STEP 2006 Solutions Thread

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  1. SimonM's Avatar
    1 05-05-2009  21:31
    • TSR Idol
    • Posts: 9,204
    STEP 2006 Solutions Thread
    Last edited by SimonM; 04-08-2011 at 21:19.
  2. Daniel Freedman's Avatar
    2 05-05-2009  22:17
    • Overlord in Training
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    Re: STEP 2006 Solutions Thread
    STEP III 2006, Question 2

    Spoiler:
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    i) \displaystyle I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{1 - \sin{\theta} \sin{2 \alpha}} \ \mathrm{d} \theta

    Let \theta = - \psi \implies \mathrm{d} \theta = - \mathrm{d}\psi

    \\ \displaystyle \therefore I = - \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\cos^2{{(-\psi)}}}{1 - \sin{{(-\psi)}} \sin{2 \alpha}} \ \mathrm{d} \psi = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{{(\psi)}}}{1 + \sin{(\psi)} \sin{2 \alpha}} \ \mathrm{d} \psi = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{1 + \sin{\theta} \sin{2 \alpha}} \ \mathrm{d} \theta , as theta and psi are just dummy variables, and sine is odd and cos is even.

    \\ \displaystyle \therefore 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{1 - \sin{\theta} \sin{2 \alpha}} + \frac{\cos^2{\theta}}{1 + \sin{\theta} \sin{2 \alpha}} \ \mathrm{d} \theta \\ \\ \\ = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{ 2 \cos^2{\theta}}{1 - \sin^2{\theta}\sin^2{2\alpha}} \ \mathrm{d} \theta = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2}{\sec^2{\theta} - \tan^2{\theta}\sin^2{2\alpha}} \ \mathrm{d} \theta \\ \\ \\ = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2}{1 + \tan^2{\theta} - \tan^2{\theta}(1 - \cos^2{2\alpha})} \mathrm{d}\theta \\ \\ \\ = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2}{1+\tan^2{\theta}\cos^2{ 2\alpha}} \mathrm{d} \theta

    as required.

    ii) \displaystyle J = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\sec^2{\theta}}{1+\tan^2{\ theta}\cos^2{2 \alpha}} \mathrm{d} \theta = \sec^2{2 \alpha} \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\sec^2{\theta}}{\sec^2{2 \alpha} + \tan^2{\theta}} \mathrm{d} \theta

    Let u = \tan{\theta} \implies \mathrm{d}\theta = \frac{\mathrm{d}u}{\sec^2{\theta }}

    \\ \displaystyle \therefore J = \sec^2{2 \alpha} \int_{-\infty}^{\infty} \frac{1}{\sec^2{2 \alpha} + u^2} \mathrm{d}u = \sec^2{2 \alpha} \big[ \cos{2\alpha} \arctan{(u\cos{2\alpha})} \big]_{-\infty}^{\infty} \\ \\ \\ = \sec{2\alpha} \big[ \frac{\pi}{2} - - \frac{\pi}{2} \big] = \pi \sec{2\alpha}

    as \cos{2\alpha} > 0 for 0 < \alpha < \frac{\pi}{4}

    iii)

    \\ \displaystyle I\sin^2{2\alpha} + J\cos^2{2\alpha} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}}{1 + \tan^2{\theta}\cos^2{2\alpha}} \mathrm{d} \theta \\ \\ \\ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}}{1 + (\sec^2{\theta}-1)\cos^2{2\alpha}} \mathrm{d} \theta \\ \\ \\ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}}{\ sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}} \mathrm{d}\theta = [\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \pi

    \\ \therefore I\sin^2{2\alpha} + \pi \sec{2\alpha}\cos^2{2\alpha} = \pi \\ \\ = \pi ( 1 - \cos{2\alpha}) \\ \\ = \pi ( 1 - (1-2\sin^2{2\alpha})) \\ \\ = \pi(2\sin^2{2\alpha})

    \\ \therefore I = \frac{\pi(2\sin^2{2\alpha})}{\si n^2{2\alpha}} = \frac{\pi(2\sin^2{2\alpha})}{4\s in^2{\alpha}\cos^2{\alpha}} = \frac{1}{2}\pi \sec^2{\alpha}

    as required.

    iv) If \frac{\pi}{4} < \alpha < \frac{\pi}{2} , then \cos{2\alpha} < 0 , which will change the value of J to -\pi \sec{2\alpha}

    \\ \therefore I\sin^2{2\alpha} - \pi \sec{2\alpha} \cos^2{2\alpha} = \pi \\ \\ \implies I \sin^2{2\alpha} = \pi(2\cos^2{\alpha}) \\ \\ \implies I = \frac{ \pi(2\cos^2{\alpha}) }{4 \sin^2{\alpha} \cos^2{\alpha}} = \frac{1}{2} \pi \mbox{cosec}^2{\alpha}

    Last edited by Daniel Freedman; 06-05-2009 at 01:37.
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  3. Daniel Freedman's Avatar
    3 05-05-2009  22:40
    • Overlord in Training
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    Re: STEP 2006 Solutions Thread
    STEP III 2006, Question 3
    Spoiler:
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    i) tan is an odd function ie tan(-x) = - tan(x) and therefore can have no even powers of x in its series expansion

