You are Here: Home >< Maths

# OCR MEI FP2 Thread - AM 27th June 2016 Watch

1. (Original post by Wunderbarr)
If you draw the lines, it can be used to explain why the w3 +w2 + w = 0 thing is true, by looking at the vector sum of the lines (if drawn as vectors going from the origin to Z).
What is the w3 +w2 + w = 0 thing?
2. (Original post by aproei)
What is the w3 +w2 + w = 0 thing?
https://youtu.be/WlM1WjmAkA0?t=14m53s

Maybe I should've written w2+ w +1 = 0 instead.
3. (Original post by Wunderbarr)
https://youtu.be/WlM1WjmAkA0?t=14m53s

Maybe I should've written w2+ w +1 = 0 instead.
Gotcha
4. anyone help on jan 06 4Cii? dont understand what to do after I intergrated by parts

Posted from TSR Mobile
5. (Original post by HFancy1997)
anyone help on jan 06 4Cii? dont understand what to do after I intergrated by parts

Posted from TSR Mobile
You get to the necessary result, then realise you need to change the 2arsinh(4/3) -1 into a natural log somehow...

So, look in the formula booklet. Page 2, bottom right corner.
6. (Original post by Wunderbarr)
You get to the necessary result, then realise you need to change the 2arsinh(4/3) -1 into a natural log somehow...

So, look in the formula booklet. Page 2, bottom right corner.
Sorry I shouldve been more clear,

I got uv but dunno what to do with the -intergral(Vdu/dx), I just dont see how went to the next step

Posted from TSR Mobile
7. (Original post by HFancy1997)
Sorry I shouldve been more clear,

I got uv but dunno what to do with the -intergral(Vdu/dx), I just dont see how went to the next step

Posted from TSR Mobile
Rewrite the integral as brackets and powers (so no fractions or square roots).

You should be able to tell what type of integration is done more clearly.

The rewritten integral should appear something like:
Spoiler:
Show
2x(9+4x2)½
8. (Original post by Wunderbarr)
Rewrite the integral as brackets and powers (so no fractions or square roots).

You should be able to tell what type of integration is done more clearly.

The rewritten integral should appear something like:
Spoiler:
Show
2x(9+4x2)½
damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

Posted from TSR Mobile
9. (Original post by HFancy1997)
damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

Posted from TSR Mobile
It's a simple integration. Think about reversing the chain rule.
10. (Original post by HFancy1997)
damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

Posted from TSR Mobile
I'm going off now, and you should too soon!

Happy hunting tomorrow everyone!
11. (Original post by HFancy1997)
damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

Posted from TSR Mobile
I just watched this video on the topic which helped but I think in the exam I will use substitution even though it takes longer.

12. Could not finish off that last question. I always struggle with actual numbers and tedious expansions. Left my answer as a long string of (3+sqrt(5))^2 and such. Think I got everything else.
13. Damn that was tough, two new styles of questions never done before too. Annoying though, forgot something very basic during the exam
14. last question you had to change sinh to exponentials then it gives you a surd numbee, not sure if its right but I did that

Posted from TSR Mobile
15. (Original post by Bunderwump)
Could not finish off that last question. I always struggle with actual numbers and tedious expansions. Left my answer as a long string of (3+sqrt(5))^2 and such. Think I got everything else.
Yes me too I just ran out of time
16. I think tha exam is the hardest ive ever done. Im just praying on low grade boundaries or im kissing uni goodbye. How did everone else find it.
17. Someone do an unofficial markscheme plzz

Here are fragments from my memory:

For Q1 I remember getting pi/6 as the integration

Q2: in part (i) it simplified to 1 - z

root arguments were pi/18, 13pi/18, -11pi/18 and modulus was root 2

Q3: M-1^n tends to infinity

Q4: show that..
x = plus/ minus ln 3+root5 /2

curve had y intercept at 2 and was symmetrical for plus and minus x

Yh thats all I can remember atm and these are just my answers so may be wrong
18. (Original post by Sophieoo1)
Someone do an unofficial markscheme plzz

Here are fragments from my memory:

For Q1 I remember getting pi/6 as the integration

Q2: in part (i) it simplified to 1 - z

root arguments were pi/18, 13pi/18, -11pi/18 and modulus was root 2

Q3: M-1^n tends to infinity

Q4: show that..
x = plus/ minus ln 3+root5 /2

curve had y intercept at 2 and was symmetrical for plus and minus x

Yh thats all I can remember atm and these are just my answers so may be wrong
got all these

Posted from TSR Mobile
19. (Original post by Sophieoo1)
Yes me too I just ran out of time
My answers, by no means right, just want to compare with people :

1)i) 1 - x^2 + x^4

ii) x - 1/3 x^3 + 1/5 x^5

iii) pi/6

iv) Draw the graph, sort of like a loop but starting at pi/4 and ending on the horizontal axis

As the angles tends to 0 r tends to infinity

v) a^2 ln(2 root2) or 1/2 a^2 ln 8

2)i) 1-z

ii) Show that C + jS thing

iii) Modulus of the cube roots was root2 the angles were pi/18 13pi/18 25pi/18

3)i) Eigenvalues were -1/6, 1 eigenvectors were (1 1) and (3 -4) I think, cant quite remember the order

ii) M^n tended towards (3 4) or something like that order might be different
................................ .(3 4)
cant quite remember

iii) Not too sure on this, I think it didnt tend to a limit (maybe infinity), as the elements of the matrix were greater than 1 but some were negative so it could have positive or negative infinity

4)i) Show that arcosh thing

ii) ln((3+root5)/2) and ln((3-root5)/2)

iii) The graph sort of looked like x^2 but started at y=2

The area bound by the curve was 5root5 / 2
The area bound by the line y = 5 was 5ln((7+3root5)/2)

Then take one away from the other.

I think that was all the questions, not sure how many are right, just what I got
20. I ****ed that up. **** **** **** ****

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 6, 2017
Today on TSR

### Anxious about my Oxford offer

What should I do?

### Am I doomed because I messed up my mocks?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE