HSC 2012 MX1 Marathon #1 (archive) (3 Viewers)

largarithmic · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Trebla said:
This only works if n is an integer. In this case, n is allowed to be any real number. Most people would fall for this trap...

Isnt it now just L'Hopital's again?

Carrotsticks · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

largarithmic said:
Isnt it now just L'Hopital's again?

Yeah if you use L'Hopital's Principle, the question is fairly straightforward.

But I don't think it's in the Extension 1 syllabus.

Unless it was given beforehand as one of those 'DO NOT PROVE' things.

Trebla · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

It can be done without L'Hopital's rule using results within the scope of even the 2 unit syllabus. The tricky bit is thinking about how to approach it.

nightweaver066 · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

deswa1 said:
Disclaimer: This is from a 4U paper but 3U students can do it.

i) Step 1: n = 1, LHS = cosx
RHS = $\frac{sin\frac{3x}{2} - sin\frac{x}{2}}{2sin\frac{x}{2}}$
= $\frac{2cosxsin\frac{x}{2}}{2sin\frac{x}{2}}$
= $cosx$
= LHS

True for n = 1.

Step 2: Assume for n = k, $cosx + cos2x + cos3x + ... + coskx = \frac{sin(k + \frac{1}{2})x - sin\frac{x}{2}}{2sin\frac{x}{2}}$ is true.
RTP for n = k + 1, i.e., $cosx + cos2x + cos3x + ... + cos(k + 1)x = \frac{sin(k + \frac{3}{2})x - sin\frac{x}{2}}{2sin\frac{x}{2}}$

Proof: LHS = $cosx + cos2x + cos3x + ... + coskx + cos(k+1)x$

$= \frac{sin(k + \frac{1}{2})x - sin\frac{x}{2}}{2sin\frac{x}{2}} + cos(k+1)x$

$= \frac{sin(k + \frac{1}{2})x - sin\frac{x}{2} + 2sin\frac{x}{2}cos(k+1)x}{2sin\frac{x}{2}}$

$= \frac{sin(k + \frac{1}{2})x - sin\frac{x}{2} + sin(k + \frac{3}{2}) + sin(-\frac{1}{2} -k)x}{2sin\frac{x}{2})}$

$= \frac{sin(k + \frac{3}{2})x - sin\frac{x}{2}}{2sin\frac{x}{2}}$

$\text{True for n = k + 1 assuming it is true for n = k}$

Step 3: As it is true for n = 1 (step 1), it must also be true for n = 1 + 1 = 2 (step 2) and so on. Hence, the result is generally true for all positive integers n

ii) Replacing x with 2x and letting n = 8,

$cos2x + cos4x + cos6x + ... + cos16x = \frac{sin(17x) - sinx}{2sinx}$

$= \frac{cos9xsin8x}{sinx}$

$= \frac{2cos9xsin4xcos4x}{sinx}$

$= \frac{4cos9xcos4xcos2xsin2x}{sinx}$

$= \frac{8cos9xcos4xcos2xcosxsinx}{sinx}$

$= 8cos9xcos4xcos2xcosx$

seanieg89 · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Trebla said:
It can be done without L'Hopital's rule using results within the scope of even the 2 unit syllabus. The tricky bit is thinking about how to approach it.

$Define $f(x):=x^r$ for positive $x$ and real $r.\\$Then $f'(x)=rx^{r-1}$ and we have:$\\ \\r=f'(1)=\lim_{x\rightarrow 1^-}\frac{1-f(x)}{1-x}=\lim_{x\rightarrow 1^-}\frac{1-x^r}{1-x}$ and the result follows.$

Of course here we are assuming that the derivative of x^r is rx^{r-1} even for irrational r, something not proven in the high school syllabus. (Indeed irrational powers are not even defined in high school.)

SpiralFlex · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

$\textup{Let}\; f(x)=x^4+2ax^2+4bx+c \; \; \textup{such that} \; a,b,c \in \mathbb{R} \; \textup{with} \; b \neq 0. \; \textup{It is also known that} \;f(x)=(x^2+2tx+r)(x^2-2tx+s) \;\textup{whereby} \; r,s,t \in \mathbb{R}$

$\textup{a) Prove that}\; t \neq 0\;.$

$\textup{b) Express} \; s \;\textup{and} \;r \; \textup{in terms of} \; a,b,t\;.$

$\textup{c) Prove that}\; 4t^6+4at^4+(a^2-c)t^2-b^2=0\;.$

nightweaver066 · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:
$\textup{Let}\; f(x)=x^4+2ax^2+4bx+c \; \; \textup{such that} \; a,b,c \in \mathbb{R} \; \textup{with} \; b \neq 0. \; \textup{It is also known that} \;f(x)=(x^2+2tx+r)(x^2-2tx+s) \;\textup{whereby} \; r,s,t \in \mathbb{R}$

$\textup{a) Prove that}\; t \neq 0\;.$

$\textup{b) Express} \; s \;\textup{and} \;r \; \textup{in terms of} \; a,b,t\;.$

