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    And does anyone rmb what Q1 G for the summation one was about? The acc question

    and what Q1J-b acc was
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    Heres my thoughts for question 1H:

    1) y = a - x^2
    2) y = x^4 - a --> x-intercepts at 4throot(a) and -4throot(a)

    set y=0 and find x-intercepts:
    1) root(a), -root(a)
    2) 4throot(a), -4throot(a)

    integrate both between limits (x-intercepts):
    1) [ax - x^3/3] = (a^1.5 - (a^1.5)/3) - (-a^1.5 - (-a^1.5)/3) = 2a^1.5 - 2a^1.5/3 = 4/3 * a^1.5 = 4a.root(a)/3
    2) [x^5/5 - ax] = ((a^1.25)/5 - a^1.25) - ((-a^1.25)/5 - (-a^1.25)) = 2a^1.25/5 - 2a^1.25 = -8/5 * a^1.25 = -8a.fourthroot(a)/5

    set the two areas equal to each other:
    4/3 * a^1.5 = 8/5 * a^1.25
    a^0.25 = 6/5
    a = (6/5)^4
    only one answer to do with (6/5)^4
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    Well I've ****ed up 🙄
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    Looking at the answers, I reckon I got somewhere between 65 and 70.

    Should that be enough to get an offer from Imperial that's solid?
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    (Original post by 17lina)
    And does anyone rmb what Q1 G for the summation one was about? The acc question

    and what Q1J-b acc was
    The summation question I think you're talking about, to quote myself:

    (Original post by ShatnersBassoon)
    There was some kind of recurrence relation like
    a1 = 1; a(n+1) = Σak [from 1 to n]
    and the question was to find Σ(1/an) [from 1 to infinity]

    Sorry about the bad formatting; in words:

    There was a sequence where each term was the sum of every previous term in the sequence, starting with 1. So it goes 1, 1, 2, 4, 8, ... Find the sum of the reciprocals of this series. The sum was 1 + 1 + 1/2 + 1/4 + 1/8 + ..., which you could use the GP formula on, or 1+1 = 2 and the rest is a fairly famous infinite sequence which sums to 1. So the answer was 3.
    1J: Might have messed up some of the options in particular because I spent more time analysing some of them than others. But this was roughly it:

    Let ∏(n) denote the number of unique prime factors in n; for example, ∏(8) = 1 and ∏(6) = 2. Let x(n) equal the last digit of n. Which of the following statements is false?

    (a) If ∏(n) = 1, there are some values of x(n) for which n must not be prime.
    (b) If ∏(n) = 1, there are some values of x(n) for which n must be prime.
    (c) If ∏(n) = 1, there are some values of x(n) which are not possible to obtain.
    (d) If ∏(n) + x(n) = 2, it is not possible to determine whether n is prime.
    (e) If ∏(n) = 2, there are some values of x(n) which are not possible to obtain.

    [I think the TSR consensus is (b), which is what I put. Although "Which is false?" instead of "Which is true?" nearly caught me out.]
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    MCQ,H: The area for x^4-a is negative right? (Provided a>0) Or does that not matter somehow?
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    (Original post by steven7891)
    MCQ,H: The area for x^4-a is negative right? (Provided a>0) Or does that not matter somehow?
    If an integral yields a negative result when evaluated, I believe you take its absolute value to find the area under the curve.
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    (Original post by Superdjman1)
    Looking at the answers, I reckon I got somewhere between 65 and 70.

    Should that be enough to get an offer from Imperial that's solid?
    If you did get 65-70, that's a pretty impressive score, so I'd say yes.
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    I think I got around 50. This seems too low for Imperial but I have strong GCSE's and AS results. Would that help me get an offer?
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    (Original post by ShatnersBassoon)
    The summation question I think you're talking about, to quote myself:



    1J: Might have messed up some of the options in particular because I spent more time analysing some of them than others. But this was roughly it:

    Let ∏(n) denote the number of unique prime factors in n; for example, ∏(8) = 1 and ∏(6) = 2. Let x(n) equal the last digit of n. Which of the following statements is false?

    (a) If ∏(n) = 1, there are some values of x(n) for which n must not be prime.
    (b) If ∏(n) = 1, there are some values of x(n) for which n must be prime.
    (c) If ∏(n) = 1, there are some values of x(n) which are not possible to obtain.
    (d) If ∏(n) + x(n) = 2, it is not possible to determine whether n is prime.
    (e) If ∏(n) = 2, there are some values of x(n) which are not possible to obtain.

    [I think the TSR consensus is (b), which is what I put. Although "Which is false?" instead of "Which is true?" nearly caught me out.]
    hmm... (c) seems correct - if ∏(n) = 1, x(n) cannot be equal to zero. If the last digit of a number is zero, we can always factor out 10 and therefore 2 and 5, rendering ∏(n)>1.
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    (Original post by Insecure)
    hmm... (c) seems correct - if ∏(n) = 1, x(n) cannot be equal to zero. If the last digit of a number is zero, we can always factor out 10 and therefore 2 and 5, rendering ∏(n)>1.
    Yes, but the question is which option is false. Your logic is correct so (c) is not the right answer.
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    (Original post by ShatnersBassoon)
    Yes, but the question is which option is false. Your logic is correct so (c) is not the right answer.
    Yup, I meant (c) is true, not correct. Hence we can eliminate it.
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    late to the party

    guess who didnt get the  \displaystyle x^2 + 1 is a factor of... MC question

    Spoiler:
    Show

    Hint: me! what a f**k up
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    Anybody remembers the answer to the MCQ about the graph of (x^2-1)+(cospix)
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    (Original post by goofyygoober)
    Anybody remembers the answer to the MCQ about the graph of (x^2-1)+(cospix)
    answer a.
    it was the one through the origin with 4 x-intercepts:
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    (Original post by boombox111)
    answer a.
    it was the one through the origin with 4 x-intercepts:
    Do you remember what answer b looks like? I think there are two looks quite similar but I forgot which one I choose.
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    (Original post by goofyygoober)
    Do you remember what answer b looks like? I think there are two looks quite similar but I forgot which one I choose.
    ah yeah they were pretty similar - i got a bit hung up between the two
    answer a had four x-intercepts, while answer b had 6
    if you set y=0 then you get (x-1)^2 = cos(pi*x)
    drawing y=(x-1)^2 and y=cos(pi*x), you get four intersections
    hence the four x-intercepts and answer a.
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    (Original post by boombox111)
    ah yeah they were pretty similar - i got a bit hung up between the two
    answer a had four x-intercepts, while answer b had 6
    if you set y=0 then you get (x-1)^2 = cos(pi*x)
    drawing y=(x-1)^2 and y=cos(pi*x), you get four intersections
    hence the four x-intercepts and answer a.
    Thanks then I think I have chosen the more complex one which is b.. sad for losing another 4 marks... btw what do you think of this year's grade boundary, greater or lower than last year??
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    (Original post by goofyygoober)
    Anybody remembers the answer to the MCQ about the graph of (x^2-1)+(cospix)
    the graph that looked lots like a molar tooth im pretty sure, thats all I could see when doing the question :laugh:
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    (Original post by goofyygoober)
    Thanks then I think I have chosen the more complex one which is b.. sad for losing another 4 marks... btw what do you think of this year's grade boundary, greater or lower than last year??
    yeah i lost plenty too aargh
    it was the same kinda standard as 2015/13 so i'd guess around 55 and 61 for the interviewed/accepted averages. could be a tad higher tho
 
 
 
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