SimonM
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STEP I:
1: Solution by Unbounded
2: Solution by Unbounded
3: Solution by nota bene
4: Solution by SimonM
5: Solution by nota bene
6: Solution by Unbounded
7: Solution by sonofdot
8: Solution by Unbounded
9: Solution by cliverlong
10: Solution by Dadeyemi
11:Solution by Unbounded
12: Solution by Robbie10538
13: Solution by Unbounded
14: Solution by brianeverit


STEP II:
1: Solution by sonofdot
2: Solution by dadeyemi
3: Solution by sonofdot
4: Solution by dadeyemi
5: Solution by dadeyemi
6: Solution by welshenglish
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by tommm
10: Solution by tommm
11: Solution by tommm
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit


STEP III:
1: Solution by SimonM
2: Solution by Dadeyemi
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Dadeyemi
6: Solution by Elongar
7: Solution by Dadeyemi
8: Solution by Dadeyemi
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by 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
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SimonM
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STEP III, Question 1

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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
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Oh I Really Don't Care
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STEP III, Question 1

a)

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 \displaystyle \int^{a}_{1} \frac{ln(x)}{x} dx = \left(\frac{[ln(x)]^2}{2}\left)^{a}_{1} = \frac{[ln(a)]^2}{2}

ln(a) tends to infinty as a tends to infinty;



b)

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 \dispalystyle V  = \pi \int^{a}_{1} y^2 dx = \pi \int^{a}_{1} \left(\frac{ln(x)}{x}\left)^2 dx

Let \dispalystyle e^u = x and it follows that the integral is transformed to;

\displaystyle \pi \int^{ln(a)}_{0} u^2e^{-u} du

And IBP twice yields;

\displaystyle \pi \left(-e^{-u}\left(u^2 + 2u - 2\left)\left)^{ln(a)}_{0} = \pi \left(\frac{-1}{a}\left((ln(a))^2 + 2ln(a) - 2\left) - 2)\left

As a tends to infinty the volume tends to  2\pi
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SimonM
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1-e^{-x} is strictly increasing, but doesn't tend to infinity
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sonofdot
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STEP II 2002 Question 3

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\displaystyle F_n = 2^{2^n} + 1 \ \ n = 0, \ 1, \ 2, \ldots

F_0 = 2^1 +1 = 3 \br

F_1 = 2^2 + 1 = 5 \br

F_2 = 2^4 + 1 = 17 \br

F_3 = 2^8 + 1 = 257

If k=1, then F_k - 2 = 3 and F_0 F_1 \ldots \F_{k-1} = F_0 = 3, therefore it is true for k=1.

If k=2, then F_k - 2 = 15 and F_0 F_1 \ldots \F_{k-1} = F_0 F_1 = 3 \times 5 = 15, therefore it is true for k=2.

If k=3, then F_k - 2 = 255 and F_0 F_1 \ldots \F_{k-1} = F_0 F_1 F_2 = 3 \times 5 \times 17 = 255, therefore it is true for k=3.

Assume, for some k F_0 F_1 \ldots F_{k-1} = F_k - 2

\begin{array}{rl}

\therefore F_0 F_1 \ldots F_{k-1} F_k & = F_k(F_k - 2) \\ \br \\

& = \left( 2^{2^k} + 1\right) \left(2^{2^k} -1 \right) \\ \br \\

& = \left( 2^{2^k} \right)^2 -1 \\ \br \\

& = 2^{2^{k+1}} -1 = F_{k+1} - 2 \end{array}

Therefore by induction, F_0 F_1 \ldots F_{k-1} = F_k - 2 for all positive integers k.

Now consider F_r and F_s with r<s. Assume that F_r and F_s have a common factor p

\begin{array}{rl}

p | F_r & \implies p | (F_0 F_1 \ldots F_r \ldots F_{s-1}) \\ \br \\

& \iff p | (F_s - 2) \end{array}

Since all Fermat numbers are odd, F_s and F_s - 2 cannot share a common factor greater than one, hence by contradiction, no two Fermat numbers have a common factor greater than 1.

