[DFranklin]

(Original post by generalebriety)
\therefore (*):\; y = Ax + B + x \sinh^{-1} x - (1+x^2)^{1/2}.[/latex]

I'm not sure this is right. Are you sure you can just add your 2 solutions?

[DFranklin]

(Original post by nota bene)
Okay, STEP I Question 2

(David, hope I've made all cases clear etc. this looks like one of those tedious ones again...)

Actually, I think you were supposed to do this by logic.

From this we can see that there are 3 possible ways of getting a sum that has eight ways of being combined. However, for this type of array we are trying to create 'e' will appear in four sums (d+e+f and b+e+h and a+e+k and c+e+g). For the sum 14 the max number of times a digit appears is three (for 1, 2, 3, 4, 5 and 7) and for 15 the max number of times a digit appears is four (for 5), lastly for 16 the max number of times a digit appears is three (for 9, 8, 7, 6, 5 and 3). This means to create an array of this type the combinations that will be present are the ones presented in the emboldened line in the list of possibilities.

Therefore follows the answers to the questions
i) Shown 15 above.
ii) Shown 5 above.
iii) Implicitly shown as we only have two sums with 9 involved when creating the sum 15. This means that the 9 must be located at a position where it only appears in two summations (therefore b, d, f or h). (On a sidenote e appears in four sums as earlier mentioned and a, c, g and k appear in three summations)
iv) Setting b=9 (knowing e=5) gives h=1 and a+c= 4 and 2 (from our table of possibilities) we get two possible arrays (one with first line 492 and one with 294).

This is almost all fine; I wouldn't bother on here, but when writing it down you should actually list the 2 magic squares for (iv) (Nothing serious, but it might cost you a mark).

v) Calculating these possibilities I think is just to multiply 4 by 2 by 4 (because there are four possible positions of the 9, b, d, f and h) to get 32 because once you fix one number all the other will fall out with two possibilities, and there are four numbers to fix (since the 5 is automatically in the middle and fixing a 9 or a 1 in opposite positions will produce the same array i.e. fixing 9=b implies h=1 and fixing h=1 implies b=9).

I don't see where the 2nd 4 comes from. (I think the correct answer is 8 magic squares).

Incidentally, I think you were "supposed" to answer the question as follows:

(i) 45 = (a+b+c)+(d+e+f)+(g+h+i) = 3n => n = 15.
(ii) So 45 +3e = (a+b+c+d+e+f+g+h+i) + 3e = (a+e+i)+(c+e+g)+(b+e+h)+(d+e+f) = 4n = 60. So 3e=15, so e = 5.
(iii) If not, then one of a,c,g,i = 9. Wlog, suppose a = 9. Then i = 1 and b+c = 6. b isn't 1 and so c is at most 4. Then as c+f+i = 15 we have f > 9, which is impossible. So one of b,d,f,h = 9.
(iv) b = 9, so a+c = 6. and h = 1. As neither a,c can be 1, one must be 2 and the other must be 4. Suppose a = 2. Then i = 8, so g = 6, so d=7 and f=3. So there's only 1 solution with a=2. By symmetry there's only one solution with c=2. So total of 2 solutions.
(v) There are 4 choices for where the 9 goes (b,d,f,h), and then 2 choices for where the 2 goes (either side of the 9). These choices fix the remaining numbers. So total number of solutions = 8.

[khaixiang]

STEP III 1992 Question 3

Sketch of C1, whose parametric equations are



C2 is a circle which passes through the origin, so C2 is of the form
Substitute into C2,

t=0 corresponds to the point of intersection at the origin and the other possible intersections, (i.e. R and S) of C1 and C2 satisfies as required.

Since (t-s) and (t-r) are factors of , is also a factor. Using long division, the required quadratic equation is . Note that we do not need to finish off the long division, that is to show that the remainder is zero since the constant "a" and "b" is used (It might not even be possible). Discriminant of this quadratic equation (whose roots are paremeters of points of further intersections) is which is always less than zero. So the only real intersections of C1 and C2 are O, R and S.

