Farhan.Hanif93
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Papers found here. (Credit to Tarquin Digby for those [and Charles98 for his photos])

STEP II:
1: Solution by mikelbird
2: Solution by LightBlueSoldier
3: Solution by Stray
4: Solution by Khallil
5: Solution by Blazy
6: Solution by DFranklin
7: Solution by cliverlong
8: Solution by LightBlueSoldier, Solution by Stray
9: Solution by Brammer
10: Solution by JosephML
11: Solution by Brammer
12: Solution by shamika
13: Solution by DJMayes


STEP III:
1: Solution by jtSketchy
2: Solution by Brammer
3: Solution by edrraa, Solution by mikelbird
4: Solution by Farhan.Hanif93
5: Solution by metaltron
6: Solution by ctrls
7: Solution by Farhan.Hanif93
8: Solution by DFranklin
9: Solution by JosephML
10: Solution by JosephML
11: Solution by Brammer
12: Solution by DJMayes
13: Solution by DJMayes
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JosephML
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II/10:

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For the first part, get the two equations for x and y and combine them:

y = u \sin \theta - \frac{1}{2}gt^2 and x=u \cos \theta t, giving:

y(x, \lambda) =\lambda x - x^2 \frac{\lambda^2 + 1}{2u^2}g

Now we wish to maximize y for a given value of x so we take what is known as the partial derivative of y with respect to \lambda i.e. ignore x and differentiate wrt to \lambda:

\displaystyle \frac{\partial y}{\partial \lambda} = - \frac{gx^2}{u^2} \lambda + x.

To get the max, we solve \frac{\partial y}{\partial \lambda} = 0 which gives  \lambda = \frac{u^2}{gx}. Putting this back into y(x, \lambda) will give us Y:

\displaystyle Y = \frac{u^2}{2g} - \frac{g}{2u^2}x^2.

The sketch of this is fairly straight forward with the set of points which can be reached being those with Y > 0 and 0\leq y \leq Y.

Finally the maximum possible distance reached by the particle is when Y = 0 which gives us x = \frac{u^2}{g}.
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shamika
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II/12:

(i)

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P(t < T < t + \delta t | T > t) = \frac{P(t < T < t + \delta t)}{P(T>t)}. The denominator is just 1 - F(t). The numerator is \frac{F(t + \delta t) - F(t)}{\delta t} \cdot \delta t = f(t) \cdot \delta t, since the first term is tends to the derivative of F(t) (i.e. f(t)), as t tends to 0.


(ii)

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In this case, f(t)=\frac{1}{a}. giving immediately h(t) = \frac{1}{a-t}.


(iii)

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Note that f(t) =\frac{dF(t)}{dt}. Therefore we want to solve \frac{1}{t}= \frac{\frac{dF(t)}{dt}}{1-F(t)}. Separating the variables gives

\int {\frac{dt}{t}} = \int{\frac{dF(t)}{1-F(t)}}, i.e.

\log t = -\log(1 - F(t)) + c. When t = a, F(t) = 0, so c = log a, and so rearranging we get F(t) = 1 - \frac{a}{t}. Differentiating, we have \frac{a}{t^2}.

As a check, we can insert F(t) and f(t) into the definition of the hazard function; we get h(t) = \frac{a}{t^2} \cdot \frac{t}{a} = \frac{1}{t}, as required.


(iv)

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We prove each implication separately. If f(t) = k\exp(-k(t-b)), then F(t)=k\int_b^t exp(-k(x-b)dx, where x is just a dummy variable. This gives F(t) = 1 - \exp(-k(t-b)). Subbing this into the formula of h(t), we get h(t) = k, a constant.

On the other hand, if h(t) = k, then if we solve k(t - F(t)) = \frac{dF(t)}{dt}, which (after separating the variables and integrating to find F(t)), gives F(t) = 1 - \exp(-k(t-b)), which differentiates to the required form of f(t).


