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The Official OCR (Not MEI) FP2 January 10th 2011 Thread

Official thread for the (non MEI) OCR FP2 exam on the 10th of January.

How are we all feeling?

I'm personally dreading polar coordinates and I hope the reduction formulae question is kind! :s-smilie:

(Goes back to revision)

Scroll to see replies

:bump:

Mr M
...


Mr M, you're probably busy right now/doing C1 answers/teaching etc etc but I trust the FP2 answers will be up soon/eventually?
Thanks!
(edited 13 years ago)
Reply 2
I thought it was quite an easy paper - still there were some really annoying questions such as the questions about y^2 = f(x) which I definitely lost marks on.
Original post by busbybabe
I thought it was quite an easy paper - still there were some really annoying questions such as the questions about y^2 = f(x) which I definitely lost marks on.


Seriously??!!! :eek: I thought it was bloody hard! :cry2: I don't think I got the last question at all. :frown: It was a fail paper seriously... :frown:
Reply 4
the last term you just had to do what would do with cos^n and make it cosh^n-1, cosh
by parts it and you get what they ask, although I did guess a bit when applying the limits to the already integrated bit.

I just mucked up on the stupid newton raphson questions aswell - ITS NOT REAL MATHS
Original post by busbybabe
the last term you just had to do what would do with cos^n and make it cosh^n-1, cosh
by parts it and you get what they ask, although I did guess a bit when applying the limits to the already integrated bit.

I just mucked up on the stupid newton raphson questions aswell - ITS NOT REAL MATHS


Oh by the last question, I meant the last part. I5... Like an idiot instead of evaluating Io I evaluated I1. :rant:
Reply 6
Original post by nikita_atikin
Oh by the last question, I meant the last part. I5... Like an idiot instead of evaluating Io I evaluated I1. :rant:


pretty sure thats what youre meant to do ?
If you evaluate I5 it gives you something in I3
If you evaluate I3 it gives you something in I1 (1/3pi or something)
then sub that value in etc.
Original post by busbybabe
pretty sure thats what youre meant to do ?
If you evaluate I5 it gives you something in I3
If you evaluate I3 it gives you something in I1 (1/3pi or something)
then sub that value in etc.


Err... oops, my brain has stopped working. Instead of I1, I launched myself into Io. Which is absolutely pointless so I didn't get very far.

I wish I could do that paper again. :frown: Reckon the grade boundaries will be low? They're always around 49-53for an A... I might need a lower one :frown:
Reply 8
honestly no idea - last years June was outragously hard but they didn't even publish them on the examiners report (which they usually do). Could be anywhere from 49-55. Just want MR M to get the answers up so I can see how ive did
Original post by busbybabe
honestly no idea - last years June was outragously hard but they didn't even publish them on the examiners report (which they usually do). Could be anywhere from 49-55. Just want MR M to get the answers up so I can see how ive did


Last year's grade boundary for an A was 52. I did it as my mock and got 52 so that's how I know lol :tongue:

Same I'm waiting for Mr M too but I think I've forgotten most of the answers I put down. Method marks :love: Help me please!
Reply 10
may aswell post mine to see if you got anything close to it

first one was the tan0.5x substitution - i got it as the integral of 1/1+t which was ln(1+tan0.5x) ?

second was a mclaurin : x + 1/3x^3

third was that f(x) graph question: assymptote was y=0 , y took all values fbigger than 5/2 and less than -5/2??
for y^2 = f(x) i wasnt sure of the max and min, and said x took all values > 0 (complete guess)

fourth was a cosh show that and then i got x = ln(0.25 + root 5/2 i think?)

5th was newton raphson = simple rearrange, get f'(x) although when i put f'(a) in it didnt get 0
and it convered to a root that started 1.8 something

6th was the series integrals - cant remember values but i remember i couldnt do the x^x one

7th i cant really remember, i think it was an integration between pi/3 and -pi/3 and i think the answer was just something in pi?

8th was the reduction formula discussed -

dyou have the question paper?
Original post by busbybabe
may aswell post mine to see if you got anything close to it

first one was the tan0.5x substitution - i got it as the integral of 1/1+t which was ln(1+tan0.5x) ?

second was a mclaurin : x + 1/3x^3

third was that f(x) graph question: assymptote was y=0 , y took all values fbigger than 5/2 and less than -5/2??
for y^2 = f(x) i wasnt sure of the max and min, and said x took all values > 0 (complete guess)

fourth was a cosh show that and then i got x = ln(0.25 + root 5/2 i think?)

5th was newton raphson = simple rearrange, get f'(x) although when i put f'(a) in it didnt get 0
and it convered to a root that started 1.8 something

6th was the series integrals - cant remember values but i remember i couldnt do the x^x one

7th i cant really remember, i think it was an integration between pi/3 and -pi/3 and i think the answer was just something in pi?

8th was the reduction formula discussed -

dyou have the question paper?


I can't remember any of my answers but nope, I don't have the question paper sorry. :frown:

1) S*** we had to put tan back into the first question :mad: I got 0.5ln(1+2t) And obviously t had to be substituted with tan0.5x so I was supposed to get 0.5ln(1+tan0.5x)

2) I got the McLaurins okay.

3) Same for y, I wasn't sure if it was -2.5 or not :s-smilie: I tried a sketch but I wasn't sure. The max and min I said it was +or- root(5/2) and I said x>=0

4) Can't remember but I think I got sinh related answers because cosh wouldn't work for -0.5 or something? Can't remember

5) Newton raphson was alright although I didn't get the very f'(alpha) with any of x^3-5x+3=0's roots.

