HSC 2016 MX2 Marathon (archive) (2 Viewers)

hedgehog_7 · Jan 23, 2016

Re: HSC 2016 4U Marathon

(b) Let C1 ≡ x2 + 3y2 − 1, C2 ≡ 4x2 + y2 − 1, and let λ be a real number.
(i) Show that C1 + λC2 = 0 is the equation of a curve through the points of intersection
of the ellipses C1 = 0 and C2 = 0.
(ii) Determine the values of λ for which C1 + λC2 = 0 is the equation of an ellipse

Paradoxica · Jan 23, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
(b) Let C1 ≡ x2 + 3y2 − 1, C2 ≡ 4x2 + y2 − 1, and let λ be a real number.
(i) Show that C1 + λC2 = 0 is the equation of a curve through the points of intersection
of the ellipses C1 = 0 and C2 = 0.
(ii) Determine the values of λ for which C1 + λC2 = 0 is the equation of an ellipse

Please write original questions. this is a former HSC question, if I do recall.

InteGrand · Jan 23, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Please write original questions. this is a former HSC question, if I do recall.

Maybe he just wants help with it. If that's the case, it might be better to create a thread for it.

Paradoxica · Jan 23, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent By considering the vectors, $e^{\frac{i\pi}{180}}, e^{\frac{2i\pi}{180}}, e^{\frac{3i\pi}{180}}, \cdots ,e^{\frac{42i\pi}{180}}, e^{\frac{43i\pi}{180}}, e^{\frac{44i\pi}{180}}\\$ evaluate $\dfrac{\displaystyle \sum_{n=1}^{44}\cos n^{\circ}}{\displaystyle \sum_{n=1}^{44}\sin n^{\circ} }$

.

Paradoxica said:
$\noindent By considering $\log(x^n -1)$ or otherwise, find $ \\ \frac{31}{2-\alpha}+\frac{31}{2-\beta}+\frac{31}{2-\gamma}+\frac{31}{2-\delta}\\$ where $\alpha,\beta,\gamma,\delta$ are complex fifth roots of unity.$ \\ $Hence, evaluate, in closed form, $ \sum_{r=1}^{n-1} \frac{2^n-1}{2-e^{\frac{2ir\pi}{n}}}$

Blast1 · Jan 23, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
.

$1. e^{ix}=cisx. \ $By pairing vectors together$, cis(\frac{\pi}{180})$ with $cis(\frac{44\pi}{180}),....cis(\frac{22\pi}{180})$ with $cis(\frac{23\pi}{180}), $The resultant vector after adding the vectors in each pair gives us $ \sqrt 2cis(\frac{\pi}{8}). $So, the resultant vector of adding all the resultant vectors together is$ 22 \sqrt2cis(\frac{\pi}{8}). $What we're after is just the real part over the imaginary part of this resultant vector. So, we just need $ \cot(\frac{\pi}{8}), $which equals$ \sqrt2 + 1. \\ 2. $Note that $z^n-1=(z-1)...(z-cis(\frac{2r\pi}{n}))...(z-cis(\frac{2\pi (n-1))}{n})). $Log both sides then differentiate both sides with respect to z, $ \frac{nz^{n-1}}{z^n-1}= \frac{1}{z-1}+\sum_{r=1}^{n-1}\frac{1}{z-cis(\frac{2r\pi}{n})}. $ Multiply through by$ z^n-1 $ then sub z=2, so $ \sum_{r=1}^{n-1}\frac{2^n-1}{2-cis(\frac{2r\pi}{n})} = n2^{n-1}-2^n+1.$

Note that cisx= e^ix for the second question as well.

