Timing in the MX2 exam (1 Viewer)

[Stefano]

what is his rule?

'If you are hurt, run to teh L'Hospital'

 

[Stefano]

SOMEONE HAD TO SAY IT!! lol..

<b>L'Hospital's rule:</b>

If f(x) and g(x) approach 0 as x approaches a, and f '(x) / g'(x) approaches L as x approaches a, then the ratio f(x) / g(x) approaches L as well.

 

[Antwan23q]

err ok.
how do you people know all this out side of the sylabus theorems? do u get taught them at school?

 

[Stefano]

err ok.
how do you people know all this out side of the sylabus theorems? do u get taught them at school?

I skip meals to study maths that's outside of the syllabus.

 

[acmilan]

SOMEONE HAD TO SAY IT!! lol..

<b>L'Hospital's rule:</b>

If f(x) and g(x) approach 0 as x approaches a, and f '(x) / g'(x) approaches L as x approaches a, then the ratio f(x) / g(x) approaches L as well.

Thats one of the many forms the rule takes

 

[noah]

With Q1(e) of the SBHS paper, even though it says "evaluate", are you justified in sketching 2-|x| and finding the sum of the areas between the curve and the x-axis between -3 and 3? In general, I'm just wondering whether you're allowed to use any available method to answer a question or whether "evaluate" implies an alegbraic method must be used.

you have to be careful though because the areas under the x axis are negative.

if you just add the areas you'll get 5 instead of 3

 

[haboozin]

you have to be careful though because the areas under the x axis are negative.

if you just add the areas you'll get 5 instead of 3

indeed,
This 4u exam was probably one of the hardest, but i've done 1-5
and my worse was 2 :'( Think i would've got 10 there..

 

[yrtherenonames]

No doubt SHS '05 was hard, but have you guys seen James Ruse 2002?
It's insanely hard; people like justchillin would be celebrating

 

[~ ReNcH ~]

ln[tan(&pi;/2 -x)] = ln(cotx)

as x--> &pi;/2 , cotx ---> 0 &there4; ln[tan(&pi;/2 -x)] ---> -&infin;

I think you need to use the integral properties for this one because the question hinges around the symmetry (I think) the graphs have around x=&pi;/4. You can't really say "ln[tan(&pi;/2 -x)] = 0 hence &int; ln(tanx) dx = 0" if you know what I mean.

For that question, I said ln(cotx) = ln(1/tanx) = ln1 - ln(tanx). You transfer the ln(tanx) to the LHS, you get 2|ln(tanx)dx = |0dx --> |ln(tanx) = 0. Is 0 the right answer?

 

[~ ReNcH ~]

you have to be careful though because the areas under the x axis are negative.

if you just add the areas you'll get 5 instead of 3

Yep. I just took any area below the x-axis as being "negative".

 

[KFunk]

For that question, I said ln(cotx) = ln(1/tanx) = ln1 - ln(tanx). You transfer the ln(tanx) to the LHS, you get 2|ln(tanx)dx = |0dx --> |ln(tanx) = 0. Is 0 the right answer?

Yup, that's pretty much how I solved it except I made ln(1/tanx) = -ln(tanx) (leading to the same conclusion). I had a look at the graph in graphmatica and zero would look to be the rigth answer.

 

[~ ReNcH ~]

SOMEONE HAD TO SAY IT!! lol..

<b>L'Hospital's rule:</b>

If f(x) and g(x) approach 0 as x approaches a, and f '(x) / g'(x) approaches L as x approaches a, then the ratio f(x) / g(x) approaches L as well.

I hate nit-picking and I'm no mathematician...far from it in fact , but I'm quite sure it's L'Hopital's rule (no "s"). The last thing you'd want is to use a rule outside of the syllabus and spell it wrong esp. if someone like Pender or Arnold were marking it

 

[ishq]

Looks like I wasn't the only one who was crucified by SBHS '05 then! Makes my brain cell feel better.

In other news, why didn't I give my HSC in 1999. Or even, accelerate in MX2 in....year...SIX? Has anyone seen that paper? It is a gift to mankind. Comparing with 2000. *ahem*

That is my mathematical contribution. I'm off to write my creative section II journey. Again. This time, a feature article journey.

 

[Antwan23q]

yeh i agree, its like totally easy. why cant they all be like 1999

 

[sikeveo]

I hate nit-picking and I'm no mathematician...far from it in fact , but I'm quite sure it's L'Hopital's rule (no "s"). The last thing you'd want is to use a rule outside of the syllabus and spell it wrong esp. if someone like Pender or Arnold were marking it

Or Hespe or Choy.

 

[Antwan23q]

Or Hespe or Choy.

... just when i start to think i might actually do well in the hsc, you people punk me out by talkin about things ive never even heard of...

 

[Slidey]

I hate nit-picking and I'm no mathematician...far from it in fact , but I'm quite sure it's L'Hopital's rule (no "s"). The last thing you'd want is to use a rule outside of the syllabus and spell it wrong esp. if someone like Pender or Arnold were marking it

Actually, "L'Hospital" is the correct English translation of his name. In French, it is: "L'Hôpital".

That is to say: "L'Hopital" is the incorrect way to spell his name.

 

[~ ReNcH ~]

Actually, "L'Hospital" is the correct English translation of his name. In French, it is: "L'Hôpital".

That is to say: "L'Hopital" is the incorrect way to spell his name.

Oh ic...I stand corrected. Thanks Slide Rule

 

[forgotten_dark]

hey what rough score in 3unit would put u in band 6? just generally a raw score of 90% or above, or is it lower like in 4unit? if someone has already asked this just direct me to the thread please...thanks

 

[yrtherenonames]

hehe time to join in on the nit-picking action
There is no word in the English language called L'hospital, Slide Rule. Sure L'hôpital in french means hospital in english, but you certainly can't just 'translate' his name. hehe and if you could it would be 'the hospital'.
Did you mean that that's the correct transliteration?