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STEP Maths I, II, III 1989 solutions

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    (Original post by DFranklin)
    Well, you can't really integrate that without ending up doing what Brian did.

    Let's write v(x) for v, to emphasise that v is a function of x.

    Then you have

    \frac{1}{2}v(0)^2 - \frac{1}{2}v(d)^2 = \int_0^d \frac{\omega (v(x)^2 +V^2)}{v(x)}dx

    So, here's the problem - at this point you have absolutely no idea what v(x) is, which is going to make finding the integral pretty tricky.

    So, let's relabel some of the variables: I'm going to write 'x' instead of d, and 't' instead of 'x'.

    \frac{1}{2}v(0)^2 - \frac{1}{2}v(x)^2 = \int_0^d \frac{\omega (v(t)^2 +V^2)}{v(t)}dt

    Now diff both sides w.r.t. x:

    v(x)\frac{dv}{dx} =   \frac{\omega (v(x)^2 +V^2)}{v(x)}

    Which is Brian's first line. So from here you can solve for v as a function of x, which will give you the desired result.

    Obviously it's quicker to just use the v dv/dx form of acceleration to start with.
    Thanks, I sort of get what you mean. My thinking was that v is the derivative of x so if we integrated something with v in it we would get something in terms of x.

    I´ll think it over.
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    You can integrate v w.r.t. x, sure. Integrating 1/v is going to be tricky...
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    (Original post by DFranklin)
    You can integrate v w.r.t. x, sure. Integrating 1/v is going to be tricky...
    Yes, that is exactly where I was running into problems.
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    STEP 1 question 8
    A solution has allready been posted using the suggested way but the question let you solve it any other way (apart from newton rapheson et al). This is the way you could do it if , like me, you haven´t done De moivre´s theorem in FP2.
    (I am using a spanish keyboard which makes typing in latex a nightmare so, for my own ease, I have let theta=x)
    LHS=cos(4x)=cos(2x+2x)=cos^2(2x)-sin^2(2x)
    using the identities:
     sin^2(2x)=1-cos^2(2x)

cos2x=2cos^2x-1
    We get:
    2(2cos^2x-1)^2-1=8cos^4-8cos^2+1 as required.

    Next part:
    cos6x=cos(3x+3x)
    By standard results:
    cos3x=4cos^3x-3cosx
    So, we do the same as in the previous part:
    2cos^23x-1=2(16cos^6x-24cos^4x+9cos^2x)-1

=32cos^6x-48cos^4x+18cos^2x-1 as required.


    Next part:
     16x^{6} - 28x^{4} + 13x^{2} -1 =0
    Whenever they give you questions like this, it is worth checking for easy roots. This one has two, x=1 and, noticing that all the powers are even, x=-1.
    We can then factorise it to get:
    16x^{6} - 28x^{4} + 13x^{2} -1 =(x+1)(x-1)(16x^4-12x^2+1)=0
    (16x^4-12x^2+1)=0 is a quadratic in terms of x^2 so all you have to do is use the quadratic formula and you are done.
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    (Original post by Farhan.Hanif93)
    STEP I - Q1
    solution
    first part
    Note that the radius of CDE is r-2h. Therefore the length HD is given by r+(r-2h) = 2r-2h.

    Considering the area of a circle of this diameter:
    A=\pi (r-h)^2.

    Consider the areas of the following semi-circles-
    AGH: A_1=\frac{1}{2}\pi r^2
    ABC and EFG: A_2 = \frac{1}{2}\pi h^2
    CDE: A_3 = \frac{1}{2}\pi (r-2h)^2

    Note that the area of ABCDEFGH is found by considering the value of A_1 - 2A_2 + A_3 = \frac{\pi}{2}(r^2+r^2-4hr+4h^2-2h^2) = \pi (r^2-2rh+h^2) = \pi (r-h)^2. Therefore, since A = A_1-2A_2+A_3 = \pi (r-h)^2, the area enclosed by ABCDEFGH is equal to the area of a circle of diameter HD, as required.

    second part
    We now seek to find the equations of the 4 parabolas and then use integration to find the areas.
    Consider an (x,y) coordinate system with an origin at A.
    The equations of the following parabolas are:
    AHG: y=x(2r-x)
    ABC: y=x(2h-x)

    Considering the area enclosed between these parabolae and the positive x-axis:
    Area under parabola AGH = I_1 = \int^{2r}_0x(2r-x)dx = \frac{4}{3}r^3.
    Area under parabola ABC = Area under parabola EFG = I_2=\int^{2h}_0x(2h-x)dx = \frac{4}{3}h^3

    Now consider another coordinate system with an origin at the midpoint of the line CE.
    Area enclosed by parabola CDE and the x-axis = I_3 = \int^{r-2h}_{2h-r}(x-(r-2h)(x+(r-2h))dx = -\frac{4}{3}(r-2h)^2

    Therefore area of ABCDEFGH = I_1-2I_2-I_3. (We subtract I_3 because it has a negative area using our coordinate systems and we're looking for an absolute value of the area enclosed by ABCDEFGH)
    Therefore, we have the following, where A_p is the area of this shape:
    A_p=\frac{4}{3}(r^3-2h^3 + (r-2h)^3) = \boxed{\frac{8}{3}(r^3-3r^2h+6rh^2-5h^3)}
    Just did the question, and got different answers to you... I diagree with you about the equations of the parabola. For example, the equation for your AHG: y=x(2r-x) does not necessary represent our curve. Although the roots are correct, the equation can be multiplied by a constant a while still holding the two roots to be true. Then we can find the actual value of a using the co-ordinates of the maximum or minimum of the parabola. Your equations just assumes that a=1, right? (correct me if I am wrong...I have just started looking at step)

    The final answer I got was, A_p=\frac{2}{3}(2r-2h)^2. This fitted with my answer for the first part, as (2r-2h), is the length HD.

    Slighty of topic, but do you think that we would not lose marks for just stating the equations of the parabolas/ having similar "jumps" in our working for STEP? Is the marking really strict, and do we get our papers back?
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    (Original post by brianeverit)
    1989 Paper 3 nos. 3-7
    Sorry but I don't understand a line in your solution to STEP III Q3.
    In it, you say:
    "if P=M then X_k=M^kX_0+Q.O=M^kX_0 \Rightarrow X_m=X_0
    But M^{k-1}+M^{k-2}+...+M^2+M+I=O when k=m not for any value of k which is what you seem to have assumed. Am I reading it wrongly?
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    (Original post by ben-smith)
    Sorry but I don't understand a line in your solution to STEP III Q3.
    In it, you say:
    "if P=M then X_k=M^kX_0+Q.O=M^kX_0 \Rightarrow X_m=X_0
    But M^{k-1}+M^{k-2}+...+M^2+M+I=O when k=m not for any value of k which is what you seem to have assumed. Am I reading it wrongly?
    M^m=I and so X^m=M^mX_0=X_0 for any m
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    Here are my attempts at STEP III numbers 11 and 12.
    I'm not at all confident of them so would welcome comments or criticisms.
    Attached Files
  1. File Type: pdf 1989.III.11.pdf (111.4 KB, 51 views)
  2. File Type: pdf 1989.III.12.pdf (53.7 KB, 36 views)
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    Here is my solution to 1989 Step 3 Question7
    Attached Files
  3. File Type: pdf Step1989Paper3Question7.pdf (68.9 KB, 51 views)
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    (Original post by squeezebox)
    STEP II Q1

    

\cos(3\theta) = \cos(2\theta + \theta) = \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) = (2cos^{2}(\theta) - 1)\cos(\theta) - 2\sin(\theta)\cos(\theta)sin(\th  eta) = 2cos^{3}(\theta) - \cos(\theta) - 2(1-\cos^{2}(\theta))\cos(\theta) = 4cos^{3}(\theta) - 3\cos(\theta).

    Substituting x= y-a into the equation:

     24x^{3} - 72x^{2} + 66x -19 = 24(y-a)^{3} - 72(y-a)^{2} + 66(y-a) -19 = 0

    Notice that the reduced form shown in the question has no z^{2} term, so we need to find the value of a which will get rid of the y^{2} term.

    Expanding out and collectiing terms we get:

     24y^{3} - 72(a+1)y^{2} + (72a^{2} + 144a +66)y - 24a^{3} - 72a^{2} -66a -19 = 0

    So the value of a to get rid of the y^{2} is -1.

    So we now have:

     24y^{3} - 72(a+1)y^{2} + (72a^{2} + 144a +66)y - 24a^{3} - 72a^{2} -66a -19 = 24t^{3} - 6y -1 = 0

     \Rightarrow 24y^{3} - 6y = 1 (*)

    Substituting y=z/b into (*);

    24(\frac{z}{b})^{3} - 6(\frac{z}{b}) = 1

    dividing both sides by 2 and multiplying by b;

    12\frac{z^{3}}{b^{2}} - 3z = \frac{b}{2} .

    And so to make this equal to;

    4z^{3}- 3z = \frac{b}{2}

    we make b = \pm \sqrt3.

    Lets take b = \sqrt3 and let z = \cos(\theta)

    so;

     4z^{3}- 3z = 4\cos^{3}(\theta) - 3\cos(\theta) = \cos(3\theta) = \frac{\sqrt3}{2}

    \Rightarrow 3\theta = \frac{pi}{6}, \frac{11\pi}{6} and  \frac{13\pi}{6} (these give distinct values of cos\theta, and hence, three distinct solutions.)


    we know that; x = y - a = (z/b) - a.