Paradoxica · Jan 23, 2016

Re: HSC 2016 4U Marathon

Blast1 said:
$1. e^{ix}=cisx. \ $By pairing vectors together$, cis(\frac{\pi}{180})$ with $cis(\frac{44\pi}{180}),....cis(\frac{22\pi}{180})$ with $cis(\frac{23\pi}{180}), $The resultant vector after adding the vectors in each pair gives us $ \sqrt 2cis(\frac{\pi}{8}). $So, the resultant vector of adding all the resultant vectors together is$ 22 \sqrt2cis(\frac{\pi}{8}). $What we're after is just the real part over the imaginary part of this resultant vector. So, we just need $ \cot(\frac{\pi}{8}), $which equals$ \sqrt2 + 1. \\ 2. $Note that $z^n-1=(z-1)...(z-cis(\frac{2r\pi}{n}))...(z-cis(\frac{2\pi (n-1))}{n})). $Log both sides then differentiate both sides with respect to z, $ \frac{nz^{n-1}}{z^n-1}= \frac{1}{z-1}+\sum_{r=1}^{n-1}\frac{1}{z-cis(\frac{2r\pi}{n})}. $ Multiply through by$ z^n-1 $ then sub z=2, so $ \sum_{r=1}^{n-1}\frac{2^n-1}{2-cis(\frac{2r\pi}{n})} = n2^{n-1}-2^n+1.$

Note that cisx= e^ix for the second question as well.

Congrats for your first post. You a long time lurker or recent?

Blast1 · Jan 23, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Congrats for your first post. You a long time lurker or recent?

Fairly new here, I discovered these forums earlier this week lol.

I'm still not quite used to the art of LaTexing

(I'm using a site called codecogs to help me out for the time being)

Drsoccerball · Jan 25, 2016

Re: HSC 2016 4U Marathon

$(i) Use a substitution to prove the AM-GM progression in the function $ f(x)=x-e^{x-1} $ that is prove:$

$(\frac{\sum_{k=1}^n{x_k}}{n})^n \geq \prod_{k=1}^{n}{x_k}$

$(ii) Using Jensen's theorem prove the AM-GM progression$

$Jensen's theorem $ = \frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \leq f(\frac{x_1+x_2+...+x_n}{n}) $ For all concave down continuous curves.$

Blast1 · Jan 25, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
$(i) Use a substitution to prove the AM-GM progression in the function $ f(x)=x-e^{x-1} $ that is prove:$

$(\frac{\sum_{k=1}^n{x_k}}{n})^n \geq \prod_{k=1}^{n}{x_k}$

$(ii) Using Jensen's theorem prove the AM-GM progression$

$Jensen's theorem $ = \frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \leq f(\frac{x_1+x_2+...+x_n}{n}) $ For all concave down continuous curves.$

$$ (i) Notice that $ e^{x-1} \geq x $ for all x$ \geq1. $ Now, motivated by the fact that we need to remove the e somehow, we make the substitution $ x_i=\frac{na_i}{a_1+...+a_n}. $ Thus $ e^{\sum_{i=1}^{n}x_i-n} \geq x_1x_2...x_n. \Rightarrow 1 \geq \frac{n^n(a_1...a_n)}{a_1+...+a_n}. $ Rearranging this completes the problem.$ \\ (ii) $The natural logarithmic function is a concave down function, so its use is valid here. $ \frac{\ln(x_1x_2...x_n)}{n} \leq \ln(\frac{x_1+...x_n}{n}). $ Raising both sides to e, $ x_1...x_n \leq (\frac{x_1+...x_n}{n})^n$

Drsoccerball · Jan 26, 2016

Re: HSC 2016 4U Marathon

Blast1 said:
$$ (i) Notice that $ e^{x-1} \geq x $ for all x$ \geq1. $ Now, motivated by the fact that we need to remove the e somehow, we make the substitution $ x_i=\frac{na_i}{a_1+...+a_n}. $ Thus $ e^{\sum_{i=1}^{n}x_i-n} \geq x_1x_2...x_n. \Rightarrow 1 \geq \frac{n^n(a_1...a_n)}{a_1+...+a_n}. $ Rearranging this completes the problem.$ \\ (ii) $The natural logarithmic function is a concave down function, so its use is valid here. $ \frac{\ln(x_1x_2...x_n)}{n} \leq \ln(\frac{x_1+...x_n}{n}). $ Raising both sides to e, $ x_1...x_n \leq (\frac{x_1+...x_n}{n})^n$

