STEP maths I, II, III 1991 solutions

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  1. gyrase's Avatar
    41 30-05-2007  18:25
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    Re: STEP maths I, II, III 1991 solutions
    II/1, Let t = lambda, alpha = z

    h(x) = ax^2+bx+c = a(x^2+bx/a+c/a) = a((x+b/2a)^2-(b/2a)^2+c/a), so we require (b/2a)^2 = c/a, so b^2 = 4ac, i.e.discriminant = 0

    3x^2+4x+t(x^2-2) = (3+t)x^2+4x-2t where a=3+t, b=4,c=-2t and we require that 16 = -8t(3+t), 0 = 3t+t^2+2, t= -1, -2.

    (1) f(x)-g(x) = 2(x+1)^2
    (2) f(x)-2g(x) = (x+2)^2

    f(x) = 2x(1)-(2) = 4(x+1)^2 - (x+2)^2
    g(x) = (1)-(2) = 2(x+1)^2 - (x+2)^2, hence A = 4, B = -1, C = 2, D = -1, m = 1, n = 2.

    3x^2+4x+t(x^2+z) = (3+t)x^2+4x+zt, so a=3+t,b=4, c=zt, we require b^2 = 4ac, so 16 = 4(3+t)zt, 0 = 3zt+zt^2-4, a=z,b=3z,c=-4 then 9z^2 = -16z, z = -16/9.
  2. Rabite's Avatar
    42 30-05-2007  19:05
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    Re: STEP maths I, II, III 1991 solutions
    Ooh, very neat solution.

    Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
  3. khaixiang's Avatar
    43 30-05-2007  19:17
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Rabite)
    Ooh, very neat solution.

    Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
    Yeah, that's the answer I got, double-checked it with Maple just now. But I really can't be bothered to substitute in, maybe you plug it back wrongly?
  4. gyrase's Avatar
    44 30-05-2007  19:19
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Rabite)
    Ooh, very neat solution.

    Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
    I'm not too sure, I breifly did it and got u = e^-x and -(e^x+e^-x)
  5. khaixiang's Avatar
    45 30-05-2007  19:22
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Decota)
    I'm not too sure, I breifly did it and got u = e^-x and -(e^x+e^-x)
    That's the first part, Rabite is talking about the second part of the question.
  6. gyrase's Avatar
    46 30-05-2007  19:29
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by khaixiang)
    That's the first part, Rabite is talking about the second part of the question.
    oh i see. 2ln(cosh½x) is indeed correct.
    Last edited by gyrase; 30-05-2007 at 19:52.
  7. Rabite's Avatar
    47 30-05-2007  21:39
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    Re: STEP maths I, II, III 1991 solutions
    Well I'll type it up :p:

    u^2 + 2u \sinh x - 1 = 0
    The quadratic formula involves fractions, and I hate fractions, so let's complete the square:
    (u+\sinh x)^2 - 1 - \sinh ^2 x=0

    u+\sinh x= ± \cosh x

    u = \cosh x -\sinh x Or -(\cosh x+\sinh x)

    coshx-sinhx is uh...e^{-x} and -(coshx+sinhx) is -e^x.
    y`>0 at x=0 so we have to choose the + solution.

    \frac{dy}{dx} = e^{-x}
    Integrate and find the +C to get
    y = 1-e^{-x}.

    For the next bit, let u = dy/dx for a start:
    u^2 \sinh x + 2u - \sinh x =0
    u^2 + 2u cosech x - 1 =0
    (u+cosech x)^2 - 1 - cosech^2 x =0
    u+cosech x = ±\coth x
    u = \coth x - cosech x Or -(\coth x + cosech x) .

    \coth x - cosech x = \frac{\cosh x - 1}{\sinh x}
    = \tanh{½x} (Consider double "angle" formulae)
    But u = dy/dx:
    \frac{dy}{dx} = \tanh{½x}
    Tanh = sinh / cosh, which is a derivative over its function:
    y = 2 \ln \cosh ½ x +C

    The solution passes through (0,0); so C =0.

    (Taking the negative solution when we square rooted earlier results in something that can't pass through (0,0))

    [edit] Is there no /cosech in Latex?
    Last edited by Rabite; 30-05-2007 at 22:22.
  8. khaixiang's Avatar
    48 30-05-2007  22:06
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Rabite)

    [edit] Is there no /cosech in Latex?
    TSR's LaTeX is messed up then, the proper code is \csch
    Nice work btw, though "1" and "coth(x)" is missing somewhere in the middle. All the STEP III questions are so unbearably long and tedious Good thing the style was revamped.
  9. Rabite's Avatar
    49 30-05-2007  22:20
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    Re: STEP maths I, II, III 1991 solutions
    I tried \csch too

    Anyway. I think I've found the missing buggers, but maybe I've still missed one...
    I kept doing stuff like \sinhx which just didn't come out as anything.

