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Timing in the MX2 exam (1 Viewer)

~ ReNcH ~

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justchillin said:
I've done the exact same thing and I'd have to agree with those marks - 2000 was bitchy... as for the papers from 95-99 they are relatively ezy...
Do you mind me asking what mark you got in the 2000 paper? I perceive you as a good benchmark coz I know that you'd be searching for ~97+/100 aligned.

justchillin said:
And id have to say for 100/120 ur looking at about 96/97, considering top of the state gets about 110-114...
...or in the case of the 1993 paper, notably the most difficult, 120/120...
 

haboozin

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justchillin said:
I've done the exact same thing and I'd have to agree with those marks - 2000 was bitchy... as for the papers from 95-99 they are relatively ezy...
And id have to say for 100/120 ur looking at about 96/97, considering top of the state gets about 110-114...
yea i agree with 2000...

that was sooo bad i couldn't do so much probably would have got what Rench got.

But i think 2003 wasnt that bad.

If we are going from difficualty from 1993 to 2004
top 4 most difficault would be:
1993 1994 2000 2003

i think 1995-1999 are all the same difficulty and they're not that bad.
2002 was just the easiest exam ever.

1994 is really weird... its all like hard 3unit questions (not really inequalities) - the question 8 of it isnt that hard just standard inequalities havnt done part b though.
i dont think 2004 was that easy but it was all the questions i was good at, so i would have really liked that test.
 

~ ReNcH ~

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I'm hoping for a 2005 HSC exam that's somewhere between the 2001, 2002 and 2004 papers in terms of difficulty. Hopefully ours is nothing like the 1993 or 2000 papers.
 

Templar

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~ ReNcH ~ said:
...or in the case of the 1993 paper, notably the most difficult, 120/120...
That's just Henderson though.
 

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~ ReNcH ~ said:
I've done papers from 2000-2004, and based on my marks the most difficult was 2000 and then 2003, followed by 2004, 2001 and 2002. Nevertheless, my marks aren't really anything to go by :rolleyes: - I'm not too sure about pre-2000 papers though.
Goes to show, different things for different people. I found 2002 harder than 2003 (2003 was pretty cruisy but 2002 tripped me up in places).
 

haboozin

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Templar said:
That's just Henderson though.
1993 was all like uni maths problems, but explained.

Henderson had probably already started on those questions so they looked generic to him...
 

ishq

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Completely off topic here - and I apologise, but I didnt want to make a thread about it - has anyone done the Grammar 2005 Trial? Are the solutions up on here anywhere?

If not, could someone PLEASE prove:
cos4x = (1 - 6tan^2x + tan^4x)/(1 + tan^2x + tan^4x)
It is Q7 (a)(ii). The first part asked to write down cos2x in terms of tanx.

I thanks you! :D

Beautiful Exam. Wish it was the HSC.
 

~ ReNcH ~

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KFunk said:
Goes to show, different things for different people. I found 2002 harder than 2003 (2003 was pretty cruisy but 2002 tripped me up in places).
Q7(c) and Q8 of the 2003 paper tripped me up. Looking back at the rest of the paper though, I made a few silly errors so hopefully that won't happen in the HSC exam.
 

KFunk

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Say you let t = tanx, then cos2x = (1 - t<sup>2</sup>)/(1+t<sup>2</sup>) and sin2x = 2t/(1 + t<sup>2</sup>)

Then cos4x = cos<sup>2</sup>2x - sin<sup>2</sup>2x

= {(1 - t<sup>2</sup>)<sup>2</sup> - (2t)<sup>2</sup>}/(1+t<sup>2</sup>)<sup>2</sup>

= (1 - 6t<sup>2</sup> + t<sup>4</sup>)/(1 + 2t<sup>2</sup> + t<sup>4</sup>)

= (1 - 6tan<sup>2</sup>x + tan<sup>4</sup>x)/(1 + 2tan<sup>2</sup>x + tan<sup>4</sup>x)

(since t = tanx)
 

ishq

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KFunk said:
Say you let t = tanx, then cos2x = (1 - t<sup>2</sup>)/(1+t<sup>2</sup>) and sin2x = 2t/(1 + t<sup>2</sup>)

Then cos4x = cos<sup>2</sup>2x - sin<sup>2</sup>2x

= {(1 - t<sup>2</sup>)<sup>2</sup> - (2t)<sup>2</sup>}/(1+t<sup>2</sup>)<sup>2</sup>

= (1 - 6t<sup>2</sup> + t<sup>4</sup>)/(1 + 2t<sup>2</sup> + t<sup>4</sup>)

= (1 - 6tan<sup>2</sup>x + tan<sup>4</sup>x)/(1 + 2tan<sup>2</sup>x + tan<sup>4</sup>x)

(since t = tanx)
Elegance.

Merci.
 

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KFunk said:
Say you let t = tanx, then cos2x = (1 - t<sup>2</sup>)/(1+t<sup>2</sup>) and sin2x = 2t/(1 + t<sup>2</sup>)

Then cos4x = cos<sup>2</sup>2x - sin<sup>2</sup>2x

= {(1 - t<sup>2</sup>)<sup>2</sup> - (2t)<sup>2</sup>}/(1+t<sup>2</sup>)<sup>2</sup>

= (1 - 6t<sup>2</sup> + t<sup>4</sup>)/(1 + 2t<sup>2</sup> + t<sup>4</sup>)

= (1 - 6tan<sup>2</sup>x + tan<sup>4</sup>x)/(1 + 2tan<sup>2</sup>x + tan<sup>4</sup>x)

(since t = tanx)
Nice method. I probably would of done it via De Moivres though.
 

KFunk

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Yeah, that was the first thing I thought of but I realised it could get messy so I came up with an easier way to do it (I say easier without having actually tried both).
 

Stefano

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KFunk said:
Yeah, that was the first thing I thought of but I realised it could get messy so I came up with an easier way to do it (I say easier without having actually tried both).
I'd say complex numbers would be easier, but probably longer. In an exam I'd do it with complex numbers, just to be on the safe side.
 

KFunk

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Due to my tendency to make stupid mistakes in a mass of working, for me longer ussually = harder.
 

Stefano

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KFunk said:
Due to my tendency to make stupid mistakes in a mass of working, for me longer ussually = harder.
Haha, yes me too. However, I can never remember those darn t-formulae!
 

Antwan23q

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yeh same, everytime i need them i have to work them all out individually.
 

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