[Doctor.]

Just about to start Jun'11 paper! It doesn't look very nice

[Doctor.]

Finished, that was NOT a nice paper

Although it did teach me, if I go much slower I would make far less mistakes

[Sem193]

(Original post by Doctor.)
Finished, that was NOT a nice paper

Although it did teach me, if I go much slower I would make far less mistakes

OMG. I had to do that test for my mock :s

[Luppy021]

June 2008 paper
Question 7i struggling to find out how the question is implying these values of the constants
Question 8i not sure how to do this, if we use an identity etc?
Question 9i Isn't the range the numbers that come out, how do we know what to do???

Thanks if anyone can help

[Sem193]

(Original post by Luppy021)
June 2008 paper
Question 7i struggling to find out how the question is implying these values of the constants
Question 8i not sure how to do this, if we use an identity etc?
Question 9i Isn't the range the numbers that come out, how do we know what to do???

Thanks if anyone can help

7i) Well, I think when it says the number of plants now that means t=0. You know N=42 so in the equation with the exponential you 42=Ae^m0 so that's 42=Ae^0 then you can get your answer for A there. Now the other parts are a little trickier and I may make some mistakes.

So you know what A is now. When it says that the number of plants is doubling every 9 years t=9 and N=84. Put that into the exponential equation to find m.

To find k use t=9 and N=84 again.

8i) Just use the cos2x identity and actually use the value for cos60.

9i) :O Erm, I have no idea how to do that because I hate functions. I think I need to revise the range a little more.

[Luppy021]

(Original post by Sem193)
7i) Well, I think when it says the number of plants now that means t=0. You know N=42 so in the equation with the exponential you 42=Ae^m0 so that's 42=Ae^0 then you can get your answer for A there. Now the other parts are a little trickier and I may make some mistakes.

So you know what A is now. When it says that the number of plants is doubling every 9 years t=9 and N=84. Put that into the exponential equation to find m.

To find k use t=9 and N=84 again.

8i) Just use the cos2x identity and actually use the value for cos60.

9i) :O Erm, I have no idea how to do that because I hate functions. I think I need to revise the range a little more.

Brilliant thank you for your help, when you say about the cos2x identity which one exactly?

[Doctor.]

God I hated that question! So badly worded

[Sem193]

(Original post by Luppy021)
Brilliant thank you for your help, when you say about the cos2x identity which one exactly?

Sorry that was a bit vague. Just the normal one when you expand it. So For 3(cosx-60) do CosACosB+SinASinB and then multiply it by 3 then do the same for the next one, but remember it's cosAcosB-sinAsinB

I'm so bad at explaining things :/

[Luppy021]

(Original post by Sem193)
Sorry that was a bit vague. Just the normal one when you expand it. So For 3(cosx-60) do CosACosB+SinASinB and then multiply it by 3 then do the same for the next one, but remember it's cosAcosB-sinAsinB

I'm so bad at explaining things :/

Ok I understand it now, thanks

[dizzeedollee]

Hello! Could someone possibly explain something to me? I've just done the June 2010 paper for C3 and there's something I don't understand about question 9 iii) and iv)

For iii) this is what I did. I found inverse g which is (x-b)/a and then equated it to ax + b so that I have (x-b)/a = ax+b. Then since the equation says this is true for all values of x (and the domain is all real numbers) I substituted x= 0 so that the equation became -b/a = b which I rearranged to give -b/b=a as in the mark scheme. However the mark scheme says that you cannot use zero as a substitution, why?

For part iv) I got -(4x^2 - 12x) + b < 5 and again the question says that this is true for all values of x so I just substituted x = 0 to get b < 5. However this is not correct, why not?

Also really confused by their mark scheme for iv) too. It says you can take the fact that 4(x-1.5)^2 - 9 = 4x^2 - 12x and substitute it into -(4x^2 - 12x). That makes -(4(x-1.5)^2 - 9) = -4(x-1.5)^2 + 9 = 0 which can then be rearranged to -4(x-1.5)^2 = -9. However the mark scheme says that -4(x-1.5)^2 = positive 9. That's not right is it? :/

Thanks a lot

[pre-med-guy]

How do you answer part ii) of this question? It's from the Jun 11 paper:


http://d.pr/i/vlob

[710]

This exam is make or break, along with Core 4, for possible A* grades...

[ihategeography]

has anyone got the jan 2012 paper?

[Killjoy-]

I would also like the jan 2012 paper, if anyone has it please upload.

[Doctor.]

(Original post by Killjoy-)
I would also like the jan 2012 paper, if anyone has it please upload.

Didn't know you're doing this exam. How are you finding It?

[Killjoy-]

(Original post by Doctor.)
Didn't know you're doing this exam

I'm not - its for a friend who doesn't have a TSR account.
We have study leave now, so unless he goes back I don't think he can get the paper.

[King_Arthur]

(Original post by Killjoy-)
I would also like the jan 2012 paper, if anyone has it please upload.

you have the paper on this thread
Jan 2012
http://www.thestudentroom.co.uk/show...xam+discussion

[King_Arthur]

Here is arseys old thread has papers tips etc........



http://www.thestudentroom.co.uk/show...an%202012%20C3

[Doctor.]

(Original post by anjelofernando)
Here is arseys old thread has papers tips etc........



http://www.thestudentroom.co.uk/show...an%202012%20C3

Thanks for the link

[Killjoy-]

(Original post by anjelofernando)
you have the paper on this thread
Jan 2012
http://www.thestudentroom.co.uk/show...xam+discussion

Thank you, but unless I'm mistaken that thread is for the Edexcel papers?
Or do you mean to say the OCR paper is in that thread?