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C3 (Not MEI) - Thursday June 14 2012, AM

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Reply 60

Doctor.

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

Reply 61

Doctor.

Finished, that was NOT a nice paper :no:

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

Reply 62

Sem193

Original post by Doctor.

Finished, that was NOT a nice paper :no:

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

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

Reply 63

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 :wink:

Reply 64

Sem193

Original post by Luppy021

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.

Reply 65

Luppy021

Original post by Sem193

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

Reply 66

Doctor.

God I hated that question! So badly worded :angry:

Reply 67

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 :smile:

then do the same for the next one, but remember it's cosAcosB-sinAsinB :smile:

I'm so bad at explaining things :/

(edited 11 years ago)

Reply 68

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 :smile:

then do the same for the next one, but remember it's cosAcosB-sinAsinB :smile:

I'm so bad at explaining things :/

Ok I understand it now, thanks

Reply 69

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 :smile:

(edited 11 years ago)

Reply 70

pre-med-guy