Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    Offline

    7
    ReputationRep:
    (Original post by m0.4444)
    i got that one wrong. But i think its 2 - sqrt2
    Ikr i made such a ****ing silly mistake. I'm kicking myself
    Offline

    2
    ReputationRep:
    Is that not the answer to B
    Posted from TSR Mobile
    Attached Images
     
    Offline

    7
    ReputationRep:
    (Original post by m0.4444)
    But how many inequalities where there with c and 0.
    For the last bit of 2 I had no idea how to approach it i was goona do wat you did but i didnt have time so i came up with some like contradiction saying 2c-1=92 did anyone get anything like this. And when are the results published
    Offline

    9
    ReputationRep:
    (Original post by louisforrest)
    Is that not the answer to B
    Posted from TSR Mobile
    I got sqrt(2) - 1 I think but I can't be sure
    Offline

    7
    ReputationRep:
    That graph was a right?
    Offline

    7
    ReputationRep:
    Any predictions for boundaries?? easier or harder than last year??
    Offline

    4
    ReputationRep:
    What did people put for the penultimate multiple choice question (about ax+by) and why? I put "max(a, b)" but it was a slight guess - I know you could get max(a, b) [from x = 1 and y = 0 or vice versa] and I ruled 2 of the other options out but does anyone have any explanation?

    (Original post by Amitbalter)
    MCQs:
    1) l^105
    2) 2-root2
    3)didn't do
    4)2 for even 3 for odd
    5) the one that had a positive value for y when X=1, and also passed through the origin
    6) didn't do
    7) 3
    8) a<(6/5)^4
    9) didn't do
    10) b
    On question 1, does anyone know where I went wrong:

    It was a GP with a = 1, r = l, n = 15
    Sum = (1-l^15)/(1-l) [which was one of the options]

    For (5) I think you're describing (a) - I checked on Wolfram Alpha after I came out the exam and I think that was the right one.

    For everything else (that you answered) I got the same, so that's good.

    (Original post by DavidBick)
    If you calculate them all carefully, they all come to 107. This rhymes with the fact that the function A^n(B^m(x))= (2^n)(3^n)x+(2^n*3^n - 1) which is used in the last question. Note that 107 = 108-1

    The latter also shows that it is impossible to reach 214x+96 (or whatever it was) since for each compound function added, the constant is > 1/2 the x coefficient. Hence, since 96 < 1/2 of 214, it is impossible with any sum.
    I agree with this. I don't remember explicitly calculating that function though - do you think that me stating without proof "the constant will always be one less than the coefficient of x" would lose me marks?

    Also for the last part, I justified it with another reason as well, that 214 = 2 * 107 [which are prime], and 214 =/= 2^m * 3^n for any integer m,n, so there was no way of combining the functions to get 214x + c. Is this true?

    (Original post by m0.4444)
    What did people get for q6?

    It was the dancers one about off-beat dances. if the sum is odd. If i remember for the very last parts i got something about there being 2 rings if n was even and 1 ring if n was odd?
    So I ran out of time on part (vi) of this but for the rest:
    (i) 0 - you can't have an off-beat dance with 4 people
    (ii) 3? or 6? - if A is 1 then that forces the numbers of the other five people; I think the pattern was 100100 so you could have 100100, 001001 or 010010 [starting with A]. Might be getting confused with a different part though.
    (iii) Explanation - has to be a multiple of 3 because we need 100 to be repeated an integer number of times.
    (iv) n is a multiple of 3? If it was even, you get two smaller groups of n/2 holding hands as in the first parts of the question, so n has to be a multiple of 3. If n was odd, you get a longer chain that starts with A, goes to the end and loops back round to B and carries on. You need n to be a multiple of 3 either way so I think that was the answer.
    (v) 8 ways (either 1 or 3 of them have a 1)
    Offline

