STEP 2005 Solutions Thread Watch

SimonM
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STEP III, Question 3

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f(x) = x^2+px+q
g(x) = x^2+rx+s

f(g(x)) = (x^2+rx+s)^2+p(x^2+rx+s)+q =
q + p s + s^2 + (p r  + 2 r s )x + (p + r^2 + 2 s )x^2 + 

 2 r x^3 + x^4

Comparing coefficients, we have

a = 2r
b = p+2s+r^2
c = pr+2rs
d = q+ps+s^2

Therefore if we can find suitable values for p,q,r,s we have:

c = \left ( b - \left ( \frac{a}{2} \right )^2 \right ) \left ( \frac{a}{2} \right )

If a,b,c satisfy those equations, then we can solve simultaneously to find p,r,s and take q to be whatever value we need.

Expanding

(x^2+vx+w)^2-k = x^4+2vx^3+(v^2+2w)x^2+2vwx+w^2-k

This clearly satisfies our equation, so we're done.

f=x^4 - 4x^3 + 10x^2 - 12x + 4

(10-2^2)(-2) =-12 so it satisfies our equation.

Therefore:

f=(x^2-2x+3)^2-5=0

Therefore we have

x^2-2x+3 = \pm  \sqrt{5}

Therefore \displaystyle x \in \left \{ 1- \sqrt{\sqrt{5}-2},   1+ \sqrt{\sqrt{5}-2}, 1- i\sqrt{\sqrt{5}+2},   1+ i\sqrt{\sqrt{5}+2} \right \}
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SimonM
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STEP III, Question 4

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\displaystyle u_{n+2} = \frac{u_{n+1}}{u_{n}} \left ( k u_n - u_{n+1} \right)

Assuming no zero terms (which they let us know)

\displaystyle \frac{u_{n+2}}{u_{n+1}} = k - \frac{u_{n+1}}{u_n}

Therefore we have the recurrence: in \displaystyle a_n = \frac{u_{n+1}}{u_n}

\displaystyle a_{n+1} = k - a_{n}

a_n = \displaystyle (\frac{b}{a}, k - \frac{b}{a}, \frac{b}{a}, k - \frac{b}{a}, \frac{b}{a}, k - \frac{b}{a}, \frac{b}{a}, \cdots) (induction if that's what floats your boat...)

This proves what we want, and c = k - \frac{b}{a}

Therefore \displaystyle u_{2n} = \left ( \frac{bc}{a} \right )^{n-1} b

and \displaystyle u_{2n-1} = \left ( \frac{bc}{a} \right )^{n-2} bc

i) Geometric \Rightarrow a_n = k, \frac{b}{a} = k - \frac{b}{a} \Rightarrow k = 2 \frac{b}{a}.

ii) Periodic, period 2: \frac{bc}{a} = 1

iii) Period, period 4: \left ( \frac{bc}{a} \right ) = -1
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SimonM
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STEP II, Question 1

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Let y = x^2 e^{-x^2}

Therefore \frac{dy}{dx} = 2x e^{-x^2} + x^2 (-2x)e^{-x^2} = e^{-x^2}2x(1-x)(1+x)

Therefore the solutions are x \in \{-1,0,1 \}

We have \displaystyle \left ( P(x)e^{-x^2} \right )' = P'(x)e^{-x^2} -2xP(x)e^{-x^2}

Therefore P'(x) - 2xP(x) = kx(x^2-a^2)(x^2-b^2)

Therefore \deg P = 4

Let P = x^4+px^3+qx^2+rx+s

P' = 4x^3+3px^2+2qx+r

Therefore comparing coefficients, we have:

-2=k
-2p = 0
-2q + 4 = -k(a^2+b^2)
-2r + 3p = 0
-2s + 2q = ka^2b^2
r = 0

Therefore k = -2, p = 0, q = -a^2-b^2 + 2, r = 0, s = a^2b^2 - a^2 - b^2+2

Therefore take P = x^4 +(-a^2-b^2+2)x^2 + (a^2b^2 - a^2-b^2 +2)
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SimonM
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STEP II, Question 3

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y = \sin x - x \cos x is an increasing function, from 0 to 1 with a slight curve

i) \displaystyle \int_0^{\pi /2} y \, dx = \left [ - \cos x - x \sin x \right ]_0^{\pi/2} + \int_0^{\pi/2} \sin x \, dx = \left [ - \cos x - x \sin x  - \cos x \right ]_0^{\pi/2} = 2 - \frac{\pi}{2}

ii) \displaystyle \int_0^{\pi/2} y^2 \, dx = \int_0^{\pi/2} \left ( \right \sin x - x \sin 2x + x^2\cos^2 x) \, dx = \frac{\pi^3}{48} - \frac{\pi}{8}

