# STEP I, II, III 1999 solutions

Watch
(Updated as far as #111) SimonM - 26.04.2009

Done

1: Solution by SimonM

2: Solution by SimonM

3: Solution in Siklos booklet

4: Solution by brianeverit

5: Solution by Aurel-Aqua

6: Solution by SimonM

7: Solution by Aurel-Aqua

8: Solution by SimonM

9: Solution by lilman91

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by lilman91 and Aurel-Aqua

1: Solution by brianeverit

2: Solution by nota bene, Solution by Square

3: Solution by nota bene, #28, etc.

4: Solution by brianeverit

5: Solution by I hate maths

6: Solution by nota bene

7: Solution by nota bene

8: Solution by brianeverit

9: Solution by brianeverit

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by SimonM And Aurel-Aqua

14: Solution by SimonM

1: Solution by Zhen Lin (PDF)

2: Solution by Agrippa (PDF)

3: Solution by Zhen Lin (PDF)

4: Solution by Zhen Lin (PDF)

5: Solution by Zhen Lin (PDF)

6: Solution by squeezebox, #61

7: Solution by Zhen Lin (PDF), Post #52

8: Solution by squeezebox

9: Solution by brianeverit

10: Solution by squeezebox

11: Solution by brianeverit

12: Solution by Aurel-Aqua

13: Solution by toasted-lion

14: Solution by brianeverit

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

Done

**STEP I:**1: Solution by SimonM

2: Solution by SimonM

3: Solution in Siklos booklet

4: Solution by brianeverit

5: Solution by Aurel-Aqua

6: Solution by SimonM

7: Solution by Aurel-Aqua

8: Solution by SimonM

9: Solution by lilman91

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by lilman91 and Aurel-Aqua

**STEP II:**1: Solution by brianeverit

2: Solution by nota bene, Solution by Square

3: Solution by nota bene, #28, etc.

4: Solution by brianeverit

5: Solution by I hate maths

6: Solution by nota bene

7: Solution by nota bene

8: Solution by brianeverit

9: Solution by brianeverit

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by SimonM And Aurel-Aqua

14: Solution by SimonM

**STEP III:**1: Solution by Zhen Lin (PDF)

2: Solution by Agrippa (PDF)

3: Solution by Zhen Lin (PDF)

4: Solution by Zhen Lin (PDF)

5: Solution by Zhen Lin (PDF)

6: Solution by squeezebox, #61

7: Solution by Zhen Lin (PDF), Post #52

8: Solution by squeezebox

9: Solution by brianeverit

10: Solution by squeezebox

11: Solution by brianeverit

12: Solution by Aurel-Aqua

13: Solution by toasted-lion

14: Solution by brianeverit

**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

Last edited by Notnek; 2 years ago

2

reply

Report

#2

I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

0

reply

Report

#4

In STEP questions like that, you can be almost certain that the second part relates to the first part in some way. In this case, put y = e^x - 1 - k tan^-1 x, and find dy/dx...

0

reply

Report

#5

Aaah. I see. I will try that now.

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)

0

reply

Report

#6

Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.

0

reply

Report

#7

I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.

0

reply

Report

#8

Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.

0

reply

Report

#9

__STEP III - Question 8__

i)

.

So the differential equation becomes:

, whose general solution is: .

Now, since when x=0, y=0 and , the particular solution is (*).

ii) Since .

So this time the differential equation is:

general solution: (**).

Now, since y and are continuous at , (**) has the same value as (*) at and also their derivatives take the same value. From this we can work out the values of C and D, (C=0 and D= ). Which gives us: .

Using the same method in the previous part, we can deduce that .

iii)

=

=

(using )

=

=

2

reply

Report

#10

(Original post by

I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

**Zhen Lin**)I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

0

reply

Report

#11

Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).

0

reply

Report

#12

Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether is increasing/decreasing faster than . I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...

0

reply

Report

#13

I dunno if what I did was correct, but I considered; g(x) = . By playing around with it, I noticed that as x , g(x) . If you want me to into anyfurther detail , I will.

1

reply

Report

#15

(Original post by

Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether is increasing/decreasing faster than . I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...

**Zhen Lin**)Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether is increasing/decreasing faster than . I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...

0

reply

Report

#17

For (ii), I don't see why you say that s is positive. Given the conditions they've given, it's pretty obvious you're supposed to use "B^2-4AC<0" on n^2(p-r)-ns+q.

0

reply

Report

#18

0

reply

Report

#20

II/7

This question seems too easy, so I've probably made a mistake:

Then

Stationary points can occur when dy/dx=0 i.e. the numerator needs to be zero <=> 0=a-x <=>x=a Hence there is at most one stationary point. The derivative is undefined if i.e. and for a>1 this has no (real) solutions and thus the function has more than zero stationary points. I.e. there is one stationary point, as desired.

i) a=2 f(0)=0 so stationary point at and and thus the point is a maximum.

Also note

ii) a=1 f(0)=0 From above we have that there are no stationary points because the denominator of the function (and it's derivative) goes to 0 as a goes to 1 i.e. vertical asymptote at x=1 Same as in i) note that

Sketches see attached. (Very rough, would put much more effort in during a real exam)

This question seems too easy, so I've probably made a mistake:

Then

Stationary points can occur when dy/dx=0 i.e. the numerator needs to be zero <=> 0=a-x <=>x=a Hence there is at most one stationary point. The derivative is undefined if i.e. and for a>1 this has no (real) solutions and thus the function has more than zero stationary points. I.e. there is one stationary point, as desired.

i) a=2 f(0)=0 so stationary point at and and thus the point is a maximum.

Also note

ii) a=1 f(0)=0 From above we have that there are no stationary points because the denominator of the function (and it's derivative) goes to 0 as a goes to 1 i.e. vertical asymptote at x=1 Same as in i) note that

Sketches see attached. (Very rough, would put much more effort in during a real exam)

2

reply

X

### Quick Reply

Back

to top

to top