MX2 Marathon (4 Viewers)

jathu123 · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

InteGrand said:
$\noindent Well done! Incidentially, another method is to consider$

$\sin( k\theta )\sin \frac{\theta}{2} = \ldots \quad (\text{use a product to difference formula})$

$\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from $k=1$ to $n$.$

http://community.boredofstudies.org...uestion-7-ekmans-compilation.html#post7213476
for the working out of integrands clever solution if anyones interested

altSwift · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

InteGrand said:
$\noindent Well done! Incidentially, another method is to consider$

$\sin( k\theta )\sin \frac{\theta}{2} = \ldots \quad (\text{use a product to difference formula})$

$\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from $k=1$ to $n$.$

Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?

Paradoxica · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
$I am never writing latex again$

Alternatively, you can use the fact that the Imaginary part of 1 is 0 to do this:

$$\noindent$ S \equiv \Im{\left( \sum_{k=0}^n z^k \right)} \equiv \Im{\left( \frac{z^{n+1}-1}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{2i\sin{\frac{(n+1)x}{2}}}{2i\sin{\frac{x}{2}}} \right )} \\\\ \equiv \Im{ \left( \left( \cos{\frac{nx}{2}}+i\sin{\frac{nx}{2}} \right ) \frac{\sin{\frac{(n+1)x}{2}}}{\sin{\frac{x}{2}}} \right )} \equiv \frac{ \sin{\frac{nx}{2} }\sin{ \frac{(n+1)x}{2}} }{\sin{ \frac{x}{2}} }$

altSwift · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

I'm not sure what you did on this line, I'm guessing its in regard to the fact that Im(1) = 0

$\Im{\left( \frac{z^{n+1}-1}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}$

Also, did you mean:

$\Im{\left( \frac{z^{n+1}-z}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}$

InteGrand · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?

One thing (probably not the only thing) you can try is searching up past years' HSC 4U marathons here on BOS (including the Advanced Level ones, though these are typically harder than HSC style Q's or would be broken into multiple steps if asked in the HSC exam).

Paradoxica · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
Also, did you mean:

$\Im{\left( \frac{z^{n+1}-z}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}$

That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.

altSwift · Feb 13, 2018

Re: HSC 2018 MX2 Marathon

Paradoxica said:
That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.

Sorry if I'm missing something obvious, but just to clarify

$$\noindent$ S = \Im{\left( \sum_{k=0}^n z^k \right)} = \Im(z^0 + z^1 + z^2 + ... + z^{n+1}) = \Im(1 + z^1 + z^2 + ... + z^{n+1})$

$= \Im(\frac{(1)(z^{n+1} - 1)}{z - 1}) = $ ... $ = \frac{ \sin{\frac{nx}{2} }\sin{ \frac{(n+1)x}{2}} }{\sin{ \frac{x}{2}} }$

When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha

InteGrand · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
Sorry if I'm missing something obvious, but just to clarify

$$\noindent$ S = \Im{\left( \sum_{k=0}^n z^k \right)} = \Im(z^0 + z^1 + z^2 + ... + z^{n+1}) = \Im(1 + z^1 + z^2 + ... + z^{n+1})$

$= \Im(\frac{(1)(z^{n+1} - 1)}{z - 1}) = $ ... $ = \frac{ \sin{\frac{nx}{2} }\sin{ \frac{(n+1)x}{2}} }{\sin{ \frac{x}{2}} }$

When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha

$\noindent Remember, adding $1$ (or any real number) to something won't change its imaginary part! (You can probably prove this fact easily if you're unsure of it.) So to compute$

$\mathrm{Im}\left(z+z^{2} +\cdots + z^{n}\right),$

$\noindent you can just compute$

$\mathrm{Im}\left(\color{red}{1}\color{black}+z+z^{2} + \cdots + z^{n}\right).$

$\noindent This is what Paradoxica did. (The advantage of doing this is it makes the resulting GP formula a bit simpler compared to if we leave out the $\color{red}{1}\color{black}$, so it's easier to compute the imaginary part of.)$

altSwift · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

InteGrand said:
$\noindent Remember, adding $1$ (or any real number) to something won't change its imaginary part! (You can probably prove this fact easily if you're unsure of it.) So to compute$

$\mathrm{Im}\left(z+z^{2} +\cdots + z^{n}\right),$

$\noindent you can just compute$

$\mathrm{Im}\left(\color{red}{1}\color{black}+z+z^{2} + \cdots + z^{n}\right).$

$\noindent This is what Paradoxica did. (The advantage of doing this is it makes the resulting GP formula a bit simpler compared to if we leave out the $\color{red}{1}\color{black}$, so it's easier to compute the imaginary part of.)$

Yeah that makes more sense now, its pretty creative. Thank you

altSwift · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

$New Question!$

$$Find the smallest \textbf{positive} value of m if $(1 + i)^m = (1 - i)^m$

fan96 · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

Is the answer $m=4$ ?

