HSC 2015 MX1 Marathon (archive) (1 Viewer)

leehuan · Jul 27, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Use the substitution $u=\frac{1}{\sqrt{3}}\tan{x}$ to evaluate $\int^{\frac{\pi}{4}}_0 \frac{\mathrm{d}x}{3-2\sin^2{x}}$

Multiply through by sec^2(x), convert the bottom sec^2(x) to 1+tan^2(x) and then the substitution can be applied immediately.

I'll try the binomial question after English

leehuan · Jul 28, 2015

Re: HSC 2015 3U Marathon

$\\ $(i)$ { \left( 1+x \right) }^{ 2 }=1+x\left( 2+x \right) \\ { \left( 1+x \right) }^{ 4n }={ \left( 1+x\left( 2+x \right) \right) }^{ 2n }\\ { \left( 1+x \right) }^{ 4n }=\sum _{ k=0 }^{ 2n }{ \left( \begin{matrix} 2n \\ k \end{matrix} \right) { x }^{ k }{ \left( 2+x \right) }^{ k } } \\ $Part (i) falls out directly from the coefficient of ${ x }^{ 2n }$ on the L.H.S. being \left( \begin{matrix} 4n \\ 2n \end{matrix} \right)$

$\\ $(ii)$ \sum _{ k=0 }^{ 2n }{ \left( \begin{matrix} 2n \\ k \end{matrix} \right) { x }^{ k }{ \left( 2+x \right) }^{ k } } \\ =\left( \begin{matrix} 2n \\ 0 \end{matrix} \right) +\dots +\left( \begin{matrix} 2n \\ n \end{matrix} \right) { x }^{ n }{ \left( 2+x \right) }^{ n }+\left( \begin{matrix} 2n \\ n+1 \end{matrix} \right) { x }^{ n+1 }{ \left( 2+x \right) }^{ n+1 }+\left( \begin{matrix} 2n \\ n+2 \end{matrix} \right) { x }^{ n+2 }{ \left( 2+x \right) }^{ n+2 }+\dots +\left( \begin{matrix} 2n \\ 2n \end{matrix} \right) { x }^{ 2n }{ \left( 2+x \right) }^{ 2n }\\ $Now, the terms involving ${ x }^{ 2n }$ can be found by taking the product of ${ x }^{ n },{ x }^{ n+1 },{ x }^{ n+2 },\dots, { x }^{ 2n }\dots $ with the term involving $ { x }^{ n },{ x }^{ n-1 },{ x }^{ n-2 },\dots , [$Constant term$] $ in $T_{ n },{ T }_{ n+1 },{ T }_{ n+2 },\dots ,{ T }_{ n } \\$

$\\ $Hence the terms involving ${ x }^{ n },{ x }^{ n-1 },{ x }^{ n-2 },\dots , [$Constant term$] $ must be \\ extracted from ${ \left( 2+x \right) }^{ k } \\ $Hence the co-efficient of $ { x }^{ 2n } $ must be: $ \\ \left( \begin{matrix} 2n \\ n \end{matrix} \right) \left( \begin{matrix} n \\ n \end{matrix} \right) +2\left( \begin{matrix} 2n \\ n+1 \end{matrix} \right) \left( \begin{matrix} n \\ n-1 \end{matrix} \right) +{ 2 }^{ 2 }\left( \begin{matrix} 2n \\ n+2 \end{matrix} \right) \left( \begin{matrix} n \\ n-2 \end{matrix} \right) +\dots +{ 2 }^{ n }\left( \begin{matrix} 2n \\ 2n \end{matrix} \right) \left( \begin{matrix} n \\ 0 \end{matrix} \right)$

$\\ $by consideration of the general term of the expansion of ${ \left( 2+x \right) }^{ k }$ \\ The result falls out from equating with part (i)$

Apologies for the rather poor wording.

