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    (Original post by souktik)
    Ah, okay. Thanks for the response.
    Even if that's not completely accurate, a relatively weak college at Oxford is still a relatively weak college at Oxford!
    Very true And if I remember well, St John's is a pretty academic college anyway, so I really wouldn't worry about that if I were you
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    (Original post by jadoreétudier)
    Does anyone want to do the 1999 and or 2002 papers?

    I posted yesterday but fear my post has been lost in the conversation.

    Okay, I'll take a look at the 2002 paper now.
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    (Original post by souktik)
    Hey, how is St John's for math? I didn't research much before applying, to be honest.
    like what bluebell said i heard st johns and merton are regarded as pretty top academically (and st johns is richerst!)

    (Original post by jadoreétudier)
    Does anyone want to do the 1999 and or 2002 papers?

    I posted yesterday but fear my post has been lost in the conversation.

    i did the 1999 paper a few days ago but the paper ive done it on has now been binned! im going to do 2002 paper today though, so ill be sure to compare answers with you later in the evening!
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    (Original post by IceKidd)
    like what bluebell said i heard st johns and merton are regarded as pretty top academically (and st johns is richerst!)

    i did the 1999 paper a few days ago but the paper ive done it on has now been binned! im going to do 2002 paper today though, so ill be sure to compare answers with you later in the evening!
    Okay, thanks, and please tell me if your Part 1 answers are the same as mine. I'm not absolutely confident about a couple, but I hardly feel like checking carefully.

    Spoiler:
    Show

    A. a
    B. b
    C. c
    D. b
    E. d
    F. b
    G. a
    H. d
    I. c
    J. a
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    (Original post by yl95)
    Merton would be one of them!

    Also, Balliol, Magdalen and New but I'm not sure how much that holds true. Noble would know better than me!

    Posted from TSR Mobile
    There's no such thing as an academic powerhouse really. Merton may have had that reputation, but from speaking to two people I've known studying maths at Merton they're more laid back than my college (New) for maths. I think I remember also the user "anyone_can_fly" on here, who studies maths at Merton, saying that last year no-one got more than 40% in collections and it wasn't much of an issue. If you get less than 40% in collections at my college the main tutor will put you on an official warning :lol:

    EDIT: I picked out Merton because that's really the only college people think is the most demanding academically, and it isn't really true at all.
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    Does anyone else feel that the "-" between c and 1/c in the first line of problem 3 actually be a "+"?

    Anyway, here are some of the answers - I'm not writing any solutions or proofs here - that I got:
    Spoiler:
    Show

    2.(i) A=2b-a2; B=b2
    2.(ii) Either use b=4, a=2.root(7) or b=-4, a=2.root(3)

    3.(iii) 32/9

    4.(iii) The four equations are x+y=1, -x+y=1, -x-y=1, x-y=1 in the 1st, 2nd, 3rd and 4th quadrants.
    4.(iv) Maximum is 1, attained at (1,0), (-1,0), (0,1), (0,-1) and minimum is 1/root(2), attained at (0.5,0.5), (-0.5,0.5), (-0.5,-0.5), (0.5,-0.5).

    5.(i) (a) 90; (b) 0; (c) 22680
    5.(ii) (a) YES; (b) NO; (c) YES; (d) NO
    5.(iii) (a) NO; (b) YES; (c) NO; (d) YES
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    Can someone help me with 2010 Q3 part i? I don't understand the bit about multiplying through the inequality

     sin x < x < tan x
    by
     cos(x)

    Shouldn't that give you
     sin x cos x < x cos x < sin x ?

    Mark scheme attached

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    (Original post by shayiqbal)
    Can someone help me with 2010 Q3 part i? I don't understand the bit about multiplying through the inequality

     sin x < x < tan x
    by
     cos(x)

    Shouldn't that give you
     sin x cos x < x cos x < sin x ?
    You only need to multiply the second part of the inequality,  x < tan x by  cos(x).
    As you have noted yourself, that gives you the result  x cos x < sin x .
    Combine this with the first part of the original inequality,  sin x < x to get the result  x cos x < sin x < x .
    Hope this helps. I'll be happy to clarify any ambiguities in my explanation.
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    (Original post by souktik)
    You only need to multiply the second part of the inequality,  x < tan x by  cos(x).
    As you have noted yourself, that gives you the result  x cos x < sin x .
    Combine this with the first part of the original inequality,  sin x < x to get the result  x cos x < sin x < x .
    Hope this helps. I'll be happy to clarify any ambiguities in my explanation.
    Thank you for the reply.

    I see now you only multiply the second part of the inequality, which makes sense, but that would give you (if I understand this right)  sinx < xcosx < sinx which makes no sense? How did you rearrange the terms to get the required result?

    Again, thanks for your time!
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    (Original post by shayiqbal)
    Thank you for the reply.

    I see now you only multiply the second part of the inequality, which makes sense, but that would give you (if I understand this right)  sinx < xcosx < sinx which makes no sense? How did you rearrange the terms to get the required result?

