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STEP Maths I,II,III 1987 Solutions

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Original post by abra-cad-abra
what is on your x axis and what on your y axis


It's a long time since I looked at this question and I do agree that there is something wrong with my solution. I will try to have another look at it.
Original post by abra-cad-abra
What do you mean by the distance measured on the x axis?

as the question says a point is chosen randomly. meaning their x coorcdinate is chosen randomly on [-1,1] so there is an equal probability for all values on this domain which is the whole domain considered int he question.


so x has uniform distribution 1/2 on [-1,1] . im not saying the distance has uniform distribution, rather the x-coord of the point has uniform distribution. which is surely correct no, thats just what you have done with theta

from this g(x)=sqrt(2-2x). and we want E[g(x)] and Var[g(x)]


The point is chosen randomly on the circle which is not the same as saying that the x co-ordinate is chosen randomly. It is theta that has a uniform distribution not the x co-ordinate.
STEP 1987 Paper I question 13
My earlier solution was incorrect. I think this one is better.
For x≤0, md2vdt2=−mμ2x⇒d2xdt2+μ2x=0⇒x=Acos⁡μt+Bsin⁡μtx \leq 0,\ m\frac{d^2v}{dt^2}=-m \mu^2 x \Rightarrow \frac{d^2x}{dt^2}+\mu^2 x=0 \Rightarrow x=A\cos \mu t +B\sin \mu t
Unparseable latex formula:

t=0 \Rightarrwo x=0 \mathrm{\ and\ }\frac{dx}{dt}=v_0\Rightarrow A=0,\ B=\frac{v_0}{\mu} \mathrm{\ so\ }x=\frac{v_0}{\mu} \sin \mu t


t=πμ⇒x=0,dxdt=−v0 t=\frac{\pi}{\mu} \Rightarrow x=0, \frac{dx}{dt}=-v_0
x now becomes negative so d2xdt2=−κdxdt⇒x=C+De−κt x \mathrm{\ now\ becomes\ negative\ so\ } \frac{d^2x}{dt^2}=-\kappa\frac{dx}{dt}\Rightarrow x=C+De^{-\kappa t}
t=πμ,dxdt=−v0⇒C+D=0,κD=v0 so x=−v0κ(1−e−κt)t=\frac{\pi}{\mu}, \frac{dx}{dt}=-v_0 \Rightarrow C+D=0, \kappa D=v_0 \mathrm{\ so\ }x=-\frac{v_0}{\kappa}(1-e^{-\kappa t})
The sketch will consist of a half period of a sine curve then exponentially tending to x=−v0κ x = -\frac {v_0} {\kappa}
1987.I.13.pdf3.6KB
Original post by abra-cad-abra
what is on your x axis and what on your y axis


I have posted a revised version of the solution on the 1987 solutions thread. Does it look o.k. now?
Original post by brianeverit
I have posted a revised version of the solution on the 1987 solutions thread. Does it look o.k. now?


yeah good
Original post by brianeverit
The point is chosen randomly on the circle which is not the same as saying that the x co-ordinate is chosen randomly. It is theta that has a uniform distribution not the x co-ordinate.


im still not quite convinced here though. say a point is chosen randomly, it is not a xcoord and y coord chosen separately as they wouldnt lie on the curve. Either the y coord is chosen randomly which determines 2 possible x coords or the x coord is chosen randomly which determines 2 possible y coords. And i explain only 1 coord needs to be considered.

I dont see why either approach are different really. btw in yours you had the pdf as 2/pi but went on to use 1/pi in the integrals. is that just a mistake?
Original post by abra-cad-abra
yeah good

im still not quite convinced here though. say a point is chosen randomly, it is not a xcoord and y coord chosen separately as they wouldnt lie on the curve. Either the y coord is chosen randomly which determines 2 possible x coords or the x coord is chosen randomly which determines 2 possible y coords. And i explain only 1 coord needs to be considered.

I dont see why either approach are different really. btw in yours you had the pdf as 2/pi but went on to use 1/pi in the integrals. is that just a mistake?


Consider the arc AP subtending an angle of 60 degrees at the centre of the circle and let N be the foot of the perpendicular to the x axis. The probability that our randomly chosen point lies on AP is clearly 1/6 but the probability that x lies in AN is 1/4
And I think the pdf should be 1/pi since theta is uniform over -pi/2 to +pi/2
for q16 how did you work out the variance for the normal approximation. from binomial it should be var=npq. what did you do? and also you require a continuity correction, no?
Original post by brianeverit
Consider the arc AP subtending an angle of 60 degrees at the centre of the circle and let N be the foot of the perpendicular to the x axis. The probability that our randomly chosen point lies on AP is clearly 1/6 but the probability that x lies in AN is 1/4
And I think the pdf should be 1/pi since theta is uniform over -pi/2 to +pi/2


see above post please. q16 step i
Original post by abra-cad-abra
for q16 how did you work out the variance for the normal approximation. from binomial it should be var=npq. what did you do? and also you require a continuity correction, no?


