The Student Room Group

STEP 2003 Solutions Thread

Scroll to see replies

Reply 40
8 Horizontal
Sorry :o:

:p:

I've done all 8 pures on II now :eek: don't think I've ever done that on a paper before.
I just logged on to type my solution up:p: I did it earlier today at work when bored - it was a nice and unusually simple question I thought.
STEP III, 2003, Question 7

Spoiler

tommm
STEP II 2003 Q2

Spoiler



Genius. I spent an hour getting two geometrical proofs, which are actually quite nice, but so much more complicated to find. I attatched one I scanned in, the other is similar. Your method is really neat though.
Reply 44
toasted-lion
Genius. I spent an hour getting two geometrical proofs, which are actually quite nice, but so much more complicated to find. I attatched one I scanned in, the other is similar. Your method is really neat though.


Very nice. In STEP, if there's two parts, you usually have to apply the method from the first part to the second part.
toasted-lion
Genius. I spent an hour getting two geometrical proofs, which are actually quite nice, but so much more complicated to find. I attatched one I scanned in, the other is similar. Your method is really neat though.


I like what you've done, though I would have never attempted to come up with that myself.

I'm pretty sure Tom's way is the way you're expected to do it.
tommm
.


There's a problem with the LaTeX in your answer for III, 10. I can't spot it though (most probably an extra / missing curly bracket)
Reply 47
Daniel Freedman
There's a problem with the LaTeX in your answer for III, 10. I can't spot it though (most probably an extra / missing curly bracket)


Yeah, I just put in an extra curly bracket and it's showing now, but it's still not right...
tommm
STEP II 2003 Q4

Spoiler



Upon using the substitution y=Rcosuy = R\cos u, and some manipulation with the cosine double-angle formula, we get

R2(u0.5sin(2u)0arccos(d/R)R^2(u - 0.5\sin(2u)|^{\arccos(d/R)}_0 (N.B. that represents limits of integration)

The difficulty in evaluating this lies in the evaluation of 0.5sin(2arccos(d/R))0.5\sin(2\arccos(d/R)). First, using the sine double angle formula, we get

0.5sin(2arccos(d/R))=sin(arccos(d/R))cos(arccos(d/R))=(d/R)sin(arccos(d/R))0.5\sin(2\arccos(d/R)) = \sin(\arccos(d/R))\cos(\arccos(d/R)) = (d/R)\sin(\arccos(d/R))

By drawing a suitable right-angled triangle, we find that
Unparseable latex formula:

\sin(2 \arccos (d/R)) = \frac{\sqrt{R^2 - d^2}{R}}

and from this we can find the required result.



Hmm... yuk! Consider the area of the sector OGH and the triangle OGH, subtracting to get the segment. Nice perseverance, but it's actually pretty easy.
tommm
STEP II 2003 Q4

Spoiler



Are you sure we want the distance to the edge and not to the centre? We had d as the distance to centre before. I got:

for (i) d=cb d = |c-b| (note the x-coord of the centre is irrelevant)
for (ii) d=cm2+1 d = \frac{|c|}{\sqrt{m^2 +1}} (perpendicular distance)
for (iii) d=ma+cbm2+1 d = \frac{|ma + c - b|}{\sqrt{m^2 +1}} (perpendicular distance again)

and I just plugged these straight into the equation from the first part. As you noted (i) and (ii) follow from (iii), but I'm sure they put them in that order for a reason (ie to give you bite-sized chunks of the overall thought-process). Please tell me if I made a mistake

P.S. You got a stray /latex tag :P
STEP III Q13

(i)

Spoiler


(ii)

Spoiler


(iii)

Spoiler

That was an impressively quick update on the original post Simon. Would you be able to fix all the dead links to other years' solutions some time maybe?
Reply 51
toasted-lion
That was an impressively quick update on the original post Simon. Would you be able to fix all the dead links to other years' solutions some time maybe?


I wasn't aware that there were any dead links, if you would care to show me the threads, I'll do it as soon as I can
SimonM
I wasn't aware that there were any dead links, if you would care to show me the threads, I'll do it as soon as I can


None of them link here, and very few link to 2002, even though the pages exist. Check a couple, you will see what I mean.
Reply 53
toasted-lion
None of them link here, and very few link to 2002, even though the pages exist. Check a couple, you will see what I mean.


I cordially invite you to slap me in the face for misreading what you said and wasting 15 minutes checking every link to a solution on 2002 and 2003 before realising you meant making the solution threads link to one another!
SimonM
I cordially invite you to slap me in the face for misreading what you said and wasting 15 minutes checking every link to a solution on 2002 and 2003 before realising you meant making the solution threads link to one another!


If it makes you feel better, have a preverbial slap in the face. Sorry if my post was ambiguous. Seriously though, much respect for organising these threads, they're invaluable.
Adje
III/1

ddxarcsin(x1x+3) \displaystyle \frac{d}{dx} \arcsin \left(\frac{x-1}{x+3}\right)

Unparseable latex formula:

\displaystyle = \frac{\sqrt{4}}{(x+3)\sqrt{2+2x}}}}



Unparseable latex formula:

\displaystyle = \frac{2}{\sqrt{2}}\frac{1}{{(x+3)\sqrt{1+x}}}}}



Unparseable latex formula:

\displaystyle \Rightarrow \int \frac{1}{{(x+3)\sqrt{x+1}}}}} \ dx = \frac{\sqrt{2}}{2}\arcsin \left(\frac{x-1}{x+3}\right) + C



This is what I got too, but it says it has to be valid for x>-1. Isn't the above only valid for x>1?
toasted-lion
This is what I got too, but it says it has to be valid for x>-1. Isn't the above only valid for x>1?


x1x+31 |\frac{x-1}{x+3}| \le 1
DeanK22
x1x+31 |\frac{x-1}{x+3}| \le 1


In the question it says "given that x + a > 0 and x + b > 0", but I am not sure whether this means it's not valid if that doesn't apply?
toasted-lion
In the question it says "given that x + a > 0 and x + b > 0", but I am not sure whether this means it's not valid if that doesn't apply?


x >= -1 is required for |(x-1)/(x+3)| <= 1

... however I will look at the question now.

edit; well the conditions are there so you don't have to start using complex numbers.

i.e.

sqrt(a-b) and it is given a > b and sqrt(2x+2) , etc

x is not => -1 but > -1 as if x = -1 you would get division by 0.

I wonder how they spot these things - it is certainly noon trivial approach to guess such functions work.
DeanK22

Spoiler



So do you think that in this particular question they just want you to do the algebra rather than worry about ranges of validity?

Quick Reply

Latest