Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    14
    ReputationRep:
    General thoughts on the difficulty? I thought it was fairly average, no obvious gift questions but quite a few questions that it was easyish to get started with.
    Offline

    14
    ReputationRep:
    I think I've got two ~fulls (Q3 and Q12), significant chunks of Q1, Q7 (I think that was the integration one?), and the differential equation question, and 2-3 marks on Q5. On the integration, I did everything up to getting an integral of 1/cos x + sin x + 1 in the last part, so hopefully that will be 15ish, and on d.e. I did everything up to the last part. Hopefully that'll be a 1.
    Offline

    0
    ReputationRep:
    (Original post by sweeneyrod)
    I got to your last line, but couldn't get any further.
    before you use the double angle formulae you can just divide through by (cos)^2 and then use sub u = tanx i think
    Offline

    18
    ReputationRep:
    (Original post by sweeneyrod)
    General thoughts on the difficulty? I thought it was fairly average, no obvious gift questions but quite a few questions that it was easyish to get started with.
    Slightly easier then last year. I guess
    95/70 for S/1.
    I was horrified tbh, no vectors and geometry set me back massively. ****inf ********.


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Q1 and 3 were trivial. Q6 was near impossible. I've heard of only 1 person managing the last part of Q6 out of 30 or so of the more able people. Any guesses for the mark distribution of Q6?

    Edit to add: you've said it was on the easier side and then given lower boundaries for a 1 than 2010-2014?
    Sorry, what was q3 again? Was that the f_n(x) one and proving that there are one or no solutions.

    Also, which one was q6? I thought q6 was the DE one but I may be completely wrong.

    On another note, how did people find the question where for the last part you had to find tan(theta) and x given y =(3+sqrt(7))/2, or something like that.
    Offline

    7
    ReputationRep:
    Q2 STEP II Solution
    Attached Images
      
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by tridianprime)
    Sorry, what was q3 again? Was that the f_n(x) one and proving that there are one or no solutions.

    Also, which one was q6? I thought q6 was the DE one but I may be completely wrong.

    On another note, how did people find the question where for the last part you had to find tan(theta) and x given y =(3+sqrt(7))/2, or something like that.
    Yep, that was Q3. And Q6 was the DE, the last part of Q6 was the bit about y_(mn).
    Offline

    14
    ReputationRep:
    (Original post by physicsmaths)
    Slightly easier then last year. I guess
    95/70 for S/1.
    I was horrified tbh, no vectors and geometry set me back massively. ****inf ********.


    Posted from TSR Mobile
    I was actually sad about no vectors, because I did some vectors questions the day before and felt super prepared.
    Offline

    1
    ReputationRep:
    For the last integration I found a neat trick, split 1/cosx(cosx+sinx) into (sinx-cosx)/cosx + 2cosx/(sinx+cosx)
    Offline

    7
    ReputationRep:
    (Original post by tridianprime)
    Sorry, what was q3 again? Was that the f_n(x) one and proving that there are one or no solutions.

    Also, which one was q6? I thought q6 was the DE one but I may be completely wrong.

    On another note, how did people find the question where for the last part you had to find tan(theta) and x given y =(3+sqrt(7))/2, or something like that.
    i liked it, what was your answer?
    Offline

    1
    ReputationRep:
    (Original post by riquix)
    Attachment 551619Attachment 551621Attachment 551623Attachment 551625Q9 from STEP 2. If someone could verify this I'd be grateful. About 90% certain I got it right.


    By a completely different method, I got the same answers for a and c, but [my b] is [your b] + [our c]
    I think you put b as the distance the bullet moves relative to the block, and I put it as the absolute distance the bullet moves.
    I remember being unsure which it wanted, because the question said b was the distance the bullet traveled before coming to rest relative to the block.
    What do people think?
    Edit: Also, did it want b and c in terms of a? Arrrgh

    And side note, anybody have a question paper?
    Offline

    2
    ReputationRep:
    Name:  IMG_1982.JPG
Views: 491
Size:  108.5 KBName:  IMG_1983.JPG
Views: 468
Size:  115.9 KBName:  IMG_1984.JPG
Views: 435
Size:  118.3 KBName:  IMG_1985.JPG
Views: 432
Size:  110.9 KBName:  IMG_1986.JPG
Views: 460
Size:  98.5 KBQ1 Solution
    Offline

    18
    ReputationRep:
    Q3 was beautiful.


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by farryharnworth)
    By a completely different method, I got the same answers for a and c, but [my b] is [your b] + [our c]
    I think you put b as the distance the bullet moves relative to the block, and I put it as the absolute distance the bullet moves.
    I remember being unsure which it wanted, because the question said b was the distance the bullet traveled before coming to rest relative to the block.
    What do people think?

    And side note, anybody have a question paper?
    Wasn't b 'the distance the bullet moves in the block'? In that case, it would be [your b] - [our c]?
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Yep, that was Q3. And Q6 was the DE, the last part of Q6 was the bit about y_(mn).
    I am a bit confused - was q6 really that difficult. I have possibly got myself confused but I thought that you just repeated the previous method and showed that the expression you obtain for applying (*) to y_n(Y_m(x)) = m^2n^2, which can be done by subbing in the dy/dx values you can obtain from (*) when applied using just y_n(y_m)) and y_m.
    Offline

    1
    ReputationRep:
    (Original post by riquix)
    Wasn't b 'the distance the bullet moves in the block'? In that case, it would be [your b] - [our c]?
    Honestly I'm not sure. I think the wording was unclear, but maybe it would turn out that I was just panicking a bit at the time if I read it again.
    Regardless, there must be some credit for either answer.
    Offline

    2
    ReputationRep:
    (Original post by gasfxekl)
    i liked it, what was your answer?
    Let's just say I didn't like it. I did a foolish thing in the first part where to show that y^2 + 1 >= I just said we can choose theta to be 0 and hence y^2 >=. I realised this was completely stupid when I then had to find tan(theta).
    I always thought deduce meant to use the previous part but I now think that is hence.

    How did you go about deducing the y^2 + 1 >= part?
    Offline

    1
    ReputationRep:
    (Original post by computerkid)
    For the last integration I found a neat trick, split 1/cosx(cosx+sinx) into (sinx-cosx)/cosx + 2cosx/(sinx+cosx)
    Yeh, that's what I did, or you could split it to sin(x)/cos(x) + (cos(x)-sin(x))/(cos(x)+sin(x))= tan(x) +tan(pi/4-x).
    Offline

    2
    ReputationRep:
    (Original post by farryharnworth)
    Honestly I'm not sure. I think the wording was unclear, but maybe it would turn out that I was just panicking a bit at the time if I read it again.
    Regardless, there must be some credit for either answer.
    Hmmm. Also, I didn't see that at the end of part (ii) it asked for b and c to be in terms of a, m, M and u (i.e. that you had to eliminate R). I left both b and c in terms of m, M, u and R. How many marks do you think I'd lose for this?
    Offline

    7
    ReputationRep:
    (Original post by tridianprime)
    Let's just say I didn't like it. I did a foolish thing in the first part where to show that y^2 + 1 >= I just said we can choose theta to be 0 and hence y^2 >=. I realised this was completely stupid when I then had to find tan(theta).
    I always thought deduce meant to use the previous part but I now think that is hence.

    How did you go about deducing the y^2 + 1 >= part?
    need to resee the question, but when you rearranged it it was equivalent to a square being greater or equal to zero - (ycostheta+sintheta)^2 or something.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.