $As $ Im(u_1) < 0, \; u_1 = -cos\frac{\theta}{2} - isin\frac{\theta}{2} - 1$ and $ u_2 = cos\frac{\theta}{2} + isin\frac{\theta}{2} -1$

$ii)\; \frac{u_2}{u_1} = -\frac{cos\frac{\theta}{2} -1 + isin\frac{\theta}{2}}{cos\frac{\theta}{2} + 1 + isin\frac{\theta}{2}} \times \frac{cos\frac{\theta}{2} + 1 - isin\frac{\theta}{2}}{cos\frac{\theta}{2} + 1 - isin\frac{\theta}{2}} \newline = -\frac{cos^2\frac{\theta}{2} - 1 -isin\frac{\theta}{2}cos\frac{\theta}{2}+ isin\frac{\theta}{2} + isin\frac{\theta}{2}cos\frac{\theta}{2} + isin\frac{\theta}{2} + sin^2\frac{\theta}{2}}}}{cos^2\frac{\theta}{2} + 2cos\frac{\theta}{2} + 1 + sin^2\frac{\theta}{2}}$

$= -\frac{2isin\frac{\theta}{2}}{2 + 2cos\frac{\theta}{2}}$

$= -\frac{4isin\frac{\theta}{4}cos\frac{\theta}{4}}{4sin^2\frac{\theta}{4}}$

$= -itan\frac{\theta}{4}$

$iii) \; (\frac{u_2}{u_1})^n = (-itan\frac{\theta}{4})^n$

$= tan^n\frac{\theta}{4}(cos(\frac{\pi}{2}) - sin(\frac{\pi}{2})^n$

$= tan^n\frac{\theta}{4}(cos\frac{n\pi}{2} - sin\frac{n\pi}{2})$

$$For it to be a real number, $\; Im(\frac{u_2}{u_1})^n = 0$

$sin\frac{n\pi}{2} = 0$

$\frac{n\pi}{2} = k\pi$ , where k = 0, 1, 2, 3...$

$n = 2k$

$As the range is $ 0 < \theta < \pi, \; $no values of theta satisfy it$

$iv) u_1 = -cos\frac{\theta}{2} - isin\frac{\theta}{2} - 1$

$= - (2cos^2\frac{\theta}{4} + 2isin\frac{\theta}{4}cos\frac{\theta}{4})$

$= -2cos\frac{\theta}{4}(cos\frac{\theta}{4} + isin\frac{\theta}{4})$

$(u_1)^{12} = (-2)^{12}cos^{12}\frac{\theta}{4}(cos\frac{\theta}{4} + isin\frac{\theta}{4})^{12}$

$= 2^{12}cos^{12}\frac{\theta}{4}(cos3\theta + isin3\theta)$

$$As $ \frac{u_2}{u_1} = -itan\frac{\theta}{4}$

$u_2 = -itan\frac{\theta}{4} \times u_1$

$\therefore u_2 = -2isin\frac{\theta}{4}(cos\frac{\theta}{4} + isin\frac{\theta}{4})$

$= -2sin\frac{\theta}{4}(cos(\frac{\pi}{2} + \frac{\theta}{4}) + isin(\frac{\pi}{2} + \frac{\theta}{4})$

$(u_2)^{12} = 2^{12}sin^{12}\frac{\theta}{4}(cos(6\pi + 3\theta) + isin(6\pi + 3\theta))$

$= 2^{12}sin^{12}\frac{\theta}{4}(cos3\theta + isin3\theta$

$(u_1)^12 - (u_2)^12 = 2^{12}(cos3\theta + isin3\theta)(cos^{12}\frac{\theta}{4} - sin^{12}\frac{\theta}{4})$

$$For $ (u_1)^{12} - (u_2)^{12}$ to be a real number, $ Im((u_1)^12 - (u_2)^12) = 0$

$\therefore sin3\theta = 0$

$3\theta = n\pi$

$\theta = \frac{n\pi}{3}$ where n = 0, 1, 2, 3...$

$\theta = \frac{\pi}{3}, \; \frac{2\pi}{3}$

For b ii) No need to rationalise.

Edit: Didn't see your last sentence

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HSC 2012 MX2 Marathon (archive) (1 Viewer)

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