Hence find the equation of the curve that passes through the point (1, 1) and intersects each of the above parabolas orthogonally. Sketch this curve.

(Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.)

Spoiler

Reply 22

Glutamic Acid

II/7:

Q is given by a rotation of OP by 30 degrees extended onto P, so

q = p + e^{i \pi/6}p = 2 + 2(\dfrac{\sqrt{3}}{2}+ \dfrac{i}{2}) = 2 + \sqrt{3} + i

r = q + (q - p)e^{i \pi/6} = 2 + \sqrt{3} + i + (\sqrt{3} + i)(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}) = 3 + \sqrt{3} + (1 + \sqrt{3})i

s = r + (r - q)e^{i \pi/6} = 3 + \sqrt{3} + (1 + \sqrt{3})i + \dfrac{1}{2}(1 + \sqrt{3}i)(\sqrt{3} + i) = 3 + \sqrt{3} + (3 + \sqrt{3})i

OPS is isosceles, and it's clear that s has an argument of

5 \pi / 12

, so it must be rotated by

7 \pi / 12

to lie on the x-axis.

So the new position of S is

[3 + \sqrt{3} + (3 + \sqrt{3})i]e^{7i \pi / 12} = (3 + \sqrt{3})(1 + i)e^{7i \pi / 12} = \sqrt{2}(3 + \sqrt{3})e^{i \pi / 4}e^{7i \pi / 12}

= \sqrt{2} (3 + \sqrt{3}) e^{5i \pi / 6} = -\dfrac{1}{2}(3 \sqrt{2} + \sqrt{6})(\sqrt{3} - i)

, as required.

Reply 23

SimonM

STEP III, Question 5

Spoiler

Reply 24

Oh I Really Don't Care

STEP III, Question 4

Spoiler

sg.jpg23.3KB

Reply 25

Glutamic Acid

II/8:

(I can't help but feel I've missed the point of this question.)

(i) f(x) = 1.

(ii) Refer to Elongar's post here: http://www.thestudentroom.co.uk/showpost.php?p=17921335&postcount=29

insane ramblings

(iii)

\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = - \text{f}(x) \Rightarrow \ln y = - \text{F}(x) + c

. When x = 0, y = 1, as F(0) = 0; c = 0.

\Rightarrow y = e^{-\text{F}(n)}

.

The ratio between y(n+1) and y(n) is

\dfrac{e^{-\text{F}(n+1)}}{e^{-\text{F}(n)}} = e^{-\text{F}(1)}

; and if F(1) =

\displaystyle \int_0^1 \text{f}(x) \, \text{d}x

then this is > 0; exp(-F(1)) < 1 and y(n) -> 0 as n -> infinity.

Reply 26

ForGreatJustice

thats what I done also :P seems too simple i know

Reply 27

SimonM

STEP III, Question 2

Spoiler

Reply 28

Elongar

For STEP II/8, part (ii),

f(x)

need not be a straight line, just consider

f(x) = \cos ( 2 \pi x ) + k, k > 1

.

However, consider,

\displaystyle \int_0^1 f(x) dx = \int_0^1 f(x+1) dx

(simply integrating both sides of the equation we're given). This gives,

F(2) = 2F(1)

Then, simply integrating between

n

and

n+1

, gives

F(n+1) - F(n) = F(n+2) - F(n+1)

.

This suggests that the area under the graph of

f(x)

between integral values of x remains the same. The result follows.

EDIT: It's simpler if you integrate between

0

and

n

. This gives,

F(n+1) - F(n) = F(1)

Reply 29

Glutamic Acid

Elongar

For STEP II/8, part (ii),

f(x)

need not be a straight line, just consider

f(x) = \cos ( 2 \pi x ) + k, k > 1

.

However, consider,

\displaystyle \int_0^1 f(x) dx = \int_0^1 f(x+1) dx

(simply integrating both sides of the equation we're given). This gives,

F(2) = 2F(1)

Then, simply integrating between

n

and

n+1

, gives

F(n+1) - F(n) = F(n+2) - F(n+1)

.

