FP2 (Not MEI) - Thursday June 14 2012, AM

Original post by NightBear

Jan 12 was such a nice paper!

Parts of it were - I got 62 but I made a daft mistake on question 4. (lost me about 6 marks! Didn't see it said n triangles rather than n-1)

Original post by Ree69

June 10 Q6ii is ridiculous. Calculating the second derivative of such a function should not be on an A-Level paper!

It's not THAT bad? Just tedious? :tongue:

Reply 86

NightBear

12

Original post by wibletg

Parts of it were - I got 62 but I made a daft mistake on question 4. (lost me about 6 marks! Didn't see it said n triangles rather than n-1)

That's well over 90% :tongue:

I'm guessing you need an A*?

Reply 87

Original post by NightBear

That's well over 90% :tongue:

I'm guessing you need an A*?

It's about 97/98 UMS, which I'd quite happily take :tongue:

I need an A* but since I got 99 and 98 in M2 and M3 I only need a low B in either this or S2 :smile:

Reply 88

SecondHand

I did Jan 12 today and thought it was quite a tricky paper in the last few questions, scored 62 though which is well over the A boundary and I only need an A overall to match my offer.

Reply 89

Original post by SecondHand

I did Jan 12 today and thought it was quite a tricky paper in the last few questions, scored 62 though which is well over the A boundary and I only need an A overall to match my offer.

The polar loops graph took some thinking about - I nearly went in and did the area of the top loop subtracted by the area of the bottom loop.

That and the very last question! (series of stuff I think?)

Apparently people struggled to finish question 9 hence the grade boundaries (I read the examiners report :tongue:

)

One thing I NEED to do tomorrow (somewhere else where I dropped a few marks) is justifying why the sign is the way it is when you work out the inverse of a cosh or sinh using a log :smile:

Reply 90

lilipop

Can someone explain Jan12 9 iv please? really don't understand the mark scheme...

Reply 91

Original post by lilipop

Can someone explain Jan12 9 iv please? really don't understand the mark scheme...

You're using the fact that In-2 - In is the same as the righthand side of the equation I believe, and going up to infinity.

This means that they cancel each other out and you're left with just I1 at the end.

Reply 92

Anon 17

^ That, and the fact that the last I_infinity would be essentially zero due to the formula for I_n.

Can anyone who's got it upload or PM me the Jan 12 mark scheme? I've marked my paper off Mr. M's unofficial answers from the Jan 12 FP2 topic... But that's not exactly a proper mark scheme =P

I found the paper pretty easy (but long) except for question 8ii), still can't figure out how to cancel everything.

Reply 93

lilipop

Original post by wibletg

You're using the fact that In-2 - In is the same as the righthand side of the equation I believe, and going up to infinity.

This means that they cancel each other out and you're left with just I1 at the end.

I got In-I1 not I1, how do you cancel In?

Reply 94

Ree69

Original post by wibletg

It's not THAT bad? Just tedious? :tongue:

Must be just me then - it took me a while, I think over 10 minutes. :s-smilie:

Reply 95

Ree69

Further Pure 2 Mark Scheme January 2012.pdf130.6KB

Original post by Anon 17

Can anyone who's got it upload or PM me the Jan 12 mark scheme? I've marked my paper off Mr. M's unofficial answers from the Jan 12 FP2 topic... But that's not exactly a proper mark scheme =P

I found the paper pretty easy (but long) except for question 8ii), still can't figure out how to cancel everything.

Here you go mate.

Reply 96

JongKey

Original post by NightBear

Oh damn, I meant Jun 06 :/ nvm, I think I've worked it out.

I still don't get it, could you tell me how you did it? It's that series question isn't it?

Reply 97

Ree69

Original post by lilipop

I got In-I1 not I1, how do you cancel In?

Well...

Notice that

\frac{1}{2}(\frac{3}{5})^{2} + \frac{1}{4}(\frac{3}{5})^{4} + \frac{1}{6}(\frac{3}{5})^{6} +... = \frac{1}{n-1}(\frac{3}{5})^{n-1}

for n = 3, 5, 7, 9, ...right up to infinity.

But

\frac{1}{n-1}(\frac{3}{5})^{n-1} = I_{n-2} - I_n

because

-\frac{1}{n-1}(\frac{3}{5})^{n-1} = I_n - I_{n-2}

as shown in ii).

So

\frac{1}{2}(\frac{3}{5})^{2} + \frac{1}{4}(\frac{3}{5})^{4} + \frac{1}{6}(\frac{3}{5})^{6} +... = (I_1 - I_3) + (I_3 - I_5) + (I_5 - I_7)... = I_1 = \ln\ \frac{5}{4}

.

(edited 11 years ago)

Reply 98

Photox

1

Banging out the past papers, getting 90+ %, feeling pretty damn good about this exam, now just a good nights sleep, and a core 3 warm up before it, then its gonna be oh so good :3 Good Luck People :biggrin:

Reply 99