(v)

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Similar to (iii) and (iv), solve \frac{\lambda}{\theta^{\lamda}} \cdot t^{\lamda -1} \cdot (1 - F(t)) = \frac{dF(t)}{dt}, to give F(t) = 1 - \exp(-(\frac{t}{\theta})^{\lambda}). Finally, differentiating, we get f(t) = \frac{\lambda}{\theta}\cdot (\frac{t}{\theta})^{\lambda - 1} \codt \exp(-(\frac{t}{\theta})^{\lambda})

Note: this is the Weibull distribution, which is an important distribution in actuarial science to model very extreme losses. Of course, there was no reason you needed to know that to complere the question. The concept of the hazard function is also a fundamental theoretical actuarial tool, but I've never used it in practice.
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LightBlueSoldier
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Some photos: II 2:

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And II 8

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JosephML
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III/9

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We have
\displaystyle \bold{r} = \frac{kt - 1 + e^{-kt}}{k^2}\bold{g} + \frac{1- e^{-kt}}{k}\bold{u},

so differentiating gives us:

\displaystyle \bold{v} = \frac{1-e^{-kt}}{k}\bold{g} + e^{-kt}\bold{u}

and again:

\displaystyle \bold{a} = e^{-kt} \bold{g} - k e^{-kt} \bold{u}.

For the initial conditions just put in t = 0.

Now we must show that this fits with \bold{F} = m \bold{a}:

\bold{F} &= m \bold{g} - mk\bold{v} \\    &= m\bold{g} - m((1-e^{-kt})\bold{g} - ke^{-kt}\bold{u}) \\ &= m(e^{-kt}\bold{g} - ke^{-kt}\bold{u})

And so we indeed have \bold{F} = m\bold{a}.

Now letting \bold{u} = u\cos \alpha \bold{i} + u \sin \alpha \bold{j} and \bold{g} = -g \bold{j} and subbing into \bold{r} we get:

\displaystyle \bold{r} = (- \frac{kt - 1 + e^{-kt}}{k^2} + \frac{1-e^{-kt}}{k} u \sin \alpha) \bold{j} + \frac{1- e^{-kt}}{k} u \cos \alpha \bold{i}

For \bold{r . j} = \bold{0} we must have the \bold{j} component of \bold{r} to be zero. Hence giving:

\displaystyle ku \sin \alpha = (\frac{kT}{1- e^{-kT}} -1)g.

Similarly writing \bold{v} in terms of \bold{i} and \bold{j} we have:

\displaystyle \bold{v} = (e^{-kt} u \sin \alpha - \frac{1- e^{-kt}}{k}g) \bold{j} + e^{-kt} u \cos \alpha \bold{i}

which then gives

\displaystyle \tan \beta = - \frac{e^{-kT} u \sin \alpha - k^{-1}(1-e^{-kT}g}{e^{-kT}u \cos \alpha}

which simplifies to the required result. Combining the two results gives us:

\displaystyle \tan \beta = \tan \alpha \left(\frac{e^{kT} -1 - kT}{kT + e^{-kT} - 1}\right)
We are given that \sinh \left(kT\right) > kT. So from the definition of \sinh we have:

\displaystyle \frac{e^{kT} - e^{-kT}}{2} > kT \\ \Rightarrow e^{kT} - kT - 1< e^{-kT} + kT -1 \Rightarrow \frac{e^{kT} - kT - 1}{e^{-kT} + kT -1} > 1

And so we get the required \tan \beta > \tan \alpha. Finally since \alphaand \beta are both acute we must have \beta > \alpha
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user2020user
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II/4:
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(i)

\begin{aligned} \text{I} = \int_{1/b}^{b} \dfrac{x\ln x}{\left( a^2 + x^2 \right) \left( a^2 x^2 + 1 \right)} \text{ d}x & \ \overset{u=1/x}= \ \int_{b}^{1/b} \dfrac{\frac{1}{u} \ln \left( \frac{1}{u} \right)}{\left( a^2 + \frac{1}{u^2} \right) \left( \frac{a^2}{u^2} + 1 \right)} \cdot \dfrac{\text{ d}u}{-u^2} \\ & = \int_{b}^{1/b} \dfrac{ - \frac{1}{u} \ln u}{- \left( a^2 u^2 +1 \right) \frac{1}{u^2} \left(a^2 + u^2 \right)} \text{ d}u \\ & \int_{b}^{1/b} \dfrac{u\ln u}{\left( a^2 u^2 +1 \right) \left(a^2 + u^2 \right)} \text{ d}u \\ & = - \int_{1/b}^{b} \dfrac{u\ln u}{\left( a^2 u^2 +1 \right) \left(a^2 + u^2 \right)} \text{ d}u \end{aligned}