    Thanks for that :p:
  10. nota bene's Avatar
    50 30-05-2007  22:26
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Speleo)
    Am I going crazy or are matrices no longer on the STEP syllabus?
    http://www.ocr.org.uk/Data/publicati...ficat98487.pdf
    *bawls* Matrices are easy:/


    (Original post by Speleo)
    Also, does anyone have the '05 papers? (II and III '06 are available if you go back a couple hundred pages to last June in this forum).
    Drop me a pm with your email, I think I have II and III from 2005, but lack I from 2005 and I from 2004, all others I think I have.
  11. Rabite's Avatar
    51 30-05-2007  22:29
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by nota bene)
    *bawls* Matrices are easy:/
    *fills a 4x4 Matrix's elements with 1=0 and drops it on nota*

    Maybe they just forgot to put it in the syllabus >>;;
    They're so tedious that they didn't deserve their own section or something.
  12. DFranklin's Avatar
    52 30-05-2007  22:49
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by nota bene)
    I'm attempting I/3 I'll see if I sort it out...


    Okay, I'll post what I have now (probably should scan a sketch tomorrow as well), but it is not finished, and likely contains a few mistakes
    Spoiler:
    Show

    Some observations I made was that if we let the first complex number have a length a the second will have a+ad etc. as long as we are in the first quadrant. This gives z_{n+1}=z_n+ad^{n-1}
    Maybe I'm missing something, but I don't see how this works; you seem to have gone directly from talking about the lengths of the vectors to their actual (complex) values. In particular, I don't understand how you're taking into account the value of theta.

    For what it's worth, I think the correct formula for the first bit is:
    Spoiler:
    Show
    \frac{z_{n+1}-z_n}{z_n-z_{n-1}} = de^{i\theta}
  13. DFranklin's Avatar
    53 30-05-2007  23:38
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Rabite)
    II, 2:
    I'm not entirely sure how to finish this one off, so someone else do that.
    We find the equations for the asymptotes meet at (1,-1), and your guess that the axes are at +/-45 degrees seems a good one.

    So we suspect that the axes have equations x+y=1, x-y =2. You could probably stop there, but partly to reassure myself, and partly for the hell of it, we can go further:

    We look for \alpha, \beta such that \alpha(x-y-2)^2-\beta(x+y)^2 resembles the given equation.

    Easiest is to look first at the coefficients of x and y, which tells us that \alpha = 4, and then the coefficient of x^2 tells us \beta = 1.

    Expanding out, we find 4(x-y-2)^2-(x+y)^2 = 3x^2-10xy+3y^2-16x+16y+16 and so using the given equation for the curve, we obtain (x+y)^2 - 4(x-y-2)^2 = 1.

    So we have explicitly obtained the equation of a hyperbola in terms of our new axes, confirming our guess.
    Last edited by DFranklin; 30-05-2007 at 23:50.
  14. generalebriety's Avatar
    54 31-05-2007  00:32
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    Re: STEP maths I, II, III 1991 solutions
    Someone let me know when stuff's finished. :p:
  15. nota bene's Avatar
    55 31-05-2007  01:25
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by DFranklin)
    Maybe I'm missing something, but I don't see how this works; you seem to have gone directly from talking about the lengths of the vectors to their actual (complex) values. In particular, I don't understand how you're taking into account the value of theta.

    For what it's worth, I think the correct formula for the first bit is:
    Spoiler:
    Show
    \frac{z_{n+1}-z_n}{z_n-z_{n-1}} = de^{i\theta}
    Yes, I have r^ncis(n\theta) towards the end, but it needs to be mentioned in the beginning. In general that whole post is totally messed up and should not be looked at:p:.

    Someone else feel free to do this question, or I'll try to get it correct later.

    generalebriety: I think my solution to I/6 is okay unless I've made a stupid mistake somewhere. (added Decota's comment, just to be on the safe side; I had seen it was not differentiated correctly, but not bothered to mention it as it was totally wrong either way.) Also take away this question as 'partially solved' until I fix it; it is horrible at the moment!
  16. justinsh's Avatar
    56 31-05-2007  03:26
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    Re: STEP maths I, II, III 1991 solutions
    STEP I question 8

    (a) Let y=\pi-x \Rightarrow dy=-dx.

