# HSC 2016 MX2 Marathon (archive) (1 Viewer)

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#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

$\bg_white \\ i) Find a closed formula for \ \int_0^{\pi /2} \sin^{2n+1}(x) \ dx \ where \ n \ is a non-negative integer \\\\ ii) Hence find \ \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}}$

EDIT: Will rework so it fits into the non-advanced marathon

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#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

$\bg_white \\ Find with proof \ \lim_{x \to 0} \frac{\sin(x^2)}{(\sin x)^2}$

#### He-Mann

##### Vexed?
Re: HSC 2016 4U Marathon

Excuse me.

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#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

Three mathematics students were seated around a table each wearing a coloured hat, either red or yellow, given to them by their teacher. The students could not see the colour of their own hat, but could see the colour of the other students. The teacher told the students that at least one among them have a yellow hat. The teacher then said that the first student to rise and mention the colour of their own hat would win a prize.

One student finally rose and correctly announced their answer, what did the student say, and how did they work it out?

#### kashkow

##### Active Member
Re: HSC 2016 4U Marathon

Three mathematics students were seated around a table each wearing a coloured hat, either red or yellow, given to them by their teacher. The students could not see the colour of their own hat, but could see the colour of the other students. The teacher told the students that at least one among them have a yellow hat. The teacher then said that the first student to rise and mention the colour of their own hat would win a prize.

One student finally rose and correctly announced their answer, what did the student say, and how did they work it out?
In all cases the answer is Yellow (by any student wearing a yellow hat).

The three cases are as follows:

Y-Y-R

Y-R-R

Y-Y-Y

For the sake of the explanation let's name the order of hat wearers as A-B-C respectively.

Case 1: Student A guesses yellow because he thinks that if he has a red hat, student B would immediately deduce he is Yellow. However due to Student B's silence, he must be Yellow.

Case 2: Student A easily deduces he is yellow due to the teachers comment that one must have a yellow hat.

Case 3: Student A can deduce the fact he has a yellow hat, due to the logic of case 1. Assuming the students are all equally rational and logical beings (ie. assuming Student B and C can just as easily come to the same conclusions as A in case 1): If Student A wore a red hat, Students B and C would be able to deduce they are wearing yellow hats. However due to their silence, Student A must be wearing a yellow hat.

Nice question, not too hard. But is this even 4U?? haha shouldn't it be in a logic thread or something?

Edit: P.S. I know a few questions like these (well I can remember one but forgot the specifics other). But I won't post the one I do know for two reasons. 1: not here (don't think this is the appropriate thread?? ahaha) and 2. because it would probably be too easy to solve after this one's been said.

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#### Drsoccerball

##### Well-Known Member
Re: HSC 2016 4U Marathon

The return of Sy123.

#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

In all cases the answer is Yellow (by any student wearing a yellow hat).

The three cases are as follows:

Y-Y-R

Y-R-R

Y-Y-Y

For the sake of the explanation let's name the order of hat wearers as A-B-C respectively.

Case 1: Student A guesses yellow because he thinks that if he has a red hat, student B would immediately deduce he is Yellow. However due to Student B's silence, he must be Yellow.

Case 2: Student A easily deduces he is yellow due to the teachers comment that one must have a yellow hat.

Case 3: Student A can deduce the fact he has a yellow hat, due to the logic of case 1. Assuming the students are all equally rational and logical beings (ie. assuming Student B and C can just as easily come to the same conclusions as A in case 1): If Student A wore a red hat, Students B and C would be able to deduce they are wearing yellow hats. However due to their silence, Student A must be wearing a yellow hat.

Nice question, not too hard. But is this even 4U?? haha shouldn't it be in a logic thread or something?

Edit: P.S. I know a few questions like these (well I can remember one but forgot the specifics other). But I won't post the one I do know for two reasons. 1: not here (don't think this is the appropriate thread?? ahaha) and 2. because it would probably be too easy to solve after this one's been said.
You've got it. I should of probably of posted it in the advanced thread, having a separate logic thread is going to separate everything too much.