Since y < 1 \Rightarrow y^2 < y

Therefore \displaystyle \frac{\pi^3}{48} - \frac{\pi}{8} < 2 - \frac{\pi}{2} \Rightarrow \pi^3 + 18\pi< 96
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JAKstriked
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GHOSH gave all the solutions I have for 1 2005:Q2,3,4,7
I'll try another one
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sonofdot
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STEP III 2005 Question 9

Hopefully my diagram makes it clear what all the symbols I use mean
Diagram
First Part
For the first collision, the coefficient of restitution is given by

\displaystyle e = \frac{b_1 - a_1}{(1-e)v+2ev} \Leftrightarrow ev(1+e) = b_1 - a_1

By conservation of momentum:

4em(1-e)v - (1-e)^2m(2ev) = 4ema_1 + (1-e)^2mb_1 \br



\Leftrightarrow 2ev(1-e^2) = 4ea_1 + (1-e)^2 b_1

Solving these simultaneously for a_1 and b_1 gives a_1 = ev(1-e), \, b_1 = 2ev
Second Part
B then collides with the wall, and, using the coefficient of restitution, e = \frac{b_2}{2ev} \Leftrightarrow b_2 = 2e^2 v

The next collision takes place at a distance d_1 from the wall. The time between the collisions is t_1 For particle A, t_1 = \frac{d - d_1}{ev(1-e)} and for B, t_1 = \frac{d}{2ev} + \frac{d_1}{2e^2 v}

So we have \frac{d - d_1}{ev(1-e)} = \frac{de + d_1}{2e^2v} which boils down to give d_1 = de
Third Part
Clearly, when this collision happens, particle A will continue in the same direction, as will particle B, which will rebound off the wall to collide with A again in a similar way. I'm going to outline a brief proof by induction that the speed of A after the nth collision is a_n = e^n v (1-e) and the speed of B after the nth collision with A and after it has hit the wall, b_{2n} = 2e^{n+1} v, though I don't know how much detail you'd have to go into on a STEP mechanics question...

So, assume that, going into the (r+1)th collision, the speed of A is a_r = e^r v (1-e) and the speed of B is b_{2r} = 2e^{r+1} v

The coefficient of restitution for this collision is given by:

\displaystyle e=\frac{b_{2r+1}-a_{r+1}}{e^r v(1-e) + 2e^{r+1} v} \Leftrightarrow e^{r+1} v (1+e) = b_{2r+1} - a_{r+1}

And, by conservation of momentum:

4em(e^r v (1-e)) - (1-e)^2m(2e^{r+1} v) = 4ema_{r+1} + (1-e)^2m b_{2r+1} \br



\Leftrightarrow 2e^{r+1}v (1-e^2) = 4ea_{r+1} + (1-e)^2 b_{2r+1}

Again, solving these simultaneously for a_{r+1} and b_{2r+1} gives a_{r+1} = e^{r+1} v (1-e), \, b_{2r+1} = 2e^{r+1} v

And then, after B strikes the wall, you get b_{2(r+1)} = 2e^{r+2} v, so our hypothesis has been "proved".

Now consider d_n, the distance from the wall at which the (n+1)th collision takes place, and t_n, the time between the nth and (n+1)th collisions. For particle A: t_n = \frac{d_{n-1} - d_n}{e^n v (1-e)} and for particle B: t_n = \frac{d_{n-1}}{2e^n v} + \frac{d_n}{2e^{n+1}v}

This gives \frac{d_{n-1} - d_n}{e^n v (1-e)} = \frac{ed_{n-1} + d_n}{2e^{n+1} v} which boils down to d_n = ed_{n-1}

Clearly, the first distance is d, so the next is de, then de^2, and so on, so d_n = de^n

The time between the (n+1)th and the nth collisions, is therefore given by:

\begin{array}{rl}

t_n & \displaystyle = \frac{d_{n-1} - d_n}{e^n v (1-e)} \\ \br \\

& \displaystyle = \frac{de^{n-1} - de^n}{e^n v (1-e)} \\ \br \\

& \displaystyle = \frac{d - de}{e v (1-e)} \\ \br \\

& \displaystyle = \frac{d}{ev} \end{array}

Since d, e and v are all constant, we have shown that the times between successive collisions are equal.
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nuodai
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Ah what the hell,
STEP I 2005 Question 1

Part (i)

For a 5-digit number to add up to 43 the digits must be either:
(1): (9, 9, 9, 9, 7) [ 5 ways of rearranging ]
(2): (9, 9, 9, 8, 8) [ 5!/3!2! = 10 ways of rearranging ]

5 + 10 = 15
Part (ii)

We can classify groupings of digits by how many 9s they have and, if no 9s. Also, since since 8*4 + 7 = 39 there can't be less than four 8s if there are no 9s.

