# Thread: HSC 2018-2019 MX2 Marathon

1. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2-2z+2=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n
Originally Posted by mrbunton
soz dudes i made a typo; z^2-2z+2=0 where a and b are roots
i apologise for any inconvenience this may have caused
$Roots (\alpha and \beta): 1 \pm i$
$Subbing in roots and y + 1 = \cot(x) into the LHS:$

$LHS = \frac{(cotx + i)^n - (cotx - i)^n}{2i}$

$= \frac{(\frac{cosx}{sinx} + i)^n - (\frac{cosx}{sinx} - i)^n}{2i}$

$= \frac{\frac{1}{sin^nx}(cosx + isinx)^n - \frac{1}{sin^nx}(cosx - isinx)^n}{2i}$

$= \frac{(cosx + isinx)^n - (\frac{(cosx - isinx)(cosx + isinx)}{cosx + sinx})^n}{2isin^nx}$

$= \frac{(cosx + isinx)^n - \frac{1}{(cosx + isinx)^n}}{2isin^nx}$

$= \frac{(cosx + isinx)^n - (cosx + isinx)^{-n}}{2isin^nx}$

$= \frac{cos{(nx)} + isin{(nx)} - (cos{(-nx)} + isin{(-nx)})}{2isin^nx}$

$= \frac{cos{(nx)} + isin{(nx)} - (cos{(nx)} - isin{(nx)})}{2isin^nx}$

$= \frac{2isin{(nx)}}{2isin^nx}$

$= \frac{sin{(nx)}}{sin^nx} = RHS$

yup

3. ## Re: HSC 2018 MX2 Marathon

Oh whoops... I read that as $\sin(x^n)$.

4. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
A classic one

$Simplify \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$
Before I write up the latex, is the answer

$\frac{sin(\frac{nx}{2})(sin(x\frac{n+1}{2}))}{sin( \frac{x}{2})} ???$

5. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
Before I write up the latex, is the answer

$\frac{sin(\frac{nx}{2})(sin(x\frac{n+1}{2}))}{sin( \frac{x}{2})} ???$
Correct!

6. ## Re: HSC 2018 MX2 Marathon

Originally Posted by jathu123
A classic one

$Simplify \sin(x)+\sin(2x) + \sin(3x) + ... + \sin(nx)$
Originally Posted by InteGrand
Correct!
$Let z = cosx + isinx$

$sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}(z + z^2 + z^3 + ... + z^n)$

$GP with a = z and r = z$

$\frac{z(1-z^n)}{1-z}$

$= \frac{(cosx + isinx)(1-(cosx + isinx)^n)}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- cosnx - isinnx))}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- (1 - 2sin^2\frac{nx}{2}) - 2isin\frac{nx}{2}cos\frac{nx}{2}))}{1- (1 - 2sin^2\frac{x}{2}) - 2isin\frac{x}{2})cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin^2\frac{nx}{2} - 2isin\frac{nx}{2}cos\frac{nx}{2})}{2sin^2\frac{x}{ 2} - 2isin\frac{x}{2}cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})}{(2sin\frac{x}{2})(sin\frac{x}{2 } - icos\frac{x}{2})}$

$= \frac{(cosx + isinx)(sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{(sin\frac{x}{2})(sin\frac{x}{2} - icos\frac{x}{2})(sin\frac{x}{2} + icos\frac{x}{2})}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{nx}{2}cos\frac{x}{2} + sin\frac{nx}{2}sin\frac{x}{2} + i(sin\frac{nx}{2}cos\frac{x}{2} - cos\frac{nx}{2}sin\frac{x}{2}))}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$Now sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}\left(\frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}\right)$

$= \frac{(sin\frac{nx}{2})(cosx)(sin\frac{x(n-1)}{2}) + (sin\frac{nx}{2})(sinx)(cos\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})\left((sin\frac{x(n-1)}{2})(cosx) + (cos\frac{x(n-1)}{2})(sinx)\right)}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{2x + xn - x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{x(n + 1)}{2})}{sin\frac{x}{2}}$

