STEP Maths I, II, III 1994 Solutions

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Reply 100

STEP I question 5

for the coordinates of the midpoint i get ( (p+q)/2 , (p^2 + q^2)/2 )

using this and pq = -1 gives the locus as 4x^2 + 1 = 2y

for part ii) i got 2 equations in p and q

y = 2px - p^2
y = 2qx - q^2

solving for x gives x = (p + q)/2

subbing back in to either equation results in y=pq=-1

Reply 101

speedy_s

STEP III q4

not convinced this is 100% correct, but here you go anyway!

dy/dx = 2sqrt(y) (+ or -)

y^(-1/2)dy= 2dx (for +)

2y^(1/2) = 2x + c

y = (x+k)^2

where k = b - a

(for - )
y = (m-x)^2

where m = a + b

is that correct?

for the graphs to go through the origin k = m = 0, hence both solutions indicate y = x^2 and so they do coincide

hence a(1) = a(2) = 1

other solutions all have repeated roots across the x axis which represent the family of curves for the DE, but on all positive side of y, since y can not be -ve since 4y = DE^2

the tangent is y=0, which satisfies the DE

Reply 102

Swayum

brooker

Agreed on pq = -1 but the midpoint surely isn't ((p+q)/2, (p^2 + q^2)/2)

The midpoint of (p, p^2) and (q, q^2) is ((p-q)/2, (p^2 - q^2)/2)

I get y = x(p+q) but I don't know what to do from there.

Reply 103

DFranklin

The midpoint of (x0,y0) and (x1, y1) is ((x0+x1)/2,(y0+y1)/2). (Think about what has to happen when x0=x1, or p=q in the actual question).

Reply 104

Swayum

Wow I am actually so stupid... why was I SUBTRACTING the coordinates? And it's not even the first time I've done this. We learnt how to find midpoints in like year 8...

God help me.

Reply 105

Glutamic Acid

zrancis

If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:

$v^2\le0$

hence:

$\frac{2g(M-msin\theta)dsin\theta}{M+m}\le0$

$sin\theta(M-msin\theta)\le0$

So:

Provided $sin\theta\ge0$

$M-msin\theta\le0$

QED (I hope)

I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great.

Hmm, I disagree. v^2 can't be < 0, and your final condition contradicts the condition found in part (i) which would suggest the ball and wagon won't move at all.

An attempt:

When B hits the ground, the string goes slack; deceleration = $g \sin \theta$ . It'll have travelled a distance of $d \sin \theta$ so if the wagon is not to collide with the pulley, $s < d - d \sin \theta$ .

Using "v^2 = u^2 + 2as" with v = 0, s = $d - d \sin \theta$ and a = $-g \sin \theta$ : $\dfrac{u^2}{2g \sin \theta} < d - d \sin \theta$ .

Substituting in u^2 from (i):
$\dfrac{2g(M - m \sin \theta) d \sin \theta}{2g \sin \theta} < d - d \sin \theta \Rightarrow M - m \sin \theta < 1 - \sin \theta$
$\Rightarrow M - m \sin \theta + \sin \theta < 1$

Reply 106

Glutamic Acid

STEP I/Q14:

I usually avoid straying into the unknown of STEP statistics questions, so there's a more than reasonable chance of a screw-up.

(i) $\displaystyle \mu = \int_0^{\infty} \lambda t e^{- \lambda t} \, \text{d}t = \left[-te^{-\lambda t} \right]^{\infty}_0 + \int_0^{\infty} e^{- \lambda t} = 0 - \frac{1}{\lambda} \left[ e^{-\lambda t}\right]^{\infty}_0 = \frac{1}{\lambda}$

$\displaystyle \sigma^2 = \int_0^{\infty} \lambda t^2 e^{-\lambda t} \, \text{d}t - \mu^2 = \left[-t^2e^{-\lambda t} \right]^{\infty}_0 + \frac{2}{\lambda} \int_0^{\infty} te^{- \lambda t} \, \text{d}t - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$

(ii) $P(U \le u)$ means $T_i \le u$ for all i = $P(T_i \le u)^n$ .

$\displaystyle = \left( \int_0^u \lambda e^{- \lambda t} \, \text{d}t \right)^n = \left( \left[ -e^{- \lambda t} \right]^u_0 \right)^n = \left(1 - e^{-\lambda u} \right)^n$

$P(U \le u)$ is the cumulative distribution function, PDF = $\dfrac{\text{d}}{\text{d}u} (1 - e^{-\lambda u})^n = n \lambda e^{- \lambda u}(1 - e^{-\lambda u})^{n-1}$

(iii) $P(T > t)$ means $T_i > t$ for all i = $P(T_i > t)^n$

$\displaystyle = \left( \int_t^{\infty} \lambda e^{- \lambda t} \, \text{d}t \right)^n = \left( \left[ -e^{-\lambda t}\right]^{\infty}_t \right)^n = e^{- \lambda t n}$

Cumulative distribution function = P(T <= t) = 1 - P(T > t) = $1 - e^{- \lambda tn}$ . PDF = derivative with respect to t = $\lambda n e^{- \lambda t n}$

(iv) The same as (i) with $\lambda n$ in place of $\lambda$ so $\mu = \dfrac{1}{\lambda n}$ and $\sigma^2 = \dfrac{1}{\lambda^2n^2}$ .

