nuodai
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39 down, 0 to go!

STEP I:
1: Solution by DeanK22, last bit by und;
2: Solution by darkness9999; Solution by nuodai
3: Solution by nuodai
4: Solution by DeanK22
5: Solution by Blackjamin
6: Solution by Reminisce
7: Solution by Give
8: Solution by My Alt
9: Solution by Felix Felicis
10: Solution by safmaster
11: Solution by brianeverit
12: Solution by nuodai
13: Solution by QED


STEP II:
1: Solution by Glutamic Acid
2: Solution by Glutamic Acid
3: Solution by Daniel Freedman
4: Solution by Daniel Freedman
5: Solution by Glutamic Acid
6: Solution by Daniel Freedman
7: Solution by Glutamic Acid
8: Solution by Daniel Freedman
9: Solution by Glutamic Acid
10: Solution by Farhan.Hanif93
11:Solution by brianeverit
12: Solution by metaltron
13: Solution by brianeverit


STEP III:
1: Solution by DeanK22
2: Solution by DeanK22
3: Solution by Daniel Freedman
4: Solution by DeanK22
5: Solution by Daniel Freedman
6: Solution by around
7: Solution by around
8: Solution by DeanK22
9: Solution by brianeverit
10: Solution by Daniel Freedman
11: Solution by brianeverit
12: Solution by miml
13: Solutions by Daniel Freedman and brianeverit


Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007 - 2008
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darkness9999
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STEP I Q2

STEP I Q2


\displaystyle t= x+ \sqrt{x^2 + 2bx + c}

\displaystyle\Rightarrow \frac{dt}{dx}= 1+ \frac{x+b}{\sqrt{x^2 + 2bx + c}}

\displaystyle\Rightarrow 1 + \frac{x+b}{t-x}

\displaystyle\Rightarrow \frac{t+b}{t-x}

\displaystyle\therefore \boxed{\frac{dx}{dt}=\frac{t-x}{t+b}}




\displaystyle\int \frac{1}{\sqrt{x^2 + 2bx + c}} \ dx

\displaystyle\Rightarrow \int \frac{(t-x)}{(t+b)(t-x)}  \ dt

\displaystyle\Rightarrow \displaystyle\int \frac{1}{t+b} \ dt

\displaystyle\Rightarrow \boxed{\ln{\left|x + \sqrt{x^2 + 2bx + c} + b\right|} + Constant}



I will update my post with the last bit ...!
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nuodai
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STEP I 2008 Question 2

Solution

Rearranging the equation gives t - x = \sqrt{x^2 + 2bx + c} \Rightarrow t^2 - 2tx + x^2 = x^2 + 2bx + c \Rightarrow t^2 - 2tx = 2bx + c

Differentiating this with respect to t then gives:
\newline \displaystyle 2t - 2t\dfrac{\mbox{d}x}{\mbox{d}t} - 2x = 2b\dfrac{\mbox{d}x}{\mbox{d}t}\newline

\Rightarrow 2x - 2t = \dfrac{\mbox{d}x}{\mbox{d}t}(-2b - 2t)\newline

\Rightarrow \dfrac{\mbox{d}x}{\mbox{d}t} = \dfrac{2x-2t}{-2b-2t}\times \dfrac{-1/2}{-1/2} = \boxed{\dfrac{t-x}{t+b}}

We can use the substitution t = x + \sqrt{x^2 + 2bx + c} in the integral to give us:
\newline \displaystyle \int \dfrac{1}{\sqrt{x^2 + bx + c}}\ \mbox{d}x\newline

 = \int \dfrac{1}{t-x}\cdot\dfrac{t-x}{t+b}\ \mbox{d}t \newline

= \int \dfrac{1}{t+b}\ \mbox{d}t = \ln | t + b | + C = \ln | x + \sqrt{x^2 + 2bx + c} + b| + C

When b^2 = c, we obtain \sqrt{x^2 + 2bx + c} = \sqrt{(x + b)^2} = \pm(x + b).

