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Official thread FP1, 1st Febraury 2010 watch

  • View Poll Results: How do you rate this paper (on a scale of 1-10 with 10 being the best)?
    1
    5
    7.25%
    2
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    1.45%
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    1.45%
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    2.90%
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    1.45%
    7
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    8.70%
    8
    6
    8.70%
    9
    15
    21.74%
    10
    32
    46.38%

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    11
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    (Original post by quietlyconfidant)
    no one fancy doing a mark scheme then????
    I think everyone's over the exam now. :p:
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    (Original post by cambo211)
    which would be surprising as the paper was out of 75...
    sorry 74/75
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    8
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    did everyone get -8 and 6 for the coordinates of the very last question?
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    (Original post by Khodu)
    [latex] z_{1} = 2+8i , z_{2} = 1-i [/l

    9. Matrices


    My first ever attempt with latex, do not laugh lol
    Would You or someone please do question 3, and break down the marks,? , also how much marks do you think i will losr for not saying , because it it ture for n=1, true for all R,

    Cheers in adv.
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    (Original post by Rough Inclined Plane)
    did everyone get -8 and 6 for the coordinates of the very last question?
    For 9e I think it was (-4(sqrt)2 , 3(sqrt)2).
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    (Original post by nadamo)
    Would You or someone please do question 3, and break down the marks,? , also how much marks do you think i will losr for not saying , because it it ture for n=1, true for all R,

    Cheers in adv.
    You could look at the mark scheme for Jan 09, there's a very similar question to the one we got.
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    well i dont think anyone is that bothered to make a mark scheme cus it was a straight forward exam
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    (Original post by Robofish)
    well i dont think anyone is that bothered to make a mark scheme cus it was a straight forward exam
    No I mean the exam from last year January 2009.
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    (Original post by Mastercard)
    For 9e I think it was (-4(sqrt)2 , 3(sqrt)2).
    :yep:
    That's what I got.
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    (1) Complex numbers z_{1} = 2+8i , z_{2} = 1-i

    Find:
    Credit to 2710 i think
    a)  \frac{z_{1}}{z_{2}}

    ans= -3 + 5i

    b)  |\frac{z_{1}}{z_{2}}|

    \sqrt 34

    c)  arg( |\frac{z_{1}}{z_{2}}|)

     2.11

    (2) Root Intervals
     f(x) = 3x^2- \frac{11}{x^2}

    (a) Write down, to 3 decimal places, the value of f(1.3) and the value of f(1.4).
    f(1.3)= -1.439

    f(1.4)= 0.268

    The equation f(x) = 0 has a root α between 1.3 and 1.4
    (b) Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width
    0.025 which contains α.

    1.375 < a < 1.4

    (c) Taking 1.4 as a first approximation to α, apply the Newton-Raphson procedure once to
    f(x) to obtain a second approximation to α, giving your answer to 3 decimal places.

    x_2= 1.384


    3. Proof by induction
    4. Co-ordinate systems- parabola
    5. Matrices

    Credit to Mastercard

    Given Matrix A, which I think looked like this \begin{pmatrix}  a & -5 \\ 2 & a + 4 \\  \end{pmatrix}. But I'm not sure about the -2 and 5.

    (a) Find Det A. Det A = a(a+4) +10 = a^2 + 4a +10

    (b) Prove that Matirix A is non-singular for all values of a.

    b^2 -4ac = -24

     -24 < 0 \Longrightarrow a\not=0 and therefore the Matrix is non-singular.

    (c) Find A^{-1} given that a=0.

    A^{-1} = \frac{1}{10}\begin{pmatrix}  4 & 5 \\ -2 & 0 \\  \end{pmatrix}

    6.Complex numbers

    Credit to Mastercard

    Given that two roots of the equation  x^3 - 12x^2 + ax + b are  2, 5+2i find:

    (a) The third root. Answer 5-2i.

    (b) Find the values of a and b. Answer a=49, b=-58.

    (c) Display the three roots on an Argand diagram.



    7.Co-ordinate systems rectangular hyperbola
    8. Proof by induction and summation

    a) \displaystyle\sum_{r=1}^n  = r^3 =\frac{1}{4}n^2(n+1)^2
    Prove true for all positive integers.

    Prove that n = 1 is true

    {1}^3  = 1

    \frac{1}{4}(1)^2(1+1)^2 = \frac{1}{4}(4)= 1

    Therefore true for n = 1

    Assume true for n = k

    \displaystyle\sum_{r=1}^k  \frac{1}{4}k^2(k+1)^2

    Prove true for n= k+1

    \displaystyle\sum_{r=1}^{k+1}  =\displaystyle\sum_{r=1}^k   + (k+1)^{th} term

    = \frac{1}{4}k^2(k+1)^2 + (k+1)^3

    = \frac{1}{4}(k+1)^2[k^2+4(k+1)]

    = = \frac{1}{4}(k+1)^2[k^2+4k+4]

    = \frac{1}{4}(k+1)^2(k+2)^2

    Therefore true for n = k+1 if n = k is true. True for n = 1, therefore by induction true for all positive integers.

