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STEP I, II, III 2002 Solutions

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    I have some additions to the solutions to some of the STEP III qu's posted by Dadeyemi that would complete them. Would they be wanted?

    Also, SimonM (or anybody else), what is the LaTeX for surrounding particular results in boxes, as you sometimes do for answers to parts of questions? That might make my solutions a little more readable...

    Meanwhile,

    Question 6, STEP III, 2002

    \displaystyle y^4  \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^4 = (y^2 - 1)^2

    Standard seperating of variables yields:

    \displaystyle \frac{y^4}{(y^2 - 1)^2} (\mathrm{d}y)^4 = 1 . (\mathrm{d}x)^4

    At this point, we would like to take roots, but notice that we must consider seperately the cases | y | \le 1, and | y | > 1, or we will be taking square roots of a negative number.

    (a) For | y | > 1:

    \displaystyle \int \frac{y}{\sqrt{y^2 - 1}} \mathrm{d}y = \int \pm 1 . \mathrm{d}x

    Resulting in:

    \displaystyle \sqrt{y^2 - 1} = \pm x + c

    \displaystyle y^2 = (\pm x + c)^2 + 1

    Notice that if x \in [- \infty, \infty], it is no longer necessary to consider \pm x, and we have:

    \displaystyle \boxed{y^2 = (x + c)^2 + 1}

    (b) For | y | \le 1:

    \displaystyle \int \frac{y}{\sqrt{1 - y^2}} \mathrm{d}y = \int \pm 1 . \mathrm{d}x

    Resulting in:

    \displaystyle - \sqrt{1 - y^2} = \pm x + c

    \displaystyle y^2 = 1 - (\pm x + c)^2

    Or simply:

    \displaystyle \boxed{y^2 = 1 - (x + c)^2}

    Now we consider parts (i) and (ii) of the question:

    (i) \displaystyle \frac{\sqrt{3}}{2} < 1, therefore we use the second formula to determine c:

    Substituting  \left(0 , \frac{\sqrt{3}}{2} \right) into \displaystyle y^2 = 1 - (x + c)^2 yields,

    \displaystyle \boxed{c = \pm \frac{1}{2}}

    (ii) \displaystyle \frac{\sqrt{5}}{2} > 1, therefore we use the first formula to determine c:

    Substituting  \left(0 , \frac{\sqrt{5}}{2} \right) into \displaystyle y^2 = (x + c)^2 + 1 yields (as if by magic),

    \displaystyle \boxed{c = \pm \frac{1}{2}}.

    Therefore, all the solution curves that pass through one, pass through the other point as well, and we have the following two functions:

    \displaystyle

y^2 = \left\{ \begin{array}{ll}

1 - (x + \frac{1}{2})^2 & \mathrm{if} ~| y | \le 1\\

 & \\

1 + (x + \frac{1}{2})^2 & \mathrm{if} ~| y | > 1\\

\end{array}

    and

    \displaystyle

y^2 = \left\{ \begin{array}{ll}

1 - (x - \frac{1}{2})^2 & \mathrm{if} ~| y | \le 1\\

 & \\

1 + (x - \frac{1}{2})^2 & \mathrm{if} ~| y | > 1\\

\end{array}

    To prove that y = \pm 1 satisfies the differential equation, we simply substitute them straight in, yielding,

    \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 0, as desired.

    It should be fairly clear that in the case | y | < 1, we have the formula for a circle:

    \displaystyle y^2 + (x + c)^2 = 1

    Specifically, for the two functions, circles with centers \displaystyle \left(\pm \frac{1}{2} , 0 \right), and radius 1.

    To sketch the functions for | y | > 1 , we simply consider what happens as x \to \pm \infty, and notice that they are symmetric in the x axis.

    I will attach a sketch as soon as I can.
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    (Original post by Elongar)
    Also, SimonM (or anybody else), what is the LaTeX for surrounding particular results in boxes, as you sometimes do for answers to parts of questions? That might make my solutions a little more readable..
    Do you mean liked this?