##### -insert title here-
Re: HSC 2016 4U Marathon

You've got it. I should of probably of posted it in the advanced thread, having a separate logic thread is going to separate everything too much.
M8

I created a logic marathon for a reason.

##### -insert title here-
Re: HSC 2016 4U Marathon

$\bg_white \\ i) Find a closed form for \ \int_0^{\pi /2} \sin^{2n+1}(x) \ dx \ where \ n \ is a non-negative integer \\\\ ii) Hence find \ \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}}$
Using the Beta and Gamma Functions we can find an explicit closed form for the integral.

$\bg_white \text{B}(p,q) = 2\int_0^\frac{\pi}{2} \sin^{2p-1}{x}\cos^{2q-1}{x} \text{d}x$

Clearly, p = n+1 and q = ½ for the integral above.

$\bg_white \text{B} \left( n+1,\frac{1}{2} \right) = \frac{\Gamma (n+1) \Gamma \left( \frac{1}{2} \right) }{\Gamma \left( n+\frac{3}{2} \right) }$

Using Legendre's Duplication Formula:

$\bg_white \Gamma (z) \Gamma \left( z+\frac{1}{2} \right) = 2^{1-2z} \Gamma \left( \frac{1}{2} \right) \Gamma (2z)$

Set z = n+1

$\bg_white \Gamma (n+1) \Gamma \left( n+\frac{3}{2} \right) = 2^{-(2n+1)} \Gamma \left( \frac{1}{2} \right) \Gamma (2n+2)$

Now we can proceed

$\bg_white \text{B} \left( n+1,\frac{1}{2} \right) = \frac{2^{2n+1} \left( \Gamma \left( n+1 \right) \right)^2 \Gamma \left( \frac{1}{2} \right)}{\Gamma \left( \frac{1}{2} \right) \Gamma \left( 2n+2 \right)} = \frac{2^{2n+1} (n!)^2}{(2n+1)!} = \frac{2^{2n+1}}{(2n+1) \binom{2n}{n}}$

Thus we conclude:

$\bg_white \int_0^\frac{\pi}{2} \sin^{2n+1}{x} \, \text{d}x = \frac{2^{2n}}{(2n+1) \binom{2n}{n}}$

Moving things around we obtain:

$\bg_white \frac{1}{ \binom{2n}{n}} = \int_0^\frac{\pi}{2} \frac{(2n+1)\sin^{2n+1}x}{2^{2n}} \text{d}x$

Using the fact that sinx≤1 for all real x, an inequality on the integral is obtained.

$\bg_white \frac{1}{ \binom{2n}{n}} = \int_0^\frac{\pi}{2} \frac{(2n+1)\sin^{2n+1}x}{2^{2n}} \text{d}x < \int_0^\frac{\pi}{2} \frac{2n+1}{2^{2n}} \text{d}x = \frac{(2n+1)\pi}{2^{2n+1}}$

And hence:

$\bg_white 0 < \sum_{n=0}^\infty \frac{1}{\binom{2n}{n}} < \sum_{n=0}^\infty \frac{(2n+1)\pi}{2^{2n+1}} = \frac{10\pi}{9}$

Thus, by the comparison test, the sum converges absolutely, since all the terms are strictly decreasing and non-negative, with strict upper and lower bounds.

Hence, the interchange of the order of summation and integration is allowed.

$\bg_white S = \sum_{k=0}^\infty (2n+1)\frac{\sin^{2n+1}{x}}{2^{2n}}$

$\bg_white S - \frac{S}{4}\sin^2{x} = \sin{x} + \sum_{k=0}^\infty \left( \frac{(2n+3)\sin^{2n+3}{x}}{2^{2n+2}} - \frac{(2n+1)\sin^{2n+3}{x}}{2^{2n+2}} \right)$

$\bg_white \frac{S}{4}(4-\sin^2{x}) = \sin{x} + \sum_{k=0}^\infty \frac{\sin^{2n+3}{x}}{2^{2n+1}} = \sin{x} + \frac{2\sin^3{x}}{4-\sin^2{x}}$

$\bg_white S = \frac{4\sin{x}}{4-\sin^2{x}} + \frac{8\sin^3{x}}{(4-\sin^2{x})^2}$

Integrating across the bounds is a trivial task, and the relevant working will be left as an exercise to the reader.

