STEP Maths I,II,III 1987 Solutions

DFranklin

I'm having a certain amount of difficulty working out what you're actually trying to do here (more me being tired than any fault on your part).

I'm equally tired, and it is my fault, because you just made me realise it is easier to get a sum in a nice closed form if I apply the limits! :rolleyes:

If you can get things down to an actual formula, I'll plug it into my computer here and see if it does at least agree numerically.
Judging by reactions, it seems this question is pretty tough for STEP I!

I'll type it up in 5 min, hold on whilst I'm latexing!

And well, the question certainly requires a lot of fiddling, and is very easy to make small slips on. But it is really a nice question.

Working: By earlier argument

\sin(\frac{22}{23}\pi)=\sin(\frac{1}{23}\pi)

and by symmetry follows that

\sin(\frac{44}{23}\pi)=- \sin(\frac{2}{23}\pi)

Putting k=10 into my above general formula we have

\displaystyle\frac{\cos( \frac{22}{23} \pi) \sin( \frac{22}{23}\pi)}{ \cos( \frac{2}{23} \pi)}

=

\frac{ \sin( \frac{44}{23} \pi)}{2 \cos( \frac{2}{23} \pi)}

[using the product formula for cos(a)sin(b)] And the top of that fraction is equal to

- \sin(\frac{2}{23}\pi)

and thus the required answer in

-\frac{1}{2}\tan(\frac{2}{23}\pi)

Reply 142

ukgea

6

nota bene

I'm equally tired, and it is my fault, because you just made me realise it is easier to get a sum in a nice closed form if I apply the limits! :rolleyes:

I'll type it up in 5 min, hold on whilst I'm latexing!

And well, the question certainly requires a lot of fiddling, and is very easy to make small slips on. But it is really a nice question.

Working: By earlier argument

\sin(\frac{22}{23})=\sin(\frac{1}{23})

and therefore follows that

\sin(\frac{44}{23})=\sin(\frac{2}{23})

Putting k=10 into my above general formula we have

\displaystyle\frac{\cos( \frac{22}{23} \pi) \sin( \frac{22}{23})}{ \cos( \frac{2}{23} \pi)}

=

\frac{ \sin( \frac{44}{23} \pi)}{2 \cos( \frac{2}{23} \pi)}

[using the product formula for cos(a)sin(b)] And the top of that fraction is equal to

\sin(\frac{2}{23}\pi)

and thus the required answer in

\tan(\frac{2}{23}\pi)

Not quite.

Reply 143

DFranklin

18

Nota, you're not going to believe what you've just done again when you recheck that!

(There's also a sign error, you've said something = sin(2pi/23) when it doesn't...)

Reply 144

ukgea

Not quite.

Damn, that factor of 1/2 again, that's the second time... :s-smilie:

Reply 145

DFranklin

Nota, you're not going to believe what you've just done again when you recheck that!

(There's also a sign error, you've said something = sin(2pi/23) when it doesn't...)

Haha, yeah - I know! I think I need someone to tell me 'eeep, not finished' when writing tests, I seem to fail to do anything 100% right...

Hopefully it's all correct by now :smile:

Reply 146

DFranklin

18

nota bene

Haha, yeah - I know! I think I need someone to tell me 'eeep, not finished' when writing tests, I seem to fail to do anything 100% right...

It's times like this you think "fortunately, I no longer have to worry about sitting STEP anymore!".

Hopefully it's all correct by now :smile:

Well, your answers agree with mine! :smile:

Reply 147

STEP III, Q4.

Spoiler

Easiest STEP question I've ever seen.

Reply 148

DFranklin

18

Dystopia

STEP III, Q4.

Spoiler

Easiest STEP question I've ever seen.

It's still not as easy as the "sum of logs" question on STEP I 1987, but it's very easy for STEP III. As I said before, the question difficulty for 1987 seems all over the place. Though later years also have the occasional surprisingly easy question making it onto STEP III. I remember in particular a "find the length of a cycloid" question that was very short (and I still managed to get wrong :rolleyes:

).

Reply 149

STEP III, Q5.

Spoiler

Reply 150

STEP III, Q6.

x' + 2x - 5y = 0 \Rightarrow x'' + 2x' - 5y' = 0

y' = 2\cos t + 2y - ax

x'' + 2x' - 10\cos t - 10y + 5ax = 0

5y = x' + 2x

x'' + 2x' - 10\cos t - 2x' - 4x + 5ax = 0

x'' + (5a - 4)x = 10\cos t

a = 1 \Rightarrow x'' + x = 10\cos t

CF:

\lambda^{2} + 1 = 0 \Rightarrow \lambda = \pm i

x = A\cos t + B\sin t

PI: Normally we would suggest a solution of the form

P\cos t + Q\sin t

However, both of these are part of the CF, so instead suppose

x = Pt\cos t + Qt\sin t

x' = P\cos t - Pt\sin t + Q\sin t + Qt\cos t

x'' = -P\sin t - P\sin t - Pt\cos t + Q\cos t + Q\cos t - Qt\cos t

Putting these into the equation we get

-2P = 0, \; 2Q= 10 \Rightarrow P = 0, \; Q=5

Therefore the solution is

x = A\cos t + (B + 5t)\sin t

Using the initial conditions, we get

A = 0, \; B=-5

Therefore the particular solution is

x = 5(t - 1)\sin t, \; y=t(2\sin t + \cos t) - (\sin t + \cos t)

If a > 1, the complementary function would be of the form

A\cos(\alpha t) + B\sin(\alpha t), \; \alpha > 1

. Therefore, the particular integral would not have to include t, and the solutions for x and y would also not include it.

Reply 151

Gonna update the first post later this evening, about to pop out just now.

Reply 152

STEP III, Q10

Spoiler

Interesting question.

DFranklin

It's still not as easy as the "sum of logs" question on STEP I 1987, but it's very easy for STEP III. As I said before, the question difficulty for 1987 seems all over the place. Though later years also have the occasional surprisingly easy question making it onto STEP III. I remember in particular a "find the length of a cycloid" question that was very short (and I still managed to get wrong :rolleyes:

).

Ah, well; haven't had a look at that one yet. :p:

Reply 153

Have solution to I/14 will type up tomorrow

edit:

Spoiler

Nice little question.

Reply 154

I've done STEP III, Q1, but I shall hold off posting as someone else might want to give it a try; as DFranklin said, it doesn't require Further Maths knowledge.

Square

Gonna update the first post later this evening, about to pop out just now.

That's good, although there still appear to be some mistakes/omissions. If you're busy I could make the corrections and PM it to you?

Reply 155

Hmm, Yeah some of the links seem to be dead, I'll sort it now.

Reply 156

If Square wants me to I can update the OP (I'm a mod, so I can edit his post)

Reply 157

Sure thing I don't mind at all.

II/6 is a dead link and I can't find the post so I'm thinking I maybe listed it there in error, but other than that the rest seem to be working. I don't think I have missed out anyone's solutions, if I have I'm sorry!

Reply 158

Square

Sure thing I don't mind at all.

II/6 is a dead link and I can't find the post so I'm thinking I maybe listed it there in error, but other than that the rest seem to be working. I don't think I have missed out anyone's solutions, if I have I'm sorry!

STEP I, Q5

STEP II, Q6

Apart from those two it looks fine.

Reply 159