Turn on thread page Beta
    Offline

    2
    ReputationRep:
    Didn't 1,0,0 work for part i of Q5?
    Offline

    1
    ReputationRep:
    (Original post by RichE)
    Powers don't work that way.
    You're right...another stupid mistake...there goes mid 80's to mid 70s...
    Offline

    4
    ReputationRep:
    (Original post by RuairiMorrissey)
    Didn't 1,0,0 work for part i of Q5?
    Someone else said this was what you got if you worked out one of the terms (2, 8, 24 etc.)* instead of the sum (2, 10, 34 etc.).

    *or something like that - not sure any more what the sequence actually was
    Offline

    2
    ReputationRep:
    (Original post by ShatnersBassoon)
    Someone else said this was what you got if you worked out one of the terms (2, 8, 24 etc.)* instead of the sum (2, 10, 34 etc.).

    *or something like that - not sure any more what the sequence actually was
    I thought that was part V?
    Offline

    2
    ReputationRep:
    (Original post by ShatnersBassoon)
    I haven't seen much discussion about question 5 yet, other than the first two parts:

    (i) and (ii) A = C = 2; B = -2.
    (iii) This was induction, or at least, induction without the wordy conclusion. Basically you found s_{k+1} = s_k + [the (k+1)th term] and showed it was equal to f(k+1).

    (iv) I split t_k up into two series: the first, the coefficients of n, was a GP - I think with powers of 2? - so it was easy to find a formula for that. The second was just s_k so sub in f(k), add the two together and you've got t_k.
    u_k was the sum of fractions in the form n/(2^n), so you put them over a common denominator (2^n), with 2^n + 2^(n-1) + ... 1 as the numerator, use the GP formula to find an expression for the numerator and then simplify the fraction.

    (v) This was meta. The sum of sums. I split it into the sum of three terms from 1 to n: the first GP I found in t_k; another weird GP; -2 [which summed to -2n]. My final answer was (something \times n - somethingelse)^{k-1} + 4.

    But I felt like I was doing part (v) in particular in at best an inefficient, and at worst a completely wrong, way. What did everyone else do?
    What do you think the first two parts of this will be in terms of the marks?
    Offline

    4
    ReputationRep:
    (Original post by RuairiMorrissey)
    I thought that was part V?
    Part (v) was the sum of the sums; that is:
    \sum_{k=1}^{n} s_k = 2 + (2+8) + (2+8+24) + ... + s_n
    Unless this is the third question I've read wrong.

    (Original post by louisforrest)
    What do you think the first two parts of this will be in terms of the marks?
    Based on the time I spent on each part, I would guess 4 marks. But if I was doing things inefficiently, as I suspect I was, then it could have been worth a little bit more. I think the main point of the question was to get onto the induction and the "recognise that this sequence relates to the one we've just shown you"-type stuff, so it was really just an introduction. Sorry if that's bad news for you - I suspect a lot of people will have just stopped after part (ii).
    Offline

    2
    ReputationRep:
    did anyone else find a<1 for the one about area of the graphs enclosed by x axis. I dont see how it is (6/5)^4
    Offline

    2
    ReputationRep:
    (Original post by ShatnersBassoon)
    Part (v) was the sum of the sums; that is:
    \sum_{k=1}^{n} s_k = 2 + (2+8) + (2+8+24) + ... + s_n
    Unless this is the third question I've read wrong.
    I've read loads wrong I wouldn't worry, we'll have to wait until the paper is out.
    Offline

    1
    ReputationRep:
    anyone got any ideas about the weightings of the marks for each part of question 3? (or any of the long questions, for that matter)
    Offline

    2
    ReputationRep:
    (Original post by ShatnersBassoon)
    Part (v) was the sum of the sums; that is:
    \sum_{k=1}^{n} s_k = 2 + (2+8) + (2+8+24) + ... + s_n
    Unless this is the third question I've read wrong.

