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STEP Maths I, II, III 1996 Solutions

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    (Original post by DFranklin)
    Step QI, P5.

    iii) Again, use the quadratic formula:
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\

    \implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
    .

    Using (ii),
     z = 1 \pm  (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.
    Just wondering, what happened to your 2a in the quadratic formula? I had a (1+i) in the bottom of the fraction, which would explain why they asked "express your answer in its simplest form", i.e multiplying through by the complex conjugate (1-i)/(1-i), I checked my solution with a ex Cambridge Maths guy and we both seemed to get the same bit. so your answer / (1+i)
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    Quite possible; I seem to recall losing the will to live with that question about half the way through...
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    What do you think the grade boundary for a 1 would be for this paper?
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    Shall we add the amendment to that then? Funny you mention that, This was a question which I got the knack of quite quickly (as opposed to many others which took me literally years)
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    (Original post by insparato)
    Typing up STEP III question 2 Now

     x + y + az = 2 - eqn 1
     x + ay + z = 2 - eqn 2
     2x + y + z = 2b - eqn 3

    eqn (2 - 1)

     ay-y + z-az = 0
     y(a-1) = z(a-1)
     y = z -eqn4

    sub eqn 4 into 3

     2x+2y = 2b
     x + y = b -eqn 5

    sub eqn 4 into 2
     x + (a+1)y = 2 -eqn6

    subtracting 6 from 5

     y + ay - y = 2-b

     y = \frac{2-b}{a}

    Sub  y = \frac{2-b}{a} into eqn 5

     x + \frac{2-b}{a} = b

     ax + 2-b = ab

     ax =  ab + b - 2

     x = \frac{ab+b-2}{a}

    Sub these back into eqn 1 just to see they are indeed satisfy the system.

     \frac{ab+b-2}{a} + \frac{2-b}{a} + \frac{2a-ab}{a} = \frac{2a}{a} = 2

    Therefore

     x = \frac{ab+b-2}{a}

     y = \frac{2-b}{a}

     z = \frac{2-b}{a}

    Not complete, the cases A = 0, A = 1 have to be investigated as the standard solution blows up when A = 0 and A = 1. Ive done alot of maths today so i wont be doing this tonight but i might have a go tomorrow or something... Anyone who feels up to finishing it can do so freely .
    Such a nice alternative solution with matricies i would type it out but i do ot know how to latex a matrix o.o sorry guys...
    If anyone requests the solution i can always do a copy and scan in just message me
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    Here is an alternative approach to STEP 1996 I Q10.
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    (Original post by themaths)
    Such a nice alternative solution with matricies i would type it out but i do ot know how to latex a matrix o.o sorry guys...
    If anyone requests the solution i can always do a copy and scan in just message me
    \text{If }a=0 \text{ there are no solutions unless }b=2
     \text{ in which case we have }x+y=x+z=2 \text{ so}x=\lambda, y=\lambda, z=2-\lambda
    \text{ If }a=1\text{ then }x+y+z=2 \text{ and }2x+y+z=2b \text{ so }x=2(b-1), y+z=4-2b
    \text{i.e. }x=2(b-1),y=\lambda,z=4-2b-\lambda
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    (Original post by DFranklin)
    Step QI, P5.

    (i) (r+s\sqrt{3})=4-2\sqrt{3} \implies r^2+3s^2+2rs\sqrt{3} = 4-2\sqrt{3}. Suppose rs \neq -1. Then
    r^2+3s^2-4 = -2(1+rs)\sqrt{3} would give us a rational expression for \sqrt{3}, contradiction.

    So rs=1, \text{ so } s = -1/r \text { and so } r^2+3/r^2 = 4 \implies r^4-4r^2+3 = 0\\

\implies r^2 = 1, 3.

    As r is rational, r^2 = 1, \text{ so } r = \pm 1, s = -r. (i.e. our solutions are \pm(1-\sqrt{3}).

    (ii) (p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i. So
    p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}. Again, we set up the quadratic and get:
    p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0. Use the quadratic formula to get:
    2p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}
    Now  (3-2\sqrt{3})^2 = 3^2+4\cdot 3 - 12\sqrt{3} = 21 - 12\sqrt{3}, \quad (1-\sqrt{3})^2 = 1 + 3 - 2 \sqrt{3} = 4-2\sqrt{3}.
    So (3-2\sqrt{3})^2+4(1-\sqrt{3}))^2 = 37 - 20 \sqrt{3}.

    As in (i), we need to find the square root of this, so look for rational \alpha, \beta with (\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}.

    We find \beta = -10/\alpha and end up with a quadratic \alpha^4 - 37\alpha^2+300=0. Solve to find \alpha^2 = 12, 25 and since we want \alpha rational deduce \alpha = \pm 5 and so the square root is \pm(5-2\sqrt{3}).

