# MX2 Marathon (1 Viewer)

#### iStudent

##### Well-Known Member
Re: HSC 2018 MX2 Marathon

RPQ and PQB are supplementary (parallel lines)

PQB = QSB (angle in alt segment)

hence RPQ + PSB = 180

since opposite angles are supplementary, RSQP is cyclic?

been 4 years hope I am right

#### pikachu975

Re: HSC 2018 MX2 Marathon

Cant seem to figure out how to do this question:
https://imgur.com/a/jRaWrmq
BSQ = PQB (alternate angle theorem)
PQB = 180-RPQ (co-interior angles in a parallelogram are supplementary)
BSQ + RPQ = 180

Hence opposite angles are supplementary so RSQP is a cyclic quad

Hope this reasoning is legit, forgot maths

EDIT: Nvm too slow

#### Drongoski

##### Well-Known Member
Re: HSC 2018 MX2 Marathon

It is easy to prove RSQP is cyclic; so that this can be easily shown in many ways.

e.g.

Let angle AQS = @; .: angle QBS = @; .: angle PRS = @ (PR // QB)

.: angle AQS = angle PRS

.: external angle AQS of quadrilateral RSQP = int opp angle PRS

.: RSQP must be cyclic

QED

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#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent For n = 0, 1, 2... let I_n = \int_{0}^{\pi/4} tan^{n}\theta d\theta$
$\bg_white \noindent a) Show that I_1 = \frac{1}{2}ln2$
$\bg_white \noindent b) Show that, for n \geq 2, I_n + I_{n-2} = \frac{1}{n-1}$

$\bg_white \noindent c) For n \geq 2, explain why I_n < I_{n-2} and deduce that$

$\bg_white \noindent \frac{1}{2(n+1)} < I_n < \frac{1}{2(n-1)}$

$\bg_white \noindent d) Use the reduction formula in part b) to find I_5, and hence deduce that$

$\bg_white \frac{2}{3} < ln2 < \frac{3}{4}$

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

a)
\bg_white \begin{aligned} I_1 &= \int^{\frac\pi4}_0 \tan \theta \,d\theta \\ &= \left[\phantom{\frac\,\,}-\log \cos\theta\right]^{\frac\pi4}_0\\&=\log 1 - \log \frac{\sqrt2}{2} \\ &= \log \frac{2}{\sqrt 2} \\ &= \frac12\log 2 \end{aligned}

b)

\bg_white \begin{aligned} I_n + I_{n+2} &= \int_0^{\frac\pi4} \tan^n\theta + \tan^{n-2}\theta\,d\theta \\ &=\int_0^{\frac\pi4} \tan^{n-2}\theta(1+\tan^2\theta)\,d\theta \\ &=\int^{\frac\pi4}_{0} \left(\tan\theta\right)^{n-2} \sec^2\theta\,d\theta \\ &= \left[\frac{1}{n-1} \left(\tan \theta\right)^{n-1}\right]^{\frac\pi4}_0\\&= \frac{1}{n-1}\end{aligned}

c)

For $\bg_white 0 \leq x \leq \pi/4$, $\bg_white 0\leq \tan \theta\leq 1$.

If $\bg_white 0\leq \tan \theta\leq 1$, then clearly $\bg_white \left(\tan\theta\right)^n >\left(\tan\theta\right)^{n+1}$.

Therefore, $\bg_white \left(\tan\theta\right)^n <\left(\tan\theta\right)^{n-2}\implies I_n < I_{n-2}$.

From b) $\bg_white I_{n-2} = \frac{1}{n-1} - I_n$. Since $\bg_white I_n < I_{n-2}$,

\bg_white \begin{aligned} I_{n} &< \frac{1}{n-1} - I_n \\ I_n&< \frac{1}{2(n-1)} \end{aligned}

Replacing $\bg_white n$ with $\bg_white n +2$, we get $\bg_white I_{n} = \frac{1}{n+1} - I_{n+2} \implies I_{n+2} = \frac{1}{n+1} - I_{n}$ with $\bg_white I_{n+2} < I_{n}$ and so

\bg_white \begin{aligned} \frac{1}{n+1}-I_{n} &< I_n \\\frac{1}{2(n+1)} &< I_n \end{aligned}

Giving, as required,

$\bg_white \frac{1}{2(n+1)} < I_n < \frac{1}{2(n-1)}$

d)

From b),

$\bg_white I_n = \frac{1}{n-1} -I_{n-2}$.

$\bg_white I_1 = \frac 12 \log 2$

$\bg_white I_3 = \frac12 -\frac 12 \log 2$

$\bg_white I_5 = \frac 14 - I_3 = \frac 12 \log 2 - \frac 14$

From c),

\bg_white \begin{aligned} \frac{1}{12} < \,&I_5 < \frac18 \\\phantom{0}\\ \frac{1}{3} <\frac12&\log2 < \frac38 \\\phantom{0}\\ \frac23 <\,&\log2<\frac34\end{aligned}

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

$\bg_white Let\, I_n = \int_0^{\frac12} \frac{(\tan^{-1}2x)^n}{4x^2+1}\,dx\,for\,n\in\mathbb{Z}^+.$

$\bg_white i) Show that\, I_n =\frac{1}{2(n+1)}\left(\frac\pi4\right)^{n+1}$

$\bg_white ii) Hence show that\, I_0 \times I_1 \times I_2 \times ... \times I_n = \frac{1}{2^{2n}(2n)!}\left(\frac\pi4\right)^{2n^2+n}$

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

yoyoyo
there was a nice question i found a while back and thought i would post it here:

given that e^x can be written as a sum of an odd and even function, find the two functions.
Is the intended answer the Taylor series for $\bg_white e^x$?

