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OCR A2 (Non-MEI) C4 Exam - 18th June 2014

There doesn't seem too be a thread for this yet. Mainly for exam day but feel free to post questions/tips etc.

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Reply 1
looking forward to this :smile:

resitting it this year to get an A*, as last year I got 95 in C3 but only 78 in C4 :frown:.
Oh well, having done further maths C4 is bliss.. For instance last year I hated implicit differentiation, but now I see it as free marks; and after doing 3D vectors in FP3, the 2D vectors in C4 which I used to hate are actually quite relaxing xD
The Vectors in C4 are still 3D but we don't really worry about that.

I really hate C4 compared to C3, but I probably got 100 in C3 so I only need 80. I would like to get as high as possible but I'm taking a break today and I'll revise hard tomorrow.
(edited 9 years ago)
Reply 3
Anyone know how to do this question by any chance? Struggling so hard with C4!

Here's the link to a picture of it

http://imgur.com/gRNg3Vj
Original post by Majeue
Anyone know how to do this question by any chance? Struggling so hard with C4!

Here's the link to a picture of it

http://imgur.com/gRNg3Vj


I can do part i but I can't do ii for some reason. I presume you seperate it iinto xtanx * tanx and then do parts again but I can't do it :frown:(

edit: worked it out, you change the tan squared into sec and work it out from there :smile:
(edited 9 years ago)
Reply 5
Original post by Sweetcorn_1
I can do part i but I can't do ii for some reason. I presume you seperate it iinto xtanx * tanx and then do parts again but I can't do it :frown:(


Could you explain part i for me please? would be very helpful!
Original post by Majeue
Could you explain part i for me please? would be very helpful!


Well it's integration by parts.

let u = x and dv/dx = sec^2(x)
du/dx =1

and since we all know that tanx differentiated is sec^2x, v = tanx.

from then on just use the rule. part ii you change tan into sec and just use your previous answer.
Reply 7
bump
Reply 8
Original post by Sweetcorn_1
Well it's integration by parts.

let u = x and dv/dx = sec^2(x)
du/dx =1

and since we all know that tanx differentiated is sec^2x, v = tanx.

from then on just use the rule. part ii you change tan into sec and just use your previous answer.


Would this be the right way to do part i?

http://imgur.com/Hcl1sVM
Original post by Majeue
Would this be the right way to do part i?

http://imgur.com/Hcl1sVM


uh.. no.. You're supposed to use integration by parts like I said.

the answer will be xtanx + lncosx
Original post by Schraderrr
help please??




(i) Expand (1+2x)^1/2

(ii) Expand (1+x)^-3 and multiply it with the answer in i. Only multiply the parts which produce co-efficients of x^3 and below because that's what the question asks.

(iii) |2x|<1
=> |x|<1/2
Reply 12
Original post by Sweetcorn_1
uh.. no.. You're supposed to use integration by parts like I said.

the answer will be xtanx + lncosx


You have to show that it equals the integral of cos^2(θ) though
(edited 9 years ago)
Original post by Majeue
You have to show that it equals the integral of cos^2(θ) though


????

Are you sure you're showing me the right question? The question you're showing me says "integration by parts" in the question with no thetas anywhere.
Reply 14
Original post by Sweetcorn_1
????

Are you sure you're showing me the right question? The question you're showing me says "integration by parts" in the question with no thetas anywhere.


Yeah somehow I got the wrong question sorry. It's 4i and ii of the first paper here - june 05

http://blogs.thegrangeschool.net/maths/files/2012/03/C4-Past-Paper-Booklet-2012.pdf
Original post by Majeue
Yeah somehow I got the wrong question sorry. It's 4i and ii of the first paper here - june 05

http://blogs.thegrangeschool.net/maths/files/2012/03/C4-Past-Paper-Booklet-2012.pdf


Then you've done it right if you've shown what they asked for, there are many ways to do it, my way is basically the same.
Original post by Majeue
Yeah somehow I got the wrong question sorry. It's 4i and ii of the first paper here - june 05

http://blogs.thegrangeschool.net/maths/files/2012/03/C4-Past-Paper-Booklet-2012.pdf


x=tanθ, therefore x^2 = (tanθ)^2
Also, dx=(secθ)^2 (the differential of tanθ)
So you then have the integral of (secθ)^2 divided by (1+(tanθ)^2 )^2

Since 1+(tanθ)^2 = (secθ)^2, you're left with the integral of / (secθ)^2 as you can do some cancelling of powers. This is equivalent to (cosθ)^2 For the second part you just have to change the limits and use the double angle formula (cosθ)^2= (1+cos2θ) / 2
Has anybody got a link to the June 2013 paper?
Original post by Sweetcorn_1

Someone posted these links in the C3 thread. Credits to them.


Thanks.:smile: Is there a mark scheme also?

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