# Thread: HSC 2018 MX2 Marathon

1. ## 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 $k = 5$.

2. ## 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.

3. ## 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 $k=-2-\sqrt{33}$.

4. ## 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.

5. ## Re: HSC 2018 MX2 Marathon

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

6. ## Re: HSC 2018 MX2 Marathon

$Let z = (\sqrt{3} + i)^m$ where $m \in \mathbb{Z}^+$

$\therefore z= \left(2\, \text{cis}\, \frac{\pi}{6}\right)^m= 2m \cos \left(\frac{m\pi}{6}\right) + i\left(2m \sin\left(\frac{m\pi}{6}\right)\right)$
__________________________________________________ ___

If $z$ is real, then $Im(z) = 0$

\begin{aligned}\therefore 2m \sin\left(\frac{m\pi}{6}\right) &= 0 \\ \sin\left(\frac{m\pi}{6}\right) &= 0 \end{aligned}

The smallest value of $m$ is $6$.
__________________________________________________ ___

If $z$ is purely imaginary, then $Re(z) = 0$

\begin{aligned}\therefore 2m \cos \left(\frac{m\pi}{6}\right) &= 0 \\ \cos \left(\frac{m\pi}{6}\right) &= 0\end{aligned}

The smallest value of $m$ is $3$.

7. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
$Let z = (\sqrt{3} + i)^m$ where $m \in \mathbb{Z}^+$

$\therefore z= \left(2\, \text{cis}\, \frac{\pi}{6}\right)^m= 2m \cos \left(\frac{m\pi}{6}\right) + i\left(2m \sin\left(\frac{m\pi}{6}\right)\right)$
__________________________________________________ ___

If $z$ is real, then $Im(z) = 0$

\begin{aligned}\therefore 2m \sin\left(\frac{m\pi}{6}\right) &= 0 \\ \sin\left(\frac{m\pi}{6}\right) &= 0 \end{aligned}

The smallest value of $m$ is $6$.
__________________________________________________ ___

If $z$ is purely imaginary, then $Re(z) = 0$

\begin{aligned}\therefore 2m \cos \left(\frac{m\pi}{6}\right) &= 0 \\ \cos \left(\frac{m\pi}{6}\right) &= 0\end{aligned}

The smallest value of $m$ is $3$.
Correct!
Here is another (I really like these questions)

$a) Show that if n is divisible by 3 then:$

$(1 + \sqrt{3}i)^{2n} + (1 - \sqrt{3}i)^{2n} = 2^{2n + 1}$

$b) Simplify the expression if n is \textbf NOT divisible by 3$

8. ## Re: HSC 2018 MX2 Marathon

The equation does not seem to hold for $n=3$.

https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

Is there supposed to be an $i$ somewhere?

9. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
The equation does not seem to hold for $n=3$.

https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

Is there supposed to be an $i$ somewhere?
Yes it should be $(1 + \sqrt3 i)^{2n} + (1 - \sqrt3 i)^{2n} = 2^{2n + 1}$

10. ## Re: HSC 2018 MX2 Marathon

$(1+i\sqrt{3})^{2n} + (1-i\sqrt{3})^{2n}$

$= \left(2 \text{ cis } \frac{\pi}{3} \right)^{2n} + \left(2 \text{ cis } \frac{-\pi}{3} \right)^{2n}$

$= 2^{2n}\left( \text{cis } \frac{2n\pi}{3} + \text{cis } \frac{-2n\pi}{3}\right)$

$= 2^{2n}\left(\cos\frac{2n\pi}{3} + i \sin \frac{2n\pi}{3} + \cos \frac{-2n\pi}{3} + i \sin \frac{-2n\pi}{3}\right)$

Since $\cos x$ is an even function and $\sin x$ is an odd function,

$= 2^{2n}\left(\cos\frac{2n\pi}{3} + i \sin \frac{2n\pi}{3} + \cos \frac{2n\pi}{3} - i \sin \frac{2n\pi}{3}\right)$

$= 2^{2n}\left(2\cos\frac{2n\pi}{3}\right)$

$=\left(2^{2n}\times 2^1\right) \left(\cos \frac{2n\pi}{3}\right)$

$= 2^{2n+1}\left(\cos \frac{2n\pi}{3}\right)$
__________________________________________________ ___________

If $3 \mid n$, then

$\cos \frac{2n\pi}{3} = \cos 2k\pi, \, k \in \mathbb{Z}$

Which is the general solution for $\cos x = 1$

$\therefore \cos \frac{2n\pi}{3} = 1$

$\therefore 2^{2n+1}\left(\cos \frac{2n\pi}{3}\right) = 2^{2n+1}$

11. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
The equation does not seem to hold for $n=3$.

https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

Is there supposed to be an $i$ somewhere?
Originally Posted by jathu123
Yes it should be $(1 + \sqrt3 i)^{2n} + (1 - \sqrt3 i)^{2n} = 2^{2n + 1}$
My bad, yeah thats the correct equation

12. ## Re: HSC 2018 MX2 Marathon

An interesting roots of unity question from the SGS notes.

i) Find the fourth roots of $-1$ in the form $a+ib$.

ii) Hence write $z^4+1$ as a product of two quadratic factors with real co-efficients.

13. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
An interesting roots of unity question from the SGS notes.

i) Find the fourth roots of $-1$ in the form $a+ib$.

ii) Hence write $z^4+1$ as a product of two quadratic factors with real co-efficients.
$Roots: cis\frac{\pi}{4} (\frac{1}{\sqrt2} + \frac{i}{\sqrt2}), cis\frac{-\pi}{4} (\frac{1}{\sqrt2} + \frac{-i}{\sqrt2}),$

$cis\frac{3\pi}{4} (\frac{-1}{\sqrt2} + \frac{i}{\sqrt2}), cis\frac{-3\pi}{4} (\frac{-1}{\sqrt2} + \frac{-i}{\sqrt2})$

$Now z^4 + 1 = (z - cis\frac{\pi}{4})(z - cis\frac{-\pi}{4})(z - cis\frac{3\pi}{4})(z - cis\frac{-3\pi}{4})$

$\noindentUsing the fact that z + \bar z = 2\mathrm{Re} (z) and that z\bar z = x^2 + y^2 (and the fact the roots are in conjugate pairs)$

$z^4 + 1 = (z^2 - 2(\frac{1}{\sqrt2})z + 1)(z^2 + 2(\frac{1}{\sqrt2})z + 1)$

$z^4 + 1 = (z^2 - \sqrt2 z + 1)(z^2 + \sqrt2 z + 1)$

14. ## Re: HSC 2018 MX2 Marathon

$\noindent If P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 , solve:$

$\noindent P(x) = 0 over C without using long division, sum/product of roots etc or the polynomial remainder theorem$

15. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
$\noindent If P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 , solve:$

$\noindent P(x) = 0 over C without using long division, sum/product of roots etc or the polynomial remainder theorem$
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.

16. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sharky564
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.
Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too

17. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sharky564
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.
Originally Posted by altSwift
Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too
$For reference anyway:$

$P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2$
$= x^2 (2x^2 + 7x + 2 - \frac{7}{x} + \frac{2}{x^2})$

$= x^2 \left(2(x^2 + \frac{1}{x^2}) + 7(x - \frac{1}{x}) + 2\right)$

$Using (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2$

$= x^2 \left(2(x - \frac{1}{x})^2 + 7(x - \frac{1}{x}) + 6\right)$

$Factorise as a quadratic from there$

$\noindent Idk if there is any point of doing it this way if you can use either the remainder theorem or equate coefficients. Still looked pretty nice when I first saw it$

18. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sharky564
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.
What made the polynomial so easy to factor?

19. ## Re: HSC 2018 MX2 Marathon

probability: how many 3 digits numbers exist where the middle digit is the average of the first and last digit.

20. ## Re: HSC 2018 MX2 Marathon

Note that only even numbers can be halved to obtain an integer, and $odd + odd = even$ and $even + even= even$. Finally, the average of any two digits is never greater than 9 or less than 0.

If the first digit is odd (1, 3, 5, 7, 9), then the last digit must be odd too (1, 3, 5, 7, 9).

If the first digit is even (2, 4, 6, 8), then the last digit must be even or zero (0, 2, 4, 6, 8).

Any two digits can only have one possible average, so the middle digit doesn't matter.

Therefore the number of digits is $5\times 5 + 4 \times 5 = 45$

yup

22. ## Re: HSC 2018 MX2 Marathon

Show that the polynomial $P(x) = x^3-3x+k$ (where $k$ is an integer such that $k > 2$) has exactly one real zero $r$ where $r < -1$.

23. ## Re: HSC 2018 MX2 Marathon

derive equation to get turning points at -1 and 1
find y values of these turning points:
(-1,2+k) and (1,k-2)
since the poly is odd it will have at least one root.
when k>2; both turning points are above the x-axis; so no extra roots are possible.

proving that r<-1 or similar value:
when letting k=2; that poly is less then 0. the poly can be factored into (x+2)(x-1)^2<0
condition is only true at x<-2 or r<-2
-2<-1 so r<-1

24. ## Re: HSC 2018 MX2 Marathon

$\noindent Given z = x + iy is a variable point and \alpha = a + ib is a fixed point on the Argand plane,$

$\noindent (i) Show that z\alpha -\bar{z}\bar{\alpha} = 0 represents a straight line through the origin O$

$\noindent (ii) Suppose that z\textsubscript{1} and z\textsubscript{2} are the solutions to the simultaneous equations: z\alpha -\bar{z}\bar{\alpha} = 0 and \mid z - \beta \mid = k , where \beta = p + iq , and where p, q and k are positive real numbers, show that \mid z\textsubscript{1} \mid \mid z\textsubscript{2} \mid = \mid p^2 + q^2 - k^2 \mid$

25. ## Re: HSC 2018 MX2 Marathon

i) the line created is bx+ay=0.since you said i) is not the locus of z; and z can be truly anything; then we can observe that bx+ay=0 only yields to purely real values; so it becomes the line y=0 on the imaginary number plane

ii) (x-p)^2+(y-q)^2= k^2
sub y=0 and expand and bring to one side
x^2 -(2p)x +(p^2+ q^2 - k^2) =0
product of roots: z1*z2= p^2 + q^2 - k^2

which is the same thing as we are meant to show. normally one would say the product of the roots are just the real parts of z1 and z2; but there is no imaginary part so condition is still true.

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