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    (Original post by n_251)
    lol I thought it was harder than most of the past papers; probably as hard as June 09

    So, going by that 80 UMS would be in the low to mid-50s but you never know!
    It was 58 in June, that is not low or mid 50s, thats high 50's
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    (Original post by mohamed4iraq)
    i got those answers
    1)a)1.386,2.290
    b)7.368
    c)i)((x*2lnx)/2)-0.25x*2+c
    5)y*2+(x*2 +lnx -4)*3 or sth like that but to the power of 3
    6) 0.299
    7)a -4 & 41
    b) the area was sth like 624
    the paper is posted in page 3
    seems quite right, got the same answers, (dont remember question 2) at all though :P )
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    (Original post by Megiddo)
    so did you multiply your answer in part a by 4pi for part b?i think i might have missed something cos the answer in part a i got was root3 - 1 i think
    well you times it by 16Pi , but since the answer to part a) was already 1/4 of what you got, 16 times 1/4 gives 4, hope that makes things clear :P
    it was (1/4)[(root3)-1] for part a), times that by 16Pi and see what you get :P
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    for the vectors question AY=27.9 (my answer)
    i did get no. 5 wrong. i integrated 9 into 9x, but differentiated 6/X into 3/X^2. so im gonna lose one mark in 5a. and probably 3 more in 5 b. and i forgot to convert 32ln4 to 64lin2 in 2.c.ii. so one more mark lost.
    i should get 69-70-ish, because im pretty sure i didnt get anything else wrong. most of the other candidates lost a minimum of 10-15 marks, and many couldnt even do no. 10. or the vectors question.
    A grade should definitely be below 60. let's just hope its around 55, then i have a shot at scoring 100
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    here are the solutions and paper
    Attached Images
  1. File Type: pdf C4 Jan 2010 Solutions.pdf (509.0 KB, 1110 views)
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    MY ANSWERS [/U]

    2)a)1.386,2.290
    b)7.368
    c)i)((x*2lnx)/2)-0.25x*2+c

    3)a) -2sin2x - 3sin3y = 0
    b)y= 1/9 pi

    4)a) A= (-6,4,-1)
    b)19/26
    c)x = (10,0,11)
    d)AX = (16,-4,12)
    e)4(26)^0.5 as given

    5)a)9x +6lnx + C
    b)y^2=(6x+4lnx -2 )^3

    6) 0.299

    7)a -4 & 41
    b) the area was sth like 624

    8)a)0.25(3^0.5 - 1 )
    b)4pi (3^0.5 - 1 )
    the paper is posted in page 3

    some answers r updated
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    btw Arsey in ur solution the last q abt volume is wrong
    bcz volume = pi * interal of y*2
    not 2pi
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    ^Just what I was abut to say. Just half your value you got in the end and that is correct.
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    (Original post by ojpowermake)
    Q2, c), (ii) did anyone get 64ln2 instead of the 4ln2?
    yes, I remember a 64ln2 in there somewhere.
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    I made quite a few stupid mistakes
    Looks like the A* is out of the window
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    (Original post by Arsey)
    here are the solutions and paper
    Much obliged sir i'll rep you again, but like someone else pointed out the integral is pi y^2, not 2pi y^2, other than that I agree with all of them

    Also I just changed swapped around the limits to get rid of the minus sign, but that's preference.
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    (Original post by Moa)
    It was 58 in June, that is not low or mid 50s, thats high 50's
    Oh was it? :confused:

    My mistake; then I reckon we'll be looking at similar boundaries this time too!
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    I made loads of stupid mistaks; will method marks be substantial or not? E.g. for question 6 or whatever where it asks you to find dr/dt, I used the surface area of a sphere formula, 4pi x r^2. Will I lose all the marks?
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    [borat] Great Success ! [/borat]

    If I got less than 68% in M1 (which I almost certainly didn't), I would have needed 67.5% in this paper, I wasn't sure about my 0.299 answer to question 6, but I think, with that I've been pushed up to around 80% in this paper!

    Fell down on question 7 though, where I could only get the x values to be 41 and... 41 not the -4, any help?

    EDIT: OH I've figured it out, two values of t, one being +/- 3 which gives 41, and one being 0, which gives -4. I should have got that, missed out on 6 marks that question.
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    (Original post by hussam1992)
    well you times it by 16Pi , but since the answer to part a) was already 1/4 of what you got, 16 times 1/4 gives 4, hope that makes things clear :P
    it was (1/4)[(root3)-1] for part a), times that by 16Pi and see what you get :P
    oh sweet i got the answer by luck haha, probably be marked wrong i guess cos i missed out the /4 from part a.
    hopefully the marking person doesn't mark too harshly, for other questions i would've got the answers if i had got the right values, but the answer i gave to the values i got, using the methods were right.
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    (Original post by 12ian34)
    [borat] Great Success ! [/borat]

    Fell down on question 7 though, where I could only get the x values to be 41 and... 41 not the -4, any help?

    EDIT: OH I've figured it out, two values of t, one being +/- 3 which gives 41, and one being 0, which gives -4. I should have got that, missed out on 6 marks that question.
    i actually wrote down the t values as 0,-3 and 3, and got -4, and the other was was -41 (with -3) when i typed in my calculator, and i couldn't understand why so i got really confused and just put 3! fml wasn't exaclty obvious when your brain is at 50%
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    (Original post by psja1322)
    I made loads of stupid mistaks; will method marks be substantial or not? E.g. for question 6 or whatever where it asks you to find dr/dt, I used the surface area of a sphere formula, 4pi x r^2. Will I lose all the marks?
    i think you should get method marks. for some of the questions, answers would vary depeding on the value you used. it wouldn't be fair to penalise someone too much for having found an incorrect value in the workings, and then using that to work out an answer.
    just my thoughts.
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    for the differential equation involving the circle did you have to give it to a certain accuracy? I worked it out exactly then also wrote down 0.3. Some of my friends say that you had to give it to 3sf's whereas others swear that it didn't specify the accuracy.

    In the last 3 marker question on the volume, i multiplied by 4pi instead of 16pi (was in a real rush!), how many marks would I expect to lose?
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    (Original post by Megiddo)
    i actually wrote down the t values as 0,-3 and 3, and got -4, and the other was was -41 (with -3) when i typed in my calculator, and i couldn't understand why so i got really confused and just put 3! fml wasn't exaclty obvious when your brain is at 50%
    Ahh unlucky! You figured more of it out in the exam than I did!

    Calculators are stupid like that, if you put -3^2 in like that, they'll do the old BIDMAS malarky on you (Brackets, Indices, Division, Multiplic... and so on) and square the 3, then (-) the answer to get -9.

    You have to put (-3)^2, only then will it come up with the answer you were looking for= 9!

    So the limits were -4 and +41, with -4 being obtained by using t=0 and +41 being obtained by using either t=3 or t=-3, as t^2 gives the same answer in either case.
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    (Original post by Megiddo)
    i think you should get method marks. for some of the questions, answers would vary depeding on the value you used. it wouldn't be fair to penalise someone too much for having found an incorrect value in the workings, and then using that to work out an answer.
    just my thoughts.
    thanks megiddo. Did that on loads of questions. i should have got above 90 percent but i wasnt in maths mode at the time and i lost around20 marks. Think I just missed out on the A*.
 
 
 
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