So, the integral reduces to \displaystyle \int \dfrac{1}{x+b}\ \mbox{d}x = \ln |x + b| + C

By use of the substitution, we have \ln |x + \sqrt{(x+b)^2} + b| = \ln |x + b + |x + b|| + C

When x + b > 0,\ |x + b| = x + b
\newline\Rightarrow \ln |x + b + |x + b|| + C = \ln (2[x+b]) + C\newline

 = \ln 2 + \ln (x + b) + C\newline

 = \boxed{ \ln (x + b) + D}
So it holds when x + b > 0.

When x + b < 0,\ |x + b| = -(x+b)
\newline \Rightarrow \ln | x + b + |x + b| | + C = \ln |x + b - (x + b) |  + C\newline

 = \boxed{\ln 0 + C}
This is undefined, so it doesn't hold when x + b < 0


EDIT: Beaten to it! Well we did it different ways so I'll put them both up :p:
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Oh I Really Don't Care
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STEP III Question 1

Using the suggested hint;

 \displaystyle \frac{(x+y)}{3} = (ax+by)(x+y) = \left(ax^2 + by^2\right) + xy(a+b) = \frac{(1+5xy)}{5}

Doing the exact idea again;

 \displaystyle \frac{(x+y)}{5} = (ax^2+by^2)(x+y) = \left(ax^3+by^3\right) + xy(ax + by) = \frac{(3+7xy)}{21}

Dividing the first equation obtained by the following one results in [after rearranging];

 xy = \frac{3}{35} along with  x+y = \frac{6}{7}

 (x - \alpha )(x - \beta) = x^2 - (\alpha + \beta)x + \alpha \beta

So x and y are the roots of the quadratic;

 p^2 - \frac{6p}{7} + \frac{3}{35} = 0

It follows;

 \displaystyle x = \frac{3}{7} \pm \frac{2\sqrt{30}}{35} \; and \; y= \frac{3}{7} \mp \frac{2\sqrt{30}}{35}

Subsituting these into equation (i) and equation (ii) gives two simulateneous eqnts in a and b that are too tedious for me to solve without being in an exam.
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Oh I Really Don't Care
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STEP III, Question 2

This is a classic example of a telescoping series and the summation is as follows:

 \displaystyle \sum_{r=0}^{n} \left((r+1)^k - r^k \right) = (n+1)^k

We then proceed by binomial expansion;

 \displaystyle \sum_{r=0}^{n} \left((r+1)^k - r^k \right) = (n+1)^k = k\sum_{r=0}^{n}r^{k-1} + \binom{k}{2}\sum_{r=0}^{n} r^{k-2} + \dots + \sum_{r=0}^{n} 1

Using the notation;

 \displaystyle S_k(n) = \sum_{r=0}^{n}r^k it follows that;

 \displaystyle kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2} S_{k-2}(n) - \dots - \binom{k}{k-1} S_1(n)

Applying the above with k = 4,5 to find S_3 and S_4 respectively quickly works.

(ii) DFranklin solutions;

Claim: S_k(n) is a poly of order k+1 for all k.

Proof: Use strong induction on k. Suppose true for k < N.

Then by (*), S_N(n) = (n+1)^{N+1} + "the sum of a load of polynomials that are of order < N+1 (by induction hypothesis)". Since the sum of an N+1 degree poly and a load of polys of degree < N+1 is a poly of degree N+1, S_N(n) is a N+1 degree poly. Since S(0) = n is a degree 1 poly, the result follows by induction.

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.
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around
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STEP III 2008 Question 6

Solution


Remember throughout that \frac{dy}{dx} = p

 y = p^2 + 2xp

Differentiate w.r.t x:

\frac{dy}{dx} = 2p\frac{dp}{dx} + 2x\frac{dp}{dx} + 2p

Or

-1 = \frac{dp}{dx}(2 + 2\frac{x}{p})

This now results in a linear differential equation; solve using the I.F:

{e}^\int 2\ln p dx = p^2

\frac{d}{dp}(xp^2) = -2p^2
xp^2 = \frac{-2}{3}p^3 + A

From the constraints we see that A is zero:

x = \frac{-2}{3} \frac{dy}{dx}

Separate variables and integrate:

y = \frac{-3}{4}x^2 + C

We can find y from the equation given earlier: when x = 2 y = -3. Hence C = 0.