Sy123 · Jul 28, 2015

Re: HSC 2015 3U Marathon

$\\ $i) Explain why$ \ (x^n-1) = (x-1)(1+ x + x^2 + \dots + x^{n-1}) \ $by expanding$$

$\\ $ii) Hence show that$ \ 1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^{n}}{1-x}$

$\\ $iii) Hence show that$ \ 1 + x + x^2 + \dots = \frac{1}{1-x} \ $for$ \ |x| < 1$

davidgoes4wce · Jul 29, 2015

Re: HSC 2015 3U Marathon

Six equal circular discs are placed so that their centres are on the circumference of a given circle and each disc touches its two neighbours. If the radius of the given circle is r, find:

a) The radius of a seventh disc that touches all the other discs.

The answers were: r/2 and 3r/2

I didn't quite understand where the '3r/2' came from.

davidgoes4wce · Jul 29, 2015

Re: HSC 2015 3U Marathon

This might be a better photo

rand_althor · Jul 29, 2015

Re: HSC 2015 3U Marathon

The two circles which touch all other discs are the blue ones.

I'm going to assume you've found the inner blue circle's radius to be r/2.

Radius of outer blue circle:
r = DE + DF
= r + r/2
= 3r/2

DatAtarLyfe · Jul 29, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
The two circles which touch all other discs are the blue ones.

I'm going to assume you've found the inner blue circle's radius to be r/2.

Radius of outer blue circle:
r = DE + DF
= r + r/2
= 3r/2

nice, didn't even see that

Drsoccerball · Jul 29, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
$\\ $i) Explain why$ \ (x^n-1) = (x-1)(1+ x + x^2 + \dots + x^{n-1}) \ $by expanding$$

$\\ $ii) Hence show that$ \ 1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^{n}}{1-x}$

$\\ $iii) Hence show that$ \ 1 + x + x^2 + \dots = \frac{1}{1-x} \ $for$ \ |x| < 1$

Should be in 2u marathon

davidgoes4wce · Aug 2, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
The two circles which touch all other discs are the blue ones.

I'm going to assume you've found the inner blue circle's radius to be r/2.

Radius of outer blue circle:
r = DE + DF
= r + r/2
= 3r/2

I initially got stumped on this as I didn't know that there was an 8th circle (the big circle as well). But this is a part b) to this question. What is the length of the outside perimeter of the given figure.

rand_althor · Aug 2, 2015

Re: HSC 2015 3U Marathon

$\Delta AA_1A_5$ is an equilateral triangle with side length $r$ and angles $60^{\circ}$ each. $\Delta A A_5 A_6$ has identical properties. This can be shown by proving the triangles are congruent (SSS). \\$\angle A_1 A_5 A_6=60^{\circ}+60^{\circ}=120^{\circ}. \\$Reflex $\angle A_1A_5A_6=360^{\circ}-120^{\circ}=240^{\circ}=\frac{4\pi}{3}. \\$Length of major arc BD$=r\theta=\frac{r}{2}\times\frac{4\pi}{3}=\frac{2\pi r}{3}$ units$ \\$Outside perimeter $=6\times$Length of major arc BD$=6\times\frac{2r\pi}{3}=4\pi r $ units$ \\$

davidgoes4wce · Aug 2, 2015

Re: HSC 2015 3U Marathon

It's cleared it up. Circle geometry is not my strength. I will have to go through another text book as I have learnt this fairly rushed.

Going through Fitzpatrick book and I think there were like 21 Theorems to remember.

davidgoes4wce · Aug 2, 2015

Re: HSC 2015 3U Marathon

Think if you learning Circle Geometry you want to use a book that does it in colour as well. You can see things a bit more clearer when you read the rules and theorems and visualise the diagrams in place.

davidgoes4wce · Aug 2, 2015

Re: HSC 2015 3U Marathon

im going to Coles to get some breakfast because I'm starving be back soon to talk more maths.

pomsky · Aug 2, 2015

Re: HSC 2015 3U Marathon

$$ Q1 Find $\int \frac{dx}{1+\sqrt{x+1}} \\$Q2 Evaluate $\int_{1}^{5} \frac{x}{(2x-1)^{\frac{3}{2} }}dx $ using the substitution $ x=\frac{1}{2}(u^2+1) \\ $Q3 Solve the differential equation $ (1+x^2) \frac{dy}{dx} + xe^{-y} =0 $, given that the curve y= f(x) passes through (3,0).$

