Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    0
    ReputationRep:
    (Original post by chemari1)
    or was question 7 the one about the roots? please anyone
    Yep
    Offline

    1
    ReputationRep:
    (Original post by thebearissquare)
    Think question 7 was about the roots, what did you get for alpha x beta?
    -1
    Offline

    1
    ReputationRep:
    (Original post by Crozzer24)
    For the inverse of B was it 1/25(A)
    Yeah I got that
    Offline

    2
    ReputationRep:
    (Original post by XenoMJ)
    I got 0.00813 3.S.F
    Oh, yes I did get that. Double zero, yes, my bad.
    Offline

    1
    ReputationRep:
    (Original post by XenoMJ)
    I got 0.00813 3.S.F
    same
    Offline

    2
    ReputationRep:
    (Original post by xX-MikeHunt-Xx)
    Yep
    in question 5 how many marks will I have got by plotting C1 right only? and what question asked about what the value converged to as x tended to infinity? pplus was the answer 0?
    Offline

    7
    ReputationRep:
    (Original post by -Gifted-)
    Not sure about the double zero. I got one zero.
    I got double zero. I got it as 8.13 to the power of 10^-3
    Offline

    3
    ReputationRep:
    My answers:

    1. M^{-1}=\dfrac{1}{8+2p} \begin{pmatrix} 1 & 2 \\ -p & 8 \end{pmatrix}

A = \dfrac{4 \times 6}{2} = 12, \, |M| = 8 + 2 \times 3 = 14, \, \therefore A' = 168

2. -\dfrac{21}{29} + \dfrac{20}{29}i

a = \dfrac{8}{29}, \, b = \dfrac{38}{29}

3. \lambda = 3, \, \mu = 25

B^{-1} = \dfrac{1}{25}A = \dfrac{1}{25}\begin{pmatrix} 3 & 6 & -4 \\ 2 & 5 & -1 \\ -1 & 4 & 3 \end{pmatrix}

4. \mathrm{proof:} \, \displaystyle\sum_{r = 1}^{n} r^{2}(2r - p)

\mathrm{p s.t. coefficient on} n^3 == n^4, \, p = \dfrac{3}{2}

5. \mathrm{Argand diagram:}

C_{1}: |z+3-4i|=5, \, C_{2}: arg(z+3-6i) = \dfrac{\pi}{2}

\mathrm{intersection:} -3+9i
    6: \mathrm{proof \, by \, induction:}

u_{1} = 8, \, u_{n+1} = 3u_{n} + 2n + 5

\mathrm{RTS:} u_{n} = 4(3^{n}) - n - 3

\mathrm{I can write this out if we really need...}

7. f(z) = 2z^{4} - 9z^{3} + Az^{2} + Bz - 26

\alpha > \beta, \, \gamma \, \mathrm{and} \, \delta \, \mathrm{are \, imaginary}, \, \gamma = 3+2i

\mathrm{show} \, \alpha + \beta = -\dfrac{3}{2}

\alpha \beta = -1

\alpha = \dfrac{1}{2}, \, \beta = -2 \,\,\, \mathrm{(I \, think \, order \, will \, matter)}

A = 6, \, B = 51

\mathrm{find roots of} \, f(\dfrac{w}{i}) = 0 \Rightarrow z = \dfrac{w}{i}, \, \therefore w = zi

w = \dfrac{1}{2}i, \, -2i, \, 2 + 3i, \, -2 + 3i

8. y = \dfrac{3x^{2} - 9}{x^{2} + 3x - 4}

\mathrm{asymptotes:} x = 1, \, x = -4, \, y = 3

\displaystyle \lim_{x \rightarrow \infty}(y) = 3^{-}

\displaystyle \lim_{x \rightarrow - \infty}(y) = 3^{+}

y \geq 0 \Leftrightarrow x < -4, \, - \sqrt{3} \leq x < 1, \, x \geq \sqrt{3}
    9. \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}=\dfrac{3}{4(2r-1)}-\dfrac{1}{2r+1}+\dfrac{1}{4(2r+3  )}

\mathrm{proof for} \, \displaystyle \sum_{1}^{n}

\displaystyle \lim_{n \rightarrow \infty} = \dfrac{2}{3}

\displaystyle \sum_{20}^{50} = 000813

    I forgot to halve the area in 1, and started writing 3/2 instead of 2/3 in 9 but apart from that it was alright
    Offline

    1
    Hardly bad, only lost at least 20 marks...
    Offline

    2
    ReputationRep:
    (Original post by Crozzer24)
    Think I've been stupid and left the answer to the last question as 0.00816 or something? Didn't realise 0's at the front classed as sig figs lol
    The zeroes don't, I got 0.00813 I think?
    Offline

    1
    ReputationRep:
    (Original post by XenoMJ)
    I got 0.00813 3.S.F
    same
    Offline

    1
    ReputationRep:
    How do you do q4, when coefficient of n^4 = n^3?
    Offline

    2
    ReputationRep:
    (Original post by lfcjatt)
    How do you do q4, when coefficient of n^4 = n^3?
    you hAD TO WORK OUT THE COEFFICIENT OF N^4, Which I think was 1/2 (sorry for the caps) then work out the coefficient of n^3 in terms of p, and solve that as an equation.
    I think my final answer was p=3/2, but I can't really remember
    Offline

    7
    ReputationRep:
    Multiply the brackets out and work out coefficients of n^3 in terms of p then work out p
    Offline

    1
    ReputationRep:
    MEI can go slip in the shower, die in a fire and burn in a hole
    Offline

    1
    ReputationRep:
    Anyone remember the mark-split for the individual parts of q7? Roots of the polynomial eq'n.

    I did alpha*beta*given root*complex conjugate = -b/a
    I got that p = 3/2 and I think another answer might have been -1 ??