$Second one is good but not to sure about your 1st one?$

$x \leq e^{x-1}$

$$ Let $ X= \frac{x_1+x_2+...+x_n}{n}$

$\frac{x_1}{X}\leq \frac{e^{\frac{x_1}{X}}}{e} ---(1)$

$\frac{x_2}{X} \leq \frac{e^{\frac{x_2}{X}}}{e} ---(2)$
...
$\frac{x_n}{X}\leq \frac{e^{\frac{x_n}{X}}}{e}---(n)$

$(1)(2)...(n) : \frac{x_1x_2...x_n}{X^n} \leq \frac{e^{\frac{1}{X}(nX)}}{e^n}$

$\therefore x_1x_2...x_n \leq X^n$

$(\frac{x_1+x_2+...+x_n}{n})^n \geq x_1x_2 ...x_n$

hedgehog_7 · Jan 30, 2016

Re: HSC 2016 4U Marathon

The tangents at two points P (x1,y1) and Q (x2,y2) are on the ellipse x^2/16 + y^2/9 = 1 intersect at A (x0,y0)

a) If the chord PQ: xx0 / 16 + yy0 / 9 = 1 touches the circle x^2 + y^2 / 9 , then by considering the distance of the chord from the origin, show that the point A (x0,y0) satisfies
9(x0)2 / 256 + (y0)^2 / 9 =1

b) Give a geometric descrpition of the locus of A

Drsoccerball · Jan 30, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent The integral $U_n$ is defined as $\int_0^\pi \frac{\cos{n x}}{5-4\cos{x}} $d$x $ for all non-negative $n. \\$a) Show $U_{n+1}+U_{n-1}-\frac{5}{2}U_n = 0 \\$b) Determine integers $A$ and $B$ such that $AU_0 -BU_1 = \pi\\$c) Evaluate $U_0$, and hence, $U_1.\\$ d) Using the reductive recurrence relation from a), determine the first few values of $U_n .\\$e) Conjecture the closed form of $U_n$. Prove your conjecture, by mathematical induction, or any other means you see fit.$

For (e) is the answer :
$U_n = \frac{\pi}{3(n+1)}$ ?

InteGrand · Jan 30, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
For (e) is the answer :
$U_n = \frac{\pi}{3(n+1)}$ ?

No

Drsoccerball · Jan 30, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
No

I guess I should use more terms rather than only the first two...

Paradoxica · Jan 30, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent By the way, for integration Q's with limits like this and a rational function of trig. functions in the integrand, we shouldn't expect the answer to have $\pi$ in it because when we find the antiderivative, it'll involve these trig. functions, so when we then sub. in the limits of integration, the $\pi$ will go away, as is the case when evaluating the trig. functions at such values.$

Um. The answer contains a linear function of pi. Just saying.

Drsoccerball · Jan 30, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Um. The answer contains a linear function of pi. Just saying.

Paradoxica= The New Integrand
Integrand= Drsoccerball

Paradoxica · Jan 30, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Paradoxica= The New Integrand
Integrand= Drsoccerball

Drsoccerball = Kaido

jks don't kill me pls

Drsoccerball · Jan 30, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Drsoccerball = Kaido

jks don't kill me pls

After what you told me in inbox I hope that's a joke

...

Why do you keep telling me not to kill you ahahah even on facebook?

InteGrand · Feb 5, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
Came across this in Fitzpatrick, my question is the book mentions about 'major axis' and 'minor axis'. Does major axis always refer to horizontal and minor axis always vertical?

First time I have also heard about semi-major axis and semi-minor axis.

No, 'major' just means the longer one, 'minor' the shorter one. It's just that for most of these HSC Ellipse Q's, they place the longer one on the horizontal axis.

The 'semi-' means half, so semi-major axis is half the major axis, like how a radius is half a diameter for a circle.

davidgoes4wce · Feb 5, 2016

Re: HSC 2016 4U Marathon

Sorry thats a better image:

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