The second part proceeds much like the first; first, differentiate w.r.t x:

\frac{dx}{dp} + \frac{2x}{p} = - \frac{ln p + 1}{p}

This has the same I.F as before:

\frac{d}{dx} (xp^2) = -p (ln p +1)

Integrate this:

xp^2 = -\frac{p^2 ln\ p}{2} - frac{p^2}{4} + B

Using the initial conditions it's pretty obvious that B is again 0.

Re-arrange this a little:

\frac{1}{\sqrt e} e^-2x = \frac{dy}{dx}

y = \frac{-1}{2\sqrt e}e ^ -2x + D

Again, use the original equation they provide to find y (-1/2) and hence D is zero



In other news I found this paper pretty tough. I think this was the only full solution I managed when I did it as a mock :<
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Oh I Really Don't Care
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STEP III, Question 4

(i) Letting y = x so I don't refer to a y or x by mistake the problem will be restated as;

show that

x > tanh(x/2) if x > 0

Let  y_1 = x \; and \; y_2 = tanh(\frac{x}{2}) when x = 0, y_1 == y_2 = 0

We then consider the derivitive and see that  (y_2)' &lt; 1 when x > 0 and the result follows.

archosh(z) > (z-1)/root(z^2-1) Let z = cosh(x) the result trivially follows from the above after rewriting (x^2-1) as (x-1)(x+1) and remembering the identity linking cosh(x) with sinh^2(x/2)

(ii) We need to that if in an interval [a,b] that f(x) => g(x) it follows that;

 \displaystyle \int_{a}^{b} f(x) dx \ge \int_{a}^{b} g(x) dx

Using IBP and splitting the integral we see that;

 \displaystyle \int_1^x arcosh(x) dx \ge \int_1^x \frac{x-1}{\sqrt{x^2-1}} dx

and the result follows.

Integrating (ii) wrt x gives the required result.
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nuodai
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STEP I Question 3

Introduction

If ab + cd \ge bc + ad, then ab + cd - bc - ad \ge 0

The proof that this is the case is as follows:
\newline ab + cd - bc - ad\newline

= a(b-d) + c(d-b)\newline

= a(b-d) - c(b-d)\newline

= (b-d)(a-c)\newline

= (d-b)(c-a)
This is greater than or equal to 0, since d \ge b,\ c \ge a
Part (i)

It follows from the introduction that if we let a = b = x and c = d = y, then it still holds that c \ge a,\ d \ge b, so we can just substitute them in to give us x^2 + y^2 \ge xy + xy \Rightarrow \boxed{ x^2 + y^2 \ge 2xy }

We are then told that x \ge z,\ y \ge z, and it is hinted that a = b = z, meaning that c = x,\ d = y. Once again, this fits the necessary conditions, so \boxed{z^2 + xy \ge xz + yz}

Combining the two inequalities gives us x^2 + y^2 + z^2 + xy \ge xz + yz + 2xy \Rightarrow \boxed{x^2 + y^2 + z^2 \ge xy + xz + yz}


Part (ii) coming when I can be bothered with tedium.
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Oh I Really Don't Care
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STEP III, question 8

1 + 2P = 0 , <---> p = -1/2

S(1+px) = (x/3) + (x*x/6) + (x*x*x/12) + ... - (x*x/6) - (x*x*x/12) - .... = x/3

It follows

 \displaystyle S_{\infty}(1-\frac{x}{2}) = \frac{x}{3} \iff S_{\infty} = \frac{2x}{3(2-x)} (as clearly we know that x =/= 2 .

