# STEP maths I, II, III 2007 solutions

Watch
Announcements

(Up-to-date as of post #115)

FINISHED!

Well done everbody!

Solutions for earlier STEP papers are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved (red above); if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

STEP I - STEP II - STEP III

STEP I:

1 - solved by generalebriety, alternative by Dystopia, also see thread

2 - part (i) solved by eponymous, part (ii) solved by eponymous.

3 - solved by insparato, partial alternative solution by calcium878

4 - solved by ukgea

5 - part (i) solved by calcium878, part (ii) solved by calcium878

6 - solved by coffeym

7 - solved by Dystopia

8 - solved by calcium878

9 - solved by coffeym

10 - part (i) solved by Square, part (ii) solved by DFranklin

11 - solved by DFranklin

12 - solved (rather sketchily) by ukgea

13 - solved by DFranklin

14 - solved by DFranklin

STEP II:

1 - solved by SsEe

2 - solved by Chewwy

3 - solved by insparato

4 - solved by generalebriety

5 - solved by generalebriety

6 - solved by Chewwy

7 - solved by ukgea

8 - solved by Dystopia

9 - solved by DFranklin

10 - solved by Dystopia

11 - solved by Dystopia

12 - solved by DFranklin

13 - solved by SsEe

14 - solved by coffeym

STEP III:

1 - solved by generalebriety, also see thread

2 - solved by Speleo

3 - solved by generalebriety

4 - solved by insparato

5 - solved by DFranklin

6 - solved by ukgea

7 - solved by generalebriety

8 - solved by generalebriety

9 - solved by DFranklin

10 - solved by DFranklin

11 - solved by DFranklin

12 - solved by SsEe

13 - solved by DFranklin

14 - solved by DFranklin

1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

FINISHED!

Well done everbody!

**Introduction/goal**Solutions for earlier STEP papers are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved (red above); if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

**Link to the papers:**STEP I - STEP II - STEP III

STEP I:

1 - solved by generalebriety, alternative by Dystopia, also see thread

2 - part (i) solved by eponymous, part (ii) solved by eponymous.

3 - solved by insparato, partial alternative solution by calcium878

4 - solved by ukgea

5 - part (i) solved by calcium878, part (ii) solved by calcium878

6 - solved by coffeym

7 - solved by Dystopia

8 - solved by calcium878

9 - solved by coffeym

10 - part (i) solved by Square, part (ii) solved by DFranklin

11 - solved by DFranklin

12 - solved (rather sketchily) by ukgea

13 - solved by DFranklin

14 - solved by DFranklin

STEP II:

1 - solved by SsEe

2 - solved by Chewwy

3 - solved by insparato

4 - solved by generalebriety

5 - solved by generalebriety

6 - solved by Chewwy

7 - solved by ukgea

8 - solved by Dystopia

9 - solved by DFranklin

10 - solved by Dystopia

11 - solved by Dystopia

12 - solved by DFranklin

13 - solved by SsEe

14 - solved by coffeym

STEP III:

1 - solved by generalebriety, also see thread

2 - solved by Speleo

3 - solved by generalebriety

4 - solved by insparato

5 - solved by DFranklin

6 - solved by ukgea

7 - solved by generalebriety

8 - solved by generalebriety

9 - solved by DFranklin

10 - solved by DFranklin

11 - solved by DFranklin

12 - solved by SsEe

13 - solved by DFranklin

14 - solved by DFranklin

**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

3

reply

Report

#2

STEP II Q6, solution by chewwy, shamelessly stolen from the STEP II thread:

http://www.thestudentroom.co.uk/atta...4&d=1183161671

http://www.thestudentroom.co.uk/atta...4&d=1183161671

2

reply

Report

#4

An excellent idea.

III/1:

--

Substituting into the given equation:

Substituting into (*):

as required.

III/1:

--

Substituting into the given equation:

Substituting into (*):

as required.

1

reply

Report

#5

STEP III Q2, solution by Speleo, shamelessly stolen from the STEP III thread:

(Original post by

2

i

as required.

Binomial expansion valid for i.e. as given.

Every successive term is multiplied by two extra negative terms, one fraction from the exponent and one power of -4x, so every term stays positive overall, so all minus signs can be removed.

Using the formula just proved and noting that there is a divisor of in total from the product of :

as required.

Verify that 2x is the first term by setting n = 1 and getting 2x.

ii

Differentiate both sides with respect to x:

Setting :

Multiply both sides by :

as required.

iii

This time integrate instead:

Set x = 0:

Set x = :

as required.

**Speleo**)2

i

as required.

Binomial expansion valid for i.e. as given.

Every successive term is multiplied by two extra negative terms, one fraction from the exponent and one power of -4x, so every term stays positive overall, so all minus signs can be removed.

Using the formula just proved and noting that there is a divisor of in total from the product of :

as required.

Verify that 2x is the first term by setting n = 1 and getting 2x.

ii

Differentiate both sides with respect to x:

Setting :

Multiply both sides by :

as required.

iii

This time integrate instead:

Set x = 0:

Set x = :

as required.

2

reply

Report

#6

STEP III Q6:

Firstly,

and similarly for q.

Thus we have

and the first result follows.

Then, note that if and are perpendicular,

or

Either way, we have

and thus

from which it follows that

as required.