$I am never writing latex again$

7. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$Let z = cosx + isinx$

$sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}(z + z^2 + z^3 + ... + z^n)$

$GP with a = z and r = z$

$\frac{z(1-z^n)}{1-z}$

$= \frac{(cosx + isinx)(1-(cosx + isinx)^n)}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- cosnx - isinnx))}{1-cosx-isinx}$

$= \frac{(cosx + isinx)(1- (1 - 2sin^2\frac{nx}{2}) - 2isin\frac{nx}{2}cos\frac{nx}{2}))}{1- (1 - 2sin^2\frac{x}{2}) - 2isin\frac{x}{2})cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin^2\frac{nx}{2} - 2isin\frac{nx}{2}cos\frac{nx}{2})}{2sin^2\frac{x}{ 2} - 2isin\frac{x}{2}cos\frac{x}{2}}$

$= \frac{(cosx + isinx)(2sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})}{(2sin\frac{x}{2})(sin\frac{x}{2 } - icos\frac{x}{2})}$

$= \frac{(cosx + isinx)(sin\frac{nx}{2})(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{(sin\frac{x}{2})(sin\frac{x}{2} - icos\frac{x}{2})(sin\frac{x}{2} + icos\frac{x}{2})}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(sin\frac{nx}{2} - icos\frac{nx}{2})(sin\frac{x}{2} + icos\frac{x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{nx}{2}cos\frac{x}{2} + sin\frac{nx}{2}sin\frac{x}{2} + i(sin\frac{nx}{2}cos\frac{x}{2} - cos\frac{nx}{2}sin\frac{x}{2}))}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$Now sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}\left(\frac{(sin\frac{nx}{2})(cosx + isinx)(cos\frac{x(n-1)}{2} + isin\frac{x(n-1)}{2})}{sin\frac{x}{2}}\right)$

$= \frac{(sin\frac{nx}{2})(cosx)(sin\frac{x(n-1)}{2}) + (sin\frac{nx}{2})(sinx)(cos\frac{x(n-1)}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})\left((sin\frac{x(n-1)}{2})(cosx) + (cos\frac{x(n-1)}{2})(sinx)\right)}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{2x + xn - x}{2})}{sin\frac{x}{2}}$

$= \frac{(sin\frac{nx}{2})(sin\frac{x(n + 1)}{2})}{sin\frac{x}{2}}$

$I am never writing latex again$
$\noindent Well done! Incidentially, another method is to consider$

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

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

8. ## Re: HSC 2018 MX2 Marathon

Originally Posted by InteGrand
$\noindent Well done! Incidentially, another method is to consider$

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

$\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from k=1 to n.$
Question 7 Of Ekman's Compilation
for the working out of integrands clever solution if anyones interested

9. ## Re: HSC 2018 MX2 Marathon

Originally Posted by InteGrand
$\noindent Well done! Incidentially, another method is to consider$

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

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

10. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$I am never writing latex again$
Alternatively, you can use the fact that the Imaginary part of 1 is 0 to do this:

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

11. ## 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:

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

12. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
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).

13. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
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.

14. ## Re: HSC 2018 MX2 Marathon

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

15. ## Re: HSC 2018 MX2 Marathon

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

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

$\noindent you can just compute$

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

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

16. ## Re: HSC 2018 MX2 Marathon

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

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

$\noindent you can just compute$

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

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

17. ## Re: HSC 2018 MX2 Marathon

$New Question!$

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

18. ## Re: HSC 2018 MX2 Marathon

Is the answer $m=4$?

(I intend to post a writeup later)

19. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
Is the answer $m=4$?

(I intend to post a writeup later)
yes it's 4

20. ## 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, $\frac{m\pi}{4} = \pi \implies m = 4$

21. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
$(1+i)^m = (1-i)^m$

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!

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

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

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

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

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

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

22. ## 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.
integral composite graph.png

23. ## 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,

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

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

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 $k = 5$.

24. ## 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.

25. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
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.

Originally Posted by mrbunton
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 $k=-2-\sqrt{33}$.

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