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

Yep, having come home and tried again I've seen what I've done wrong, and been slightly horrified haha. For some reason that exam really did not go my way. Thanks though :smile:

Reply 181

Femto

8

Now for STEP... Eugh I hate my life.

Reply 182

Ree69

14

Original post by tomp99

Interesting. The question read "Find the maximum value." I might be more worried if it had said "show that the max value is this." Plus they'd just asked us to find the tangents at the pole so we've already found the minimum values.

4 marks: Attempts to find dr/dtheta and =0 M1
Uses product rule M1 maybe?
...and an intermediary answer mark and a final answer mark? Hm, you might be right.

Original post by CJAW

I thought about this too and I actually just tried to double differentiate sin2thetacostheta and it gets quite intricate so from that I reckon that they wouldn't have expected you to do it.

Also, because theta was only between 0 and 1/2 pie, would it be possible to have a max and min in that range? I'm thinking that it wouldn't so therefore it would have to be max.

Yeah...knowing OCR mark schemes :frown:

. I don't think it needs to be as hard as calculating the second derivative; when solving

\frac{dr}{d\theta} = 0

, the only possible solutions were

\cos\theta = 0

and

\sin\theta = \pm\frac{\sqrt{3}}{3}

. We know the first one gives a minimum from i). So the second must be a maximum.

Reply 183

Qwertyfoo

Unparseable latex formula:

\text{f}(n+1)-\text{f}(n)=\left(1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}+\frac{1}{n}-\ln(n)\right)=\frac{1}{n+1} - \ln(n+1)+\ln(n)=\frac{1}{n+1}-\left(\ln(n+1)-\ln(n)\right)=\frac{1}{n+1}-\ln\left(1+\frac{1}{n}\right) \approx\frac{1}{n+1}-\left(\frac{1}{n}-\frac{1}{2n^2}\right)=-\frac{n-1}{2n^2(n+1)}[br]\end{align*}[br]

Reply 184

CJAW

9

Original post by theWildman

Counting, I dropped at least 14 marks, probably not too many more though. Does anyone have any idea what the UMS scaling will be on this?

Obviously no-one knows, but usually 50 give or take is 80ums. So 56-58 would be 90.
This was a hard paper so maybe take rather than give.

Reply 185

tomp99

Original post by theWildman

Hey everyone, how did you manage that value for r as a maximum? I differentiated r but only managed to get theta=pi/4 and subsequently had Rmax = (rt2)/2.

I also couldn't do the last part of the last question, I was under the impression that the only possible solutions to y^2=y would be y=0 or y=1, neither of which produced the same value of x for both equations (C2 and the line).

8iii) I'd already found y in the form y=x/2+1+8/(x+2) from part (8i), which helped a lot in this part.

y^2 = x/2 + 1 + 8/(x+2) and y = x/2 + 1 (which also means that x+2 = 2y)

Therefore y^2 - y - 8/(x+2) = 0 and so y^2 - y - 8/2y = 0

=> y^3 - y^2 - 4 = 0

By inspection (factor theorem, looking at factors of 4), y=2 is a solution to this - in fact, by what they tell you, it's the only solution. If y=2, then by x+2=2y, x must also be 2. So the intersection is (2,2).

Reply 186

cocopops93

on january it was 49 for an 80 UMS but this one was a bit better then january o i'm thinking it will be mid 50's for 80 UMS judging by other people's comments too. I reckon i got 38 make for definite and maybe 20 ish ones so fingers crossed for my B :smile:

Reply 187

theWildman

Original post by Qwertyfoo

Unparseable latex formula:

\text{f}(n+1)-\text{f}(n)=\left(1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}+\frac{1}{n}-\ln(n)\right)=\frac{1}{n+1} - \ln(n+1)+\ln(n)=\frac{1}{n+1}-\left(\ln(n+1)-\ln(n)\right)=\frac{1}{n+1}-\ln\left(1+\frac{1}{n}\right) \approx\frac{1}{n+1}-\left(\frac{1}{n}-\frac{1}{2n^2}\right)=-\frac{n-1}{2n^2(n+1)}[br]\end{align*}[br]

If you don't brackets the logs with a -ve you can get:

f(n+1)-f(n)= 1/(n+1) + lnn - ln(n+1) = 1/(n+1) + ln(1 - 1/(1+n))
=1/(n+1) - ((-1/(n+1)) + 0.5(-1/(n+1))^2)
=1/(2(n+1)^2)

This is what I did, so obviously I didn't end up with the desired result, but I couldn't see what was wrong with it. Is this an alternative and correct solution (although not what they want) or is my maths at fault?

