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# STEP Maths I, II, III 1988 solutions

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1. (Updated as far as #68) SimonM - 15.06.2009

Seeing as people seem to have asked for a thread for this for quite a while I thought I might as well make one.

I guess you're all familiar what these threads are for by now; post your solutions to the older STEP papers as there are no solutions available freely on the net - plus it's good revision Feel free to post alternative solutions, and please point out when you see a weakness or mistake in a solution.

If you look in the STEP megathread you will find a link to the papers - if you still have problems accessing them, PM me or someone else who has them.

STEP I:
1: Solution by nota bene
2: Solution by nota bene
3: Solution by Glutamic Acid
4: Solution by brianeverit
5: Solution by Swayum
6: Solution by brianeverit
7: Solution by Glutamic Acid
8: Solution by brianeverit
9: Solution by nota bene
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
15: Solution by nota bene
16: Solution by brianeverit

STEP II:
1: Solution by kabbers
2: Solution by SimonM
3: Solution by Glutamic Acid
4: Solution by brianeverit
5: Solution by Squeezebox
6: Solution by SimonM
7: Solution by Square
8: Solution by brianeverit
9: Solution by SimonM
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 kabbers
2: Solution by Squeezebox
3: Solution by mikelbird and brianeverit
4: Solution by mikelbird
5: Solution by kabbers
6: Solution by kabbers
7: Solution by squeezebox
8: Solution by SimonM
9: Solution by SimonM
10: Solution by ben-smith
11: Solution by Jkn
12. Solution by ben-smith
13: Solution by ben-smith
14: Solution by ben-smith
15:
16: Solution by ben-smith

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
2. STEP I Q1

For max/min set h'(x)=0 i.e.
To find the nature of the stationary point, consider e.g. h'(e-0.2)=positive and h'(e+0.2)=negative hence a (global) maximum. For sketching purposes might be worth noting that and . Graph see attached (Mathematica because of lack of scanner).

Solving is equivalent of solving (last step valid as neither m nor n can be 0).
m=n is the trivial solution.
Consider the line f=c (i.e. a constant). For there are two solutions. There's one solution when or . And for there are no solutions.

Now if then ().

Thus and .

(What am I missing here, this looks trivial for a STEP question...)

----

STEP I Q2

()

Let

Now WLOG

Has solutions cos(x)=0 and sin(x)=1/2
i.e.

where
By quadratic formula the solutions are i.e. u=-1, 2
has no real solutions and has the root (which we have already found).

where
Solutions to this are and as sin(x)=u will have no real solutions.

, so a (local)min
, so a (global) min
, so a (global) max
, so a (local) max
Attached Images

3. I haven't bothered to download the paper, so this may not be a valid objection, but I feel I need to point out that 2^4 = 4^2...
4. Yeah, I certainly need to! Had forgotten that one, which is by inspection.
5. High, quick question: What programme are you using to write out the Mathematics? I've seen such quite alot here but have no idea to do so myself. :|

Thanks.
6. Nota: I think you would be expected to give at least some justification why (2,4) and (4,2) are the only 'non-trivial' solutions. (Without giving you a hard time for the slipup: if you missed (2,4), how do you know you didn't miss anything else?)
7. (Original post by The Lyceum)
High, quick question: What programme are you using to write out the Mathematics? I've seen such quite alot here but have no idea to do so myself. :|

Thanks.
LaTeX. It's installed on the forum software - see the thread in the maths forum called "PLEASE READ BEFORE POSTING !!" for an explanation of how to use it.
8. (Original post by generalebriety)
LaTeX. It's installed on the forum software - see the thread in the maths forum called "PLEASE READ BEFORE POSTING !!" for an explanation of how to use it.
Thanks.
9. Im about half way through II/7.

So if someone could refrain from spoiling my fun and posting the solution before I wake up tomorrow and finish the question please!
10. Good idea Nota . I'll have a look at the paper(s) tomorrow and see if I get anything out.

Also....1989 has many unsolved questions. [/shameless plug]

11. (Original post by DFranklin)
Nota: I think you would be expected to give at least some justification why (2,4) and (4,2) are the only 'non-trivial' solutions. (Without giving you a hard time for the slipup: if you missed (2,4), how do you know you didn't miss anything else?)
Ah, yes. *feels stupid*

Consider the line f=c (i.e. a constant). For there are two solutions. There's one solution when or . And for there are no solutions.

Now if then ().

Thus and .
12. Sorry, but I'm not seeing how the above shows there's only two solutions.

(I would give a very simple solution based on the graph and the location of the turning point).
13. III/1:

Sketch

I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

Differentiating (quotient rule is your friend), we get

So we have turning points at x = 0, and x =

Differentiating again, we get

Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.

Noticing the denominator of y, we find that there is an asymptote at x = -1.

And because of the exponential properties of e^-x, y tends to zero as

y tends to as since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.

So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14

Prove

First note that

Hence we may split the integral into

Consider

Now consider

I posit the inequality for x > 0

So our inequality holds.

So:

Thus

And therefore,

Now notice that the graph of is greater than 0 for x > 0, and hence so is the infinite integral.

So,

14. k here's the graph from the above q. if i've made any mistakes, please feel free to point them out
Attached Images

15. You've made a mistake integrating (x-1)e^{-x}. (Still right answer, but some of what you've written is wrong).
16. (Original post by DFranklin)
You've made a mistake integrating (x-1)e^{-x}. (Still right answer, but some of what you've written is wrong).
ok thanks, corrected
17. I/9

i) Recall that , now we can use this to integrate by parts.

Putting in the limits gives So

ii) Let
So the integral becomes
Going back to x we have
18. I've got III/3 out. Will post later. Suprisingly easy, its more like a FP3 question to be fair. Might need a check over because my complex transformations are more than a little rusty.
19. Done rough "workings out" for I/9 again two surprisingly easy integrals.
20. (Original post by insparato)
Done rough "workings out" for I/9 again two surprisingly easy integrals.
Agreed, took me 10 minutes on the tube this morning when on my way to the library

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