# PC2 Maths, AQA 24/05/10Watch

8 years ago
#81
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)
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#82
I think the last log quesitnon was 3, i remember it being stupidly low!
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8 years ago
#83
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.
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8 years ago
#84
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.
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#85
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.
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8 years ago
#86
the last part of the last question was a killlerrrrrrr uffffffffffff i lost 4 marks already
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8 years ago
#87
(Original post by 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
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8 years ago
#88
(Original post by 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
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8 years ago
#89
(Original post by 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.
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8 years ago
#90
(Original post by 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
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8 years ago
#91
Does anyone remember what they got for question seven when you had to solve cos(2x)=1/2 ?
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8 years ago
#92
(Original post by *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
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8 years ago
#93
Yeah i think i did

2(pi)x=27.2
(pi)x=13.6
x=13.6/pi
x=4.33 to 3sf
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8 years ago
#94
(Original post by 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
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8 years ago
#95
(Original post by *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
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8 years ago
#96
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 wwsnt in any pasp papers. feel like i got everyhting wrong ..
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8 years ago
#97
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
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8 years ago
#98
(Original post by Phalange)
0.4 and like 2.6
Arghhh, you sure it wasn't 0.5 something? cause the 2x values were like 1. something and 5.something then you halve them
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8 years ago
#99
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
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8 years ago
#100
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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