We can do a similar thing by considering  \frac{px^{n+1}}{2^{n+1}} and it follows:

 \displaystyle S_{n+1} = \left(1 - \frac{x^{n+1}}{2^{n+1}} \right) \frac{2x}{3(2-x)}

(ii) Do the same and we obtain: q = 1 , p = -2.5

 \displaystyle T_{\infty} (1+px+qx^2) = 2 + 3x \iff T_{\infty} = \frac{4+6x}{(2-x)(1-2x)}

We then proceed by partial fractions:

 \displaystyle \frac{4+6x}{2(1-\frac{x}{2})(1-2x)} = \frac{A}{1-\frac{x}{2}} + \frac{B}{1-2x}

B = 14/3 and A = -8/3

Then proceed as above;

 \displaystyle T_{n+1} = \frac{14}{3} \left(\frac{1-(2x)^{n+1}}{1-2x}\right) - \frac{4}{3} \left(\frac{1-(\frac{x}{2})^{n+1}}{2-x}\right)
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DFranklin
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(Original post by DeanK22)
(ii) To me this is blatantly obviously true. It has basically been shown by the working above. I do not know what the eaminer wants from this - just look at the above epression? I guess you could talk about the relationship between integration and summation? For a STEP III I am not going to tell you that saying "well it is kin of obvious" is the thing to do but this seems like the case here.
I think you'd be expected to do a proof by induction for the first bit.

e.g. Claim: S_k(n) is a poly of order k+1 for all k.

Proof: Use strong induction on k. Suppose true for k < N.

Then by (*), S_N(n) = (n+1)^{N+1} + "the sum of a load of polynomials that are of order < N+1 (by induction hypothesis)". Since the sum of an N+1 degree poly and a load of polys of degree < N+1 is a poly of degree N+1, S_N(n) is a N+1 degree poly. Since S(0) = n is a degree 1 poly, the result follows by induction.

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.

To be honest, your "it is kind of obvious" comment looks dubious here. There are some pretty clear steps that you need to go through. And waffling about links between integration and summation indicates to me that you are missing the point of the question.
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SimonM
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nuodai do you think you could use the same format for the first post as the other years (ie using "Solution by name" as the hyperlink. The only exception to this is 2007 which I intend to fix in the near future)
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Oh I Really Don't Care
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(Original post by DFranklin)
For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.
:o: I just read the part about proving it would be a polynomial of degree k+1 - these other bits look abit more involved. Do you think it would be neccessary to prove the bit about the degree of the polnomial or would it be OK to say that from the previous working it can be seen?
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nuodai
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(Original post by SimonM)
nuodai do you think you could use the same format for the first post as the other years (ie using "Solution by name" as the hyperlink. The only exception to this is 2007 which I intend to fix in the near future)
Consider it done.
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DFranklin
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(Original post by DeanK22)
:o: I just read the part about proving it would be a polynomial of degree k+1 - these other bits look abit more involved. Do you think it would be neccessary to prove the bit about the degree of the polnomial or would it be OK to say that from the previous working it can be seen?
No, it wouldn't be OK.
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Glutamic Acid
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II/7:.

(i) y = u(1+x^2)^{1/2} \Rightarrow \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}u}{\text{d}x}(1+x^2)^{1/2} + xu(1+x^2)^{-1/2}

\Rightarrow \dfrac{(1+x^2)^{-1/2}}{u} \left[ \dfrac{\text{d}u}{\text{d}x}(1+x^2)^{1/2} + xu(1+x^2)^{-1/2} \right] = xu(1+x^2)^{1/2} + \dfrac{x}{1+x^2}

\Rightarrow \dfrac{1}{u^2}\dfrac{\text{d}u}{\text{d}x} = x(1+x^2)^{1/2} \Rightarrow \dfrac{-1}{u} = \dfrac{1}{3}(1+x^2)^{3/2} + c when x = 0, y = 1 so u = 1 so c = -4/3.

Solution: - \dfrac{(1+x^2)^{1/2}}{y} = \dfrac{1}{3}(1+x^2)^{3/2} - \dfrac{4}{3}.

