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PC2 Maths, AQA 24/05/10

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Reply 80
I found it odd they asked about the circumference too... its 2*pie*r ?
Never saw that in a paper

Also, how did show there was no maximum point? (It said "hence" and was 2 marks)
Reply 81
I think the last log quesitnon was 3, i remember it being stupidly low!
Reply 82
Guys, I don't understand where you are coming from. In the last 2 log questions, you had to get an answer given in the exam. The very last question, you had to get 1/(4log2) as the answer.
I bet the examiners are just sitting in their armchairs chuckling to themselves as we speak. I didn't even know we were meant to use our answers from the previous question, they usually say 'hence...'
This sucks.
Reply 84
I think for the circumfrence it said the circumfrence of another circle or radius x m, is the same as the area calculated in part b, whats x?

so one takes the area (27.2? or 72.7?) nd dividews by 2*pie i think.
Reply 85
the last part of the last question was a killlerrrrrrr uffffffffffff i lost 4 marks already
Reply 86
rob...
I think for the circumfrence it said the circumfrence of another circle or radius x m, is the same as the area calculated in part b, whats x?

so one takes the area (27.2? or 72.7?) nd dividews by 2*pie i think.


Yes, that is correct. You must make the following equation

27.2=2pie r


and solve for r
Reply 87
rob...
I think for the circumfrence it said the circumfrence of another circle or radius x m, is the same as the area calculated in part b, whats x?

so one takes the area (27.2? or 72.7?) nd dividews by 2*pie i think.


Phew yeah it was perimeter I think. Something like 27.2 yeah. I remember r for the sector was 8 and theta 1.4
rob...
I think for the circumfrence it said the circumfrence of another circle or radius x m, is the same as the area calculated in part b, whats x?

so one takes the area (27.2? or 72.7?) nd dividews by 2*pie i think.


Yup, that is what one should have done. :yep:
Reply 89
Phalange
I found it odd they asked about the circumference too... its 2*pie*r ?
Never saw that in a paper

Also, how did show there was no maximum point? (It said "hence" and was 2 marks)


if u find the second deffrential then substitue that x=1 value , it worked out to be positive , (greater than 0) so it wasnt a maximum i hope dats right becuase thats what i wrote
Does anyone remember what they got for question seven when you had to solve cos(2x)=1/2 ?
Reply 91
*A_S_H*
if u find the second deffrential then substitue that x=1 value , it worked out to be positive , (greater than 0) so it wasnt a maximum i hope dats right becuase thats what i wrote


LOL NOOOO I solved it for =0 and it was positive so I said no maximum as its a minimum
********
grrrr
Reply 92
Yeah i think i did

2(pi)x=27.2
(pi)x=13.6
x=13.6/pi
x=4.33 to 3sf
Reply 93
Averagegirl:)
Does anyone remember what they got for question seven when you had to solve cos(2x)=1/2 ?


0.4 and like 2.6
something like that... answers in radians i think
Reply 94
*A_S_H*
if u find the second deffrential then substitue that x=1 value , it worked out to be positive , (greater than 0) so it wasnt a maximum i hope dats right becuase thats what i wrote


You had to let dy/dx equal zero and find the x value and substitute into d2y/dx2. Then you get a positive answer hence the stationary point is a minima
Reply 95
For questions seven i kept getting 0 and it wouldn't equal half! and i forgot the bludy circumference of a circle thats not fair :frown: wwsnt in any pasp papers. feel like i got everyhting wrong ..
Reply 96
lifeadd1ct
You had to let dy/dx equal zero and find the x value and substitute into d2y/dx2. Then you get a positive answer hence the stationary point is a minima


It was only 2 marks I didn't think you'd have to go through all of that
ah well
Reply 97
Phalange
0.4 and like 2.6
something like that... answers in radians i think


Arghhh, you sure it wasn't 0.5 something? cause the 2x values were like 1. something and 5.something then you halve them
Reply 98
lifeadd1ct
You had to let dy/dx equal zero and find the x value and substitute into d2y/dx2. Then you get a positive answer hence the stationary point is a minima


ohhhh i substituted x=1 :frown:
Reply 99
lifeadd1ct
You had to let dy/dx equal zero and find the x value and substitute into d2y/dx2. Then you get a positive answer hence the stationary point is a minima

How many marks were on that question? Seems like that's the onlt one i got correct at the moment..

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