FOUR NINES:
(9, 9, 9, 9, 3) [ 5 ways of rearranging]

THREE NINES:
(9, 9, 9, 8, 4) [ 5!/3! = 20 ways of rearranging ]
(9, 9, 9, 7, 5) [ 5!/3! = 20 ways of rearranging ]
(9, 9, 9, 6, 6) [ 5!/3!2! = 10 ways of rearranging ]

TWO NINES:
(9, 9, 8, 8, 5) [ 5!/2!2! = 30 ways of rearranging ]
(9, 9, 8, 7, 6) [ 5!/2! = 60 ways of rearranging ]
(9, 9, 7, 7, 7) [ 5!/2!3! = 10 ways of rearranging ]

ONE NINE:
(9, 8, 8, 8, 6) [5!/3! = 20 ways of rearranging ]
(9, 8, 8, 7, 7) [5!/2!2! = 30 ways of rearranging ]

NO NINES:
(8, 8, 8, 8, 7) [5 ways of rearranging ]

Doing some good ol' addition gives you 5 + 20 + 20 + 10 + 30 + 60 + 10 + 20 + 30 + 5 = 210 ways of rearranging the numbers. And no LaTeX required!
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Unbounded
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Question 5, STEP I, 2005
(i)
Case 1:  k\not= 0

 \displaystyle\int_0^1 (x+1)^{k-1} \ \mathrm{d}x = \left [ \dfrac{(x+1)^k}{k} \right ]_0^1 = \boxed{\dfrac{2^k-1}{k}}

Case 2: k = 0

 \displaystyle\int_0^1 \dfrac{1}{x+1} \ \mathrm{d}x = \left [ \ln |x+1| \right ]_0^1 = \boxed{\ln 2}

As  k \to 0, \dfrac{2^k-1}{k} \to \ln 2  \implies k \approx 0, \dfrac{2^k-1}{k} \approx \ln 2

I wasn't exactly sure what to put for that last part
(ii)
With the substitution  u = x+1

 I = \displaystyle\int_0^1 x(x+1)^m \ \mathrm{d}x = \displaystyle\int_1^2 \left(u^{m+1} - u^m\right) \ \mathrm{d}u

Case 1:  m \not= -1, -2

 I = \left [ \dfrac{u^{m+2}}{m+2} - \dfrac{u^{m+1}}{m+1} \right ]_1^2 = \left[\dfrac{2^{m+2}}{m+2} - \dfrac{2^{m+1}}{m+1}\right] - \left [ \dfrac{1}{m+2} - \dfrac{1}{m+1} \right ]

 = \dfrac{2^{m+2} - 1}{m+2} + \dfrac{1 - 2^{m+1}}{m+1} = \boxed{\dfrac{2^{m+1}m+1}{(m+1)(  m+2)}}

Case 2:  m = -1

I = \displaystyle\int_1^2 \left(1 - \dfrac{1}{u}\right) \ \mathrm{d}u = \left [ u - \ln u \right ]_1^2 = \left [ 2 - \ln 2 \right ] - \left [ 1 - \ln 1 \right ]

 = \boxed{1 - \ln 2}

Case 2:  m = -2

 I = \displaystyle\int_1^2 \left(\dfrac{1}{u} - \dfrac{1}{u^2}\right) \ \mathrm{d}u = \left [ \ln u + \dfrac{1}{u} \right ]_1^2

 = \left [ \ln 2 + \dfrac{1}{2} \right ] - \left [ \ln 1 + 1 \right ] = \boxed{\ln 2 - \dfrac{1}{2}}
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nuodai
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STEP I 2005 Question 9

Solution

The rod is non-uniform so the weight can be anywhere. I said it acted at distance x from B. In the instance when there are two strings holding it up, taking moments about B gives you Wx = 3lT \Rightarrow x = \dfrac{3lT}{W}.

When there is a pivot (call this P) at a distance l from A and a string at B, the distance from P from which the weight acts is therefore 2l -x . Therefore, taking moments about P gives you W(2l - x) = 2lT. Rearranging this with the substitution x = \dfrac{3lT}{W} gives you 2lW - \dfrac{3lTW}{W} = 2lT \Rightarrow 2W - 3T = 2T \Rightarrow \boxed{2W = 5T}, as required.