All Fermat numbers are either prime or made up of prime factors. Since no Fermat numbers share a common factor greater than one, no Fermat numbers can have the same prime factor. Since there are infinitely many Fermat numbers, there must therefore be infinitely many prime numbers.
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Unbounded
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Question 1, STEP I, 2002
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We need to find the points of intersection between the two ellipses:

 (x+2)^2 + 2y^2 = 18 - ellipse 1
9(x-1)^2 + 16y^2 = 25 - ellipse 2

making y^2 the subject in the equation for ellipse 1:

 y^2 = \frac{18-(x+2)^2}{2}

and substitution into the equation for ellipse 2:

 9(x-1)^2 + 8(18-(x+2)^2) = 25
 \implies 9x^2-18x+9 + 144-8x^2-32x-32 = 25
 \implies x^2 - 50x +96 = 0
\implies (x-48)(x-2) = 0
 \implies x = 48, 2

Subbing back into the equation for y^2:

 2y^2 = 18-(50)^2 \implies y \ \mathrm{is \ not \ real}
 \therefore x \not= 48

 2y^2 = 18-(4)^2
 \implies 2y^2 = 2 \implies y^2 = 1 \implies y = \pm 1

Therefore the coordinates of intersection are (2,1) and (2,-1)

From a sketch, we can see that this means that the centre of a circle passing through those two points must lie on the x-axis.

Letting the coordinates of the centre of the circle be (a,0):

 (x-a)^2+y^2=r^2 where r is the radius of the circle.

Substitute the point (2,1) in:

 (2-a)^2 + 1 = r^2 \implies r^2 = 5 - 4a + a^2

So the equation of the circle now becomes:

 (x-a)^2 + y^2 = 5-4a+a^2

 \implies \boxed{x^2 -2ax + y^2 = 5-4a} \ \ \ \square
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Dadeyemi
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Some attached

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(the end of the question 2 is the reason why I should not do maths late at night :/
Attached files
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Dadeyemi
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Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
Attached files
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Unbounded
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Question 6, STEP I, 2002
First Part
Let the height of the equilateral triangle base be x.

From pythag:  x^2 + \frac{a^2}{4} = a^2

 \implies x^2 = \frac{3a^2}{4}

 \implies x = \frac{a\sqrt{3}}{2}

Therefore the area of this equilateral triangle base is given by:

 \frac{ax}{2} = \frac{a^2\sqrt3}{4}

Drawing a line down from the apex of the pyramid to the triangle base, letting h be the height, from pythagoras, we can see that:

 h^2 = b^2 - (\frac{2x}{3})^2

\implies h^2 = b^2 - \frac{a^2}{3}

\implies h^2 = \frac{1}{3} (3b^2-a^2)

 \implies h = \frac{\sqrt3}{3} (3b^2-a^2)^{\frac{1}{2}}

The volume of the cone, V is given by,  V = \frac{1}{3} Ah

 \therefore V = \frac{1}{3} \times \frac{a^2\sqrt3}{4} \times \frac{\sqrt{3}}{3} (3b^2-a^2)^{\frac{1}{2}}

 \implies \boxed{V = \frac{1}{12} a^2 (3b^2-a^2)^{\frac{1}{2}}} \ \ \ \square
Second Part
The area of the new base is to be found:

The height of the isoceles base, y, is given by:

 b^2 = y^2 + (\frac{a}{2})^2

 \implies y^2 = \frac{1}{4} (4b^2-a^2)

 \implies y = \frac{1}{2} (4b^2-a^2)^{\frac{1}{2}}

And so the area of this isoceles base is  \frac{1}{4} a(4b^2-a^2)^{\frac{1}{2}}

Now  V = \frac{1}{3} Ah

Let the new height of the pyramid be k

 \therefore V = \frac{1}{3} \times \frac{1}{4} a(4b^2-a^2)^{\frac{1}{2}} \times k

 \implies V = \frac{k}{12} a(4b^2-a^2)^{\frac{1}{2}}

\implies \frac{1}{12} a^2 (3b^2-a^2)^{\frac{1}{2}} = \frac{k}{12} a(4b^2-a^2)^{\frac{1}{2}}

 \implies k = \dfrac{a(3b^2-a^2)^{\frac{1}{2}}}{(4b^2-a^2)^{\frac{1}{2}}}

 \implies \boxed{k^2 = \dfrac{a^2(3b^2-a^2)}{4b^2-a^2}} \ \ \ \square
Third Part
Considering the right angled triangle formed now with the height of this pyramid and the height of the equilateral triangle. Let  \theta denote the angle between the equilateral triangle and the horizontal.

