HSC 2012 MX1 Marathon #1 (archive) (2 Viewers)

SpiralFlex · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
For the record, to indicate inverse tan, we use arctan (x)

ie: arctan (pi/4) = 1

Laugh out loud.

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:
Laugh out loud.

wait wtf, did I just say that arctan (pi/4) = 1?

Oh dear this is what 2 hours of sleep gets you.

I meant arctan (1) = pi/4

Oh dear me.

Trebla · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Given that - 1 < x < 1 and n is an arbitrary real number, find

$\displaystyle\lim_{x \rightarrow 1^-}(1-x^n)(1 + x + x^2 + ... )$

This one is deceptively simple.

Carrotsticks · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Find the shortest distance between the two branches of the curve y=(x^2+1) / (x+1)

SpiralFlex · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
wait wtf, did I just say that arctan (pi/4) = 1?

Oh dear this is what 2 hours of sleep gets you.

I meant arctan (1) = pi/4

Oh dear me.

Clearly someone needs sugar or sleep.

Trebla · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks 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...

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.

tywebb · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Next questions: http://4unitmaths.com/polynomials-lc.pdf

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.

tywebb · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

Yes. Ta.

I fixed it now. It works now.

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....

tywebb · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

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.

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

tywebb · Jan 5, 2012

Re: 2012 HSC MX1 Marathon

I don't think the HSC is easier than the Leaving Certificate (1916-1966). In fact the extra year makes the HSC harder.

The questions I posted are hard but can still be done by Extension 1 students.

Nevertheless I agree that the current Extension 2 course is easier than the Level 1 course (1966-1980).

HSC 2012 MX1 Marathon #1 (archive) (2 Viewers)

Well-Known Member

Retired

Administrator

Retired

Well-Known Member

Administrator

Member

Retired

Administrator

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

dangerman

Retired

dangerman

Retired

dangerman

Retired

dangerman

Users Who Are Viewing This Thread (Users: 0, Guests: 2)