    Prove \cot{x} - \tan{x} = 2\cot{2x}

    \mbox{LHS} = \frac{1}{\tan{x}} - \frac{\tan{x}}{1} = \frac{1 - \tan^2{x}}{\tan{x}} = 2 \left( \frac{1-\tan^2{x}}{2\tan{x}} \right) = 2 \left( \frac{1}{\frac{2\tan{x}}{1-\tan^2{x}}} \right) = \frac{2}{\tan{2x}} = 2\cot{2x} = \mbox{RHS}

    \displaystyle \therefore \frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n - \sum_{n=0}^{\infty} a_n x^n = 2 \left( \frac{1}{2x} + \sum_{n=0}^{\infty} b_n (2x)^n \right) \implies \sum_{n=0}^{\infty} \left( b_n x^n - a_n x^n \right) = 2^{n+1} \sum_{n=0}^{\infty} b_n x^n \implies b_n - a_n = 2^{n+1}b_n \ (x \neq 0 ) \implies a_n = b_n (1 - 2^{n+1})

    as required.

    ii)

    \frac{1}{\tan{x}} + \frac{\tan{x}}{1} = \frac{1 + \tan^2{x}}{\tan{x}} = 2 \left( \frac{1+\tan^2{x}}{2\tan{x}} \right) = 2 \left( \frac{1}{\frac{2\tan{x}}{1+\tan^ 2{x}}} \right) = \frac{2}{\sin{2x}} = 2\mbox{cosec}{2x}


    \therefore \cot{x} + \tan{x} = 2\mbox{cosec}{2x}

    \\ \displaystyle \implies \frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^n = 2 \left( \frac{1}{2x} + \sum_{n=0}^{\infty} c_n (2x)^n \right) \implies b_n + a_n = 2^{n+1}c_n \ \ (x \neq 0) \implies c_n = b_n (2^{-n} - 1)

    as required.

    iii)

    \displaystyle \left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)^2 + x^2 = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n \right)^2 \Leftrightarrow (x \cot{x})^2 + x^2 = (x \mbox{cosec}{x})^2 \Leftrightarrow x^2 ( \cot^2{x} + 1 ) = x^2 \mbox{cosec}^2{x} \Leftrightarrow \cot^2{x} + 1 = \mbox{cosec}^2{x} \ (x \neq 0 ) \Leftrightarrow \cos^2{x} + \sin^2{x} = 1

    which is true by Pythagoras'.

    \\ \displaystyle \therefore \displaystyle \left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)^2 + x^2 = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n \right)^2 \implies \left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)\left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right) = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n - x \right) \left( 1 + x \sum_{n=0}^{\infty} c_n x^n + x \right) \implies \left( 1 + x ( b_0 + b_1x + b_2x^2 + ... ) \right)^2 = (1 + x(c_0 + c_1x + c_2x^2 + ... ) - x)(1 + x(c_0 + c_1x + c_2x^2 + ... ) + x) \implies \left( 1 + x ( b_0 + b_1x + b_2x^2 + ... ) \right)^2= (1 + x((c_0 - 1) + c_1x + c_2x^2 + ... ))(1 + x((c_0+1) + c_1x + c_2x^2 + ... ) )

    EDIT: There was no point in factorising this way. I did the question whilst typing it up, and it seemed an obvious thing to do.

    Equating coefficients gives:

    \\ x: \ \ 2b_0 = c_0 + 1 + c_0 - 1 \implies b_0 = c_0 = 0 \implies a_0 = 0

    \\ x^2 : \ \ 2b_1 + b_0^2 = 2c_1 + (c_0+1)(c_0-1) \implies 2b_1 = -b_1 - 1 \implies 3b_1 = -1 \implies b_1 = -\frac{1}{3} \implies a_1 = (1 - 2^2)(- \frac{1}{3}) = 1

    as required.

    \\ x^4: \ \ 2b_3 + 2b_0 b_2 + b_1^2 = 2c_3 + (c_0 - 1)c_2 + c_1^2 + (c_0 + 1)c_2 \implies 2b_3 + \frac{1}{9} = 2c_3 + \frac{1}{36} \implies 2b_3 + \frac{1}{12} = 2(2^{-3}-1)b_3 \implies b_3 = -\frac{1}{45} \implies a_3 = (1-2^4)(-\frac{1}{45}) = \frac{1}{3}
    Last edited by Daniel Freedman; 05-05-2012 at 12:06.
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  4. Daniel Freedman's Avatar
    4 06-05-2009  02:12
    • Overlord in Training
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    Re: STEP 2006 Solutions Thread
    STEP III 2006, Question 1

    Spoiler:
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    y = \frac{2x(x^2-5)}{x^2 - 4} = \frac{2x(x+\sqrt{5})(x-\sqrt{5})}{(x+2)(x-2)}

    Therefore the curve has vertical asymptotes with equations x = 2 and x = -2 .

    The curve has roots x = - \sqrt{5}, \ 0, \ \sqrt{5} .

    As x \to \pm \infty, \ y \to 2x , so the curve has an oblique asymptote with equation y = 2x .