Yews n=40,p=q=1/4 so Var=npq=-10 If you mean for X then since alpha is constant, Var 60(1-alpha)+beta=Var beta.
I'll have another think about the conitinuity correction
Original post by brianeverit
Yews n=40,p=q=1/4 so Var=npq=-10 If you mean for X then since alpha is constant, Var 60(1-alpha)+beta=Var beta.
I'll have another think about the conitinuity correction


oh i see thank you
Original post by brianeverit
Yews n=40,p=q=1/4 so Var=npq=-10 If you mean for X then since alpha is constant, Var 60(1-alpha)+beta=Var beta.
I'll have another think about the conitinuity correction


wait i dont understand. if p is 1/4 then isnt mean=np=10 ??
Original post by abra-cad-abra
wait i dont understand. if p is 1/4 then isnt mean=np=10 ??


That was just my silly mistake. N=40 p=q=1/2 !
Original post by Mazzacre
Step II, Q1

(i)(x−a)2+(2x−b)2+(3x−c)2=x2−2ax+a2+4x2−4bx+b2+9x2−6c+c2 (\mathrm{i}) (x-\mathrm{a})^2 + (2x - \mathrm{b})^2 + (3x - \mathrm{c})^2 = x^2 - 2\mathrm{a}x + \mathrm{a}^2 + 4x^2 - 4\mathrm{b}x + \mathrm{b}^2 + 9x^2 - 6\mathrm{c} + \mathrm{c}^2

=14x2−x(2a+4b+6c)+a2+b2+c2 = 14x^2 - x(2\mathrm{a} + 4\mathrm{b} + 6\mathrm{c}) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

=14x2−x(14x)+a2+b2+c2  since  7x=a+2b+3c = 14x^2 - x(14x) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 \,\, \mathrm{since} \,\,7x = \mathrm{a} + 2\mathrm{b} + 3\mathrm{c}

=a2+b2+c2 = \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2



(ii)(2x−a)2+(3x−b)2+(3x−c)2=4x2−4ax+a2+9x2−6bx+b2+9x2−6c+c2 (\mathrm{ii}) (2x-\mathrm{a})^2 + (3x - \mathrm{b})^2 + (3x - \mathrm{c})^2 = 4x^2 - 4\mathrm{a}x + \mathrm{a}^2 + 9x^2 - 6\mathrm{b}x + \mathrm{b}^2 + 9x^2 - 6\mathrm{c} + \mathrm{c}^2

=22x2−x(4a+6b+6c)+a2+b2+c2 = 22x^2 - x(4\mathrm{a} + 6\mathrm{b} + 6\mathrm{c}) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

=22x2−x(22x)+a2+b2+c2  since  11x=2a+3b+3c = 22x^2 - x(22x) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 \,\, \mathrm{since} \,\,11x = 2\mathrm{a} + 3\mathrm{b} + 3\mathrm{c}

=a2+b2+c2 = \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

if  αa+βb+γc=α2+β2+γ22x \mathrm{if}\,\, \alpha\mathrm{a} + \beta\mathrm{b} + \gamma\mathrm{c} = \frac{\alpha^2 + \beta^2 + \gamma^2}{2}x

then a2+b2+c2=(αx−a)2+(βx−b)2+(γx−c)2 \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 = (\alpha x - \mathrm{a})^2 + (\beta x - \mathrm{b})^2 + (\gamma x - \mathrm{c})^2

Apologies if there are errors, rather rushed it.


I see how the first part is done, but not the last part of the question.
It asks:
'Give a general result of which (i) and (ii) are special cases,'
I am so dim witted that i cant see the obvious point if there is any in this question, and was wondering exactly what the examiners are asking, and how you've done it.
Thanks in advance.
(edited 10 years ago)
Original post by SimonM
STEP III, Question 8

Spoiler




this is not step iii q8?!?!!?
Reply 294
Avatar for davros
davros
16
Original post by Yung_ramanujan
this is not step iii q8?!?!!?


Looks like Q10 to me :smile:
Original post by davros
Looks like Q10 to me :smile:


yeah i realised when i got to 10!


is there a solution to 8 here?
Reply 296
Avatar for davros
davros
16
Original post by Yung_ramanujan
yeah i realised when i got to 10!


is there a solution to 8 here?


Not sure without going through the whole thing! :frown:

Pretty sure I've found a few broken links in the STEP solutions thread before now :smile:
Original post by Yung_ramanujan
yeah i realised when i got to 10!


is there a solution to 8 here?


Here is my solution for III.8
1987.III.8.pdf34.4KB
how can you write npi should it not be 2npi to leave the angle unchanged
Original post by brianeverit
STYEP 1987 Fma
Questions 2,3 and 4


For question 2,

A+-B+-C = kpi
therefore, sin^2(A)=sin^2(B+C) or sin^2(A)=sin^2(B-C)

why is this true?

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