    Based on the time I spent on each part, I would guess 4 marks. But if I was doing things inefficiently, as I suspect I was, then it could have been worth a little bit more. I think the main point of the question was to get onto the induction and the "recognise that this sequence relates to the one we've just shown you"-type stuff, so it was really just an introduction. Sorry if that's bad news for you - I suspect a lot of people will have just stopped after part (ii).
    Nevermind you were right.

    Please end me, i've miss-read SO many : (
    Offline

    9
    ReputationRep:
    (Original post by peted0herty)
    anyone got any ideas about the weightings of the marks for each part of question 3? (or any of the long questions, for that matter)
    I suspect that the last parts of q3 will be 2-3 marks each but I can't remember exactly what the question structure was
    Offline

    2
    ReputationRep:
    **** already down to 5 in the multiple choice
    Offline

    2
    ReputationRep:
    (Original post by boombox111)
    for what values of a is the area between y = a^2 - x^2 and the x-axis greater than the area between y = x^4 - a and the x-axis?
    something like this.
    Tbh i got that one wrong. But i worked it out after. The x^4 graph has x-intercepts wider than the x^2 for x>1. Therefore the x^4 has a greater area for a>1. However, for x<1 the x^4 graph has a smaller gap between the x-intercepts and so has a smaller area.
    Offline

    2
    ReputationRep:
    (Original post by louisforrest)
    did anyone else find a<1 for the one about area of the graphs enclosed by x axis. I dont see how it is (6/5)^4
    I think it was one of the a<1 or a>1. I cant remember the details. I got it wrong but in hindsight that was the right answer i think.
    Offline

    2
    ReputationRep:
    (Original post by m0.4444)
    I think it was one of the a<1 or a>1. I cant remember the details. I got it wrong but in hindsight that was the right answer i think.
    No I've realised I'm wrong its (6/5)^4
    Offline

    13
    ReputationRep:
    Predicted grade boundaries anyone?
    Offline

    2
    ReputationRep:
    (Original post by louisforrest)
    No I've realised I'm wrong its (6/5)^4
    Why? I know i got it wrong i put all values of a.
    Offline

    4
    ReputationRep:
    (Original post by peted0herty)
    anyone got any ideas about the weightings of the marks for each part of question 3? (or any of the long questions, for that matter)
    There were 7 parts to this question, right? I don't think I've ever seen that many before in a MAT question. I'd expect the marks to have been fairly homogeneous - i.e. 2 marks for each of them, with a third mark for whichever part they thought was trickiest.

    It was basically all "show that" and "explain that", from what I can remember, so I don't think there's much to talk about, but:

    Can't remember the first parts.

    For \int_a^b x^n dx = -\int_b^a x^n dx, I think you just needed to evaluate them and show they were both equal to \frac{x^{b+1}-x^{a+1}}{n+1}.

    The subsequent "show that" seemed to me more like an "explain that", but I said
    p(x) = a_0 + a_1x + a_2x^2 + ... + a_kx^k
    Then I set its indefinite integral, a_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + ... + \frac{a_kx^k}{k} equal to a new function, F(x).
    Then the LHS of their equation was F(b) - F(a), and the RHS was -(F(a) - F(b)) = F(b) - F(a).

    Then we had an "explain that" with 2α-t and so on as the limits of the integral. This was because the function was bilateral; I don't think I explained it very well but basically the value of f(t) at any point from t to α corresponded 1-to-1 to a point from 2α-t to 2α, so the symmetry meant the areas were equal.

    G(x) - there was some kind of show that I can't quite recall requiring the \int_a^b x^n dx = -\int_b^a x^n dx rule from earlier.

    Finally, you had to show G(x) = 0 given that it was bilateral. I equated some stuff to find G(x) = -G(x) ∀ x, so G(x) = 0.
    Offline

    2
    ReputationRep:
    I'm not sure how you'd work it out but that's what I put in the calculator and that's when they're both equal

    Posted from TSR Mobile
    Attached Images
      
    Offline

    3
    ReputationRep:
    Looking at this I'm like How the hell does everyone remember the questions and answers? I literally dont rmb a thing😂
 
 
 
Turn on thread page Beta
Updated: October 30, 2017
Poll
Do you think parents should charge rent?

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.