I suppose this question is on the easy side, hope I've done it correctly.

[*bobo*]

STEP III
11) particle: constant v=V and time before collision=t distance travellled= s(1)
s(1)=Vt
barrier: u=4V a=-4V/T, time =t, distance travelled=s(2)
s(2)=4Vt- 2Vt^2/T

s(1)+s(2)= 75VT/32 as this is their initial distance apart.

=>2Vt^2- 5Vt + 7(5VT^2)/32=0
=> t= (5VT - (25V^2T^2 - (75(VT)^2/4))^0.5)/4V
= 5T/4- 5T/8= 5T/8

0.5= (v-(3V/2))/((3V/2)+V) where v is the speed of particle after collision
v= 11V/4
distance between particle and wall at first collision with barrier=5VT/8
so time taken for particle to reach wall after P collides with barrier= (5VT/8)/(11V/4)= 5T/22
after collsion with wall,speed of P= 11V/8 (e=0.5 and wall is stationary)
total time taken for P to collide with wall= 5T/22 +5T/8= 75T/88

s is the distance between the wall and where P and barrier collide for second time, when P is travelling at a speed of 11V/8, and t is the time between the collision of P and the wall and P and the barrier for the second time:

s= 11Vt/8

((75VT/32) –s)=distance travelled by wall before 2nd collision

((75VT/32) –s)= 4V( t + (75T/88)) – (2V/T)(t +(75T/88))^2

Sort into a quadratic of t, solve for t which equals 3T/11, then total time= 3T/11 + 75T/88= 99T/88= 9T/8

[ad absurdum]

STEP II question 7 attached.

When I started this question, I thought there was no chance it would have fitted on one page

[ad absurdum]

STEP I question 5 attached

In hindsight, I really should have multplied out the original expression when getting the polynomial in standard polynomial form. No harm done though. Also, for the last part, I wasn't sure if they wanted you to actually do it - I would have done it in the exam but the thought of another page of working didn't appeal to me too much right now. (And yes, I don't know my alphabet, incase you were wondering about me naming my polynomials).

[DFranklin]

(Original post by ad absurdum)
STEP I question 5 attached

In hindsight, I really should have multplied out the original expression when getting the polynomial in standard polynomial form. No harm done though. Also, for the last part, I wasn't sure if they wanted you to actually do it - I would have done it in the exam but the thought of another page of working didn't appeal to me too much right now.

You might have been able to skip some working, but I don't think what you did was enough.

In fact, it's not true to say "no harm done", because although you get the same answer, by doing it your way it is not at all obvious why (1-x)(1-x^2)(1-x^4) works out to have every coefficient of x^k be +/-1. It does, but it's not obvious that it's not a piece of luck that won't generalise.
Whereas if do it like:

(1-x)(1-x^2) = (1-x)-x^2(1-x) = 1-x-x^2+x^3
(1-x-x^2+x^3)(1-x^4) = (1-x-x^2+x^3) - x^4(1-x-x^2+x^3) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7

it's a lot more obvious. (And it's also a lot less work, which is why I suspect you were actually supposed to actually do the calculations for the n=15 case - it's not that infeasible I think. You can just verify that the same approach works rather than deriving it like you did in the first case).

Meh: stop sitting on the fence David and see how long it takes...

We have

Deduce that our partition will be: {1,2,4,7,8,11,13,14} = {3,5,6,9,10,12,15}.

5 mins to do it, if that (mainly writing) and 5 mins to verify this works with a calculator (allowed back then) (or prove it, by same arguments).