    Hence the solutions of the equation are:

     x = \frac{\cos(\frac{\pi}{18})}{\sqr  t3} + 1 , \frac{\cos(\frac{11\pi}{18})}{\s  qrt3} +1 and  \frac{\cos(\frac{13\pi}{18})}{\s  qrt3} + 1


    The question also wanted a counterexample to show that not all cubics can be solved by this method and trivially
    x^3-1=0 is such a counterexample
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    (Original post by Dystopia)
    STEP III, Q8

    \frac{\mathrm{d}^{2}x}{\mathrm{d  }t^{2}} = 4\frac{\mathrm{d}x}{\mathrm{d}t} - 4\frac{\mathrm{d}y}{\mathrm{d}t}

    \frac{\mathrm{d}^{2}x}{\mathrm{d  }t^{2}} - 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 48e^{2t} + 48e^{-2t}

    Let u = \frac{\mathrm{d}x}{\mathrm{d}t} - 2x

    Then \frac{\mathrm{d}u}{\mathrm{d}t} - 2u = 48e^{2t} + 48e^{-2t}

    Using an integrating factor, e^{-2t}, we get

    ue^{-2t} = \int 48 + 48e^{-4t} \; \mathrm{d}t = 48t - 12e^{-4t} + c

    u = e^{2t}(48t + c) - 12e^{-2t}

    \frac{\mathrm{d}x}{\mathrm{d}t} - 2x = e^{2t}(48t + c) - 12e^{-2t}

    xe^{-2t} = \int 48t + c - 12e^{-4t} \; \mathrm{d}t = 24t^{2} + ct + 3e^{-4t} + d

    x = e^{2t}(24t^{2} + ct + d) + 3e^{-2t}

    Applying boundary conditions:

    0 = d + 3 \Rightarrow d = -3

    \frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t}(24t^{2} + ct + d) + e^{2t}(48t + c) - 6e^{-2t} = 4(x-y)

    2d + c - 6 = 0 \Rightarrow c = 6 - 2d = 6 + 6 = 12

    x = e^{2t}(24t^{2} + 12t - 3) + 3e^{-2t}

    4y = 4x - \frac{\mathrm{d}x}{\mathrm{d}t} = e^{2t}(48t^{2} - 4t - 18) + 18e^{-2t} = 4te^{2t}(12t - 1) - 36\sinh 2t

    y = te^{2t}(12t - 1) - 9\sinh 2t

    ---

    I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.
    Is there a slight error in your result for y, I make
    y=\left(12t^2-24t-\dfrac{9}{2} \right)\text{e}^{-2t}-9\sinh 2t
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    Here's my solution for Q10, STEP I 1989.
    Sorry for dodgy presentation, I am practicing using my new graphics tablet
    I realise that LaTeX would be easier to read

    If I have understood this question, I don't like it at all... not a good STEP question.
    On the other hand I may be wrong in some important way, if so please correct!
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    Here is my solution for Q11 STEP II 1989. (sorry I called it Q10 in the pic!)
    I did it because I thought it hadn't been done (because I thought it was Q10, doh!), but anyway you might find my slightly simpler method interesting.

    It's a very easy question I think, although it helps a lot to consider relative acceleration.

    One last comment: the second part (why the lift comes to rest) I described as "obvious" .. do you see why?
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    Here is my solution to STEP II 1989 Q12.

    Quite easy using Lami's theorem; I suspect the first part might be a bit messy otherwise.
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    Here's my answer for STEP III 1989 Q12.
    Note to Brian Everit: I was pleased to see after I did it that we modeled it almost exactly the same way.
    The only difference is in the end; my reading of the question suggested they didn't want an explicit solution for \phi but just to deduce from the form of \frac{\textrm{d}\phi}{\textrm{d}  t}.

    But however you look at it, I think you did a great job (That's assuming neither of missed something..)
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    Here is my answer for STEP II 1989 Q4.

    After comparing with Brian Everit's solution on post 65, I see that we have exactly the same answer. Only thing different is I have a quicker way of doing the first part.

    Not the best question ever - asking students to "tell me how you would prove it, but don't prove it" and then asking them to draw 4 different quadratic/quadratic graphs is pretty evil...
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    (Original post by waxwing)
    Here is my answer for STEP II 1989 Q4.

    After comparing with Brian Everit's solution on post 65, I see that we have exactly the same answer. Only thing different is I have a quicker way of doing the first part.

    Not the best question ever - asking students to "tell me how you would prove it, but don't prove it" and then asking them to draw 4 different quadratic/quadratic graphs is pretty evil...
    Well done finding that method for the first part. I wish I had thought of it myself.
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    (Original post by brianeverit)
    Well done finding that method for the first part. I wish I had thought of it myself.
    Yes, it is clever isn't it, except for the tiny detail that I've just realised it doesn't work! :mad: :mad:
    I can't take  \ln{(f(x))} if f(x) is negative!
    Well, I guess I can if I'm allowing complex answers.
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    (Original post by waxwing)
    Yes, it is clever isn't it, except for the tiny detail that I've just realised it doesn't work! :mad: :mad:
    I can't take  \ln{(f(x))} if f(x) is negative!
    Well, I guess I can if I'm allowing complex answers.
    I don't have the question in front of me, but if you're only interested in the behaviour when |x| is large then f(x) won't be negative then.
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    (Original post by DFranklin)
    I don't have the question in front of me, but if you're only interested in the behaviour when |x| is large then f(x) won't be negative then.
    Thanks for the suggestion, it would be OK if it were so, but unfortunately the first part of this monstrously long question asks to prove that \frac{xf'(x)}{f(x)}= \ldots , given only that a,b,c,d are real and x not equal to c and d. Only then does it say "... and deduce that if |x| is large..."

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