Thus we conclude:

$\bg_white \sum_{n=0}^\infty \frac{1}{\binom{2n}{n}} = \frac{4}{3} + \frac{2\pi}{9\sqrt{3}}$

#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

$\bg_white \\ i) Find a closed formula for \ \int_0^{\pi /2} \sin^{2n+1}(x) \ dx \ where \ n \ is a non-negative integer \\\\ ii) Hence find \ \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}}$

EDIT: Will rework so it fits into the non-advanced marathon
Mostly just algebra and following instructions, it is a fine exercise

$\bg_white \\ Let \ I_n = \int_{0}^{ \frac{\pi}{2} } \sin^{2n+1}(x) \ dx \ for integers \ n \geq 0 \\\\ i) Show that \ I_n = \frac{2n}{2n+1}I_{n-1} \ using integration by parts \\\\ ii) Hence show that \ \frac{1}{\binom{2n}{n}} = \frac{2n+1}{4^n}I_n$

$\bg_white \\ iii) Explain why the infinite sum \ \sum_{n=0}^{\infty} (2n+1)z^{2n+1} \ defined for \ |z| < 1 \ can be written as \ \sum_{n=0}^{\infty} (2n+1)z^{2n+1} = \frac{z}{1-z^2} + \frac{2z^3}{1-z^2} + \frac{2z^5}{1-z^2} + \frac{2z^7}{1-z^2} + \dots \\\\ iv) Hence show that \ \sum_{n=0}^{\infty} (2n+1)z^{2n+1} = \frac{z(1+z^2)}{(1-z^2)^2}$

$\bg_white \\ v) Now by using part (ii) show that \ \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}} = 4 \int_0^{\pi /2} \frac{\sin x (4 + \sin^2x)}{(4 -\sin^2 x)^2} \ dx \\\\ vi) Show that \ \int \frac{1}{(1+x^2)^2} \ dx = \frac{1}{2} \left(\frac{x}{x^2+1} + \tan^{-1}x \right) + c \\\\ vii) Evaluate the integral in part (v) using the substitution \ u = \cos x \\\\ viii) Hence show that \ \sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}} = \frac{2}{27}(18 + \pi \sqrt{3})$

#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

Using the Beta and Gamma Functions we can find an explicit closed form for the integral.

$\bg_white \text{B}(p,q) = 2\int_0^\frac{\pi}{2} \sin^{2p-1}{x}\cos^{2q-1}{x} \text{d}x$

Clearly, p = n+1 and q = ½ for the integral above.

$\bg_white \text{B} \left( n+1,\frac{1}{2} \right) = \frac{\Gamma (n+1) \Gamma \left( \frac{1}{2} \right) }{\Gamma \left( n+\frac{3}{2} \right) }$

Using Legendre's Duplication Formula:

$\bg_white \Gamma (z) \Gamma \left( z+\frac{1}{2} \right) = 2^{1-2z} \Gamma \left( \frac{1}{2} \right) \Gamma (2z)$

Set z = n+1

$\bg_white \Gamma (n+1) \Gamma \left( n+\frac{3}{2} \right) = 2^{-(2n+1)} \Gamma \left( \frac{1}{2} \right) \Gamma (2n+2)$

Now we can proceed

$\bg_white \text{B} \left( n+1,\frac{1}{2} \right) = \frac{2^{2n+1} \left( \Gamma \left( n+1 \right) \right)^2 \Gamma \left( \frac{1}{2} \right)}{\Gamma \left( \frac{1}{2} \right) \Gamma \left( 2n+2 \right)} = \frac{2^{2n+1} (n!)^2}{(2n+1)!} = \frac{2^{2n+1}}{(2n+1) \binom{2n}{n}}$

Thus we conclude:

$\bg_white \int_0^\frac{\pi}{2} \sin^{2n+1}{x} \, \text{d}x = \frac{2^{2n}}{(2n+1) \binom{2n}{n}}$

Moving things around we obtain:

$\bg_white \frac{1}{ \binom{2n}{n}} = \int_0^\frac{\pi}{2} \frac{(2n+1)\sin^{2n+1}x}{2^{2n}} \text{d}x$

Using the fact that sinx≤1 for all real x, an inequality on the integral is obtained.

