SimonM
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(Updated as far as #96) SimonM - 23.03.2009

For explanation, scroll down:

STEP I:
1: Solution by nota bene, also solved in p 21 of Siklos' booklet, also DFranklin
2: Solution by datr
3: Solution by DFranklin
4: Solved in p 17 of Siklos' booklet
5: Solution by ukgea
6: Solution by DFranklin
7: Solution by DFranklin
8: Solution by ukgea
9: Solved in p 107 of Siklos' booklet
10: Solution by DFranklin
11: Solution by ukgea
12: Solved in p 147 of Siklos' booklet
13: Solution by nota bene
14: Solution by ukgea


STEP II:
1: Solution by Koyla
2: Solved in p 25 of Siklos' booklet
3: Solution by insparato
4: Solution by DFranklin
5: Solution by dvs
6: Solved in p 27 of Siklos' booklet
7: Solution by waxwing
8: Solution by DFranklin
9: Solved in p 127 of Siklos' booklet
10: Solution by DFranklin
11: Solution by ad absurdum
12: Solution by datr
13: Solution by DFranklin
14: Solution by datr


STEP III:
1: Solution by dvs
2: Solution by khaixiang
3: Solution by ukgea
4: Solved in p 57 of Siklos' booklet
5: Solution by DFranklin
6: Solution by DFranklin, also by khaixiang
7: Solved in p 77 of Siklos' booklet
8: Solution by dvs
9: Solution by ukgea
10: Solution by DFranklin
11: Solution by DFranklin
12: Solution by DFranklin
13: Solution by DFranklin
14: Solution by DFranklin


Explanation:
In this thread, we will try to collectively provide solutions for the questions from the STEP Maths I, II and III papers from 1997, seeing as no solutions are available on the net.

If you want to contribute, simply find a question which is marked with "unsolved" above (you might want to check for new solutions after the time the list was last updated as well), solve it, and post your solution in the thread. I will try to update the list as often as possible. If you have a solution to a question that uses a different idea from one already posted, submit it as well, as it's often useful to see different approaches to the same problem.

If you find an error in one of the solutions, simply say so, and hope that whoever wrote the solution in the first place corrects it.

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
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nota bene
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I
Q1

Possibilities to obtain 10p by using coins of 1, 2, 5 and 10p

10
5+5
5+2+2+1
5+2+1+1+1
5+1+1+1+1+1
2+2+2+2+2
2+2+2+2+1+1
2+2+2+1+1+1+1
2+2+1+1+1+1+1+1
2+1+1+1+1+1+1+1+1
1+1+1+1+1+1+1+1+1+1

Possibilities to obtain 20p by using 1, 2, 5 and 10p coins:
20=10+10 Hence we can see that there are 11 possibilities to obtain it if added with a 10.

Now we will have;
5+5+5+5
5+5+5+2+2+1 etc. 2 times more
5+5+2+2+2+2+2, 5+5+2+2+2+2+1+1 etc. 4 times more
5+2+2+2+2+2+2+2+1 etc. 7 times more
2+2+2+2+2+2+2+2+2+2 etc. 10 times more

i.e. total answer =41
And, as DFranklin pointed out later, I missed the obvious 20p:p:

Edited: corrected stupid thinking mistake, should not just work on a STEP question for 5min:p:

Edit2: Now it shall be OK, and when checking in Siklos I find he writes something that is wrong (only finds 6ways to get 5+....) but the final answer in there is the same.
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insparato
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Ive got STEP II Question 3 to write up. Ill do it abit later on .

I didnt see you posting this oops! Ill get a mod to delete the one i started.
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dvs
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Where are the papers?

You guys are going to eat up all my free time. :p:
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ukgea
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Papers coming, upload is sloooow... :p:

insparato: Actually yours was first, so I should be the one to say sorry. But I just thought that you might want some relief from having to update the thread, I mean I just happen to got too much free time right now so I might just as well do it, you concentrate on doing the questions instead. :p:
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insparato
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Ukgea can attach them or you could pm me and ill give you the link.

I dont think im allowed to provide you it on TSR threads.

