HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

Sy123 · Dec 10, 2014

Re: HSC 2015 4U Marathon - Advanced Level

SilentWaters said:
If $p(x)$ has integer coefficients, then the product of its roots must be of a rational form (the quotient of two integers, the leading coefficient and the constant). To rationalize such a product with some irrational $p+\sqrt{q}$ we must have its conjugate as a necessary root.

What about for example a polynomial with roots $\sqrt{2}$ and $\sqrt{8}$ ?

Just knowing about the product of roots alone is not enough to conclude that each irrational pair needs its conjugate

SilentWaters · Dec 10, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Alternatively, set:

$p(x) = \left[x-(p+\sqrt{q})\right]\left[x+(p-\sqrt{q})\right]\cdot q(x)+ax+b$
$=\left[(x-p)^{2} - q\right]\cdot q(x) +ax+b$

where the coefficients of $q(x)$ , together with a and b, are rational. Since $p+\sqrt{q}$ is a root,

$p(p+\sqrt{q}) = a(p+\sqrt{q}) + b = 0$

But here we'd need an equality of a rational and irrational side for non-zero $a$ and $b$ :

$a(p+\sqrt{q}) = -b$

This forces us to have $a,b = 0$ , which makes our conjugate pairs indeed roots of $p(x)$ without remainder.

SilentWaters · Dec 10, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Sy, empty your PM box

Sy123 · Dec 10, 2014

Re: HSC 2015 4U Marathon - Advanced Level

SilentWaters said:
Sy, empty your PM box

Done

Sy123 · Dec 10, 2014

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Let,$ \ n \ $be a positive even integer, and let$ \ p(x) \ $be a polynomial of $ \ n$-th degree, such that, $ \ p(k) = p(-k) \ $, for $ \ k=1,2,3, \dots, n \ \\ $Prove that there exists a polynomial,$ \ q(x) \ $such that,$ \ p(x) = q(x^2)$

seanieg89 · Dec 11, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Writing e=(f+(-f))/2, o=(f-(-f))/2 we can write any function as the sum of an odd and even part.

p(k)=p(-k) for all k implies that the odd part of p must vanish at 1,2,...,n. As any odd function also vanishes at zero, this odd part must be the zero polynomial. (It has more roots than its degree).

By direct calculation though, the odd part of a real polynomial is just the sum of terms with odd degree. Saying this is zero is just saying that all terms in p have even degree, from which the conclusion follows.

Sy123 · Dec 11, 2014

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
Writing e=(f+(-f))/2, o=(f-(-f))/2 we can write any function as the sum of an odd and even part.

p(k)=p(-k) for all k implies that the odd part of p must vanish at 1,2,...,n. As any odd function also vanishes at zero, this odd part must be the zero polynomial. (It has more roots than its degree).

By direct calculation though, the odd part of a real polynomial is just the sum of terms with odd degree. Saying this is zero is just saying that all terms in p have even degree, from which the conclusion follows.

Nice!

Here is my solution:

$\\ $The function$ \ f(x) = p(x) - p(-x) \ $has$ \ n \ $roots, at$ \ x=1,2,3, \dots, n \\ \\ $Since$ \ f \ $is the sum of polynomials it is itself a sum of polynomials, the leading term of$ \ p(x) \ $is identical to the leading term of$ \ p(-x) \ $so the leading term of$ \ f \ $must be$ \ 2a_{n-1}x^{n-1} \ $where$ \ a_{n-1} \ $is the coefficient of$ \ x^{n-1} \ $in$ \ p(x) \\ \\ $So, by the Fundamental Theorem of Algebra$ \ f \ $has$ \ (n-1) \ $roots, but$ \ f \ $must at least have$ \ n \ $roots, so$ \ a_{n-1} = a_{n-3} = \dots = 0 \\ \\ $That is, all odd degree coefficients must be zero, and so then we must have$ \ f(x) = 0 \\ \\ $Therefore, there are only even degree terms in$ \ p \ $so there must be a polynomial$ \ q(x^2) = p(x)$

Sy123 · Dec 11, 2014

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $A monic cubic has roots$ \ \alpha, \beta, \gamma \ $show that all 3 roots are real if and only if$ \ (\alpha - \beta)^2(\beta - \gamma)^2(\alpha - \gamma)^2 > 0$

seanieg89 · Dec 11, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Here is the most elementary proof I know of the fundamental theorem of algebra.

(You may assume that every continuous function attains a minimum on any disk of the form $D_r=\{z\in\mathbb{C}:|z|\leq r\}$ . This means that there exists $w\in D_r$ such that $f(z)\geq f(w)$ for all $z\in D_r$ .)

Let P be a degree n polynomial with no complex roots.

1. Show that

$|P(z)|=|\sum_{k=0}^n a_k z^k| > |a_0|$

for all z with

$|z|>\max\{\frac{\sum_{k=0}^{n-1}|a_k|+|a_0|}{|a_n|},1\}.$

2. Deduce that $|P(z)|$ attains a nonzero minimum over all of $\mathbb{C}$ at some $z_0\in\mathbb{C}$ .

3. Let

$Q(z):=\frac{P(z_0-z)}{P(z_0)}.$

Show that the constant term of Q is 1. Let $az^k$ be the nonzero term of least positive degree.