     \boxed{\mathrm{Whatever \ you \ want \ in \ here.}}

    Code: \boxed{Whatever you want in here.}

    Next time you're unsure of a code for a symbol, hover your cursor over someone else's LateX which has coded such a symbol, and the LateX coding will appear. :yy:
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    Many thanks . I'll bear that in mind.
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    Question 8, STEP I, 2002
    First Part
    Writing out the first few  C_k:
     C_1 = C

     C_2 = C(1+\alpha) - R_1

     C_3 = (C(1+\alpha) - R_1)(1+\alpha) - R_2 = C(1+\alpha)^2 - R_1(1+\alpha) - R_2

     C_4 = (C(1+\alpha)^2 - R_1(1+\alpha) - R_2)(1+\alpha) - R_3

     C_4 = C(1+\alpha)^3 - R_1(1+\alpha)^2 - R_2(1+\alpha) - R_3

    Generalising this:

     \boxed{C_{n+1} = C(1+\alpha)^n - \displaystyle\sum_{k=1}^n R_k (1+\alpha)^{n-k}}


    Substituting in  R_k = (1+\alpha)^kr

     C_{n+1} = C(1+\alpha)^n - \displaystyle\sum_{k=1}^n (1+\alpha)^k(1+\alpha)^{n-k}}r

     \iff C_{n+1} = C(1+\alpha)^n - \displaystyle\sum_{k=1}^n (1+\alpha)^nr

     \iff C_{n+1} = C(1+\alpha)^n - n(1+\alpha)^nr

    If the loan is payed off in N years, then  C_{N+1} = 0

     \implies C(1+\alpha)^N - N(1+\alpha)^Nr = 0

     \iff (C-Nr)(1+\alpha)^N = 0

     \implies C-Nr = 0 \iff \boxed{r = \dfrac{C}{N}} \ \ \ \square
    Second Part
    Now all  R_k are the same, given by  R_k = R

     \implies C_{n+1} = C(1+\alpha)^n - \displaystyle\sum_{k=1}^n R(1+\alpha)^{n-k} \ \ \ (\ast )

    Looking at the sum, we see that it is a geometric series:

     \displaystyle\sum_{k=1}^n R(1+\alpha)^{n-k} = R + R(1+\alpha) + R(1+\alpha)^2 + \cdots + R(1+\alpha)^{n-1}

     \iff \displaystyle\sum_{k=1}^n R(1+\alpha)^{n-k} = \dfrac{R((1+\alpha)^n -1)}{\alpha}

    So the equation  (\ast ) now becomes:

     C_{n+1} = C(1+\alpha)^n - \dfrac{R((1+\alpha)^n -1)}{\alpha}

    Again, using the fact that the loan is payed off in N years, and so  C_{N+1} = 0

     \implies C(1+\alpha)^N - \dfrac{R((1+\alpha)^N -1)}{\alpha} = 0

     \iff C(1+\alpha)^N = \dfrac{R((1+\alpha)^N -1)}{\alpha}

     \iff \boxed{\dfrac{R}{C} = \dfrac{\alpha(1+\alpha)^N}{(1+\a  lpha)^N -1}} \ \ \ \square
    Third Part
     \alpha = \frac{1}{50} and N = 4, we can approximate  (1+\alpha)^N :

    (1+\frac{1}{50})^4  = 1 + \frac{2}{25} + \frac{3}{1250} + \cdots

    As the terms become very small, we can approximate it to the 3rd term:

    (1+\frac{1}{50})^4 \approx 1 + \frac{2}{25} + \frac{3}{1250}

     \iff (1+\frac{1}{50})^4 \approx \dfrac{1353}{1250}

     \implies \dfrac{R}{C} \approx \frac{1}{50} \times \frac{1353}{1250} \div \frac{103}{1250}

     \implies \dfrac{R}{C} \approx \dfrac{1353}{103 \times 50} = \dfrac{27 \times 50 + 3}{103 \times 50}

    As \frac{3}{103\times 50} is very small, we can ignore it.