This suggests that the area under the graph of

f(x)

between integral values of x remains the same. The result follows.

Ah, very nice. I can see how (iii) relates to (ii) now.

Reply 30

qgujxj39

STEP III 2001, Q11

A uniform cylinder of radius

a

rotates freely about its axis, which is fixed and horizontal. The moment of inertia of the cylinder about its axis is

I

. A light string is wrapped around the cylinder and supports a mass

m

which hangs freely. A particle of mass M is fixed to the surface of the cylinder. The system is held at rest with the particle vertically below the axis of the cylinder, and then released. Find, in terms of

I, a, M, m, g

and

\theta

, the angular velocity of the cylinder when it has rotated through angle

\theta

Spoiler

Show that the cylinder will rotate without coming to a halt if

m/M > \sin\alpha

, where

\alpha

satisfies

\alpha = \tan\frac{1}{2}\alpha

and

0 < \alpha < \pi

Spoiler

Reply 31

SimonM

STEP III, Question 14

Spoiler

Reply 32

Mark13

DFranklin

It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

I think the fact the number of plates broken by each student obeys a Poisson distribution, and the fact that the number of plates broken by a student is independent of the number broken by the other students, shows that the breakages are completely independent.

Reply 33

DFranklin

Mark13

I think to be convincing, you would want an argument that uses the properties of a Poisson distribution in a slightly more fundamental way than that. (But heck, it's STEP I, you're probably right).

Reply 34

Aurel-Aqua

STEP II, Question 14

T = \lambda X + \frac{1}{2}(1-\lambda)Y

E[X] = 1*p = p

E[Y] = 1*2p = 2p

E[X^2] = 1*p = p

E[Y^2] = 1*2p = 2p

We may use

E[XY] = E[X]E[Y]

since X and Y are independent.

First part:

E[T] = E[\lambda X + \frac{1}{2}(1-\lambda)Y] = \frac{1}{2}(\lambda (2E[X] - E[Y])+ E[Y]) = \frac{1}{2}(\lambda(2p-2p) + 2p) = p

Second part:

E[T^2] = E[(\frac{1}{2}(\lambda(2X-Y)+Y)^2] = \frac{1}{4}E[\lambda^2(2X-Y)^2 + 2\lambda Y(2X-Y) + Y^2] =

= \frac{1}{4}(\lambda^2E[4X^2-4XY+Y^2]+2\lambda E[2XY - Y^2] + E[Y^2] =

=\frac{1}{4}(\lambda^2(4p-8p^2+2p) + 2\lambda(4p^2-2p) + 2p =

=\frac{1}{2}(\lambda^2(3p-4p^2)+2\lambda(2p^2-p)+p)

Var[T] = E[T^2]-E[T]^2 = \frac{1}{2}(\lambda^2(3p-4p^2)+2\lambda(2p^2-p)+p-2p^2)

For minimum variance,

\frac{\mathrm{d}Var[T]}{\mathrm{d}\lambda} = \frac{1}{2}(2\lambda(3p-4p^2)+2\lambda(2p^2-p)) = 0 \implies \displaystyle \lambda = \frac{1-2p}{3-4p}

, as required.

Third part:
Remember:

Var[\overline{X}] = \frac{1}{n}Var[X]

and for a normal variable,

\displaystyle P(X\leq z) = \Phi\left(\frac{z-\mu}{\sigma}\right)

\displaystyle Var[A+b]=Var[A]

.

Since n is large, we will make the use of the Normal distribution. In this case,

\displaystyle \mu = E[\overline{T} - p] = p - E[p] = p - p = 0

.

If the normal curve is symmetric (ie.

\mu = 0

), then we can say that

\displaystyle P(|X|<x) = 2P(X<x)-1

.

Thus,

\displaystyle P(\left|\overline{T}-p\right| < b) = 2\Phi\left(\frac{b-\mu}{Var[\overline{T}-b]}\right) - 1 = 2\Phi\left(\frac{b}{s^2}\right) - 1

, where

\displaystyle s^2 = Var[\overline{T}-b] = \frac{1}{n}Var[T]

.