(ii) Let y = u(1+x^3)^{1/3} \Rightarrow \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}u}{\text{d}x}(1+x^3)^{1/3} + ux^2(1+x^3)^{-2/3}

\Rightarrow \dfrac{(1+x^3)^{-1/3}}{u} \left[ \dfrac{\text{d}u}{\text{d}x} (1+x^3)^{1/3} + ux^2(1+x^2)^{-2/3} \right] = ux^2(1+x^2)^{1/3} + \dfrac{x^2}{1 + x^3}

\Rightarrow \dfrac{1}{u^2} \dfrac{\text{d}u}{\text{d}x} = x^2(1+x^3)^{1/3} \Rightarrow \dfrac{-1}{u} = \dfrac{1}{4}(1+x^3)^{4/3} + c
u = 1, again, so c = - 5/4.

Solution: - \dfrac{(1+x^3)^{1/3}}{y} = \dfrac{1}{4}(1+x^3)^{4/3} - \dfrac{5}{4}.

(iii) I conjecture the solution to be - \dfrac{(1+x^n)^{1/n}}{y} = \dfrac{1}{n+1}(1+x^n)^{1 + 1/n} - \dfrac{n+2}{n+1}. To prove it, one would make the substitution y = u(1+x^n)^(1/n).
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Glutamic Acid
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II/5:

\displaystyle \int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x} \, \text{d}x = \int_0^{\pi/2} \frac{2 \cos x \sin x}{1 + \sin^2 x}  \, \text{d}x Letting u = sin x, this becomes
\displaystyle \int_0^1 \frac{2u}{1 + u^2}  \, \text{d}u = \left[ \ln(1 + u^2) \right]^1_0 = \ln 2.

\displaystyle \int_0^{\pi/2} \frac{\sin x}{1 + \sin^2 x}  \, \text{d}x = \int_0^{\pi/2} \frac{- \sin x}{\cos^2 x - 2}  \, \text{d}x u = cos x, this becomes
\displaystyle \int_1^0 \frac{\text{d}u}{u^2 -2} = \frac{1}{2 \sqrt{2}} \left[ \ln \left| \frac{x - \sqrt{2}}{x + \sqrt{2}} \right| \right]^0_1 = \frac{1}{2 \sqrt{2}} \ln \frac{\sqrt{2} + 1}{\sqrt{2} - 1}

(1 + \sqrt{2})^5 = 1 + 5 \sqrt{2} + 20 + 20 \sqrt{2} + 20 + 4 \sqrt{2} = 41 + 29 \sqrt{2} &lt; 99 \Leftrightarrow 29 \sqrt{2} &lt; 58 \Leftrightarrow \sqrt{2} &lt; 2 \Leftrightarrow 2 &lt; 4.
\sqrt{2} &gt; 1.4 \Leftrightarrow 2 &gt; 1.96

As 2^{\sqrt{2}} &gt; 2^{1.4} (exponential functions are strictly increasing), sufficient to show that 2^{7/5} &gt; 1 + \sqrt{2} \Leftrightarrow 2^7 &gt; (1 + \sqrt{2})^5, and 128 > 99 is enough to show this holds.

Suppose \ln 2 &gt; \dfrac{1}{2 \sqrt{2}} \ln \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} \Leftrightarrow 2 \sqrt{2} \ln 2 &gt; \ln (1 + \sqrt{2})^2 \Leftrightarrow \sqrt{2} \ln 2 &gt; \ln (\sqrt{2} + 1) \Leftrightarrow \ln 2^{\sqrt{2}} &gt; \ln(1 + \sqrt{2}) \Leftrightarrow 2^{\sqrt{2}} &gt; \sqrt{2} + 1, previously established, so the first integral is greater. And we are done, forever.
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Glutamic Acid
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II/9:

-h = 40t \sin \alpha - \frac{1}{2} \cdot 10t^2 and 20 = 40t \cos \alpha therefore -h = \dfrac{40 \sin \alpha}{2 \cos \alpha} - \dfrac{5}{4} \sec^2 \alpha \Rightarrow -h = 20 \tan \alpha - \dfrac{5}{4}(\tan^2 \alpha + 1)
 \Rightarrow \tan \alpha = \dfrac{20 \pm \sqrt{400 - 5(5/4 -h)}}{5/2}.
The larger root means the ball will loop up in the air for ages and ages and ages like some sort of stupid duck so we choose the smaller root.
As t = sec a / 2, and cos a approx 1, t is approx 1/2.

