HSC 2016 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Nov 3, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
My teacher said it's allowed....
.-.

Never ever use it in the HSC unless you are wiling to explain what a taylor series is and how to derive the result....

dan964 · Nov 3, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$I= \int_0^{2\pi}{sin(nx)(sin(2x)}dx$

$I= \int_0^{2 \pi}{\frac{z^n -\frac{1}{z^n}}{2i} \cdot \frac{z^2-\frac{1}{z^2}}{2i}}dx$

$I= \frac{-1}{4}\int_0^{2\pi}{(z^{n+2} +\frac{1}{z^{n+2}} -(z^{n-2}+\frac{1}{z^{n-2}}}))dx$

$I= \frac{-1}{4} \int_0^{2 \pi}{cos(n+2)x -cos(n-2)x}dx$

The rest is straight forward...

So yes yours seems right.

Method is good for HSC but...
Your 4th line is missing something, when you reapply the formula proven in the previous part; a constant comes out (a 1/2 be exact). This happens twice but since you can factorise, so the 1/4 out the front should be a 1/8.
$I= \frac{-1}{8} \int_0^{2 \pi}{cos(n+2)x -cos(n-2)x}dx$

Paradoxica · Nov 3, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Never ever use it in the HSC unless you are wiling to explain what a taylor series is and how to derive the result....

What about the differential equation solution, which both of these satisfy? (or is there some difficulty in elementarily justifying that if two functions are both solutions to a second-order differential equation, they are identical?)

InteGrand · Nov 3, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
What about the differential equation solution, which both of these satisfy? (or is there some difficulty in elementarily justifying that if two functions are both solutions to a second-order differential equation, they are identical?)

Just because two functions satisfy the same second-order ODE, it doesn't mean they are identical.

dan964 · Nov 3, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
What about the differential equation solution, which both of these satisfy? (or is there some difficulty in elementarily justifying that if two functions are both solutions to a second-order differential equation, they are identical?)

not worth it

Drsoccerball · Nov 3, 2015

Re: HSC 2016 4U Marathon

dan964 said:
Method is good for HSC but...
Your 4th line is missing something, when you reapply the formula proven in the previous part; a constant comes out (a 1/2 be exact). This happens twice but since you can factorise, so the 1/4 out the front should be a 1/8.
$I= \frac{-1}{8} \int_0^{2 \pi}{cos(n+2)x -cos(n-2)x}dx$

Yes sorry about that. The working out was rushed.

Paradoxica · Nov 3, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
Just because two functions satisfy the same second-order ODE, it doesn't mean they are identical.

Ah yes, thanks for clearing that. One of my peers claimed that was so (for the same situation I described), and I instantly doubted him, but did not have any merit to my claim. Also, I forgot to mention the differing of constants, but that probably doesn't change the matter of the fact.

Sy123 · Nov 4, 2015

Re: HSC 2016 4U Marathon

$\\ $The polynomial$ \ ax^3 + bx^2 + bx + a \ $is such that$ \ a,b \ $are positive and all the roots are real, prove that$ \ b \geq 3a$

Paradoxica · Nov 4, 2015

Re: HSC 2016 4U Marathon

Sy123 said:
$\\ $The polynomial$ \ ax^3 + bx^2 + bx + a \ $is such that$ \ a,b \ $are positive and all the roots are real, prove that$ \ b \geq 3a$

$If the roots exist, then there must be two stationary points, except in the doubly degenerate case where they are identical. So $3ax^2 + 2bx + b$ has real (not necessarily distinct) roots. Taking the discriminant to be non-negative, we have the inequality $4b^2 - 4(3a)(b) \geq 0 \Rightarrow b^2 \geq 3ab$. As a and b are given to be positive, we can safely divide b out of the inequality. This leads to $b \geq 3a$ and we are done.$

leehuan · Nov 4, 2015

Re: HSC 2016 4U Marathon

Until some fresh 2016'ers come in, and topic order becomes more important in my eyes, here's a question.

