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AQA Core 4 MPC4 -- 16th June 2011

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Reply 140
Avatar for harky 1
harky 1
Original post by lamin.sessay

Original post by lamin.sessay
Dear Readers,

Can you please explain in detail how you did the final 9 mark question? I am totally bemused. I tried to solve it repeatedly but to no avail.

Many thanks,
Lamin Sessay


Yes lamin Pwd 2k11 BRAAAAP
Original post by bish93
i couldn't get it to equal 0 either :/
another exam board cock up? or is that just wisful thinking cos i couldnt do it? :P


Well do you think our Constant of 2/e (which nearly everyone ive spoke to about the exam said they got it as!!) was incorrect? in which case i would say the possibility of an exam board cock up is very likely !!
I differentiated it twice, got the exact same thing both times, and i dont see how it could be anything else :s-smilie:. :frown:
I thought this paper was the hardest core 4 paper I've ever seen in my life, I resat Core 3 on Monday and it was really nice, and Further Maths was a nice paper too, I had high hopes for this, I felt like ripping it up! :frown:
Reply 143
Avatar for Crowaled
Crowaled
Original post by charlotte7474
Hiya, after differentiating it, i got C = 2/e, but then when you have to prove it was a stationary point, i couldnt get it to = 0. :s-smilie:
Quite a lot of people in my class couldnt either, did anyone actually get it to equal 0??!


2y + e^(2x)y² = x² + C

so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

(dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

Stationary point when (1,1/e)

2(1/e) + e²(1/e²) = 1² + C
2/e + 1 = 1 + C
2/e = C

Prove that it's a stationary point or something it said,

So dy/dx = 0

0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

0 = (2x - 2e^(2x)y²)

0 = 2 - 2e²(1/e²)

0 = 2 - 2

0 = 0 Proven.
Reply 144
Avatar for Misiak
Misiak
Original post by Crowaled
2y + e^(2x)y² = x² + C

so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

(dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

Stationary point when (1,1/e)

2(1/e) + e²(1/e²) = 1² + C
2/e + 1 = 1 + C
2/e = C

Prove that it's a stationary point or something it said,

So dy/dx = 0

0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

0 = (2x - 2e^(2x)y²)

0 = 2 - 2e²(1/e²)

0 = 2 - 2

0 = 0 Proven.


yeah i did that
Reply 145
Avatar for j04millj
j04millj
Original post by bish93
i couldn't get it to equal 0 either :/
another exam board cock up? or is that just wisful thinking cos i couldnt do it? :P


Wishful thinking, I'm afraid, it was right. The constant was 2/e though.
Original post by Crowaled
2y + e^(2x)y² = x² + C

so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

(dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

Stationary point when (1,1/e)

2(1/e) + e²(1/e²) = 1² + C
2/e + 1 = 1 + C
2/e = C

Prove that it's a stationary point or something it said,

So dy/dx = 0

0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

0 = (2x - 2e^(2x)y²)

0 = 2 - 2e²(1/e²)

0 = 2 - 2

0 = 0 Proven.



2e²(1/e²) = 2?

oh crap yeah ofcourse it does,
thanks for clearing that up ! :smile:
As I said earlier, I'm quite sure that the year is 2016. I came across a question in a past paper similar to this one and I also didn't get it but I did once my teacher explained. He said to think of it like this - when a baby is 3 months old, it's in its first year. You don't round down and say it's 0.
(edited 12 years ago)
Reply 148
Avatar for JontyC
JontyC
Original post by Crowaled
2y + e^(2x)y² = x² + C

so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

(dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

Stationary point when (1,1/e)

2(1/e) + e²(1/e²) = 1² + C
2/e + 1 = 1 + C
2/e = C

Prove that it's a stationary point or something it said,

So dy/dx = 0

0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

0 = (2x - 2e^(2x)y²)

0 = 2 - 2e²(1/e²)

0 = 2 - 2

0 = 0 Proven.


Think I got that. I was unsure about the differentiating, but i got the stationary point equal to 0, so I guess I did do it correctly!
can any1 just remember the last question? I want to do it now again!!
Original post by Crowaled
2y + e^(2x)y² = x² + C

so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

(dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

Stationary point when (1,1/e)

2(1/e) + e²(1/e²) = 1² + C
2/e + 1 = 1 + C
2/e = C

Prove that it's a stationary point or something it said,

So dy/dx = 0

0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

0 = (2x - 2e^(2x)y²)

0 = 2 - 2e²(1/e²)

0 = 2 - 2

0 = 0 Proven.


I was just wondering whether you thought my method was right for proving it was a stationary point. I got dy/dx on its own and then subsituted the given values for x and y (1,1/e) and this equalled zero which shows that dy/dx=0, therefore it is a SP. Would that still get the marks
Reply 151
Avatar for harky 1
harky 1
how many marks will i get for the question where we had to solve A, B and C if i only got 2 of those values correct?
Reply 152
Avatar for DdotT
DdotT
17
Original post by charlotte7474
Hiya, after differentiating it, i got C = 2/e, but then when you have to prove it was a stationary point, i couldnt get it to = 0. :s-smilie:
Quite a lot of people in my class couldnt either, did anyone actually get it to equal 0??!


yh i did thats how i am confident i was correct but just wondered still or perhaps i completely messed up and i am wrong :P
Reply 153
Avatar for JC1993
JC1993
"I was just wondering whether you thought my method was right for proving it was a stationary point. I got dy/dx on its own and then subsituted the given values for x and y (1,1/e) and this equalled zero which shows that dy/dx=0, therefore it is a SP. Would that still get the marks "

That's what I did :smile: ^
Reply 154
Avatar for Brister
Brister
7
Original post by harky 1

Original post by harky 1
Messed that up then =( u think the grade boundries will get lowered?


I would think so, because the people I talked to said it was difficult for them. Then again, there are always plenty of people who say that. We'll just have to wait and see, and hope for the best!
Reply 155
Avatar for asem93
asem93
The year is 2015...
When is the year when it will first exceed 100,000... Of course 2015
(edited 12 years ago)
Reply 156
Avatar for JontyC
JontyC
Original post by haaroon321
I was just wondering whether you thought my method was right for proving it was a stationary point. I got dy/dx on its own and then subsituted the given values for x and y (1,1/e) and this equalled zero which shows that dy/dx=0, therefore it is a SP. Would that still get the marks


That's exactly what I did, it's the way I've always proved a stationary point in the past, and got the marks!

I guess theres more than one way of proving it, I guess our way was simpler. :biggrin:
Reply 157
Avatar for shabir77
shabir77
The answer to that question was 2015.... 1st of August 2015!!
Original post by Signed Solution
As I said earlier, I'm quite sure that the year is 2016. I came across a question in a past paper similar to this one and I also didn't get it but I did once my teacher explained. He said to think of it like this - when a baby is 3 months old, it's in its first year. You don't round down and say it's 0.


I hope you're right. :redface:

I put 2016 because I thought about those questions when they ask if I have x items and each box holds x how many boxes do you need. If you get and answer like 4.3 then you say 5 because 3 won't be enough.
Reply 159
Avatar for Zing
Zing
To be honest, it seems that half have put 2015 and half have put 2016 so they'll probably just accept either because the question was difficult to understand. At least I hope they do :smile:

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