As Nicholas has tried to do above .
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insparato
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(Original post by ukgea)
Papers coming, upload is sloooow... :p:

insparato: Actually yours was first, so I should be the one to say sorry. But I just thought that you might want some relief from having to update the thread, I mean I just happen to got too much free time right now so I might just as well do it, you concentrate on doing the questions instead. :p:
Hehe no prob :p: sure if you've got the free time. I managed to get Question 3 on STEP II out today so gonna type that out now.
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ukgea
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Meh!

"STEP I 1997.pdf:
Your file of 1.32 MB bytes exceeds the forum's limit of 609.7 KB for this filetype.
STEP II 1997.pdf:
Your file of 1.17 MB bytes exceeds the forum's limit of 609.7 KB for this filetype.
STEP III 1997.pdf:
Your file of 1.33 MB bytes exceeds the forum's limit of 609.7 KB for this filetype."

How were your files so small, insparato?
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insparato
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There were all under 600kb just lucky i guess.
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nota bene
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shall we upload them to rapidshare or something instead? (Although that would mean we had to download the files to the harddrive...)
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Kolya
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STEP II Question 1

Note that the first term + last term, 2000+5777 = 7777. The second term + last-1'th term, 2002+5775 = 7777, and so on.

We only have to consider the number of terms in the 2000's and 5000's. As we are adding the terms up, we need to find the number of terms in the 2000's.

Looking at the progression:
2000
2002
2005
2007

2020
2022
2025
2027

2050
2052
2055
2057

2070
2072
2075
2077
etc...

There are 16 terms with the first two digits, and this occurs 4 times with each different hundreds number, so there are 64 "sets" of 7777.

Therefore:
64\cdot 7777 = 2^6 \cdot 7 \cdot 1111 = 2^6 \cdot 7 \cdot 11 \cdot 101

Done.
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ukgea
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Perhaps. If you want to do it, then fine. Otherwise we could just have people PMing around the links, besides I think most people who are actually doing STEP revision know where to find them anyway.
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nota bene
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(Original post by ukgea)
Perhaps. If you want to do it, then fine. Otherwise we could just have people PMing around the links, besides I think most people who are actually doing STEP revision know where to find them anyway.
Yes, I think so too, but it seemed like dvs wanted them, I'll see if I upload them to rapidshare soon, otherwise someone could PM him, and others wanting it (if that isn't done...)
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dvs
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(Original post by nota bene)
Yes, I think so too, but it seemed like dvs wanted them, I'll see if I upload them to rapidshare soon, otherwise someone could PM him, and others wanting it (if that isn't done...)
I was sent the link. Will look over them after I have some lunch.
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DFranklin
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Nota Bene - I think you're just missing the "obvious" solution of just using the 20p! As I happen to have an approach I think can be useful in similar problems, I hope no-one minds if I give it here:

Step I, Q1, 1997.

Let P(n,k) be the no. of ways of making n using coins <=k. We make the following observations:

P(n,1) = 1 for all n.

P(1,2) = 1, P(2,2) = 2, P(n,2)=floor(n/2)+1 (using 0,1,2,...,floor(n/2) 2's).

P(10,5) = P(10,2)+P(5,2)+1 = 6 + 3 + 1 = 10 (first term is using no 5's, second term is using one 5, last term is using 2 fives. We will use a similar construction repeatedly later).
P(10,10) = P(10,5)+1 = 11. (first term using no 10s, 2nd term using 1 10).

For the 2nd part:

P(20,5)=P(20,2)+P(15,2)+P(10,2)+P(5,2)+1 = 11 + 8 + 6 + 3 + 1 = 29
P(20,10) = P(20,5)+P(10,5)+1 = 29 + 10 + 1 = 40
P(20,20) = P(20,10)+1 = 41.
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DFranklin
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I'll do Paper I Q3,6,7.
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nota bene
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(Original post by DFranklin)
Nota Bene - I think you're just missing the "obvious" solution of just using the 20p!
Yeah, LOL
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insparato
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STEP II Question 3