     \therefore \boxed{\dfrac{R}{C} \approx \dfrac{27}{103}} \ \ \ \square
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    Question 13, STEP I, 2002
    (i)
    For the roots to be real, the discriminant must be greater than or equal to zero, ie:

     U^2 \geq 4V

    Case 1: U = 1
    For the inequality to hold, we require V = -1

    And the probability for this case is  \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}

    Case 2: U = 0
    For the inequality to hold, we require V = -1

    The probability for this case is  \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}

    Case 3: U = -1
    For the inequality to hold, again we require V = -1

    The probability for this case is  \frac{1}{4} \times \frac{1}{3} = \frac{1}{9}

    Adding up the probabilities, we get:

     \frac{2}{9} + \frac{1}{6} + \frac{1}{9} = \frac{4}{18} + \frac{3}{18} + \frac{2}{18}

     = \frac{9}{18} = \boxed{\frac{1}{2}} \ \ \ \square
    (ii)
    Going back to cases (and using the fact that we know V must be -1)

    Case 1: U = 1

     x = \dfrac{-1 \pm \sqrt{5}}{2}

    The larger root is  \dfrac{\sqrt{5}-1}{2} and has a probability of  \frac{2}{9}

    Case 2: U = 0

     x = \pm 1

    The larger root is 1, and this has a probability of  \frac{1}{6}

    Case 3: U = -1

     x = \dfrac{1\pm \sqrt{5}}{2}

    The larger root is  \dfrac{1+\sqrt{5}}{2} and this has a probability of  \frac{1}{9} .

    The expected value of the large root E(R) is now given by:

     \dfrac{E(R)}{2} = \dfrac{\sqrt{5}-1}{2} \times \dfrac{2}{9} + 1 \times \dfrac{1}{6} + \dfrac{1+\sqrt{5}}{2} \times \dfrac{1}{9}

     \dfrac{E(R)}{2} = \dfrac{1}{3} \left [ \dfrac{2\sqrt{5}-2}{6} + \dfrac{3}{6} + \dfrac{1+\sqrt{5}}{6} \right ]

     \iff \dfrac{E(R)}{2} = \dfrac{1}{3} \left [ \dfrac{3\sqrt{5}+2}{6} \right ]

     \iff \boxed{E(R) = \dfrac{3\sqrt{5}+2}{9}}
    (iii)
     x^3 + (U-2V)x^2 + (1-2UV)x + U = 0

     \iff (x+U)(x^2 - 2Vx + 1) = 0

     x = -U \implies \boxed{U = -1}

     x^2 - 2Vx + 1 = 0 \implies x = \dfrac{2V \pm \sqrt{4V^2-4}}{2} = V \pm \sqrt{V^2-1}

    We require the root to be real  \therefore V = \pm 1

     \implies x = V \pm 0 = V \implies \boxed{V = 1}

    Therefore we require U = -1 and V = 1

    Therefore the probability that the roots of the equation are positive is:

     \dfrac{1}{3} \times \dfrac{2}{3} = \boxed{\dfrac{2}{9}}
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    STEP II 2002 Q9

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    There's loads of ugly algebraic manipulation involved in this question, so apologies if I don't type out full working in getting from one stage to the next.

    Considering horizontal motion:
    u = V\cos\alpha
    a = 0
    s = s_H = horizontal displacement after time t
    t = t

    Using s = ut + 0.5at^2 we obtain t = \displaystyle\frac{s_H}{V\cos\al  pha}

    Considering vertical motion:
    u = V\sin\alpha
    a = -g
    s = s_v
    t = \displaystyle\frac{s_H}{V\cos\al  pha}

    which gives

    s_v = s_H\tan\alpha - \frac{gs^2_H}{2V^2}\sec^2\alpha (*)

    when s_H = a or b, s_V = h, because the particle just clears the walls

    Therefore h = a\tan\alpha - \frac{ga^2}{2V^2}\sec^2\alpha = b\tan\alpha - \frac{gb^2}{2V^2}\sec^2\alpha

    rearranging, we obtain \displaystyle\frac{a\tan\alpha - h}{a^2} = \displaystyle\frac{b\tan\alpha - h}{b^2}

    which rearranges to give

    \tan\alpha = \displaystyle\frac{h(b + a)}{ab}

    substituting this into (*), and using the identity \tan^2x + 1 = \sec^2x, we obtain

    \frac{2V^2}{g} = \displaystyle\frac{(\frac{a^2b^2 + h^2(b+a)^2}{b^2})}{\frac{h(b+a)}  {ab} - h}

    which rearranges to give the required result.

