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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

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 8 years ago)

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

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

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)

STEP II 2005, Question 6

Spoiler

sonofdot

What happens when n = 2 in part ii)?

STEP II 2005, Question 2

Spoiler

STEP I: Question 12

Part 1

Part 2

Part 1

Spoiler

Part 2

Spoiler

II/4:

1st part

2nd part

3rd part

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

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.

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.

GHOSH gave all the solutions I have for 1 2005:Q2,3,4,7

I'll try another one

I'll try another one

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