(ii) h > 5/4.

(iii)\tan \alpha = \dfrac{20 - \sqrt{400 + 25/4}}{5/2} = \dfrac{20 - 20(1 + 1/64)^{1/2}}{5/2} \approx \dfrac{20 - 20(1 + 1/128)}{5/2} = -1/16 (first term binomial expansion). Using the tan a = a approximation and that 1 radian is 57 degres, alpha is approximately 57/16 degrees.
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Daniel Freedman
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STEP III 2008, Question 5

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Assume for induction that  T_n^2 -T_{n-1}T_{n+1} = f(x)

We wish to show that  T_{n+1}^2 -T_{n}T_{n+2} = f(x)

 \\ T_n^2 -T_{n-1}T_{n+1} = f(x) \\

\\ \implies T_n^2 - (2xT_n - T_{n+1})T_{n+1} = f(x), \ \mbox{by} \ (*)  \\

\\ \implies T_{n+1}^2 + T_n^2 - 2xT_nT_{n+1} = f(x) \\

\\ \implies T_{n+1}^2 + T_n( T_n - 2xT_{n+1}) = f(x) \\

\\ \implies T_{n+1}^2 + T_n( - T_{n+2}) = f(x), \ \mbox{by} \ (*)  \\

\\ \implies T_{n+1}^2 -T_{n}T_{n+2} = f(x)

as required. The base case  f(x) = T_1^2 - T_0 T_2 is given, thus completing the proof by induction.

In the case  f(x) \equiv 0 , we have

 \\ T_n^2 -T_{n-1}T_{n+1} = 0 \\

\\ \implies \frac{T_{n+1}}{T_n} = \frac{T_n}{T_{n-1}} = ... = \frac{T_1}{T_0} \\

\\ \implies T_n = r^n T_0, \ \mbox{where } r = \frac{T_1}{T_0}

(multiplying all the terms together (apart from the term in T_(n+1)) gives  \frac{T_n}{T_0} , which is also equivalent to  \left(\frac{T_1}{T_0}\right)^n )

Substituting into (*) gives,

\\ r^{n+1} T_0 - 2xr^n T_0 + r^{n-1}T_{0} = 0 \\

\\ \implies r^{n-1} (r^2 - 2xr  + 1) = 0 \\ 

\\ \implies r^2 - 2xr + 1 = 0

as T_0 is assumed to be non-zero.

 \therefore r = x \pm \sqrt{x^2 - 1}

by the quadratic formula.

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Daniel Freedman
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STEP III 2008, Question 3

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The line SP is given by:

 \\ \frac{y - b\sin{\theta}}{ - b\sin{\theta}} = \frac{x - a\cos{\theta}}{-ae - a\cos{\theta}} \\

\\ \implies y = \frac{b\sin{\theta}(x - a\cos{\theta})}{ae + a\cos{\theta}} + b\sin{\theta}

To find the gradient of the tangent at P:

 \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{b\cos{\theta}}{a\sin{\theta}} = - \frac{b}{a} \cot{\theta}

ON is perpendicular to the tangent at P, which means

 \\ \frac{y}{x}(-\frac{b}{a} \cot{\theta}) = 1 \\

\\ \implies y = \frac{a}{b} \tan{\theta} x

is the equation of the line through ON.