    As x \to 0, \ y \to \frac{-10x}{-4} = \frac{5}{2}x . This is the equation of the tangent to the curve at the origin (can be verified by differentiating)

    [SKETCH ATTACHED]

    i)

    3x(x^2-6) = (x^2-4)(x+3) \implies \frac{2x(x^2-5)}{x^2 - 4} = \frac{2}{3}x + 2

    [SKETCH ATTACHED]

    The graph shows that the two curves have three points of intersection, which means the above equation has three real roots

    ii)

    4x(x^2-5) = (x^2 - 4)(5x - 2) \implies \frac{2x(x^2-5)}{x^2 - 4} = \frac{5}{2}x - 1

    [SKETCH ATTACHED]

    The graph shows the two curves only have one point of intersection (using the fact that the line has the same gradient as the tangent to the curve at the origin). This means the above equation has one real root.

    iii)

    \\ 4x^2(x^2-5)^2 = (x^2 - 4)^2 (x^2+1) \implies \left(\frac{2x(x^2-5)}{x^2 - 4} \right)^2 = x^2 + 1 \implies \frac{2x(x^2-5)}{x^2 - 4} = \pm \sqrt{x^2+1}

    [SKETCH ATTACHED]

    The graph shows the curves have six points of intersection, which means the above equation has six real roots (the curves don't meet again because there are only two solutions of \sqrt{x^2+1} = 2x
    Attached Thumbnails
    Click image for larger version.  Name: graph 1.GIF  Views: 167  Size: 9.1 KB  ID: 69287   Click image for larger version.  Name: graph 2.GIF  Views: 127  Size: 9.9 KB  ID: 69288   Click image for larger version.  Name: graph 3.GIF  Views: 120  Size: 10.4 KB  ID: 69289   Click image for larger version.  Name: graph 4.GIF  Views: 157  Size: 10.4 KB  ID: 69290  
    Last edited by Daniel Freedman; 06-05-2009 at 02:21.
  5. Oh I Really Don't Care's Avatar
    5 10-05-2009  02:50
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    Re: STEP 2006 Solutions Thread
    STEP III 8


    Spoiler:
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    \displaystyle \delta(1\times1) = \delta(1) + \delta(1) by property four. It follows \delta(1) = 0 By property three we see that \delta \lambda = 0 for any lambda in R

    \displaystyle \delta(x^2) = \delta(x \times x) = x\delta(x) + x\delta(x) = 2x

    \displaystyle \delta(x\timesx^2) = x\delta(x^2) + x^2\delta(x) = 3x^2


    We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; \displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

    We will now prove property 1
    Spoiler:
    Show

    \displaystyle \frac{d}{dx}(x^n) = \lim_{dx \to 0}\frac{ (x+dx)^n - x^n}{dx} = \lim_{dx \to 0} \frac{x^n + nx^{n-1}dx - x^n + \phi}{dx} = (*) phi represents terms in the binomial expansion whose power of dx is greater then or equal to 2. It follow that (*) is equivalent to;

    \displaystyle \lim_{dx \to 0} \left(nx^{n-1} + \frac{\phi}{dx} \right) = nx^{n-1}

    Setting n = 0 we have \frac{d}{dx}(x^0) = 0 x^{-1} = 0


    We now prove property 2 is met by this operation

    Spoiler:
    Show


    \displaystyle \frac{d}{dx}\left(f(x) + g(x)\right)

    \displaystyle = \lim_{dx \to 0} \frac{ f(x+dx) + g(x+dx) - f(x) - g(x)}{dx}

    \displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)-f(x)}{dx} + \lim_{dx \to 0} \frac{ g(x+dx)-g(x)}{dx}

    \displaystyle = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)



    We prove property three is met by the above operation.

    Spoiler:
    Show

    Let g(x) = \lambda f(x)

    \displaystyle \frac{d}{dx}g(x) = \lim_{dx \to 0} \frac{g(x+dx)-g(x)}{dx}

    \displaystyle = \lim_{dx \to 0} \frac{\lambda f(x+dx) - \lambda f(x)}{dx}

    \displaystyle = \lambda\left(\lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}\right)

    \displaystyle = \lambda\left(\frac{d}{dx} f(x) \right)




    We now prove the final property;
    Spoiler:
    Show

    \displaystyle \frac{d}{dx} f(x) g(x)

    \displaystyle= \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) - f(x)g(x)}{dx}

    \displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) + g(x)f(x+dx) - g(x)f(x+dx) - f(x)g(x) }{dx}

    \displaystyle = \lim_{dx \to 0} \left( g(x) \frac{f(x+dx)-f(x)}{dx} + f(x+dx) \frac{ g(x+dx)-g(x)}{dx}\right)


    We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

    Spoiler:
    Show
    We proceed by induction.

    Assume that for some k in N that delta(x^k) = kx^(k-1). Assume that the result holds for k + 1.

    \displaystyle \delta(x^{k+1}) = \delta(x^k \times x) = x\delta(x^k) + x^k\delta(x) = kx^{k-1} \times x + x^k = (k+1)x^k

    P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains N. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.


    As required
    Last edited by Oh I Really Don't Care; 10-05-2009 at 03:49.
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  6. Glutamic Acid's Avatar
    6 10-05-2009  09:38
    • TSR Legend
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    Re: STEP 2006 Solutions Thread
    I/8:

    (i) Let the base be OAC, so area = 1/2ca. Height is b, so the area is b*1/3*(1/2ca) = abc/6.

