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Mr M's OCR (not OCR MEI) Core 2 Answers May 2013

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For the binomial expansion second part C=-11 I got it as 24 realised what I did wrong but would I lose all three marks or just 1 or 2??


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Original post by missypink<3
Hi Mr M,
Do you have a rough idea how many UMS I would get for 68/72?


I'd say around 94.
Reply 222
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HayyanN
image.jpgDon't ask me why, but for some stupid reason this is what I put for 7 ii. Is it worth any marks ? :/
thanks
Reply 223
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Zakee
Original post by eggfriedrice
I'd say around 94.



Are you predicting that 72/72 is 100 UMS for this paper then?
Original post by Zakee
Are you predicting that 72/72 is 100 UMS for this paper then?


Well put it this way, if we predict the worse case scenario (72 for 100 UMS), then we know what we have AT LEAST got, so from there it can stay the same or get better. :wink: I mean it's not like it'll be far off anyways.
Reply 225
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Namige
17
Original post by Zakee
Are you predicting that 72/72 is 100 UMS for this paper then?
What is it usually for C2?
This seemed like a particularly easy paper so probably 72/72 for 100 UMS.
But for some context here are the minimum 100% marks from the last few years (i.e. 2*(AB boundary) + A boundary)

Jan 2013 68/72
Jun 2012 69/72
Jan 2012 72/72 (maxed out)
Jun 2011 72/72 (maxed out)
Jan 2011 72/72 (maxed out)
Jun 2010 69/72

Mean = 70.3

IMO this is probably 72/72 for 100
Reply 227
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Zakee
Original post by Namige
What is it usually for C2?



Generally, 71/72 is 100 UMS for Core 2 (the average over the past 4 years).
Reply 228
Avatar for Zakee
Zakee
Original post by dakenSoren
This seemed like a particularly easy paper so probably 72/72 for 100 UMS.
But for some context here are the minimum 100% marks from the last few years (i.e. 2*(AB boundary) + A boundary)

Jan 2013 68/72
Jun 2012 69/72
Jan 2012 72/72 (maxed out)
Jun 2011 72/72 (maxed out)
Jan 2011 72/72 (maxed out)
Jun 2010 69/72

Mean = 70.3

IMO this is probably 72/72 for 100



When it is maxed out, what would dropping one mark below 100 UMS mean? 99 UMS?
Reply 229
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aran0909
Hi, how do you do Q5 i and ii?
Thanks
Original post by Zakee
When it is maxed out, what would dropping one mark below 100 UMS mean? 99 UMS?


I'm not sure and wondering the same thing,
does anyone else know?
Original post by Mr M
Mr M's OCR (not OCR MEI) Core 2 Answers May 2013


1. 6.39 (4 marks)


2. (i) 254 and 106 degrees (3 marks)

(ii) 71.6 and 252 degrees (3 marks)


3. (i) 64+960x+6000x264+960x+6000x^2 (4 marks)

(ii) c = - 11 (3 marks)


4. (a) 54x4−3x2+x+k\frac{5}{4}x^4 -3x^2+x+k (3 marks)

(b) (i) −12x−2+k-12 x^{-2} + k (2 marks)

(ii) a = 2 (3 marks)


5. (i) 62.2 (4 marks)

(ii) 34.0 (4 marks)


6. (i) 963 (3 marks)

(ii) 17 (6 marks)


7. (i) Show ... (4 marks)

(ii) 3730\frac{37}{30} (5 marks)


8. (i) (a) (0, 1) (1 mark)

b) (0, 4) (1 mark)

c) Any value where a > 1 and any value where 0 < b < 1 (2 marks)

(ii) Show ... (5 marks)


9. (i) 15 (2 marks)

(ii) Show f(-0.5)=0 and f(x)=(2x+1)(2x−3)(x+1)f(x)=(2x+1)(2x-3)(x+1) (6 marks)

(iii) θ=2π3\theta = \frac{2\pi}{3} and θ=4π3\theta =\frac{4\pi}{3} and θ=π\theta=\pi (4 marks)


I'll be around a bit this afternoon and then part of this evening to answer questions. Please post in this thread rather than sending me private messages.


for question 4, if i had +c on the ends would that still be okay
and for the trapezium rule one (1) if i used the values of x=5,7,11 (forgot 9) and done everything else right and came out with 4.56, how many would i drop ?
Does anyone have the question paper?
Reply 233
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Shuster
Hi,
For the question with the tangent, i worked out the gradient of the tangent to be 4, instead of 3. I then proceeded correctly, found the equation of the tangent ect.. How many method marks do you think I will get? Thanks.
Reply 234
Avatar for Shuster
Shuster
For the very last question, iii) i didn't have time to finish so i only wrote down 2 of the possible 3 values. Will i only lose one mark?
Original post by benitashrestha
Hey Mr M,
you know for the logs question I did.
a^x=4b^x
xloga=2 +xlogb
xloga-xlogb=2
x(loga/b) = 2

then I got lost.
how many marks do you think I will gain out of 5?


2 I think
Original post by studentro
How many marks do you think I would lose for the following mistakes:
1. h = (11 - 5) / 2 = 3, Area = 3/2 * (8/5 + 8/11 + 2 ( 8/8 ) ) = 6.49 - Did 3 ordinates instead of strips
6.i. a = 6, d = 1.8, S24 = 1/2 * 24 * ( 2(6) + 1.8 ( 24 - 1 ) ) = 640.8 - Did sum of first 24 instead of 30
8.c. a = 2, b = -2 - Did a negative value of b instead of 0 < b < 1
9.i. Remainder = 15/(x-2) - Divided remainder by the quotient rather than leaving it as the answer obtained using the factor theorem

Thanks in advance


1) drop 1 mark

6i) drop 1 or 2 marks

8c) drop 1 mark

9i) This is fine
Original post by smithstefan1995
For the binomial expansion second part C=-11 I got it as 24 realised what I did wrong but would I lose all three marks or just 1 or 2??


This was posted from The Student Room's iPhone/iPad App


Probably 2 marks
Original post by HayyanN
image.jpgDon't ask me why, but for some stupid reason this is what I put for 7 ii. Is it worth any marks ? :/
thanks


I don't think so
Original post by hayhayhay1
for question 4, if i had +c on the ends would that still be okay
and for the trapezium rule one (1) if i used the values of x=5,7,11 (forgot 9) and done everything else right and came out with 4.56, how many would i drop ?


No problem with + c

drop 2 marks for trapezium rule

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