We can now also simplify x, to get x = \dfrac{3lT}{\frac{5T}{2}} = \dfrac{6l}{5}; this is the distance along the rod from B that the weight acts.

To find \theta it's best just to take moments about B again. Doing so gives (W\cos \theta)x = \dfrac{3lT}{2}. Using our result from before that x = \dfrac{6l}{5}, this simplifies to  3lT\cos\theta = \dfrac{3lT}{2}, which simplifies to \cos \theta = \dfrac{1}{2} \Rightarrow \boxed{\theta = \dfrac{\pi}{3}}

Now we've found \theta, we need to find the coefficient of friction. Resolving vertically to find the normal contact force gives us R = W - \dfrac{T}{2}\cos \theta. Using the information we have already found, we have W = \dfrac{5T}{2}, \cos \theta = \dfrac{1}{2} so this simplifies to R = \dfrac{9T}{4}.

In order to find \mu we need to resolve horizontally to find the frictional force \mu R acting horizontally "inwards" (towards the rest of the rod) at B. We are told that the rod is not slipping, so the frictional force must be greater than opposing forces, so we therefore get:
\newline \displaystyle \mu R \ge \dfrac{T}{2}\sin \theta \Rightarrow \mu \ge \dfrac{\frac{\sqrt{3}T}{4}}{\fra  c{9T}{4}}\newline

\Rightarrow \boxed{\mu \ge \dfrac{\sqrt{3}}{9}}
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sonofdot
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STEP II 2005 Question 8

First Part
\displaystyle \frac{dy}{dx} = \frac{x^3 y^2}{(1+x^2)^{5/2}} for x \geq 0 with y=1 when x=0

\displaystyle\Rightarrow \int \frac{1}{y^2} \, dy = \int \frac{x^3}{(1+x^2)^{5/2}} \, dx

The RHS can be integrated by parts, with u=x^2 \Rightarrow \frac{du}{dx} = 2x and \frac{dv}{dx} = \frac{x}{(1+x^2)^{5/2}} \Leftarrow v = -\frac{1}{3(1+x^2)^{3/2}}

\displaystyle\Rightarrow -\frac{1}{y} = - \frac{x^2}{3(1+x^2)^{3/2}} + \int \frac{2x}{3(1+x^2)^{3/2}} \, dx \br 



\br



\displaystyle\Rightarrow -\frac{1}{y} = - \frac{x^2}{3(1+x^2)^{3/2}} - \frac{2}{3(1+x^2)^{1/2}} + C \br 



\br



\displaystyle\Rightarrow \frac{1}{y} = \frac{x^2}{3(1+x^2)^{3/2}} + \frac{2(1+x^2)}{3(1+x^2)^{3/2}} + C \br 



\br



\displaystyle\Rightarrow \frac{1}{y} = \frac{2+3x^2}{3(1+x^2)^{3/2}} + C

Substituting in x=0 and y=1 gives C=1/3, so we get \displaystyle\boxed{\frac{1}{y} = \frac{2+3x^2}{3(1+x^2)^{3/2}} + \frac13} as required.

For large x, we can say 1+x^2 \approx x^2 so (1+x^2)^{3/2} \approx x^3

\displaystyle\therefore \frac{1}{y} \approx \frac{x^2}{x^3} + \frac13 \approx \frac13 \left(1+\frac{3}{x} \right)

Looking at the first two terms of the expansion of (1+\frac{3}{x})^{-1} gives, for large x:

\displaystyle y \approx 3\left( 1 - \frac{3}{x} \right) \approx \boxed{3-\frac{9}{x}}
Sketch One
I was lazy and didn't sketch it myself, but some points to note:

1) dy/dx is 0 only when x=0, since y cannot be 0 or the above equation blows up, so the only turning point is at (0,1)

2) The graph has an asymptote x=3, and looks like the graph of y=3-9/x for large x

Sketch Two
Note that if we separate the variables again, we will get:
\displaystyle\frac{1}{z^2} = -\int \frac{x^3}{(1+x^2)^{5/2}} \, dx
So this graph is just graph of y^2 (where the above was a sketch of y)

So it is symmetrical in the x-axis (which gives the two curves requested, passing through (0,1) and (0,-1)) and it has asymptotes y=\pm \sqrt3

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roubiliac
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Great thread!
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nuodai
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STEP I 2005 Question 11

Part (i)

The path of the particle can be represented by parametric equations x = \sin 2t,\ y = 2\cos t. A the origin, x = y = 0, so using the double-angle formulae:
\newline \sin 2t = 2\cos t\newline

\Rightarrow 2\sin t \cos t = 2\cos t\newline

\Rightarrow 2\cos t(\sin t - 1) = 0
\newline\Rightarrow \cos t = 0,\ \sin t = 1 \Rightarrow \boxed{ t = \frac{\pi}{2}, \frac{3\pi}{2}} , as required.
Part (ii)

Differentiating to find the velocity components gives us \dot{x} = 2\cos 2t,\ \dot{y} = -2\sin t. If this is perpendicular to the displacement, then \begin{pmatrix}x\\y\end{pmatrix} \cdot \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = 0, giving us 2\sin 2t \cos 2t - 4\sin t \cos t = 0.

