4 Unit Revising Marathon HSC '10 (1 Viewer)

ninetypercent · Jan 13, 2010

conics

The normal at a point P on the curve xy=c^2 meets the asymptotes in Q, R. Show that the locus of the midpoint of QR has the equation $4x^3y^3 + c^2(x^2 - y^2)^2 = 0$

kaz1 · Jan 13, 2010

lol at 09ers revising for HSC '10.

addikaye03 · Jan 13, 2010

kaz1 said:
lol at 09ers revising for HSC '10.

Not many '10ers or '11ers posting ay. Get on here Cazic, out smart us all lol

xXmuffin0manXx · Jan 13, 2010

addikaye03 said:
Not many '10ers or '11ers posting ay. Get on here Cazic, out smart us all lol

probably because most have only just finished complex numbers

shaon0 · Jan 13, 2010

ninetypercent said:
conics

The normal at a point P on the curve xy=c^2 meets the asymptotes in Q, R. Show that the locus of the midpoint of QR has the equation $4x^3y^3 + c^2(x^2 - y^2)^2 = 0$

Can we make a new rule? No conics questions

gurmies · Jan 13, 2010

shaon0 said:
Can we make a new rule? No conics questions

Bear in mind that it's only early for '10ers - they would most probably be up to conics now =] There do exist some interesting conics questions, there was one I posted a solution for a while back which was quite unusual.

Trebla · Jan 13, 2010

untouchablecuz said:
$R=\cos(t\sqrt{\alpha\beta}) $ and $J=-\frac{\alpha}{\sqrt{\alpha\beta}}\sin(t\sqrt{\alpha\beta})$

not sure about their fate, tho i do know that $\frac{J^2}{(\frac{\alpha}{\beta})}+R^2=1$ - their love forms an ellipse

Well basically, their level love will always be out of phase with each other if you consider the behaviour of a sine curve relative to a cosine curve i.e. their romance is doomed to fail!

Also, although this came from a first year university course, the way the question is worded makes it perfectly acceptable as a potential Ext2 question (as Harder Ext1) because it uses tools that are not new to you. You don't even need to know about second order ODEs to solve this question because it guides you into solving it...

shaon0 · Jan 13, 2010

gurmies said:
Bear in mind that it's only early for '10ers - they would most probably be up to conics now =] There do exist some interesting conics questions, there was one I posted a solution for a while back which was quite unusual.

Yeah, i solved a question in 4u a long time ago which basically used similar triangles and geometry opposed to algebra bashing.

jet · Jan 14, 2010

Keep the focus on the game people

shaon0 · Jan 14, 2010

equ of normal:
px-y/p=c(p^2-(1/p^2))
At x=0:
y=-cpk where k=p^2-p^-2
At y=0:
x=ck/p

Mid QR: (ck/2p,-cpk/2)
x=ck/2p and y= -cpk/2
p^2=-y/x

2px=ck where k=y^2/x^2-x^2/y^2=y^4-x^4/(xy)^2
4p^2x^2=c^2k^2
4x^4.y^2(-y/x)=c^2(y^2-x^2)^2
4x^4.y^2(y/x)+c^2(y^2-x^2)^2=0
4x^3.y^3+c^2(x^2-y^2)^2=0

Someone post up a question.

addikaye03 · Jan 14, 2010

Unless someone wants to do the above tedious Conics Q.

New Questions:

1. (Pretty easy):
If a+b=-3 and ab=1, find rt(a/b)+rt(b/a).

2. (Bonus Q, i can't think of a soln):
∫cosx dx/(7+cos2x)

[Probably something really simple, just blonde moment lol]

untouchablecuz · Jan 14, 2010

addikaye03 said:
Unless someone wants to do the above tedious Conics Q.

New Questions:

1. (Pretty easy):
If a+b=-3 and ab=1, find rt(a/b)+rt(b/a).

