Edexcel A2 C4 Mathematics June 2016 - Official Thread

Original post by Supermanxxxxxx

ImageUploadedByStudent Room1464640913.197247.jpg

Spent so long on a still couldn't do it

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Did you divide through by (X-6)(X-3) then do partial fractions to put it as A/(x-6) + B/(X-3) ?

Reply 2162

Supermanxxxxxx

5

Original post by Middriver

Did you divide through by (X-6)(X-3) then do partial fractions to put it as A/(x-6) + B/(X-3) ?

Nope I didn't even think of that, thanks

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Reply 2163

Original post by Supermanxxxxxx

Spent so long on a still couldn't do it

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Edit:nevermind

Reply 2164

jovdawesome

Original post by Clovers

Hi,
Can someone please explain why, when solving a differential equation, you don't put constants of integration on both sides? Thanks in advance :smile:

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hi look at this video to understand it http://www.examsolutions.net/maths-revision/core-maths/integration/differential-equations/seperable/tutorial-1a.php

Reply 2165

jovdawesome

Original post by Supermanxxxxxx

Nope I didn't even think of that, thanks

Posted from TSR Mobile

yh use partial fractions

Reply 2166

ryandaniels2015

11

Anybody know how I should go about learning vectors, I think I understand the topic but can't do any pp questions! :frown:

Reply 2167

Original post by metrize

Can someone help on jan 13 c4 question 8?

https://googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/January%202013%20QP%20-%20C4%20Edexcel.pdf

This is my method:
-ln(3-theta all over A)=t/125
So 3-theta over A=e^(-t/125)
So 3-theta = Ae^(-t/125)
So theta = 3-Ae^(-t/125)

But the answer is meant to be theta = 3+Ae^(-t/125)

Not sure why you start with -ln(3-theta all over A) I thought the A would be a constant on the dt side, so it would start -ln(3-theta)= t/125 +c ?

Reply 2168

Original post by Middriver

Not sure why you start with -ln(3-theta all over A) I thought the A would be a constant on the dt side, so it would start -ln(3-theta)= t/125 +c ?

Yeah i called the constant ln A so then when i took it away and with logs rules i made it -ln( 3-theta all over A)

Reply 2169

Original post by metrize

Yeah i called the constant ln A so then when i took it away and with logs rules i made it -ln( 3-theta all over A)

Oh right, did you manage to get it right?

Reply 2170

imran_

guys for c3, when drawing modulus graphs and equating it to eachother how would you know if they intercept twice or not?

https://07a69ccf283966549a9350d1a66951a7bc96e2dc.googledrive.com/host/0B1ZiqBksUHNYZ0JQM1NRcmdHdXM/June%202014%20(R)%20QP%20-%20C3%20Edexcel.pdf

heres an example, question 7

Reply 2171

Original post by Middriver

Oh right, did you manage to get it right?

I did the steps I had said at first but fot the signs wrong in the end so im not sure where i went wrong

Reply 2172

Original post by metrize

I did the steps I had said at first but fot the signs wrong in the end so im not sure where i went wrong

Hmh shall I send my workings might be able to see where sign change went wrong? I didn't use lnA though so don't want to confuse you or anything.

Reply 2173

Original post by Middriver

Hmh shall I send my workings might be able to see where sign change went wrong? I didn't use lnA though so don't want to confuse you or anything.

Yeah please do, thanks

Reply 2174

jovdawesome

Original post by metrize

Can someone help on jan 13 c4 question 8?

https://googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/January%202013%20QP%20-%20C4%20Edexcel.pdf

This is my method:
-ln(3-theta all over A)=t/125
So 3-theta over A=e^(-t/125)
So 3-theta = Ae^(-t/125)
So theta = 3-Ae^(-t/125)

But the answer is meant to be theta = 3+Ae^(-t/125)

hello, your first step is wrong.. why are you dividing by A? thats not how its done.. the constant youre meant to be Adding is C and not A. Heres how i did it:

1) Dtheta/dt = (3-theta) / 125
2) 1/(3-theta) with respect to dtheta = 1/125 with respect to dt
3) -ln(3-theta) = t/125 + 'C'
4) ln(3-theta) = -t/125 - C
5) (3-theta) = e^-0.008t-c
6) theta= 3 - e^-0.008t-c
7) theta= 3 - (e^-0.008t e^-c)
8) theta = -e^-ce^-0.008t+3
9) theta =Ae^-0.008t+3
where A= -e^-cit says A is a constant and -e^-cis a constant

Reply 2175

Original post by jovdawesome

hello, your first step is wrong.. why are you dividing by A? thats not how its done.. the constant youre meant to be Adding is C and not A. Heres how i did it:

1) Dtheta/dt = (3-theta) / 125
2) 1/(3-theta) with respect to dtheta = 1/125 with respect to dt
3) -ln(3-theta) = t/125 + 'C'
4) ln(3-theta) = -t/125 - C
5) (3-theta) = e^-0.008t-c
6) theta= 3 - e^-0.008t-c
7) theta= 3 - (e^-0.008t e^-c)
8) theta = -e^-ce^-0.008t+3
9) theta =Ae^-0.008t+3
where A= -e^-cit says A is a constant and -e^-cis a constant

I did that because at the beginning i said the constant is ln A and so i took away ln A and with log rules it divides by A inside the log, its the way that our teacher had taught it us, will learn how to do it with c, thanks for posting the working out

Reply 2176

ryandaniels2015

11

Original post by ryandaniels2015

Anybody know how I should go about learning vectors, I think I understand the topic but can't do any pp questions! :frown:

any help?

Nvm he's done it^

Original post by metrize

I did that because at the beginning i said the constant is ln A and so i took away ln A and with log rules it divides by A inside the log, its the way that our teacher had taught it us, will learn how to do it with c, thanks for posting the working out

When adding the constant, we must use 'C' in most cases. We must only add lnC if it makes it easier for us to solve it. For example if a question was ln(x-2)= ln5.. we can add lnC here because it will be easier for us to solve as we can cancel the ln on both sides. However in all questions ur simply meant to add C and NOT lnC..
I recommend you go over videos on examsolutions and learn this topic ASAP as we have just above 2 weeks left for the exam .. Good luck :smile:

Reply 2179