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

    2
    ReputationRep:
    Jordan\ Youshould also add the solutions for the downing college one for more practice

    http://www.dow.cam.ac.uk/documents/t...Maths_test.pdf
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by FutureOxford3)
    Jordan\ Youshould also add the solutions for the downing college one for more practice

    http://www.dow.cam.ac.uk/documents/t...Maths_test.pdf
    Yeah, of course! I'll add it to the OP.
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Very much so, the questions test general maths and will be useful practice for any college including Oxford.
    Oxford is not a college tho :P
    Offline

    18
    ReputationRep:
    (Original post by Krollo)
    Two pleasingly different solutions. What maths is all about, perhaps

    Posted from TSR Mobile
    The final answer is correct but after spotting that pattern it has been assumed it holds for all n, I would prove it by induction then let n=1000.

    I can't found who did the A one, but I can't find so maybe Krollo can make me find the solution-er.
    Posted from TSR Mobile
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by physicsmaths)
    The final answer is correct but after spotting that pattern it has been assumed it holds for all n, I would prove it by induction then let n=1000.

    I can't found who did the A one, but I can't find so maybe Krollo can make me find the solution-er.
    Posted from TSR Mobile
    Hence why I deleted it
    Offline

    18
    ReputationRep:
    (Original post by Jordan\)
    Hence why I deleted it
    I thought i just could not find it. Assume it holds for. K, show it holds for k+1 (by multiplying by A) and that's it.


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    Did that Z(root2) question on paper, I will write it up soon whilst I learn latex..


    Posted from TSR Mobile
    Offline

    8
    ReputationRep:
    (Original post by physicsmaths)
    Did that Z(root2) question on paper, I will write it up soon whilst I learn latex..


    Posted from TSR Mobile
    Pell equation!

    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    (Original post by Renzhi10122)
    Pell equation!

    Posted from TSR Mobile
    Yep! Amazing stuff that is


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by FutureOxford3)
    Jordan\ Youshould also add the solutions for the downing college one for more practice

    http://www.dow.cam.ac.uk/documents/t...Maths_test.pdf
    We figured those were far too elementary to provide solutions for. You can check your answers with a quick WA computation.
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    We figured those were far too elementary to provide solutions for. You can check yoir answers with a quick WA computation.
    What is the answer to 5 and 6 then?
    Offline

    22
    ReputationRep:
    (Original post by FutureOxford3)
    What is the answer to 5 and 6 then?
    Don't have the paper with me, but the question is a straightforward sketch and calculus problem. You can check your answers using WA.

    Six is admittedly hard, but it's not worth providing 8 solutions if we only really need it for 1 question.
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Don't have the paper with me, but the question is a straightforward sketch and calculus problem. You can check your answers using WA.

    Six is admittedly hard, but it's not worth providing 8 solutions if we only really need it for 1 question.
    Yea, the only one I can't do is number 6. I will just ask Jordan or physicsmath to help me then.
    Offline

    1
    ReputationRep:
    (Original post by Krollo)
    Specimen 2, Q7

    Hint
    Spoiler:
    Show

    (x-1) is a factor of x^n -1 always.
    When is (x+1) a factor of x^n+1?

    Solution
    Spoiler:
    Show

    Unless otherwise stated, assume all new constants are positive integers.


    Suppose p = a^n - 1.
    Then p = (a-1)(a^n-1 + ... + 1)

    So p is composite unless a-1 = 1 (as the second bracket is greater or equal to the first), so a=2.

    So p = 2^n - 1.

    Suppose n is composite, i.e. n=cd, with neither c or d less than 2.

    Then p =l 2^(cd) -1 = (2^c)^d -1 = (2^c -1)(2^(cd-c) + ... +1)

    As c>1, p is thus composite. So n cannot be composite - in other words, n is prime.

    - - -

    Let q = a^n + 1. We seek the conditions in which q is prime.

    If a is 1, then q is clearly always prime.

    If a is an odd integer greater than 1, then a^n is odd, so q is even (and not 2) hence it is never prime in this case.

    If a is an even integer, then if n is odd and greater than 1, q=(a+1)(a^n-1 - ... +1). q is thus never prime as a+1 > 1.

