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

Examine · Mar 9, 2012

Re: 2012 HSC MX1 Marathon

Solve the equation 2^y^2+3-9^2y=0

bleakarcher · Mar 9, 2012

Re: 2012 HSC MX1 Marathon

^ could u use parentheses?

bleakarcher · Mar 9, 2012

Re: 2012 HSC MX1 Marathon

its quite unclearly written.

Kingportable · Mar 9, 2012

Re: 2012 HSC MX1 Marathon

Can anyone help me? I am working on a question from John Fitzpatrick 3unit Text book. It is in Chapter 29a) question 50. please help

dulip · Mar 11, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
Also adding on to part iii), find the percentage error of the approximate and actual area.

Do you still need this? Should i write up my soln

nightweaver066 · Mar 11, 2012

Re: 2012 HSC MX1 Marathon

dulip said:
Do you still need this? Should i write up my soln

I don't need this, i'm posting it up for people to do for practise.. lol.

Scroll down a bit further after that question to find Carrotsticks addition to the question.

Drongoski · Mar 11, 2012

Re: 2012 HSC MX1 Marathon

Kingportable said:
Can anyone help me? I am working on a question from John Fitzpatrick 3unit Text book. It is in Chapter 29a) question 50. please help

Solution already posted to this question in another thread.

seanieg89 · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

deswa1 said:
What about now:
$\textup{It is known that the series }1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{diverges}$

With a bit of thinking, a good 3U student (I think) should be able to get it but this proof might not be fully rigourous...

Comparison to which geometric series?

Here is a cute overkill proof that uni students may find amusing:

$If the harmonic series converges to some value $H$, then the alternating harmonic series $L:=1-1/2+1/3-\ldots$ is absolutely convergent. Hence we can rearrange the terms of the sum $L$ and obtain: \\$L=1-1/2+1/3-1/4+1/5-\ldots=\\(1+1/2+1/3+1/4+1/5+\ldots)-2(1/2+1/4+1/6+1/8+\ldots)=H-H=0.\\ $But on the other hand, we have: $\\L=(1-1/2)+(1/3-1/4)+\ldots>0\\ $since each bracketed term is positive. This contradiction shows that the harmonic series CANNOT converge :).$

Sidekickk · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Log x base a can be defined as a:
A. base
B. index
C. operator
D. none of the above

Srs question

Carrotsticks · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Did you send this question to Math Man earlier this morning?

Sidekickk · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Did you send this question to Math Man earlier this morning?

uhh no? who the hell is math man

Carrotsticks · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Sidekickk said:
uhh no? who the hell is math man

Never mind. He was asked the exact same question this morning.

According to the teacher who wrote the exam, the answer is index.

Sidekickk · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Aha yep correct. Stupid question though rite? :/ I'm guessing (hoping) for over 3/4 of the grade to get it wrong.

Carrotsticks · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Sidekickk said:
Aha yep correct. Stupid question though rite? :/ I'm guessing (hoping) for over 3/4 of the grade to get it wrong.

Indeed a bit of a poorly worded question.

Perhaps your teachers will decide to omit it from the entire exam, or award everybody a mark for that question regardless of their answer.

Sidekickk · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Fingers crossed >.<

SpiralFlex · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

Hola, Spiral is hungry. GROWL

$The diagram above shows a British 50 pence coin. The seven circular arcs AB, BC, ... , GA are of equal length and their centres are E, F,....,D respectively. Each arc has radius a.$

$$(i) Show that the sector AEB has area$ \; \frac{1}{14} \pi a^2$

$$(ii) Hence, or otherwise, show that the face of the coin has area $ \; \frac{1}{2} a^2(\pi-7\tan \frac{\pi}{14})$

Timske · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:
Hola, Spiral is hungry. GROWL

$The diagram above shows a British 50 pence coin. The seven circular arcs AB, BC, ... , GA are of equal length and their centres are E, F,....,D respectively. Each arc has radius a.$

$$(i) Show that the sector AEB has area$ \; \frac{1}{14} \pi a^2$

$$(ii) Hence, or otherwise, show that the face of the coin has area $ \; \frac{1}{2} a^2(\pi-7\tan \frac{\pi}{14})$

I don't understand the question

so AE = AB = a ?

SpiralFlex · Mar 12, 2012

Re: 2012 HSC MX1 Marathon

AB has a centre of E. BC has a centre of F and so forth.

Examine · Mar 14, 2012

Re: 2012 HSC MX1 Marathon

Differentiate y=3sqrroot2 using the First Principles method.

Carrotsticks · Mar 14, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
Differentiate y=3sqrroot2 using the First Principles method.

Are you missing an x somewhere, or is it intended to be a constant..?

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

same

Active Member

Active Member

Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Retired

Member

Retired

Member

Retired

Member

Well-Known Member

Sequential

Well-Known Member

same

Retired

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