 \implies \sin \theta = \frac{k}{x}

 \implies \sin \theta = \dfrac{a(3b^2-a^2)^{\frac{1}{2}}}{4b^2-a^2}^{\frac{1}{2}}} \times \frac{2}{a\sqrt{3}}

 \implies \sin \theta = 2 \sqrt{\dfrac{(3b^2-a^2}{3(4b^2-a^2)}}

 \implies \boxed{\theta = \mathrm{arcsin} \ \left(2 \sqrt{\dfrac{(3b^2-a^2}{3(4b^2-a^2)}}\right)}
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sonofdot
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STEP II 2002 Question 1

Part One
\begin{array}{rl}

\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{1-\cos 2\theta} \, d\theta

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{1-(1-2\sin^2 \theta)} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{2\sin^2 \theta} \, d\theta \\ \br \\

& \displaystyle = \frac12 \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \mathrm{cosec}^2 \theta \, d\theta \\ \br \\

& \displaystyle = \frac12 \left[ -\cot \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ \br \\

& \displaystyle = \frac12 \left( \frac{1}{\tan \frac{\pi}{6}} - \frac{1}{\tan \frac{\pi}{4}} \right) \\ \br \\

& \displaystyle = \frac12 \left( \sqrt3 - 1 \right) \\ \br \\

& \displaystyle = \boxed{\frac{\sqrt3}{2} - \frac{1}{2}}

\end{array}
Part Two
\displaystyle\int_{\frac{\sqrt3}{2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx

x=\sin 2\theta \Rightarrow dx = 2\cos 2 \theta \, d\theta

\begin{array}{rl}

\therefore \displaystyle\int_{\frac{\sqrt3}{2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 2\theta}} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{2 - 4\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\

& \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\

& \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}
Part Three
\displaystyle\int_1^{\frac{2}{\sqrt3}} \frac{1}{y(y-\sqrt{y^2 - 1})} \, dy

\displaystyle y = \frac1x \Rightarrow dy = -\frac{1}{x^2} \, dx

\begin{array}{rl}

\displaystyle\therefore \int_1^{\frac{2}{\sqrt3}} \frac{1}{y(y-\sqrt{y^2 - 1})} \, dy

& \displaystyle = \int_1^{\frac{\sqrt3}{2}} \frac{x}{\frac1x - \sqrt{\frac{1}{x^2}-1}} \times - \frac{dx}{x^2} \\ \br \\

& \displaystyle = \int_1^{\frac{\sqrt3}{2}} -\frac{x}{x - x^2 \sqrt{\frac{1-x^2}{x^2}}} \, dx \\ \br \\

& \displaystyle = \int_1^{\frac{\sqrt3}{2}} -\frac{x}{x - x \sqrt{1-x^2}} \, dx \\ \br \\

& \displaystyle = \int_1^{\frac{\sqrt3}{2}} -\frac{1}{1 - \sqrt{1-x^2}} \, dx \\ \br \\

& \displaystyle = \int^1_{\frac{\sqrt3}{2}} \frac{1}{1 - \sqrt{1-x^2}} \, dx  = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}
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sonofdot
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STEP I 2002 Question 7

\displaystyle I = \int_0^a \frac{\cos x}{\sin x + \cos x} \, dx and \displaystyle J = \int_0^a \frac{\sin x}{\sin x + \cos x} \, dx with 0 \leq a &lt; \frac{3\pi}{4}

(Note that in the interval 0 \leq x &lt; \frac{3\pi}{4}, \sin x + \cos x \not= 0)

\begin{array}{rl}

I+J & \displaystyle = \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \\ \br \\

& \displaystyle = \int_0^a 1 \, dx \\ \br \\

& \displaystyle = \left[ x \right]_0^a = a \end{array}

\begin{array}{rl}

I-J & \displaystyle = \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \\ \br \\

& \displaystyle = \left[ \ln (\sin x + \cos x) \right]_0^a \\ \br \\

& \displaystyle = \ln (\sin a + \cos a)\end{array}

\displaystyle (I+J)+(I-J) = a + \ln (\sin a + \cos a) \Rightarrow \boxed{2I = a + \ln (\sin a + \cos a)}
(i)
Let \displaystyle I_1 = \int_0^{\frac{\pi}{2}} \frac{\cos x}{p \sin x + q \cos x} \, dx