Using the double-angle formula for \sin 2t again gives us:
\newline 2\sin 2t\cos 2t = 2\sin 2t\newline

\Rightarrow 2\sin 2t(\cos 2t - 1) = 0\newline

\Rightarrow \sin 2t = 0,\ \cos 2t = 1 \Rightarrow 2t = 0, \pi, 2\pi, 3\pi, ... \Rightarrow \boxed{t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} }
Part (iii)

I maybe cheated a bit here by using a matrix... there are other ways to do this, but this was by far the simplest.

If the lines are parallel, then the matrix \begin{pmatrix} x & \dot{x} \\ y & \dot{y} \end{pmatrix} is singular, so \begin{vmatrix} \sin 2t & 2\cos 2t \\ 2\cos t & -2\sin t \end{vmatrix} = 0

That gives me -2\sin t \sin 2t - 4\cos 2t \cos t = 0

Once again using the double-angle formula for sine, we get:
\newline -4\sin^2 t \cos t - 4\cos 2t\cos t = 0\newline

\Rightarrow \cos t(\sin^2 t + \cos 2t) = 0

Now using \cos 2t = 1 - 2\sin^2 t, we get:
\newline \cos t(1 - \sin^2 t) = 0 \Rightarrow \cos t = 0,\ \sin t = \pm 1 \Rightarrow t = \frac{\pi}{2}, \frac{3\pi}{2}

Subbing these into the equations gives \boxed{ \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} }, which is the origin, as required.
Part (iv)

If the particle is at a maximum distance from the origin, then \dfrac{ \dot{y} }{ \dot{x} } = 0, so \dfrac{-2\sin t}{2\cos 2t} = 0. This happens when \sin t = 0 and \cos 2t \ne 0, which gives us t = 0, \pi, ..., \ t \ne \frac{\pi}{4}, \frac{3\pi}{4}, ..., so t = 0, \pi. Subbing this back in to the displacement equations gives us \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix} \mbox{ or } \begin{pmatrix} 0 \\ -2 \end{pmatrix}, so the maximum distance is 2. [We know it's a maximum and not a minimum because we've already shown that it goes through the origin.]

The sketch looks like a figure-of-8, with the part where the loops cross at the origin, and crossing the y-axis at \pm 2... I'd draw a diagram but I have no means to do so!
To be honest, this question could just as well have been in the Pure section!
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nuodai
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Christ, I'm normally a Pure man, but these applied questions are great fun :p:

STEP II 2005 Question 12
Part (i)

To help I drew a table that looked something like:
\newline \displaystyle \begin{matrix}  & \mbox{Anna} & \mbox{Bella} \\ \mbox{Online} & p & q \\ \mbox{Truth} & a & b \end{matrix}

If it is heads, then EITHER [Anna is at the computer AND telling the truth] OR [Bella is at the computer AND telling the truth], so ap + bq = \dfrac{1}{2} \Rightarrow \boxed{ 2(ap+bq) = 1 }, as required.
Part (ii)

Either the first correspondence is true and the second is false (1) or the first is false and the second is true (2). The probability that the coin came down heads requires (1), so the probability is (1)/[(1) + (2)].

(1) can be expressed by, (ap + bq)([1-a]p + [1-b]q). Letting ab + bq = X, this simplfies to X(p + q - X) = X(1-X). Similarly, (2) can be expressed as ([1-a]p + [1-b]q)(ap + bq), meaning that it is the same as (1), so (1) + (2) = 2(1).

This means that the probability that the coin came down heads is \dfrac{X(1+X)}{2X(1+X)} = \boxed{\dfrac{1}{2}}, as required.
Part (iii)

In a similar fashion, either both times whoever we speak to is telling the truth (3) or both times they are lying (4).