    9. Matrices
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    (Original post by Khodu)
    (1) Complex numbers z_{1} = 2+8i , z_{2} = 1-i

    Find:
    Credit to 2710 i think
    a)  \frac{z_{1}}{z_{2}}

    ans= -3 + 5i

    b)  |\frac{z_{1}}{z_{2}}|

    \sqrt 34

    c)  arg( |\frac{z_{1}}{z_{2}}|)

     2.11

    (2) Root Intervals
     f(x) = 3x^2- \frac{11}{x^2}

    (a) Write down, to 3 decimal places, the value of f(1.3) and the value of f(1.4).
    f(1.3)= -1.439

    f(1.4)= 0.268

    The equation f(x) = 0 has a root α between 1.3 and 1.4
    (b) Starting with the interval [1.3, 1.4], use interval bisection to find an interval of width
    0.025 which contains α.

    1.375 < a < 1.4

    (c) Taking 1.4 as a first approximation to α, apply the Newton-Raphson procedure once to
    f(x) to obtain a second approximation to α, giving your answer to 3 decimal places.

    x_2= 1.384


    3. Proof by induction
    4. Co-ordinate systems- parabola
    5. Matrices

    Credit to Mastercard

    Given Matrix A, which I think looked like this \begin{pmatrix}  a & -5 \\ 2 & a + 4 \\  \end{pmatrix}. But I'm not sure about the -2 and 5.

    (a) Find Det A. Det A = a(a+4) +10 = a^2 + 4a +10

    (b) Prove that Matirix A is non-singular for all values of a.

    b^2 -4ac = -24

     -24 < 0 \Longrightarrow a\not=0 and therefore the Matrix is non-singular.

    (c) Find A^{-1} given that a=0.

    A^{-1} = \frac{1}{10}\begin{pmatrix}  4 & 5 \\ -2 & 0 \\  \end{pmatrix}

    6.Complex numbers

    Credit to Mastercard

    Given that two roots of the equation  x^3 - 12x^2 + ax + b are  2, 5+2i find:

    (a) The third root. Answer 5-2i.

    (b) Find the values of a and b. Answer a=49, b=-58.

    (c) Display the three roots on an Argand diagram.



    7.Co-ordinate systems rectangular hyperbola
    8. Proof by induction and summation

    a) \displaystyle\sum_{r=1}^n  = r^3 =\frac{1}{4}n^2(n+1)^2
    Prove true for all positive integers.

    Prove that n = 1 is true

    {1}^3  = 1

    \frac{1}{4}(1)^2(1+1)^2 = \frac{1}{4}(4)= 1

    Therefore true for n = 1

    Assume true for n = k

    \displaystyle\sum_{r=1}^k  \frac{1}{4}k^2(k+1)^2

    Prove true for n= k+1

    \displaystyle\sum_{r=1}^{k+1}  =\displaystyle\sum_{r=1}^k   + (k+1)^{th} term

    = \frac{1}{4}k^2(k+1)^2 + (k+1)^3

    = \frac{1}{4}(k+1)^2[k^2+4(k+1)]

    = = \frac{1}{4}(k+1)^2[k^2+4k+4]

    = \frac{1}{4}(k+1)^2(k+2)^2

    Therefore true for n = k+1 if n = k is true. True for n = 1, therefore by induction true for all positive integers.

    9. Matrices
    i think 6 is wrong....
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    (Original post by fredbraty)
    i think 6 is wrong....
    Why do you say that?
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    (Original post by 2710)
    Why do you say that?
    a was 49 but b was 100 and something i think
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    can some do question 9
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    (Original post by fredbraty)
    a was 49 but b was 100 and something i think
    No. Given that one of the roots is 2, if you sub in 2 for x, and 49 and -58, you get 0, which is correct. On the other hand, if you sub in 100 something....

    Sorry mate
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    (Original post by 2710)
    No. Given that one of the roots is 2, if you sub in 2 for x, and 49 and -58, you get 0, which is correct. On the other hand, if you sub in 100 something....

    Sorry mate
    hm...as i said i am not sure
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    (Original post by quietlyconfidant)
    can some do question 9
    9a) Find the transformation represented by Matrix \begin{pmatrix} \frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}  \\\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}  \end{pmatrix}.

    Answer: Anticlockwise rotation through 45 degrees about (0,0)

    b) \begin{pmatrix} \frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}  \\\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}  \end{pmatrix}\begin{pmatrix}  p  \\\ q\end{pmatrix} = \begin{pmatrix}  3\sqrt 2  \\\ 4\sqrt 2  \end{pmatrix}

    Therefore, p - q = 6 and p+q=8
    Which gives p=7, q=1.

    c) Distance between (7,1) and the origin = \sqrt 50 = 5\sqrt2

    d) Initial Matrix squared. Answer \begin{pmatrix} 0 & -1  \\1 & 0  \end{pmatrix}

    e) Transform (3\sqrt2, 4\sqrt2) using matrix squared.

    Answer = \begin{pmatrix} 0 & -1  \\1 & 0  \end{pmatrix}\begin{pmatrix}  3\sqrt 2  \\\ 4\sqrt 2  \end{pmatrix} = (-4\sqrt2, 3\sqrt2)

    I hope I haven't made any stupid errors here. Had to do it from memory since the paper's been taken down.
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    thanks very much. i think thats what i got. repped bam!
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    thank youuu nice paper i thought
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    :tumble:

    any maths people around... Getting scared. Going to have to do S1,S2,M1,C4 in June. And that's the minimum.
 
 
 
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Updated: February 8, 2010
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