Reply 107

Glutamic Acid

III/13:

Unconfident about the first part. Let T be the total time, t₀ be the time the rope has been in use and t be the time the artist can continue to use it.

$\displaystyle P(T > t_0 + t) = \int_{t_0 + t}^{\infty} \lambda e^{- \lambda t} \, \text{d}t = \left[-e^{-\lambda t} \right]^{\infty}_{t_0 + t} = e^{-\lambda(t_0 + t)}$

$\displaystyle P(T > t_0) = \int_{t_0}^{\infty} \lambda e^{- \lambda t} \, \text{d}t = \left[-e^{-\lambda t} \right]^{\infty}_{t_0} = e^{-\lambda t_0}$

Consider $P(T > t_0 + t | T > t_0) = \dfrac{P(T > t_0 + t)}{P(T > t_0)} = \dfrac{e^{-\lambda (t_0 + t)}}{e^{-\lambda t_0}} = e^{-\lambda t}$

Define a variable $X = T - t_0 \Rightarrow P(X > t) = e^{-\lambda t} \Rightarrow P(X \le t) = 1 - e^{-\lambda t} \Rightarrow \text{PDF} = \lambda e^{-\lambda t}$ (probability density function is the derivative of the cumulative distribution function.)

This is the same as the original distribution, so regardless of how long the rope has been in use the probabilities for how long it can continue to be used are the same?

Both ropes fail: $\displaystyle P(T \le t) = \left( \int_0^t \lambda e^{-\lambda t} \, \text{d}t \right)^2$ ; this is the cumulative distribution function.

$\displaystyle \text{f}(t) = \frac{\text{d}}{\text{d}t} \left( \int_0^t \lambda e^{-\lambda t} \, \text{d}t \right)^2 = 2 \lambda e^{- \lambda t} \int_0^t \lambda e^{-\lambda t} \, \text{d}t = 2 \lambda e^{- \lambda t}(1 - e^{-\lambda t})$ , with zero probabilities otherwise.

Start of performance at (n-1)h, end at nh. Both fail =
$\displaystyle \left( \int_{(n-1)h}^{nh} \lambda e^{-\lambda t} \, \text{d}t \right)^2 = \left( \left[-e^{-\lambda t} \left]^{nh}_{(n-1)h} \right)^2 = (-e^{-nh \lambda} + e^{-\lambda (n-1)h})^2 = e^{-2n \lambda h} (e^{\lambda h} - 1)^2$ .

Probability of them bothfailing during the same performance = P(both fail on 1st) + P(both fail on 2nd) + ...

$= \displaystyle \sum_{n=1}^{\infty} (e^{\lambda h} - 1)^2 e^{-2 n \lambda h} = (e^{\lambda h} - 1)^2 \frac{e^{-2 \lambda h}}{1 - e^{-2 \lambda h}} = \frac{1 - 2e^{-\lambda h} + e^{- 2 \lambda h}}{1 - e^{-2 \lambda h}}$

$\displaystyle = \frac{(1 - e^{- \lambda h})^2}{1 - e^{-2 \lambda h}} = \frac{1 - e^{-\lambda h}}{1 + e^{-\lambda h}} = \frac{e^{\lambda h/2} - e^{-\lambda h/2}}{e^{\lambda h/2} + e^{-\lambda h/2}} = \tanh \left(\frac{\lambda h}{2}\right)$ .

Reply 108

Oh I Really Don't Care

Only managed to obtain the first paper for 1994 but completed it - so if any working is required because it has not been posted - then will post.

However there are some II and III questions that certainly have no solutions - so if anyone would PM them - will post them [if can do them that is :wink:

Reply 109

Swayum

DeanK22

http://www.thestudentroom.co.uk/showpost.php?p=16034501&postcount=9

Reply 110

SimonM

STEP I, Question 2

Spoiler

Reply 111

SimonM

I think we have a contender for easiest/fastest STEP question

Reply 112

Oh I Really Don't Care

Swayum

http://www.thestudentroom.co.uk/showpost.php?p=16034501&postcount=9

hmm. it might be this pc but those links are not opening.

Reply 113

Swayum

DeanK22

hmm. it might be this pc but those links are not opening.

Yeah, my host was letting me upload my files for free because I knew him quite well off another website. Unfortunately, he can no longer afford it, so he took down the server.

Reply 114

DFranklin

SimonM

I think we have a contender for easiest/fastest STEP question

Q3, STEP I, 1997 is a challenger...

Reply 115

SimonM

STEP III, Question 12

Spoiler

Reply 116

SimonM

STEP III, Question 10

I suck at these relative motion questions :frown:

Spoiler

Reply 117

maltodextrin

Rabite

STEP II:
Here's a half-baked Q2 (the stupid one with cos^{2n+1} and what not) - I would probably bet money that it's wrong though.
[edit] Oops, first PDF was crapped up since it went over a page. I should really learn how to use Latex tags so I dont' have to use this crappy program.

[edit again] Also, I think Q5 is straightforward (attached), but I may have overlooked something...

In question 5, just to be certain could anyone confirm that there are infinitely many solutions when c = 2? (Rabite left a question mark by it in his solution)

Reply 118

brianeverit

maltodextrin

In question 5, just to be certain could anyone confirm that there are infinitely many solutions when c = 2? (Rabite left a question mark by it in his solution)

Yes. Any value of x between 1 and 3 is a solution.

Reply 119

brianeverit

1994 Paper 3 number 11

1994PAPERIII.no.11.pdf46.3KB

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