InteGrand · Aug 2, 2015

Re: HSC 2015 3U Marathon

pomsky said:
$$ Q1 Find $\int \frac{dx}{1+\sqrt{x+1}} \\$Q2 Evaluate $\int_{1}^{5} \frac{x}{(2x-1)^{\frac{3}{2} }}dx $ using the substitution $ x=\frac{1}{2}(u^2+1) \\ $Q3 Solve the differential equation $ (1+x^2) \frac{dy}{dx} + xe^{-y} =0 $, given that the curve y= f(x) passes through (3,0).$

$1. $I = \int \frac{1}{1+\sqrt{x+1}}\text{ d}x$. Let $u=1+\sqrt{x+1}$, so $\sqrt{x+1} = u-1\Rightarrow x = (u-1)^2 -1$. So $\mathrm{d}x = 2(u-1)\text{ d}u$. Therefore,$

$I=\int \frac{2(u-1)\text{ d}u}{u}$

$=2\int \left(1-\frac{1}{u} \right)\text{ d}u$

$=2(u-\ln u)+c$

$=2\left[1+\sqrt{x+1}-\ln \left( 1+\sqrt{x+1} \right)\right]+c$

$=2+2\left[\sqrt{x+1}-\ln \left( 1+\sqrt{x+1} \right)\right]+c$

$$=2\left[\sqrt{x+1}-\ln \left( 1+\sqrt{x+1} \right)\right]+C$, where $C=c+2$ is an arbitrary constant.$

InteGrand · Aug 2, 2015

Re: HSC 2015 3U Marathon

pomsky said:
$$ Q1 Find $\int \frac{dx}{1+\sqrt{x+1}} \\$Q2 Evaluate $\int_{1}^{5} \frac{x}{(2x-1)^{\frac{3}{2} }}dx $ using the substitution $ x=\frac{1}{2}(u^2+1) \\ $Q3 Solve the differential equation $ (1+x^2) \frac{dy}{dx} + xe^{-y} =0 $, given that the curve y= f(x) passes through (3,0).$

$2. $x=\frac{1}{2}(u^2 + 1)\Rightarrow u^2 + 1 = 2x\Rightarrow u = \pm \sqrt{2x-1}$. In substitutions for integrals, we need the substitution to be one-to-one, so we need to choose a branch here. We shall choose $u=\sqrt{2x-1}$, as it is more convenient.$

$So $\mathrm{d}u = \frac{2}{2\sqrt{2x-1}}\text{ d}x\Rightarrow \mathrm{d}u = \frac{1}{\sqrt{2x-1}}\text{ d}x$. Also, when $x=1,u=\sqrt{2\times 1 -1}=1$, and when $x=5,u=\sqrt{2\times 5 -1}=3$. So,$

$\int_{1}^{5} \frac{x}{(2x-1)^{\frac{3}{2} }}\text{ d}x=\int_{1}^{5} \frac{x}{2x-1}\times\frac{1}{\sqrt{2x-1}}\text{ d}x$

$$=\int _1 ^3 \frac{\frac{1}{2}(u^2 + 1)}{u^2}\text{ d}u$ (implementing the substitutions (note $2x-1=u^2$))$

$=\frac{1}{2}\int _1 ^3 \left(1+\frac{1}{u^2} \right)\text{ d}u$

$=\frac{1}{2} \left[u-\frac{1}{u} \right]_1 ^3$

$=\frac{1}{2} \left[\left(3-\frac{1}{3}\right)- \left(1-\frac{1}{1}\right)\right]$

$=\frac{1}{2} \times 2\tfrac{2}{3}$

$$=1\tfrac{1}{3}$.$

InteGrand · Aug 2, 2015

Re: HSC 2015 3U Marathon

pomsky said:
$$ Q1 Find $\int \frac{dx}{1+\sqrt{x+1}} \\$Q2 Evaluate $\int_{1}^{5} \frac{x}{(2x-1)^{\frac{3}{2} }}dx $ using the substitution $ x=\frac{1}{2}(u^2+1) \\ $Q3 Solve the differential equation $ (1+x^2) \frac{dy}{dx} + xe^{-y} =0 $, given that the curve y= f(x) passes through (3,0).$

$$(1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} + x\mathrm{e}^{-y}=0$, $y(3)=0$.$