(Note again that for 0 \leq x \leq \frac{\pi}{2} sin(x) and cos(x) are never negative, so psin(x) + qcos(x) is never equal to 0)

Let \displaystyle J_1 = \int_0^{\frac{\pi}{2}} \frac{\sin x}{p \sin x + q \cos x} \, dx

\begin{array}{rl}q^2 I_1 + pqJ_1

& \displaystyle = q \int_0^{\frac{\pi}{2}} \frac{q\cos x + p\sin x}{p \sin x + q \cos x} \, dx \\ \br \\

& \displaystyle = q \int_0^{\frac{\pi}{2}} 1 \, dx \\ \br \\

& \displaystyle = \frac{q\pi}{2}\end{array}

\begin{array}{rl}p^2 I_1 - pq J_1

& \displaystyle = p \int_0^{\frac{\pi}{2}} \frac{p\cos x - q\sin x}{p \sin x + q \cos x} \, dx \\ \br \\

& \displaystyle = p \left[ \ln (p \sin x + q\cos x) \right]_0^{\frac{\pi}{2}} \\ \br \\

& \displaystyle = p\ln \frac{p}{q}\end{array}

\displaystyle\therefore (p^2 + q^2)I_1 = \frac{q\pi}{2} + p\ln\frac{p}{q} \Rightarrow \boxed{I_1 = \frac{1}{p^2 + q^2} \left( \frac{q\pi}{2} + p\ln \frac{p}{q} \right)}
(ii)
Let \displaystyle I_2 = \int_0^{\frac{\pi}{2}} \frac{\cos x+4}{3\sin x + 4 \cos x +25} \, dx and let \displaystyle J_2 = \int_0^{\frac{\pi}{2}} \frac{\sin x+3}{3\sin x + 4 \cos x +25} \, dx

\begin{array}{rl}4(4I_2 + 3J_2)

& \displaystyle = 4\int_0^{\frac{\pi}{2}} \frac{4\cos x+3\sin x + 25}{3\sin x + 4 \cos x +25} \, dx \\ \br \\

& \displaystyle = 4\int_0^{\frac{\pi}{2}} 1 \, dx = 2\pi \end{array}

\begin{array}{rl}3(3I_2 - 4J_2)

& \displaystyle = 3\int_0^{\frac{\pi}{2}} \frac{3\cos x+ 12 - 4\sin x - 12}{3\sin x + 4 \cos x +25} \, dx \\ \br \\

& \displaystyle = 3\int_0^{\frac{\pi}{2}} \frac{3\cos x - 4\sin x}{3\sin x + 4 \cos x +25} \, dx \\ \br \\

& \displaystyle = 3\left[ \ln ( 3\sin x + 4 \cos x + 25) \right]_0^{\frac{\pi}{2}} \\ \br \\

& \displaystyle = 3\ln \frac{28}{29}\end{array}

\displaystyle\therefore 16I_2 + 12J_2 + 9I_2 - 12J_2 = 2\pi + 3\ln \frac{28}{29} \Rightarrow \boxed{I_2 = \frac{1}{25} \left( 2\pi + 3 \ln \frac{28}{29} \right)}
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Glutamic Acid
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II/8:

For x < 0, dy/dx = -y ==> 1/y dy/dx = -1 ==> ln y = -x +c ==> y = Ae^(-x); as y = a when x = -1, a = Ae so A = a/e; therefore y = ae^{-(x+1)} for x < 0.

For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore y = be^{x-1} for x > 0.

If a = b, there is no jump.

Note that e^x - 1 is positive if x > 0, and negative if x < 0.

\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = e^x - 1 \Rightarrow \ln y = e^x - x + k \Rightarrow y = Ae^{e^x - x} for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e; y = e^{e^x - x + 1}

\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = 1 - e^x \Rightarrow \ln y = x - e^x + c \Rightarrow y = Be^{x - e^x} for x < 0.