6) Series and integrals, sneaky but I think I gave it a good go. Managed to show the bounds successfully. I took the ln of y=x^x so you let lny=xlnx and then differentiate both sides. I then said 1/y=0 (don't know why I said so) and fiddled it a bit to get e^-1. Don't think I'm right though because I "cheated" the way it worked :s-smilie: If you get my drift.

7) What were the tangents? And the line of symmetry? I said the tangents were +or- pi/3.
The integration was 0.25pi?

8) Discussed. It was crap.
:bump: Just so other FP2ers can see and join the suck fest... :smile:
Reply 13
This is what I got :

1. ln(1+tan0.5x) + c
2i. (show)
ii. x + (x^3)/6 (I stupidly wrote (x^3)/6)
3i. a) y=0 , b) between/equal to 2.5 and -2.5
ii. a) y=0, b) 2.5, -2.5 c) x greater than or equal to 0
4i. (show)
ii. x=ln (rt5 +/- 1)/2
5i. (show)
ii. (show)
iii. I did it wrong but the answer I worked out after was 1.86... or something
6i. (show)
ii. (show)
iii. 0.438
iv. I drew 3 rectangles on two seperate diagrams - upper and lower bounds with the stationary point as being a value of a length on both diagrams
7i. (show) - showed values of theta on either side of 0
ii. pi, pi/3, -pi/3
iii. area = pi/2
8i. (show) - I showed ln (2+rt 3) on both sides
ii. (show)
iii. (24rt3)/5

Most of them are right I think

EDIT - Ive got the paper aswell if anyone wants to know about certain questions. I would scan it in but I've scribbled all over and its extremely messy.
(edited 13 years ago)
Original post by UFK
This is what I got :

1. ln(1+tan0.5x) + c
2i. (show)
ii. x + (x^3)/6 (I stupidly wrote (x^3)/6)
3i. a) y=0 , b) between/equal to 2.5 and -2.5
ii. a) y=0, b) 2.5, -2.5 c) x greater than or equal to 0
4i. (show)
ii. x=ln (rt5 +/- 1)/2
5i. (show)
ii. (show)
iii. I did it wrong but the answer I worked out after was 1.86... or something
6i. (show)
ii. (show)
iii. 0.438
iv. I drew 3 rectangles on two seperate diagrams - upper and lower bounds with the stationary point as being a value of a length on both diagrams
7i. (show) - showed values of theta on either side of 0
ii. pi, pi/3, -pi/3
iii. area = pi/2
8i. (show) - I showed ln (2+rt 3) on both sides
ii. (show)
iii. (24rt3)/5

Most of them are right I think

EDIT - Ive got the paper aswell if anyone wants to know about certain questions. I would scan it in but I've scribbled all over and its extremely messy.


7iii) It's 0.5pi??? I spend bloody ages on it and got 0.25pi. I'm defo failing this exam :frown:
Reply 15
Original post by nikita_atikin
7iii) It's 0.5pi??? I spend bloody ages on it and got 0.25pi. I'm defo failing this exam :frown:


I think thats correct. I integrated between limits pi/3 and 0, then multiplied by 2 because the line theta=0 was a line of symmetry. Therefore the answer would've been (pi/4 x 2) = pi/2. I may be wrong though, thats what I did anyway.
Original post by UFK
I think thats correct. I integrated between limits pi/3 and 0, then multiplied by 2 because the line theta=0 was a line of symmetry. Therefore the answer would've been (pi/4 x 2) = pi/2. I may be wrong though, thats what I did anyway.


How is it different to integrating between pi/3 and -pi/3? Surely it should give the same answer? :confused:

Also, your McLaurins should be x + x^3/3 and not x^3/6. :smile:
Reply 17
Original post by UFK
This is what I got :

1. ln(1+tan0.5x) + c
2i. (show)
ii. x + (x^3)/6 (I stupidly wrote (x^3)/6)
3i. a) y=0 , b) between/equal to 2.5 and -2.5
ii. a) y=0, b) 2.5, -2.5 c) x greater than or equal to 0
4i. (show)
ii. x=ln (rt5 +/- 1)/2
5i. (show)
ii. (show)
iii. I did it wrong but the answer I worked out after was 1.86... or something
6i. (show)
ii. (show)
iii. 0.438
iv. I drew 3 rectangles on two seperate diagrams - upper and lower bounds with the stationary point as being a value of a length on both diagrams
7i. (show) - showed values of theta on either side of 0
ii. pi, pi/3, -pi/3
iii. area = pi/2
8i. (show) - I showed ln (2+rt 3) on both sides
ii. (show)
iii. (24rt3)/5

Most of them are right I think

EDIT - Ive got the paper aswell if anyone wants to know about certain questions. I would scan it in but I've scribbled all over and its extremely messy.



please scan it in mate, it would be really useful for me, ta.
Reply 18
Original post by nikita_atikin
How is it different to integrating between pi/3 and -pi/3? Surely it should give the same answer? :confused:

Also, your McLaurins should be x + x^3/3 and not x^3/6. :smile:


I got that all the sin's cancel as sin pi/3 = 0 and that all it was was 1/3pi + 1/3pi

and the 7 part ii) should have 1/3pi, -1/3pi, pi and -pi ?
(edited 13 years ago)
Original post by busbybabe
I got that all the sin's cancel as sin pi/3 = 0 and that all it was was 1/3pi + 1/3pi

and the 7 part ii) should have 1/3pi, -1/3pi, pi and -pi ?


So was your answer for the area question 2pi/3?

I guess Mr M will confirm once and for all but what UKF seems to make sense because they wouldn't get us to find a line of symmetry if it wasn't going to be any use to us in the next few parts :s-smilie:

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