    Plugging this back in, we have

    2p^2 = (3-2\sqrt{3}) \pm (5-2\sqrt{3}) = -2, 8-4\sqrt{3}.

    Since p is real, we take the latter root, dividing by 2 and using (i) we find p = \pm (1-\sqrt{3}). Use pq = (1-\sqrt{3}) to get final solutions \pm (1-\sqrt{3}+i).

    (iii) Again, use the quadratic formula:
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\

    \implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
    .

    Using (ii),
     z = 1 \pm  (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.
    Sorry, ignore this post .
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    (Original post by DFranklin)
    Step II, Q3:

    F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.

    F0.F2-F1^2 = -1, F1.F3-F2^2 = 1, F2.F4-F3^2 = -1.

    Claim F_{n+1}F_{n-1}-F_n^2 = (-1)^n. Proof by induction on n.
    By above, true for n<=3. Assume true for n = k. Then
    F_{k+2}F_k - F_{k+1}^2 \\

= (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ 

= F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\

= [F_{k+1}F_{k-1}+(-1)^{k+1}]  - F_{k+1}(F_{k+1} - F_k) \\

= (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\

= (-1)^{k+1}F_{k+1}^2

    So true for n=k+1 and so true for all n by induction.

    Want to show F_{n+k} = F_kF_{n+1}+F_{k-1}F_n. Assume true for 1 < k <=m. Then
    F_{n+k+1} = F_{n+k}+F_{n+k-1}\\

= F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\

=(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\

=F_{k+1}F_{n+1}+F_k F_n.

    So true for k=m+1.

    Explictly when k=1 we have F_{n+1} = F_1F_{n+1}+F_0F_n = F_{n+1}. So true for k=1, so true for all k.
    This question has been annoying me this afternoon, in the end I settled with what you have here, but I am not wholly satisfied.
    For the second part, you have assumed that F_{n+k-1} takes the same form as F_{n+k} isn't that defeating the whole point of the induction?
    We assume n=k, but in the proof it seems n=k-1 is assumed too, am I talking a load of rubbish?
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    The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

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    (Original post by BabyMaths)
    The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

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    Sorry i didn't specify i was talking about the last part, but I have been sorted out on the step prep thread
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    (Original post by DFranklin)
    Step QI, P5.

    (i) (r+s\sqrt{3})=4-2\sqrt{3} \implies r^2+3s^2+2rs\sqrt{3} = 4-2\sqrt{3}. Suppose rs \neq -1. Then
    r^2+3s^2-4 = -2(1+rs)\sqrt{3} would give us a rational expression for \sqrt{3}, contradiction.

    So rs=1, \text{ so } s = -1/r \text { and so } r^2+3/r^2 = 4 \implies r^4-4r^2+3 = 0\\

\implies r^2 = 1, 3.

    As r is rational, r^2 = 1, \text{ so } r = \pm 1, s = -r. (i.e. our solutions are \pm(1-\sqrt{3}).

    (ii) (p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i. So
    p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}. Again, we set up the quadratic and get:
    p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0. Use the quadratic formula to get:
    2p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}
    Now  (3-2\sqrt{3})^2 = 3^2+4\cdot 3 - 12\sqrt{3} = 21 - 12\sqrt{3}, \quad (1-\sqrt{3})^2 = 1 + 3 - 2 \sqrt{3} = 4-2\sqrt{3}.
    So (3-2\sqrt{3})^2+4(1-\sqrt{3}))^2 = 37 - 20 \sqrt{3}.

    As in (i), we need to find the square root of this, so look for rational \alpha, \beta with (\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}.

    We find \beta = -10/\alpha and end up with a quadratic \alpha^4 - 37\alpha^2+300=0. Solve to find \alpha^2 = 12, 25 and since we want \alpha rational deduce \alpha = \pm 5 and so the square root is \pm(5-2\sqrt{3}).

    Plugging this back in, we have

    2p^2 = (3-2\sqrt{3}) \pm (5-2\sqrt{3}) = -2, 8-4\sqrt{3}.

    Since p is real, we take the latter root, dividing by 2 and using (i) we find p = \pm (1-\sqrt{3}). Use pq = (1-\sqrt{3}) to get final solutions \pm (1-\sqrt{3}+i).

    (iii) Again, use the quadratic formula:
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\

    \implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
    .