    From our expression for tan(alpha), we get

    \tan(\alpha + \delta\alpha) = (h + \delta h)\frac{a+b}{ab}

    Using the formula for tan(A+B) and rearranging:

    \frac{a+b}{ab}\delta h = \displaystyle\frac{\delta\alpha\  sec^2\alpha}{1 - \delta\alpha\tan\alpha}

    Because \delta\alpha is small, we can disregard its term in the denominator, giving the required result.

    Now we similarly obtain:

    \frac{2(V + \delta V)^2}{g} = \frac{ab}{h + \delta h} + \frac{(a+b)^2(h + \delta h)}{ab}

    replacing the part of the last fraction that equals 2V^2/g and rearranging:

    \frac{2}{g}((v + \delta v)^2 - v^2) = \frac{ab}{h + \delta h} + \frac{(a+b)^2\delta h}{ab} - \frac{ab}{h} (A)

    Now the LHS = \frac{2}{g}(2v\delta v + \delta v^2), of which we can disregard the \delta v^2 term because it is very small. Therefore LHS = \frac{4}{g}V\delta v.

    From this, we can conclude that \delta v has the same sign as the RHS of equation (A). Therefore, \delta v > 0 iff

    \frac{ab}{h + \delta h} + \frac{(a+b)^2\delta h}{ab} - \frac{ab}{h} > 0

    This rearranges to h^2 + h\delta h > \frac{a^2b^2}{(a + b)^2}

    we can disregard the h\delta h term because it is much smaller than all the others, and then we can square root to give the required result. The other result can easily be shown by reversing the > sign.
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    STEP II 2002 Q10

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    after t hours, the competitor has (42\frac{3}{8} - 13t)km left to run

    the remainder is run at (14+\frac{2t}{T})\mathrm{kmh}^{-1}, in a time (T-t) hours

    using speed = distance/time:

    14 + 2t/T = \displaystyle\frac{42\frac{3}{8} - 13t}{T - t}

    which rearranges to

    (2T)t^2 - t + (42\frac{3}{8} - 14T) = 0

    This is a quadratic in t. Clearly t is real, therefore it has real roots, so b^2 \geq 4ac

    \implies 1 \geq \frac{8}{T}(42\frac{3}{8} - 14T)

    which rearranges to give T \geq 3 for all values of t.

    If T = 3, then

    \frac{2}{3}t^2 - t + \frac{3}{8} = 0

    which has the (repeated) solution t = 3/4.

    Therefore competitor one runs at 13km/h for 3/4 hours
    then (29/2)km/h for the remains 9/4 hours

    Therefore distance run by competitor one after time x
    = 13x, x \leq 3/4
    = \frac{29}{2}x - \frac{9}{8}, x \geq 3/4

    (The 9/8 is obtained by observing that the two expressions must agree when x = 3/4.)

    Now the speed of competitor 2 after time x
    = 16 - kx
    we can integrate to find the distance:
    distance = 16x - \frac{1}{2}kx^2 + C
    from considering the distance travelled initially (0) and after 3 hours (42+3/8), we find
    c = 0, k = 5/4

    Therefore distance = 16x - \frac{5}{8}x^2

    therefore the distance between the two
    = |16x - \frac{5}{8}x^2 - 13x|, x \leq 3/4
    = |16x - \frac{5}{8}x^2 + \frac{9}{8} - \frac{29}{2}|, x \geq 3/4

    Via differentiation, we find that the first expression has no maxima in the required range, but the second expression has a maximum at x = 6/5.

    Substituting this in, we find that the maximum distance is 71/40km.

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    STEP II 2002 Q11

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    Consider the beam to be made up of tiny sections of length \delta x and distance x from B.

    Weight of one of these sections = \frac{\alpha W}{l}(\frac{x}{l})^{\alpha - 1} \delta x

    now distance of CoM from B = \displaystyle\frac{\sum x(\mathrm{weight})}{W}

    = \frac{1}{W} \displaystyle\sum^{x=l}_{x=0} x\frac{\alpha W}{l}(\frac{x}{l})^{\alpha - 1} \delta x

    Taking the limit as \delta x \rightarrow 0, it becomes an integral:

    = \displaystyle\int^l_0 \displaystyle\frac{\alpha x^{\alpha}}{l^{\alpha}} \mathrm{d}x

    = \displaystyle\frac{\alpha x^{\alpha + 1}}{l^{\alpha}(\alpha + 1)}|^l_0

    = \displaystyle\frac{\alpha l}{\alpha + 1}

    We now draw a diagram of the rod against the wall. There are normal reactions R_A, R_B at A and B respectively, with corresponding frictional forces \mu R_A, \mu R_B (acting towards the base of the wall and vertically upwards respectively.) There is also the weight of the rod, W, at the centre of mass.

