You can let someone else take control of the OP whenever.
STEP I:
1: Solution by Horizontal 8
2: Solution by Horizontal 8
3: Solution by Horizontal 8
4: Solution by Horizontal 8
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by Unbounded
10: Solution by Unbounded
11: Solution by Unbounded
12: Solution by Unbounded
13: Solution by AnonyMatt and Jkn
14: Solution by nota bene
STEP II:
1: Solution by tommm
2: Solution by tommm
3: Solution by Horizontal 8
4: Solution by tommm
5: Solution by Dadeyemi
6: Solution by Horizontal 8
7: Solution by SimonM
8: Solution by tommm
9: Solution by Elongar
10: Solution by TwoTwoOne
11: Solution by Rocious
12: Solution by SimonM
13: Solution by SimonM
14: Solution by TwoTwoOne
STEP III:
1: Solution by Adje
2: Solution by SimonM
3: Solution by Adje
4: Solution by tommm
5: Solution by SCE1912
6: Solution by SimonM
7: Solution by Daniel Freedman
8: Solution by Adje
9: Solution by brianeverit
10: Solution by tommm
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by toastedlion
14: Solution by SimonM
Solutions written by TSR members:
1987  1988  1989  1990  1991  1992  1993  1994  1995  1996  1997  1998  1999  2000  2001  2002  2003  2004  2005  2006  2007
STEP 2003 Solutions Thread
Announcements  Posted on 



III/1

STEP III, Question 14

Given that's what I've written down on my whiteboard, I'll assume that was a typo

III, 3

STEP I question 4
(Not 100% confident, since I always make mistakes with intervals)

III/8

STEP II Q3
First Part:
Spoiler:Show
Second part:
Spoiler:Show
Suppose statement is true for n=k.
By the inductive hypothesis is irrational and it follow from our result in the first part of the question that the cube root of this must be irrational therefore it must be true for n=k+1.
The case where n=1 is true (from the question) therefore the result is true for all positive integral value of n.
Last part:

I did III/Q4 yesterday, I'll type it up soon.

Question 5, STEP I, 2003

STEP I question 3
Let x= theta for notational purposes
(i)
(ii)
(iii)

Question 7, STEP I, 2003

Question 9, STEP I, 2003

Question 8, STEP I, 2003
Spoiler:ShowLet the B = yV. the rate at which B increases with respect to time is:
From the question, we see that:
where C is some constant
where
When A completely turns into B, y = 1.
and we have a contradiction and can conclude that A never completely turns into B.
I have a very strong feeling I've missed something, in the question 
STEP III 2003, Q4
Spoiler:Show
, therefore
Using , we obtain the equation of the tangent:
This meets the curve again when . Substituting this in and rearranging to find a quadratic in , we obtain:
By the factor theorem, we find that is a root. We can factorise and divide by (because ) to obtain
.
Using the quadratic formula, we obtain
One of these gives , which we can disregard.
Therefore, as required.
The double angle formula for tangent is:
Therefore,
So, if , then
Therefore, if , then
We can also apply the formula similarly to the cotangent function, so
Using the identities and , we get
as required.
Another value of which satisfies the required properties is .

Question 10, STEP I, 2003

STEP III 2003 Q10
Spoiler:Show
n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.
Integrating gives
and using the initial conditions gives
Separating variables gives
Using the initial conditions gives
Therefore, upon rearranging a bit, we get
As , the argument of tangent and hence as required.
In this second case, our first integration combined with the initial conditions gives
Now, for simplicity, let .
Separating variables we obtain
which becomes
Rearranging a lot, we get the rather nasty answer:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.x = \sqrt{d^2  \frac{2U}{k}} \displaystyle\frac{1 + \frac{d  \sqrt{d^2  \frac{2U}{k}}}}{d + \sqrt{d^2  \frac{2U}{k}}} e^{\sqrt{d^2  \frac{2U}{k}}kt}{1  \frac{d  \sqrt{d^2  \frac{2U}{k}}}{d + \sqrt{d^2  \frac{2U}{k}}} e^{\sqrt{d^2  \frac{2U}{k}}kt}
As , the fraction
Therefore .
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