2. (Bonus Q, i can't think of a soln):
∫cosx dx/(7+cos2x)

[Probably something really simple, just blonde moment lol]

$\\1. \ \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=\frac{a+b}{\sqrt{ab}}=-3\\2. \ \int \frac{\cos x}{7+\cos2x}dx=\int \frac{\cos x}{7+1-2\sin^2x}dx=\int \frac{\cos x}{8-2\sin^2x}dx=\frac{1}{2}\int \frac{1}{4-\sin^2x}d(\sin x), $ proceed with partial fractions$$

addikaye03 · Jan 14, 2010

untouchablecuz said:
$\\1. \ \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=\frac{a+b}{\sqrt{ab}}=-3\\2. \ \int \frac{\cos x}{7+\cos2x}dx=\int \frac{\cos x}{7+1-2\sin^2x}dx=\int \frac{\cos x}{8-2\sin^2x}dx=\frac{1}{2}\int \frac{1}{4-\sin^2x}d(\sin x), $ proceed with partial fractions$$

God i'm retarded, thanks untouchable, really shoulda got that lol.. I did Q.1 a little differently.

Since ab=1, and a+b=-3 therefore a and b are negative. rt(a^2)=-a. a=1/b, and b=1/a.

rt(a^2)+rt(b^2)=(-a)+(-b)=-(a+b)=3. LOL Yours makes mine look tedious. Wouldn't your solution suggest the answer is -3? (answer: a,b < 0. On the denominator rt (a) x rt (b).)

gurmies · Jan 14, 2010

shaon0 · Jan 14, 2010

gurmies said:
$\left ( a \right )\text{Show that} \\\\ \frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta}=i\cot\frac{\theta}{2} \\\\ \left ( b \right )\text{Let n be a positive integer. Show that all the roots of the equation}\left ( z-1 \right )^{n}+\left ( z+1 \right )^{n}=0 \,\, \text{can be written as} \,\, i\alpha_{k}, \,\, \text{where} \,\, \alpha_{k}\in \mathbb{R}, \,\, k=0,1,...,n-1. \\\\ \left ( c \right )\text{Using the relations between roots and coefficients, show that }\sum_{k=0}^{n-1}\alpha_{k}^{2}=n\left ( n-1 \right )$ $\left ( d \right )\text{Let} \,\, P_{0}, P_{1},...,P_{n-1} \,\, \text{be the points in an Argand plane representing the roots of the equation in (b)}.\,\, \\\\ Q \,\, \text{is the point representing} \,\, r\left ( \cos\beta+i\sin\beta \right ) \,\, \text{where}\,\,r\geq 0 \,\,\text{and} \,\, \beta \in \mathbb{R}. \,\,\\\\ \text{If} \,\, d_{k} \,\, \text{is the distance between} \,\, P_{k} \,\, \text{and} \,\, Q, \,\, \text{show that} \,\, \sum_{k=0}^{n-1}d^{2}_{k} \,\, \text{is independent of} \,\, \beta.$

(1+cos@+isin@)/(1-cos@-isin@)
=(1+2(cos*)^2-1+2sin*cos*)/(1-1+2(sin*)^2-2isin*cos*) where *=@/2
=icos*/sin*.(cis*/cis*)
=icot*

(z-1)^n+(z+1)^n=0
1+((z+1)/(z-1))^n=0
1+(-icot*)^n=0
cot*=i^(2/n-3)

{sum} a0a1+a0a2+...+a{n-2}a{n-1}=(n n-2)
Since roots are of form ia{k}:
2a0^2+2a1^2+...+2a{n-1}^2=(n n-2)
{sum} a{k}^2=0.5 (n n-2)
{sum} a{k}^2=n(n-1)

(n)(r^2+1) maybe the answer for the last part but i'll try the last part after lunch

cyl123 · Jan 14, 2010

shaon0 said:
1+((z+1)/(z-1))^n=0
1+(-icot*)^n=0
cot*=i^(2/n-3)

This part requires you assume z=cosx+isinx which assumes |z|=1

Rather:
((z+1)/(z-1))^n=-1
Taking mod from both sides gives:
|z+1|^n/|z-1|^n=1
|z+1|/|z-1|=1
|z+1|=|z-1|

Solving for z geometrically implies z is purely imaginary or 0, and hence can be expressed as z=ia where a is real.