[*bobo*]

STEP II
13)i) T-mg(mu)=ma
mg-T=ma
a= g(1- mu)/2
P(1): a=0.5g(1- mu), u=0, s=b
v (when P(2) reaches floor)= (gb(1-mu))^0.5

P(1): a=-g(mu), u=(gb(1-mu))^0.5, s=(d-b), v^2>0 for P(1) to reach edge table
gb(1-mu)-2g(mu)(d-b)>0
(mu)<(b/(2d-b))

ii) P(1): a=-g(mu) ,u=(gb(1-mu))^0.5, v=0 time=t(1)
using a= (v-u)/t
t(1)= (b(1- mu)/g)^0.5/(mu)= (1/mu).(b(1-mu)/g)^0.5

P(2): velocity after bouncing off ground> (gb(1-mu)/2(mu^2))^0.5, v=0, a=-g time=t(2)
using a= (v-u)/t
u= v- at
gt(2)>(gb(1-mu)/2(mu^2))^0.5
t(2)> (b(1-mu)/(2g(mu^2)))^0.5

time taken (=2t(2))to bounce for 2nd time> ((2^0.5)/mu)(b(1-mu)/g)^0.5

so 2t(2)> (2^0.5)t(1)
so P(1) comes to rest before P(2) bounces for the 2nd time

[ad absurdum]

(Original post by DFranklin)
You might have been able to skip some working, but I don't think what you did was enough.

In fact, it's not true to say "no harm done", because although you get the same answer, by doing it your way it is not at all obvious why (1-x)(1-x^2)(1-x^4) works out to have every coefficient of x^k be +/-1. It does, but it's not obvious that it's not a piece of luck that won't generalise.
Whereas if do it like:

(1-x)(1-x^2) = (1-x)-x^2(1-x) = 1-x-x^2+x^3
(1-x-x^2+x^3)(1-x^4) = (1-x-x^2+x^3) - x^4(1-x-x^2+x^3) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7

it's a lot more obvious. (And it's also a lot less work, which is why I suspect you were actually supposed to actually do the calculations for the n=15 case - it's not that infeasible I think. You can just verify that the same approach works rather than deriving it like you did in the first case).

Meh: stop sitting on the fence David and see how long it takes...

We have

Deduce that our partition will be: {1,2,4,7,8,11,13,14} = {3,5,6,9,10,12,15}.

5 mins to do it, if that (mainly writing) and 5 mins to verify this works with a calculator (allowed back then) (or prove it, by same arguments).

Yeah, I didn't realise I'd done it a stupid way until I went to scan it in, so I was thinking it was more work than it actually was, I agree that it's actually not much. I'm going out just now but I'll do the other half properly tomorrow (assuming I'm feeling up to it that is ).

[nota bene]

(Original post by DFranklin)
Actually, I think you were supposed to do this by logic.

Yes, I realised quite quickly this was not the method sought for, although it only took me 35 minutes so wasn't that bad.

(Original post by DFranklin)
This is almost all fine; I wouldn't bother on here, but when writing it down you should actually list the 2 magic squares for (iv) (Nothing serious, but it might cost you a mark).

Hehe, yes, had them on paper but was tired of latexing matrices after that other question...

(Original post by Dfranklin)
I don't see where the 2nd 4 comes from. (I think the correct answer is 8 magic squares).

Yes, don't know what I was thinking it is quite obvious that once you fix one number in all its possible positions and multiply by two all other possibilities will have been covered. Edited that in the post.

(Original post by DFranklin)
Incidentally, I think you were "supposed" to answer the question as follows:

(i) 45 = (a+b+c)+(d+e+f)+(g+h+i) = 3n => n = 15.
(ii) So 45 +3e = (a+b+c+d+e+f+g+h+i) + 3e = (a+e+i)+(c+e+g)+(b+e+h)+(d+e+f) = 4n = 60. So 3e=15, so e = 5.
(iii) If not, then one of a,c,g,i = 9. Wlog, suppose a = 9. Then i = 1 and b+c = 6. b isn't 1 and so c is at most 4. Then as c+f+i = 15 we have f > 9, which is impossible. So one of b,d,f,h = 9.
(iv) b = 9, so a+c = 6. and h = 1. As neither a,c can be 1, one must be 2 and the other must be 4. Suppose a = 2. Then i = 8, so g = 6, so d=7 and f=3. So there's only 1 solution with a=2. By symmetry there's only one solution with c=2. So total of 2 solutions.
(v) There are 4 choices for where the 9 goes (b,d,f,h), and then 2 choices for where the 2 goes (either side of the 9). These choices fix the remaining numbers. So total number of solutions = 8.