    Using (ii),
     z = 1 \pm  (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.
    Hi, I think you forgot to divide by 2(1+i) to obtain the final answer in part III, and then you might want to simplify it
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    For III Q2.
    You can show that those are the answers are correct with the exception of a=1,0. Here's how:
    The three equation can be rewritten as three matrices as shown below:
     \begin{bmatrix} 1 & 1 & a \\ 1 & a & 1 \\ 2 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 2b \end{bmatrix}
    We can now solve it because of the result
    Ainverse (A)R = Ainverse B
    which goes to :
    R = Ainverse B
    now this means that the equations have no solution when A is non-invertible.
    A is invertible when  det A =0
    so when  2a-2a^2 =0
    so when  a=0 or  a=1
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    (Original post by DFranklin)
    Quite possible; I seem to recall losing the will to live with that question about half the way through...
    I think I've got a nicer solution to this problem which I've posted below:

    Q5 STEP I:

    Part i)

    Spoiler:
    Show
     (r + s\sqrt3)^2 = 3 -2\sqrt3 + 1 = (\sqrt3 -1)^2

     r + s\sqrt3 = \pm(\sqrt3 -1)

    So the only combination which results in rational r and s is:

     r = \pm1 , s= \mp1


    Part ii)

    Spoiler:
    Show
     (p+qi)^2 = i^2 + 2(1-\sqrt3)i + 4 - 2\sqrt3 = i^2 + 2(1/\sqrt3)i + (\sqrt3 -1)^2 = (i+1-\sqrt3)^2

     p + qi = \pm(i+1-\sqrt3)

     p = \pm(1-\sqrt3) , q = \pm1


    Part iii)

    Spoiler:
    Show
    By the quadratic formula:

     z = \frac{2\pm\sqrt{4-4(i+1)(2\sqrt3-2)}}{2(1+i)} = \frac{1\pm\sqrt{1 - (1+i)(2\sqrt3 - 2)}}{1+i} = \frac{1\pm\sqrt{(3-2\sqrt3)+2i(1-\sqrt3)}}{1+i} = \frac{1\pm(i+1-\sqrt3)}{1+i}

     z = \frac{\sqrt3 - i}{1+i} = \frac{(\sqrt3 - 1) - (\sqrt3 + 1)i}{2}

    or

     z = \frac{2+i-\sqrt3}{1+i} = \frac{(3-\sqrt3) + (\sqrt3 - 1)i}{2}
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    (Original post by DFranklin)
    Solved it, yes, written it up, no - nice job. I didn't see: did you manage the last bit? I thought it surprisingly tricky; needing to find the value of tan(3pi/8) was a tad annoying...
    You missed a trick if you needed the value of tan(3pi/8).
    Notice that you're letting a=cos(3pi/4), use t substitutions with t=tan(3pi/8) and it all cancels down very quickly to 2arctan(1)
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    (Original post by DFranklin)
    Step II, Q3:

    F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.

    F0.F2-F1^2 = -1, F1.F3-F2^2 = 1, F2.F4-F3^2 = -1.

    Claim F_{n+1}F_{n-1}-F_n^2 = (-1)^n. Proof by induction on n.
    By above, true for n<=3. Assume true for n = k. Then
    F_{k+2}F_k - F_{k+1}^2 \\

= (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ 

= F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\

= [F_{k+1}F_{k-1}+(-1)^{k+1}]  - F_{k+1}(F_{k+1} - F_k) \\

= (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\

= (-1)^{k+1}F_{k+1}^2

    So true for n=k+1 and so true for all n by induction.

    Want to show F_{n+k} = F_kF_{n+1}+F_{k-1}F_n. Assume true for 1 < k <=m. Then
    F_{n+k+1} = F_{n+k}+F_{n+k-1}\\

= F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\

=(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\

=F_{k+1}F_{n+1}+F_k F_n.

    So true for k=m+1.

    Explictly when k=1 we have F_{n+1} = F_1F_{n+1}+F_0F_n = F_{n+1}. So true for k=1, so true for all k.
    In your final proof by induction the proof for k=m+1 assumes the result for k=m and k=m-1. Doesn't that mean that it can not be used to prove the result for k=2, since that would rely on the result for k=0 ,for which the result is undefined? Therefore you have to explicitly show the result to hold for k=1 and k=2 to complete the proof?
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    (Original post by Nick_)
    ..
    I explicitly show the result for n<=3 in the first few lines of the proof.
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    (Original post by DFranklin)
    I explicitly show the result for n<=3 in the first few lines of the proof.
    I meant the second proof by induction
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    Fair point.
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    (Original post by Rainingshame)
    For III Q2.
    You can show that those are the answers are correct with the exception of a=1,0. Here's how:
    The three equation can be rewritten as three matrices as shown below:
     \begin{bmatrix} 1 & 1 & a \\ 1 & a & 1 \\ 2 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 2b \end{bmatrix}
    We can now solve it because of the result
    Ainverse (A)R = Ainverse B
    which goes to :
    R = Ainverse B
    now this means that the equations have no solution when A is invertible.
    A is invertible when  det A =0
    so when  2a-2a^2 =0
    so when  a=0 or  a=1
    You mean non-invertible.

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