# STEP Maths I,II,III 1987 Solutions

Watch
Announcements

(Updated as far as #214) SimonM - 07.05.2009

1: Solution by Square & DFranklin

2: Solution by DFranklin

3: Solution by coffeym

4: Solution by DeathAwaitsU

5: Solution by Coffeym

6: Solution by brianeverit

7: Solution by ukgea & Solution by nota bene

8: Solution by kabbers

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by nota bene

15: Solution by brianeverit

16: Solution by brianeverit

1: Solution by Mazzacre

2: Solution by brianeverit

3: Solution by brianeverit

4: Solution by brianeverit

5: Solution by Dystopia

6: Solution by Dystopia

7: Solution by Dystopia

8: Solution by brianeverit

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

15: Solution by brianeverit

16: Solution by brianeverit

1: Solution by DeanK22 Solution by SimonM

2: Solution by SimonM

3: Solution by Dystopia

4: Solution by Dystopia

5: Solution by Dystopia

6: Solution by Dystopia

7: Solution by ukgea

8: Solution by SimonM

9: Solution by SimonM

10: Solution by Dystopia

11: Solution by ben-smith

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

15: Solution by brianeverit

16: Solution by SimonM

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 Square & DFranklin

2: Solution by DFranklin

3: Solution by coffeym

4: Solution by DeathAwaitsU

5: Solution by Coffeym

6: Solution by brianeverit

7: Solution by ukgea & Solution by nota bene

8: Solution by kabbers

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by nota bene

15: Solution by brianeverit

16: Solution by brianeverit

**STEP II:**1: Solution by Mazzacre

2: Solution by brianeverit

3: Solution by brianeverit

4: Solution by brianeverit

5: Solution by Dystopia

6: Solution by Dystopia

7: Solution by Dystopia

8: Solution by brianeverit

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

15: Solution by brianeverit

16: Solution by brianeverit

**STEP III:**1: Solution by DeanK22 Solution by SimonM

2: Solution by SimonM

3: Solution by Dystopia

4: Solution by Dystopia

5: Solution by Dystopia

6: Solution by Dystopia

7: Solution by ukgea

8: Solution by SimonM

9: Solution by SimonM

10: Solution by Dystopia

11: Solution by ben-smith

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

15: Solution by brianeverit

16: 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

0

reply

Report

#3

(Original post by

Is there a thread for 1987 STEP solutions? I cant find it if there is.

If not we should make one

**Square**)Is there a thread for 1987 STEP solutions? I cant find it if there is.

If not we should make one

0

reply

Report

#4

Well I'd need some contributions from other regulars, especially with II/III.

Got some of the STEP I questions written up though.

Got some of the STEP I questions written up though.

0

reply

Report

#6

I'll just make it this thread then, post your solutions gogo.

Gonna copy the format from the other threads and gonna put it into my opening post, will change the thread name also.

Gonna copy the format from the other threads and gonna put it into my opening post, will change the thread name also.

0

reply

Report

#8

Go for it, I'll do some questions this night - gonna go out to the cinema and then go to the pub for dinner, but will be back before midnight

edit: I've PMd the link

edit: I've PMd the link

0

reply

Report

#9

Bugger, think there might actually be a mistake in my solution, taken it down for now.

0

reply

Report

#10

STEP II, Q7.

The trapezium rule approximates the area y=f(x) under the curve by trapezia, all having a base of width h. The side lengths are f(n), f(n+h) for n from the lower limit to h less than the upper limit. The area of each trapezium is therefore h/2(f(n) + f(n+1)), which, when added together, results in the expression given.

Therefore ln x is a convex function, so the area under the curve is underestimated by trapezia.

As required. This is closer to than g(kn) because the intervals are closer together for the second approximation and so it is more accurate.

The trapezium rule approximates the area y=f(x) under the curve by trapezia, all having a base of width h. The side lengths are f(n), f(n+h) for n from the lower limit to h less than the upper limit. The area of each trapezium is therefore h/2(f(n) + f(n+1)), which, when added together, results in the expression given.

Therefore ln x is a convex function, so the area under the curve is underestimated by trapezia.

As required. This is closer to than g(kn) because the intervals are closer together for the second approximation and so it is more accurate.

0

reply

Report

#11

Differentiating

Finding x-ordinates of turning points:

Consider any turning point:

Expand out using addition formula sin terms go to zero:

Now if n is even the expression is going to be:

And if n is odd the expression will be:

Now consider the next turning point directly after the one shown above ie. (n+1)pi

This expression will be of the form:

Which is going to equal:

So, depending on from the last part if n is even or odd, it is now the opposite here, which means that this next turning point will have the opposite sign from the previous one.

