# 2004 a STEP thread... Watch

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(Updated as far as #55) SimonM - 10.05.2009

1: Solution by DeanK22

2: Solution by sonofdot

3: Solution by SimonM

4: Solution by Unbounded

5: Solution by Unbounded

6: Solution by brianeverit

7: Solution by SimonM

8: Solution by Daniel Freedman

9: Solution by Unbounded

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by Unbounded

13: Solution by brianeverit

14: Solution by brianeverit

1: Solution by Unbounded

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solutiom by Aurel-Aqua

5: Solution by Daniel Freedman

6: Solution by SimonM

7: Solution by Daniel Freedman

8: Solution by Daniel Freedman

9: Solution by brianeverit

10: Solution by Daniel Freedman

11: Solution by Zuzuzu

12: Solution by Aurel-Aqua

13: Solution by brianeverit

14: Solution by brianeverit

1: Solution by sonofdot

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solution by SimonM

5: Solution by DeanK22

6: Solution by SimonM

7: Solution by DeanK22

8: Solution by DeanK22

9: Solution by brianeverit

10: Solution by Aurel-Aqua

11: Solution by SimonM

12: Solution by brianeverit

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

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 DeanK22

2: Solution by sonofdot

3: Solution by SimonM

4: Solution by Unbounded

5: Solution by Unbounded

6: Solution by brianeverit

7: Solution by SimonM

8: Solution by Daniel Freedman

9: Solution by Unbounded

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by Unbounded

13: Solution by brianeverit

14: Solution by brianeverit

**STEP II:**1: Solution by Unbounded

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solutiom by Aurel-Aqua

5: Solution by Daniel Freedman

6: Solution by SimonM

7: Solution by Daniel Freedman

8: Solution by Daniel Freedman

9: Solution by brianeverit

10: Solution by Daniel Freedman

11: Solution by Zuzuzu

12: Solution by Aurel-Aqua

13: Solution by brianeverit

14: Solution by brianeverit

**STEP III:**1: Solution by sonofdot

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solution by SimonM

5: Solution by DeanK22

6: Solution by SimonM

7: Solution by DeanK22

8: Solution by DeanK22

9: Solution by brianeverit

10: Solution by Aurel-Aqua

11: Solution by SimonM

12: Solution by brianeverit

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

**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

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#2

Sequel to "2001: A Space Odyssey"?

Sounds cool, but the solutions will be on Meikleriggs. No reason we can't have a separate set, though.

Sounds cool, but the solutions will be on Meikleriggs. No reason we can't have a separate set, though.

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#3

(Original post by

... to do the papers - solutions only...

...

**DeanK22**)... to do the papers - solutions only...

...

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#5

(Original post by

... a laughable comment.

**DeanK22**)... a laughable comment.

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#6

__STEP II, 2004, Question 2__
Spoiler:

Show

No solutions to

When

due to symmetry.

where

This interval has length 2t the sum of the lengths of the intervals is 4t due to symmetry as before.

as

Graph attached (slightly dodgy)

EDIT: In the first part, you'd probably need to mention why the discriminant of this particular quadratic is still b^2 - 4ac (consider x < 0 and x > 0).

When

due to symmetry.

where

This interval has length 2t the sum of the lengths of the intervals is 4t due to symmetry as before.

as

Graph attached (slightly dodgy)

EDIT: In the first part, you'd probably need to mention why the discriminant of this particular quadratic is still b^2 - 4ac (consider x < 0 and x > 0).

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#9

__STEP I, 2004, Question 5__A:

B:

C:

D:

E:

where

Statement: The sum of any term in B with any term in C is a term in E

as for all b, c there exists an e such that

Statement: The square of any term in B is in D.

as for all b, there exists a d such that

Statement: The square of any term in C is also in D.

as for all c, there exists a d such that

(i) Considering the equation with modulus 5:

Looking at quadratic residues with modulus 5:

Therefore, there exists no integer such that it squares to have a residue of 3 (mod 5).

(ii) Again looking at the equation with modulus 5:

Similarly, looking at quartic residues with modulus 5:

Considering positive integers , we have pairs with quartic residues (mod 5) of either (1,1), (0,0), (1,0), or (0,1), and so the residues modulo 5 of the pairs are either (1,2), (0,0), (1,0) or (0,2). and summing the pairs in the brackets, none sum to 4 (mod 5), so there exist no pairs of integers x and y, such that

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#11

__STEP I Question 2__i) Graph attached (I was lazy and didn't draw it myself...)

From graph:

ii) Similarly,

(sum of a GP)

iii)

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#12

__STEP II, 2004, Question 10__
Spoiler:

i)

To find the time after which it stops,

After it has stopped,

Note that it is necessary that , as at , P has returned to its point of projection.

When they collide,

We want smallest positive t, therefore as required.

ii)

At

After 2s, the total distance travelled by P is equal to .

There is a constant resistive force of 4N, so the total work done is 24J.

Show

i)

To find the time after which it stops,

After it has stopped,

Note that it is necessary that , as at , P has returned to its point of projection.

When they collide,

We want smallest positive t, therefore as required.

ii)

At

After 2s, the total distance travelled by P is equal to .

There is a constant resistive force of 4N, so the total work done is 24J.

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#14

__STEP II, 2004, Question 3__
Spoiler:

as required.

At a stationary point,

Therefore the coordinates of the stationary points are

At

At

(graph attached)

i)

Therefore curve has same roots as C, is symmetrical about x-axis, and is not defined were -1 < x < 0

(graph attached)

ii)

This is an even function, and so is symmetrical about the y axis. It has roots at

Let , which is curve C.

Therefore it has stationary points at

(graph attached)

Show

as required.

At a stationary point,

Therefore the coordinates of the stationary points are

At

At

(graph attached)

i)

Therefore curve has same roots as C, is symmetrical about x-axis, and is not defined were -1 < x < 0

(graph attached)

ii)

This is an even function, and so is symmetrical about the y axis. It has roots at

Let , which is curve C.

Therefore it has stationary points at

(graph attached)

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#15

Someone do II question 1. I can't be arsed typing it up (and it's a rubbish question).

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#16

__STEP I, 2004, Question 9__the vertical velocity of the particle upwards is initially . noting that at the greatest height 'h', v = 0

As the particle doesn't hit the top,

__at point P__, let the time be T:

Considering it horizontally:

Considering it vertically:

Substitution back into :

Substitution into :

considering roots of

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#18

(Original post by

Someone do II question 1. I can't be arsed typing it up (and it's a rubbish question).

**Daniel Freedman**)Someone do II question 1. I can't be arsed typing it up (and it's a rubbish question).

__STEP II, 2004, Question 1__
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#20

I know it says answers only at the top of this thread but I'd like to ask a question. Sorry.

I just did 4i from paper 2 and out popped the required result but I didn't use/need the condition that .

Any comments on why this condition is necessary?

I just did 4i from paper 2 and out popped the required result but I didn't use/need the condition that .

Any comments on why this condition is necessary?

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