$\bg_white \frac{1}{ \binom{2n}{n}} = \int_0^\frac{\pi}{2} \frac{(2n+1)\sin^{2n+1}x}{2^{2n}} \text{d}x < \int_0^\frac{\pi}{2} \frac{2n+1}{2^{2n}} \text{d}x = \frac{(2n+1)\pi}{2^{2n+1}}$

And hence:

$\bg_white 0 < \sum_{n=0}^\infty \frac{1}{\binom{2n}{n}} < \sum_{n=0}^\infty \frac{(2n+1)\pi}{2^{2n+1}} = \frac{10\pi}{9}$

Thus, by the comparison test, the sum converges absolutely, since all the terms are strictly decreasing and non-negative, with strict upper and lower bounds.

Hence, the interchange of the order of summation and integration is allowed.

$\bg_white S = \sum_{k=0}^\infty (2n+1)\frac{\sin^{2n+1}{x}}{2^{2n}}$

$\bg_white S - \frac{S}{4}\sin^2{x} = \sin{x} + \sum_{k=0}^\infty \left( \frac{(2n+3)\sin^{2n+3}{x}}{2^{2n+2}} - \frac{(2n+1)\sin^{2n+3}{x}}{2^{2n+2}} \right)$

$\bg_white \frac{S}{4}(4-\sin^2{x}) = \sin{x} + \sum_{k=0}^\infty \frac{\sin^{2n+3}{x}}{2^{2n+1}} = \sin{x} + \frac{2\sin^3{x}}{4-\sin^2{x}}$

$\bg_white S = \frac{4\sin{x}}{4-\sin^2{x}} + \frac{8\sin^3{x}}{(4-\sin^2{x})^2}$

Integrating across the bounds is a trivial task, and the relevant working will be left as an exercise to the reader.

Thus we conclude:

$\bg_white \sum_{n=0}^\infty \frac{1}{\binom{2n}{n}} = \frac{4}{3} + \frac{2\pi}{9\sqrt{3}}$
See my expanded question for a more 4U approach

##### -insert title here-
Re: HSC 2016 4U Marathon

See my expanded question for a more 4U approach
You didn't formally justify the interchange of summation and integration, so nah

#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

$\bg_white \\ Find with proof \ \lim_{x \to 0} \frac{\sin(x^2)}{(\sin x)^2}$
Bump, and another question (suitable for this thread):

$\bg_white \\ Show that \ 2^{n+2} + 3^{2n+1} \ is divisible by \ 7 \ for all integers \ n \geq 1$

#### KingOfActing

##### lukewarm mess
Re: HSC 2016 4U Marathon

Bump, and another question (suitable for this thread):

$\bg_white \\ Show that \ 2^{n+2} + 3^{2n+1} \ is divisible by \ 7 \ for all integers \ n \geq 1$
$\bg_white \lim_{x\to 0} \frac{\sin{(x^2)}}{\sin^2{x}} = \lim_{x\to 0} \frac{\sin{(x^2)}}{x^2}\left( \frac{x}{\sin{x}}\right )^2 = 1 \times 1 = 1$

Honestly, I'd consider that 3U level Unless by "With Proof" you meant "By squeeze theorem" in which case I wouldn't see it as a suitable 4u question without more guidance.

\bg_white 2^{1+2} + 3^{2\times1+1} = 35=7\times 5\\Suppose the result holds for some integer n (i.e. 2^{n+2} + 3^{2n+1} = 7m,\quad m \in \mathbb N) Then for n+1 we have:\\\begin{align*}2^{n+3} + 3^{2n+3} &= 2\times 2^{n+2} + 9\times3^{2n+1} \\&=2\times2^{n+2} +2 \times 3^{2n+1} + 7 \times 3^{2n+1} \\&= 7(2m + 3^{2n+1})\end{align*}

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##### -insert title here-
Re: HSC 2016 4U Marathon

Bump, and another question (suitable for this thread):