So this is going to equal:

Since this is true for n and n+1 then all the turning points must be part of a geometric progression with common ratio: as required.

If anyone can think of a better way of doing the explaining bit at the end they are more than welcome. I've missed out quite abit of the individual working stages but have labelled what I have done so hopefully you can follow it .

Maybe there is a tidier way to do it with GP formulae I dont know.

Finding x-ordinates of turning points:

Consider any turning point:

Expand out using addition formula sin terms go to zero:

Now if n is even the expression is going to be:

And if n is odd the expression will be:

Now consider the next turning point directly after the one shown above ie. (n+1)pi

This expression will be of the form:

Which is going to equal:

So, depending on from the last part if n is even or odd, it is now the opposite here, which means that this next turning point will have the opposite sign from the previous one.

So this is going to equal:

Since this is true for n and n+1 then all the turning points must be part of a geometric progression with common ratio: as required.

If anyone can think of a better way of doing the explaining bit at the end they are more than welcome. I've missed out quite abit of the individual working stages but have labelled what I have done so hopefully you can follow it .

Maybe there is a tidier way to do it with GP formulae I dont know.

0

reply

Report

#12

STEP I, question 4.

Let

Take each term to logarithm base 2 (change the base rule)

Split the second brackets into two parts.

P = 1 + 1/4 + 1/16...

N = -1/2 - 1/8 - 1/32...

So

P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

Change the base to e, so we have the natural logarithm.

Using the power rule

(multiplying top and bottom by 1/2)

WWWWW.

Let

Take each term to logarithm base 2 (change the base rule)

Split the second brackets into two parts.

P = 1 + 1/4 + 1/16...

N = -1/2 - 1/8 - 1/32...

So

P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

Change the base to e, so we have the natural logarithm.

Using the power rule

(multiplying top and bottom by 1/2)

WWWWW.

0

reply

Report

#13

(Original post by

If anyone can think of a better way of doing the explaining bit at the end they are more than welcome.

**Square**)If anyone can think of a better way of doing the explaining bit at the end they are more than welcome.

Then , so

But , so .

So and so the values of f at the turning points form a GP with ratio as required.

1

reply

Report

#14

STEP I Q3

Our differential equation, referred to from here on as (*) is:

Let

Hence in (*):

After cancelling an and rearranging, (*) reduces to:

Hence, after dealing with partial fractions (which should be routine)

Thus

Now

Hence as

This completes the question, although I would like someone to have a quick scan for errors as I did this quite roughly.

Our differential equation, referred to from here on as (*) is:

Let

Hence in (*):

After cancelling an and rearranging, (*) reduces to:

Hence, after dealing with partial fractions (which should be routine)

Thus

Now

Hence as

This completes the question, although I would like someone to have a quick scan for errors as I did this quite roughly.

0

reply

Report

#15

III/7

(using formula for sum of geometric progression with common ratio -x^2)

as required.

Rearranging this, we get

But for

and

.

The last three statements together prove

as required.

Let us set in

Since the RHS of this inequality clearly approches zero as n grows towards positive infinity, we have by squeezing that

approaches as , and the next required result immediately follows.

Let the error in the approximation of be . The error is given by

,

which is exactly four times the middle term in

For we have

Thus we have

as required.

(using formula for sum of geometric progression with common ratio -x^2)

as required.

Rearranging this, we get

But for

and

.

The last three statements together prove

**(*)**as required.

Let us set in

**(*)**. Using the right half, we getSince the RHS of this inequality clearly approches zero as n grows towards positive infinity, we have by squeezing that

approaches as , and the next required result immediately follows.

Let the error in the approximation of be . The error is given by

,

which is exactly four times the middle term in

**(*)**for . Thus from the left part of**(*)**it followsFor we have

Thus we have

as required.

1

reply

Report

#16

On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?

0

reply

Report

#17

(Original post by

On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?

**DeathAwaitsU**)On a side note, is it bad if I can only do 1 question per STEP 1 at this stage (the stage being week before start of upper sixth)?

0

reply

Report

#18

(Original post by

Depends. Are you talking about 1987? No. If you're talking about 2004-present, then maybe. But you'll improve.

**generalebriety**)Depends. Are you talking about 1987? No. If you're talking about 2004-present, then maybe. But you'll improve.

0

reply

Report

#19

(Original post by

what is good if you have only done c1 and c2?

**eponymous**)what is good if you have only done c1 and c2?

0

reply

Report

#20

(Original post by

The 1992 STEP II paper had quite a few nice questions, I seem to remember.

**generalebriety**)The 1992 STEP II paper had quite a few nice questions, I seem to remember.

0

reply

X

### Quick Reply

Back

to top

to top