(3): (ap + bq)(ap + bq) = X^2
(4): ([1-a]p + [1-b]q)([1-a]p + [1-b]q) = (1-X)^2

So, the probability is \dfrac{X^2}{X^2 + (1-X)^2}. Since we are given that 2X = 1 \Rightarrow X = \dfrac{1}{2}, this simplifies to \dfrac{X^2}{X^2 + (1-X)^2} = \dfrac{1/4}{1/4 + 1/4} = \dfrac{1/4}{1/2} = \boxed{\dfrac{1}{2}}
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Unbounded
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Question 6, STEP I, 2005
(i)
 \mathrm{AP}^2 = (x-5)^2 + (y-16)^2

 \mathrm{BP}^2 = (x+4)^2 + (y-4)^2

 \therefore (x-5)^2 + (y-16)^2 = 4(x+4)^2 + 4(y-4)^2

 \iff x^2 -10x + 25 + y^2 - 32y + 256 = 4x^3 + 32x + 64 + 4y^2 -32y + 64

 \iff 153 = 3x^2 + 42x + 3y^2

 \iff 51 + 49 = x^2 + 14x + 49 + y^2

 \iff 100 = (x+7)^2 + y^2 \ \ \ \square

Or:  51 = x^2 + 14x + y^2 \ \ \ (\ast )
(ii)
Let \mathrm{Q} = (x,y)

 \mathrm{QC}^2 = (x-a)^2 + y^2

 \mathrm{QD}^2 = (x-b)^2 + y^2

 \therefore (x-a)^2 + y^2 = k^2(x-b)^2 + k^2y^2

 \iff a^2-k^2b^2 = x^2(k^2-1) + x(2a-2bk^2) + y^2(k^2-1)  \ \ \ (\ast \ast )

Comparing coefficients of different terms of  (\ast ) with  (\ast \ast ) , the ratio between the coefficients must be the same for the path to be the same.

 \therefore \dfrac{a^2-k^2b^2}{51} = \dfrac{k^2-1}{1}

 \iff a^2 - b^2k^2 = 51k^2-51

 \iff k^2 (b^2+51) = a^2 + 51 \iff k^2 = \dfrac{a^2+51}{b^2+51}

And similarly:

 \dfrac{k^2-1}{1} = \dfrac{2a-2bk^2}{14}

 \iff 7k^2 - 7 = a-bk^2

 \iff k^2(b+7) = a+7 \iff k^2 = \dfrac{a+7}{b+7}

 \therefore \dfrac{a+7}{b+7} = \dfrac{a^2+51}{b^2+51} \ \ \ \square

 \iff (a+7)(b^2+51) = (b+7)(a^2+51)

 \iff ab^2 + 51a + 7b^2 = a^2b + 51b + 7a^2

 \iff a^2b - b^2a + 7a^2 - 7b^2 + 51b-51a = 0

 \iff ab(a-b) + 7(a+b)(a-b) - 51(a-b) = 0

We can divide by (a-b) , for the case  a \not= b

\therefore ab + 7(a+b) = 51

 \iff (a+7)(b+7) = 100 \ \ \ \square
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Unbounded
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(Original post by nuodai)
STEP I 2005 Question 11
(ii) \Rightarrow \sin 2t = 0,\ \boxed{\cos 2t = 1 \Rightarrow 2t = 0, \pi, 2\pi, 3\pi,} ... \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}
Not sure about the bit I've boxed.  \cos \theta = 1 \iff \theta = 2n\pi for some integer n.

edit: for (iii) I think as a different approach: the direction of the velocity and displacement will be the same when  \dfrac{\dot{y}}{\dot{x}} = \dfrac{y}{x}
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So we end up solving:

 \dfrac{-2\sin t}{2\cos 2t} = \dfrac{2\cos t}{\sin 2t} \iff \dfrac{-\sin t}{\cos 2t} = \dfrac{1}{\sin t}

 \iff \sin^2 t = 1-2\cos^2 t \iff \cos^2 t = 0 \implies t = \frac{\pi}{2}, \frac{3\pi}{2} which are the times when it is at the origin, as found in part (i)
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Daniel Freedman
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#36
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STEP III 2005, Question 5

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 \\ y = ax^2 + bx + c \\

\\ \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b = m \\

\\ \implies x = \frac{m-b}{2a}

This gives the x coordinate when the gradient of the tangent to the curve is m. The y coordinate is given by

 y = a\left(\frac{m-b}{2a}\right)^2 + b\left(\frac{m-b}{2a}\right) + c

The equation of a line is given by  y - y_1 = m(x-x_1)

 \\ \therefore y - \left(a\left(\frac{m-b}{2a}\right)^2 + b\left(\frac{m-b}{2a}\right) + c\right) = m\left(x-  \frac{m-b}{2a}\right)  \\

\\ \implies y - mx = \left(a\left(\frac{m-b}{2a}\right)^2 + b\left(\frac{m-b}{2a}\right) + c\right) - \frac{m(m-b)}{2a} \\

\\ \implies y - mx = c \\

\\ \implies y - mx = c  - \frac{(m-b)^2}{4a}

as required.