\begin{aligned} 2\text{I} & = \int_{1/b}^{b} \dfrac{x\ln x}{\left( a^2 + x^2 \right) \left( a^2 x^2 + 1 \right)} \text{ d}x + \underbrace{\left( - \int_{1/b}^{b} \dfrac{u\ln u}{\left( a^2 u^2 +1 \right) \left(a^2 + u^2 \right)} \text{ d}u \right)}_{\text{Rewriting } u \text{ as } x} \\ & = \int_{1/b}^{b} \dfrac{x\ln x}{\left( a^2 + x^2 \right) \left( a^2 x^2 + 1 \right)} \text{ d}x - \int_{1/b}^{b} \dfrac{x\ln x}{\left( a^2 x^2 +1 \right) \left(a^2 + x^2 \right)} \text{ d}x = 0 \end{aligned} \\ \\ \text{ I} \ = 0 \end{aligned}
(ii)

\displaystyle \begin{aligned} \text{J} = \int_{1/b}^{b} \dfrac{\arctan x}{x} \text{ d}x & \ \overset{u=1/x}= \ \int_{b}^{1/b} \dfrac{\arctan \left( \frac{1}{u} \right)}{\left( \frac{1}{u} \right)} \cdot \dfrac{\text{ d}u}{-u^2} \\ & = \int_{b}^{1/b} \dfrac{ -\arctan \left( \frac{1}{u} \right)}{u} \text{ d}u \\ & = \int_{1/b}^{b} \dfrac{ \arctan \left( \frac{1}{u} \right)}{u} \text{ d}u \end{aligned}

\begin{aligned} 2\text{J} & = \int_{1/b}^{b} \dfrac{ \arctan x}{x} \text{ d}x + \underbrace{\int_{1/b}^{b} \dfrac{ \arctan \left( \frac{1}{u} \right)}{u} \text{ d}u}_{\text{ Rewriting } u \text{ as } x} \\ & = \int_{1/b}^{b} \dfrac{ \arctan x + \arctan \left( \frac{1}{x} \right)}{x} \text{ d}x \ = \ \dfrac{\pi}{2} \int_{1/b}^{b} \dfrac{\text{d}x}{x} \ = \ \dfrac{\pi}{2} \ln (b^2) \end{aligned}

\text{ J} \ = \dfrac{\pi \ln b}{2}
(iii)

 \begin{aligned} I & = \int_{0}^{\infty} \dfrac{1}{\left( a^2 + x^2 \right)^2} \text{ d}x \\ & \overset{u = a{^2}/x}= \dfrac{1}{a^{2}} \int_{0}^{\infty} \dfrac{u^{2}}{(u^2 + a^2)^{2}} \text{ d}u \\ & \overset{+a^{2}-a^{2}}= \dfrac{1}{a^{2}} \left( \int_{0}^{\infty} \dfrac{1}{u^{2}+a^{2}} \text{ d}u - a^{2} I \right) \implies 2I = \dfrac{1}{a^{2}} \underbrace{ \left( \int_{0}^{\infty} \dfrac{1}{u^{2}+a^{2}} \text{ d}u \right)}_{= \pi/2a} \end{aligned}
I'll edit the last part into this post in a short while.
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III/9
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I only got up as far as the result involving uksin - how many marks do you think I'd have got?
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Kasel
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III/9
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
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III/6

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By letting h(t)=f''(t) with the given inequality and noting that the inequality f''(t)<0 holds for 0<t<x_0 where 0<x_0<x, you get,

\int_0^{x_0} f''(t) dt = f'(x_0) - f(0) = f'(x_0)>0.

Since f'(0)=0.Repeating the process by setting f'(x)=h(t) gives the required inequality.

(i) Choosing f(x)=1 - \cos{x} \cosh{x}, if we can prove that f(x)>0 for all 0<x<\frac{\pi}{2}, then the result follows. To do this, we differentiate it twice, so

f'(x) = \sin{x}\cosh{x} - \cos{x}\sinh{x},
f'(x) = \cos{x}\cosh{x} + \sin{x}\sinh{x} + \sin{x}\sinh{x} - \cos{x}\cosh{x} = 2\sin{x}\sinh{x}.

Noting that for 0<x<\frac{\pi}{2}, \sin{x}>0 and \sinh{x}>0 so it follows that f''(x)>0 for this range of x. Since f(0)=f'(0)=0, the conditions for the inequality in the original question is met and we can deduce f(x)>0 for the given range of x and the result follows.