$\textup{c) Prove that}\; 4t^6+4at^4+(a^2-c)t^2-b^2=0\;.$

a) $(x^2 + 2tx + r)(x^2 - 2tx + s)$

$= x^4 -2tx^3 + sx^2 + 2tx^3 - 4t^2x^2 + 2tsx + rx^2 - 2rtx + rs$

$= x^4 + (s -4t^2 + r)x^2 + (2ts - 2rt)x + rs \equiv x^4 + 2ax^2 + 4bx + c$

Equating coefficient of x

$\therefore st - rt = 2b ..... (1)$

$b = \frac{t(s - r)}{2}$

$\therefore \text{As b } \neq 0, t \neq 0$

b) Equating coefficient of x^2,

$s - 4t^2 + r = 2a$

$s = 2a + 4t^2 - r ..... (2)$

Subbing (2) in (1), $2at + 4t^3 - rt - rt = 2b$

$r = \frac{2t^3 + at - b}{t}$

Also, from (2), $r = 4t^2 - s + 2a ..... (3)$

Subbing (3) in to (1), $st - 4t^3 + st - 2at = 2b$

$s = \frac{2t^3 + at + b}{t}$

c) Equating constants,

$c = rs$

$rs = \frac{2t^3 + at - b}{t} \times \frac{2t^3 + at + b}{t}$

$\therefore c = \frac{(2t^3 + at)^2 - b^2}{t^2}$

$ct^2 = 4t^6 + 4at^4 + a^2t^2 - b^2$

$4t^6 + 4at^4 + (a^2 - c)t^2 - b^2 = 0$

Parametrics Qn.

Carrotsticks · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
Next questions: http://4untimaths.com/polynomials-lc.pdf

Might want to swap the 't' and the 'i' in the link so then it works.

Carrotsticks · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

lol tywebb, I think those questions are sorta pushing it for the average 4U student, let alone 3U....

Carrotsticks · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
In those days there was no year 12. 1967 was the first year 12.

So it was the equivalent of our year 11 students who did those questions.

lol indeed.

But you seemed to have forgotten that the 'HSC' from 1980 onwards is much easier than the leaver's exam from beforehand

cutemouse · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
In those days there was no year 12. 1967 was the first year 12.

So it was the equivalent of our year 11 students who did those questions.

lol indeed.

Just a question, if don't mind answering... How did you find the Level 1 (or whatever maths course you did) when you did it? And how did you go in it? Just curious -- that's all.

Carrotsticks · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

I'm flicking through the papers from 1960 onwards, and they have:

- Linear Algebra (including matrices and curve rotations)

- Convergence/Divergence tests

- Number Theory

- Series expansions.

The above are not taught until University nowadays.

HOWEVER, I think from the student's perspective (from that time period), it would have been normal, since they have been learning it for several years.

These topics would seem to be alien to the HSC student nowadays, since they are not taught.

Though I'm quite convinced that if you somehow acquired a time machine and stole a 'last year of highschool' student from the 1950's, then put them up against a 4U student from 2011, the 1950's person would destroy the 2011.

cutemouse · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
The main thing to remember when comparing leaving certificate to HSC is (regardless of content) it was the equivalent of year 11 students who sat the leaving certificate.

Year 12 was only added in the year of the first HSC exams in 1967 as a result of the Wyndham Report of 1957 and the Education Act 1961.

Did you do the Level 1 course? If so, did you find it difficult at the time when you did it?

nightweaver066 · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Another question in the case that people have not done polynomials or feel intimidated by the nature of those questions..

SpiralFlex · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
Another question in the case that people have not done polynomials or feel intimidated by the nature of those questions..

Agreed, I think lowered key questions will encourage more people to post.

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

Being within the 3U syllabus does not necessarily equate to being okay for 3U.

For example, the 4U 'Harder 3U' topic. Why else is it put into the Extension 2 course?

Upper and Lower bound approximation for integrals are most surely within the 3U course. In fact, it can even be done using methods from 2U.

However, they are never put into 2U exams because they are simply too difficult. It may be within the scope of the 2U syllabus, but the skill of a person doing 2U is not enough to be able to competently do such questions.

Trebla · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

This one is slightly challenging, but not hardcore difficult (though it may require you sit there and think carefully) so I think it is more suitable for Ext1 level given that it only uses 2 unit results.

Find the value of the following limiting sum

$\dfrac{2}{3}+\dfrac{5}{3^2}+\dfrac{8}{3^3}+...$

Alternatively, a relatively easier problem:

$\hspace{-1cm} $Show that the quadratic polynomial $kx^2 + 3kx + 6$ can never be negative definite.$$

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

$\textup{Let} \; S(n)=\frac{2}{3}+\frac{5}{3^2}+\frac{8}{3^3}+\frac{11}{3^4}...(1)$

$3S(n)=2+\frac{5}{3}+\frac{8}{3^2}+\frac{11}{3^3}+....(2)$

$(2)-(1),$

$2S(n)=(2+\frac{5}{3}+\frac{8}{3^2}+\frac{11}{3^3}+...)-(\frac{2}{3}+\frac{5}{3^2}+\frac{8}{3^3}+\frac{11}{3^4})$

$\therefore S(n)=\frac{1}{2}(2+\frac{3}{3}+\frac{3}{3^2}+\frac{3}{3^3}+....)$

$\textup{Applying limiting sum}:$

$S_{\infty }=\frac{1}{2}(3+\frac{\frac{1}{3}}{1-\frac{1}{3}})$

$\therefore S_{\infty }=1 \frac{3}{4}$

Am I correct Treblie?

Carrotsticks · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

The trend you demonstrate is undeniable (with a few exceptions such as 2009 HSC, where all of Q8 was just Harder 3U).

What you showed us is what should theoretically be happening in the HSC.

However in practice, it may not be so.

I have done and seen several HSC/Trial papers, but I have very rarely encountered questions of that difficulty in the exam.

Those questions are excellent for any 4U student practising Polynomials (since such questions are everywhere!), but I think they would be too difficult for someone doing purely 3U.

Perhaps they are good for a student who does 3U/4U, but not for 2U/3U.

SpiralFlex · Jan 6, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
Mostly.

But answer is 5/4 because you should have

$S_{\infty }=\frac{1}{2}(2+\frac{\frac{1}{3}}{1-\frac{1}{3}})$

Dear lord...

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