Edit: nevermind, I found some others

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#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Marathon

Is the intended answer the Taylor series for $\bg_white e^x$?

Edit: nevermind, I found some others

its Hyperbolic function of sin and cos
so e^x = sinhx+coshx
Its legit like deriving cosx and sinx in complex field

The taylor expansion is just an approximate for e^x for values less than 1 or close to 0 but can get quite close. You would just factorise the even and odd parts of the function. However, that does not mean you cant use it, I tried it with Maclaurin series and it seems to look solid.

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#### InteGrand

##### Well-Known Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent By the way, you can do this with any function, not just e^x (provided that it makes sense to input both x and -x into your function). We can write a given f(x) as a sum of an \color{red}{even }\color{black} function and an \color{blue}{odd }\color{black} function as \boxed{f(x) = \color{red}\frac{1}{2}(f(x) + f(-x))\color{black} + \color{blue}\frac{1}{2}(f(x) - f(-x))\color{black}}, and in fact this is the \emph{only} way to make such a decomposition.$

#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Marathon

Yeah it clicked in when i was thinking of expressing these interms of Complex version of cos and sin and basically simultaneous

#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Marathon

More qtns

#### Sy123

##### This too shall pass
Re: HSC 2018 MX2 Marathon

$\bg_white \\ i) A point P moves on the rectangular hyperbola x^2 - y^2 = a^2. Let M be the point on the tangent to the hyperbola at P be the point closest to the origin. \\\\Show that this curve is given by the locus \ (x^2 + y^2)^2 = a^2(x^2 - y^2). \\\\ This curve is called the \textit{Lemniscate of Bernoulli}$

$\bg_white \\ ii) Show that if F_1\left(\frac{a}{\sqrt{2}}, 0 \right) and F_2\left(\frac{-a}{\sqrt{2}}, 0\right) are fixed points in the plane. Then for any point Q on the Lemniscate of Bernoulli, then |QF_1| \cdot |QF_2| = 1$

#### InteGrand

##### Well-Known Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent For each real number \sigma, let f(\sigma) be the maximum value of |\log_{2}(1 + x) - x- \sigma| for 0\leq x \leq 1. Find the value of \sigma that minimises f(\sigma).$

##### -insert title here-
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent For each real number \sigma, let f(\sigma) be the maximum value of |\log_{2}(1 + x) - x- \sigma| for 0\leq x \leq 1. Find the value of \sigma that minimises f(\sigma).$
tsk tsk trying to introduce multivariable optimisation in a sidestepping manner

#### InteGrand

##### Well-Known Member
Re: HSC 2018 MX2 Marathon

tsk tsk trying to introduce multivariable optimisation in a sidestepping manner
Haha, that wasn't the motivation at all.

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

$\bg_white \\ i) A point P moves on the rectangular hyperbola x^2 - y^2 = a^2. Let M be the point on the tangent to the hyperbola at P be the point closest to the origin. \\\\Show that this curve is given by the locus \ (x^2 + y^2)^2 = a^2(x^2 - y^2). \\\\ This curve is called the \textit{Lemniscate of Bernoulli}$

$\bg_white \\ ii) Show that if F_1\left(\frac{a}{\sqrt{2}}, 0 \right) and F_2\left(\frac{-a}{\sqrt{2}}, 0\right) are fixed points in the plane. Then for any point Q on the Lemniscate of Bernoulli, then |QF_1| \cdot |QF_2| = 1$
The equation of the tangent at the hyperbola is given by

$\bg_white y-a\tan\theta = \csc \theta(x-a\sec\theta)$

Because the shortest distance between a point and a line is the perpendicular distance, $\bg_white M$ is the intersection between the tangent line and the line perpendicular to the tangent which also passes through the origin.

That line is

$\bg_white y = -x \sin \theta$

Solving these two equations gives

$\bg_white x = \frac{a \cos \theta}{1+\sin^2\theta} \qquad y=- \frac{a \sin\theta \cos \theta}{1+\sin^2\theta}$

Now,

$\bg_white a^2(x^2-y^2)=a^4\times \frac{\cos^2 \theta - \sin^2\theta \cos^2\theta}{(1+\sin^2\theta)^2}=\frac{a^4\cos^4 \theta}{(1+\sin^2\theta)^2}$

$\bg_white (x^2+y^2)^2 = \frac{a^4}{(1+\sin^2\theta)^4}\times(\cos^2\theta + \sin^2\theta\cos^2\theta)^2=\frac{a^4\cos^4 \theta}{(1+\sin^2\theta)^2}$

Hence the locus of $\bg_white M$ is given by the equation $\bg_white (x^2+y^2)^2=a^2(x^2-y^2)$.