At x = 0, y = e^{1 - 0 + 1} = e^2 using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so
e^2 = Be^{-1} \Rightarrow B = e^3

y = e^{e^x - x + 1} \text{for} x &gt; 0
y = e^{3 + x - e^x} \text{for} x &lt; 0


(i) \displaystyle \lim_{x \to + \infty} \exp(e^x - x + 1) \exp(-e^x) = \lim_{x \to + \infty} \exp(1 - x) = 0

(ii) \displaystyle \lim_{x \to - \infty} e^{3 + x - e^x} e^{-x} = \lim_{x \to -\infty} e^{3 - e^x} = e^3.
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Glutamic Acid
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II/7: (Scary scary vectors.)
??????????

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

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

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

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

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

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

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

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

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

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

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

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

The second equation gives \alpha = \dfrac{2}{2 - 3 \lambda}. Substituting into the third gives \beta = \dfrac{2}{2 - 3 \lambda}. Substituting these into the first:
1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1); multiplying out gives 2 \lambda = 0 \Rightarrow \lambda = 0, so no non-zero solutions.
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Question 2, STEP I, 2002
First Part
f(x) = x^m(x-1)^n

 \implies f'(x) = mx^{m-1}(x-1)^n + nx^m(x-1)^{n-1}

 = x^{m-1}(x-1)^{n-1}(m(x-1)+nx)

At stationary points, f'(x) = 0

 \implies x^{m-1}(x-1)^{n-1}(m(x-1)+nx) = 0

 \implies x = 0, 1, \frac{m}{m+n}

 0 &lt; \dfrac{m}{m+n} &lt; 1 which is the point that satisfies the conditions.
Second Part
 f'(x) = x^{m-1}(x-1)^{n-1}(m(x-1)+nx)

 \implies f''(x) = (m-1)x^{m-2}(x-1)^{n-1}(m(x-1)+nx)
+ (n-1)x^{m-1}(x-1)^{n-2}(m(x-1)+nx)
+ x^{m-1}(x-1)^{n-1}(m+n)

 \implies f''\left(\dfrac{m}{m+n}\right) = \left(\dfrac{m}{m+n}\right)^{m-1}\cdot \left(\dfrac{m}{m+n} -1\right)^{n-1} \cdot(m+n)

Now the first and third brackets are always positive. The middle bracket alternates signs. When n is even, the middle bracket is negative, which implies f''(x) < 0 which implies it is a maximum. When n is odd, the middle bracket is positive, which implies it is a minimum. Q.E.D.
Third Part
Sketches to follow shortly
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Aurel-Aqua
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I'm not checking if anyone posted the solutions, cause I completed these not long ago.

2002 II Question 1

\displaystyle \int_{\pi/6}^{\pi/4} \frac{1}{1-\cos{\2\theta}} \,\mathrm{d}\theta =

\displaystyle = \int_{\pi/6}^{\pi/4} \frac{1}{2\sin^2\theta} \,\mathrm{d}\theta =
\displaystyle =\frac{1}{2}\left[-\cot^2\theta\right]_{\pi/6}^{\pi/4} =
\displaystyle = \frac{1}{2}\left(\frac{\sqrt{3}/2}{1/2} - 1\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} as required.

\displaystyle x = \sin{2\theta} \implies \theta = \frac{1}{2}\arcsin{x}
and \implies \mathrm{d}x =
\displaystyle = 2\cos{2\theta}\mathrm{d}\theta

\displaystyle \implies \int_{\sqrt{3}/2}^1 \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x =
\displaystyle =\int_{\pi/6}^{\pi/4} \frac{1-2\sin^2\theta}{\sin^2\theta}\,\mathrm{d}\theta =
\displaystyle = \sqrt{3}-1-\int_{\pi/6}^{\pi/4}2\,\mathrm{d}\theta = \sqrt{3} - 1 - \frac{\pi}{6}, as required.