Trebla · Jan 14, 2010

Here's an interesting one (in my opinion anyway):

$$Suppose that $:\\\\S_{n}=\sum_{j=0}^{n}\dfrac{\sin[(2j+1)x]}{2j+1}\\\\(a)$ Show that $S_{n} = \displaystyle\int_{0}^{x} \left (\sum_{j=0}^{n} \cos [(2j+1)y] \right )\,dy\\\\(b)$ Use the fact that $e^{i\theta}=\cos \theta + i\sin \theta$ to show that $\\\\ (i) \cos [(2j+1)y] = \dfrac{1}{2}\left ( e^{iy(2j+1)}+e^{-iy(2j+1)} \right )\\\\$and$\\\\(ii) \sin [(2j+1)y] = \dfrac{1}{2i}\left ( e^{iy(2j+1)}-e^{-iy(2j+1)} \right )\\\\(c) $ Use the result from (b)(i) in (a) to show that$\\\\S_{n}=\dfrac{1}{2}\displaystyle\int_{0}^{x}\left [e^{iy}\left (\dfrac{e^{2iy(n+1)}-1}{e^{2iy}-1} \right )+e^{-iy}\left (\dfrac{e^{-2iy(n+1)}-1}{e^{-2iy}-1} \right ) \right ]\,dy\\\\(d) $ Hence, use the result in (b)(ii) to deduce that $\\\\ \sum_{j=0}^{n}\dfrac{\sin[(2j+1)x]}{2j+1} = \dfrac{1}{2} \int_{0}^{x}\dfrac{\sin[2y(n+1)]}{\sin y}\,dy$

jet · Jan 15, 2010

Ok guys, new guideline:

Please use LaTeX. It is much easier to read, and it just looks pretty!

I'll add it to the rules at the start. If you absolutely hate LaTeX with a passion and refuse to use it, I will edit your post once you do your solution, and turn it into LaTeX. Does that sound fair?

gurmies · Jan 15, 2010

Beautiful. Also I see there's no solution to that problem yet. I will post one if there's none by 11:30 - started writing yesterday, but there was tons of code.

Koroneos · Jan 16, 2010

$(a)\,\,S_n=\sum_{j=0}^{n}\int_{0}^{x}\cos[(2j+1)y]dy\\=\sum_{j=0}^{n}\frac{\sin[(2j+1)x]}{2j+1}$

$(b)(i)\,\, 2\cos\theta=cis\theta+cis(-\theta)\\2\cos\theta=e^{i\theta}+e^{-i\theta}\\\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})\\\cos[(2j+1)y]=\frac{1}{2}(e^{iy(2j+1)}+e^{-iy(2j+1)})$by letting$\,\,\theta=[(2j+1)y]$

$(ii)\,\,2i\sin\theta=cis\theta-cis(-\theta)\\2i\sin\theta=e^{i\theta}-e^{-i\theta}\\\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\\\sin[(2j+1)y]=\frac{1}{2i}(e^{iy(2j+1)}-e^{-iy(2j+1)})$by letting$\,\,\theta=[(2j+1)y]$

$(c)\,\,S_n=\int_{0}^{x}\sum_{j=0}^{n}\cos[(2j+1)y]dy\\=\frac{1}{2}\int_{0}^{x}(\sum_{j=0}^{n}e^{iy(2j+1)}-\sum_{j=0}^{n}e^{-iy(2j+1)})dy\\=\frac{1}{2}\int_{0}^{x}\left [ e^{iy}\left ( \frac{e^{2iy(n+1)}-1}{e^{2iy}-1} \right )+e^{-iy}\left ( \frac{e^{-2iy(n+1)}-1}{e^{-2iy}-1} \right ) \right ]dy\,\,$using sum of GP$$

$(d)\,\,\frac{1}{2}\int_{0}^{x}\frac{\sin[2y(n+1)]}{\sin y}dy\\=\frac{1}{2}\int_{0}^{x}\frac{e^{2iy(n+1)}-e^{-2iy(n+1)}}{e^{iy}-e^{-iy}}dy\\=\frac{1}{2}\int_{0}^{x}\frac{e^{2iny+3iy}-e^{-2iny-iy}}{e^{2iy}-1}dy\\=\frac{1}{2}\int_{0}^{x}(\frac{e^{2iny+3iy}-e^{iy}}{e^{2iy}-1}-\frac{e^{-2iny-iy}-e^{iy}}{e^{2iy}-1})dy\\=\frac{1}{2}\int_{0}^{x}\left [ e^{iy}\left ( \frac{e^{2iy(n+1)}-1}{e^{2iy}-1} \right )+e^{-iy}\left ( \frac{e^{-2iy(n+1)}-1}{e^{-2iy}-1} \right ) \right ]dy\\=S_n\,\,$from part c$\\=\sum_{j=0}^{n}\frac{\sin[(2j+1)x]}{2j+1}$

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4 Unit Revising Marathon HSC '10 (1 Viewer)

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