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#1
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STEP I:
1: Solution by DeanK22, last bit by und;
2: Solution by darkness9999; Solution by nuodai
3: Solution by nuodai
4: Solution by DeanK22
5: Solution by Blackjamin
6: Solution by Reminisce
7: Solution by Give
8: Solution by My Alt
9: Solution by Felix Felicis
10: Solution by safmaster
11: Solution by brianeverit
12: Solution by nuodai
13: Solution by QED

STEP II:
1: Solution by Glutamic Acid
2: Solution by Glutamic Acid
3: Solution by Daniel Freedman
4: Solution by Daniel Freedman
5: Solution by Glutamic Acid
6: Solution by Daniel Freedman
7: Solution by Glutamic Acid
8: Solution by Daniel Freedman
9: Solution by Glutamic Acid
10: Solution by Farhan.Hanif93
11:Solution by brianeverit
12: Solution by metaltron
13: Solution by brianeverit

STEP III:
1: Solution by DeanK22
2: Solution by DeanK22
3: Solution by Daniel Freedman
4: Solution by DeanK22
5: Solution by Daniel Freedman
6: Solution by around
7: Solution by around
8: Solution by DeanK22
9: Solution by brianeverit
10: Solution by Daniel Freedman
11: Solution by brianeverit
12: Solution by miml
13: Solutions by Daniel Freedman and 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 - 2008
0
12 years ago
#2
STEP I Q2

STEP I Q2

I will update my post with the last bit ...!
0
#3
STEP I 2008 Question 2

Solution

Rearranging the equation gives

Differentiating this with respect to then gives:

We can use the substitution in the integral to give us:

When , we obtain .

So, the integral reduces to

By use of the substitution, we have

When

So it holds when .

When

This is undefined, so it doesn't hold when

EDIT: Beaten to it! Well we did it different ways so I'll put them both up
0
12 years ago
#4
STEP III Question 1

Using the suggested hint;

Doing the exact idea again;

Dividing the first equation obtained by the following one results in [after rearranging];

along with

So x and y are the roots of the quadratic;

It follows;

Subsituting these into equation (i) and equation (ii) gives two simulateneous eqnts in a and b that are too tedious for me to solve without being in an exam.
0
12 years ago
#5
STEP III, Question 2

This is a classic example of a telescoping series and the summation is as follows:

We then proceed by binomial expansion;

Using the notation;

it follows that;

Applying the above with k = 4,5 to find S_3 and S_4 respectively quickly works.

(ii) DFranklin solutions;

Claim: S_k(n) is a poly of order k+1 for all k.

Proof: Use strong induction on k. Suppose true for k < N.

Then by (*), "the sum of a load of polynomials that are of order < N+1 (by induction hypothesis)". Since the sum of an N+1 degree poly and a load of polys of degree < N+1 is a poly of degree N+1, S_N(n) is a N+1 degree poly. Since S(0) = n is a degree 1 poly, the result follows by induction.

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.
1
12 years ago
#6
STEP III 2008 Question 6

Solution

Remember throughout that

Differentiate w.r.t x:

Or

This now results in a linear differential equation; solve using the I.F:

From the constraints we see that A is zero:

Separate variables and integrate:

We can find y from the equation given earlier: when x = 2 y = -3. Hence C = 0.

The second part proceeds much like the first; first, differentiate w.r.t x:

This has the same I.F as before:

Integrate this:

Using the initial conditions it's pretty obvious that B is again 0.

Re-arrange this a little:

Again, use the original equation they provide to find y (-1/2) and hence D is zero

In other news I found this paper pretty tough. I think this was the only full solution I managed when I did it as a mock :<
2
12 years ago
#7
STEP III, Question 4

(i) Letting y = x so I don't refer to a y or x by mistake the problem will be restated as;

show that

x > tanh(x/2) if x > 0

Let when x = 0, y_1 == y_2 = 0

We then consider the derivitive and see that when x > 0 and the result follows.

archosh(z) > (z-1)/root(z^2-1) Let z = cosh(x) the result trivially follows from the above after rewriting (x^2-1) as (x-1)(x+1) and remembering the identity linking cosh(x) with sinh^2(x/2)

(ii) We need to that if in an interval [a,b] that f(x) => g(x) it follows that;

Using IBP and splitting the integral we see that;

and the result follows.

Integrating (ii) wrt x gives the required result.
0
#8
STEP I Question 3

Introduction

If , then

The proof that this is the case is as follows:

This is greater than or equal to 0, since
Part (i)

It follows from the introduction that if we let and , then it still holds that , so we can just substitute them in to give us

We are then told that , and it is hinted that , meaning that . Once again, this fits the necessary conditions, so

Combining the two inequalities gives us

Part (ii) coming when I can be bothered with tedium.
2
12 years ago
#9
STEP III, question 8

1 + 2P = 0 , <---> p = -1/2

S(1+px) = (x/3) + (x*x/6) + (x*x*x/12) + ... - (x*x/6) - (x*x*x/12) - .... = x/3

It follows

(as clearly we know that x =/= 2 .