$\bg_white \\ Show that \ 2^{n+2} + 3^{2n+1} \ is divisible by \ 7 \ for all integers \ n \geq 1$

$\bg_white \noindent Observe that 3 \equiv -4 \mod{7} \\\\ Observe that 2^3 \equiv 1 \mod{7} \\\\ Then 2^{3n} \equiv 1 \mod{7} \\\\ 2^{n+2} + 3^{2n+1} \equiv 2^{n+2} - 4^{2n+1} \mod{7} \\\\ 2^{n+2} + 3^{2n+1} \equiv 2^{n+2} \times 2^{3n} - 2^{4n+2} \mod{7} \\\\ 2^{n+2} + 3^{2n+1} \equiv 2^{4n+2} - 2^{4n+2} \mod{7} \\\\ 2^{n+2} + 3^{2n+1} \equiv 0 \mod{7}$

##### -insert title here-
Re: HSC 2016 4U Marathon

$\bg_white \lim_{x\to 0} \frac{\sin{(x^2)}}{\sin^2{x}} = \lim_{x\to 0} \frac{\sin{(x^2)}}{x^2}\left( \frac{x}{\sin{x}}\right )^2 = 1 \times 1 = 1$

Honestly, I'd consider that 3U level Unless by "With Proof" you meant "By squeeze theorem" in which case I wouldn't see it as a suitable 4u question without more guidance.
Yeah. Needs to be something harder, like this:

$\bg_white \lim_{x \to 0} \frac{\sqrt{1-\cos{x^2}}}{1-\cos{x}}$

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#### Sy123

##### This too shall pass
Re: HSC 2016 4U Marathon

$\bg_white \\ P(a \cos \alpha, b\sin \alpha) \ and \ Q(a \cos \beta, b \sin \beta) \ are points that lie on the ellipse \\\\ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\\\ It is given that \ \alpha + \beta = \frac{\pi}{2} \\\\ i) Find the coordinates of the midpoint \ M \ of the chord \ PQ \ in terms of \ \theta \\\\ ii) Show that the locus of \ M \ is a straight line through the origin$

##### -insert title here-
Re: HSC 2016 4U Marathon

$\bg_white \\ P(a \cos \alpha, b\sin \alpha) \ and \ Q(a \cos \beta, b \sin \beta) \ are points that lie on the ellipse \\\\ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\\\ It is given that \ \alpha + \beta = \frac{\pi}{2} \\\\ i) Find the coordinates of the midpoint \ M \ of the chord \ PQ \ in terms of \ \theta \\\\ ii) Show that the locus of \ M \ is a straight line through the origin$
Consider the conjugate points dilated by a constant factor perpendicular to the major axis of the ellipse onto the auxiliary circle which exists alongside the ellipse.

Clearly, these points are reflected about the line y=x, due to the condition on the sum of the angular parameters.

So obviously, the midpoint of the Auxiliary points is the line y=x, inside the auxiliary circle.

Reverse the dilation on the points to bring them back onto the ellipse. Perpendicular dilation from a line does not change the fact that the locus of the midpoints is a line, so the statement is true.

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#### DatAtarLyfe

##### Booty Connoisseur
Re: HSC 2016 4U Marathon

For what values of n is
a) (root(3)+i)^n purely imaginary
Im confused about the npi/6=pi/2 + kpi part. I thought the general solution was 2kpi(plusminus)pi/2?
Converting to mod-arg form and expanding by DMT, you then make the real part = 0
so cos(npi/6) = 0
if you solve, you realise that npi/6 = +-(pi/2, 3pi/2, 5pi/2.....) = pi/2 + kpi

#### InteGrand

##### Well-Known Member
Re: HSC 2016 4U Marathon

For what values of n is
a) (root(3)+i)^n purely imaginary
Im confused about the npi/6=pi/2 + kpi part. I thought the general solution was 2kpi(plusminus)pi/2?
$\bg_white \noindent Note that the set of values of \pm \frac{\pi}{2} + 2k\pi as k runs through the integers is the same as that of \frac{\pi}{2} +k \pi as k runs through the integers, since \frac{\pi}{2} and -\frac{\pi}{2} are distance \pi apart.$

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