    If someone could post the question+answers would be v grateful
    Offline

    1
    ReputationRep:
    (Original post by TillSoft)
    you hAD TO WORK OUT THE COEFFICIENT OF N^4, Which I think was 1/2 (sorry for the caps) then work out the coefficient of n^3 in terms of p, and solve that as an equation.
    I think my final answer was p=3/2, but I can't really remember
    How did you get the coefficient of n^4? and was the coefficient of n^3 = 2?

    i got p=+/-sqrt(2) or something like that :/

    not sure what i did then lol
    Offline

    2
    ReputationRep:
    (Original post by Duskstar)
    My answers:

    1. M^{-1}=\dfrac{1}{8+2p} \begin{pmatrix} 1 & 2 \\ -p & 8 \end{pmatrix}

A = \dfrac{4 \times 6}{2} = 12, \, |M| = 8 + 2 \times 3 = 14, \, \therefore A' = 168

2. -\dfrac{21}{29} + \dfrac{20}{29}i

a = \dfrac{8}{29}, \, b = \dfrac{38}{29}

3. \lambda = 3, \, \mu = 25

B^{-1} = \dfrac{1}{25}A = \dfrac{1}{25}\begin{pmatrix} 3 & 6 & -4 \\ 2 & 5 & -1 \\ -1 & 4 & 3 \end{pmatrix}

4. \mathrm{proof:} \, \displaystyle\sum_{r = 1}^{n} r^{2}(2r - p)

\mathrm{p s.t. coefficient on} n^3 == n^4, \, p = \dfrac{3}{2}

5. \mathrm{Argand diagram:}

C_{1}: |z+3-4i|=5, \, C_{2}: arg(z+3-6i) = \dfrac{\pi}{2}

\mathrm{intersection:} -3+9i
    6: \mathrm{proof \, by \, induction:}

u_{1} = 8, \, u_{n+1} = 3u_{n} + 2n + 5

\mathrm{RTS:} u_{n} = 4(3^{n}) - n - 3

\mathrm{I can write this out if we really need...}

7. f(z) = 2z^{4} - 9z^{3} + Az^{2} + Bz - 26

\alpha > \beta, \, \gamma \, \mathrm{and} \, \delta \, \mathrm{are \, imaginary}, \, \gamma = 3+2i

\mathrm{show} \, \alpha + \beta = -\dfrac{3}{2}

\alpha \beta = -1

\alpha = \dfrac{1}{2}, \, \beta = -2 \,\,\, \mathrm{(I \, think \, order \, will \, matter)}

A = 6, \, B = 51

\mathrm{find roots of} \, f(\dfrac{w}{i}) = 0 \Rightarrow z = \dfrac{w}{i}, \, \therefore w = zi

w = \dfrac{1}{2}i, \, -2i, \, 2 + 3i, \, -2 + 3i

8. y = \dfrac{3x^{2} - 9}{x^{2} + 3x - 4}

\mathrm{asymptotes:} x = 1, \, x = -4, \, y = 3

\displaystyle \lim_{x \rightarrow \infty}(y) = 3^{-}

\displaystyle \lim_{x \rightarrow - \infty}(y) = 3^{+}

y \geq 0 \Leftrightarrow x < -4, \, - \sqrt{3} \leq x < 1, \, x \geq \sqrt{3}
    9. \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}=\dfrac{3}{4(2r-1)}-\dfrac{1}{2r+1}+\dfrac{1}{4(2r+3  )}

\mathrm{proof for} \, \displaystyle \sum_{1}^{n}

\displaystyle \lim_{n \rightarrow \infty} = \dfrac{2}{3}

\displaystyle \sum_{20}^{50} = 000813

    I forgot to halve the area in 1, and started writing 3/2 instead of 2/3 in 9 but apart from that it was alright
    the asymptote was 1
    Offline

    0
    ReputationRep:
    For Qs 7 i think, how did you calculate alpha x beta. I didn't get that and so couldn't really do the rest of the qs and find A and B and then the w/j bit. I've done all the fp1 past papers and there has been nothing like this qs before. Also the series method of differences qs (i think it was near the end, honestly can't remember anymore) was a toughie, i think I may have gotten partly there but still hard
    Offline

    3
    ReputationRep:
    (Original post by lfcjatt)
    How do you do q4, when coefficient of n^4 = n^3?
    You had something like

    \dfrac{1}{6}n(n+1)(3n^{2}+(3-2p)n-3) = \dfrac{1}{6}(n^{2}+n)(3n^{2}+(3-2p)n-3)

    as the formula I think. The coefficient on n^{4} is \dfrac{1}{2} if you multiply it out, and the coefficient on n^{3} is given by multiplying \dfrac{1}{6} \times (1 \times 3 + 1 \times (3 - 2p)) by matching up terms in either bracket to get n^{3} terms.

    Set \dfrac{1}{2} = \dfrac{1}{6} \times (1 \times 3 + 1 \times (3 - 2p)) and solve for p.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.