$\bg_white |QF_1| =|a|\sqrt{\left(\frac{\cos^2\theta}{1+\sin^2\theta}-\frac{1}{\sqrt{2}}\right)^2+\left(\frac{\sin\theta\cos\theta}{1+\sin^2\theta}\right)^2}$

$\bg_white = |a| \sqrt{\frac{\cos^2\theta+ \sin^2\theta\cos^2\theta}{(1+\sin^2\theta)^2} + \frac{-\sqrt 2 \cos \theta + \frac 12 + \frac 12 \sin^2 \theta}{1+\sin^2\theta}}$

$\bg_white = |a| \sqrt{\frac{(\cos \theta - \frac{1}{\sqrt{2}})^2 + \frac 12 \sin^2\theta}{1+\sin^2\theta}}$

And similarly,

$\bg_white |QF_2| = |a| \sqrt{\frac{(\cos \theta + \frac{1}{\sqrt{2}})^2 + \frac 12 \sin^2\theta}{1+\sin^2\theta}}$

So,

$\bg_white |QF_1|\cdot|QF_2|=\frac{a^2}{|1+\sin^2\theta |}\sqrt{\left(\cos^2\theta-\frac 1 2\right)^2+\frac 12 \sin^2\theta\left(\left(\cos\theta+\frac{1}{\sqrt{2}}\right)^2+\left(\cos\theta-\frac{1}{\sqrt{2}}\right)^2\right)+\frac 14 \sin^4\theta }$

$\bg_white =\frac{a^2}{|1+\sin^2\theta|}\sqrt{\cos^4\theta-\cos^2\theta +\sin^2\theta \cos^2\theta+\frac 14 + \frac 12 \sin^2\theta+\frac 14 \sin^4\theta }$

$\bg_white =\frac{a^2}{|1+\sin^2\theta|}\sqrt{\cos^4\theta-\cos^2\theta+(1-\cos^2\theta)\cos^2\theta+\frac 14 \left(1+\sin^2\theta\right)^2}$

$\bg_white =\frac{a^2}{|1+\sin^2\theta|}\sqrt{\frac 14 \left(1+\sin^2\theta\right)^2}$

$\bg_white = \frac{a^2}{2}$

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#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent Let T_n be a positive sequence such that T_n=2\sqrt{T_1+T_2+...+T_n}-1. Show by mathematical induction that T_1+T_2+...+T_n=n^2.$

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#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent Let T_n be a sequence such that T_1=1, T_2=2 and T_{n+2}=T_{n+1}+T_n. Show by mathematical induction that T_nT_{n+2}-(T_{n+1})^2=\left(-1\right)^n.$

#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent Let T_n be a positive sequence such that T_n=2\sqrt{T_1+T_2+...+T_n}-1. Show by mathematical induction that T_1+T_2+...+T_n=n^2.$
This one is not that too tough, right?

$\bg_white T_1=2\sqrt{T_1}-1$
$\bg_white Let x=\sqrt{T_1}$
$\bg_white x^2=2x-1$
$\bg_white (x-1)^2=0$
$\bg_white x=1$
$\bg_white T_1=x^2=1^2$
$\bg_white The statement is true for n=1.$

$\bg_white Assume the statement is true for n=k.$
$\bg_white T_{k+1}=2\sqrt{T_1+T_2+...+T_k+T_{k+1}}-1$
$\bg_white Let y=\sqrt{T_1+T_2+...+T_k+T_{k+1}}$
$\bg_white y^2-k^2=2y-1$
$\bg_white (y-1)^2=k^2$
$\bg_white y=k+1 or y=-k+1 (rejected because the sequence is positive)$
$\bg_white T_1+T_2+...+T_k+T_{k+1}=y^2=(k+1)^2$
$\bg_white The statement is true for n=k+1.$

$\bg_white \\By mathematical induction, the statement is true for all positive integers n.$

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#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Marathon

This one is not that too tough, right?

$\bg_white T_1=2\sqrt{T_1}-1$
$\bg_white Let x=\sqrt{T_1}$
$\bg_white x^2=2x-1$
$\bg_white (x-1)^2=0$
$\bg_white x=1$
$\bg_white T_1=x^2=1^2$
$\bg_white The statement is true for n=1.$

$\bg_white Assume the statement is true for n=k.$
$\bg_white T_{k+1}=2\sqrt{T_1+T_2+...+T_k+T_{k+1}}-1$
$\bg_white Let y=\sqrt{T_1+T_2+...+T_k+T_{k+1}}$
$\bg_white y^2-k^2=2y-1$
$\bg_white (y-1)^2=k^2$
$\bg_white y=k+1 or y=-k+1 (rejected because the sequence is positive)$
$\bg_white T_1+T_2+...+T_k+T_{k+1}=y^2=(k+1)^2$
$\bg_white The statement is true for n=k+1.$

$\bg_white \\By mathematical induction, the statement is true for all positive integers n.$
Stupid_girl can you mentor me sensei, no joke