4. Explain why we can find complex c with $c^k=-a^{-1}.$

5. Show that $R(z):=Q(cz)=1-z^k+z^{k+1}S(z)$ for some polynomial S.

6. Show that $|R(x)|<1$ for some small positive x.

7. Deduce a contradiction, and make your conclusion.

InteGrand · Dec 11, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $A monic cubic has roots$ \ \alpha, \beta, \gamma \ $show that all 3 roots are real if and only if$ \ (\alpha - \beta)^2(\beta - \gamma)^2(\alpha - \gamma)^2 > 0$

I think we need to add the condition that the polynomial has only real coefficients, since otherwise, we could have non-real roots not coming in conjugate pairs, which could satisfy the given inequality.

Also, the greater than sign should be a greater-than-or-equal-to sign I think, since there could be a real triple root, which would make the LHS equal to 0.

seanieg89 · Dec 12, 2014

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
I think we need to add the condition that the polynomial has only real coefficients, since otherwise, we could have non-real roots not coming in conjugate pairs, which could satisfy the given inequality.

Also, the greater than sign should be a greater-than-or-equal-to sign I think, since there could be a real triple root, which would make the LHS equal to 0.

If there are any multiple roots at all (real or not), equality holds, so we have no iff.

The statement should probably be:

For real polys, show that all three roots are distinct and real iff blah > 0.

dan964 · Dec 20, 2014

Re: HSC 2015 4U Marathon - Advanced Level

a polynomial of order n
has roots sin and cosine of Pi/(4n) respectively
for n>0
(I) Deduce using Demouivre's Theorem where necessary that n is of the form 4k + 1
(II) Prove by mathematical induction
that the product of the roots is always -sqrt(0.5)

part (I) is easy
part (II) hard

turntaker · Dec 20, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Euler's>demoives

dan964 · Dec 20, 2014

Re: HSC 2015 4U Marathon - Advanced Level

turntaker said:
Euler's>demoives

Euler's is barely in the scope then again neither are most of my q.

trapizi · Dec 22, 2014

Re: HSC 2015 4U Marathon - Advanced Level

Solve for x
$y=\frac{ln(\frac{x}{m}-sa)}{r^2}$

integral95 · Dec 22, 2014

Re: HSC 2015 4U Marathon - Advanced Level

trapizi said:
Solve for x
$y=\frac{ln(\frac{x}{m}-sa)}{r^2}$

$Me^{rry} = x-mas$

Sy123 · Jan 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Let$ \ P \ $be a point on the plane$ \ p \ $and$ Q \ $is a fixed point in space that does not lie on$ \ p \\ \\ $Find all points$ \ R \ $on the plane$ \ p \ $such that$ \\ \\ \frac{PR + PQ}{QR} \ $is a maximum$$

FrankXie · Jan 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ P \ $be a point on the plane$ \ p \ $and$ Q \ $is a fixed point in space that does not lie on$ \ p \\ \\ $Find all points$ \ R \ $on the plane$ \ p \ $such that$ \\ \\ \frac{PR + PQ}{QR} \ $is a maximum$$

$$ My answer is that $ \max\frac{PR+PQ}{QR}={\rm cosec}\frac{\alpha}{2} $ (where $ \alpha $ is the angle between the line $ PQ $ and the plane $ p $ ) and that the maximum is attained when $ R $ lies in the line $ PH $ such that $ PR=PQ. $ Here $ H $ is the foot of perpendicular from $ Q $ to the plane $ p.$

Sy123 · Jan 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

FrankXie said:
$$ My answer is that $ \max\frac{PR+PQ}{QR}={\rm cosec}\frac{\alpha}{2} $ (where $ \alpha $ is the angle between the line $ PQ $ and the plane $ p $ ) and that the maximum is attained when $ R $ lies in the line $ PH $ such that $ PR=PQ. $ Here $ H $ is the foot of perpendicular from $ Q $ to the plane $ p.$

The angle alpha is variable remember!

So we need to minimise the angle alpha, so R is only 1 point on the plane (rather than a circle on the plane p), the specific point that lies on the line on the plane p that has the smallest angle between it and PQ

Sy123 · Jan 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

FrankXie said:
$$ My answer is that $ \max\frac{PR+PQ}{QR}={\rm cosec}\frac{\alpha}{2} $ (where $ \alpha $ is the angle between the line $ PQ $ and the plane $ p $ ) and that the maximum is attained when $ R $ lies in the line $ PH $ such that $ PR=PQ. $ Here $ H $ is the foot of perpendicular from $ Q $ to the plane $ p.$

The angle alpha is variable remember!

So we need to minimise the angle alpha, so R is only 1 point on the plane (rather than a circle on the plane p), the specific point that lies on the line on the plane p that has the smallest angle between it and PQ

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