$\therefore (1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = -x\mathrm{e}^{-y}$

$$\Rightarrow \frac{1}{1+x^2}\frac{\mathrm{d}x}{\mathrm{d}y} = -\frac{\mathrm{e}^y}{x}$ (inverting both sides)$

$$\Rightarrow \frac{x}{1+x^2}\frac{\mathrm{d}x}{\mathrm{d}y} = -\mathrm{e}^y$ (rearranging)$

$$\Rightarrow \int \frac{x}{1+x^2}\frac{\mathrm{d}x}{\mathrm{d}y} \text{ d}y = \int -\mathrm{e}^y\text{ d}y$ (integrating both sides with respect to $y$)$

$\Rightarrow \int \frac{x}{1+x^2}\text{ d}x = -\mathrm{e}^y +c$

$$\Rightarrow \frac{1}{2}\ln \left( 1+x^2\right) = -\mathrm{e}^y +c$ (we don't need to introduce a constant $c_2$ on the L.H.S., because we could just bring it to the R.H.S. and get an arbitrary constant and have $c-c_2$, which is itself an arbitrary constant, so it is enough just to have $+c$ on the R.H.S.).$

$Since $y=0$ when $x=3$, we have $\frac{1}{2}\ln \left( 1+3^2\right) = -\mathrm{e}^0 +c$$

$$\Rightarrrow \frac{1}{2}\ln 10 = -1+c\Rightarrow c = 1+\frac{1}{2}\ln 10$.$

$Now, $\frac{1}{2}\ln \left( 1+x^2\right) = -\mathrm{e}^y +c \Rightarrow \mathrm{e}^y =-\frac{1}{2}\ln \left( 1+x^2\right) +c $$

$\Rightarrow y = \ln \left[ -\frac{1}{2}\ln \left( 1+x^2\right) +c \right]$

$\Rightarrow y = \ln \left[ -\frac{1}{2}\ln \left( 1+x^2\right) +1+\frac{1}{2}\ln 10 \right]$

$\Rightarrow y = \ln \left\{ \frac{1}{2}\left[\ln 10 -\ln \left( 1+x^2\right)\right] +1 \right\}$

$\Rightarrow y = \ln \left\[ \frac{1}{2}\ln \left( \frac{10}{1+x^2}\right) +1 \right]$

rand_althor · Aug 2, 2015

Re: HSC 2015 3U Marathon

$\begin{align*}\text{2. }&\int^5_1 \frac{x}{(2x-1)^{\frac{3}{2}}}\, \mathrm{d}x \\\text{Let }x&=\frac{1}{2}(u^2+1)\to u^2=2x-1 \\\frac{\mathrm{d}x}{\mathrm{d}u}&=u\to\mathrm{d}x=u\mathrm{d}u \\\text{When }x&=5\to u=3,\, x=1\to u=1\\\int^3_1\frac{\frac{1}{2}(u^2+1)}{(u^2)^\frac{3}{2}}\,u\mathrm{d}u&=\frac{1}{2}\int^3_1\frac{u^3+u}{u^3} \, \mathrm{d}u \\&=\frac{1}{2} \int^3_1 1+\frac{1}{u^2} \, \mathrm{d}u \\&=\frac{1}{2} \left [u-\frac{1}{u} \right ]^3_1 \\&=\frac{1}{2} \left [(3-\frac{1}{3})-(1-1) \right ] \\&=\frac{1}{2} \times \frac{8}{3} \\&=\frac{4}{3}\end{align*}$

davidgoes4wce · Aug 2, 2015

Re: HSC 2015 3U Marathon

Find the cartesian equation of the curves whose parametric equations are:

rand_althor · Aug 2, 2015

Re: HSC 2015 3U Marathon

$$Let $t=\tan{\frac{\theta}{2}}. \therefore \sin{\theta}=\frac{2t}{1+t^2}, \cos{\theta}=\frac{1-t^2}{1+t^2}. $ We can write the parametric forms as$: x=\sin{\theta}, y=\cos{\theta}. $ Since $\sin^2{\theta}+\cos^2{\theta}=1$, the cartesian form is$ :x^2+y^2=1.$

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