From the above, the equations of the tangent with gradient m to each curve can be written:

 \\ y - mx = c_1 - \frac{(m-b_1)^2}{4a_1}, \mbox{ and } y - mx = c_2 - \frac{(m-b_2)^2}{4a_2} . Equating the RHS of each equation (as these two equations describe the same tangent):

 \\ c_1 - \frac{(m-b_1)^2}{4a_1} = c_2 - \frac{(m-b_2)^2}{4a_2} \ \ \left(a_1, a_2 \neq 0 \right) \\

\\ \Leftrightarrow 4c_1a_1a_2 - a_2(m^2-2b_1m +b_1^2) = 4c_2a_1a_2 - a_1(m^2 - 2mb_2 + b_2^2) \\

\\ \Leftrightarrow (a_2 - a_1)m^2 + 2(a_1 b_2 - a_2 b_1)m + 4a_1a_2(c_2-c_1) + a _2 b_1^2 - a_1b_2^2 = 0

as required.

For the two curves to have exactly one common tangent, this quadratic in m must have equal roots, meaning that its discriminant must be equal to 0.

 \\ \Leftrightarrow 4(a_1 b_2 - a_2 b_1)^2 - 4(a_2-a_1)(4a_1a_2(c_2-c_1) + a _2 b_1^2 - a_1b_2^2) = 0 \\

 \\ \Leftrightarrow a_1^2b_2^2 - 2a_1a_2b_1b_2 + a_2^2b_1^2 - (a_2-a_1)(4a_1a_2c_2 - 4a_1a_2c_1 + a_2b_1^2 - a_1b_2^2) \\  \\ \Leftrightarrow 4a_1a_2^2c_1 - 4a_1a_2^2c_2 + 4a_1^2a_2c_2 - 4a_1^2a_2c_1 + a_1a_2b_1^2 - 2a_1a_2b_1b_2  + a_1a_2b_2^2= 0  \\

\\ \Leftrightarrow a_1a_2 \left( (b_1-b_2)^2 - 4(a_1-a_2)(c_1-c_2) \right) = 0 \\

\\ \Leftrightarrow (b_1-b_2)^2 - 4(a_1-a_2)(c_1-c_2) = 0 \mbox{ as } a_1, a_2 \neq 0

For the curves to touch each other, the quadratic in x that results from eliminating y must have equal roots ie it's discriminant must be 0.

This quadratic is  x^2(a_1-a_2) + x(b_1-b_2) + (c_1-c_2) = 0 , whose discriminant is the expression above equal to 0 (provided  a_1 \neq a_2 ). Hence, in the case  a_1 \neq a_2 , the two curves have exactly one common tangent if and only if they touch each other.

If  a_1 = a_2 = a , the condition on the tangent becomes

 \\ 2(b_2 - b_1)m + 4a(c_2-c_1) + b_1^2 - b_2^2 = 0 \\

\\ \implies m = \frac{2a(c_1-c_2)}{b_2-b_1} + b_1 + b_2

Which gives the condition  b_1 \neq b_2 (if they were equal, this would imply  c_1 = c_2 ie. the curves wouldn't be distinct).
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Daniel Freedman
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#37
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(Original post by DeanK22)
STEP III

Question 1


If cos(B) = Sin(A) this implies A = (4n+1)pi/2 (+/-) B

Let B = A + x for some x in R

It follows the problem is now Cos(A+x) = Sin(A) *

Epanding using the addition formulae we see that * is transformed to;

Cos(A)Cos(x) - Sin(A)Sin(x) = Sin(A)

<->  Sin(A)(1+Sin(x))=CosA)Cos(x)
<-> tan(A) = \frac{Cos(x)}{1+Sin(x)}

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It is relativel easy to show that  \frac{cos(x)}{1+sin(x)} = tan(\frac{\pi}{4} - \frac{x}{2}) if after searching for GA formula / derivation it has not been found, we use the identity sin(x) = cos(pi/2 - x) = cos(2(pi/4 - x/2)) and expand using the double angle formulae and the result follows fairly soon.