(ii) Here you can prove each bound separately. Since for 0<x<\frac{\pi}{2} you can show that x>0, \sin{x}>0 and \sinh{x}>0, we can multiply over the denominators without worrying about flipping the inequality. This gives,

\frac{1}{\cosh{x}} < \frac{\sin{x}}{x} \Leftrightarrow \sin{x}\cosh{x}-x>0.

So we set f(x)=\sin{x}\cosh{x}-x and prove it is positive in the given domain using the inequality again. Differentiating twice gives,

f'(x)=\cos{x}\cosh{x} + \sin{x}\sinh{x} - 1,
f''(x)=-\sin{x}\cosh{x} + \cos{x}\sinh{x} + \cos{x}\sinh{x} + \cos{x}\cosh{x} = 2\cos{x}\sinh{x}.

Noting again the the two functions are positive in the domain and that f(x)=f'(x)=0, the result follows.

Finally for the upper bound, rearranging the inequality as,

\frac{\sin{x}}{x} < \frac{x}{\sinh{x}} \Leftrightarrow x^2 - \sin{x}\sinh{x}>0,

And setting f(x)=x^2 - \sin{x}\sinh{x}>0, again it is sufficient to show that f(x)>0 for the given domain to establish the inequality. Differentiating twice,

f'(x)=2x - \cos{x}\sinh{x} - \sin{x}\cosh{x},
f''(x)=2+\sin{x}\sinh{x} - \cos{x}\cosh{x} - \cos{x}\cosh{x} - \sin{x}\sinh{x} = 2\left(1-\cos{x}\cosh{x}\right).

Note that this gives the same inequality from part (i), so we can quickly deduce that f''(x)>0 for the given domain. Since f(0)=f'(0)=0, we can deduce that f(x)>0 and the result follows.

Note: Most of this is just tedious manipulation to show the initial inequalities are met, so I've omitted some of the details of showing certain functions are positive and that f(0) and f'(0) are equal to zero in the various cases. If there's an faster way to do this question that I've missed, I would be interested to hear about it.
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JosephML
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(Original post by Kasel)
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
Ah, you're right. I missed that bit. I'll correct the solution, thanks.

(Original post by CD315)
I only got up as far as the result involving uksin - how many marks do you think I'd have got?
I should suspect somewhere between 8 and 12 marks.
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Kasel
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II/4
Do you think we would have had to prove arctan(x) + arctan(1/x)=pi/2?
There are a few short proofs but I forgot to include any in my solution.
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Farhan.Hanif93
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STEP III, Q4

(i)
Note, expanding the square in the integrand of I_1:

I - I_1  =  -\displaystyle\int_0^1 y^2(1+\tan^2 x) + 2yy'\tan x dx

=  - \displaystyle\int_0^1 y^2\sec^2x +\left(\dfrac{d}{dx}(y^2)\right) \tan x dx

=  - \displaystyle\int_0^1 \dfrac{d}{dx}[y^2\tan x] dx

=  y(0)^2\underbrace{\tan (0)}_{= \ 0} - \underbrace{y(1)^2}_{= \ 0}   \tan (1)

\Rightarrow \boxed{I-I_1 = 0}, as required.

Hence I = \displaystyle\int_0^1 \underbrace{(y' + y\tan x)^2}_{\geq 0} dx, which can interpreted as the area under a curve that is always above the x-axis. It follows that \boxed{I\geq 0} with I=0 \iff y'+y\tan x = 0 over the interval of integration i.e. for every x s.t. 0\leq x\leq 1. Solving:

\iff \displaystyle\int\dfrac{y'}{y} dx= \displaystyle\int -\tan x dx

\iff y= A\cos x.

Note y(1) = 0 \Rightarrow A=0 so that \boxed{I=0 \iff y\equiv 0} on 0\leq x\leq 1, as desired.


(ii)
Let J_1 = \displaystyle\int_0^1 (y' + ay\tan b x)^2 dx and observe that J_1 \geq 0 by the argument in (i). Moreover, we require b<\dfrac{\pi}{2} in order to avoid integrating over the discontinuity of \tan bx at x= \dfrac{\pi}{2b} i.e. for our integral to make sense.

Furthermore:

J-J_1 = - a\displaystyle\int_0^1 a y^2 \sec^2 bx + 2yy'\tan bx dx

Setting b=a allows us to write the integrand as the derivative of a product, provided that a<\pi /2 for the above reasons.