    (ii) \cos \theta is the angle between BC and AC. BC = \left( \begin{array}{c} 0 \\ -b \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right)

    AC = \left( \begin{array}{c} -a \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right)

    \left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right) . \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right) = \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \cos \theta \Rightarrow \cos \theta = \dfrac{c^2}{\sqrt{(a^2 + c^2)(b^2 + c^2)}}

    Area of triangle = "\dfrac{1}{2}ab \sin \theta". Note that \sin^2 \theta = 1 - \cos^2 \theta, so \sin^2 \theta = 1 - \dfrac{c^4}{(a^2 + c^2)(b^2 + c^2)} = \dfrac{a^2b^2 + a^2c^2 + b^2c^2}{(a^2 + c^2)(b^2 + c^2)}

    \text{area} = \dfrac{1}{2} \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \dfrac{\sqrt{a^2b^2 + b^2c^2 +a^2c^2}}{\sqrt{(a^2 + c^2)(b^2 + c^2)}} = \dfrac{1}{2}\sqrt{a^2b^2 + b^2c^2 +a^2c^2}

    The area will clearly be equal to abc/6, only that the height of the new triangle is d.

    \dfrac{1}{6}abc = \dfrac{1}{6}(a^2b^2 + b^2c^2 +a^2c^2)^{1/2}d \Rightarrow a^2b^2c^2 = d^2(a^2b^2 + b^2c^2 +a^2c^2)

    \Rightarrow \dfrac{1}{d^2} = \dfrac{a^2b^2 + b^2c^2 +a^2c^2}{a^2b^2c^2} \Rightarrow \dfrac{1}{d^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}
  7. Daniel Freedman's Avatar
    7 10-05-2009  11:01
    • Overlord in Training
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    Re: STEP 2006 Solutions Thread
    (Original post by DeanK22)
    STEP III 8


    Spoiler:
    Show

    \displaystyle \delta(1\times1) = \delta(1) + \delta(1) by property four. It follows \delta(1) = 0 By property three we see that \delta \lambda = 0 for any lambda in R

    \displaystyle \delta(x^2) = \delta(x \times x) = x\delta(x) + x\delta(x) = 2x

    \displaystyle \delta(x\timesx^2) = x\delta(x^2) + x^2\delta(x) = 3x^2


    We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; \displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}

    We will now prove property 1
    Spoiler:
    Show

    \displaystyle \frac{d}{dx}(x^n) = \lim_{dx \to 0}\frac{ (x+dx)^n - x^n}{dx} = \lim_{dx \to 0} \frac{x^n + nx^{n-1}dx - x^n + \phi}{dx} = (*) phi represents terms in the binomial expansion whose power of dx is greater then or equal to 2. It follow that (*) is equivalent to;

    \displaystyle \lim_{dx \to 0} \left(nx^{n-1} + \frac{\phi}{dx} \right) = nx^{n-1}

    Setting n = 0 we have \frac{d}{dx}(x^0) = 0 x^{-1} = 0


    We now prove property 2 is met by this operation

    Spoiler:
    Show


    \displaystyle \frac{d}{dx}\left(f(x) + g(x)\right)

    \displaystyle = \lim_{dx \to 0} \frac{ f(x+dx) + g(x+dx) - f(x) - g(x)}{dx}

    \displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)-f(x)}{dx} + \lim_{dx \to 0} \frac{ g(x+dx)-g(x)}{dx}

    \displaystyle = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)



    We prove property three is met by the above operation.

    Spoiler:
    Show

    Let g(x) = \lambda f(x)

    \displaystyle \frac{d}{dx}g(x) = \lim_{dx \to 0} \frac{g(x+dx)-g(x)}{dx}

    \displaystyle = \lim_{dx \to 0} \frac{\lambda f(x+dx) - \lambda f(x)}{dx}

    \displaystyle = \lambda\left(\lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}\right)

    \displaystyle = \lambda\left(\frac{d}{dx} f(x) \right)




    We now prove the final property;
    Spoiler:
    Show

    \displaystyle \frac{d}{dx} f(x) g(x)

    \displaystyle= \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) - f(x)g(x)}{dx}

    \displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) + g(x)f(x+dx) - g(x)f(x+dx) - f(x)g(x) }{dx}

    \displaystyle = \lim_{dx \to 0} \left( g(x) \frac{f(x+dx)-f(x)}{dx} + f(x+dx) \frac{ g(x+dx)-g(x)}{dx}\right)


    We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

    Spoiler:
    Show
    We proceed by induction.

    Assume that for some k in N that delta(x^k) = kx^(k-1). Assume that the result holds for k + 1.

    \displaystyle \delta(x^{k+1}) = \delta(x^k \times x) = x\delta(x^k) + x^k\delta(x) = kx^{k-1} \times x + x^k = (k+1)x^k

    P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains N. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.


