STEP I, II, III 2002 Solutions

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  1. ben-smith's Avatar
    181 26-05-2012  09:27
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Rahul.S)
    you did that ques when you were in y12 :O
    good times
    back before I had to worry about this university business.
  2. Rahul.S's Avatar
    182 26-05-2012  12:13
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    Re: STEP I, II, III 2002 Solutions
    (Original post by ben-smith)
    good times
    back before I had to worry about this university business.
    :lol: dont tell me you have done most of the TRIPOS already -_-
  3. mikelbird's Avatar
    183 03-06-2012  10:12
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    Re: STEP I, II, III 2002 Solutions
    (Original post by z0tx)
    I find the solution above relies less on guessing and is closer to the techniques hinted at in the exercise than the other one posted on this thread.
    I think this solution is actually more in the spirit of the question....
    Attached Files
  4. File Type: pdf Step2002Paper1Question5.pdf (59.3 KB, 21 views)
  5. z0tx's Avatar
    184 03-06-2012  13:26
    • Exalted and Worshipped Member
    Re: STEP I, II, III 2002 Solutions
    (Original post by mikelbird)
    I think this solution is actually more in the spirit of the question....
    Hm... Yes you're right.
  6. desijut's Avatar
    185 04-06-2012  13:03
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Dadeyemi)
    Some more;

    Did these quite a while ago I'm afraid some may be partial solutions.
    on q2, the last part says "hence" what i did was spot that, it was the same as the sum above it but with r=0 which led me straight to pi/2
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  7. adrienne_om's Avatar
    186 04-06-2012  19:38
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    Re: STEP I, II, III 2002 Solutions
    (Original post by toasted-lion)
    Question 5, I'll pick up where Dadeymi left off:

    So \left(ab^2 + (a-1)b +(a-1)\right)=0

    \implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} - \frac{a^2 - 2a +1}{4a^2}

    ]
    Above from --- http://www.thestudentroom.co.uk/show...1#post19333071

    Maybe a correction here -- I'm not sure -- but I got a positive sign in the RHS of the second equation ---

    ab^2 + (a-1)b +(a-1)=0

    \iff b^2 + \dfrac{a-1}{a}b = \dfrac{1 - a}{a}

    \iff b^2 + \dfrac{a-1}{a}b + \left(\dfrac{a - 1}{2a})^2\right = \dfrac{1 - a}{a} + \left(\dfrac{a - 1}{2a}\right)^2

    \iff \left( b + \dfrac{a-1}{2a} \right)^2 = \dfrac{1-a}{a} + \dfrac{a^2 - 2a +1}{4a^2}
    Last edited by adrienne_om; 04-06-2012 at 19:42.
  8. brianeverit's Avatar
    187 11-06-2012  09:14
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    Re: STEP I, II, III 2002 Solutions
    (Original post by adrienne_om)
    Above from --- http://www.thestudentroom.co.uk/show...1#post19333071

    Maybe a correction here -- I'm not sure -- but I got a positive sign in the RHS of the second equation ---

    ab^2 + (a-1)b +(a-1)=0

    \iff b^2 + \dfrac{a-1}{a}b = \dfrac{1 - a}{a}

    \iff b^2 + \dfrac{a-1}{a}b + \left(\dfrac{a - 1}{2a})^2\right = \dfrac{1 - a}{a} + \left(\dfrac{a - 1}{2a}\right)^2

    \iff \left( b + \dfrac{a-1}{2a} \right)^2 = \dfrac{1-a}{a} + \dfrac{a^2 - 2a +1}{4a^2}
    \text{I agree with you. So we finally have }

    b=\dfrac{1}{2a}\left[1-a\pm \sqrt{(a-1)^2-4a(a-1)} \text{ or }k=\dfrac{b}{1-a}=\dfrac{1}{2a} \pm\dfrac{\sqrt{1+2a-3a^2}}{2a(1-a)}
  9. Xero Xenith's Avatar
    188 11-06-2012  20:17
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    Re: STEP I, II, III 2002 Solutions
    Seems Paper II Q1 has a bad solution here:

    http://www.thestudentroom.co.uk/show...9#post18023769

    No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

    The last few lines should be

    Part Two
    \begin{array}{rl} \therefore \displaystyle\int_{\frac{\sqrt3} {2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 \theta}} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos^2 - 2\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\ & \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\ & \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}