    If a is an even integer, then if n is even but has an odd factor r where r>1, and n=rs, then q = a^(rs) +1 = (a^s)^r + 1 = (a^s+1)(a^rs-s... +1) and so q is never prime.

    This leaves the only possibility as when a is even, and n is even and has no odd factors > 1, i.e. n is a non-negative integer power of two. Examples of this would be
    2^4 +1 =17, 6^1+1 = 7, 6^2 +1 =37.

    Posted from TSR Mobile
    Um, didn't the question ask for 2^n+1 not a^n+1? Also, when a = 8 n =2 you get 65.
    Offline

    18
    ReputationRep:
    (Original post by ForeignStudent34)
    Um, didn't the question ask for 2^n+1 not a^n+1? Also, when a = 8 n =2 you get 65.
    Well, they tell you to look at a^n -1 first and see what you can notice, they then want you to notice a form of a^n+1
    And then factor and see what you can see when a=2. I have not read through Krollos solution but i trust it is correct as his structure/idea looks fine imo.


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    I think in general they want to see 2^n+1=2^n-(-1)=(3)(some integer) if n is odd 2^n+1 is prime only when n=1 then do a similar thing for n is even, which krollo has done but to a greater depth.


    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    Specimen Test 2, Question 5:

    Hint:
    Spoiler:
    Show
    Use the equation $\mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}
    Solution:
    Spoiler:
    Show
    a) Let A be the event the final coin selected is biased and let B be the event the coin comes up heads three times

    $\mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=\frac{\frac{1  }{6}(\frac{3}{4})^3}{\frac{5}{6}  (\frac{1}{2})^3+\frac{1}{6}( \frac{3}{4})^3}=\frac{27}{67}$

    b) Let A be the event the biased coin is among the three selected and let B be the event the coin comes up heads three times

    $\mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}=\frac{\frac{1  }{2}(\frac{2}{3}(\frac{1}{2})^3+  \frac{1}{3}(\frac{3}{4})^3)}{ \frac{5}{6}(\frac{1}{2})^3+\frac  {1}{6}(\frac{3}{4})^3}=\frac{43}  {67}$
    Offline

    8
    ReputationRep:
    Downing Admissions Paper:
    Question 2:
    i)
    Spoiler:
    Show
    4 - x < 3 - 2x

    2x- x < 3 - 4

    x< -1
    ii)
    Spoiler:
    Show
    |x- 3| = 8

    x= -5
    or
    x= 11
    Offline

    8
    ReputationRep:
    Downing Admissions Paper:
    Question 3:
    Hint:
    Spoiler:
    Show
    Think about how taking -1 as a factor in a modulus function affects the function
    Solution:
    Spoiler:
    Show
    Because both inequalities are less than E/2 we may safely add them:

    |x - x_0 |+| y - y_0| < \frac{E}{2} + \frac{E}{2}

    |x - x_0 + y - y_0| < E

    |(x+ y) - (x_0 + y_0)| < E (1)

    |(x_0+ y_0) - (x + y)| < E

    |(x_0- x) + (y_0 - y)| < E

    |(-1)(x- x_0)| + |y_0 - y| < E

    |x- x_0| + |y_0 - y| < E

    |(x- y) + (y_0 - x_0)| < E

    |(x- y) - (x_0 - y_0)| < E (2)
    Offline

    8
    ReputationRep:
    Downing Admissions Paper:
    Question 4:
    Spoiler:
    Show
    Let V denote the volume:

    V= (1 - 2x)(1 - 2x)(x)

    V= (1 - 4x + 4x^2)(x)

    V= (x - 4x^2 + 4x^3)

    \frac{dV}{dx}=1 - 8x + 12x^2

    \frac{dV}{dx}= (6x - 1)(2x - 1)

    x= \frac{1}{6} or x = \frac{1}{2}

    x cannot be \frac{1}{2} so x must be \frac{1}{6}

    Check:
    \frac{d^2V}{d x^2} = 24x - 8

    When x = \frac{1}{6}:

    \frac{d^2V}{d x^2} = 4 - 8 = - 4

    Therefore V is a maximum at x = \frac{1}{6}
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 18, 2017
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?

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.