So, in the case a<\pi /2:

J - J_1 = -a \displaystyle\int_0^1 \dfrac{d}{dx}[y^2 \tan ax] dx

 =a[y(0)^2\underbrace{\tan 0}_{= \ 0} - \underbrace{y(1)^2}_{= \ 0}\tan a]

\Rightarrow \boxed{J = J_1 \geq 0}, as desired, with the argument being valid for \boxed{a< \dfrac{\pi}{2}}.

In the case a=\pi / 2:

Integrating the first term in the integrand of J = \displaystyle\int_0^1 \left(y'^2 - \left(\dfrac{\pi}{2}\right)^2 y^2\right) dx by parts with u=y', we obtain:

J = y(0)y'(0) - \displaystyle\int_0^1 y(y''+(\pi /2 )^2 y) dx

So it is sufficient to search for a function y\not\equiv 0 s.t. y''+(\pi /2)^2y =0 with either y(0) or y'(0) = 0.

Seeing that the general solution of the second order differential equation is y = A\cos \dfrac{\pi x}{2}  + B\sin \dfrac{\pi x}{2}, y(1) = 0 forces us to take B=0 and observing that \dfrac{d}{dx}[A\cos \dfrac{\pi x}{2} ]|_{x=0} = 0, it follows that  y= \cos \dfrac{\pi x}{2} is s.t. y(1) = 0, y'(0)=0 and y''+(\pi /2)^2 y = 0 on 0<x<1.

Hence y=\cos \dfrac{\pi x}{2} \Rightarrow J = 0 - \displaystyle\int_0^1 y\cdot 0 dx = 0 so \boxed{y=\cos \dfrac{\pi x}{2} } will do the trick.



It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.
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user2020user
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...
I did but I left it out of the solution above for brevity. The one I wrote in the paper was very similar to this, but it looks really forced with regards to the 1-1 in the denominator.

\begin{aligned} \phi = \arctan b + \arctan \frac{1}{b} \implies \tan \phi & = \dfrac{\tan \left( \arctan b \right) + \tan \left( \arctan \frac{1}{b} \right)}{1 - \tan \left( \arctan b \right)\tan \left( \arctan \frac{1}{b} \right)} \\ & = \dfrac{b + \frac{1}{b}}{1 - 1} \end{aligned}

\therefore \ \tan \phi = \displaystyle \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \implies \phi = \arctan \left( \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \right) = \dfrac{\pi}{2}
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To avoid limits you can say let y=arctan(x), then tan(y)=x
and
cot(y)=(1/x)=tan(pi/2 - y)
which gives arctan(x)+arctan(1/x)= y+pi/2-y=pi/2
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Ah, of course! :facepalm2:
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STEP III - Q12

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i)

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Note that  y = e^x is increasing. Thus:

 P(x \leq x_m ) = \frac{1}{2} \iff P(y \leq e^{x_m} ) = \frac{1}{2}

So  y_m = e^{x_m}



ii)

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 P(Y \leq y ) = P(e^X \leq y) = P(X \leq lny)

Differentiating, we deduce a p.d.f of  \dfrac{f(lny)}{y} , as required.

The mode occurs at a stationary point of the p.d.f (not at an endpoint; this is not possible as the range is infinite) so occurs when the derivative of the p.d.f is 0:

 \Rightarrow \dfrac{f'(lny) - f(lny)}{y^2} = 0

From which  f'(ln\lambda) = f(ln\lambda) follows.



iii)

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Our integral is merely the integral of the p.d.f of the random variable  N(\mu + \sigma^2, \sigma) , so equals 1. Then:

 E(Y) = E(e^X) = \dfrac{1}{\sigma\sqrt{2\pi}} \displaystyle\int_{-\infty}^{\infty} exp(-\dfrac{1}{2\sigma^2}((x-\mu)^2-2\sigma^2x)) \ dx

\dfrac{1}{\sigma\sqrt{2\pi}} \displaystyle\int_{-\infty}^{\infty} exp(-\dfrac{1}{2\sigma^2}((x-\mu-\sigma^2)^2-2\mu \sigma^2 - \sigma^4)) \ dx

 = exp(\mu + \frac{1}{2}\sigma^2) I

Where I is the integral mentioned above equalling one, from which our result follows.



iv):

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Using our p.d.f. X has median  \mu so Y has median  e^{\mu}