 \frac{ax+b}{x^2+2x+2} + \frac{cx+d}{x^2-2x+2} = \frac{1}{x^4+4}

Therefore

 (ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) = 1

Expand out

 ax^3 -2ax^2 + 2ax + bx^2 -2bx + 2b + cx^3 +2cx^2 + 2cx + dx^2 + 2dx + 2d = 1

 (a+c)x^3 (-2a+b+2c+d)x^2 + (2a-2b+2c+2d)x + 2b + 2d = 1

Comparing coefficients

 a + c = 0 - 1

 -2a+b+2c+d = 0 - 2

 2a-2b+2c+2d = 0 - 3

 2b+2d = 1 - 4

 a = -c - Substituting this into 2 and 3

 b + 4c + d = 0

 b = d - Substituting this into 4

 4b = 1

 b = \frac{1}{4}

 4d = 1

 d = \frac{1}{4}

Substituting values of b and d into 2

 \frac{1}{2} + 4c = 0

 c = -\frac{1}{8} - Substituting back into 1

 a = \frac{1}{8}

So

 a = \frac{1}{8}

 b = \frac{1}{4}

 c = -\frac{1}{8}

 d = \frac{1}{4}

Therefore

 \frac{1}{x^4+4} = \frac{\frac{1}{8}x+\frac{1}{4}}{x^2+2x+2} + \frac{\frac{1}{4}-\frac{1}{8}x}{x^2-2x+2}

 = \frac{x+2}{8x^2+16x+16} + \frac{2-x}{8x^2-16x+16}

 \displaystyle \int \frac{1}{x^4+4} = \int \frac{x+2}{8x^2+16x+16} + \frac{2-x}{8x^2-16x+16}

\displaystyle = \frac{1}{8}\int \frac{x+2}{x^2+2x+2} + \frac{2-x}{x^2-2x+2}

Consider \displaystyle \int \frac{x+2}{x^2+2x+2}

\displaystyle \int \frac{x+2}{x^2+2x+2} = \int \frac{2x+2}{x^2+2x+2} - \frac{x}{x^2+2x+2}

 = ln(x^2+2x+2) - \int \frac{x}{x^2+2x+2}

Consider  \displaystyle \int \frac{x}{x^2+2x+2}

 \displaystyle \int \frac{x}{x^2+2x+2} = \int \frac{2x+2}{x^2+2x+2} - \frac{2}{x^2+2x+2} - \frac{x}{x^2+2x+2}

 \displaystyle 2\int \frac{x}{x^2+2x+2} = ln (x^2+2x+2) - \int \frac{2}{x^2+2x+2}

\displaystyle = ln (x^2+2x+2) - \int \frac{2}{(x+1)^2+1}

let u = x+1

 \frac{du}{dx} = 1

 du = dx

 = ln (x^2+2x+2) - 2\int\frac{1}{u^2+1}

 = ln(x^2+2x+2) - 2arctanu

 = ln(x^2+2x+2) - 2arctan(x+1)


 \displaystyle 2\int \frac{x}{x^2+2x+2} = ln(x^2+2x+2) - 2arctan(x+1)

\displaystyle \int \frac{x}{x^2+2x+2} = \frac{1}{2}ln(x^2+2x+2) - arctan(x+1)

Therefore

\displaystyle \int \frac{x+2}{x^2+2x+2} = ln(x^2+2x+2) - [\frac{1}{2}ln(x^2+2x+2) - arctan(x+1)]

 = \frac{1}{2}ln(x^2+2x+2) + arctan(x+1)

Consider  \displaystyle \int \frac{2-x}{x^2-2x+2}

 \displaystyle \int \frac{2-x}{x^2-2x+2} = \int \frac{2x-2}{x^2-2x+2} +\int \frac{4}{x^2-2x+2} - \int \frac{3x}{x^2-2x+2}

\displaystyle = ln (x^2-2x+2) + 4\int \frac{1}{(x-1)^2+1} - 3 \int \frac{x}{x^2-2x+2}

Consider \displaystyle \int \frac{1}{(x-1)^2+1}

 u = x-1

 du = dx

\displaystyle \int \frac{1}{(x-1)^2+1} = \int \frac{1}{u^2+1}

 = arctan(x-1)

 \displaystyle \int \frac{2-x}{x^2-2x+2} = ln(x^2-2x+2) +4arctan(x-1) - \int \frac{3x}{x^2-2x+2}

Consider \displaystyle \int \frac{3x}{x^2-2x+2}

\displaystyle \int \frac{3x}{x^2-2x+2} = \int \frac{2x-2}{x^2-2x+2} + \int \frac{x}{x^2-2x+2} + \int \frac{2}{x^2-2x+1}

 2\displaystyle \int \frac{x}{x^2-2x+2} = ln (x^2-2x+2) + 2arctan(x-1)

