HSC 2014 MX2 Marathon (archive) (1 Viewer)

[Axio]

Re: HSC 2014 4U Marathon

I did it using complex numbers, I think this works:

a+11i rotated pi/3 = b+37i.

So (a+11i) x cispi/3 = b+37i,

then solving the real and imaginary parts makes ab = 21root3 x 5root3 = 315

 

[Sy123]

Re: HSC 2014 4U Marathon

I did it using complex numbers, I think this works:

a+11i rotated pi/3 = b+37i.

So (a+11i) x cispi/3 = b+37i,

then solving the real and imaginary parts makes ab = 21root3 x 5root3 = 315

Yep that's the idea

 

[awesome-0_4000]

Re: HSC 2014 4U Marathon

I am sure there is a more efficient method, but here is my attempt:


Edit: both the plus signs in the square roots for the final expression should be minuses.

 

[Chlee1998]

Re: HSC 2014 4U Marathon

Though looking back on it, there really isn't much difference in flipping it or not lol.

Which was why I was confused and asked if what you wrote was a typo

 

[braintic]

Re: HSC 2014 4U Marathon

Answer: cos(pi/10)

Done. 4U is so easy.

 

[Axio]

Re: HSC 2014 4U Marathon

(2000 Q5i) Consider the polynomial: p(x) = ax^4 + bx^3 + cx^2 + dx + e, where a, b, c, d and e are integers. Suppose n is an integer such that p(n) = 0. Prove that n divides e.

 

[Chlee1998]

Re: HSC 2014 4U Marathon

(2000 Q5i) Consider the polynomial: p(x) = ax^4 + bx^3 + cx^2 + dx + e, where a, b, c, d and e are integers. Suppose n is an integer such that p(n) = 0. Prove that n divides e.

p(n) = an^4 + bn^3+ .......+ e = 0

-e = an^4 + bn^3 + .....+dn
= n(an^3+bn^2+.......d)
Since factor in brackets is integer, -e/n must be integral hence n divides -e and hence also divides e

 

[Sy123]

Re: HSC 2014 4U Marathon

 

[FrankXie]

Re: HSC 2014 4U Marathon

Solve simultaneous equations to find points of intersection. Let the circle be $(x-h)^2+(y-k)^2=r^2$. Then sub $y=\frac{1}{x}$. Thine resulting quartic equation should have solutions $x_1, x_2, x_3, x_4$. Finally use product of roots.

 

[dan964]

Re: HSC 2014 4U Marathon

Here is a question:
(i) Prove that e^x is lim (n--> infinity) [x^0/0! + x^1/1! ... x^n/n!] by integration
/ 1
(ii) By assuming Euler's formula (e^iq = cis q) derive expressions for cos x and sin x and hence deduce that d/dx sin (x) = cos x
/ 2

edit: don't attempt...

 

[Carrotsticks]

Re: HSC 2014 4U Marathon

Here is a question:
(i) Prove that e^x is lim (n--> infinity) [x^0/0! + x^1/1! ... x^n/n!] by integration
/ 1
(ii) By assuming Euler's formula (e^iq = cis q) derive expressions for cos x and sin x and hence deduce that d/dx sin (x) = cos x
/ 2

The first question is not suitable for this thread. It belongs more so to the advanced thread (and then even then, the standard of method of constructing an integral, using the recurrence and showing that the error vanishes is by no means something that even the best of HSC students can be expected to do).

The second question has no place in any high school section.

 

[Carrotsticks]

Re: HSC 2014 4U Marathon

second question is actually not that hard. just substitute itheta in part (i) and evaluate real and complex parts.
(the first part is needed)

Thank you, I am aware of the methodology you intended.

Now, I'd like you to answer a question of mine because I feel that you're just throwing higher level machinery around without really knowing how it works.

Your proof of part (i) is to prove the Taylor Series for the exponential function. The proof you are requesting is via real integration. In other words, once you've deduced this, you may only assume that the identity holds for real values of x. How is it then that we can "just substitute itheta"??

 

[Carrotsticks]

Re: HSC 2014 4U Marathon

well I though this was the advanced thread
whoops

(the same way you prove Euler's formula, by differentiating as well?)

Start from the ground up.

Integer exponents are okay, intuitive.

Rational exponents are still okay (though to a slightly lesser degree of intuition).

Irrational exponents start getting tricky but can still be taken care of since any irrational exponent can be approximated as closely as we like by some rational exponent.

Complex exponents ??????? Does such a thing even exist???

Differentiating complex exponents ?????? How can we even think about this unless we know what complex exponents are??

 

[Carrotsticks]

Re: HSC 2014 4U Marathon

http://mathfaculty.fullerton.edu/mathews/c2003/ComplexIntegralMod.html

No, I do not want to see what your Google search yielded.

I want you to understand what kinds of questions you are asking, and thus why they are not suitable for the high school level. It's all too easy for us to ask questions on results in higher levels of Mathematics, asking students to assume these results blindly unless absolutely necessary, and then brush over the machinery behind it.

 

[seanieg89]

Re: HSC 2014 4U Marathon

Lel, was planning on tearing into that post when I finished up with my essential work for today. Am glad Carrotsticks held down the fort.

You need to really know what you are talking about to properly formulate questions that guide students through out-of-syllabus results, otherwise you are usually doing them more harm than good.

(Here out of syllabus means in terms of content, not difficulty.)

 

[RealiseNothing]

Re: HSC 2014 4U Marathon

second question is actually not that hard. just substitute itheta in part (i) and evaluate real and complex parts.
(the first part is needed)

For all real x:

Substituting in x=i gives

We have proven that

See your problem?

 

[dan964]

Re: HSC 2014 4U Marathon

yeah I have.
for some reason
the edit function doesn't work
nevermind switched to firefox.
for the benefit I have removed all comments.

here is a more appropriate question for yous...

 

[Chlee1998]

Re: HSC 2014 4U Marathon

yeah I have.
for some reason
the edit function doesn't work
nevermind switched to firefox.
for the benefit I have removed all comments.




here is a more appropriate question for yous...

part i, ii are pretty self explanatory. for part iii) just expand (z - 1/z)^5 using binomial which will give pretty much give the required result.

for the second part of part iii), just replace sin 3theta with 3 sin theta - 4sin^3 theta and rearrange to get sin5 theta in terms of powers of sin theta.
More specifically, 5sin theta - 20 sin ^3 theta + 16 sin^5 theta.

 

[Chlee1998]

Re: HSC 2014 4U Marathon

Here's a nice Q (well in the syllabus carrot and Sean lol)

ABC is a triangle with AB = 360, BC = 240 and AC = 180. The internal and external bisectors of CAB meet BC and BC produced at P and q respectively. Find the radius of the circle which passess through A, P and Q.

 

[RealiseNothing]

Re: HSC 2014 4U Marathon

Here's a nice Q (well in the syllabus carrot and Sean lol)

ABC is a triangle with AB = 360, BC = 240 and AC = 180. The internal and external bisectors of CAB meet BC and BC produced at P and q respectively. Find the radius of the circle which passess through A, P and Q.

The problem wasn't so much it wasn't in the syllabus, but more so that the maths was really flawed.