Using \displaystyle x = \frac{1}{y}, \displaystyle \mathrm{d}y = -y^2\mathrm{d}x, we obtain:

\displaystyle \int_{1}^{2/\sqrt{3}} \frac{1}{y(y-\sqrt{y^2-1^2})} \,\mathrm{d}y =

\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-xy^2}{\frac{1}{x}-\sqrt{\frac{1}{x^2}-1}} \,\mathrm{d}x =
\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-x^2y^2}{1-\sqrt{1-x^2}} \,\mathrm{d}x =
\displaystyle = \int_{\sqrt{3}{2}}^{1} \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x = \sqrt{3}-1-\frac{\pi}{6}
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Aurel-Aqua
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2002 II Question 11

Total weight (to check our calculus) \displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}\,\mathrm{d}x = aWl^{-\alpha}\left[\frac{1}{\alpha}x^\alpha \right]_0^l = Wl^{-\alpha}l^\alpha = W

Total moment \displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}x\,\mathrm{d}x = \alphaWl^{-\alpha}\int_0^l x^\alpha \,\mathrm{d}x
\displaystyle = \frac{\alpha}{\alpha+1}Wl.

We know that total moment \displaystyle = W\overline{x}, so \displaystyle\overline{x} = \frac{\alpha l}{\alpha+1}, as required.

Draw out a diagram: A touching the ground, with R_A going up, F_A going right. B touching the wall, with F_B going up, R_B going left. Somewhere ebtween A and B, we have the centre of mass. We denote \displaystyle \overline{x_A} = l - \overline{x} = \frac{l}{\alpha+1} to be the distance of c.o.m. from A. Angle \displaystyle\theta going from the horizontal to the line AB.

Equating the horizontal forces: \displaystyle F_A = R_B. Noting that \displaystyle F_A = \mu R_A, we get \displaystyle\mu R_a = r_B. Noting that \displaystyle f_B=\mu R_B, we get \displaystyle F_B = \mu^2 R_A
Equating the vertical forces: \displaystyle W = R_A+F_B = R_A(1+\mu^2).
Equating the moments: \displaystyle W\overline{x_A}\cos\theta = R_Bl\sin\theta + F_Bl\cos\theta \implies R_Bl\tan\theta = W\overline{x_A} - F_Bl = \mu R_al\tan\theta = R_A(1+\mu^2)\overline{x_A} - \mu^2R_Al
\displaystyle \implies \mu l\tan\theta = (1+\mu^2)\overline{x_A} - \mu^2l = (1+\mu^2)\frac{l}{\alpha+1}-\mu^2l
\displaystyle \implies \mu\tan\theta = \frac{1+\mu^2}{\alpha+1} - \mu^2 = \frac{1-\alpha\mu^2}{\alpha+1}
\displaystyle \implies \tan\theta = \frac{1-\alpha\mu^2}{(1_\alpha)\mu} as required.

Not sure if I'm right, but I'll give it a go:
\displaystyle \alpha = \frac{3}{2}, and \displaystyle \theta = \frac{\pi}{4} \implies (1+\alpha)\mu = 1-\alpha\mu^2 \implies 3\mu^2+5\mu-2 = 0 \implies \mu = \frac{2}{6} = \frac{1}{3}
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Aurel-Aqua
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2002 II Question 12

\displaystyle P(\text{k of K days, something with probability of q happens}) = \binom{K}{k}q^k(1-q)^{K-k}
q = probability such that more than l coins of the L coins will produce less than m heads

Estimating q: Q \~ B(L,z), where z is the probability of a coin producing less than m heads.
We can say that \displaystyle Q' \~ N(Lz,Lz(1-z)), which gives:
\displaystyle q = P(Q&gt;l) \approx 1 - \Phi\left(\frac{(l-\frac{1}{2})-Lz}{\sqrtLz(1-z)}\right) = \Phi\left(\frac{2Lz-2l+1}{2\sqrt{Lz(1-z)}}\right) \approx \Phi\left(\frac{2Lz-2l+1}{2\sqrt{Lz}}\right) for small z.
We then substitute \displaystyle h = Lz to get \displaystyle q = \Phi\left(\frac{2h-2l+1}{2\sqrt{h}}\right) as required.

z is the probability that less than m heads are made after M throws.
\displaystyle Z ~ B(M,p) \implies Z \approx N (Mp, Mp(1-p))
\displaystyle z = P(Z&lt;m) = \Phi\left(\frac{(m-\frac{1}{2})-Mp}{\sqrt{Mp(1-p)}}\right) = \Phi\left(\frac{2m-1-Mp}{2\sqrt{Mp(1-p)}}\right).

\displaystyle h = Lz = L\Phi\left(\frac{2m-1-Mp}{2\sqrt{Mp(1-p)}}\right) as required.