 \displaystyle \int \frac{x}{x^2-2x+2} = \frac{1}{2}ln (x^2-2x+2) + arctan(x-1)

Therefore

 \displaystyle \int \frac{2-x}{x^2-2x+2} = ln(x^2-2x+2) +4arctan(x-1) -3[\frac{1}{2}ln (x^2-2x+2) + arctan(x-1)]

Finally

 \displaystyle \int \frac{1}{x^4+4} = \frac{1}{8} \int \frac{x+2}{x^2+2x+2} +  \frac{2-x}{x^2-2x+2}

\\ =  [\frac{1}{8}ln(x^2+2x+2) -\frac{1}{8}(\frac{1}{2}ln(x^2+2x+2)-arctan(x+1)) + \frac{1}{8}ln(x^2-2x+2) + \frac{1}{2}arctan(x-1) - \frac{3}{8}(\frac{1}{2}(ln(x^2-2x+2) + arctan(x-1)]_0^1

\\ =  [\frac{1}{8}ln5 - \frac{1}{8}(\frac{1}{2}ln5 - arctan2)+ \frac{1}{8}ln1 + \frac{1}{2}ln0 - \frac{3}{8}(\frac{1}{2}ln1+arctan0)] - [\frac{1}{8}ln2 - \frac{1}{8}(\frac{1}{2}ln2 - arctan(1))+\frac{1}{8}ln2 + \frac{1}{2}arctan(-1) -\frac{3}{8}(\frac{1}{2}ln2+arctan(-1))

\\=  [\frac{1}{16}ln5 +\frac{1}{8}arctan(2)] - [\frac{1}{16}ln2 + \frac{\pi}{32} + \frac{1}{8}ln2 -\frac{\pi}{8} - \frac{3}{16}ln2 +\frac{3\pi}{32}]

=  \frac{1}{16}ln5 + \frac{1}{8}arctan2

This has been a nightmare typing this up as i dont exactly write in a logical order when its a big question on paper. Ill read through it just to check it. Although it does get the answer.
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ukgea
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STEP I question 8

Clearly, the equation has no non-positive roots (as the RHS is always positive for real x).

Then we can take logs of both sides for the equivalent

\ln x = x\ln a

\ln x - x\ln a = 0.

Let f(x) = \ln x - x\ln a. Then

\displaystyle f'(x) = \frac{1}{x} - \ln a

Note that f'(x) is strictly decreasing (well for positive x anyway, which are the only ones we're worrying about).

Now, setting f'(x) = 0 we get

\displaystyle \frac{1}{x} = \ln a

\displaystyle x = \frac{1}{\ln a}

This is a maximum since f'(x) was decreasing. The value at the maximum is

\displaystyle f\left(\frac{1}{\ln a}\right) = - \ln {(\ln a)} - 1

For there to be zeroes to f(x), we must have that the maximum is larger than or equal to zero:

- \ln{(\ln a)} - 1 \geq 0

\ln{(\ln a)} \leq -1

Exponentiating:

\ln a \leq e^{-1}

and again:

a \leq e^{1/e},

and so there are no real roots if  a &gt; e^{1/e}.

If 0 &lt; a &lt; 1, then \ln a &lt; 0 and so f'(x) is strictly positive (again for x>0). This means that there can be at most 1 solution to the equation. Consider now f(a)

f(a) = \ln a - a\ln a = (1 - a)\ln a

which is negative since 1 - a &gt; 0 and \ln a &lt; 0. Then consider f(1):

f(1) = \ln 1 - \ln a = - \ln a

which is positive since \ln a &lt; 0. Because f(x) is continuos, there must then be one point x, with a &lt; x &lt; 1 such that f(x) = 0, i.e. the equation f(x) = 0 has at least one solution.

But now f(x) = 0 must have exactly one root. Since this equation is equivalent (for positive x) to the given equation x = a^x, it follows that this latter one also has exactly one solution.
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DFranklin
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Inspirato: does it come out simpler if you write:

\displaystyle \int \frac{x+2}{x^2+2x+2} dx = \int \frac{x+1}{x^2+2x+2} + \frac{1}{x^2+2x+2}dx as the 1st bit is just \frac{1}{2}\frac{d}{dx} ln(x^2+2x+2).

(And similarly for the other part).

I actually set essentially the same question as this for some students, once upon a while...
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