    Resolving horizontally: R_B = \mu R_A
    Resolving vertically: W = \mu R_B + R_A
    Therefore W = (1 + \mu^2)R_A, which we will use later.

    Taking moments about B:

    \frac{W\alpha l}{\alpha + 1}\cos\theta = l\mu R_A \sin\theta = lR_A\cos\theta

    which rearranges eventually to give \tan\theta = \displaystyle\frac{\alpha + 1 - (\frac{W\alpha}{R_A})}{(1 + \alpha)\mu}

    Substituting in W = (1 + \mu^2)R_A gives the required result.

    If \alpha = 3/2, then \tan\theta = \displaystyle\frac{2 - 3\mu^2}{5\mu}

    If the beam slides for any \theta < \pi/4, then the greatest possible value of \mu is given when \theta = \pi/4 gives limiting equilibrium

    Therefore \tan(\pi/4) = \displaystyle\frac{2 - 3\mu^2}{5\mu}

    \implies 3\mu^2 + 5\mu - 2 = 0

    \implies \mu = \displaystyle\frac{-5 + \sqrt{25 + 24}}{6} (ignoring the negative root)

    \implies \mu = \frac{1}{3}



    Yay, all three mechanics.
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    f(x)=x^n+a_1x^{n-1}+...+a_{n-1}x+a_n=(x+k_1)(x+k_2)\cdots(x+k  _n)
    Looking at f(0) gives that f(0)=a_n=\displaystyle\prod_{i=1  }^n k_i as desired
    Now, consider f(1) and f(1)=1+\displaystyle\sum_{i=1}^n a_i=\displaystyle\prod_{i=1}^n (1+k_i) as required.
    If you consider f(-1) you on one hand get \displaystyle\prod_{i=1}^n (k_i-1) and on the other hand (-1)^n+\displaystyle\sum_{i=1}^n(-1)^{n-1}a_i
    So, for even n 1-a_1+a_2-\cdots-a_{n-1}+a_n=(k_1-1)(k_2-1)\cdots(k_n-1)
    and for odd n -1+a_1-a_2+a_3-\cdots+a_{n-1}-a_n=(k_1-1)(k_2-1)\cdots(k_n-1)


    Polynomial: x^4+22x^3+172x^2+552x+576=0 Guess the root x=-2 f(-2)=16-176+688-1104+576=0 We can now factorise it to (x+2)(x^3+20x^2+132x+288) study f(-6)=-216+720-792+288=1008-1008=0. Thus (x+2)(x+6)(x^2+14x+48)=x^4+22x^3  +172x^2+552x+576 Solving the resulting quadratic gives x=\frac{-14\pm\sqrt{196-192}}{2}=-6 \text{and}-8
    Thus the roots to the polynomial are x=-2, -6 (double) and -8.
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    0\le (a+b)^2=a^2+b^2+2ab \implies 2ab\le a^2+b^2 So (a+b)^2=a^2+b^2+2ab \le a^2+b^2+a^2+b^2=2(a^2+b^2) as desired.

