You are Here: Home

# STEP I, II, III 1999 solutions Tweet

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

Announcements Posted on
Enter our travel-writing competition for the chance to win a Nikon 1 J3 camera 21-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Sign in to Reply
1. STEP I, II, III 1999 solutions
Last edited by SimonM; 05-05-2009 at 21:46.
2. Re: STEP I, II, III 1999 solutions
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...
Last edited by Zhen Lin; 24-03-2008 at 04:17.
3. Re: STEP I, II, III 1999 solutions
Err... which paper?

Edit: ah, you edited. I'll have a look.
Last edited by generalebriety; 24-03-2008 at 04:19.
4. Re: STEP I, II, III 1999 solutions
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...
5. Re: STEP I, II, III 1999 solutions
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.)
Attached Files
6. III-1999-Q01.pdf (33.7 KB, 673 views)
7. III-1999-Q03.pdf (38.9 KB, 507 views)
8. III-1999-Q04.pdf (23.1 KB, 331 views)
9. III-1999-Q05.pdf (25.9 KB, 439 views)
10. III-1999-Q07.pdf (28.2 KB, 297 views)
11. Re: STEP I, II, III 1999 solutions
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.
12. Re: STEP I, II, III 1999 solutions
I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.
Last edited by squeezebox; 24-03-2008 at 12:11.
13. Re: STEP I, II, III 1999 solutions
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.
14. Re: STEP I, II, III 1999 solutions
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 )

=

=
Last edited by squeezebox; 06-04-2008 at 00:30.
15. Re: STEP I, II, III 1999 solutions
(Original post by 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...
I would find the stationary points by differentiating, then see where there is a +ve/-ve sign change between stationary points in the original equation, if that makes sense.
Last edited by ukstudent2011; 25-03-2008 at 01:31.
16. Re: STEP I, II, III 1999 solutions
Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).
17. Re: STEP I, II, III 1999 solutions
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...
Last edited by Zhen Lin; 25-03-2008 at 02:09.
18. Re: STEP I, II, III 1999 solutions
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.
19. Re: STEP I, II, III 1999 solutions
I'm currently struggling my way through II/3, will type up when done.
under construction

II/3

Then let's compute some derivatives

so and we can see that for all S_n we will get e^(x^3) and e^(-x^3) cancelling out.
So
And similarily
So

To see the pattern I computed S_4-S_6 as well, won't type up the calculations but they are as follows:

From this we can form the following closed form where denotes ceiling[b] i.e. the nearest integer above, or equal to b.
Last edited by nota bene; 25-03-2008 at 03:00. Reason: solution
20. Re: STEP I, II, III 1999 solutions
(Original post by 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...
You know the roots must be between turning points, before the first turning point, or after the last one, so if there is a sign change between two turning points there must be one root between the turning points. It is quite easy to work out if there is another root before or after the last turning point.
21. Re: STEP I, II, III 1999 solutions
II/2
solution

i) Where p=3 q=50 r=2 and s=15 find the set of values for n where the equation above has no real roots.

For non-real roots we need
Inserting the given values of p, q, r and s:

Solving for the positive case we have and the negative case
I.e. the equation lacks real roots when and

ii) Prove that if p<r and then the above equation has no real roots fo any n.

Recall from i) that for non-real roots
Solving for n gives and is obviously complex when and thus there are no real solutions for any n.

iii) Find when the above equation has real roots when n=1, (p-r)=1 and
For real roots , and with the above values this transfers to finding where
Solving the positive case gives as desired and the negative case gives as desired.

Last edited by nota bene; 02-05-2008 at 23:33.
22. Re: STEP I, II, III 1999 solutions
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.
23. Re: STEP I, II, III 1999 solutions
I'll fix that - I thought I'd take a shortcut but turns out I didn't check carefully as obviously
24. Re: STEP I, II, III 1999 solutions
II/6
solution
then

Now, this can be used to find integrals such as Substitute and then we have
This transfers our integral to

as desired.

Next integral:
Let thus which gives

And similarily let thus which gives

Evaluating these two integrals, and adding together gives

Now to the last integral:

which both have been evaluated in the part above.

25. Re: STEP I, II, III 1999 solutions
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)
Attached Thumbnails

Last edited by nota bene; 29-03-2008 at 10:59.
Sign in to Reply
Share this discussion:
Useful resources

## Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominations

## Quick Link:

Unanswered Maths Exams Threads

## Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:
Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.