    As required
    I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that \Delta(x^n) = nx^{n-1} , using the four rules. Then let \displaystyle h(x) = \sum_{k=0}^n a_k x^k , and apply the operation to this function, making use of rules ii) and iii) to show that \displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1} .
  8. Dadeyemi's Avatar
    8 10-05-2009  13:08
    • Overlord in Training
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    Re: STEP 2006 Solutions Thread
    STEP III 2006, Question 4
    initial part
    let y = x
    \Rightarrow 2f(x) \equiv f(2x) \\ \Rightarrow 2f'(x)\equiv 2f'(2x) \\ \Rightarrow f''(x) \equiv 2f''(2x) \\ \Rightarrow f^{(n)} \equiv 2^{n-1}f^{(n)}(2x)
    Letting x = 0 we have f(0) = 0 and f^{n}(0) = 0 for n\neq 0
    so by Maclaurin f(x) \equiv f'(0)x \equiv kx for some constant k
    (i)
    G(x+y) \equiv ln(g(x+y)) \equiv ln(g(x)g(y)) \equiv ln(g(x)) + ln(g(y)) \equiv G(x)+G(y)
    so G(x) \equiv f(x) = kx
    giving g(x) = e^{kx}
    (ii)
    H(\ln (x+y)) \equiv h(xy) \equiv h(x)+h(y) \equiv H(\ln x)+G(\ln y)
    so h(x) \equiv H(\ln x) \equiv f(\ln x) \equiv k\ln x
    (iii)
    let x = \tan \theta , y = \tan \phi
    \displaystyle \Rightarrow t(\tan \theta) + t(\tan \phi) = t( \frac{\tan \theta +\tan \phi}{1- \tan \theta \tan \phi }) = t(\tan (\theta+ \phi))
    so t (x) = t (\tan \theta) = f(\theta) = k\theta = k \arctan x
    Last edited by Dadeyemi; 10-05-2009 at 13:13.
  9. Aurel-Aqua's Avatar
    9 10-05-2009  14:21
    • Adored and Respected Member
    Re: STEP 2006 Solutions Thread
    2006 I Question 13
    Part (i)
    The number of diamonds in one kilogram has a Poisson distribution of D \sim \text{Poisson}(1). We find that per 100 grams, D \sim \text{Poisson}(\lambda = 0.1). For 100T grams, D_{T} \sim \text{Poisson}(\lambda = 0.1T). T is the distribution of scores of the die.

    Drawing 1, we have P(T = 1 \cap D_{1}=0) = \frac{1}{6}(e^{-0.1}).
    Similarly, \displaystyle P((T=1 \cap D_{1})\cup(T=2\cap D_{2}=0)\cup\ldots\cup(T=6\cap D_{6}=0)) = \sum_{t=1}^6 \frac{1}{6}e^{-0.1t}=
    \displaystyle = \frac{1}{6}e^{-0.1}\sum_{t=1}^6 e^{(-0.1)(t-1)} = \frac{e^{-0.1}}{6}\frac{1-e^{-0.6}}{1-e^{-0.1}} (recognising a finite geometric sum).

    Part (i) - expectation
    \displaystyle E[D_T] = \sum_{t=1}^6 \frac{1}{6}E[D_t] = \frac{1}{6}\sum_{t=1}^6 0.1t = \frac{1}{6}(0.1+0.2+0.3+0.4+0.5+ 0.6) = \frac{2.1}{6} = \frac{0.7}{2} = 0.35

    Part (ii)
    This time, T is a geometric distribution with p = \frac{5}{6}. Thus, P(T=t)=\frac{1}{6}\left(\frac{5} {6}\right)^{t-1}.

    This time, our probability is:
    \displaystyle P((T=1 \cap D_{1}=0)\cup(T=2\cap D_{2}=0)\cup\ldots) = \sum_{t=1}^{\infty}P(T=t\cap D_{t}=0) =
    \displaystyle = \sum_{t=1}^{\infty} \left(\frac{5}{6}\right)^{t-1}\frac{1}{6}e^{-0.1t} = \frac{e^{-0.1}}{6}\sum_{t=1}^{\infty} \left(e^{-0.1}\frac{5}{6}\right)^{t-1} =
    \displaystyle = \frac{e^{-0.1}}{6}\frac{1}{1-\frac{5}{6}e^{-0.1}} = \frac{e^{-0.1}}{6-5e^{-0.1}}, as required.

    Part (ii) - expectation
    \displaystyle E[D_T] = \sum_{t=1}^\infty E[D_t]P(T=t) = \sum_{t=1}^\infty 0.1t\frac{1}{6}\left(\frac{5}{6} \right)^{t-1} =\frac{1}{60}\sum_{t=1}^\infty t\left(\frac{5}{6}\right)^{t-1}.

    Consider: \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ((1-x)^{-1}) = (1-x)^{-2} = \frac{\mathrm{d}}{\mathrm{d}x} \sum_{t=0}^\infty x^t = \sum_{t=1}^\infty tx^{t-1}.

    Thus, \displaystyle E[D_T] = \frac{1}{60}\left(1-\frac{5}{6}\right)^{-2} = \frac{36}{60} = 0.6.
  10. Aurel-Aqua's Avatar
    10 10-05-2009  14:40
    • Adored and Respected Member
    Re: STEP 2006 Solutions Thread
    2006 I Question 14
    Either I missed something crucial, or this question is incredibly easy. I'll write up the incredibly-easy interpretation of this question:

    Part (i)
    This is a geometric progression, where \displaystyle P(\text{Red on rth attempt}) = pq^{r-1}, where \displaystyle p=\frac{1}{n}, \displaystyle q = \frac{n-1}{n}, so \displaystyle P(\text{Red on rth attempt}) = \left(\frac{n-1}{n}\right)^{r-1}\frac{1}{n} = \frac{(n-1)^{r-1}}{n^r}.
    Differentiating, we get: \displaystyle \frac{\mathrm{d}P}{\mathrm{d}n} = \frac{n^r(r-1)(n-1)^{r-2} - rn^{r-1}(n-1)^{r-1}}{n^{2r}}. Equating to zero, n^r(r-1)(n-1)^{r-2} = rn^{r-1}(n-1)^{r-1} \iff n(r-1) = r(n-1) \iff n(r-1-r) = -r \iff -n=-r \iff n = r. We know that this is the maximum as r tends to infinity, P tends to zero.