(I intend to post a writeup later)

jathu123 · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
Is the answer $m=4$ ?

(I intend to post a writeup later)

yes it's 4

fan96 · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

$(1+i)^m = (1-i)^m$

Converting to polar form,

$m\sqrt{2}\left(\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4} \right)=m\sqrt{2}\left(\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4} \right)$

Since $m \neq 0$ ,

$\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4}=\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4}$

The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,

$\sin \frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

Since the sine function is odd, $\sin x = - \sin -x$ for all $x$ .

$-\sin -\frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

And clearly this is only true when $\sin \frac{m\pi}{4} = 0$ .

So therefore, $\frac{m\pi}{4} = \pi \implies m = 4$

altSwift · Feb 14, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
$(1+i)^m = (1-i)^m$

Converting to polar form,

$m\sqrt{2}\left(\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4} \right)=m\sqrt{2}\left(\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4} \right)$

Since $m \neq 0$ ,

$\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4}=\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4}$

The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,

$\sin \frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

Since the sine function is odd, $\sin x = - \sin -x$ for all $x$ .

$-\sin -\frac{m\pi}{4}=\sin -\frac{m\pi}{4}$

And clearly this is only true when $\sin \frac{m\pi}{4} = 0$ .

So therefore, $\frac{m\pi}{4} = \pi \implies m = 4$

Nice!

$$Alternatively $ (1+i)^m = (1-i)^m$

$\noindent \frac{(1+i)^m}{(1-i)^m} = 1$

$\frac{m\sqrt{2}(cis \frac{m\pi}{4})}{m\sqrt{2}(cis \frac{-m\pi}{4})} = 1$

$cis \frac{m\pi}{2} = cis0$

$\frac{m\pi}{2} = 0 + 2k\pi $ where k = 0, -1, 1, -2, 2...$$

$m = 4k \therefore $ smallest positive value when k = 1 (ie m=4)$$

mrbunton · Feb 16, 2018

Re: HSC 2018 MX2 Marathon

2u integration+circle geo
i made this question myself in paint so dont mind the bad production quality.
edit: look for the question which has the circle with one edge on -5 instead of -3. ignore the question where the circle touches -3.

fan96 · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

i) define the graph as

$f(x) = \begin{cases} 2, \,x \geq 1 \\ \frac{1}{2}x + 1, \, x < 1\end{cases}$

Let the lower bound be $k$ such that $k < 1$ .

Now,

$\int^5_1 2 \, dx = 8$

So for the integral to be zero,

$\begin{aligned}\int^1_k \frac{1}{2}x + 1\,dx &= -8 \\ \left[ \frac{1}{4}x^2 + x\right]^1_k &= -8 \\ \frac{5}{4} - \left(\frac{1}{4}k^2 + k\right) &= -8 \end{aligned}$

Rearranging,

$k^2 + 4k + 37 = 0$

Using the quadratic formula,

$k = \frac{-4 \pm \sqrt{164}}{2} = -2 \pm \sqrt{41}$

Since $k < 1$ , $k = -2 - \sqrt{41}$

So

$\int^5_{-2-\sqrt{41}}\, f(x)\, dx = 0$

(One other solution is also $k = 5$ )
_______________________________________________

ii) The area of the semicircle is $\frac{\pi}{2}$ , and $\int^5_{-1}1\, dx = 6$

The graph doesn't seem to be defined for $x < -3$ .

Clearly $6 > \frac{\pi}{2}$ , so if one of the bounds is $5$ , the only bound that will make the integral zero is $k = 5$ .

mrbunton · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1
also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.

noticing k=5 was impressive although.

fan96 · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1

It's a semicircle with diameter from -3 to -1 isn't it? So the diameter has length 2 and radius 1.

mrbunton said:
also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.

Oops. The line is $\frac{2}{3}\left(x+2\right)$ .

So the new equation is

$\int^1_k \frac{2}{3}x + \frac{4}{3}\, dx = -8$

And the quadratic is now $k^2+4k-29 = 0$ which gives $k=-2-\sqrt{33}$ .

mrbunton · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

rip the circle was meant to have radius 2; i changed it now so the circle touches -5 to -1.
edit: i keep writing questions incorrectly; i check next time btw.

altSwift · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

$Find the minimum value of the positive integer m for which $ (\sqrt3 + i)^m $ is:$
$a) real$
$b) purely imaginary$

MX2 Marathon (4 Viewers)

Active Member

Member

-insert title here-

Member

Well-Known Member

-insert title here-

Member

Well-Known Member

Member

Member

617 pages

Active Member

617 pages

Member

Member

Attachments

617 pages

Member

617 pages

Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)