    Again, thanks for your time!
    No problem, happy to help. See,  sinx < x < tanx actually stands for two separate inequalities which have been combined:
     sinx < x (1) and  x < tanx (2). You take just the second one, multiply it by  cosx . That gives you the inequality  xcosx < sinx (3). Now combine (1) and (3)
    to get the final result:
     xcosx < sinx < x .
    Is this okay?
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    (Original post by souktik)
    No problem, happy to help. See,  sinx < x < tanx actually stands for two separate inequalities which have been combined:
     sinx < x (1) and  x < tanx (2). You take just the second one, multiply it by  cosx . That gives you the inequality  xcosx < sinx (3). Now combine (1) and (3)
    to get the final result:
     xcosx < sinx < x .
    Is this okay?
    Ah I see! Thank you ever so much
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    (Original post by shayiqbal)
    Ah I see! Thank you ever so much
    You're welcome.
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    guys when it says label a turning point. e.g 2006 q3

    do you have to write coordinates- its a messy surd which youd have to cube to get the y value which makes even more messy.
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    2012 q5
    Spoiler:
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    Name:  uploadfromtaptalk1383311982138.jpg
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    Could someone explain the underlined part?


    Sent from my GT-N7100 using Tapatalk 4
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    (Original post by seohyun)
    2012 q5
    Spoiler:
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    Name:  uploadfromtaptalk1383311982138.jpg
Views: 136
Size:  60.5 KB

    Could someone explain the underlined part?


    Sent from my GT-N7100 using Tapatalk 4
    Yeah, sure, I'll try.

    m0 is 3, right?

    That means:
    m_1=2.m_0=3.2; m_2=2.m_1=3.2^2; m_3=2.m_2=3.2^3... m_n=2.m_(n-1)=3.2^n

    [You start with 3 at m_0 and multiply by 2 to go from m_k to m_(k+1) for k=0,1,2...(n-1). So you multiply by 2 a total of n times to go from m_0 to m_n.]

    Is that okay?

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    (Original post by souktik)
    Yeah, sure, I'll try.

    m0 is 3, right?

    That means:
    m_1=2.m_0=3.2; m_2=2.m_1=3.2^2; m_3=2.m_2=3.2^3... m_n=2.m_(n-1)=3.2^n

    [You start with 3 at m_0 and multiply by 2 to go from m_k to m_(k+1) for k=0,1,2...(n-1). So you multiply by 2 a total of n times to go from m_0 to m_n.]

    Is that okay?

    Posted from TSR Mobile
    Thanks, it somewhat clearer now.

    Sent from my GT-N7100 using Tapatalk 4
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    (Original post by souktik)
    Okay, thanks, and please tell me if your Part 1 answers are the same as mine. I'm not absolutely confident about a couple, but I hardly feel like checking carefully.

    Spoiler:
    Show

    A. a
    B. b
    C. c
    D. b
    E. d
    F. b
    G. a
    H. d
    I. c
    J. a
    i got exact same for errything except for I. i put d) but i wernt too sure on that. i didnt get how to use the ii and iv bit.

    if ur pretty sure ur right, can u explain it to me please
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    (Original post by jadoreétudier)
    I've done 1999 and 2002 from this website:
    http://www.mathshelper.co.uk/oxb.htm
    Although there aren't solutions. Would anyone else like to do them too and then we can compare our answers?
    My answers for both papers:
    Spoiler:
    Show
    1999

    a i
    b iii
    c i couldn't get this one
    d iv
    e iii
    f i
    g ii
    h iii
    j i
    k couldn't get this one either! my answer was 4*49*48/(52*51*50) -= 0.07 which isn't an option

    2. a) (x+3)(x-2)
    b) a is greater than or equal to -1/4.
    x= - 1 plus or minus the square root of (1+4a) all over 2
    (just the quadratic formula)
    c) when x=1, the whole thing equals zero, no matter what b is.. you get 1 - b - 1 + b
    b=-1/4 which I've said is the only value for b

    3. I don't think I knew what I was doing in this question, especially c.

    a) y=-x+3
    b) subbing in x=1, gives you y=2, no matter the value of a. just like question 2
    y= 2 + 1/a - x/a
    c) x=1 is one line. I haven't been able to come up with a general formula for the equations of the lines, but I think y=(4-root3)x + root 3 -2, might be one of them

    4. Quite confused with this question.
    a) for both integrals I got 2/3
    b)
    c) I got zero, but it says the area should be measured positively and I don't think zero counts as positive. Maybe because I've done top curve minus the bottom curve in my integration I'm showing that I'm measuring it positively?
    d) I don't know! It's difficult to see graphically that the difference in areas above and below the x axis will be the same as it is for the quadratic.

    5. a) 16*15*14*13= 43680
    b)4^4 = 256
    c) 4*3*2*1=24

    2002

    a a
    b a
    c c
    d b
    e d
    f b
    g a
    h d
    i c
    j a

    Question 2 appeared in a later paper, so I have the answers for that. 3 didn't work out well for me. 4:Attachment 250205

    5. i) i haven't written down an answer, for a guess I'll go for 18
    ii) a yes b no c yes d no
    iii) a yes b yes c no d yes
    for no.2 obviously thats in another paper. for number 3 you havent done yet,

    for number 4 i agree with ur parts i,ii,iii. on the last part i agree wit ur coord wer the max value occurs and that its 1. but u also need to find the minimum value which i got to be (1/sqrt2) at ({+or-}0.5,{+or-}0.5)

    numberr 5 i dont get -.-
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    (Original post by IceKidd)
    i got exact same for errything except for I. i put d) but i wernt too sure on that. i didnt get how to use the ii and iv bit.

    if ur pretty sure ur right, can u explain it to me please
    Err, can you please remind me what the question was?

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    IceKidd, are you an international applicant?

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