$\\$A particle moves in uniform circular motion. Prove that the centripetal force acting on this particle is $F=\frac { m{ v }^{ 2 } }{ r } $ towards the centre of the circle it traces.$

Paradoxica · Nov 4, 2015

Re: HSC 2016 4U Marathon

$$a) Expand $(x^2 + 3x + 1)^2$
$b) Hence, prove $\frac{5^{125} -1}{5^{25} -1}$ is a composite number.$

calamebe · Nov 5, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
$$a) Expand $(x^2 + 3x + 1)^2$
$b) Hence, prove $\frac{5^{125} -1}{5^{25} -1}$ is a composite number.$

At school so no pic:

a) x^4 + 6x^3 + 11x^2 + 6x + 1

b) (5^125-)/(5^25-1)=5^100 + 5^75 + 5^50 + 5^25 + 1
I let 5^25 = x
x^4 + x^3 + x^2 + x + 1 = x^4 + 6x^3 + 11x^2 + 6x + 1 -5x^3 -10x^2 - 5x
=(x+3x+1)^2 -5x(x+1)^2
=(x+3x+1-(x+1)root(5x))(x+3x+1+(x+1)root(5x))
Root(5x)=root(x^26)=x^13
Therefore it's factors are rational, so it is a composite number

Paradoxica · Nov 5, 2015

Re: HSC 2016 4U Marathon

calamebe said:
At school so no pic:

a) x^4 + 6x^3 + 11x^2 + 6x + 1

b) (5^125-)/(5^25-1)=5^100 + 5^75 + 5^50 + 5^25 + 1
I let 5^25 = x
x^4 + x^3 + x^2 + x + 1 = x^4 + 6x^3 + 11x^2 + 6x + 1 -5x^3 -10x^2 - 5x
=(x+3x+1)^2 -5x(x+1)^2
=(x+3x+1-(x+1)root(5x))(x+3x+1+(x+1)root(5x))
Root(5x)=root(x^26)=x^13
Therefore it's factors are integral, so it is a composite number

FTFY

calamebe · Nov 5, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
FTFY

Please explain

InteGrand · Nov 5, 2015

Re: HSC 2016 4U Marathon

calamebe said:
Please explain

First of all, "factors" are by definition integers (which implies they are rational). To prove a positive integer N ≥ 2 is composite, we are trying to show that it can be written as N = ab, where a and b are both integers with 1 < a, b < N (i.e. both strictly between 1 and N).

leehuan · Nov 5, 2015

Re: HSC 2016 4U Marathon

Also, recall that a rational number is any number that can be expressed in the form p/q where p and q are integers, and q=/=0.

This means a rational number could be say 4/151

calamebe · Nov 5, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Also, recall that a rational number is any number that can be expressed in the form p/q where p and q are integers, and q=/=0.

This means a rational number could be say 4/151

InteGrand said:
First of all, "factors" are by definition integers (which implies they are rational). To prove a positive integer N ≥ 2 is composite, we are trying to show that it can be written as N = ab, where a and b are both integers with 1 < a, b < N (i.e. both strictly between 1 and N).

Ah, so does integral pretty much just mean they are integers? Yeah I was confused why it was wrong that I wrote rational, now it's clear, thanks!

InteGrand · Nov 5, 2015

Re: HSC 2016 4U Marathon

calamebe said:
Ah, so does integral pretty much just mean they are integers? Yeah I was confused why it was wrong that I wrote rational, now it's clear, thanks!

Yeah, integral is another word for integer (as an 'adjective').

leehuan · Nov 5, 2015

Re: HSC 2016 4U Marathon

calamebe said:
Ah, so does integral pretty much just mean they are integers? Yeah I was confused why it was wrong that I wrote rational, now it's clear, thanks!

$\int { \left( confusing \right) d\left( terminology \right) }$

leehuan · Nov 6, 2015

Re: HSC 2016 4U Marathon

$\\ $Show that $\frac { \cos { a } +i\sin { a } }{ \cos { b } +i\sin { b } } =\cos { \left( a-b \right) } +i\sin { \left( a-b \right) }$

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