    The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

    EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
    Last edited by Xero Xenith; 11-06-2012 at 21:00.
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  10. DFranklin's Avatar
    189 11-06-2012  21:03
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Xero Xenith)
    I think the result only holds for complex numbers modulus 1.
    Not even for all numbers of modulus 1. e.g. z = i, n = 2.
  11. sonofdot's Avatar
    190 12-06-2012  11:12
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Xero Xenith)
    Seems Paper II Q1 has a bad solution here:

    http://www.thestudentroom.co.uk/show...9#post18023769

    No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

    The last few lines should be

    Part Two
    \begin{array}{rl} \therefore \displaystyle\int_{\frac{\sqrt3} {2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 \theta}} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos^2 - 2\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\ & \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\ & \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}


    The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

    EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
    Good spot, fixed
  12. Extricated's Avatar
    191 12-06-2012  14:28
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Xero Xenith)
    Seems Paper II Q1 has a bad solution here:

    http://www.thestudentroom.co.uk/show...9#post18023769

    No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

    The last few lines should be

    Part Two
    \begin{array}{rl} \therefore \displaystyle\int_{\frac{\sqrt3} {2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 \theta}} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{2\cos^2 - 2\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\ & \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\ & \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\ & \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}


    The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

    EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
    Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally. Ofcourse doesn't affect final answer, but could lead to a little confusion
    Last edited by Extricated; 12-06-2012 at 14:38.
  13. Xero Xenith's Avatar
    192 12-06-2012  14:35
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Extricated)
    Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally.
    Good spot yourself, pretty sure you're right!


    (Original post by sonofdot)
    Good spot, fixed
    Seems like there's still just a minor thing on the third line (the one with the three dots) - square root sign should be 1- \cos^2 2 \theta not just 1- \cos^2 \theta.
  14. Rahul.S's Avatar
    193 12-06-2012  15:09
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Extricated)
    Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally. Ofcourse doesn't affect final answer, but could lead to a little confusion
    nah you are being stupid here :lol:
  15. yukki0822's Avatar
    194 22-02-2013  18:30
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Glutamic Acid)
    II/7: (Scary scary vectors.)
    ??????????

    Let the lines have direction vector \left( \begin{array}{c} a \\ b \\ c \end{array} \right) which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;

    \Rightarrow \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) . \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \sqrt{2} \cos \dfrac{\pi}{4} \Rightarrow a + b = 1

    \Rightarrow \left( \begin{array}{c} a \\ b \\ c \end{array} \right) . \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right) = \sqrt{2} \cos \dfrac{\pi}{4} \Rightarrow a + c = 1.

    b = 1 - a; c = 1 - a so a^2 + (1 - a)^2 + (1 - a)^2 = 1 \Rightarrow (3a - 1)(a - 1) = 0
    The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.

    m3 has direction \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right); m4 direction \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right).

    Considering \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) . \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right) gives \cos \theta = 1/3.

    (i) A has position vector \left( \begin{array}{c} \lambda \\ \lambda \\ 0 \end{array} \right); B \left( \begin{array}{c} \lambda \\ 0 \\ \lambda \end{array} \right).
    Let P = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right), and Q = \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right)

    AQ . BP = 0 so \left( \begin{array}{c} \lambda - 1/3 \\ \lambda - 2/3 \\ -2/3 \end{array} \right) . \left( \begin{array}{c} \lambda - 1 \\ 0 \\ \lambda \end{array} \right) = 0

    (\lambda - 1)(\lambda - 2/3) - 2/3 \times \lambda = 0 \text{giving} \lambda = 1 \pm \dfrac{\sqrt{6}}{3}. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.

    (ii) AQ = \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right) + \alpha \left( \begin{array}{c} \lambda - 1/3 \\ \lambda - 2/3 \\ -2/3 \end{array} \right)

    BP = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) + \beta \left( \begin{array}{c} \lambda - 1 \\ 0 \\ \lambda \end{array} \right)

    So 1/3 + \alpha(\lambda - 1/3) = 1 + \beta(\lambda - 1)
    2/3 + \alpha(\lambda - 2/3) = 0
    2/3 - \alpha 2/3 = \lambda \beta

    The second equation gives \alpha = \dfrac{2}{2 - 3 \lambda}. Substituting into the third gives \beta = \dfrac{2}{2 - 3 \lambda}. Substituting these into the first:
    1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1); multiplying out gives 2 \lambda = 0 \Rightarrow \lambda = 0, so no non-zero solutions.
    sorry im really confused that why does the modulus of r equal to 1?
  16. BabyMaths's Avatar
    195 22-02-2013  20:03
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    Re: STEP I, II, III 2002 Solutions
    (Original post by yukki0822)
    sorry im really confused that why does the modulus of r equal to 1?
    We're only interested in the direction.
  17. DJMayes's Avatar
    196 17-03-2013  21:33
    • Overlord in Training
    Re: STEP I, II, III 2002 Solutions
    For II, Q10, I believe there's an error in the final part - the -\dfrac{29}{2} should have an x attached, which makes the final answer 81/40. I also get this answer through an entirely different approach to the question:

    Spoiler:
    Show


    13t + (T-t)(\dfrac{14T+2t}{T}) = \dfrac{339}{8}

    13Tt + 14T^2+2Tt-14Tt-2t^2 = \dfrac{339}{8}T

    14T^2 + Tt - 2t^2 = \dfrac{339}{8}T

    Differentiate implicitly:

    28T\dfrac{dT}{dt}+T+t\dfrac{dT}{ dt}-4t = \dfrac{339}{8}\dfrac{dT}{dt}

    When T is stationary, \dfrac{dT}{dt} = 0 \Rightarrow T - 4t = 0 , so t = \dfrac{1}{4}T

    Plug this back into the original equation:

    14T^2+\dfrac{1}{4}T^2-\dfrac{1}{8}T^2 = \dfrac{339}{8}T

    113T^2 - 339T = 0

    This gives maxima/minima at T = 0 and T = 3. T = 0 is obviously impossible, so T must equal 3, as required.

    For the second part, use a trapezium to calculate the final speed of the second competitor, and use this to work out the acceleration:

    \frac{3}{2}(16+v) = \frac{339}{8} \Rightarrow V = \frac{49}{4}

    And so the acceleration is then -1.25 kph.

    T = 3 \Rightarrow t = 0.75 , which means that the speed in the second part is 14.5. Next, find the time when they are moving at equal speed, as this is when maximum displacement occurs:

    16-1.25t = 14.5 \Rightarrow t = 1.2

    This is greater than 0.75 so is fine. Then, calculate the distances for each:

    x_1 = 16(1.2)-\frac{5}{8}(1.2)^2 = \frac{183}{10}

    x_2 = 0.75(13)+0.45(14.5) = \frac{651}{40}

    Subtracting one from the other yields the final answer \frac{81}{40}

    Last edited by DJMayes; 17-03-2013 at 21:36.
  18. Generic Name's Avatar
    197 4 Weeks Ago
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    Re: STEP I, II, III 2002 Solutions
    (Original post by SimonM)
    STEP I, Question 4

    Spoiler:
    Show


    \displaystyle y' = - \frac{2x}{(1+x^2)^2}

    Therefore the equation of the tangent is \displaystyle y = - \frac{2a}{(1+a^2)^2} x + \frac{1}{1+a^2} +\frac{2a^2}{(1+a^2)^2}

    Since (0,1) is a solution

    \displaystyle 1 = \frac{1}{1+a^2} +\frac{2a^2}{(1+a^2)^2} \Rightarrow 1+2a^2+a^4 = 1+a^2 + 2a^2 \Rightarrow a^4 -a^2 = 0 \Rightarrow a \in \{-1,0,1\} \Rightarrow a = 1

    This gives us the equation, \displaystyle y = -\frac{1}{2} x+1

    We have \displaystyle 1 = \left (1 - \frac{1}{2} x \right) (1+x^2) \Rightarrow 0 = x(x-1)^2 so the only intersections are 0 and 1.

    We have

    \displaystyle \int_0^1 y \, dx = \frac{\pi}{4} (given)

    The area under the tangent is a trapezium with area

    \displaystyle \frac{1}{2} \left ( 1+ \frac{1}{2} \right ) \cdot 1 = \frac{3}{4}

    Since the graph is always above it, we get \frac{3}{4} < \frac{\pi}{4} which is what we want to show.

    Considering the half closest to the y axis, and revolving it around that axis we get:

    \displaystyle \pi \int_{1/2}^1 x^2 \, dy = \pi \left [ \ln y - y\right ]_{1/2}^1 = \pi \left ( \ln 2 - \frac{1}{2} \right )
    and

    \displaystyle \pi \int_{1/2}^1 (2(y-1))^2\, dy = 4\pi \left [ \frac{y^3}{3} \right ]_{-1/2}^0 = \pi \frac{1}{6}

    Therefore \ln 2 > \frac{1}{2} + \frac{1}{6} = \frac{2}{3}
    Does this graph have an oblique asymptote?
  19. hypercube's Avatar
    198 3 Weeks Ago
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    Re: STEP I, II, III 2002 Solutions
    (Original post by Glutamic Acid)
    II/8:

    y = e^{3 + x - e^x} \text{for} x < 0
    I think this should be:

    y = e^{2e -1 + x - e^x} ​ instead?
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