Reply 188

Anon 17

Original post by Femto

Now for STEP... Eugh I hate my life.

This, this so much.

Reply 189

Qwertyfoo

I_n=\int_0^\pi x^n \sin{x}\;\mathrm{d}x=\left[-x^n \cos{x}\right]_0^\pi+n\int_0^\pi x^{n-1} \cos{x}\;\mathrm{d}x

=\pi^n+n\left(\left[x^{n-1} \sin{x}\right]_0^\pi-(n-1)\int_0^\pi x^{n-2} \sin{x}\;\mathrm{d}x\right)

=\pi^n-n(n-1)I_{n-2}

I_1=\int_0^\pi x \sin{x}\;\mathrm{d}x=\left[-x \cos{x}\right]_0^\pi+\left[\sin x\right]_0^\pi=\pi

I_3=\pi^3-(3)(2)(\pi)=\pi^3-6\pi

I_5=\pi^5-(5)(4)(\pi^3-6\pi)=\pi^5-20\pi^3+120\pi

Reply 190

tomp99

Original post by theWildman

If you don't brackets the logs with a -ve you can get:

f(n+1)-f(n)= 1/(n+1) + lnn - ln(n+1) = 1/(n+1) + ln(1 - 1/(1+n))
=1/(n+1) - ((-1/(n+1)) + 0.5(-1/(n+1))^2)
=1/(2(n+1)^2)

This is what I did, so obviously I didn't end up with the desired result, but I couldn't see what was wrong with it. Is this an alternative and correct solution (although not what they want) or is my maths at fault?

Your maths seems to be sound, apart from the exclusion of a minus sign on the bottom line which I presume is an oversight and that you know it should be there. If you manipulate it slightly, I get your answer to be 1/(2n^2(n+1)^2) less than the actual answer, a very small amount. This might be explained by the fact that we're dealing with estimations, and your estimation is only very slightly different from ours. Obviously, you haven't shown the answer you were supposed to show, but you certainly deserve some credit.

Of course, my working might just be wrong. More eye-hurting non-jsMath working to follow...

-1/(2(n+1)^2)
=-(n^2/(n+1))/(2n^2(n+1)) Let the n^2 on top be (n^2-1) +1 and divide by the n+1 to get n-1 + 1/(n+1).
=-(n-1+1/(n+1))/(2n^2(n+1))
...which, as I say, is only a very small distance away.

Actually, when n is large as they stipulate, that deviation (the reciprocal of a fourth degree polynomial in n) very quickly becomes 0, meaning your estimate tends towards the one given. If you had in fact gone all the way through to say that, they would have had to have given you a lot of marks.

(edited 11 years ago)

Reply 191

Qwertyfoo

Paper Cap A* A B C D E
Jan 2006 72 65 59 51 43 36 29
Jun 2006 68 61 54 47 40 33 27
Jan 2007 70 62 54 46 39 32 25
Jun 2007 72 66 60 53 46 39 33
Jan 2008 61 55 49 43 37 31 25
Jun 2008 61 55 49 43 37 31 25
Jan 2009 59 54 49 44 39 34 30
Jun 2009 67 60 53 46 40 34 28
Jan 2010 67 60 53 46 39 32 25
Jun 2010 62 57 52 47 42 37 32
Jan 2011 72 64 56 48 40 32 25
Jun 2011 64 58 52 46 40 35 30
Jan 2012 63 56 49 42 35 28 21

(edited 11 years ago)

Reply 192

wibletg

3

Original post by Qwertyfoo

Paper Cap A* A B C D E
Jan 2006 72 65 59 51 43 36 29
Jun 2006 68 61 54 47 40 33 27
Jan 2007 70 62 54 46 39 32 25
Jun 2007 72 66 60 53 46 39 33
Jan 2008 61 55 49 43 37 31 25
Jun 2008 61 55 49 43 37 31 25
Jan 2009 59 54 49 44 39 34 30
Jun 2009 67 60 53 46 40 34 28
Jan 2010 67 60 53 46 39 32 25
Jun 2010 62 57 52 47 42 37 32
Jan 2011 72 64 56 48 40 32 25
Jun 2011 64 58 52 46 40 35 30
Jun 2011 64 58 52 46 40 35 30
Jan 2012 63 56 49 42 35 28 21

I'm going to go ahead and say this was slightly more difficult than Jun 11 and significantly more difficult than Jun 11.