Applying ii) with f the p.d.f. given above, we find that  \lambda = e^{\mu - \sigma^2} :

 f(x) = \dfrac{1}{\sqrt{2 \pi \sigma^2}} exp(-\dfrac{(x-\mu)^2}{2\sigma^2})

 f'(x) = -\dfrac{(x-\mu)}{\sigma^2 \sqrt{2 \pi \sigma^2}}  exp(-\dfrac{(x-\mu)^2}{2\sigma^2})

 f(x) = f'(x) \Rightarrow  -\dfrac{(x-\mu + \sigma^2)}{\sigma^2 \sqrt{2 \pi \sigma^2}}  exp(-\dfrac{(x-\mu)^2}{2\sigma^2}) = 0

 \Rightarrow x = \mu - \sigma^2

Noting that  x = log\lambda , the result follows.

Then the inequality follows easily from the fact the exponential function is increasing.



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STEP II - Q8 (LaTeX)
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\{n, a, b, c_r\} \mathbb{Z}+
c_r, (0 \leq r \leq n) = \texttt{coefficient of } x^r \texttt{ in } (a + bx)^n

c_r = ^n\mathrm{C}_r  \cdot a^{n-r} b^r= \frac{n!}{r!(n-r)!} \cdot a^{n-r} b^r

c_m \geq c_r \texttt{ for } 0 \leq r \leq n

If c_m is a maximum, it must be \geq c_{m+1} and \geq c_{m-1}


Taking first, c_{m+1} :

\frac{n!a^{n-(m+1)} b^{(m+1)}}{(m+1)!(n-(m+1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}

Divide both sides by the common factor

\frac{n!a^{n-(m+1)}b^m}{m!(n-(m+1))!}

Giving

\frac{b}{m+1} \leq \frac{a}{n-m}

bn-bm \leq am+a

bn-a \leq am+bm

\frac{bn-a}{a+b} \leq m

This is almost the LHS of what we are required to show, just needs a little fiddling

\frac{bn+b-(a+b)}{a+b} \leq m

\frac{b(n+1)}{a+b} -1 \leq m  (**)

Now repeat the process for c_{m-1}:

\frac{n!a^{n-(m-1)} b^{(m-1)}}{(m-1)!(n-(m-1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}

Divide both sides by the common factor

\frac{n!a^{n-m}b^{m-1}}{(m-1)!(n-m)!}

Giving

\frac{a}{n-(m-1)} \leq \frac{b}{m}

am \leq bn + b - bm

m(a+b) \leq b(n+1)

m \leq \frac{b(n+1)}{a+b}

Which combines with (**) to give the required solution

\frac{b(n+1)}{a+b} -1 \leq m \leq \frac{b(n+1)}{a+b}

Considering the possible values of m

\texttt{Let } y=\frac{b(n+1)}{a+b}

y-1 \leq m \leq y

Either y and y-1 are both integers, or y and y-1 are non-integer values either side of an integer, so if m is an integer, it is either y and y-1 (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of m where there is 1 value, or the larger value if there are 2. This means we can consider the function G(n,a,b) to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is \lfloor.

G(n,a,b) =  \lfloor{\frac{b(n+1)}{a+b}}

i)
G(9,1,3) =  \lfloor{\frac{3(9+1)}{1+3}} = \lfloor\frac{30}{4} = \floor{7.5} = 7

G(9,2,3) =  \lfloor{\frac{3(9+1)}{2+3}} = \lfloor\frac{30}{5} = 6

ii)
G(2k,a,a) = \lfloor{\frac{a(2k+1)}{a+a}} = \lfloor\frac{2k+1}{2} = \lfloor{(k + 0.5)} = k

G(2k-1,a,a) = \lfloor{\frac{a(2k-1+1)}{a+a}} = \lfloor\frac{2k}{2} = k

iii)
If we fix n and b, the only part of G that can alter is the denominator containing a, so for large G we want small a, so G_{max} is when a=1. (Zero is not a positive integer).

iv)
For fixed a=1 and (unknown) fixed n, the part of the function varying is:

\frac{b}{b+1}

For positive integer values of b, \frac{1}{2} \leq \frac{b}{b+1} < 1.

So \frac{1}{2} \leq G(n,1,b) < n+1
giving G(n,1,b)_{max}=n

LightBlueSoldier points out below that my original b = \infty is wrong... of course the constraint isn't nearly as harsh as that, regardless of whether infinity is acceptable...