    Part (ii)

    P(\text{Red on 1st}) = \frac{1}{n}
    P(\text{Red on 2nd}) = \frac{n-1}{n}\cdot\frac{1}{n-1} = \frac{1}{n}
    P(\text{Red on rth}) = \frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-(r-2)}{n-(r-3)}\cdot\frac{n-(r-1)}{n-(r-2)}\cdot\frac{1}{n-(r-1)}=\frac{1}{n} . Clearly maximum at n = 1.

    Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.

    Edit: see GHOSH-5's post here: http://www.thestudentroom.co.uk/show...3&postcount=41
    Last edited by Aurel-Aqua; 13-06-2009 at 10:53.
  11. DFranklin's Avatar
    11 10-05-2009  14:42
    • TSR Royalty
    • Location: London
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    Re: STEP 2006 Solutions Thread
    (Original post by Daniel Freedman)
    I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that \Delta(x^n) = nx^{n-1} , using the four rules. Then let \displaystyle h(x) = \sum_{k=0}^n a_k x^k , and apply the operation to this function, making use of rules ii) and iii) to show that \displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1} .
    Actually, it's worse than that.

    They've said "given that the operator \Delta has these properties; show that...".

    Dean has said "A particular operator \delta has those properties, and for \delta, we have ...."

    But he hasn't shown that \delta is the only possible operator with the properties.

    This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).
  12. Oh I Really Don't Care's Avatar
    12 10-05-2009  14:52
    • TSR Demigod
    • Location: Mind your bleeding own you two bob ****
    Re: STEP 2006 Solutions Thread
    (Original post by DFranklin)
    Actually, it's worse than that.

    They've said "given that the operator \Delta has these properties; show that...".

    Dean has said "A particular operator \delta has those properties, and for \delta, we have ...."

    But he hasn't shown that \delta is the only possible operator with the properties.

    This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).
    Thanks for pointing that out - certainly a mistake that will not be made this Summer (hopefully). Unfortunately I do not really know how to prove the Uniquness of this operator - could you please shed some light on to how you do that? Thanks.
  13. DFranklin's Avatar
    13 10-05-2009  15:35
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    Re: STEP 2006 Solutions Thread
    Well, that's effectively what the question is asking you to show. I would:

    (1) Prove by induction that \Delta x^n = n x^{n-1} (using (i) and (iv)).
    (2) Then use (iii) to prove \Delta a_n x^n = n a_n x^{n-1}
    (3) Then use (ii) to prove \Delta \sum_0^N a_n x^n = \sum_0^N n a_n x^{n-1}.

    [Which effectively proves uniqueness of \Delta for polynomials].

    Then simply observe this is the same as the derivative in the case of polynomials.
  14. darkness9999's Avatar
    14 12-05-2009  13:29
    • Exalted and Worshipped Member
    • Location: London
    • Posts: 1,139
    Re: STEP 2006 Solutions Thread
    I have different answers to STEP I Q 14 Im not to sure who is wrong...

    Part (i)
    Spoiler:
    Show


    From the diagram it is obvious that on the (r-1)^{th} attempt you would have had to picked a blue one before the red one ...

    So
    P(R=r)=\left(\frac{n}{n+1} \right)^{r-1} \times \frac{1}{n+1}

    \Rightarrow \boxed{\frac{n^{r-1}}{(n+1)^r}}

    Now to find the maximum value of A we need to differentiate P(R=r) wtr n.

    Let f(x)= \frac{n^{r-1}}{(n+1)^r}

    Using the quatient rule:


    f‘(x)= \frac{(n+1)^r(r-1)n^{r-2}- n^{r-1}r(n+1)^{r-1}}{(n+1)^{2r}}


    \Rightarrow n^{r-2}(n+1)^{r-2}\frac{(n+1)(r-1) - nr}{(n+1)^{2r}}

    \Rightarrow n^{r-2}\frac{r-n-1}{(n+1)^{r+1}}

    Therefore it is maximised when \boxed{n=r-1}


    Part (ii)
    Spoiler:
    Show




    P(R=1) = \frac{1}{n+1}


    P(R=2) = \frac{n}{n+1}\times \frac{1}{n}

    \Rightarrow \frac{1}{n+1}


    P(R=3) = \frac{n}{n+1}\times \frac{n-1}{n}\times \frac{1}{n-1}

    \Rightarrow \frac{1}{n+1}


    Hence
    \boxed{P(R=r)= \frac{1}{n+1}}

    This probability is decreasing as n increases therefore it is again maximised when \boxed{n=r-1}

    Last edited by darkness9999; 12-05-2009 at 20:46.
  15. darkness9999's Avatar
    15 12-05-2009  16:34
    • Exalted and Worshipped Member
    • Location: London
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    Re: STEP 2006 Solutions Thread
    STEP I: Q5