    y=(a^2\cos^2(\theta)+b^2\sin^2(\  theta))^{\frac{1}{2}}+(b^2\cos^2  (\theta)+a^2\sin^2(\theta))^{\fr  ac{1}{2}} So
    \frac{dy}{dx}=\frac{1}{2}(\frac{  (b^2-a^2)\sin(2\theta)}{(a^2\cos^2(\t  heta)+b^2\sin^2( \theta))^{\frac{1}{2}}}}+ \frac{(a^2-b^2)\sin(2\theta)}{b^2\cos^2(\th  eta)+a^2\sin^2( \theta))^{\frac{1}{2}}}) = \frac{\sin(2\theta)}{2}(\frac{b^  2-a^2}{(a^2\cos^2(\theta)+b^2\sin^  2(\theta))^{\frac{1}{2}}}+\frac{  a^2-b^2}{(b^2\cos^2(\theta)+a^2\sin^  2(\theta))^{\frac{1}{2}}})
    For stationary points dy/dx=0, thus one solution is \frac{sin(2\theta)}{2}=0 \Leftrightarrow \sin(2\theta)=0 \implies \theta=k\frac{\pi}{2}
    All other solutions will be when:
    \frac{b^2-a^2}{(a^2\cos^2(\theta)+b^2\sin^  2(\theta))^{\frac{1}{2}}}= - \frac{a^2-b^2}{(b^2\cos^2(\theta)+a^2\sin^  2(\theta))^{\frac{1}{2}}} \Leftrightarrow (a^2\cos^2(\theta)+b^2\sin^2(\th  eta))^{\frac{1}{2}}=(b^2\cos^2(\  theta)+a^2\sin^2(\theta))^{\frac  {1}{2}}) \implies a^2\cos^2(\theta)+b^2\sin^2(\the  ta)=b^2\cos^2(\theta)+a^2\sin^2(  \theta) \Leftrightarrow a^2(\cos^2(\theta)-\sin^2(\theta))=b^2(\cos^2(\thet  a)-\sin^2(\theta)) \Leftrightarrow a^2\cos(2\theta)=b^2\cos(2\theta  ) \Leftrightarrow (a^2-b^2)\cos(2\theta)=0 \newline \implies a^2-b^2=0 \text{or} \cos(2\theta)=0 (\Leftrightarrow 2\theta=\frac{\pi}{2}+k\pi \Leftrightarrow \theta=\frac{\pi}{4}+k\frac{\pi}  {2}) \newline a^2-b^2=(a-b)(a+b)
    Thus a=b or a=-b

    Now we have four solutions to dy/dx=0, check these to verify which point is max/min.
    a=b and a=-b give the same value for the function y, namely y=a+b=2a=2b
    \theta=0 \implies y=|a|+|b| \newline \theta=\frac{\pi}{4}\implies y=\sqrt{2}(a^2+b^2)^{\frac{1}{2}  }
    Thus, if you recall the result proved in the beginning ((a+b)^2\le2(a^2+b^2)\implies a+b\le\sqrt{2}\sqrt{a^2+b^2}, \theta=0 produces a minimum and \theta=\frac{\pi}{4} produces a maximum.

    As global max is \sqrt{2}\sqrt{a^2+b^2}=(2(a^2+b^  2))^{\frac{1}{2}} then y\le(2a^2+2b^2)^{\frac{1}{2}}. Moreover, global minimum is |a|+|b| thus |a|+|b|\le (a^2\cos^2(\theta)+b^2\sin^2(\th  eta))^{\frac{1}{2}}+(b^2\cos^2(\  theta)+a^2\sin^2(\theta))^{\frac  {1}{2}} \le(2a^2+2b^2)^{\frac{1}{2}}
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    (Original post by nota bene)
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    For the roots, I'm pretty sure this is what you were supposed to do:

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    Note 576 = 600 - 24 = 25 x 24 - 24 = 24 x 24 = 3^2.2^6.

    Suppose the roots are ABCD. Then we know ABCD = 3^2 . 2^6

    Also, calculate 1 + 22 + 172 + 552 + 576 = 599 + 172 + 552 = 771 + 552 = 1323 = 3 x 441 = 3 x 3 x 3 x 7 x 7.

    So we know (A-1)(B-1)(C-1)(D-1) = 3 x 3 x 3 x 7 x 7. (*)

    And finally calculate 1 - 22 + 172 - 552 + 576 = 175 = 7 x 5 x 5.

    So we know (A+1)(B+1)(C+1)(D+1) = 7 x 5 x 5. (**)

    Looking at (*), to get the '7' factors, we must have two roots that are either 8 (since 8-1 = 7) or -6 (since -6-1 = -7). Or we could have an unexpectedly large root, but that isn't likely.

    But then looking at (**), we see the '8' case can't work.
    So we must have 2 roots that = -6, WLOG, C = D = -6.

    Then from (*) we get (A-1)(B-1) = 27
    and from (**) we get (A+1)(B+1) = 7

    So either {A+1, B+1} = {1,7} or {-1, -7}.