We can do a similar thing by considering and it follows:

(ii) Do the same and we obtain: q = 1 , p = -2.5

We then proceed by partial fractions:

B = 14/3 and A = -8/3

Then proceed as above;

1
12 years ago
#10
(Original post by DeanK22)
(ii) To me this is blatantly obviously true. It has basically been shown by the working above. I do not know what the eaminer wants from this - just look at the above epression? I guess you could talk about the relationship between integration and summation? For a STEP III I am not going to tell you that saying "well it is kin of obvious" is the thing to do but this seems like the case here.
I think you'd be expected to do a proof by induction for the first bit.

e.g. Claim: S_k(n) is a poly of order k+1 for all k.

Proof: Use strong induction on k. Suppose true for k < N.

Then by (*), "the sum of a load of polynomials that are of order < N+1 (by induction hypothesis)". Since the sum of an N+1 degree poly and a load of polys of degree < N+1 is a poly of degree N+1, S_N(n) is a N+1 degree poly. Since S(0) = n is a degree 1 poly, the result follows by induction.

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.

To be honest, your "it is kind of obvious" comment looks dubious here. There are some pretty clear steps that you need to go through. And waffling about links between integration and summation indicates to me that you are missing the point of the question.
0
12 years ago
#11
nuodai do you think you could use the same format for the first post as the other years (ie using "Solution by name" as the hyperlink. The only exception to this is 2007 which I intend to fix in the near future)
0
12 years ago
#12
(Original post by DFranklin)
For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.
I just read the part about proving it would be a polynomial of degree k+1 - these other bits look abit more involved. Do you think it would be neccessary to prove the bit about the degree of the polnomial or would it be OK to say that from the previous working it can be seen?
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#13
(Original post by SimonM)
nuodai do you think you could use the same format for the first post as the other years (ie using "Solution by name" as the hyperlink. The only exception to this is 2007 which I intend to fix in the near future)
Consider it done.
0
12 years ago
#14
(Original post by DeanK22)
I just read the part about proving it would be a polynomial of degree k+1 - these other bits look abit more involved. Do you think it would be neccessary to prove the bit about the degree of the polnomial or would it be OK to say that from the previous working it can be seen?
No, it wouldn't be OK.
0
12 years ago
#15
II/7:.

(i)

when x = 0, y = 1 so u = 1 so c = -4/3.

Solution: .

(ii) Let

u = 1, again, so c = - 5/4.

Solution: .

(iii) I conjecture the solution to be . To prove it, one would make the substitution y = u(1+x^n)^(1/n).
0
12 years ago
#16
II/5:

Letting u = sin x, this becomes
.

u = cos x, this becomes

.

As (exponential functions are strictly increasing), sufficient to show that , and 128 > 99 is enough to show this holds.

Suppose , previously established, so the first integral is greater. And we are done, forever.
1
12 years ago
#17
II/9:

and therefore
.
The larger root means the ball will loop up in the air for ages and ages and ages like some sort of stupid duck so we choose the smaller root.
As t = sec a / 2, and cos a approx 1, t is approx 1/2.

(ii) h > 5/4.

(iii) (first term binomial expansion). Using the tan a = a approximation and that 1 radian is 57 degres, alpha is approximately 57/16 degrees.
0
12 years ago
#18
STEP III 2008, Question 5

Spoiler:
Show

Assume for induction that

We wish to show that

as required. The base case is given, thus completing the proof by induction.

In the case , we have

(multiplying all the terms together (apart from the term in T_(n+1)) gives , which is also equivalent to )

Substituting into (*) gives,

as T_0 is assumed to be non-zero.

0
12 years ago
#19
STEP III 2008, Question 3

Spoiler:
Show
The line SP is given by:

To find the gradient of the tangent at P:

ON is perpendicular to the tangent at P, which means

is the equation of the line through ON.

Rearranging this for x, and substituting into the equation of the line SP:

as required.

For T to lie on the circle centre S, radius a, it must satisfy:

0
12 years ago
#20
STEP III 2008, Question 10

Spoiler:
Show
Resolving forces vertically on the rth particle,

Resolving forces vertically on the bottom particle,

which gives

The length L of the long string is given by

The total elastic potential energy E stored in the long string is given by

Let us divide the new heavy rope of mass M and natural length L_0, into n small ropes of mass m and natural length l. Clearly,

and

Let n, the number of small ropes the large rope is divided into, tend to infinity to find the length L of the heavy rope.

as required.

Similarly to find the energy stored in the heavy rope,

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