    Part (i)

    \int \frac{3}{(x-4)\sqrt {2x+1}} \, dx

    Substitution:

    u^2=2x+1

    \frac{du}{dx} 2u = 2

    u\times du = dx

    Therefore the integral becomes

    \int \frac{3u}{(x-4)\sqrt {2x+1}} \, du

    \Rightarrow \int \frac{3u}{u \frac{u^2-9}{2}} \, du

    \Rightarrow \int \frac{6}{(u-3)(u+3)} \, du

    Simplifying this further using partial fractions gives us
    Spoiler:
    Show

    \frac{6}{(u-3)(u+3)} \equiv \frac{A}{u-3} + \frac{B}{u+3}

    6= A(u+3) +b(u-3)

    u_3: 6= A6 \Rightarrow A=1

    u_{-3}: 6= b(-6) \Rightarrow B=-1

    \Rightarrow \frac{6}{(u-3)(u+3)} \Leftrightarrow \frac{1}{u-3} - \frac{1}{u+3}


    \int \frac{1}{u-3} - \frac{1}{u+3} \, du

    \Rightarrow \ln {u-3} - \ln {u+3} + C

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \Rightarrow \ln { \frac{u-3}{u+3} + C


    If we now sub in the substitution we used we get

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    {\ln {\frac{\sqrt {2x+1}-3}{\sqrt {2x+1}+3}+C



    Part (ii)

    \displaystyle\int^{\ln 8}_{\ln 3} \frac{2}{e^x\sqrt {e^x+1}} \, dx

    I used this substitution:

    u^2=e^x+1

    \frac{du}{dx} 2u = e^x

    \frac{2u\times du}{e^x} = dx

    So the integral becomes:


    \displaystyle\int^{3}_{2} \frac{4u}{e^{2x}\sqrt {e^x+1}} \, du

    \Rightarrow \displaystyle\int^{3}_{2} \frac{4u}{(u^2-1)(u^2-1)u} \, du

    \Rightarrow \displaystyle\int^{3}_{2} \frac{4}{(u+1)^2(u-1)^2} \, du

    Simplifying this further using partial fractions gives us
    Spoiler:
    Show

    \frac{4}{(u+1)^2(u-1)^2} \equiv \frac{A}{(u+1)^2} + \frac{B}{(u-1)^2} + \frac{C}{u-1} + \frac{D}{u+1}

    4 = A(u-1)^2 +B(u+1)^2 +C(u+1)^2(u-1) +D(u-1)^2(u+1)

    u_{-1}: 4 = A4 \Rightarrow A=1

    u_{1}: 4 = B4 \Rightarrow B=1

    u_{0}: 2 = -C +D

    u_{1}: 0 = -C -D

    \Rightarrow C= -1 & D= 1

    \Rightarrow \frac{4}{(u+1)^2(u-1)^2} \Leftrightarrow \frac{1}{(u+1)^2} + \frac{1}{(u-1)^2} - \frac{1}{u-1} + \frac{1}{u+1}


    \displaystyle\int^{3}_{2} \frac{1}{(u+1)^2} + \frac{1}{(u-1)^2} - \frac{1}{u-1} + \frac{1}{u+1} \, du

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \Rightarrow \left[-\frac{1}{u+1} - \frac{1}{u-1} + \ln {\frac{u+1}{u-1} \right]^{3}_{2}


    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \Rightarrow \left( (-\frac{1}{4} -\frac{1}{2} + \ln {\frac{4}{2}) -(-\frac{1}{3} -1 + \ln 3) \right)


    \Rightarrow \boxed {\frac{7}{12} + \ln {\frac{2}{3}}}
    Last edited by darkness9999; 12-05-2009 at 16:48.
  16. Elongar's Avatar
    16 20-05-2009  00:05
    • Respected Member
    • Location: Cambridge
    • Posts: 240
    Re: STEP 2006 Solutions Thread
    STEP III 2006, Q5

    We wish to show that \alpha, \beta, \gamma form an equilateral iff

    \displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0

    Multiplying by two and factorising, we wish to show that,

    \displaystyle (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 0

    holds for all equilateral triangles.

    Note that \alpha - \beta, \beta - \gamma, \gamma - \alpha are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is \frac{\pi}{3}. Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

    Using this, we reconstruct the equation:

    \displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = 0

    or, more simply,

    \displaystyle r^2 \left( e^{2 i \theta} + e^{2 i (\theta +\frac{\pi}{3})} + e^{2 i (\theta +\frac{2 \pi}{3})} \right) = 0

    \displaystyle r^2 e^{2 i \theta} \left( 1 + e^{i \frac{\pi}{3}} + e^{i \frac{2 \pi}{3}} \right) = 0

    The bracket evaluates to 0, so we are done.

    For the next part, we take the roots of the cubic to be \alpha, \beta, \gamma, and expand, yielding Vieta's formula:

    \displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0

    Algebra shows that:

    \displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = (\alpha + \beta + \gamma)^2 - 3 (\alpha \beta + \beta \gamma + \gamma \alpha) = a^2 - 3b = 0

    and the result follows immediately.