This certainly is neater.
Just out of interest, do you think my 'method' would gain full marks (well provided I write 8 and not 32!) or is it considered 'bad' ?

[DFranklin]

(Original post by nota bene)
Just out of interest, do you think my 'method' would gain full marks (well provided I write 8 and not 32!) or is it considered 'bad' ?

I don't think you chose the best method, but it's a valid one, so it should still get full marks. You might lose a mark or 2 for the explanation at the end which is a little confusing, but I suspect you'd have done better on paper when it's easy to show the actual squares.

As Siklos says in the booklet, sometimes you just need to solve the question any way you can. It's easy to post mortem after the event, and while we're discussing it on here, it makes sense for people to say "actually, you could have done that bit better", because that all helps the learning process. But in an exam, if you can see "I can do this question by listing 200 combinations, and that will only take 25 minutes", then for heaven's sake don't worry about whether it's an elegant solution. (Any solution in <30 mins is an elegant solution in my book...)

[khaixiang]

I've done 4 more on STEP III, but I need to revise tomorrow's A2 chemistry practical! Wish I have got more time on STEP. So I will be posting question 1, 6, 8 and 10 tomorrow if no one else has done it.

[generalebriety]

(Original post by DFranklin)
I'm not sure this is right. Are you sure you can just add your 2 solutions?

I have no idea. I don't see why not...

Edit: wait a second, I think you're right. Will have a look.

Would it be right to say that the solutions were y = Ax + B or y = x arsinh x - root(1+x^2) + C? No, it wouldn't...

[DFranklin]

(Original post by generalebriety)
I have no idea. I don't see why not...

Basically because .

[DFranklin]

(Original post by generalebriety)
Would it be right to say that the solutions were y = Ax + B or y = x arsinh x - root(1+x^2) + C? No, it wouldn't...

Having had a look, subbing in for y=Ax+B you end up finding B = -cosh A, so y=Ax-coshA is one family of solutions.

I haven't looked at how the 2nd (nastier) solution behaves.

As a general comment: If you have a first order DE (no d2y/dx2 terms), then you don't expect more than 1 arbitrary constant. But you may get more than one family of solutions if there are some many-to-one functions in there.

E.g. (dy/dx)^2 = 1 has two families of solutions y=A+x and y = A - x.

[generalebriety]

II/9:

Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 647x558

\displaystyle (1-\lambda -\mu )\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c} = \mathbf{a} - \lambda\mathbf{a} - \mu\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c} \\

\mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) + \mu (\mathbf{c} - \mathbf{a}) \longleftarrow \text{Plane through }\mathbf{a}. \\

\text{Let } \mathbf{n} = (\mathbf{b} -\mathbf{a})\cross (\mathbf{c} -\mathbf{a}) = \mathbf{b}\times\mathbf{c} - \mathbf{a}\times\mathbf{c} - \mathbf{b}\times\mathbf{a} = \mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}. \\

\text{Then }\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} . \\ \\



[(1-\lambda -\mu )\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}]\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{a}\cdot (\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a})\\

(1-\lambda -\mu )\mathbf{a}\cdot\mathbf{b}\times \mathbf{c} + \lambda\mathbf{b}\cdot\mathbf{c} \times\mathbf{a} + \mu\mathbf{c}\cdot\mathbf{a}\tim es\mathbf{b} = \mathbf{a}\cdot\mathbf{b}\times\ mathbf{c} \text{ and first result follows}. \\