    Since we know no root = 0 (since the product of ABCD isn't 0), we must have the {-1, -7} case and so 2 additional roots -2, -8.

    Note: since we didn't rule out the "unexpectedly large root test", we should still check these actually are the 4 roots.
    Note that this is actually a lot longer than your solution, but I suspect you had some failed "root guesses" that you didn't bother writing up.

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    In practical terms, I suspect using a hybrid technique to make educated guesses at the roots would be better than either pure guessing, or trying to deduce the roots solely by using the first part of the question.
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    (Original post by nota bene)
    \theta=0 \implies y=a+b

    * I think I'm missing something obvious here? I shouldn't rely on the function being non-negative...[/spoiler]
    I think this is your problem:
    Spoiler:
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    When \theta = 0,\, y = \sqrt{a^2} + \sqrt{b^2} = |a| + |b|, not a+b.
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    (Original post by DFranklin)
    For the roots, I'm pretty sure this is what you were supposed to do:

    Note 576 = 600 - 24 = 25 x 24 - 24 = 24 x 24 = 3^2.2^6.
    I did use this whilst searching for my roots, and clearly -1 doesn't work, -2 does, -3 and -4 fail after quick mental calculation, -6 is a hit. Then I just solved the quadratic. I suspected all way along that I was supposed to do something using more of the previous result, but as I knew that bashing it out by guessing roots would be quick I didn't think more about it. (I did these two questions last week, so just typed them up now)
    (Original post by DFranklin)
    I think this is your problem:
    Indeed, corrected.
    Thanks!
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    (Original post by Dadeyemi)
    Some more;

    Did these quite a while ago I'm afraid some may be partial solutions.
    RE Q3: You lost an a squared very early on and so came to get a minimum on the x-axis. In fact the minimum occurs when  x = \frac{a^2 b}{a^2 - 1} , and at this point f(x)>0. Also, your axes are very inappropriately placed seeing as b>0. Thankfully you then gave up, because if you had carried on you would have got into quite a mess!

    For anyone looking for a solution to this question I got:

    c^2 < b(a^2-1) gives 0 solutions
    c^2 > ba^2 gives 1 solution
    ba^2 \geq c^2 > b(a^2-1) gives 2 solutions
    c^2 = b(a^2-1) also gives 1 solution

    For part (i) note that we have the case where c^2 = b(a^2-1), and require  x = \frac{a^2 b}{a^2 - 1} = \frac{9}{4} . This is the only solution.

    For part (ii) it isn't quite as neat, we have the case where ba^2 \geq c^2 > b(a^2-1) so we're looking for two solutions. One is x=4 which you see quickly on inspection, for the other we have to trawl through some algebra to get x=49/16. Even if the first part of the question isn't to taste I think part two is worth doing to check you know what to do to transform the equation into something managable, namely a standard quadratic.
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    (Original post by Dadeyemi)
    Some more;

    Did these quite a while ago I'm afraid some may be partial solutions.
    I think you're missing an a^2, i get the minimum to be (b(a^2)/(a^2 - 1), (b(a^2 - 1))^1/2. To finish the question off i get:

    By looking at the graph it is clear that:
    Conditions for exactly one solution are c = b(a^2)/(a^2 - 1) or c > a(b^1/2)
    For exactly two solutions b(a^2 - 1) < c^2 < (a^2)b

    i) Squaring once gives 8x - 14 = 8(x - 2)^1/2, squaring again gives 16(x^2) - 72x + 81 = 0, by the quadratic formula, x = 9/4

    ii) Squaring once 8x - 22 = 10(x - 3)^1/2 squaring again 16(x^2) - 113x + 196 = 0 by the quadratic formula x = 4 or x = 49/16

    Seems too easy so i've probably gone wrong. I can't see how the first bit helps with the second, it tells you how many solutions the equations have but you find that out for yourself when you solve them. Any feedback welcome.
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    (Original post by toasted-lion)
    RE Q3: You lost an a squared very early on and so came to get a minimum on the x-axis. In fact the minimum occurs when  x = \frac{a^2 b}{a^2 - 1} , and at this point f(x)>0. Also, your axes are very inappropriately placed seeing as b>0. Thankfully you then gave up, because if you had carried on you would have got into quite a mess!