    Now we write,

    p = Ae^{i \phi}

    q = B + Ci

    The transformation z = p w + q has the effect of rotating our equilateral triangle by the angle \phi, magnifying it by a factor of A and translating it through the vector represented by q (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.
  17. nuodai's Avatar
    17 29-05-2009  16:11
    • PS Helper
    • TSR Legend
    Re: STEP 2006 Solutions Thread
    STEP I 2006 Question 2

    Solution

    The barn is square, so it doesn't matter which side the goat is on. Because the lengths are of side 2a and the rope is of length 4a, it makes sense that we should consider the case where the rope is between the middle of a side and the edge of the side

    In this, let P be the position of the goat, O be the position where the rope is tethered, and ABCD be the barn, with O being along AB and A, B, C, D marking the corners of the barn, going clockwise. Bear in mind that this is much more easily explained with diagrams, so try drawing it if you don't follow.

    Let's say that O is located along AB at distance x from A, where 0 < x < a (so that it is between the midpoint of AB and the point A)

    The distance OA is x, so the distance OA is 2a - x and so on. This means that if the goat walks as far as it can clockwise, P will be located at a distance x from C along the line CD. So, the goat can make:
    - A quarter-circle with radius CP = x
    - A quarter-circle with radius BP = 2a + x
    - A half-circle with radius OP = 4a
    - A quarter-circle with radius AP = 4a - x
    - A quarter-circle with radius DP = 2a - x

    So the area would be:
    \newline \displaystyle \dfrac{\pi}{4}\Big[ x^2 + (2a+x)^2 + 2(4a)^2 + (4a - x)^2 + (2a - x)^2 \Big] \newline = \dfrac{\pi}{4} \Big[ x^2 + 4a^2 + 4ax + x^2 + 32a^2 + 16a^2 - 8ax + x^2 + 4a^2 - 4ax + x^2 \Big] \newline = \dfrac{\pi}{4} \Big[ 56a^2 - 8ax + 4x^2 \Big] = \pi(14 a^2 - 2ax + x^2)\newline = \pi([x - a]^2 + 13a^2)

    This therefore gives us the maximum and minimum values. The maximum value must occur when x = 0 since 0 \le x \le a, and the maximum value must occur when x = a, therefore \boxed{ 13\pi a^2 \le A \le 14\pi a^2 }
    Last edited by nuodai; 12-06-2009 at 14:10.
  18. Oh I Really Don't Care's Avatar
    18 29-05-2009  16:13
    • TSR Demigod
    • Location: Mind your bleeding own you two bob ****
    Re: STEP 2006 Solutions Thread
    RE; Noudai

    opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.
  19. nuodai's Avatar
    19 29-05-2009  16:21
    • PS Helper
    • TSR Legend
    Re: STEP 2006 Solutions Thread
    (Original post by DeanK22)
    RE; Noudai

    opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.
    I realised that a bit too late :p: I've chopped it out of the answer though.
    Post rating:
    1
  20. tommm's Avatar
    20 29-05-2009  16:34
    • TSR Idol
    Re: STEP 2006 Solutions Thread
    STEP III 2006 Q7

    Spoiler:
    Show
    Part i):

    Using the quadratic formula we obtain

    u = \displaystyle\frac{-2\sinh x \pm \sqrt{4\sinh^2x + 4}}{2}

    =-\sinh x \pm \cosh x

    If we let u = \frac{dy}{dx}, we obtain the differential equation they're asking about, so:

    \frac{dy}{dx} = -\sinh x \pm \cosh x

    Using the condition that \frac{dy}{dx} &gt; 0 at x = 0, we see that the \pm must be a +, hence

    \frac{dy}{dx} = \cosh x - \sinh x

    Integrating we obtain

    y = \sinh x - \cosh x + c

    and using y = 0 when x = 0 we find that c = 1.


    Part ii):

    Using the quadratic formula again to find dy/dx, we obtain (apologies for not writing it out in full):

    \frac{dy}{dx} = \displaystyle\frac{-1 \pm \cosh y}{\sinh y}

    \implies \displaystyle\int\displaystyle\f rac{\sinh y}{1 \pm \cosh y}\mathrm{d}y = \int - \mathrm{d}x

    We can integrate the left hand side using recognition: the top is the derivative of the bottom, so:

    \pm \ln (1 \pm \cosh y) = c - x

    Now, if the \pm sign were a minus, then the condition x = 0 when y = 0 would be impossible, as this would mean that the arbitrary constant were infinite. We can therefore deduce that the \pm must be a +. Therefore:

    \ln (1 + \cosh y) = c - x

    Using x = 0 when y = 0, we find that c = ln2. Therefore

    \ln (1 + \cosh y) = \ln 2 - x

    \implies \cosh y = 2e^{-x} -1, which is the solution of the differential equation.

    Writing the cosh in full exponentials and multiplying by 2 gives:

    e^y + e^{-y} = 4e^{-x} - 2

    To find the asymptotes, we consider two cases:

    1) when y \rightarrow \infty, we can disregard e^{-y} and -2 because they are comparatively small, so e^y \approx 4e^{-x}. Rearranging gives the equation of the asymptote, y = \ln 4 - x.

    1) when y \rightarrow -\infty, we can disregard e^{y} and -2 because they are comparatively small, so e^{-y} \approx 4e^{-x}. Rearranging gives the equation of the other asymptote, y = x - \ln 4.
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