For the next part, we have in the case : (let represent in the Argand plane, with the circle centred in the origin and with radius a)

Multiplying the first and the third equations, and then dividing by the second, we get

there are obviously two distinct that satisfy this, as required.

For , it's instead

In this case, any can satisfy this, becayse if you choose a , you can always choose a so that the first equation is satisfied, and then a so that the second one is satisfied, and then a so that the third is satisfied, and then, and here's the catch, equation 4 follows from the other three: (multiply the first and third equations and divide by the fourth) and so the fourth equation will automatically satified.

For larger n, we have exactly the same sitution as in n = 3 when n is odd (you can use the same reasoning, just alternatingly multiply and divide all the equations together) and as in n = 4 when n is even (again the same reasoning, the last equation can always be obtained by alternatingly multiplying and dividing the first ones together.

Firstly,

and similarly for q.

Thus we have

and the first result follows.

Then, note that if and are perpendicular,

or

Either way, we have

and thus

from which it follows that

as required.

For the next part, we have in the case : (let represent in the Argand plane, with the circle centred in the origin and with radius a)

Multiplying the first and the third equations, and then dividing by the second, we get

there are obviously two distinct that satisfy this, as required.

For , it's instead

In this case, any can satisfy this, becayse if you choose a , you can always choose a so that the first equation is satisfied, and then a so that the second one is satisfied, and then a so that the third is satisfied, and then, and here's the catch, equation 4 follows from the other three: (multiply the first and third equations and divide by the fourth) and so the fourth equation will automatically satified.

For larger n, we have exactly the same sitution as in n = 3 when n is odd (you can use the same reasoning, just alternatingly multiply and divide all the equations together) and as in n = 4 when n is even (again the same reasoning, the last equation can always be obtained by alternatingly multiplying and dividing the first ones together.

0

reply

Report

#8

(Original post by

Don't suppose the papers are up on the 'net yet are they?

**Square**)Don't suppose the papers are up on the 'net yet are they?

0

reply

Report

#10

Ooooooh. III/7. Mine.

(i)

Choose x = 1 and result follows.

(ii)

Performing this substitution:

Again, choose x = 1 and result follows.

(iii)

(detail omitted!)

where

Then

and so

(i)

Choose x = 1 and result follows.

(ii)

Performing this substitution:

Again, choose x = 1 and result follows.

(iii)

(detail omitted!)

where

Then

and so

1

reply

Report

#11

**STEP I Question 3**

Right this stinks of I + J adding and substracting these integrals using the identities. But scrap that theres sin^6x so im going for reduction formula.

By parts

Let

2

reply

Report

#12

I'm also claiming III/8. Might be my last contribution of the night then. Urgh, this will be messy to latex.

(i) Putting u = x:

Putting u = exp(-x):

Adding:

General solution:

Applying substitution,

which has general solution

i.e.

y(0) = 0:

(ii)

Applying the substitution gives (detail omitted):

to which "obvious" solutions are u = x, u = exp x, and so the general solution is This gives:

y(0) = 2:

(i) Putting u = x:

Putting u = exp(-x):

Adding:

General solution:

Applying substitution,

which has general solution

i.e.

y(0) = 0:

(ii)

Applying the substitution gives (detail omitted):

to which "obvious" solutions are u = x, u = exp x, and so the general solution is This gives:

y(0) = 2:

0

reply

Report

#13

Alternative I/3. As inspirato for the beginning.

Now, (*) + (**) gives so that

And, (*)-(**) gives

So,

So that

The one with is very nearly the same, just a sign change here and there (skipping the intermediate rearrangements):

Then a traditional mash through the trig for the 6-th power ones, which I won't type up

Now, (*) + (**) gives so that

And, (*)-(**) gives

So,

So that

The one with is very nearly the same, just a sign change here and there (skipping the intermediate rearrangements):

Then a traditional mash through the trig for the 6-th power ones, which I won't type up

0

reply

Report

#14

Yep i did it that way originally on paper. But i thought im going to have to do reduction formula anyways so might as well follow through. Btw I was going to post that but the Latex was a ***** anyways so i couldnt be bothered lol.

0

reply

Report

#18

__Step I Q2__

ok, so this is my first time EVER using latex , here goes...

ok, so i cant master latex , see fatuous philomath's post

0

reply

Report

#19

(Original post by

ok, so this is my first time EVER using latex , here goes...

now

so from the first part;

rearranging a bit;

now, or

so (p-1) = 1 or 2 and (q-1) = 2 or 1

hence p = 2 or 3 and q = 3 or 2

(p-1) and (q-1) cannot be equal to -1, since this would make p or q equal to zero and p and q are non zero integers.

**eponymous**)__Step I Q2__ok, so this is my first time EVER using latex , here goes...

now

so from the first part;

rearranging a bit;

now, or

so (p-1) = 1 or 2 and (q-1) = 2 or 1

hence p = 2 or 3 and q = 3 or 2

(p-1) and (q-1) cannot be equal to -1, since this would make p or q equal to zero and p and q are non zero integers.

0

reply

Report

#20

(Original post by

I think you mean...

**fatuous_philomath**)I think you mean...

thank you so much!!

i really can't get latex right

0

reply

X

### Quick Reply

Back

to top

to top