K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6. \displaystyle z = \Phi\left(\frac{98-1-60}{2\sqrt{100\times 0.6\times 0.4}}\right) = \Phi\left(\frac{37}{2\sqrt{24}}\right)\approx\Phi(3.7) , which is a large z, so we are not allowed to approximate as needed.
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Aurel-Aqua
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2002 II Question 13

F(a) = 0
f(b) = 1, a &gt; 0

\displaystyle G(y) = \frac{F(y)}{2-F(y)}


\displaystyle F(a) = \frac{F(a)}{2-F(a)} = \frac{0}{2} = 0

\displaystyle G(b) = \frac{F(b)}{2-F(b)} = \frac{1}{2-1} = 1

\displaystyle G'(y) = \frac{(2-F(y))F'(y)-F(y)(-F'(y)}{(2-F(y))^2} = \frac{2F'(y)}{(2-F(y))^2}  \geq 0 since \displaystyle F'(y) \geq 0 since it's a cumulative distribution function.

We know \displaystyle 1 \leq (2-F(x))^2 \leq 4 \implies \frac{1}{4} \leq \frac{1}{(2-F(x))^2} \leq 1 \implies \frac{1}{2} \leq \frac{2}{(2-F(x))^2} \leq 2

We know \displaystyle E[Y] = \int_a^b xG'(x)\,\mathrm{d}x = \int_a^b \frac{2xF'(x)}{(2-F(x))^2}\, \mathrm{d}x = \int_a^b xF'(x) \times \frac{2}{(2-F(x))^2}\, \mathrm{d}x

Using the condition found before, we deduce that \displaystyle \int_a^b \frac{1}{2}xF'(x)\, \mathrm{d}x \leq \int_a^b xF'(x) \times \frac{2}{(2-F(x))^2}\, \mathrm{d}x \leq \int_a^b 2xF'(x)\, \mathrm{d}x

\displaystyle \implies \frac{1}{2}E[X] \leq E[Y] \leq 2E[Y]

\displaystyle Var[Y] = E[Y^2] - E[Y]^2.

We know \displaystyle E[Y^2]\leq 2E[X^2], so E[Y^2] \leq 2E[X^2]  = 2E[X]^2+2Var[X].

We also know that \displaystyle (E[Y])^2 \geq \frac{1}{4}E[X]^2 \implies -E[Y]^2 \leq -\frac{1}{4}E[X]^2

So \displaystyle Var[Y] \leq 2Var[X] + 2E[X]^2 - E[Y]^2 \leq 2Var[X] + 2E[X]^2 - \frac{1}{4}E[X]^2 =
 \displaystyle = 2Var[X] + \frac{7}{4}E[X]^2, as required.
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Unbounded
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Question 10, STEP I, 2002
First Part
By Newton's law of restitution, for the first collision:

 \dfrac{v+u}{v_1-u} = 1

 \iff v_1 = v+2u

Assume for some n=k it is true, ie:

 v_k = v+2ku

Looking at the (k+1)th collision:

 v_k+u = v_{k+1} - u \iff v_{k+1} = v_k + 2u = v+2(k+1)u

And so by induction, we have shown it to be true:

 \boxed{v_n = v + 2nu} \ \ \ \square
Second Part
[With a diagram] we can see that:

 \boxed{d_n - d_{n+1} = ut_n} \ \ \ (\ast )


[Also with a diagram] we can see that:

 v_nt_n = d_n + d_{n+1}

substituting for  t_n into  (\ast ), we get:

 v_nd_n-v_nd_{n+1} = ud_n + ud_{n+1}

 \iff d_{n+1} (v_n+u) = d_n (v_n-u)

 \iff d_{n+1} = \dfrac{v_n-u}{v_n+u} d_n

 \iff \boxed{d_{n+1} = \dfrac{v+(2n-1)u}{v+(2n+1)u} d_n} \ \ \ \square
Third Part
 d_n = \dfrac{v+(2n-3)u}{v+(2n-1)u} \times \dfrac{v+(2n-5)u}{v+(2n-3)u} \times \cdots \times \dfrac{v+u}{v+3u}\times d_1