We also note that multiples of pi can be added to the tangent functions argument without altering its value to obtain;

 Tan(A) = Tan(\frac{\pi}{4} - \frac{x}{2} + n\pi) \; n \in \mathbb{Z}

<->  A = \pi(n+\frac{1}{4}) - \frac{x}{2}

and the result

 A = \frac{(4n+1)\pi}{2} + B follows - although interestingly we do not have  \pm B wip is obviously needed to obtain this.
Bit of an odd method?

 \sin{A} = \cos{B} \implies \cos{(\frac{\pi}{2} - A)} = \cos{B} .

From looking at the cos graph, we can see that  \cos{x} = \cos{y} \implies x = y + 2k\pi or  x = -y + 2k \pi for some integer k, which gives

 \\ \frac{\pi}{2} - A = B + 2k \pi \implies A = (-4k+1)\frac{\pi}{2} - B \\

\\ \mbox{or} \\

\\ \frac{\pi}{2}- A = -B + 2k\pi \implies A = (-4k+1)\frac{\pi}{2} + B

which we can write as  A = (4n+1)\frac{\pi}{2} \pm B for some integer n.
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Dadeyemi
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#38
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(Original post by GHOSH-5)
(ii)
With the substitution  u = x+1

 I = \displaystyle\int_0^1 x(x+1)^m \ \mathrm{d}x = \displaystyle\int_1^2 \left(u^{m+1} - u^m\right) \ \mathrm{d}u

Case 1:  m \not= -1, -2

 I = \left [ \dfrac{u^{m+2}}{m+2} - \dfrac{u^{m+1}}{m+1} \right ]_1^2 = \left[\dfrac{2^{m+2}}{m+2} - \dfrac{2^{m+1}}{m+1}\right] - \left [ \dfrac{1}{m+2} - \dfrac{1}{m+1} \right ]

 = \dfrac{2^{m+2} - 1}{m+2} + \dfrac{1 - 2^{m+1}}{m+1} = \boxed{\dfrac{2^{m+1}m+1}{(m+1)(  m+2)}}

Case 2:  m = -1

I = \displaystyle\int_1^2 \left(1 - \dfrac{1}{u}\right) \ \mathrm{d}u = \left [ u - \ln u \right ]_1^2 = \left [ 2 - \ln 2 \right ] - \left [ 1 - \ln 1 \right ]

 = \boxed{1 - \ln 2}

Case 2:  m = -2

 I = \displaystyle\int_1^2 \left(\dfrac{1}{u} - \dfrac{1}{u^2}\right) \ \mathrm{d}u = \left [ \ln u + \dfrac{1}{u} \right ]_1^2

 = \left [ \ln 2 + \dfrac{1}{2} \right ] - \left [ \ln 1 + 1 \right ] = \boxed{\ln 2 - \dfrac{1}{2}}
It might be a little neater just to do:

 I = \displaystyle\int_0^1 x(x+1)^m \ \mathrm{d}x 

\\= \int_0^1 (x+1-1)(x+1)^m \ \mathrm{d}x  

\\= \int_0^1 (x+1)^{m+1}- \int_0^1 (x+1)^m \ \mathrm{d}x

then the results follow immediately from part (i)
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nota bene
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#39
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(Original post by Adje)
III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation


 \displaystyle \frac{dy}{dx} = -\frac{xy}{x^2 + a^2}

Separating variables and integrating:

 \displaystyle \int \frac{1}{y} \ dy = - \int \frac{x}{x^2 + a^2} \ dx

 \displaystyle \ln y = -\frac{1}{2} \ln (x^2 + a^2) + K = -\frac{1}{2} \ln [B(x^2 + a^2)] with  B = - \frac{1}{2}\ln K

 \displaystyle y = \frac{1}{\sqrt{B}\sqrt{x^2 + a^2}}

 \displaystyle y \sqrt{x^2 + a^2} = c with  c = \frac{1}{\sqrt{B}}

 \displaystyle y^2(x^2 + a^2) = c^2
I don't agree with your constant after the integration -\frac{1}{2} \ln (x^2 + a^2) + K You say this equals -\frac{1}{2} \ln [B(x^2 + a^2)] with  B = - \frac{1}{2}\ln K However, substituting that B into it gives -\frac{1}{2}(\ln\ln(k^{-\frac{1}{2}})+\ln(x^2+a^2)) and -0.5ln(ln(1/sqrt(k))) =/ k.

I got:
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\ln y=-\frac{1}{2} \ln (x^2 + a^2) + k= \ln(\frac{1}{\sqrt{x^2+a^2}})+k So y=e^k\frac{1}{\sqrt{x^2+a^2}} which also gives the desired form, with c=e^k
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Adjective
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#40
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Yes, I mean K = -1/2 ln B, don't I? I was just shifting around the constant. Sorries.
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