    7
    ReputationRep:
    for that polynomial i got n is even
    Offline

    2
    ReputationRep:
    (Original post by 11234)
    That graph was a right?
    what did it look like
    Offline

    7
    ReputationRep:
    (Original post by RuairiMorrissey)
    what did it look like
    one that had positive value of y for x=1 and through origin and was symmetirc
    Offline

    1
    ReputationRep:
    (Original post by louisforrest)
    Is that not the answer to B
    Posted from TSR Mobile
    You worked out a but u need to find x
    Offline

    7
    ReputationRep:
    for the max value of ax + by it was sqrt(a^2+b^2).
    =
    Proof.
    By cauchy schwarz (a2+b2)(x2+y2)>(ax+by)2 and x2+y2<1 so less than sqrt a2+b2
    Offline

    1
    ReputationRep:
    (Original post by ShatnersBassoon)
    What did people put for the penultimate multiple choice question (about ax+by) and why? I put "max(a, b)" but it was a slight guess - I know you could get max(a, b) [from x = 1 and y = 0 or vice versa] and I ruled 2 of the other options out but does anyone have any explanation?
    The inequality was x^2 + y^2 <= 1 which basically says (x, y) lies within the unit circle. To maximise (ax+by), it is simple to note that it is essentially 45 degrees in the unit circle i.e. x = y = root(2)/2.

    Therefore the max you can get is root(a^2 + b^2)
    Offline

    7
    ReputationRep:
    I would say it was similar to last yhears paper. So maybe similar boundaries??
    Offline

    1
    ReputationRep:
    (Original post by ShatnersBassoon)

    On question 1, does anyone know where I went wrong:

    It was a GP with a = 1, r = l, n = 15
    Sum = (1-l^15)/(1-l) [which was one of the options]

    ...

    I agree with this. I don't remember explicitly calculating that function though - do you think that me stating without proof "the constant will always be one less than the coefficient of x" would lose me marks?

    Also for the last part, I justified it with another reason as well, that 214 = 2 * 107 [which are prime], and 214 =/= 2^m * 3^n for any integer m,n, so there was no way of combining the functions to get 214x + c. Is this true?
    For question 1, sadly it wasn't a GP... they asked for the product of the terms (nearly caught me).

    The tricky thing with the last part of that question is it's not a single function, it's an arbitrary sum of them. So your first statement would be right for a solitary compound function of that form, but for a sum the difference between the x coeff. and constant will be equal to the number of functions added. You might still get a mark though!

    Equally, the (semi) primality of 214 is irrelevant in light of it being a sum. (2^1)(3^0)+(2^0)(3^1) is a basic example which produces the prime 5.

    I'm sure you'll still get marks - sorry don't want to be downer. You seem to have more of an intuition on it than others on this thread, so I imagine you're a high scorer overall.
    Offline

    7
    ReputationRep:
    (Original post by collmr)
    The inequality was x^2 + y^2 <= 1 which basically says (x, y) lies within the unit circle. To maximise (ax+by), it is simple to note that it is essentially 45 degrees in the unit circle i.e. x = y = root(2)/2.

    Therefore the max you can get is root(a^2 + b^2)
    yea boi
    Offline

    2
    ReputationRep:
    (Original post by 11234)
    one that had positive value of y for x=1 and through origin and was symmetirc
    Was it not the squigily one?
    Offline

    1
    ReputationRep:
    (Original post by louisforrest)
    Is that not the answer to B
    Posted from TSR Mobile
    You solved for a, not x. x is 1 - a so answer is 2 - sqrt2
    Offline

    7
    ReputationRep:
    (Original post by RuairiMorrissey)
    Was it not the squigily one?
    nvm. got any predictions??
    Offline

    7
    ReputationRep:
    For Q4 the geometry one: I got (cot alpa, 1) and then alpha = pi/6. Then I got the radius of the big cirle is 9. Then something like 4sqrt3-11pi/6 cant remember the exacts
 
 
 
Turn on thread page Beta
Updated: October 30, 2017
Poll
“Yanny” or “Laurel”

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.