 \iff \boxed{d_n = \dfrac{v+u}{v+(2n-1)u} d_1}

u = v,

\therefore d_n = \dfrac{2ud_1}{2nu} = \dfrac{d_1}{n}

Similarly,  d_{n+1} = \dfrac{d_1}{n+1}

By  (\ast ) we have:

 ut_n = \dfrac{d_1}{n} - \dfrac{d_1}{n+1}

 \iff \boxed{ut_n = \dfrac{d_1}{n(n+1)}} \ \ \ \square
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Unbounded
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Question 11, STEP I, 2002
First Part
Let the initial speed of  P_1 be u and let  P_1 and  P_2 leave the collision at speeds  v_1 and  v_2 respectively.

Looking at momentum in the direction of  P_1 initially:

 mu = mv_1\cos \theta + kmv_2\cos \phi

 \iff u = v_1\cos \theta + kv_2\cos \phi \ \ \ (\ast )

Looking at momentum perpendicular to the initial direction of P_1 :

 v_1\sin \theta = kv_2\sin \phi  \ \ \ (\ast \ast )

And considering the energy of the system:

 \frac{1}{2} mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}kmv_2^2

 \implies u^2 = v_1^2 + kv_2^2 \ \ \ (\ast \ast \ast )

Squaring  (\ast ) we get:

 u^2 = v_1^2 \cos^2 \theta + k^2v_2^2\cos^2 \phi + 2kv_1v_2\cos \theta \cos \phi

And substituting in  (\ast \ast \ast )

 v_1^2 + kv_2^2 = v_1^2 \cos^2 \theta + k^2v_2^2\cos^2 \phi + 2kv_1v_2\cos \theta \cos \phi

 \iff v_1^2(1-\cos^2 \theta) = kv_2^2 (k\cos^2 \phi -1) + 2kv_1v_2\cos \theta \cos \phi

 \iff v_1^2\sin^2 \theta = kv_2^2 (k\cos^2 \phi -1) + 2kv_1v_2\cos \theta \cos \phi \ \ \ (\star )

Making  v_1 the subject in  (\ast \ast ) we get:

 v_1 = \dfrac{kv_2\sin \phi}{\sin \theta}

Substituting this result into  (\star ) we get:

 k^2 v_2^2 \sin^2 \phi =  kv_2^2 (k\cos^2 \phi -1) + \dfrac{2k^2v_2^2\cos \theta \cos \phi \sin \phi}{\sin \theta}

 \iff k\sin^2 \phi = k\cos^2 \phi -1 + \dfrac{2k\cos \theta \cos \phi \sin \phi}{\sin \theta}

 \iff k(\sin^2 - \cos^2 \phi) + 1 = \dfrac{k\cos \theta \sin 2\phi}{\sin \theta}

 \iff 1 - k\cos 2\phi = \dfrac{k\cos \theta \sin 2\phi}{\sin \theta}

 \iff \sin \theta = k \left [ \cos 2 \phi \sin \theta + \sin 2\phi \cos \theta \right ]

 \iff \sin \theta = k\sin (\theta + 2\phi)

 \iff \sin^2 \theta = k \sin \theta \sin (\theta + 2\phi)

Applying product-sum formulae:

 \iff 2\sin^2 \theta = k\cos 2\phi - k\cos 2(\theta + \phi)

By double angle formulae:

 \iff 2\sin^2 \theta = 2k\sin^2 (\theta + \phi) - 2k\sin^2 \phi

 \iff \boxed{\sin^2 \theta + k\sin^2 \phi = k\sin^2 (\theta + \phi)} \ \ \ \square

I fear I've taken some unnecessarily long route
Second Part
If the angle between the particles after the collision is a right angle:

 \theta + \phi = \frac{\pi}{2}

And also  \sin \theta = \cos \phi and  \cos \theta = \sin \phi

Applying this to the equation  \sin^2 \theta + k\sin^2 \phi = k\sin^2 (\theta + \phi)

we get  \cos^2 \phi + k \sin^2 \phi = k \sin^2 (\frac{\pi}{2})

 \iff \cos^2 \phi + k \sin^2 \phi = k

 \iff  k(1-\sin^2 \phi) = \cos^2 \phi

 \iff k\cos^2 \phi = \cos^2 \phi

 \implies \boxed{k = 1} :qed:
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