    For anyone looking for a solution to this question I got:

    c^2 &lt; b(a^2-1) gives 0 solutions
    c^2 &gt; ba^2 gives 1 solution
    ba^2 \geq c^2 &gt; b(a^2-1) gives 2 solutions
    c^2 = b(a^2-1) also gives 1 solution

    For part (i) note that we have the case where c^2 = b(a^2-1), and require  x = \frac{a^2 b}{a^2 - 1} = \frac{9}{4} . This is the only solution.

    For part (ii) it isn't quite as neat, we have the case where ba^2 \geq c^2 &gt; b(a^2-1) so we're looking for two solutions. One is x=4 which you see quickly on inspection, for the other we have to trawl through some algebra to get x=49/16. Even if the first part of the question isn't to taste I think part two is worth doing to check you know what to do to transform the equation into something managable, namely a standard quadratic.
    I left this page open overnight and didn't refresh, well i guess 2 solutions is better than none lol!
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    (Original post by SimonM)
    STEP III, Question 1

    Spoiler:
    Show
    The area is \displaystyle \int_1^a \frac{\ln x}{x} \, dx = \left [ \frac{(\ln x)^2}{2} \right ]_1^a = \frac{(\ln a)^2}{2}

    As a \to \infty, \ln a \to \infty so the area tends to infinity as well

    Volume of the solid of revolution is

    \displaystyle \pi \int_1^a \left ( \frac{\ln x}{x} \right )^2 \, dx = \pi \left ( \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \int_1^a \frac{2 \ln x}{x^2} \, dx \right) =

    \displaystyle \pi \left [ - \frac{(\ln x)^2}{x} \right ]_1^a  + \pi \left [ - \frac{2 \ln x}{x} \right ]_1^a +\int_1^a \frac{2}{x^2} \, dx =

    \displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{\ln x}{x} - \frac{2}{x} \right ]_1^a =

    \displaystyle \pi \left ( 2 - \frac{(\ln a)^2}{a} - \frac{\ln a}{a} - \frac{2}{a} \right )

    As  a \to \infty, the volume tends to 2\pi
    I might be missing something obvious but how did you determine that (lna)^2/a and (lna)/a tend to zero as a tends to infinity? I knew that (lna)/a tended to zero but proved it using l'hopital because I thought they'd want to see proof, i didn't know that (lna)^2/a did though so also ended up showing it using l'hopital.
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    (Original post by maltodextrin)
    I left this page open overnight and didn't refresh, well i guess 2 solutions is better than none lol!
    At least you know you didn't go wrong now!
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    (Original post by maltodextrin)
    I might be missing something obvious but how did you determine that (lna)^2/a and (lna)/a tend to zero as a tends to infinity? I knew that (lna)/a tended to zero but proved it using l'hopital because I thought they'd want to see proof, i didn't know that (lna)^2/a did though so also ended up showing it using l'hopital.
    How's this:  \frac {ln(x)}{x} tends to 0 as x tends to infinity, which we know and seems intuitively obvious (but please post your proof as I've never heard of l'hopital) so, letting  x = \sqrt a ,  \frac {ln(\sqrt a )}{\sqrt a} tends to 0 and also  \frac {2ln(\sqrt a )}{\sqrt a} = \frac {ln(\sqrt a ^2 )}{\sqrt a} = \frac {ln( a )}{\sqrt a} tends to 0, so squaring,  \frac {(ln(a))^2}{a} tends to 0. That seems ok to me, now your "l'hopital"?
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    (Original post by toasted-lion)
    How's this:  \frac {ln(x)}{x} tends to 0 as x tends to infinity, which we know and seems intuitively obvious (but please post your proof as I've never heard of l'hopital) so, letting  x = \sqrt a ,  \frac {ln(\sqrt a )}{\sqrt a} tends to 0 and also  \frac {2ln(\sqrt a )}{\sqrt a} = \frac {ln(\sqrt a ^2 )}{\sqrt a} = \frac {ln( a )}{\sqrt a} tends to 0, so squaring,  \frac {(ln(a))^2}{a} tends to 0. That seems ok to me, now your "l'hopital"?
    http://en.wikipedia.org/wiki/L'hopital's_rule

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