# STEP I, II, III 1999 solutions

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#1
Last edited by Sir Cumference; 1 year ago
2
12 years ago
#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...
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12 years ago
#3
Err... which paper?

Edit: ah, you edited. I'll have a look.
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12 years ago
#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...
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12 years ago
#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.)
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12 years ago
#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.
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12 years ago
#7
I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.
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12 years ago
#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.
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12 years ago
#9
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12 years ago
#10
(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.
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12 years ago
#11
Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).
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12 years ago
#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...
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12 years ago
#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.
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12 years ago
#14
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.
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12 years ago
#15
(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.
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12 years ago
#16
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.

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12 years ago
#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.
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12 years ago
#18
I'll fix that - I thought I'd take a shortcut but turns out I didn't check carefully as obviously 0
12 years ago
#19
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12 years ago
#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)
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