HSC 2018-2019 MX2 Marathon

**pikachu975** · 28 May 2018, 9:21 PM

Originally Posted by CapitalSwine

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** · 28 May 2018, 9:51 PM

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

**altSwift** · 9 Jun 2018, 3:19 PM

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

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

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

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

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

**fan96** · 9 Jun 2018, 6:23 PM

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

b)

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

c)

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

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

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

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

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

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

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

Giving, as required,

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

d)

From b),

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

$I_1 = \frac 12 \log 2$

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

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

From c),

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

**fan96** · 16 Jun 2018, 2:39 PM

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

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

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

**mrbunton** · 20 Nov 2018, 11:01 AM

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.

**fan96** · 20 Nov 2018, 11:34 AM

Originally Posted by mrbunton

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 $e^x$ ?

Edit: nevermind, I found some others

**HeroWise** · 20 Nov 2018, 2:01 PM

Originally Posted by fan96

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

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.

**InteGrand** · 20 Nov 2018, 3:45 PM

$\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** · 20 Nov 2018, 5:05 PM

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

**mrbunton** · 20 Nov 2018, 10:08 PM

Originally Posted by InteGrand

$\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.$

ya. You can generate that through utilising the definition of even and odd functions which was the intent of the question.

**HeroWise** · 20 Nov 2018, 10:43 PM

More qtns

**Sy123** · 25 Nov 2018, 11:42 AM

$\\ $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}$

$\\ $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** · 30 Dec 2018, 7:24 PM

$\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 $f(\sigma)$.$

**Paradoxica** · 30 Dec 2018, 8:37 PM

Originally Posted by InteGrand

$\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** · 31 Dec 2018, 9:36 AM

Originally Posted by Paradoxica

tsk tsk trying to introduce multivariable optimisation in a sidestepping manner

Haha, that wasn't the motivation at all.

**fan96** · 31 Dec 2018, 6:44 PM

Originally Posted by Sy123

$\\ $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}$

$\\ $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

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

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

That line is

$y = -x \sin \theta$

Solving these two equations gives

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

Now,

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

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

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

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

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

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

And similarly,

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

So,

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

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

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

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

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

**stupid_girl** · 16 Feb 2019, 7:50 PM

$$\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 $T_1+T_2+...+T_n=n^2$.$

**stupid_girl** · 17 Feb 2019, 8:10 PM

$$\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 $T_nT_{n+2}-(T_{n+1})^2=\left(-1\right)^n$.$

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