STEP 2005 Solutions Thread

You can let someone else take control of the OP whenever.

STEP I:
1: Solution by nuodai
2: Solution by Unbounded
3: Solution by Unbounded
4: Solution by Unbounded
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by nuodai
10: Solution by Sk1lz
11: Solution by nuodai
12: Solution by darkness9999
13: Solution by brianeverit
14: Solution Farhan.Hanif93

STEP II:
1: Solution by SimonM
2: Solution by Daniel Freedman
3: Solution by SimonM
4: Solution by Glutamic Acid
5: Solution by SimonM
6: Solution by Daniel Freedman
7: Solution by SimonM
8: Solution by sonofdot
9: Solution by Farhan.Hanif93
10: Solution by Glutamic Acid
11: Solution by Farhan.Hanif93
12: Solution by nuodai
13: Solution by brianeverit
14: Solution by brianeverit

STEP III:
1: Solution by DeanK22
2: Solution by Adje
3: Solution by SimonM
4: Solution by SimonM
5: Solution by Daniel Freedman
6: Solution by sonofdot
7: Solution by sonofdot
8: Solution by SimonM
9: Solution by sonofdot
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by sweeneyrod
13: Solution by brianeverit
14: Solution by sonofdot

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

(edited 9 years ago)

Scroll to see replies

Reply 1

SimonM

I'm not going to type it up, but STEP III Question 7 is one of my favourite STEP questions

Reply 2

sonofdot

STEP III 2005 Question 7

This is one of my favourites too

We have $\displaystyle\int \frac{1}{u f(u)} \, du = F(u) +c$

Making a substitution $u=x^m \Rightarrow du = mx^{m-1} dx$

$\displaystyle\therefore \int \frac{mx^{m-1}}{x^m f(x^m)} \, dx = F(x^m) + c \iff \int \frac{m}{x f(x^m)} \, dx = F(x^m) + c$

(i)

(ii)

Reply 3

Daniel Freedman

STEP II 2005, Question 6

Spoiler

Reply 4

Daniel Freedman

sonofdot

.

What happens when n = 2 in part ii)?

Reply 5

Daniel Freedman

STEP II 2005, Question 2

Spoiler

Reply 6

sonofdot

Daniel Freedman

What happens when n = 2 in part ii)?

Good point, edited

Reply 7

darkness9999

STEP I: Question 12
Part 1

Spoiler

Part 2

Spoiler

Reply 8

sonofdot

STEP III 2005 Question 14

:eek:

that was a long one...

Spoiler

Reply 9

Glutamic Acid

II/4:

1st part

2nd part

3rd part

Reply 10

Oh I Really Don't Care

STEP III

Question 1

1st part

2nd part

Cheating with the graphs.

stthrthsrtrhrsth.jpg28.6KB

Reply 11

sonofdot

STEP III 2005 Question 6

Spoiler

Reply 12

Adjective

III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation

d/dx and d²/dx² of (x² + y²)

Graph

0 Point of minimum distance to origin with < c a^2

If c > a²

step052.jpg10.8KB

Reply 13

Hashshashin

For this part, is it not far simpler to note that as the gradient of PR is $\frac{1}{p}$ and the gradient of SQ is $-q$ , and as $pq = -1$ it's clear that $p = -\frac{1}{q}$ and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.