Since we are to find the maximum value of the probability, we are looking for the minimum value of variance. Therefore,

\displaystyle Var\left[T_\lambda\right] = p(\frac{(1-2p)^2}{(3-4p)^2}(3-4p)+2\frac{1-2p}{3-4p}(2p-1)+1-2p) = \frac{p(1-2p)(1-p)}{(3-4p)}

.

Therefore

\displaystyle s^2=\frac{p(1-p)(1-2p)}{n(3-4p)}

, as required.

Reply 35

qgujxj39

STEP III 2001, Q9

Spoiler

Reply 36

Aurel-Aqua

STEP II, Question 13

P(T>t) = (1+kt)^{-\alpha} = 1-F(t)

Finding the median:

P(T>M) = \frac{1}{2} = 1-F(M) \implies F(M) = \frac{1}{2} \implies 2 = (1+kM)^\alpha \implies M = \frac{\sqrt[\alpha]{2}-1}{k}

Finding the mean: we need to find the PDF:

f(t) = F'(t) = (1-(1+kt)^{-\alpha})' = k\alpha(1+kt)^{-\alpha-1}

.
Then,

\displaystyle m = E[T] = \int_0^\infty \alpha kt(1+kt)^{-\alpha-1}\, \mathrm{d}t = \left[ -\frac{(kx+1)^{-\alpha}(\alpha kx + 1)}{k(\alpha-1)} \right]_0^\infty = \frac{1}{k(\alpha-1)}

Assuming that we can take the 5th bulb to have the median time (= 1000 hours), and the 9 bulbs to have the mean time (= 2400 hours), we form two equations:

(1+1000k)^\alpha = 2

and

2400 = \frac{1}{k(\alpha-1)}

. Substituting for k, we get

2.4(\sqrt[\alpha]{2}-1) = \frac{1}{\alpha-1}

. Making

\alpha = 2

, we see that LHS is almost equal to 1, so

\alpha = 2

is a good approximation. Using the second equation, we obtain that

k \approx \frac{1}{2400}

.

Last part:

We have to find, given

M < m

\displaystyle P(T>m|T>M) = \frac{P(T>m)}{P(T>M}) = \frac{(1+km)^{-\alpha}}{(1+kM)^{-\alpha}} =

\displaystyle = \left(\frac{1+kM}{1+km}\right)^\alpha = \left(\frac{1+kM}{1+\frac{1}{\alpha-1}}\right)^\alpha \approx \frac{1}{4}\left(1+\frac{1}{2400}M\right)^2

Reply 37

Dadeyemi

STEP III, Question 6

Spoiler

PS: I enjoyed this question quite a lot, it gives a very interesting way of looking at areal coordinates.

Reply 38

Glutamic Acid

I/8:

y = x, y' = 1, y'' = 0
y = 1 - x^2, y' = -2x, y'' = -2

So p(x) + xq(x) = 0 therefore p(x) = -xq(x) and
-2 - 2xp(x) + q(x)(1 - x^2) = 0

\Rightarrow -2 + 2x^2 \text{q}(x) + \text{q}(x)(1 - x^2) = 0 \Rightarrow \text{q}(x)(2x^2 + 1 - x^2) = 2 \Rightarrow \text{q}(x) = 2(1 + x^2)^{-1}

And

\text{p}(x) = -2x(1 + x^2)^{-1}

.

Letting y = ax + b(1 - x^2) gives y' = a - 2xb and y'' = -2b, so it relies on

-2b - \dfrac{2x}{1 + x^2}(a - 2xb) + \dfrac{2}{1+x^2}(ax + b(1 - x^2)) = 0

; which if you expand out, it does.

If y =

\cos^2 \frac{x}{2}

and y =

\sin^2 \frac{x}{2}

satisfy the equation then y =

a \cos^2 \frac{x}{2} + b \sin^2 \frac{x}{2}

will satisfy it for any a and b. Letting a = 1, b = 1 gives y = 1, y' = 0, y'' = 0 so q(x) = 0.
Letting a = 1, b = -1 gives y = cos x, y' = -sin x, y'' = -cos x so -cos(x) - p(x)sin x = 0 so p(x) = -cot x.