The fact they didn't give you the quadratic and linear functions in the last question screwed over a lot of people I know who sat the exam - they won't have got any accuracy marks if they got the function wrong (even if they continued onto the last parts!).

That and the last parts of 7.

Reply 193

Qwertyfoo

mean cap = 66
mean A* = 59.461...
mean A = 53

(edited 11 years ago)

Reply 194

corpuscallosum

10

Anyone remember the question that wanted the answer in terms of 5^(1/3)?

Reply 195

Wahrheit

18

Original post by corpuscallosum

Anyone remember the question that wanted the answer in terms of 5^(1/3)?

That was last year?

Reply 196

corpuscallosum

10

Original post by Wahrheit

That was last year?

Haha ok, my bad, I kind of fell asleep doing a paper last night

Answers are here:

http://www.thestudentroom.co.uk/showthread.php?t=2031752

I'll do Core 3 later but I have an afterschool meeting first.

Reply 198

bubblegum3000

Original post by Qwertyfoo

Paper Cap A* A B C D E
Jan 2006 72 65 59 51 43 36 29
Jun 2006 68 61 54 47 40 33 27
Jan 2007 70 62 54 46 39 32 25
Jun 2007 72 66 60 53 46 39 33
Jan 2008 61 55 49 43 37 31 25
Jun 2008 61 55 49 43 37 31 25
Jan 2009 59 54 49 44 39 34 30
Jun 2009 67 60 53 46 40 34 28
Jan 2010 67 60 53 46 39 32 25
Jun 2010 62 57 52 47 42 37 32
Jan 2011 72 64 56 48 40 32 25
Jun 2011 64 58 52 46 40 35 30
Jan 2012 63 56 49 42 35 28 21

Not sure if those are actually right... at least the Jan 2012 is wrong:

A B C D E
49 42 35 28 21

I think that this one will be pretty similar to the Jan 2012... it was pretty hard!

Reply 199

Mr M (jr)

1

Ok, here's my mark scheme folks for what it's worth, I thought this was definitely harder than average).

Q1.

Unparseable latex formula:

tan^-^1e^2^x + c

(5 marks)

Q2.
(i)

\theta=\frac{\pi}{2}

or

\theta=0

(2 marks)
(ii)

\frac{4}{9}\sqrt3

(0.770 is ok)(4 marks)
(iii)

(x^2+y^2)^2=2x^2y

(3 marks)

Q3.
(i) prove... (2 marks)
(ii)

x=\frac{1}{2}ln2

or

\frac{1}{2}ln3

(6 marks)

Q4.
(i) staircase diagram converging up to

\alpha

; x₂=1.3869, x₃=1.3939 (3 marks)
(ii) steps diverging away (down and left) from

\alpha

(2 marks)
(iii) 1.3953 (4 marks)

Q5.
(i) x=1; show... (5 marks)
(ii) graph reflected in y-axis then x-axis (whole function is odd);

f(x)\geq 2ln(1+\sqrt2)

or

f(x)\leq -2ln(1+\sqrt2)

(3 marks)

Q6.
(i) Prove... (5 marks)
(ii)

\pi^5 - 20\pi^3 + 120\pi

(4 marks)

Q7.
(i) a=2, c=1, b=n, d=n-1 (3 marks)
(ii)

\frac{1}{n} < f(n) < 1

(4 marks)
(iii) Show... (5 marks)

Q8.
(i)

\frac{\frac{1}{2}x^2 + 2x + 10}{x + 2}

(4 marks)
(ii)

y \geq 4

or

y \leq -4

(4 marks)
(iii) (2,2) (4 marks)

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

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