\frac{b(n+1)}{b+1} \geq n

bn+b \geq nb+n

b \geq n



This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
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LightBlueSoldier
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(Original post by Stray)
STEP II - Q8 (LaTeX)

\{n, a, b, c_r\} \mathbb{Z}+
c_r, (0 \leq r \leq n) = \texttt{coefficient of } x^r \texttt{ in } (a + bx)^n

c_r = ^n\mathrm{C}_r  \cdot a^{n-r} b^r= \frac{n!}{r!(n-r)!} \cdot a^{n-r} b^r

c_m \geq c_r \texttt{ for } 0 \leq r \leq n

If c_m is a maximum, it must be \geq c_{m+1} and \geq c_{m-1}


Taking first, c_{m+1} :

\frac{n!a^{n-(m+1)} b^{(m+1)}}{(m+1)!(n-(m+1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}

Divide both sides by the common factor

\frac{n!a^{n-(m+1)}b^m}{m!(n-(m+1))!}

Giving

\frac{b}{m+1} \leq \frac{a}{n-m}

bn-bm \leq am+a

bn-a \leq am+bm

\frac{bn-a}{a+b} \leq m

This is almost the LHS of what we are required to show, just needs a little fiddling

\frac{bn+b-(a+b)}{a+b} \leq m

\frac{b(n+1)}{a+b} -1 \leq m  (**)

Now repeat the process for c_{m-1}:

\frac{n!a^{n-(m-1)} b^{(m-1)}}{(m-1)!(n-(m-1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}

Divide both sides by the common factor

\frac{n!a^{n-m}b^{m-1}}{(m-1)!(n-m)!}

Giving

\frac{a}{n-(m-1)} \leq \frac{b}{m}

am \leq bn + b - bm

m(a+b) \leq b(n+1)

m \leq \frac{b(n+1)}{a+b}

Which combines with (**) to give the required solution

\frac{b(n+1)}{a+b} -1 \leq m \leq \frac{b(n+1)}{a+b}

Considering the possible values of m

\texttt{Let } y=\frac{b(n+1)}{a+b}

y-1 \leq m \leq y

Either y and y-1 are both integers, or y and y-1 are non-integer values either side of an integer, so if m is an integer, it is either y and y-1 (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of m where there is 1 value, or the larger value if there are 2. This means we can consider the function G(n,a,b) to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is \lfloor.

G(n,a,b) =  \lfloor{\frac{b(n+1)}{a+b}}

i)
G(9,1,3) =  \lfloor{\frac{3(9+1)}{1+3}} = \lfloor\frac{30}{4} = \floor{7.5} = 7

G(9,2,3) =  \lfloor{\frac{3(9+1)}{2+3}} = \lfloor\frac{30}{5} = 6

ii)
G(2k,a,a) = \lfloor{\frac{a(2k+1)}{a+a}} = \lfloor\frac{2k+1}{2} = \lfloor{(k + 0.5)} = k

G(2k-1,a,a) = \lfloor{\frac{a(2k-1+1)}{a+a}} = \lfloor\frac{2k}{2} = k

iii)
If we fix n and b, the only part of G that can alter is the denominator containing a, so for large G we want small a, so G_{max} is when a=1. (Zero is not a positive integer).

iv)
For fixed a=1 and (unknown) fixed n, the part of the function varying is:

\frac{b}{b+1} (***)

This tends to 1 as b tends to infinity, but will never actually reach 1, so (***)(n+1) tends towards but never reaches n+1 so for fixed n
G(n,1,b)_{max} =n \texttt{ when } b = \infty

---

This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
The last part is incorrect. In particular the value of infinity does not exist.


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Blazy
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STEP II, Q5: (somebody quickly check through this please though I think it should be fine)

(i)
Let \displaystyle y = ux , so \displaystyle \frac{dy}{dx} = u + x\frac{du}{dx} by Product Rule.

Using this substitution, the question becomes:

Solve
\displaystyle u + x\frac{du}{dx} = \frac{2ux+x}{ux-2x} =  \frac{2u+1}{u-2} .