    \displaystyle I=\int_0^\pi xf(\sin x)dx\\
    \displaystyle=-\int_\pi^0 (\pi-y) f(\sin y)dy\\
    \displaystyle=\pi\int_0^\pi f(\sin x)dx-\int_0^\pi xf(\sin x)dx\\
    \displaystyle=\pi\int_0^{\pi/2} f(\sin x)dx+\pi\int_{\pi/2}^\pi f(\sin x)dx-I\\
    \displaystyle=\pi\int_0^{\pi/2} f(\sin x)dx-\pi\int_{\pi/2}^0 f(\sin y)dy-I\\
    \displaystyle=\pi\int_0^{\pi/2} f(\sin x)dx+\pi\int_0^{\pi/2} f(\sin x)dx-I\\
    \displaystyle=2\pi\int_0^{\pi/2} f(\sin x)dx-I\\
    \displaystyle\therefore I=\pi\int_0^{\pi/2} f(\sin x)dx

    A shorter alternative due to DFranklin:

    \displaystyle I = \int_0^{\pi/2} x f(\sin x)dx + \int_{\pi/2}^\pi x f(\sin x)dx
    \displaystyle = \int_0^{\pi/2} x f(\sin x)dx -\int_{\pi/2}^0 (\pi-y)f(\sin(\pi-y))dy (by substitution)
    \displaystyle = \int_0^{\pi/2} (x+(\pi-x)) f(\sin x)dx (recombining)
    \displaystyle = \pi \int_0^{\pi/2} f(\sin x)dx

    Using this last result:

    \displaystyle J=\int_0^\pi \frac{x}{2+\sin x}dx\\
    \displaystyle =\pi\int_0^{\pi/2} \frac{1}{2+\sin x}dx\\

    Let \displaystylet=\tan\frac{x}{2} \Rightarrow dt=\frac{1}{2}(1+\tan^2\frac{x}{ 2})dx=(1+t^2)dx. Also, \displaystyle\sin x=\frac{2t}{1+t^2}. Hence:

    \displaystyle J=\pi\int_0^1 \frac{1}{2+\frac{2t}{1+t^2}}\fra c{2dt}{t^2+1}\\
    \displaystyle=\pi\int_0^1 \frac{dt}{t^2+t+1}
    \displaystyle=\pi\int_0^1 \frac{dt}{(t+\frac{1}{2})^2+\fra c{3}{4}}
    \displaystyle=\pi\int_0^1 \frac{dt}{(\frac{2t}{\sqrt{3}}+\ frac{1}{\sqrt{3}})^2+1}
    \displaystyle=\pi\frac{2}{\sqrt{ 3}}\int_{1/\sqrt{3}}^{3/\sqrt{3}} \frac{du}{u^2+1}
    \displaystyle=\pi\frac{2}{\sqrt{ 3}}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)
    \displaystyle=\frac{\sqrt{3}\pi^ 2}{9}

    (b) Let \displaystyle u=(\sin^{-1}t)^2 \Rightarrow du=\frac{2\sin^{-1}t}{\sqrt{1-t^2}}

    Boundaries: \displaystyle(\sin^{-1}0)^2=0, (\sin^{-1}1)^2=\left(\frac{\pi}{2}\right )^2=\frac{\pi^2}{4}

    \displaystyle\int_0^1\frac{(\sin ^{-1}t)\cos[(\sin^{-1}t)^2]}{\sqrt{1-t^2}}dt\\
    \displaystyle=\frac{1}{2}\int_0^ {\pi^2/4}\cos u du=\frac{1}{2}\sin\frac{\pi^2}{4 }
    Last edited by justinsh; 31-05-2007 at 08:23.
  17. DFranklin's Avatar
    57 31-05-2007  07:45
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by justinsh)
    My part (a) is rather long, is there a shorter way?
    I = \int_0^{\pi/2} x f(\sin x) \,dx + \int_{\pi/2}^\pi x f(\sin x) \,dx. Sub y = \pi-x in the 2nd integral to get

    I = \int_0^{\pi/2} x f(\sin x) \,dx -\int_{\pi/2}^0 (\pi-y)f(\sin(\pi-y))\,dy. Recombine the two integrals to get

    I = \int_0^{\pi/2} (x+(\pi-x)) f(\sin x) \, dx

    I = \pi \int_0^{\pi/2} f(\sin x) \, dx
  18. justinsh's Avatar
    58 31-05-2007  08:27
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    Re: STEP maths I, II, III 1991 solutions
    STEP I questions 4

    Please tell me what you think, I'm not certain of what I did.
    Attached Thumbnails
    Click image for larger version.  Name: 2.jpeg  Views: 88  Size: 336.9 KB  ID: 40650   Click image for larger version.  Name: 1.jpeg  Views: 85  Size: 261.5 KB  ID: 40651  
    Last edited by justinsh; 02-06-2007 at 06:48.
  19. gyrase's Avatar
    59 01-06-2007  21:47
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    Re: STEP maths I, II, III 1991 solutions
    *Bump*

    I will rep anyone who posts a solution for II/4 in its entirety. (the trigonometry one)

    kthnx.
    Last edited by gyrase; 01-06-2007 at 21:48.
  20. Rabite's Avatar
    60 01-06-2007  22:10
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    Re: STEP maths I, II, III 1991 solutions
    Hmm...
    I can do the first bit and the last bit.

    But not the middle bit (unless we resort to complex numbers, but that's foul play right?)
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