Reply 39

Aurel-Aqua

2001 III Question 13

A: TTH
B: THH
C: HHT
D: HTT

The first two have 4 ways of happening, so probability of a win is

=\frac{1}{4}\left(p_{TT}+p_{TH}+p_{HH}+p_{HT}\right)

.

Part (i): A against B, p is for A's win:
If we start with TT, we cannot end up with anything else than TTH, ad infinitum, so

p_{TT} = 1

. (1)
If we start with TH, then half of that is a loss, and half of that is

p_{HH}

, so

p_{TH} = \frac{1}{2}p_{HH} \implies 2p_{TH} = p_{HH}

. (2)
If we start with HH, then half of that is HT, and the other half is HH, so:

p_{HH} = \frac{1}{2}p_{HT}+\frac{1}{2}p_{HH} \implies p_{HH} = p_{HT}

. (3)
If we start with HT, then half of that is TH, and another half is TT, so:

p_{HT} = \frac{1}{2}p_{TH} + \frac{1}{2}

. (4)

(2) and (3) imply

2p_{TH} = p_{HT}

, and thus with (4), it gives

p_{HT} = \frac{1}{4}p_{HT} + \frac{1}{2} \implies p_{HT} = \frac{2}{3}

. By (3),

p_{HH} = \frac{2}{3}

. By (2),

p_{TH} = \frac{1}{3}

.

So probability of A winning against B is

\frac{1}{4}\left(1+ \frac{2}{3} + \frac{2}{3} + \frac{1}{3}\right) = \frac{1}{4}\cdot\frac{8}{3} = \frac{2}{3}

.

Part (ii): B against C, p is for B's win:
By the same procedure, we obtain that

p_{HH} = 0

and

p_{TH} = 1

p_{TT} = \frac{1}{2}(p_{TT}+p_{TH}) \implies p_{TT} = p_{TH} \implies p_{TT} = 1

p_{HT} = \frac{1}{2}(p_{TT}+p_{TH}) = 1

. Probability of B winning against C is

\frac{3}{4}

.

Part (iii): C against D, p is for C's win:
Again, we find that

p_{HH} = 1

because it'll always end with HHT.

p_{TT} = \frac{1}{2}p_{TT} + \frac{1}{2}p_{TH} \implies p_{TT} = p_{TH}

p_{HT} = \frac{1}{2}p_{TH}

p_{TH} = \frac{1}{2}p_{HT} + \frac{1}{2}p_{HH} = \frac{1}{4}p_{TH} + \frac{1}{2}p_{HH} \implies \frac{3}{4}p_{TH} = \frac{1}{2} \implies p_{TH} = \frac{2}{3} \implies p_{TT} = \frac{2}{3} \implies p_{HT} = \frac{1}{3}

. Therefore the probability of C winning against D is equal to

\frac{1}{4}\left(\frac{2}{3}+\frac{2}{3}+\frac{1}{3}+\frac{3}{3}\right) = \frac{2}{3}

.

Part (iii): D against A, p is for D's win:
Starting with TT:

p_{TT} = \frac{1}{2}p_{TT} \implies p_{TT} = 0

.
HT:

p_{HT} = \frac{1}{2} + \frac{1}{2}p_{TH}

.
HH:

p_{HH} = \frac{1}{2}p_{HT} + \frac{1}{2}p_{HH} \implies p_{HH} = p_{HT}

.
TH:

p_{TH} = \frac{1}{2}p_{HT} + \frac{1}{2}p_{HH} = p_{HT}

.

Thus

p_{HT} = \frac{1}{2}+\frac{1}{2}p_{HT} \implies p_{HT} = 1 \implies p_{TH} = 1 \implies p_{HH} = 1

.

Therefore the probability of D winning against A is

\frac{3}{4}

.

Please check the last part, I might have mixed something up. But this is generally one of the easiest STEP III Statistics questions I've seen so far.