Rearranging:  \displaystyle x\frac{du}{dx} = \frac{2u+1}{u-2} - u = \frac{u^2 - 4u - 1 }{2-u}

Separating the variables and integrating:
 \displaystyle \int \frac{2-u}{u^2 - 4u - 1 } du = \int  \frac{1}{x} dx

 \displaystyle \Rightarrow -\frac{\log(u^2 - 4u - 1)}{2} = \log x + A

 \displaystyle \Rightarrow \log(u^2 - 4u - 1)= -2\log x + B

 \displaystyle \Rightarrow u^2 - 4u - 1= Cx^{-2}

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

 \displaystyle \Rightarrow y^2 - 4xy - x^2= C

The solution curve passes through (1,1) so C = -4

Hence:  \boxed{x^2 + 4xy - y^2 = 4}


(ii)
We look for a and b such that:  a - 2b = 4 and  2a + b = 3 in order to make the DE homogenous.  a = 2 and  b = -1 works.

Then substituting  x = X + 2 and  y = Y - 1 :

 \displaystyle \frac{dY}{dX} = \frac{X-2Y}{2X+Y}

Making the same substitution  Y = UX and rearranging:

 \Rightarrow \displaystyle X\frac{dU}{dX} = \frac{-U^2 - 4U + 1}{2+U}

 \Rightarrow \displaystyle \int \frac{2+U}{-U^2 - 4U + 1} dU = \int \frac{1}{X}dX

 \Rightarrow \displaystyle -\frac{\log(-U^2 - 4U + 1)}{2} = \log X + A

 \Rightarrow \displaystyle \log(-U^2 - 4U + 1) = -2\log X + B

 \Rightarrow \displaystyle -U^2 - 4U + 1 = CX^{-2}

 \Rightarrow \displaystyle -Y^2 - 4XY + X^2 = C

 \Rightarrow \displaystyle -(y+1)^2 - 4(x-2)(y+1) + (x-2)^2 = C

The solution curve goes through (1,1), so C = 5. Hence:

 \boxed {-(y+1)^2 - 4(x-2)(y+1) + (x-2)^2 = 5}


The only real difficulty here was spotting how to get rid of the -4 and -3.
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metaltron
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STEP III Q5

Beginning

Spoiler:
Show


If ABCD is a parallelogram, then  \displaystyle \vec{AB} =b - a = \vec {DC} = c  - d . So  \displaystyle a + c = b + d .
If  \displaystyle a + c = b+ d then  \displaystyle b-a = c-d so  \displaystyle \vec{AB} = \vec {DC} so ABCD is a parallelogram.

If ABCD is a parallelogram then it is a square if and only if its diagonals are perpendicular. That is  \displaystyle i(a-c) = b-d .


Part i)

Spoiler:
Show


Half the diagonal of a square is  \displaystyle \frac{\sqrt{2}}{2} times the side length of a square, and the diagonal makes an angle of 45 degrees with the side. So by a spiral enlargement:

 \displaystyle \frac{\sqrt{2}}{2}e^{i\frac{\pi}{4}} (p-q) = x - q

So  \displaystyle x = q +  \frac{\sqrt{2}}{2} \times  \frac{\sqrt{2}}{2} (1+i)(p-q)

So  \displaystyle x = q + \frac{1}{2}(p(1+i) - q(1+i)) = \frac{1}{2}(p(1+i) + q(1-i))


Part ii)

Spoiler:
Show


By symmetry:

 \displaystyle x=  \frac{1}{2}(p(1+i) + q(1-i))

 \displaystyle y=  \frac{1}{2}(q(1+i) + r(1-i))

 \displaystyle z=  \frac{1}{2}(r(1+i) + s(1-i))

 \displaystyle t=  \frac{1}{2}(s(1+i) + p(1-i))

So  \displaystyle 2(y - t) = (q-s)(1+i) + (r-p)(1-i) = (q+r-s-p) + (q-s+p-r)i

 \displaystyle 2i(x-z) = i((p-r)(1+i) + (q-s)(1-i)) = i((p-r+q-s) + (p-r+s-q)i)

So  i(x-z) = y - t so diagonals of quadrilateral are perpendicular.

So XYZT is a square if and only if  \displaystyle \vec{XY} = \vec{TZ} .

That is  \displaystyle (q-p)(1+i) + (r-q)(1-i) = (r-s)(1+i) + (s-p)(1-i) .

Looking at the co-efficient of i (which is complex itself)  \displaystyle r - p = r - p which is always true.

Equating the other parts  \displaystyle 2q - p - r = r + p - 2s .

So XYZT is a square if and only if  \displaystyle q +s = p + r which is the condition for PQRS to be a parallelogram.
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