HSC 2015 MX2 Marathon (archive) (6 Viewers)

braintic · Feb 10, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\tex${Describe and show working (sketching on the complex plane not necessary) satisfying \\ \mid z-i|=|z-1|.}$

If this were an HSC question, working would not be expected. In fact, they would probably ONLY ask for a sketch.

Drsoccerball · Feb 11, 2015

Re: HSC 2015 4U Marathon

Guys the forum says all questions must be answered before posting new ones..

Carrotsticks · Feb 11, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Guys the forum says all questions must be answered before posting new ones..

Sometimes questions get forgotten, we don't need to exercise that rule with an iron fist.

Drsoccerball · Feb 11, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
Sometimes questions get forgotten, we don't need to exercise that rule with an iron fist.

Even so there was a question 1 post before when he posted it i don't think it is that easily forgotten

Drsoccerball · Feb 11, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\tex${Describe and show working (sketching on the complex plane not necessary) satisfying \\ \mid z-i|=|z-1|.}$

the equation is y=x

VBN2470 · Feb 11, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Anyways new question:
$arg(z-2) = arg(z+2) + \frac{\pi}{3} $Find the center, radius and area of this circle$$
I added the area bit in lets see how this goes

$Using properties of Complex Numbers and standard Geometry, the locus is the major arc of a circle with radius $\frac{4}{\sqrt3}$ and centre $(0, \frac{2}{\sqrt3})$. Area of the entire circle is therefore $\frac{16\pi}{3}$. \\ NOTE that the points $(\pm2,0)$ are not included in the locus since $\arg(0)$ is undefined.$

VBN2470 · Feb 11, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION

$$ Show that the remainder when the polynomial $P(x)$ is divided by $x^2-a^2$ is $\frac{1}{2a}[P(a)-P(-a)]x+\frac{1}{2}[P(a)+P(-a)]$. \\ Find the remainder when $P(x)=x^n-a^n$ is divided by $x^2-a^2$ for non-negative values of $n$.$

Bump

Librah · Feb 11, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using properties of Complex Numbers and standard Geometry, the locus is the major arc of a circle with radius $\frac{4}{\sqrt3}$ and centre $(\frac{2}{\sqrt3}, 0)$. Area of the circle is therefore $\frac{16\pi}{3}$. \\ NOTE that the points $(\pm2,0)$ are not included in the locus since $\arg(0)$ is undefined.$

Supposed to be (0, 2/root 3). Also i don't know if he was asking for the area of part of the circle described by the locus or the entire circle itself.

VBN2470 · Feb 11, 2015

Re: HSC 2015 4U Marathon

YunLi said:
Supposed to be (0, 2/root 3). Also i don't know if he was asking for the area of part of the circle described by the locus or the entire circle itself.

Thanks, just fixed it, yeah I am also not sure if the area was referring to the whole circle or just that major segment of it described by the locus. Will just have to wait and see what the OP of the question says.

Drsoccerball · Feb 11, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Thanks, just fixed it, yeah I am also not sure if the area was referring to the whole circle or just that major segment of it described by the locus. Will just have to wait and see what the OP of the question says.

its the part thats visible

Ekman · Feb 11, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
its the part thats visible

Yeah just do:

$\frac{16\pi}{3} \times \frac{2}{3} = \frac{32\pi}{9}$

To get the area of the visible part.

Librah · Feb 11, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Yeah just do:

$\frac{16\pi}{3} \times \frac{2}{3} = \frac{32\pi}{9}$

To get the area of the visible part.

I got (32pi+12root3)/9

Sy123 · Feb 11, 2015

Re: HSC 2015 4U Marathon

$\\ $Rigorously explain why polynomials with real co-efficients with an odd degree have at least one real root$$

VBN2470 · Feb 11, 2015

Re: HSC 2015 4U Marathon

$Consider the polynomial $P(z)=a_0+a_1z+...+a_nz^n$, where $a_1, a_2,...,a_n$ are all real and $n$ is an odd integer. Suppose that $\alpha$ is a root of $P(z)$ so that $P(\alpha)=0$. \\ Then $P(\bar{\alpha})=\displaystyle\sum_{k=0}^{n}a_k (\overline{\alpha})^k=\displaystyle\sum_{k=0}^{n} \overline{a_k\alpha^k}=\overline{\displaystyle\sum_{k=0}^{n}{a_k\alpha^k}}=\overline{P(\alpha)}=0.$ \\ Then, by the Fundamental Theorem of Algebra, since the degree of the polynomial is odd, there exists an odd number of roots for $P(z)$. But since all roots of $P(z)$ exist in complex conjugate pairs, this implies the existence of at least one root, say $\beta$, such that $\beta$ = $\bar{\beta}$ \Rightarrow \, $\beta$ is real.\\Thus, it follows that for every odd degree polynomial with real coefficients, there must exist at least one real root.$

VBN2470 · Feb 11, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEW QUESTION

$$ Show that the remainder when the polynomial $P(x)$ is divided by $x^2-a^2$ is $\frac{1}{2a}[P(a)-P(-a)]x+\frac{1}{2}[P(a)+P(-a)]$. \\ Find the remainder when $P(x)=x^n-a^n$ is divided by $x^2-a^2$ for non-negative values of $n$.$

.

Drsoccerball · Feb 12, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
.

Does the question start from "find" or "show" ?

VBN2470 · Feb 12, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Does the question start from "find" or "show" ?

First part is to 'show', then second part is to 'find' i.e. use the first part to do the second, so it is like a 'Hence, find...' sort of question.

FrankXie · Feb 12, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
.

$$ let $ P(x)=(x^2-a^2)Q(x)+Ax+B $ (the division transformation). sub. $ x=\pm a, P(a)=aA+B, P(-a)=-aA+B. $ solving simultaneous equations for $ A, B, A=\frac{1}{2a}[P(a)-P(-a)], B=\frac{1}{2}[P(a)+P(-a)].$
$$ when $ P(x)=x^n-a^n, A=\frac{1-(-1)^n}{2}a^{n-1}, B=\frac{(-1)^n-1}{2}a^n$
that is, when n is even, p(x) is divisible by x^2-a^2; when n is odd, the remainder is a^{n-1}x-a^n.

FrankXie · Feb 12, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

Each of a set of equal cubes has its six faces painted red, orange, yellow, green, blue and white respectively so that no two cubes are alike and every arrangement of colours is used. How many cubes are there in the set?

braintic · Feb 12, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

Each of a set of equal cubes has its six faces painted red, orange, yellow, green, blue and white respectively so that no two cubes are alike and every arrangement of colours is used. How many cubes are there in the set?

Just checking - is the answer is 30?
(No working, so that someone else can give it a go)

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