Rearranging this for x, and substituting into the equation of the line SP:

 \\ y = \frac{b\sin{\theta}(\frac{b}{a}\cot{\theta}y - a\cos{\theta})}{ae + a\cos{\theta}} + b\sin{\theta} \\

\\ \implies y( ae + a\cos{\theta} - \frac{b^2}{a} \cos{\theta}) = bae \sin{\theta} \\

\\ \implies y = \frac{bae \sin{\theta}}{ae + a\cos{\theta} - \frac{b^2}{a} \cos{\theta}} \\

\\ \implies y = \frac{bae \sin{\theta}}{ae + a\cos{\theta} - \frac{a^2(1-e^2)}{a}\cos{\theta}} \\

\\ \implies y = \frac{be\sin{\theta}}{e + e^2 \cos{\theta}} = \frac{b\sin{\theta}}{1 + e\cos{\theta}}

as required.

For T to lie on the circle centre S, radius a, it must satisfy:

 (x+ae)^2 + y^2 = a^2

 \\ LHS = \left(\frac{b^2\cos{\theta}}{a(1+e\cos{\theta})} + \frac{ae}{1}\right)^2 + \frac{b^2 \sin^2{\theta}}{(1+e\cos{\theta})^2} \\

\\ = \frac{(b^2\cos{\theta} + a^2e(1+e\cos{\theta}))^2 + a^2b^2\sin^2{\theta}}{a^2(1+e\cos{\theta})^2} \\

\\ = \frac{a^2 ( (1-e^2) \cos{\theta} + e(1+e\cos{\theta})^2)^2 + a^2(1-e^2)(1-\cos^2{\theta})}{(1+e\cos{\theta})^2} \\

\\ = \frac{a^2( (e + \cos{\theta})^2 + (1-e^2)(1-\cos^2{\theta}))}{(1+e\cos{\theta})^2} \\

\\ = \frac{a^2(e^2\cos^2{\theta} + 2e\cos{\theta} + 1)}{(1+e\cos{\theta})^2} = a^2  \\

\\ = RHS
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Daniel Freedman
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STEP III 2008, Question 10

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Resolving forces vertically on the rth particle,

 T_{r} = T_{r-1} + mg

Resolving forces vertically on the bottom particle,

 T_1 = m_g which gives  T_r = rmg = \frac{\lambda x_r}{l}

The length L of the long string is given by

\\  \displaystyle L = \sum_{r=1}^n \left( l + x_r \right) =  \sum_{r=1}^n \left( l + \frac{rmgl}{\lambda} \right) \\ \\

\\ = nl + \frac{mgl}{\lambda} \sum_{r=1}^n r \\ \\

\\ = nl +  \frac{n(n+1)mgl}{2\lambda}

The total elastic potential energy E stored in the long string is given by

 \\ \displaystyle E = \sum_{r=1}^n \frac{\lambda x_r^2}{2l}  = \sum_{r=1}^n \frac{ r^2 m^2 g^2 l}{2\lambda} = \frac{m^2 g^2 l}{2\lambda} \sum_{r=1}^n r^2 \\ \\

\\ = \frac{n(n+1)(2n+1) m^2 g^2 l}{12\lambda}

Let us divide the new heavy rope of mass M and natural length L_0, into n small ropes of mass m and natural length l. Clearly,

 M = nm and  L_0 = nl

Let n, the number of small ropes the large rope is divided into, tend to infinity to find the length L of the heavy rope.

 \\ \displaystyle L = \lim_{n\to\infty} \left( L_0 + \frac{L_0}{n} \frac{M}{n} \frac{g}{2\lambda} n(n+1) \right) = L_0 \left(1 + \frac{Mg}{2\lambda} \right)

as required.

Similarly to find the energy stored in the heavy rope,

 \\ \displaystyle E = \lim_{n\to\infty} \left( \frac{M^2}{n^2} g^2 \frac{L_0}{n} \frac{1}{12\lambda} n(n+1)(2n+1) \right) \\

\\ = \frac{M^2g^2L_0}{12\lambda} \lim_{n\to\infty} \frac{n(n+1)(2n+1)}{n^3} \\

\\ = \frac{M^2g^2L_0}{6\lambda} \\

\\ = \frac{(2\lambda(L-L_0))^2g^2L_0}{(gL_0)^2 6\lambda} \\ \\

\\ = \frac{2(L-L_0)^2\lambda}{3L_0}
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