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Edexcel A2 C4 Mathematics June 2016 - Official Thread

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Because when you integrate it to one version, you get (1/3)lnx + C. If you integrate it to the other version, you get (1/3)ln3x + D. Where C and D are some constants. Taking the second version and applying log rules, we can expand

(1/3)ln3x + D

to (via logarithm product rule)

(1/3)lnx + (1/3)ln3 + D

In the first version we saw that we got (1/3)lnx + C.

So by comparing the two, C = (1/3)ln3 + D.

As you can see C and D are not equal, so basically when you take the indefinite integral, you get a different constant depending on how you integrate it.


Now when you take the definite integral, the constants always cancel each other out anyway, so it doesn't make a difference which version you use.
Original post by Don Pedro K.
@13 1 20 8 42 Is this even C4?

wut.JPG


Isn't that one of TeeEm's hard/very hard papers
Original post by Don Pedro K.
@13 1 20 8 42 Is this even C4?

wut.JPG


That is definitely C4. It can be done with further methods, but C4 teaches you all the information you need to solve that
Original post by 1 8 13 20 42
Isn't that one of TeeEm's hard/very hard papers


Yeah paper s.

Original post by Student403
That is definitely C4. It can be done with further methods, but C4 teaches you all the information you need to solve that


Ah okay...well I guess I'll have a got at it :s! Never seen anything like it before though!
Original post by Don Pedro K.
Yeah paper s.



Ah okay...well I guess I'll have a got at it :s! Never seen anything like it before though!


Good luck. The solution's pretty simple if you take it step by step
Original post by Don Pedro K.
Yeah paper s.



Ah okay...well I guess I'll have a got at it :s! Never seen anything like it before though!


I wouldn't worry about S and T. But that question is one of the easier ones I think. Just think about what you know about minimums and variables..
question 7a
http://qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2013/Exam-materials/6666_01_que_20150616.pdf
it doesnt say whether to do radians or in degrees unless being stupid but can some one help
Ln 3x and ln x both differentiate to 1/x
Reply 3848
Avatar for taysc
taysc
2
Original post by NotoriousS
Screen Shot 2016-06-23 at 21.22.38.png

Can someone please help me on part b?



it tells you that x<8/9.

So you can take out 1000 from under the cube root sign leaving you with
10 cube root 7.1

Then 7.1= 8-9x
Solve for x=0.1
Sub 0.1 into your previous expansion and times by 10

In summary you need your x to be small, so you make that possible using the method above. P.S the mark scheme for this question has an error
Original post by shamk123
question 7a
http://qualifications.pearson.com/content/dam/pdf/A-Level/Mathematics/2013/Exam-materials/6666_01_que_20150616.pdf
it doesnt say whether to do radians or in degrees unless being stupid but can some one help


In calculus you always use radians
Reply 3850
Avatar for taysc
taysc
2
Original post by Student403
Good luck. The solution's pretty simple if you take it step by step


What is the solution, I'm not sure how to do this?

Thank you
i know this sounds stupid but ive got a mind blank on the easiest question
Can someone please differentiate 5/2t for my stupid ar*e :wink:
Original post by taysc
What is the solution, I'm not sure how to do this?

Thank you


The shortest line joining two lines, l1 and l2, lies perpendicular to both l1 and l2. Try to take it from here
Original post by coolguy123456
phew

But then again this is edextrick.. They gave a messed up C4 q in fp2
𝐬𝐢𝐧𝟓 𝒙 𝒄𝒐𝒔 𝒙 How do you integrate this?
Original post by celessi
𝐬𝐢𝐧𝟓 𝒙 𝒄𝒐𝒔 𝒙 How do you integrate this?


sin5xcosx or sin5xcosx?
Original post by Kingsyboy
i know this sounds stupid but ive got a mind blank on the easiest question
Can someone please differentiate 5/2t for my stupid ar*e :wink:


if you mean that t is the denominator:
5/2*ln(2t)
Original post by Slimewizard
if you mean that t is the denominator:
5/2*ln(2t)


he said differentiate, not integrate
Original post by SeanFM
Yes.. although your number doesn't seem to be directly useful :tongue:

I was thinking more 480-275 = 205, so you only need 205 UMS across 3 modules.


Thanks very much. Since 3 modules =300 ums, Ive got quite a lot of UMS to play with before I drop down to a B :smile:
Original post by Student403
he said differentiate, not integrate


i s2g i'll probably mess up like this on the exam as well cause im such a ditz

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