\bigg[\mathbf{r}\cdot\mathbf{n} =\bigg] \mathbf{a}\cdot\mathbf{n} = \mathbf{b}\cdot\mathbf{n} = \mathbf{c}\cdot\mathbf{n} \\

\mathbf{a}\cdot(\mathbf{a}\times \mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{b}\cdot(\mathbf{a}\times \mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times \mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a} ). \\

\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c}) = \mathbf{b}\cdot(\mathbf{c}\times \mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times \mathbf{b}) \text{ as the dot product of two perpendicular vectors is }\mathbf{0}. \\

A, B, C, O \text{ all in same plane }\Rightarrow \mathbf{a}\cdot (\mathbf{b}\times\mathbf{c}) = \mathbf{b}\cdot (\mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot (\mathbf{a}\times\mathbf{b}) = \mathbf{0}.

[*bobo*]

STEP I
10)
P: u=o, a=P/m, time to exit barrel=t, distance travelled to exit barrel=s(1)
s(1)=(P/2m)t
B:u=0, a=P/M, time taken=t, distance travelled= s(2)
s(2)=(P/2M)t
s(1)+s(2)=l

=> t= (2mMl)/(P(m+M))
so s(1)= Ml/(m+M)

P: s= Ml/(m+M), a= P/m, u=0 using v^2=u^2 + 2as
v= ((2PMl)/(m(m+M))^0.5

[generalebriety]

(Original post by DFranklin)
Having had a look, subbing in for y=Ax+B you end up finding B = -cosh A, so y=Ax-coshA is one family of solutions.

I haven't looked at how the 2nd (nastier) solution behaves.

As a general comment: If you have a first order DE (no d2y/dx2 terms), then you don't expect more than 1 arbitrary constant. But you may get more than one family of solutions if there are some many-to-one functions in there.

E.g. (dy/dx)^2 = 1 has two families of solutions y=A+x and y = A - x.

Yeah - after making that post above, I got about as far as you said with the first solution, but then wasn't sure whether I had found every solution. Will check the other solution and amend it after doing the question above. Thanks.

[*bobo*]

STEP I
11)i)
modulus of elasticity=A
tension in: AB=x,BC=y,CD=z
extension in: AB=e(1), BC=e(2), CD=e(3)
where e(1) + e(2) + e(3)= 3(d-a)

vertical equilibrium at two hanging particles:
x= mg+y
y=mg+z
=> 2mg=x-z => 2amg/A= e(1)- e(3)= 2e(1) + e(2) +3(a-d)

=>e(1)= (amg/A) + d -a
=>e(2)=d-a
=>e(3)= d-a -(amg/A)
so height of B= 2d- (amg/A)
height of C= d- (amg/A)

ii) no tension in CD, CD=a:
e(3)=0
d= a( 1 + (mg/A))

[*bobo*]

STEP I
13) d^2x/dt^2= v dv/dx= -Gx^-2
0.5v^2= Gx^-1 + 0.5V^2- Gh^-1
v=(2Gx^-1 + V^2 -2Gh^-1)^0.5
when x is very large x^-1 approaches 0
so v=(V^2 - 2Gh^-1)^0.5
for the comet to reach very large x, v must exist.
so V^2- 2Gh^-1=>0
so V^2 >= 2Gh^-1

at a great distance from the sun, the comet is motionles:
hence V^2=2Gh^-1

so 0.5v^2=Gx^-1
=> (dx/dt)= (2Gx^-1)^0.5
=> (2/3)x^1.5= t(2G)^0.5 +C
when t=0 x=h
=> x^1.5=1.5t(2G)^0.5 + h^1.5
when x=0, t=- (2h^1.5)/(3(2G)^0.5) (which is then put as positive because motion was modelled away from sun initially and now it is moving towards)